Related
I had this question on an Algorithms test yesterday, and I can't figure out the answer. It is driving me absolutely crazy, because it was worth about 40 points. I figure that most of the class didn't solve it correctly, because I haven't come up with a solution in the past 24 hours.
Given a arbitrary binary string of length n, find three evenly spaced ones within the string if they exist. Write an algorithm which solves this in O(n * log(n)) time.
So strings like these have three ones that are "evenly spaced": 11100000, 0100100100
edit: It is a random number, so it should be able to work for any number. The examples I gave were to illustrate the "evenly spaced" property. So 1001011 is a valid number. With 1, 4, and 7 being ones that are evenly spaced.
Finally! Following up leads in sdcvvc's answer, we have it: the O(n log n) algorithm for the problem! It is simple too, after you understand it. Those who guessed FFT were right.
The problem: we are given a binary string S of length n, and we want to find three evenly spaced 1s in it. For example, S may be 110110010, where n=9. It has evenly spaced 1s at positions 2, 5, and 8.
Scan S left to right, and make a list L of positions of 1. For the S=110110010 above, we have the list L = [1, 2, 4, 5, 8]. This step is O(n). The problem is now to find an arithmetic progression of length 3 in L, i.e. to find distinct a, b, c in L such that b-a = c-b, or equivalently a+c=2b. For the example above, we want to find the progression (2, 5, 8).
Make a polynomial p with terms xk for each k in L. For the example above, we make the polynomial p(x) = (x + x2 + x4 + x5+x8). This step is O(n).
Find the polynomial q = p2, using the Fast Fourier Transform. For the example above, we get the polynomial q(x) = x16 + 2x13 + 2x12 + 3x10 + 4x9 + x8 + 2x7 + 4x6 + 2x5 + x4 + 2x3 + x2. This step is O(n log n).
Ignore all terms except those corresponding to x2k for some k in L. For the example above, we get the terms x16, 3x10, x8, x4, x2. This step is O(n), if you choose to do it at all.
Here's the crucial point: the coefficient of any x2b for b in L is precisely the number of pairs (a,c) in L such that a+c=2b. [CLRS, Ex. 30.1-7] One such pair is (b,b) always (so the coefficient is at least 1), but if there exists any other pair (a,c), then the coefficient is at least 3, from (a,c) and (c,a). For the example above, we have the coefficient of x10 to be 3 precisely because of the AP (2,5,8). (These coefficients x2b will always be odd numbers, for the reasons above. And all other coefficients in q will always be even.)
So then, the algorithm is to look at the coefficients of these terms x2b, and see if any of them is greater than 1. If there is none, then there are no evenly spaced 1s. If there is a b in L for which the coefficient of x2b is greater than 1, then we know that there is some pair (a,c) — other than (b,b) — for which a+c=2b. To find the actual pair, we simply try each a in L (the corresponding c would be 2b-a) and see if there is a 1 at position 2b-a in S. This step is O(n).
That's all, folks.
One might ask: do we need to use FFT? Many answers, such as beta's, flybywire's, and rsp's, suggest that the approach that checks each pair of 1s and sees if there is a 1 at the "third" position, might work in O(n log n), based on the intuition that if there are too many 1s, we would find a triple easily, and if there are too few 1s, checking all pairs takes little time. Unfortunately, while this intuition is correct and the simple approach is better than O(n2), it is not significantly better. As in sdcvvc's answer, we can take the "Cantor-like set" of strings of length n=3k, with 1s at the positions whose ternary representation has only 0s and 2s (no 1s) in it. Such a string has 2k = n(log 2)/(log 3) ≈ n0.63 ones in it and no evenly spaced 1s, so checking all pairs would be of the order of the square of the number of 1s in it: that's 4k ≈ n1.26 which unfortunately is asymptotically much larger than (n log n). In fact, the worst case is even worse: Leo Moser in 1953 constructed (effectively) such strings which have n1-c/√(log n) 1s in them but no evenly spaced 1s, which means that on such strings, the simple approach would take Θ(n2-2c/√(log n)) — only a tiny bit better than Θ(n2), surprisingly!
About the maximum number of 1s in a string of length n with no 3 evenly spaced ones (which we saw above was at least n0.63 from the easy Cantor-like construction, and at least n1-c/√(log n) with Moser's construction) — this is OEIS A003002. It can also be calculated directly from OEIS A065825 as the k such that A065825(k) ≤ n < A065825(k+1). I wrote a program to find these, and it turns out that the greedy algorithm does not give the longest such string. For example, for n=9, we can get 5 1s (110100011) but the greedy gives only 4 (110110000), for n=26 we can get 11 1s (11001010001000010110001101) but the greedy gives only 8 (11011000011011000000000000), and for n=74 we can get 22 1s (11000010110001000001011010001000000000000000010001011010000010001101000011) but the greedy gives only 16 (11011000011011000000000000011011000011011000000000000000000000000000000000). They do agree at quite a few places until 50 (e.g. all of 38 to 50), though. As the OEIS references say, it seems that Jaroslaw Wroblewski is interested in this question, and he maintains a website on these non-averaging sets. The exact numbers are known only up to 194.
Your problem is called AVERAGE in this paper (1999):
A problem is 3SUM-hard if there is a sub-quadratic reduction from the problem 3SUM: Given a set A of n integers, are there elements a,b,c in A such that a+b+c = 0? It is not known whether AVERAGE is 3SUM-hard. However, there is a simple linear-time reduction from AVERAGE to 3SUM, whose description we omit.
Wikipedia:
When the integers are in the range [−u ... u], 3SUM can be solved in time O(n + u lg u) by representing S as a bit vector and performing a convolution using FFT.
This is enough to solve your problem :).
What is very important is that O(n log n) is complexity in terms of number of zeroes and ones, not the count of ones (which could be given as an array, like [1,5,9,15]). Checking if a set has an arithmetic progression, terms of number of 1's, is hard, and according to that paper as of 1999 no faster algorithm than O(n2) is known, and is conjectured that it doesn't exist. Everybody who doesn't take this into account is attempting to solve an open problem.
Other interesting info, mostly irrevelant:
Lower bound:
An easy lower bound is Cantor-like set (numbers 1..3^n-1 not containing 1 in their ternary expansion) - its density is n^(log_3 2) (circa 0.631). So any checking if the set isn't too large, and then checking all pairs is not enough to get O(n log n). You have to investigate the sequence smarter. A better lower bound is quoted here - it's n1-c/(log(n))^(1/2). This means Cantor set is not optimal.
Upper bound - my old algorithm:
It is known that for large n, a subset of {1,2,...,n} not containing arithmetic progression has at most n/(log n)^(1/20) elements. The paper On triples in arithmetic progression proves more: the set cannot contain more than n * 228 * (log log n / log n)1/2 elements. So you could check if that bound is achieved and if not, naively check pairs. This is O(n2 * log log n / log n) algorithm, faster than O(n2). Unfortunately "On triples..." is on Springer - but the first page is available, and Ben Green's exposition is available here, page 28, theorem 24.
By the way, the papers are from 1999 - the same year as the first one I mentioned, so that's probably why the first one doesn't mention that result.
This is not a solution, but a similar line of thought to what Olexiy was thinking
I was playing around with creating sequences with maximum number of ones, and they are all quite interesting, I got up to 125 digits and here are the first 3 numbers it found by attempting to insert as many '1' bits as possible:
11011000011011000000000000001101100001101100000000000000000000000000000000000000000110110000110110000000000000011011000011011
10110100010110100000000000010110100010110100000000000000000000000000000000000000000101101000101101000000000000101101000101101
10011001010011001000000000010011001010011001000000000000000000000000000000000000010011001010011001000000000010011001010011001
Notice they are all fractals (not too surprising given the constraints). There may be something in thinking backwards, perhaps if the string is not a fractal of with a characteristic, then it must have a repeating pattern?
Thanks to beta for the better term to describe these numbers.
Update:
Alas it looks like the pattern breaks down when starting with a large enough initial string, such as: 10000000000001:
100000000000011
10000000000001101
100000000000011011
10000000000001101100001
100000000000011011000011
10000000000001101100001101
100000000000011011000011010000000001
100000000000011011000011010000000001001
1000000000000110110000110100000000010011
1000000000000110110000110100000000010011001
10000000000001101100001101000000000100110010000000001
10000000000001101100001101000000000100110010000000001000001
1000000000000110110000110100000000010011001000000000100000100000000000001
10000000000001101100001101000000000100110010000000001000001000000000000011
1000000000000110110000110100000000010011001000000000100000100000000000001101
100000000000011011000011010000000001001100100000000010000010000000000000110100001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011000000001100100000000100100000000000010000000010000100000100100010010000010000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000110000010000000000000000000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011000000001100100000000100100000000000010000000010000100000100100010010000010000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000110000010000000000000000000001001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011000001000000000000000000000100100000000000000000000000000000000000011
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011000001000000000000000000000100100000000000000000000000000000000000011001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001100100000000000000000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001100100000000000000000000001001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001100100000000000000000000001001000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011000001000000000000000000000100100000000000000000000000000000000000011001000000000000000000000010010000010000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011000000001100100000000100100000000000010000000010000100000100100010010000010000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000110000010000000000000000000001001000000000000000000000000000000000000110010000000000000000000000100100000100000011
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001100100000000000000000000001001000001000000110000000000001
I suspect that a simple approach that looks like O(n^2) will actually yield something better, like O(n ln(n)). The sequences that take the longest to test (for any given n) are the ones that contain no trios, and that puts severe restrictions on the number of 1's that can be in the sequence.
I've come up with some hand-waving arguments, but I haven't been able to find a tidy proof. I'm going to take a stab in the dark: the answer is a very clever idea that the professor has known for so long that it's come to seem obvious, but it's much too hard for the students. (Either that or you slept through the lecture that covered it.)
Revision: 2009-10-17 23:00
I've run this on large numbers (like, strings of 20 million) and I now believe this algorithm is not O(n logn). Notwithstanding that, it's a cool enough implementation and contains a number of optimizations that makes it run really fast. It evaluates all the arrangements of binary strings 24 or fewer digits in under 25 seconds.
I've updated the code to include the 0 <= L < M < U <= X-1 observation from earlier today.
Original
This is, in concept, similar to another question I answered. That code also looked at three values in a series and determined if a triplet satisfied a condition. Here is C# code adapted from that:
using System;
using System.Collections.Generic;
namespace StackOverflow1560523
{
class Program
{
public struct Pair<T>
{
public T Low, High;
}
static bool FindCandidate(int candidate,
List<int> arr,
List<int> pool,
Pair<int> pair,
ref int iterations)
{
int lower = pair.Low, upper = pair.High;
while ((lower >= 0) && (upper < pool.Count))
{
int lowRange = candidate - arr[pool[lower]];
int highRange = arr[pool[upper]] - candidate;
iterations++;
if (lowRange < highRange)
lower -= 1;
else if (lowRange > highRange)
upper += 1;
else
return true;
}
return false;
}
static List<int> BuildOnesArray(string s)
{
List<int> arr = new List<int>();
for (int i = 0; i < s.Length; i++)
if (s[i] == '1')
arr.Add(i);
return arr;
}
static void BuildIndexes(List<int> arr,
ref List<int> even, ref List<int> odd,
ref List<Pair<int>> evenIndex, ref List<Pair<int>> oddIndex)
{
for (int i = 0; i < arr.Count; i++)
{
bool isEven = (arr[i] & 1) == 0;
if (isEven)
{
evenIndex.Add(new Pair<int> {Low=even.Count-1, High=even.Count+1});
oddIndex.Add(new Pair<int> {Low=odd.Count-1, High=odd.Count});
even.Add(i);
}
else
{
oddIndex.Add(new Pair<int> {Low=odd.Count-1, High=odd.Count+1});
evenIndex.Add(new Pair<int> {Low=even.Count-1, High=even.Count});
odd.Add(i);
}
}
}
static int FindSpacedOnes(string s)
{
// List of indexes of 1s in the string
List<int> arr = BuildOnesArray(s);
//if (s.Length < 3)
// return 0;
// List of indexes to odd indexes in arr
List<int> odd = new List<int>(), even = new List<int>();
// evenIndex has indexes into arr to bracket even numbers
// oddIndex has indexes into arr to bracket odd numbers
List<Pair<int>> evenIndex = new List<Pair<int>>(),
oddIndex = new List<Pair<int>>();
BuildIndexes(arr,
ref even, ref odd,
ref evenIndex, ref oddIndex);
int iterations = 0;
for (int i = 1; i < arr.Count-1; i++)
{
int target = arr[i];
bool found = FindCandidate(target, arr, odd, oddIndex[i], ref iterations) ||
FindCandidate(target, arr, even, evenIndex[i], ref iterations);
if (found)
return iterations;
}
return iterations;
}
static IEnumerable<string> PowerSet(int n)
{
for (long i = (1L << (n-1)); i < (1L << n); i++)
{
yield return Convert.ToString(i, 2).PadLeft(n, '0');
}
}
static void Main(string[] args)
{
for (int i = 5; i < 64; i++)
{
int c = 0;
string hardest_string = "";
foreach (string s in PowerSet(i))
{
int cost = find_spaced_ones(s);
if (cost > c)
{
hardest_string = s;
c = cost;
Console.Write("{0} {1} {2}\r", i, c, hardest_string);
}
}
Console.WriteLine("{0} {1} {2}", i, c, hardest_string);
}
}
}
}
The principal differences are:
Exhaustive search of solutions
This code generates a power set of data to find the hardest input to solve for this algorithm.
All solutions versus hardest to solve
The code for the previous question generated all the solutions using a python generator. This code just displays the hardest for each pattern length.
Scoring algorithm
This code checks the distance from the middle element to its left- and right-hand edge. The python code tested whether a sum was above or below 0.
Convergence on a candidate
The current code works from the middle towards the edge to find a candidate. The code in the previous problem worked from the edges towards the middle. This last change gives a large performance improvement.
Use of even and odd pools
Based on the observations at the end of this write-up, the code searches pairs of even numbers of pairs of odd numbers to find L and U, keeping M fixed. This reduces the number of searches by pre-computing information. Accordingly, the code uses two levels of indirection in the main loop of FindCandidate and requires two calls to FindCandidate for each middle element: once for even numbers and once for odd ones.
The general idea is to work on indexes, not the raw representation of the data. Calculating an array where the 1's appear allows the algorithm to run in time proportional to the number of 1's in the data rather than in time proportional to the length of the data. This is a standard transformation: create a data structure that allows faster operation while keeping the problem equivalent.
The results are out of date: removed.
Edit: 2009-10-16 18:48
On yx's data, which is given some credence in the other responses as representative of hard data to calculate on, I get these results... I removed these. They are out of date.
I would point out that this data is not the hardest for my algorithm, so I think the assumption that yx's fractals are the hardest to solve is mistaken. The worst case for a particular algorithm, I expect, will depend upon the algorithm itself and will not likely be consistent across different algorithms.
Edit: 2009-10-17 13:30
Further observations on this.
First, convert the string of 0's and 1's into an array of indexes for each position of the 1's. Say the length of that array A is X. Then the goal is to find
0 <= L < M < U <= X-1
such that
A[M] - A[L] = A[U] - A[M]
or
2*A[M] = A[L] + A[U]
Since A[L] and A[U] sum to an even number, they can't be (even, odd) or (odd, even). The search for a match could be improved by splitting A[] into odd and even pools and searching for matches on A[M] in the pools of odd and even candidates in turn.
However, this is more of a performance optimization than an algorithmic improvement, I think. The number of comparisons should drop, but the order of the algorithm should be the same.
Edit 2009-10-18 00:45
Yet another optimization occurs to me, in the same vein as separating the candidates into even and odd. Since the three indexes have to add to a multiple of 3 (a, a+x, a+2x -- mod 3 is 0, regardless of a and x), you can separate L, M, and U into their mod 3 values:
M L U
0 0 0
1 2
2 1
1 0 2
1 1
2 0
2 0 1
1 0
2 2
In fact, you could combine this with the even/odd observation and separate them into their mod 6 values:
M L U
0 0 0
1 5
2 4
3 3
4 2
5 1
and so on. This would provide a further performance optimization but not an algorithmic speedup.
Wasn't able to come up with the solution yet :(, but have some ideas.
What if we start from a reverse problem: construct a sequence with the maximum number of 1s and WITHOUT any evenly spaced trios. If you can prove the maximum number of 1s is o(n), then you can improve your estimate by iterating only through list of 1s only.
This may help....
This problem reduces to the following:
Given a sequence of positive integers, find a contiguous subsequence partitioned into a prefix and a suffix such that the sum of the prefix of the subsequence is equal to the sum of the suffix of the subsequence.
For example, given a sequence of [ 3, 5, 1, 3, 6, 5, 2, 2, 3, 5, 6, 4 ], we would find a subsequence of [ 3, 6, 5, 2, 2] with a prefix of [ 3, 6 ] with prefix sum of 9 and a suffix of [ 5, 2, 2 ] with suffix sum of 9.
The reduction is as follows:
Given a sequence of zeros and ones, and starting at the leftmost one, continue moving to the right. Each time another one is encountered, record the number of moves since the previous one was encountered and append that number to the resulting sequence.
For example, given a sequence of [ 0, 1, 1, 0, 0, 1, 0, 0, 0, 1 0 ], we would find the reduction of [ 1, 3, 4]. From this reduction, we calculate the contiguous subsequence of [ 1, 3, 4], the prefix of [ 1, 3] with sum of 4, and the suffix of [ 4 ] with sum of 4.
This reduction may be computed in O(n).
Unfortunately, I am not sure where to go from here.
For the simple problem type (i.e. you search three "1" with only (i.e. zero or more) "0" between it), Its quite simple: You could just split the sequence at every "1" and look for two adjacent subsequences having the same length (the second subsequence not being the last one, of course). Obviously, this can be done in O(n) time.
For the more complex version (i.e. you search an index i and an gap g>0 such that s[i]==s[i+g]==s[i+2*g]=="1"), I'm not sure, if there exists an O(n log n) solution, since there are possibly O(n²) triplets having this property (think of a string of all ones, there are approximately n²/2 such triplets). Of course, you are looking for only one of these, but I have currently no idea, how to find it ...
A fun question, but once you realise that the actual pattern between two '1's does not matter, the algorithm becomes:
scan look for a '1'
starting from the next position scan for another '1' (to the end of the array minus the distance from the current first '1' or else the 3rd '1' would be out of bounds)
if at the position of the 2nd '1' plus the distance to the first 1' a third '1' is found, we have evenly spaces ones.
In code, JTest fashion, (Note this code isn't written to be most efficient and I added some println's to see what happens.)
import java.util.Random;
import junit.framework.TestCase;
public class AlgorithmTest extends TestCase {
/**
* Constructor for GetNumberTest.
*
* #param name The test's name.
*/
public AlgorithmTest(String name) {
super(name);
}
/**
* #see TestCase#setUp()
*/
protected void setUp() throws Exception {
super.setUp();
}
/**
* #see TestCase#tearDown()
*/
protected void tearDown() throws Exception {
super.tearDown();
}
/**
* Tests the algorithm.
*/
public void testEvenlySpacedOnes() {
assertFalse(isEvenlySpaced(1));
assertFalse(isEvenlySpaced(0x058003));
assertTrue(isEvenlySpaced(0x07001));
assertTrue(isEvenlySpaced(0x01007));
assertTrue(isEvenlySpaced(0x101010));
// some fun tests
Random random = new Random();
isEvenlySpaced(random.nextLong());
isEvenlySpaced(random.nextLong());
isEvenlySpaced(random.nextLong());
}
/**
* #param testBits
*/
private boolean isEvenlySpaced(long testBits) {
String testString = Long.toBinaryString(testBits);
char[] ones = testString.toCharArray();
final char ONE = '1';
for (int n = 0; n < ones.length - 1; n++) {
if (ONE == ones[n]) {
for (int m = n + 1; m < ones.length - m + n; m++) {
if (ONE == ones[m] && ONE == ones[m + m - n]) {
System.out.println(" IS evenly spaced: " + testBits + '=' + testString);
System.out.println(" at: " + n + ", " + m + ", " + (m + m - n));
return true;
}
}
}
}
System.out.println("NOT evenly spaced: " + testBits + '=' + testString);
return false;
}
}
I thought of a divide-and-conquer approach that might work.
First, in preprocessing you need to insert all numbers less than one half your input size (n/3) into a list.
Given a string: 0000010101000100 (note that this particular example is valid)
Insert all primes (and 1) from 1 to (16/2) into a list: {1, 2, 3, 4, 5, 6, 7}
Then divide it in half:
100000101 01000100
Keep doing this until you get to strings of size 1. For all size-one strings with a 1 in them, add the index of the string to the list of possibilities; otherwise, return -1 for failure.
You'll also need to return a list of still-possible spacing distances, associated with each starting index. (Start with the list you made above and remove numbers as you go) Here, an empty list means you're only dealing with one 1 and so any spacing is possible at this point; otherwise the list includes spacings that must be ruled out.
So continuing with the example above:
1000 0101 0100 0100
10 00 01 01 01 00 01 00
1 0 0 0 0 1 0 1 0 1 0 0 0 1 0 0
In the first combine step, we have eight sets of two now. In the first, we have the possibility of a set, but we learn that spacing by 1 is impossible because of the other zero being there. So we return 0 (for the index) and {2,3,4,5,7} for the fact that spacing by 1 is impossible. In the second, we have nothing and so return -1. In the third we have a match with no spacings eliminated in index 5, so return 5, {1,2,3,4,5,7}. In the fourth pair we return 7, {1,2,3,4,5,7}. In the fifth, return 9, {1,2,3,4,5,7}. In the sixth, return -1. In the seventh, return 13, {1,2,3,4,5,7}. In the eighth, return -1.
Combining again into four sets of four, we have:
1000: Return (0, {4,5,6,7})
0101: Return (5, {2,3,4,5,6,7}), (7, {1,2,3,4,5,6,7})
0100: Return (9, {3,4,5,6,7})
0100: Return (13, {3,4,5,6,7})
Combining into sets of eight:
10000101: Return (0, {5,7}), (5, {2,3,4,5,6,7}), (7, {1,2,3,4,5,6,7})
01000100: Return (9, {4,7}), (13, {3,4,5,6,7})
Combining into a set of sixteen:
10000101 01000100
As we've progressed, we keep checking all the possibilities so far. Up to this step we've left stuff that went beyond the end of the string, but now we can check all the possibilities.
Basically, we check the first 1 with spacings of 5 and 7, and find that they don't line up to 1's. (Note that each check is CONSTANT, not linear time) Then we check the second one (index 5) with spacings of 2, 3, 4, 5, 6, and 7-- or we would, but we can stop at 2 since that actually matches up.
Phew! That's a rather long algorithm.
I don't know 100% if it's O(n log n) because of the last step, but everything up to there is definitely O(n log n) as far as I can tell. I'll get back to this later and try to refine the last step.
EDIT: Changed my answer to reflect Welbog's comment. Sorry for the error. I'll write some pseudocode later, too, when I get a little more time to decipher what I wrote again. ;-)
I'll give my rough guess here, and let those who are better with calculating complexity to help me on how my algorithm fares in O-notation wise
given binary string 0000010101000100 (as example)
crop head and tail of zeroes -> 00000 101010001 00
we get 101010001 from previous calculation
check if the middle bit is 'one', if true, found valid three evenly spaced 'ones' (only if the number of bits is odd numbered)
correlatively, if the remained cropped number of bits is even numbered, the head and tail 'one' cannot be part of evenly spaced 'one',
we use 1010100001 as example (with an extra 'zero' to become even numbered crop), in this case we need to crop again, then becomes -> 10101 00001
we get 10101 from previous calculation, and check middle bit, and we found the evenly spaced bit again
I have no idea how to calculate complexity for this, can anyone help?
edit: add some code to illustrate my idea
edit2: tried to compile my code and found some major mistakes, fixed
char *binaryStr = "0000010101000100";
int main() {
int head, tail, pos;
head = 0;
tail = strlen(binaryStr)-1;
if( (pos = find3even(head, tail)) >=0 )
printf("found it at position %d\n", pos);
return 0;
}
int find3even(int head, int tail) {
int pos = 0;
if(head >= tail) return -1;
while(binaryStr[head] == '0')
if(head<tail) head++;
while(binaryStr[tail] == '0')
if(head<tail) tail--;
if(head >= tail) return -1;
if( (tail-head)%2 == 0 && //true if odd numbered
(binaryStr[head + (tail-head)/2] == '1') ) {
return head;
}else {
if( (pos = find3even(head, tail-1)) >=0 )
return pos;
if( (pos = find3even(head+1, tail)) >=0 )
return pos;
}
return -1;
}
I came up with something like this:
def IsSymetric(number):
number = number.strip('0')
if len(number) < 3:
return False
if len(number) % 2 == 0:
return IsSymetric(number[1:]) or IsSymetric(number[0:len(number)-2])
else:
if number[len(number)//2] == '1':
return True
return IsSymetric(number[:(len(number)//2)]) or IsSymetric(number[len(number)//2+1:])
return False
This is inspired by andycjw.
Truncate the zeros.
If even then test two substring 0 - (len-2) (skip last character) and from 1 - (len-1) (skip the first char)
If not even than if the middle char is one than we have success. Else divide the string in the midle without the midle element and check both parts.
As to the complexity this might be O(nlogn) as in each recursion we are dividing by two.
Hope it helps.
Ok, I'm going to take another stab at the problem. I think I can prove a O(n log(n)) algorithm that is similar to those already discussed by using a balanced binary tree to store distances between 1's. This approach was inspired by Justice's observation about reducing the problem to a list of distances between the 1's.
Could we scan the input string to construct a balanced binary tree around the position of 1's such that each node stores the position of the 1 and each edge is labeled with the distance to the adjacent 1 for each child node. For example:
10010001 gives the following tree
3
/ \
2 / \ 3
/ \
0 7
This can be done in O(n log(n)) since, for a string of size n, each insertion takes O(log(n)) in the worst case.
Then the problem is to search the tree to discover whether, at any node, there is a path from that node through the left-child that has the same distance as a path through the right child. This can be done recursively on each subtree. When merging two subtrees in the search, we must compare the distances from paths in the left subtree with distances from paths in the right. Since the number of paths in a subtree will be proportional to log(n), and the number of nodes is n, I believe this can be done in O(n log(n)) time.
Did I miss anything?
This seemed liked a fun problem so I decided to try my hand at it.
I am making the assumption that 111000001 would find the first 3 ones and be successful. Essentially the number of zeroes following the 1 is the important thing, since 0111000 is the same as 111000 according to your definition. Once you find two cases of 1, the next 1 found completes the trilogy.
Here it is in Python:
def find_three(bstring):
print bstring
dict = {}
lastone = -1
zerocount = 0
for i in range(len(bstring)):
if bstring[i] == '1':
print i, ': 1'
if lastone != -1:
if(zerocount in dict):
dict[zerocount].append(lastone)
if len(dict[zerocount]) == 2:
dict[zerocount].append(i)
return True, dict
else:
dict[zerocount] = [lastone]
lastone = i
zerocount = 0
else:
zerocount = zerocount + 1
#this is really just book keeping, as we have failed at this point
if lastone != -1:
if(zerocount in dict):
dict[zerocount].append(lastone)
else:
dict[zerocount] = [lastone]
return False, dict
This is a first try, so I'm sure this could be written in a cleaner manner. Please list the cases where this method fails down below.
I assume the reason this is nlog(n) is due to the following:
To find the 1 that is the start of the triplet, you need to check (n-2) characters. If you haven't found it by that point, you won't (chars n-1 and n cannot start a triplet) (O(n))
To find the second 1 that is the part of the triplet (started by the first one), you need to check m/2 (m=n-x, where x is the offset of the first 1) characters. This is because, if you haven't found the second 1 by the time you're halfway from the first one to the end, you won't... since the third 1 must be exactly the same distance past the second. (O(log(n)))
It O(1) to find the last 1 since you know the index it must be at by the time you find the first and second.
So, you have n, log(n), and 1... O(nlogn)
Edit: Oops, my bad. My brain had it set that n/2 was logn... which it obviously isn't (doubling the number on items still doubles the number of iterations on the inner loop). This is still at n^2, not solving the problem. Well, at least I got to write some code :)
Implementation in Tcl
proc get-triplet {input} {
for {set first 0} {$first < [string length $input]-2} {incr first} {
if {[string index $input $first] != 1} {
continue
}
set start [expr {$first + 1}]
set end [expr {1+ $first + (([string length $input] - $first) /2)}]
for {set second $start} {$second < $end} {incr second} {
if {[string index $input $second] != 1} {
continue
}
set last [expr {($second - $first) + $second}]
if {[string index $input $last] == 1} {
return [list $first $second $last]
}
}
}
return {}
}
get-triplet 10101 ;# 0 2 4
get-triplet 10111 ;# 0 2 4
get-triplet 11100000 ;# 0 1 2
get-triplet 0100100100 ;# 1 4 7
I think I have found a way of solving the problem, but I can't construct a formal proof. The solution I made is written in Java, and it uses a counter 'n' to count how many list/array accesses it does. So n should be less than or equal to stringLength*log(stringLength) if it is correct. I tried it for the numbers 0 to 2^22, and it works.
It starts by iterating over the input string and making a list of all the indexes which hold a one. This is just O(n).
Then from the list of indexes it picks a firstIndex, and a secondIndex which is greater than the first. These two indexes must hold ones, because they are in the list of indexes. From there the thirdIndex can be calculated. If the inputString[thirdIndex] is a 1 then it halts.
public static int testString(String input){
//n is the number of array/list accesses in the algorithm
int n=0;
//Put the indices of all the ones into a list, O(n)
ArrayList<Integer> ones = new ArrayList<Integer>();
for(int i=0;i<input.length();i++){
if(input.charAt(i)=='1'){
ones.add(i);
}
}
//If less than three ones in list, just stop
if(ones.size()<3){
return n;
}
int firstIndex, secondIndex, thirdIndex;
for(int x=0;x<ones.size()-2;x++){
n++;
firstIndex = ones.get(x);
for(int y=x+1; y<ones.size()-1; y++){
n++;
secondIndex = ones.get(y);
thirdIndex = secondIndex*2 - firstIndex;
if(thirdIndex >= input.length()){
break;
}
n++;
if(input.charAt(thirdIndex) == '1'){
//This case is satisfied if it has found three evenly spaced ones
//System.out.println("This one => " + input);
return n;
}
}
}
return n;
}
additional note: the counter n is not incremented when it iterates over the input string to construct the list of indexes. This operation is O(n), so it won't have an effect on the algorithm complexity anyway.
One inroad into the problem is to think of factors and shifting.
With shifting, you compare the string of ones and zeroes with a shifted version of itself. You then take matching ones. Take this example shifted by two:
1010101010
1010101010
------------
001010101000
The resulting 1's (bitwise ANDed), must represent all those 1's which are evenly spaced by two. The same example shifted by three:
1010101010
1010101010
-------------
0000000000000
In this case there are no 1's which are evenly spaced three apart.
So what does this tell you? Well that you only need to test shifts which are prime numbers. For example say you have two 1's which are six apart. You would only have to test 'two' shifts and 'three' shifts (since these divide six). For example:
10000010
10000010 (Shift by two)
10000010
10000010 (We have a match)
10000010
10000010 (Shift by three)
10000010 (We have a match)
So the only shifts you ever need to check are 2,3,5,7,11,13 etc. Up to the prime closest to the square root of size of the string of digits.
Nearly solved?
I think I am closer to a solution. Basically:
Scan the string for 1's. For each 1 note it's remainder after taking a modulus of its position. The modulus ranges from 1 to half the size of the string. This is because the largest possible separation size is half the string. This is done in O(n^2). BUT. Only prime moduli need be checked so O(n^2/log(n))
Sort the list of modulus/remainders in order largest modulus first, this can be done in O(n*log(n)) time.
Look for three consecutive moduli/remainders which are the same.
Somehow retrieve the position of the ones!
I think the biggest clue to the answer, is that the fastest sort algorithms, are O(n*log(n)).
WRONG
Step 1 is wrong as pointed out by a colleague. If we have 1's at position 2,12 and 102. Then taking a modulus of 10, they would all have the same remainders, and yet are not equally spaced apart! Sorry.
Here are some thoughts that, despite my best efforts, will not seem to wrap themselves up in a bow. Still, they might be a useful starting point for someone's analysis.
Consider the proposed solution as follows, which is the approach that several folks have suggested, including myself in a prior version of this answer. :)
Trim leading and trailing zeroes.
Scan the string looking for 1's.
When a 1 is found:
Assume that it is the middle 1 of the solution.
For each prior 1, use its saved position to compute the anticipated position of the final 1.
If the computed position is after the end of the string it cannot be part of the solution, so drop the position from the list of candidates.
Check the solution.
If the solution was not found, add the current 1 to the list of candidates.
Repeat until no more 1's are found.
Now consider input strings strings like the following, which will not have a solution:
101
101001
1010010001
101001000100001
101001000100001000001
In general, this is the concatenation of k strings of the form j 0's followed by a 1 for j from zero to k-1.
k=2 101
k=3 101001
k=4 1010010001
k=5 101001000100001
k=6 101001000100001000001
Note that the lengths of the substrings are 1, 2, 3, etc. So, problem size n has substrings of lengths 1 to k such that n = k(k+1)/2.
k=2 n= 3 101
k=3 n= 6 101001
k=4 n=10 1010010001
k=5 n=15 101001000100001
k=6 n=21 101001000100001000001
Note that k also tracks the number of 1's that we have to consider. Remember that every time we see a 1, we need to consider all the 1's seen so far. So when we see the second 1, we only consider the first, when we see the third 1, we reconsider the first two, when we see the fourth 1, we need to reconsider the first three, and so on. By the end of the algorithm, we've considered k(k-1)/2 pairs of 1's. Call that p.
k=2 n= 3 p= 1 101
k=3 n= 6 p= 3 101001
k=4 n=10 p= 6 1010010001
k=5 n=15 p=10 101001000100001
k=6 n=21 p=15 101001000100001000001
The relationship between n and p is that n = p + k.
The process of going through the string takes O(n) time. Each time a 1 is encountered, a maximum of (k-1) comparisons are done. Since n = k(k+1)/2, n > k**2, so sqrt(n) > k. This gives us O(n sqrt(n)) or O(n**3/2). Note however that may not be a really tight bound, because the number of comparisons goes from 1 to a maximum of k, it isn't k the whole time. But I'm not sure how to account for that in the math.
It still isn't O(n log(n)). Also, I can't prove those inputs are the worst cases, although I suspect they are. I think a denser packing of 1's to the front results in an even sparser packing at the end.
Since someone may still find it useful, here's my code for that solution in Perl:
#!/usr/bin/perl
# read input as first argument
my $s = $ARGV[0];
# validate the input
$s =~ /^[01]+$/ or die "invalid input string\n";
# strip leading and trailing 0's
$s =~ s/^0+//;
$s =~ s/0+$//;
# prime the position list with the first '1' at position 0
my #p = (0);
# start at position 1, which is the second character
my $i = 1;
print "the string is $s\n\n";
while ($i < length($s)) {
if (substr($s, $i, 1) eq '1') {
print "found '1' at position $i\n";
my #t = ();
# assuming this is the middle '1', go through the positions
# of all the prior '1's and check whether there's another '1'
# in the correct position after this '1' to make a solution
while (scalar #p) {
# $p is the position of the prior '1'
my $p = shift #p;
# $j is the corresponding position for the following '1'
my $j = 2 * $i - $p;
# if $j is off the end of the string then we don't need to
# check $p anymore
next if ($j >= length($s));
print "checking positions $p, $i, $j\n";
if (substr($s, $j, 1) eq '1') {
print "\nsolution found at positions $p, $i, $j\n";
exit 0;
}
# if $j isn't off the end of the string, keep $p for next time
push #t, $p;
}
#p = #t;
# add this '1' to the list of '1' positions
push #p, $i;
}
$i++;
}
print "\nno solution found\n";
While scanning 1s, add their positions to a List. When adding the second and successive 1s, compare them to each position in the list so far. Spacing equals currentOne (center) - previousOne (left). The right-side bit is currentOne + spacing. If it's 1, the end.
The list of ones grows inversely with the space between them. Simply stated, if you've got a lot of 0s between the 1s (as in a worst case), your list of known 1s will grow quite slowly.
using System;
using System.Collections.Generic;
namespace spacedOnes
{
class Program
{
static int[] _bits = new int[8] {128, 64, 32, 16, 8, 4, 2, 1};
static void Main(string[] args)
{
var bytes = new byte[4];
var r = new Random();
r.NextBytes(bytes);
foreach (var b in bytes) {
Console.Write(getByteString(b));
}
Console.WriteLine();
var bitCount = bytes.Length * 8;
var done = false;
var onePositions = new List<int>();
for (var i = 0; i < bitCount; i++)
{
if (isOne(bytes, i)) {
if (onePositions.Count > 0) {
foreach (var knownOne in onePositions) {
var spacing = i - knownOne;
var k = i + spacing;
if (k < bitCount && isOne(bytes, k)) {
Console.WriteLine("^".PadLeft(knownOne + 1) + "^".PadLeft(spacing) + "^".PadLeft(spacing));
done = true;
break;
}
}
}
if (done) {
break;
}
onePositions.Add(i);
}
}
Console.ReadKey();
}
static String getByteString(byte b) {
var s = new char[8];
for (var i=0; i<s.Length; i++) {
s[i] = ((b & _bits[i]) > 0 ? '1' : '0');
}
return new String(s);
}
static bool isOne(byte[] bytes, int i)
{
var byteIndex = i / 8;
var bitIndex = i % 8;
return (bytes[byteIndex] & _bits[bitIndex]) > 0;
}
}
}
I thought I'd add one comment before posting the 22nd naive solution to the problem. For the naive solution, we don't need to show that the number of 1's in the string is at most O(log(n)), but rather that it is at most O(sqrt(n*log(n)).
Solver:
def solve(Str):
indexes=[]
#O(n) setup
for i in range(len(Str)):
if Str[i]=='1':
indexes.append(i)
#O((number of 1's)^2) processing
for i in range(len(indexes)):
for j in range(i+1, len(indexes)):
indexDiff = indexes[j] - indexes[i]
k=indexes[j] + indexDiff
if k<len(Str) and Str[k]=='1':
return True
return False
It's basically a fair bit similar to flybywire's idea and implementation, though looking ahead instead of back.
Greedy String Builder:
#assumes final char hasn't been added, and would be a 1
def lastCharMakesSolvable(Str):
endIndex=len(Str)
j=endIndex-1
while j-(endIndex-j) >= 0:
k=j-(endIndex-j)
if k >= 0 and Str[k]=='1' and Str[j]=='1':
return True
j=j-1
return False
def expandString(StartString=''):
if lastCharMakesSolvable(StartString):
return StartString + '0'
return StartString + '1'
n=1
BaseStr=""
lastCount=0
while n<1000000:
BaseStr=expandString(BaseStr)
count=BaseStr.count('1')
if count != lastCount:
print(len(BaseStr), count)
lastCount=count
n=n+1
(In my defense, I'm still in the 'learn python' stage of understanding)
Also, potentially useful output from the greedy building of strings, there's a rather consistent jump after hitting a power of 2 in the number of 1's... which I was not willing to wait around to witness hitting 2096.
strlength # of 1's
1 1
2 2
4 3
5 4
10 5
14 8
28 9
41 16
82 17
122 32
244 33
365 64
730 65
1094 128
2188 129
3281 256
6562 257
9842 512
19684 513
29525 1024
I'll try to present a mathematical approach. This is more a beginning than an end, so any help, comment, or even contradiction - will be deeply appreciated. However, if this approach is proven - the algorithm is a straight-forward search in the string.
Given a fixed number of spaces k and a string S, the search for a k-spaced-triplet takes O(n) - We simply test for every 0<=i<=(n-2k) if S[i]==S[i+k]==S[i+2k]. The test takes O(1) and we do it n-k times where k is a constant, so it takes O(n-k)=O(n).
Let us assume that there is an Inverse Proportion between the number of 1's and the maximum spaces we need to search for. That is, If there are many 1's, there must be a triplet and it must be quite dense; If there are only few 1's, The triplet (if any) can be quite sparse. In other words, I can prove that if I have enough 1's, such triplet must exist - and the more 1's I have, a more dense triplet must be found. This can be explained by the Pigeonhole principle - Hope to elaborate on this later.
Say have an upper bound k on the possible number of spaces I have to look for. Now, for each 1 located in S[i] we need to check for 1 in S[i-1] and S[i+1], S[i-2] and S[i+2], ... S[i-k] and S[i+k]. This takes O((k^2-k)/2)=O(k^2) for each 1 in S - due to Gauss' Series Summation Formula. Note that this differs from section 1 - I'm having k as an upper bound for the number of spaces, not as a constant space.
We need to prove O(n*log(n)). That is, we need to show that k*(number of 1's) is proportional to log(n).
If we can do that, the algorithm is trivial - for each 1 in S whose index is i, simply look for 1's from each side up to distance k. If two were found in the same distance, return i and k. Again, the tricky part would be finding k and proving the correctness.
I would really appreciate your comments here - I have been trying to find the relation between k and the number of 1's on my whiteboard, so far without success.
Assumption:
Just wrong, talking about log(n) number of upper limit of ones
EDIT:
Now I found that using Cantor numbers (if correct), density on set is (2/3)^Log_3(n) (what a weird function) and I agree, log(n)/n density is to strong.
If this is upper limit, there is algorhitm who solves this problem in at least O(n*(3/2)^(log(n)/log(3))) time complexity and O((3/2)^(log(n)/log(3))) space complexity. (check Justice's answer for algorhitm)
This is still by far better than O(n^2)
This function ((3/2)^(log(n)/log(3))) really looks like n*log(n) on first sight.
How did I get this formula?
Applaying Cantors number on string.
Supose that length of string is 3^p == n
At each step in generation of Cantor string you keep 2/3 of prevous number of ones. Apply this p times.
That mean (n * ((2/3)^p)) -> (((3^p)) * ((2/3)^p)) remaining ones and after simplification 2^p.
This mean 2^p ones in 3^p string -> (3/2)^p ones . Substitute p=log(n)/log(3) and get
((3/2)^(log(n)/log(3)))
How about a simple O(n) solution, with O(n^2) space? (Uses the assumption that all bitwise operators work in O(1).)
The algorithm basically works in four stages:
Stage 1: For each bit in your original number, find out how far away the ones are, but consider only one direction. (I considered all the bits in the direction of the least significant bit.)
Stage 2: Reverse the order of the bits in the input;
Stage 3: Re-run step 1 on the reversed input.
Stage 4: Compare the results from Stage 1 and Stage 3. If any bits are equally spaced above AND below we must have a hit.
Keep in mind that no step in the above algorithm takes longer than O(n). ^_^
As an added benefit, this algorithm will find ALL equally spaced ones from EVERY number. So for example if you get a result of "0x0005" then there are equally spaced ones at BOTH 1 and 3 units away
I didn't really try optimizing the code below, but it is compilable C# code that seems to work.
using System;
namespace ThreeNumbers
{
class Program
{
const int uint32Length = 32;
static void Main(string[] args)
{
Console.Write("Please enter your integer: ");
uint input = UInt32.Parse(Console.ReadLine());
uint[] distancesLower = Distances(input);
uint[] distancesHigher = Distances(Reverse(input));
PrintHits(input, distancesLower, distancesHigher);
}
/// <summary>
/// Returns an array showing how far the ones away from each bit in the input. Only
/// considers ones at lower signifcant bits. Index 0 represents the least significant bit
/// in the input. Index 1 represents the second least significant bit in the input and so
/// on. If a one is 3 away from the bit in question, then the third least significant bit
/// of the value will be sit.
///
/// As programed this algorithm needs: O(n) time, and O(n*log(n)) space.
/// (Where n is the number of bits in the input.)
/// </summary>
public static uint[] Distances(uint input)
{
uint[] distanceToOnes = new uint[uint32Length];
uint result = 0;
//Sets how far each bit is from other ones. Going in the direction of LSB to MSB
for (uint bitIndex = 1, arrayIndex = 0; bitIndex != 0; bitIndex <<= 1, ++arrayIndex)
{
distanceToOnes[arrayIndex] = result;
result <<= 1;
if ((input & bitIndex) != 0)
{
result |= 1;
}
}
return distanceToOnes;
}
/// <summary>
/// Reverses the bits in the input.
///
/// As programmed this algorithm needs O(n) time and O(n) space.
/// (Where n is the number of bits in the input.)
/// </summary>
/// <param name="input"></param>
/// <returns></returns>
public static uint Reverse(uint input)
{
uint reversedInput = 0;
for (uint bitIndex = 1; bitIndex != 0; bitIndex <<= 1)
{
reversedInput <<= 1;
reversedInput |= (uint)((input & bitIndex) != 0 ? 1 : 0);
}
return reversedInput;
}
/// <summary>
/// Goes through each bit in the input, to check if there are any bits equally far away in
/// the distancesLower and distancesHigher
/// </summary>
public static void PrintHits(uint input, uint[] distancesLower, uint[] distancesHigher)
{
const int offset = uint32Length - 1;
for (uint bitIndex = 1, arrayIndex = 0; bitIndex != 0; bitIndex <<= 1, ++arrayIndex)
{
//hits checks if any bits are equally spaced away from our current value
bool isBitSet = (input & bitIndex) != 0;
uint hits = distancesLower[arrayIndex] & distancesHigher[offset - arrayIndex];
if (isBitSet && (hits != 0))
{
Console.WriteLine(String.Format("The {0}-th LSB has hits 0x{1:x4} away", arrayIndex + 1, hits));
}
}
}
}
}
Someone will probably comment that for any sufficiently large number, bitwise operations cannot be done in O(1). You'd be right. However, I'd conjecture that every solution that uses addition, subtraction, multiplication, or division (which cannot be done by shifting) would also have that problem.
Below is a solution. There could be some little mistakes here and there, but the idea is sound.
Edit: It's not n * log(n)
PSEUDO CODE:
foreach character in the string
if the character equals 1 {
if length cache > 0 { //we can skip the first one
foreach location in the cache { //last in first out kind of order
if ((currentlocation + (currentlocation - location)) < length string)
if (string[(currentlocation + (currentlocation - location))] equals 1)
return found evenly spaced string
else
break;
}
}
remember the location of this character in a some sort of cache.
}
return didn't find evenly spaced string
C# code:
public static Boolean FindThreeEvenlySpacedOnes(String str) {
List<int> cache = new List<int>();
for (var x = 0; x < str.Length; x++) {
if (str[x] == '1') {
if (cache.Count > 0) {
for (var i = cache.Count - 1; i > 0; i--) {
if ((x + (x - cache[i])) >= str.Length)
break;
if (str[(x + (x - cache[i]))] == '1')
return true;
}
}
cache.Add(x);
}
}
return false;
}
How it works:
iteration 1:
x
|
101101001
// the location of this 1 is stored in the cache
iteration 2:
x
|
101101001
iteration 3:
a x b
| | |
101101001
//we retrieve location a out of the cache and then based on a
//we calculate b and check if te string contains a 1 on location b
//and of course we store x in the cache because it's a 1
iteration 4:
axb
|||
101101001
a x b
| | |
101101001
iteration 5:
x
|
101101001
iteration 6:
a x b
| | |
101101001
a x b
| | |
101101001
//return found evenly spaced string
Obviously we need to at least check bunches of triplets at the same time, so we need to compress the checks somehow. I have a candidate algorithm, but analyzing the time complexity is beyond my ability*time threshold.
Build a tree where each node has three children and each node contains the total number of 1's at its leaves. Build a linked list over the 1's, as well. Assign each node an allowed cost proportional to the range it covers. As long as the time we spend at each node is within budget, we'll have an O(n lg n) algorithm.
--
Start at the root. If the square of the total number of 1's below it is less than its allowed cost, apply the naive algorithm. Otherwise recurse on its children.
Now we have either returned within budget, or we know that there are no valid triplets entirely contained within one of the children. Therefore we must check the inter-node triplets.
Now things get incredibly messy. We essentially want to recurse on the potential sets of children while limiting the range. As soon as the range is constrained enough that the naive algorithm will run under budget, you do it. Enjoy implementing this, because I guarantee it will be tedious. There's like a dozen cases.
--
The reason I think that algorithm will work is because the sequences without valid triplets appear to go alternate between bunches of 1's and lots of 0's. It effectively splits the nearby search space, and the tree emulates that splitting.
The run time of the algorithm is not obvious, at all. It relies on the non-trivial properties of the sequence. If the 1's are really sparse then the naive algorithm will work under budget. If the 1's are dense, then a match should be found right away. But if the density is 'just right' (eg. near ~n^0.63, which you can achieve by setting all bits at positions with no '2' digit in base 3), I don't know if it will work. You would have to prove that the splitting effect is strong enough.
No theoretical answer here, but I wrote a quick Java program to explore the running-time behavior as a function of k and n, where n is the total bit length and k is the number of 1's. I'm with a few of the answerers who are saying that the "regular" algorithm that checks all the pairs of bit positions and looks for the 3rd bit, even though it would require O(k^2) in the worst case, in reality because the worst-case needs sparse bitstrings, is O(n ln n).
Anyway here's the program, below. It's a Monte-Carlo style program which runs a large number of trials NTRIALS for constant n, and randomly generates bitsets for a range of k-values using Bernoulli processes with ones-density constrained between limits that can be specified, and records the running time of finding or failing to find a triplet of evenly spaced ones, time measured in steps NOT in CPU time. I ran it for n=64, 256, 1024, 4096, 16384* (still running), first a test run with 500000 trials to see which k-values take the longest running time, then another test with 5000000 trials with narrowed ones-density focus to see what those values look like. The longest running times do happen with very sparse density (e.g. for n=4096 the running time peaks are in the k=16-64 range, with a gentle peak for mean runtime at 4212 steps # k=31, max runtime peaked at 5101 steps # k=58). It looks like it would take extremely large values of N for the worst-case O(k^2) step to become larger than the O(n) step where you scan the bitstring to find the 1's position indices.
package com.example.math;
import java.io.PrintStream;
import java.util.BitSet;
import java.util.Random;
public class EvenlySpacedOnesTest {
static public class StatisticalSummary
{
private int n=0;
private double min=Double.POSITIVE_INFINITY;
private double max=Double.NEGATIVE_INFINITY;
private double mean=0;
private double S=0;
public StatisticalSummary() {}
public void add(double x) {
min = Math.min(min, x);
max = Math.max(max, x);
++n;
double newMean = mean + (x-mean)/n;
S += (x-newMean)*(x-mean);
// this algorithm for mean,std dev based on Knuth TAOCP vol 2
mean = newMean;
}
public double getMax() { return (n>0)?max:Double.NaN; }
public double getMin() { return (n>0)?min:Double.NaN; }
public int getCount() { return n; }
public double getMean() { return (n>0)?mean:Double.NaN; }
public double getStdDev() { return (n>0)?Math.sqrt(S/n):Double.NaN; }
// some may quibble and use n-1 for sample std dev vs population std dev
public static void printOut(PrintStream ps, StatisticalSummary[] statistics) {
for (int i = 0; i < statistics.length; ++i)
{
StatisticalSummary summary = statistics[i];
ps.printf("%d\t%d\t%.0f\t%.0f\t%.5f\t%.5f\n",
i,
summary.getCount(),
summary.getMin(),
summary.getMax(),
summary.getMean(),
summary.getStdDev());
}
}
}
public interface RandomBernoulliProcess // see http://en.wikipedia.org/wiki/Bernoulli_process
{
public void setProbability(double d);
public boolean getNextBoolean();
}
static public class Bernoulli implements RandomBernoulliProcess
{
final private Random r = new Random();
private double p = 0.5;
public boolean getNextBoolean() { return r.nextDouble() < p; }
public void setProbability(double d) { p = d; }
}
static public class TestResult {
final public int k;
final public int nsteps;
public TestResult(int k, int nsteps) { this.k=k; this.nsteps=nsteps; }
}
////////////
final private int n;
final private int ntrials;
final private double pmin;
final private double pmax;
final private Random random = new Random();
final private Bernoulli bernoulli = new Bernoulli();
final private BitSet bits;
public EvenlySpacedOnesTest(int n, int ntrials, double pmin, double pmax) {
this.n=n; this.ntrials=ntrials; this.pmin=pmin; this.pmax=pmax;
this.bits = new BitSet(n);
}
/*
* generate random bit string
*/
private int generateBits()
{
int k = 0; // # of 1's
for (int i = 0; i < n; ++i)
{
boolean b = bernoulli.getNextBoolean();
this.bits.set(i, b);
if (b) ++k;
}
return k;
}
private int findEvenlySpacedOnes(int k, int[] pos)
{
int[] bitPosition = new int[k];
for (int i = 0, j = 0; i < n; ++i)
{
if (this.bits.get(i))
{
bitPosition[j++] = i;
}
}
int nsteps = n; // first, it takes N operations to find the bit positions.
boolean found = false;
if (k >= 3) // don't bother doing anything if there are less than 3 ones. :(
{
int lastBitSetPosition = bitPosition[k-1];
for (int j1 = 0; !found && j1 < k; ++j1)
{
pos[0] = bitPosition[j1];
for (int j2 = j1+1; !found && j2 < k; ++j2)
{
pos[1] = bitPosition[j2];
++nsteps;
pos[2] = 2*pos[1]-pos[0];
// calculate 3rd bit index that might be set;
// the other two indices point to bits that are set
if (pos[2] > lastBitSetPosition)
break;
// loop inner loop until we go out of bounds
found = this.bits.get(pos[2]);
// we're done if we find a third 1!
}
}
}
if (!found)
pos[0]=-1;
return nsteps;
}
/*
* run an algorithm that finds evenly spaced ones and returns # of steps.
*/
public TestResult run()
{
bernoulli.setProbability(pmin + (pmax-pmin)*random.nextDouble());
// probability of bernoulli process is randomly distributed between pmin and pmax
// generate bit string.
int k = generateBits();
int[] pos = new int[3];
int nsteps = findEvenlySpacedOnes(k, pos);
return new TestResult(k, nsteps);
}
public static void main(String[] args)
{
int n;
int ntrials;
double pmin = 0, pmax = 1;
try {
n = Integer.parseInt(args[0]);
ntrials = Integer.parseInt(args[1]);
if (args.length >= 3)
pmin = Double.parseDouble(args[2]);
if (args.length >= 4)
pmax = Double.parseDouble(args[3]);
}
catch (Exception e)
{
System.out.println("usage: EvenlySpacedOnesTest N NTRIALS [pmin [pmax]]");
System.exit(0);
return; // make the compiler happy
}
final StatisticalSummary[] statistics;
statistics=new StatisticalSummary[n+1];
for (int i = 0; i <= n; ++i)
{
statistics[i] = new StatisticalSummary();
}
EvenlySpacedOnesTest test = new EvenlySpacedOnesTest(n, ntrials, pmin, pmax);
int printInterval=100000;
int nextPrint = printInterval;
for (int i = 0; i < ntrials; ++i)
{
TestResult result = test.run();
statistics[result.k].add(result.nsteps);
if (i == nextPrint)
{
System.err.println(i);
nextPrint += printInterval;
}
}
StatisticalSummary.printOut(System.out, statistics);
}
}
# <algorithm>
def contains_evenly_spaced?(input)
return false if input.size < 3
one_indices = []
input.each_with_index do |digit, index|
next if digit == 0
one_indices << index
end
return false if one_indices.size < 3
previous_indexes = []
one_indices.each do |index|
if !previous_indexes.empty?
previous_indexes.each do |previous_index|
multiple = index - previous_index
success_index = index + multiple
return true if input[success_index] == 1
end
end
previous_indexes << index
end
return false
end
# </algorithm>
def parse_input(input)
input.chars.map { |c| c.to_i }
end
I'm having trouble with the worst-case scenarios with millions of digits. Fuzzing from /dev/urandom essentially gives you O(n), but I know the worst case is worse than that. I just can't tell how much worse. For small n, it's trivial to find inputs at around 3*n*log(n), but it's surprisingly hard to differentiate those from some other order of growth for this particular problem.
Can anyone who was working on worst-case inputs generate a string with length greater than say, one hundred thousand?
An adaptation of the Rabin-Karp algorithm could be possible for you.
Its complexity is 0(n) so it could help you.
Take a look http://en.wikipedia.org/wiki/Rabin-Karp_string_search_algorithm
Could this be a solution? I', not sure if it's O(nlogn) but in my opinion it's better than O(n²) because the the only way not to find a triple would be a prime number distribution.
There's room for improvement, the second found 1 could be the next first 1. Also no error checking.
#include <iostream>
#include <string>
int findIt(std::string toCheck) {
for (int i=0; i<toCheck.length(); i++) {
if (toCheck[i]=='1') {
std::cout << i << ": " << toCheck[i];
for (int j = i+1; j<toCheck.length(); j++) {
if (toCheck[j]=='1' && toCheck[(i+2*(j-i))] == '1') {
std::cout << ", " << j << ":" << toCheck[j] << ", " << (i+2*(j-i)) << ":" << toCheck[(i+2*(j-i))] << " found" << std::endl;
return 0;
}
}
}
}
return -1;
}
int main (int agrc, char* args[]) {
std::string toCheck("1001011");
findIt(toCheck);
std::cin.get();
return 0;
}
I think this algorithm has O(n log n) complexity (C++, DevStudio 2k5). Now, I don't know the details of how to analyse an algorithm to determine its complexity, so I have added some metric gathering information to the code. The code counts the number of tests done on the sequence of 1's and 0's for any given input (hopefully, I've not made a balls of the algorithm). We can compare the actual number of tests against the O value and see if there's a correlation.
#include <iostream>
using namespace std;
bool HasEvenBits (string &sequence, int &num_compares)
{
bool
has_even_bits = false;
num_compares = 0;
for (unsigned i = 1 ; i <= (sequence.length () - 1) / 2 ; ++i)
{
for (unsigned j = 0 ; j < sequence.length () - 2 * i ; ++j)
{
++num_compares;
if (sequence [j] == '1' && sequence [j + i] == '1' && sequence [j + i * 2] == '1')
{
has_even_bits = true;
// we could 'break' here, but I want to know the worst case scenario so keep going to the end
}
}
}
return has_even_bits;
}
int main ()
{
int
count;
string
input = "111";
for (int i = 3 ; i < 32 ; ++i)
{
HasEvenBits (input, count);
cout << i << ", " << count << endl;
input += "0";
}
}
This program outputs the number of tests for each string length up to 32 characters. Here's the results:
n Tests n log (n)
=====================
3 1 1.43
4 2 2.41
5 4 3.49
6 6 4.67
7 9 5.92
8 12 7.22
9 16 8.59
10 20 10.00
11 25 11.46
12 30 12.95
13 36 14.48
14 42 16.05
15 49 17.64
16 56 19.27
17 64 20.92
18 72 22.59
19 81 24.30
20 90 26.02
21 100 27.77
22 110 29.53
23 121 31.32
24 132 33.13
25 144 34.95
26 156 36.79
27 169 38.65
28 182 40.52
29 196 42.41
30 210 44.31
31 225 46.23
I've added the 'n log n' values as well. Plot these using your graphing tool of choice to see a correlation between the two results. Does this analysis extend to all values of n? I don't know.
For context, this is to control multiple stepper motors simultaneously in a high-accuracy application.
Problem statement
Say I have a loop that will run i iterations. Over the course of those iterations, expression E_x should evaluate to true x times (x <= i is guaranteed).
Requirements
- E_x must evaluate to true exactly x times
- E_x must evaluate to true at more or less evenly spaced intervals*
* "evenly spaced intervals" means that the maximum interval size is minimized
Examples
For: i = 10, x = 7
E_x will be true on iterations marked 1: 1101101101
For: i = 10, x = 3
E_x will be true on iterations marked 1: 0010010010
For: i = 10, x = 2
E_x will be true on iterations marked 1: 0001000100
What is the best (or even "a good") way to have E_x evaluate to true at evenly spaced intervals while guaranteeing that it is true exactly x times?
This question is close to mine, however it assumes that E_x will always evaluate to true in the 1st and last iterations, which does not meet my requirements (see 2nd example above).
I'll use a bit different naming convention: let's there by T intervals [1..T] and N events to be fired. Also let's solve the problem as a cyclic one. To do the let's add one fake step at the end that we are guaranteed to fire event at (and this will be also the event at time 0 i.e. before the cycle). So my T is your i+1 and my N is your x+1.
If you divide T by N with reminder you'll get T = w*N + r. If r=0 the case is trivial. If r != 0 the best you can achieve is r intervals of size w+1 and (N-r) intervals of size w. The fast and simple but good enough solution would be something like this (pseudocode):
events = []
w = T / N
r = T % N
current = 0
for(i = 1; i<=N; i++) {
current += w;
if (i <= r)
current += 1;
events[i] = current;
}
You can see that the last value in the array will be T as was promised by our re-statement as a cyclic problem. It will be T because over the cycle we'll add w to current N times and add r times 1, so the sum will be w*N+r which is T.
The main drawback of this solution is that all the "long" intervals will be at the start while all the "short" interval will be at the end.
You can spread intervals more evenly if you are a bit smarter. And the resulting logic will be essentially the same as it is behind Bresenham's line algorithm referenced in comments. Imagine you are drawing a line on a plane, where X-axis represents time and Y-axis represents events, from (0,0) (which is the 0-th event, before your timeframe) to (i+1, x+1) (which is the x+1-th event, just after your timeframe). The moment to raise an event is when you switch to the next Y i.e. draw the first pixel at a given Y.
If you want to do x increments over n iterations, you can do it like this:
int incCount = 0;
int iterCount = 0;
boolean step() {
++iterCount;
int nextCount = (iterCount*x + n/2) / n; // this is rounding division
if (nextCount > incCount) {
++incCount;
return true;
}
else {
return false;
}
}
That's the easy-to-understand way. If you're on an embedded CPU where division is more expensive, you can accomplish exactly the same thing like this:
int accum = n/2;
boolean step() {
accum+=x;
if (accum >= n) {
accum-=n;
return true;
}
else {
return false;
}
}
The total amount added to accum here is iterCount*x + n/2 just like the first example, but the division is replaced with an incremental repeated subtraction. This is the way that Bresenham's line drawing algorithm works.
Is there anyway to ensure the that the fewest number of turns heuristic is met by anything except a breadth first search? Perhaps some more explanation would help.
I have a random graph, much like this:
0 1 1 1 2
3 4 5 6 7
9 a 5 b c
9 d e f f
9 9 g h i
Starting in the top left corner, I need to know the fewest number of steps it would take to get to the bottom right corner. Each set of connected colors is assumed to be a single node, so for instance in this random graph, the three 1's on the top row are all considered a single node, and every adjacent (not diagonal) connected node is a possible next state. So from the start, possible next states are the 1's in the top row or 3 in the second row.
Currently I use a bidirectional search, but the explosiveness of the tree size ramps up pretty quickly. For the life of me, I haven't been able to adjust the problem so that I can safely assign weights to the nodes and have them ensure the fewest number of state changes to reach the goal without it turning into a breadth first search. Thinking of this as a city map, the heuristic would be the fewest number of turns to reach the goal.
It is very important that the fewest number of turns is the result of this search as that value is part of the heuristic for a more complex problem.
You said yourself each group of numbers represents one node, and each node is connected to adjascent nodes. Then this is a simple shortest-path problem, and you could use (for instance) Dijkstra's algorithm, with each edge having weight 1 (for 1 turn).
This sounds like Dijkstra's algorithm. The hardest part would lay in properly setting up the graph (keeping track of which node gets which children), but if you can devote some CPU cycles to that, you'd be fine afterwards.
Why don't you want a breadth-first search?
Here.. I was bored :-) This is in Ruby but may get you started. Mind you, it is not tested.
class Node
attr_accessor :parents, :children, :value
def initialize args={}
#parents = args[:parents] || []
#children = args[:children] || []
#value = args[:value]
end
def add_parents *args
args.flatten.each do |node|
#parents << node
node.add_children self unless node.children.include? self
end
end
def add_children *args
args.flatten.each do |node|
#children << node
node.add_parents self unless node.parents.include? self
end
end
end
class Graph
attr_accessor :graph, :root
def initialize args={}
#graph = args[:graph]
#root = Node.new
prepare_graph
#root = #graph[0][0]
end
private
def prepare_graph
# We will iterate through the graph, and only check the values above and to the
# left of the current cell.
#graph.each_with_index do |row, i|
row.each_with_index do |cell, j|
cell = Node.new :value => cell #in-place modification!
# Check above
unless i.zero?
above = #graph[i-1][j]
if above.value == cell.value
# Here it is safe to do this: the new node has no children, no parents.
cell = above
else
cell.add_parents above
above.add_children cell # Redundant given the code for both of those
# methods, but implementations may differ.
end
end
# Check to the left!
unless j.zero?
left = #graph[i][j-1]
if left.value == cell.value
# Well, potentially it's the same as the one above the current cell,
# so we can't just set one equal to the other: have to merge them.
left.add_parents cell.parents
left.add_children cell.children
cell = left
else
cell.add_parents left
left.add_children cell
end
end
end
end
end
end
#j = 0, 1, 2, 3, 4
graph = [
[3, 4, 4, 4, 2], # i = 0
[8, 3, 1, 0, 8], # i = 1
[9, 0, 1, 2, 4], # i = 2
[9, 8, 0, 3, 3], # i = 3
[9, 9, 7, 2, 5]] # i = 4
maze = Graph.new :graph => graph
# Now, going from maze.root on, we have a weighted graph, should it matter.
# If it doesn't matter, you can just count the number of steps.
# Dijkstra's algorithm is really simple to find in the wild.
This looks like same problem as this projeceuler http://projecteuler.net/index.php?section=problems&id=81
Comlexity of solution is O(n) n-> number of nodes
What you need is memoization.
At each step you can get from max 2 directions. So pick the solution that is cheaper.
It is something like (just add the code that takes 0 if on boarder)
for i in row:
for j in column:
matrix[i][j]=min([matrix[i-1][j],matrix[i][j-1]])+matrix[i][j]
And now you have lest expensive solution if you move just left or down
Solution is in matrix[MAX_i][MAX_j]
If you can go left and up too, than the BigO is much higher (I can figure out optimal solution)
In order for A* to always find the shortest path, your heuristic needs to always under-estimate the actual cost (the heuristic is "admissable"). Simple heuristics like using the Euclidean or Manhattan distance on a grid work well because they're fast to compute and are guaranteed to be less than or equal to the actual cost.
Unfortunately, in your case, unless you can make some simplifying assumptions about the size/shape of the nodes, I'm not sure there's much you can do. For example, consider going from A to B in this case:
B 1 2 3 A
C 4 5 6 D
C 7 8 9 C
C e f g C
C C C C C
The shortest path would be A -> D -> C -> B, but using spatial information would probably give 3 a lower heuristic cost than D.
Depending on your circumstances, you might be able to live with a solution that isn't actually the shortest path, as long as you can get the answer sooner. There's a nice blogpost here by Christer Ericson (progammer for God of War 3 on PS3) on the topic: http://realtimecollisiondetection.net/blog/?p=56
Here's my idea for an nonadmissable heuristic: from the point, move horizontally until you're even with the goal, then move vertically until you reach it, and count the number of state changes that you made. You can compute other test paths (e.g. vertically then horizontally) too, and pick the minimum value as your final heuristic. If your nodes are roughly equal size and regularly shaped (unlike my example), this might do pretty well. The more test paths you do, the more accurate you'd get, but the slower it would be.
Hope that's helpful, let me know if any of it doesn't make sense.
This untuned C implementation of breadth-first search can chew through a 100-by-100 grid in less than 1 msec. You can probably do better.
int shortest_path(int *grid, int w, int h) {
int mark[w * h]; // for each square in the grid:
// 0 if not visited
// 1 if not visited and slated to be visited "now"
// 2 if already visited
int todo1[4 * w * h]; // buffers for two queues, a "now" queue
int todo2[4 * w * h]; // and a "later" queue
int *readp; // read position in the "now" queue
int *writep[2] = {todo1 + 1, 0};
int x, y, same;
todo1[0] = 0;
memset(mark, 0, sizeof(mark));
for (int d = 0; ; d++) {
readp = (d & 1) ? todo2 : todo1; // start of "now" queue
writep[1] = writep[0]; // end of "now" queue
writep[0] = (d & 1) ? todo1 : todo2; // "later" queue (empty)
// Now consume the "now" queue, filling both the "now" queue
// and the "later" queue as we go. Points in the "now" queue
// have distance d from the starting square. Points in the
// "later" queue have distance d+1.
while (readp < writep[1]) {
int p = *readp++;
if (mark[p] < 2) {
mark[p] = 2;
x = p % w;
y = p / w;
if (x > 0 && !mark[p-1]) { // go left
mark[p-1] = same = (grid[p-1] == grid[p]);
*writep[same]++ = p-1;
}
if (x + 1 < w && !mark[p+1]) { // go right
mark[p+1] = same = (grid[p+1] == grid[p]);
if (y == h - 1 && x == w - 2)
return d + !same;
*writep[same]++ = p+1;
}
if (y > 0 && !mark[p-w]) { // go up
mark[p-w] = same = (grid[p-w] == grid[p]);
*writep[same]++ = p-w;
}
if (y + 1 < h && !mark[p+w]) { // go down
mark[p+w] = same = (grid[p+w] == grid[p]);
if (y == h - 2 && x == w - 1)
return d + !same;
*writep[same]++ = p+w;
}
}
}
}
}
This paper has a slightly faster version of Dijsktra's algorithm, which lowers the constant term. Still O(n) though, since you are really going to have to look at every node.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.54.8746&rep=rep1&type=pdf
EDIT: THE PREVIOUS VERSION WAS WRONG AND WAS FIXED
Since a Djikstra is out. I'll recommend a simple DP, which has the benefit of running in the optimal time and not having you construct a graph.
D[a][b] is the minimal distance to x=a and y=b using only nodes where the x<=a and y<=b.
And since you can't move diagonally you only have to look at D[a-1][b] and D[a][b-1] when calculating D[a][b]
This gives you the following recurrence relationship:
D[a][b] = min(if grid[a][b] == grid[a-1][b] then D[a-1][b] else D[a-1][b] + 1, if grid[a][b] == grid[a][b-1] then D[a][b-1] else D[a][b-1] + 1)
However doing only the above fails on this case:
0 1 2 3 4
5 6 7 8 9
A b d e g
A f r t s
A z A A A
A A A f d
Therefore you need to cache the minimum of each group of node you found so far. And instead of looking at D[a][b] you look at the minimum of the group at grid[a][b].
Here's some Python code:
Note grid is the grid that you're given as input and it's assumed the grid is N by N
groupmin = {}
for x in xrange(0, N):
for y in xrange(0, N):
groupmin[grid[x][y]] = N+1#N+1 serves as 'infinity'
#init first row and column
groupmin[grid[0][0]] = 0
for x in xrange(1, N):
gm = groupmin[grid[x-1][0]]
temp = (gm) if grid[x][0] == grid[x-1][0] else (gm + 1)
groupmin[grid[x][0]] = min(groupmin[grid[x][0]], temp);
for y in xrange(1, N):
gm = groupmin[grid[0][y-1]]
temp = (gm) if grid[0][y] == grid[0][y-1] else (gm + 1)
groupmin[grid[0][y]] = min(groupmin[grid[0][y]], temp);
#do the rest of the blocks
for x in xrange(1, N):
for y in xrange(1, N):
gma = groupmin[grid[x-1][y]]
gmb = groupmin[grid[x][y-1]]
a = (gma) if grid[x][y] == grid[x-1][y] else (gma + 1)
b = (gmb) if grid[x][y] == grid[x][y-1] else (gma + 1)
temp = min(a, b)
groupmin[grid[x][y]] = min(groupmin[grid[x][y]], temp);
ans = groupmin[grid[N-1][N-1]]
This will run in O(N^2 * f(x)) where f(x) is the time the hash function takes which is normally O(1) time and this is one of the best functions you can hope for and it has a lot lower constant factor than Djikstra's.
You should easily be able to handle N's of up to a few thousand in a second.
Is there anyway to ensure the that the fewest number of turns heuristic is met by anything except a breadth first search?
A faster way, or a simpler way? :)
You can breadth-first search from both ends, alternating, until the two regions meet in the middle. This will be much faster if the graph has a lot of fanout, like a city map, but the worst case is the same. It really depends on the graph.
This is my implementation using a simple BFS. A Dijkstra would also work (substitute a stl::priority_queue that sorts by descending costs for the stl::queue) but would seriously be overkill.
The thing to notice here is that we are actually searching on a graph whose nodes do not exactly correspond to the cells in the given array. To get to that graph, I used a simple DFS-based floodfill (you could also use BFS, but DFS is slightly shorter for me). What that does is to find all connected and same character components and assign them to the same colour/node. Thus, after the floodfill we can find out what node each cell belongs to in the underlying graph by looking at the value of colour[row][col]. Then I just iterate over the cells and find out all the cells where adjacent cells do not have the same colour (i.e. are in different nodes). These therefore are the edges of our graph. I maintain a stl::set of edges as I iterate over the cells to eliminate duplicate edges. After that it is a simple matter of building an adjacency list from the list of edges and we are ready for a bfs.
Code (in C++):
#include <queue>
#include <vector>
#include <iostream>
#include <string>
#include <set>
#include <cstring>
using namespace std;
#define SIZE 1001
vector<string> board;
int colour[SIZE][SIZE];
int dr[]={0,1,0,-1};
int dc[]={1,0,-1,0};
int min(int x,int y){ return (x<y)?x:y;}
int max(int x,int y){ return (x>y)?x:y;}
void dfs(int r, int c, int col, vector<string> &b){
if (colour[r][c]<0){
colour[r][c]=col;
for(int i=0;i<4;i++){
int nr=r+dr[i],nc=c+dc[i];
if (nr>=0 && nr<b.size() && nc>=0 && nc<b[0].size() && b[nr][nc]==b[r][c])
dfs(nr,nc,col,b);
}
}
}
int flood_fill(vector<string> &b){
memset(colour,-1,sizeof(colour));
int current_node=0;
for(int i=0;i<b.size();i++){
for(int j=0;j<b[0].size();j++){
if (colour[i][j]<0){
dfs(i,j,current_node,b);
current_node++;
}
}
}
return current_node;
}
vector<vector<int> > build_graph(vector<string> &b){
int total_nodes=flood_fill(b);
set<pair<int,int> > edge_list;
for(int r=0;r<b.size();r++){
for(int c=0;c<b[0].size();c++){
for(int i=0;i<4;i++){
int nr=r+dr[i],nc=c+dc[i];
if (nr>=0 && nr<b.size() && nc>=0 && nc<b[0].size() && colour[nr][nc]!=colour[r][c]){
int u=colour[r][c], v=colour[nr][nc];
if (u!=v) edge_list.insert(make_pair(min(u,v),max(u,v)));
}
}
}
}
vector<vector<int> > graph(total_nodes);
for(set<pair<int,int> >::iterator edge=edge_list.begin();edge!=edge_list.end();edge++){
int u=edge->first,v=edge->second;
graph[u].push_back(v);
graph[v].push_back(u);
}
return graph;
}
int bfs(vector<vector<int> > &G, int start, int end){
vector<int> cost(G.size(),-1);
queue<int> Q;
Q.push(start);
cost[start]=0;
while (!Q.empty()){
int node=Q.front();Q.pop();
vector<int> &adj=G[node];
for(int i=0;i<adj.size();i++){
if (cost[adj[i]]==-1){
cost[adj[i]]=cost[node]+1;
Q.push(adj[i]);
}
}
}
return cost[end];
}
int main(){
string line;
int rows,cols;
cin>>rows>>cols;
for(int r=0;r<rows;r++){
line="";
char ch;
for(int c=0;c<cols;c++){
cin>>ch;
line+=ch;
}
board.push_back(line);
}
vector<vector<int> > actual_graph=build_graph(board);
cout<<bfs(actual_graph,colour[0][0],colour[rows-1][cols-1])<<"\n";
}
This is just a quick hack, lots of improvements can be made. But I think it is pretty close to optimal in terms of runtime complexity, and should run fast enough for boards of size of several thousand (don't forget to change the #define of SIZE). Also, I only tested it with the one case you have provided. So, as Knuth said, "Beware of bugs in the above code; I have only proved it correct, not tried it." :).
I had this question on an Algorithms test yesterday, and I can't figure out the answer. It is driving me absolutely crazy, because it was worth about 40 points. I figure that most of the class didn't solve it correctly, because I haven't come up with a solution in the past 24 hours.
Given a arbitrary binary string of length n, find three evenly spaced ones within the string if they exist. Write an algorithm which solves this in O(n * log(n)) time.
So strings like these have three ones that are "evenly spaced": 11100000, 0100100100
edit: It is a random number, so it should be able to work for any number. The examples I gave were to illustrate the "evenly spaced" property. So 1001011 is a valid number. With 1, 4, and 7 being ones that are evenly spaced.
Finally! Following up leads in sdcvvc's answer, we have it: the O(n log n) algorithm for the problem! It is simple too, after you understand it. Those who guessed FFT were right.
The problem: we are given a binary string S of length n, and we want to find three evenly spaced 1s in it. For example, S may be 110110010, where n=9. It has evenly spaced 1s at positions 2, 5, and 8.
Scan S left to right, and make a list L of positions of 1. For the S=110110010 above, we have the list L = [1, 2, 4, 5, 8]. This step is O(n). The problem is now to find an arithmetic progression of length 3 in L, i.e. to find distinct a, b, c in L such that b-a = c-b, or equivalently a+c=2b. For the example above, we want to find the progression (2, 5, 8).
Make a polynomial p with terms xk for each k in L. For the example above, we make the polynomial p(x) = (x + x2 + x4 + x5+x8). This step is O(n).
Find the polynomial q = p2, using the Fast Fourier Transform. For the example above, we get the polynomial q(x) = x16 + 2x13 + 2x12 + 3x10 + 4x9 + x8 + 2x7 + 4x6 + 2x5 + x4 + 2x3 + x2. This step is O(n log n).
Ignore all terms except those corresponding to x2k for some k in L. For the example above, we get the terms x16, 3x10, x8, x4, x2. This step is O(n), if you choose to do it at all.
Here's the crucial point: the coefficient of any x2b for b in L is precisely the number of pairs (a,c) in L such that a+c=2b. [CLRS, Ex. 30.1-7] One such pair is (b,b) always (so the coefficient is at least 1), but if there exists any other pair (a,c), then the coefficient is at least 3, from (a,c) and (c,a). For the example above, we have the coefficient of x10 to be 3 precisely because of the AP (2,5,8). (These coefficients x2b will always be odd numbers, for the reasons above. And all other coefficients in q will always be even.)
So then, the algorithm is to look at the coefficients of these terms x2b, and see if any of them is greater than 1. If there is none, then there are no evenly spaced 1s. If there is a b in L for which the coefficient of x2b is greater than 1, then we know that there is some pair (a,c) — other than (b,b) — for which a+c=2b. To find the actual pair, we simply try each a in L (the corresponding c would be 2b-a) and see if there is a 1 at position 2b-a in S. This step is O(n).
That's all, folks.
One might ask: do we need to use FFT? Many answers, such as beta's, flybywire's, and rsp's, suggest that the approach that checks each pair of 1s and sees if there is a 1 at the "third" position, might work in O(n log n), based on the intuition that if there are too many 1s, we would find a triple easily, and if there are too few 1s, checking all pairs takes little time. Unfortunately, while this intuition is correct and the simple approach is better than O(n2), it is not significantly better. As in sdcvvc's answer, we can take the "Cantor-like set" of strings of length n=3k, with 1s at the positions whose ternary representation has only 0s and 2s (no 1s) in it. Such a string has 2k = n(log 2)/(log 3) ≈ n0.63 ones in it and no evenly spaced 1s, so checking all pairs would be of the order of the square of the number of 1s in it: that's 4k ≈ n1.26 which unfortunately is asymptotically much larger than (n log n). In fact, the worst case is even worse: Leo Moser in 1953 constructed (effectively) such strings which have n1-c/√(log n) 1s in them but no evenly spaced 1s, which means that on such strings, the simple approach would take Θ(n2-2c/√(log n)) — only a tiny bit better than Θ(n2), surprisingly!
About the maximum number of 1s in a string of length n with no 3 evenly spaced ones (which we saw above was at least n0.63 from the easy Cantor-like construction, and at least n1-c/√(log n) with Moser's construction) — this is OEIS A003002. It can also be calculated directly from OEIS A065825 as the k such that A065825(k) ≤ n < A065825(k+1). I wrote a program to find these, and it turns out that the greedy algorithm does not give the longest such string. For example, for n=9, we can get 5 1s (110100011) but the greedy gives only 4 (110110000), for n=26 we can get 11 1s (11001010001000010110001101) but the greedy gives only 8 (11011000011011000000000000), and for n=74 we can get 22 1s (11000010110001000001011010001000000000000000010001011010000010001101000011) but the greedy gives only 16 (11011000011011000000000000011011000011011000000000000000000000000000000000). They do agree at quite a few places until 50 (e.g. all of 38 to 50), though. As the OEIS references say, it seems that Jaroslaw Wroblewski is interested in this question, and he maintains a website on these non-averaging sets. The exact numbers are known only up to 194.
Your problem is called AVERAGE in this paper (1999):
A problem is 3SUM-hard if there is a sub-quadratic reduction from the problem 3SUM: Given a set A of n integers, are there elements a,b,c in A such that a+b+c = 0? It is not known whether AVERAGE is 3SUM-hard. However, there is a simple linear-time reduction from AVERAGE to 3SUM, whose description we omit.
Wikipedia:
When the integers are in the range [−u ... u], 3SUM can be solved in time O(n + u lg u) by representing S as a bit vector and performing a convolution using FFT.
This is enough to solve your problem :).
What is very important is that O(n log n) is complexity in terms of number of zeroes and ones, not the count of ones (which could be given as an array, like [1,5,9,15]). Checking if a set has an arithmetic progression, terms of number of 1's, is hard, and according to that paper as of 1999 no faster algorithm than O(n2) is known, and is conjectured that it doesn't exist. Everybody who doesn't take this into account is attempting to solve an open problem.
Other interesting info, mostly irrevelant:
Lower bound:
An easy lower bound is Cantor-like set (numbers 1..3^n-1 not containing 1 in their ternary expansion) - its density is n^(log_3 2) (circa 0.631). So any checking if the set isn't too large, and then checking all pairs is not enough to get O(n log n). You have to investigate the sequence smarter. A better lower bound is quoted here - it's n1-c/(log(n))^(1/2). This means Cantor set is not optimal.
Upper bound - my old algorithm:
It is known that for large n, a subset of {1,2,...,n} not containing arithmetic progression has at most n/(log n)^(1/20) elements. The paper On triples in arithmetic progression proves more: the set cannot contain more than n * 228 * (log log n / log n)1/2 elements. So you could check if that bound is achieved and if not, naively check pairs. This is O(n2 * log log n / log n) algorithm, faster than O(n2). Unfortunately "On triples..." is on Springer - but the first page is available, and Ben Green's exposition is available here, page 28, theorem 24.
By the way, the papers are from 1999 - the same year as the first one I mentioned, so that's probably why the first one doesn't mention that result.
This is not a solution, but a similar line of thought to what Olexiy was thinking
I was playing around with creating sequences with maximum number of ones, and they are all quite interesting, I got up to 125 digits and here are the first 3 numbers it found by attempting to insert as many '1' bits as possible:
11011000011011000000000000001101100001101100000000000000000000000000000000000000000110110000110110000000000000011011000011011
10110100010110100000000000010110100010110100000000000000000000000000000000000000000101101000101101000000000000101101000101101
10011001010011001000000000010011001010011001000000000000000000000000000000000000010011001010011001000000000010011001010011001
Notice they are all fractals (not too surprising given the constraints). There may be something in thinking backwards, perhaps if the string is not a fractal of with a characteristic, then it must have a repeating pattern?
Thanks to beta for the better term to describe these numbers.
Update:
Alas it looks like the pattern breaks down when starting with a large enough initial string, such as: 10000000000001:
100000000000011
10000000000001101
100000000000011011
10000000000001101100001
100000000000011011000011
10000000000001101100001101
100000000000011011000011010000000001
100000000000011011000011010000000001001
1000000000000110110000110100000000010011
1000000000000110110000110100000000010011001
10000000000001101100001101000000000100110010000000001
10000000000001101100001101000000000100110010000000001000001
1000000000000110110000110100000000010011001000000000100000100000000000001
10000000000001101100001101000000000100110010000000001000001000000000000011
1000000000000110110000110100000000010011001000000000100000100000000000001101
100000000000011011000011010000000001001100100000000010000010000000000000110100001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011000000001100100000000100100000000000010000000010000100000100100010010000010000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000110000010000000000000000000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011000000001100100000000100100000000000010000000010000100000100100010010000010000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000110000010000000000000000000001001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011000001000000000000000000000100100000000000000000000000000000000000011
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011000001000000000000000000000100100000000000000000000000000000000000011001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001100100000000000000000000001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001100100000000000000000000001001
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001100100000000000000000000001001000001
100000000000011011000011010000000001001100100000000010000010000000000000110100001001000001000000110001000000001000000000000000000000000000000000000000010000001000000000000001100000000110010000000010010000000000001000000001000010000010010001001000001000000000000010000000000000000000000000000000000000000000000000000000000000000000000000000000000001000000000000000011000001000000000000000000000100100000000000000000000000000000000000011001000000000000000000000010010000010000001
1000000000000110110000110100000000010011001000000000100000100000000000001101000010010000010000001100010000000010000000000000000000000000000000000000000100000010000000000000011000000001100100000000100100000000000010000000010000100000100100010010000010000000000000100000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000110000010000000000000000000001001000000000000000000000000000000000000110010000000000000000000000100100000100000011
10000000000001101100001101000000000100110010000000001000001000000000000011010000100100000100000011000100000000100000000000000000000000000000000000000001000000100000000000000110000000011001000000001001000000000000100000000100001000001001000100100000100000000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000001100000100000000000000000000010010000000000000000000000000000000000001100100000000000000000000001001000001000000110000000000001
I suspect that a simple approach that looks like O(n^2) will actually yield something better, like O(n ln(n)). The sequences that take the longest to test (for any given n) are the ones that contain no trios, and that puts severe restrictions on the number of 1's that can be in the sequence.
I've come up with some hand-waving arguments, but I haven't been able to find a tidy proof. I'm going to take a stab in the dark: the answer is a very clever idea that the professor has known for so long that it's come to seem obvious, but it's much too hard for the students. (Either that or you slept through the lecture that covered it.)
Revision: 2009-10-17 23:00
I've run this on large numbers (like, strings of 20 million) and I now believe this algorithm is not O(n logn). Notwithstanding that, it's a cool enough implementation and contains a number of optimizations that makes it run really fast. It evaluates all the arrangements of binary strings 24 or fewer digits in under 25 seconds.
I've updated the code to include the 0 <= L < M < U <= X-1 observation from earlier today.
Original
This is, in concept, similar to another question I answered. That code also looked at three values in a series and determined if a triplet satisfied a condition. Here is C# code adapted from that:
using System;
using System.Collections.Generic;
namespace StackOverflow1560523
{
class Program
{
public struct Pair<T>
{
public T Low, High;
}
static bool FindCandidate(int candidate,
List<int> arr,
List<int> pool,
Pair<int> pair,
ref int iterations)
{
int lower = pair.Low, upper = pair.High;
while ((lower >= 0) && (upper < pool.Count))
{
int lowRange = candidate - arr[pool[lower]];
int highRange = arr[pool[upper]] - candidate;
iterations++;
if (lowRange < highRange)
lower -= 1;
else if (lowRange > highRange)
upper += 1;
else
return true;
}
return false;
}
static List<int> BuildOnesArray(string s)
{
List<int> arr = new List<int>();
for (int i = 0; i < s.Length; i++)
if (s[i] == '1')
arr.Add(i);
return arr;
}
static void BuildIndexes(List<int> arr,
ref List<int> even, ref List<int> odd,
ref List<Pair<int>> evenIndex, ref List<Pair<int>> oddIndex)
{
for (int i = 0; i < arr.Count; i++)
{
bool isEven = (arr[i] & 1) == 0;
if (isEven)
{
evenIndex.Add(new Pair<int> {Low=even.Count-1, High=even.Count+1});
oddIndex.Add(new Pair<int> {Low=odd.Count-1, High=odd.Count});
even.Add(i);
}
else
{
oddIndex.Add(new Pair<int> {Low=odd.Count-1, High=odd.Count+1});
evenIndex.Add(new Pair<int> {Low=even.Count-1, High=even.Count});
odd.Add(i);
}
}
}
static int FindSpacedOnes(string s)
{
// List of indexes of 1s in the string
List<int> arr = BuildOnesArray(s);
//if (s.Length < 3)
// return 0;
// List of indexes to odd indexes in arr
List<int> odd = new List<int>(), even = new List<int>();
// evenIndex has indexes into arr to bracket even numbers
// oddIndex has indexes into arr to bracket odd numbers
List<Pair<int>> evenIndex = new List<Pair<int>>(),
oddIndex = new List<Pair<int>>();
BuildIndexes(arr,
ref even, ref odd,
ref evenIndex, ref oddIndex);
int iterations = 0;
for (int i = 1; i < arr.Count-1; i++)
{
int target = arr[i];
bool found = FindCandidate(target, arr, odd, oddIndex[i], ref iterations) ||
FindCandidate(target, arr, even, evenIndex[i], ref iterations);
if (found)
return iterations;
}
return iterations;
}
static IEnumerable<string> PowerSet(int n)
{
for (long i = (1L << (n-1)); i < (1L << n); i++)
{
yield return Convert.ToString(i, 2).PadLeft(n, '0');
}
}
static void Main(string[] args)
{
for (int i = 5; i < 64; i++)
{
int c = 0;
string hardest_string = "";
foreach (string s in PowerSet(i))
{
int cost = find_spaced_ones(s);
if (cost > c)
{
hardest_string = s;
c = cost;
Console.Write("{0} {1} {2}\r", i, c, hardest_string);
}
}
Console.WriteLine("{0} {1} {2}", i, c, hardest_string);
}
}
}
}
The principal differences are:
Exhaustive search of solutions
This code generates a power set of data to find the hardest input to solve for this algorithm.
All solutions versus hardest to solve
The code for the previous question generated all the solutions using a python generator. This code just displays the hardest for each pattern length.
Scoring algorithm
This code checks the distance from the middle element to its left- and right-hand edge. The python code tested whether a sum was above or below 0.
Convergence on a candidate
The current code works from the middle towards the edge to find a candidate. The code in the previous problem worked from the edges towards the middle. This last change gives a large performance improvement.
Use of even and odd pools
Based on the observations at the end of this write-up, the code searches pairs of even numbers of pairs of odd numbers to find L and U, keeping M fixed. This reduces the number of searches by pre-computing information. Accordingly, the code uses two levels of indirection in the main loop of FindCandidate and requires two calls to FindCandidate for each middle element: once for even numbers and once for odd ones.
The general idea is to work on indexes, not the raw representation of the data. Calculating an array where the 1's appear allows the algorithm to run in time proportional to the number of 1's in the data rather than in time proportional to the length of the data. This is a standard transformation: create a data structure that allows faster operation while keeping the problem equivalent.
The results are out of date: removed.
Edit: 2009-10-16 18:48
On yx's data, which is given some credence in the other responses as representative of hard data to calculate on, I get these results... I removed these. They are out of date.
I would point out that this data is not the hardest for my algorithm, so I think the assumption that yx's fractals are the hardest to solve is mistaken. The worst case for a particular algorithm, I expect, will depend upon the algorithm itself and will not likely be consistent across different algorithms.
Edit: 2009-10-17 13:30
Further observations on this.
First, convert the string of 0's and 1's into an array of indexes for each position of the 1's. Say the length of that array A is X. Then the goal is to find
0 <= L < M < U <= X-1
such that
A[M] - A[L] = A[U] - A[M]
or
2*A[M] = A[L] + A[U]
Since A[L] and A[U] sum to an even number, they can't be (even, odd) or (odd, even). The search for a match could be improved by splitting A[] into odd and even pools and searching for matches on A[M] in the pools of odd and even candidates in turn.
However, this is more of a performance optimization than an algorithmic improvement, I think. The number of comparisons should drop, but the order of the algorithm should be the same.
Edit 2009-10-18 00:45
Yet another optimization occurs to me, in the same vein as separating the candidates into even and odd. Since the three indexes have to add to a multiple of 3 (a, a+x, a+2x -- mod 3 is 0, regardless of a and x), you can separate L, M, and U into their mod 3 values:
M L U
0 0 0
1 2
2 1
1 0 2
1 1
2 0
2 0 1
1 0
2 2
In fact, you could combine this with the even/odd observation and separate them into their mod 6 values:
M L U
0 0 0
1 5
2 4
3 3
4 2
5 1
and so on. This would provide a further performance optimization but not an algorithmic speedup.
Wasn't able to come up with the solution yet :(, but have some ideas.
What if we start from a reverse problem: construct a sequence with the maximum number of 1s and WITHOUT any evenly spaced trios. If you can prove the maximum number of 1s is o(n), then you can improve your estimate by iterating only through list of 1s only.
This may help....
This problem reduces to the following:
Given a sequence of positive integers, find a contiguous subsequence partitioned into a prefix and a suffix such that the sum of the prefix of the subsequence is equal to the sum of the suffix of the subsequence.
For example, given a sequence of [ 3, 5, 1, 3, 6, 5, 2, 2, 3, 5, 6, 4 ], we would find a subsequence of [ 3, 6, 5, 2, 2] with a prefix of [ 3, 6 ] with prefix sum of 9 and a suffix of [ 5, 2, 2 ] with suffix sum of 9.
The reduction is as follows:
Given a sequence of zeros and ones, and starting at the leftmost one, continue moving to the right. Each time another one is encountered, record the number of moves since the previous one was encountered and append that number to the resulting sequence.
For example, given a sequence of [ 0, 1, 1, 0, 0, 1, 0, 0, 0, 1 0 ], we would find the reduction of [ 1, 3, 4]. From this reduction, we calculate the contiguous subsequence of [ 1, 3, 4], the prefix of [ 1, 3] with sum of 4, and the suffix of [ 4 ] with sum of 4.
This reduction may be computed in O(n).
Unfortunately, I am not sure where to go from here.
For the simple problem type (i.e. you search three "1" with only (i.e. zero or more) "0" between it), Its quite simple: You could just split the sequence at every "1" and look for two adjacent subsequences having the same length (the second subsequence not being the last one, of course). Obviously, this can be done in O(n) time.
For the more complex version (i.e. you search an index i and an gap g>0 such that s[i]==s[i+g]==s[i+2*g]=="1"), I'm not sure, if there exists an O(n log n) solution, since there are possibly O(n²) triplets having this property (think of a string of all ones, there are approximately n²/2 such triplets). Of course, you are looking for only one of these, but I have currently no idea, how to find it ...
A fun question, but once you realise that the actual pattern between two '1's does not matter, the algorithm becomes:
scan look for a '1'
starting from the next position scan for another '1' (to the end of the array minus the distance from the current first '1' or else the 3rd '1' would be out of bounds)
if at the position of the 2nd '1' plus the distance to the first 1' a third '1' is found, we have evenly spaces ones.
In code, JTest fashion, (Note this code isn't written to be most efficient and I added some println's to see what happens.)
import java.util.Random;
import junit.framework.TestCase;
public class AlgorithmTest extends TestCase {
/**
* Constructor for GetNumberTest.
*
* #param name The test's name.
*/
public AlgorithmTest(String name) {
super(name);
}
/**
* #see TestCase#setUp()
*/
protected void setUp() throws Exception {
super.setUp();
}
/**
* #see TestCase#tearDown()
*/
protected void tearDown() throws Exception {
super.tearDown();
}
/**
* Tests the algorithm.
*/
public void testEvenlySpacedOnes() {
assertFalse(isEvenlySpaced(1));
assertFalse(isEvenlySpaced(0x058003));
assertTrue(isEvenlySpaced(0x07001));
assertTrue(isEvenlySpaced(0x01007));
assertTrue(isEvenlySpaced(0x101010));
// some fun tests
Random random = new Random();
isEvenlySpaced(random.nextLong());
isEvenlySpaced(random.nextLong());
isEvenlySpaced(random.nextLong());
}
/**
* #param testBits
*/
private boolean isEvenlySpaced(long testBits) {
String testString = Long.toBinaryString(testBits);
char[] ones = testString.toCharArray();
final char ONE = '1';
for (int n = 0; n < ones.length - 1; n++) {
if (ONE == ones[n]) {
for (int m = n + 1; m < ones.length - m + n; m++) {
if (ONE == ones[m] && ONE == ones[m + m - n]) {
System.out.println(" IS evenly spaced: " + testBits + '=' + testString);
System.out.println(" at: " + n + ", " + m + ", " + (m + m - n));
return true;
}
}
}
}
System.out.println("NOT evenly spaced: " + testBits + '=' + testString);
return false;
}
}
I thought of a divide-and-conquer approach that might work.
First, in preprocessing you need to insert all numbers less than one half your input size (n/3) into a list.
Given a string: 0000010101000100 (note that this particular example is valid)
Insert all primes (and 1) from 1 to (16/2) into a list: {1, 2, 3, 4, 5, 6, 7}
Then divide it in half:
100000101 01000100
Keep doing this until you get to strings of size 1. For all size-one strings with a 1 in them, add the index of the string to the list of possibilities; otherwise, return -1 for failure.
You'll also need to return a list of still-possible spacing distances, associated with each starting index. (Start with the list you made above and remove numbers as you go) Here, an empty list means you're only dealing with one 1 and so any spacing is possible at this point; otherwise the list includes spacings that must be ruled out.
So continuing with the example above:
1000 0101 0100 0100
10 00 01 01 01 00 01 00
1 0 0 0 0 1 0 1 0 1 0 0 0 1 0 0
In the first combine step, we have eight sets of two now. In the first, we have the possibility of a set, but we learn that spacing by 1 is impossible because of the other zero being there. So we return 0 (for the index) and {2,3,4,5,7} for the fact that spacing by 1 is impossible. In the second, we have nothing and so return -1. In the third we have a match with no spacings eliminated in index 5, so return 5, {1,2,3,4,5,7}. In the fourth pair we return 7, {1,2,3,4,5,7}. In the fifth, return 9, {1,2,3,4,5,7}. In the sixth, return -1. In the seventh, return 13, {1,2,3,4,5,7}. In the eighth, return -1.
Combining again into four sets of four, we have:
1000: Return (0, {4,5,6,7})
0101: Return (5, {2,3,4,5,6,7}), (7, {1,2,3,4,5,6,7})
0100: Return (9, {3,4,5,6,7})
0100: Return (13, {3,4,5,6,7})
Combining into sets of eight:
10000101: Return (0, {5,7}), (5, {2,3,4,5,6,7}), (7, {1,2,3,4,5,6,7})
01000100: Return (9, {4,7}), (13, {3,4,5,6,7})
Combining into a set of sixteen:
10000101 01000100
As we've progressed, we keep checking all the possibilities so far. Up to this step we've left stuff that went beyond the end of the string, but now we can check all the possibilities.
Basically, we check the first 1 with spacings of 5 and 7, and find that they don't line up to 1's. (Note that each check is CONSTANT, not linear time) Then we check the second one (index 5) with spacings of 2, 3, 4, 5, 6, and 7-- or we would, but we can stop at 2 since that actually matches up.
Phew! That's a rather long algorithm.
I don't know 100% if it's O(n log n) because of the last step, but everything up to there is definitely O(n log n) as far as I can tell. I'll get back to this later and try to refine the last step.
EDIT: Changed my answer to reflect Welbog's comment. Sorry for the error. I'll write some pseudocode later, too, when I get a little more time to decipher what I wrote again. ;-)
I'll give my rough guess here, and let those who are better with calculating complexity to help me on how my algorithm fares in O-notation wise
given binary string 0000010101000100 (as example)
crop head and tail of zeroes -> 00000 101010001 00
we get 101010001 from previous calculation
check if the middle bit is 'one', if true, found valid three evenly spaced 'ones' (only if the number of bits is odd numbered)
correlatively, if the remained cropped number of bits is even numbered, the head and tail 'one' cannot be part of evenly spaced 'one',
we use 1010100001 as example (with an extra 'zero' to become even numbered crop), in this case we need to crop again, then becomes -> 10101 00001
we get 10101 from previous calculation, and check middle bit, and we found the evenly spaced bit again
I have no idea how to calculate complexity for this, can anyone help?
edit: add some code to illustrate my idea
edit2: tried to compile my code and found some major mistakes, fixed
char *binaryStr = "0000010101000100";
int main() {
int head, tail, pos;
head = 0;
tail = strlen(binaryStr)-1;
if( (pos = find3even(head, tail)) >=0 )
printf("found it at position %d\n", pos);
return 0;
}
int find3even(int head, int tail) {
int pos = 0;
if(head >= tail) return -1;
while(binaryStr[head] == '0')
if(head<tail) head++;
while(binaryStr[tail] == '0')
if(head<tail) tail--;
if(head >= tail) return -1;
if( (tail-head)%2 == 0 && //true if odd numbered
(binaryStr[head + (tail-head)/2] == '1') ) {
return head;
}else {
if( (pos = find3even(head, tail-1)) >=0 )
return pos;
if( (pos = find3even(head+1, tail)) >=0 )
return pos;
}
return -1;
}
I came up with something like this:
def IsSymetric(number):
number = number.strip('0')
if len(number) < 3:
return False
if len(number) % 2 == 0:
return IsSymetric(number[1:]) or IsSymetric(number[0:len(number)-2])
else:
if number[len(number)//2] == '1':
return True
return IsSymetric(number[:(len(number)//2)]) or IsSymetric(number[len(number)//2+1:])
return False
This is inspired by andycjw.
Truncate the zeros.
If even then test two substring 0 - (len-2) (skip last character) and from 1 - (len-1) (skip the first char)
If not even than if the middle char is one than we have success. Else divide the string in the midle without the midle element and check both parts.
As to the complexity this might be O(nlogn) as in each recursion we are dividing by two.
Hope it helps.
Ok, I'm going to take another stab at the problem. I think I can prove a O(n log(n)) algorithm that is similar to those already discussed by using a balanced binary tree to store distances between 1's. This approach was inspired by Justice's observation about reducing the problem to a list of distances between the 1's.
Could we scan the input string to construct a balanced binary tree around the position of 1's such that each node stores the position of the 1 and each edge is labeled with the distance to the adjacent 1 for each child node. For example:
10010001 gives the following tree
3
/ \
2 / \ 3
/ \
0 7
This can be done in O(n log(n)) since, for a string of size n, each insertion takes O(log(n)) in the worst case.
Then the problem is to search the tree to discover whether, at any node, there is a path from that node through the left-child that has the same distance as a path through the right child. This can be done recursively on each subtree. When merging two subtrees in the search, we must compare the distances from paths in the left subtree with distances from paths in the right. Since the number of paths in a subtree will be proportional to log(n), and the number of nodes is n, I believe this can be done in O(n log(n)) time.
Did I miss anything?
This seemed liked a fun problem so I decided to try my hand at it.
I am making the assumption that 111000001 would find the first 3 ones and be successful. Essentially the number of zeroes following the 1 is the important thing, since 0111000 is the same as 111000 according to your definition. Once you find two cases of 1, the next 1 found completes the trilogy.
Here it is in Python:
def find_three(bstring):
print bstring
dict = {}
lastone = -1
zerocount = 0
for i in range(len(bstring)):
if bstring[i] == '1':
print i, ': 1'
if lastone != -1:
if(zerocount in dict):
dict[zerocount].append(lastone)
if len(dict[zerocount]) == 2:
dict[zerocount].append(i)
return True, dict
else:
dict[zerocount] = [lastone]
lastone = i
zerocount = 0
else:
zerocount = zerocount + 1
#this is really just book keeping, as we have failed at this point
if lastone != -1:
if(zerocount in dict):
dict[zerocount].append(lastone)
else:
dict[zerocount] = [lastone]
return False, dict
This is a first try, so I'm sure this could be written in a cleaner manner. Please list the cases where this method fails down below.
I assume the reason this is nlog(n) is due to the following:
To find the 1 that is the start of the triplet, you need to check (n-2) characters. If you haven't found it by that point, you won't (chars n-1 and n cannot start a triplet) (O(n))
To find the second 1 that is the part of the triplet (started by the first one), you need to check m/2 (m=n-x, where x is the offset of the first 1) characters. This is because, if you haven't found the second 1 by the time you're halfway from the first one to the end, you won't... since the third 1 must be exactly the same distance past the second. (O(log(n)))
It O(1) to find the last 1 since you know the index it must be at by the time you find the first and second.
So, you have n, log(n), and 1... O(nlogn)
Edit: Oops, my bad. My brain had it set that n/2 was logn... which it obviously isn't (doubling the number on items still doubles the number of iterations on the inner loop). This is still at n^2, not solving the problem. Well, at least I got to write some code :)
Implementation in Tcl
proc get-triplet {input} {
for {set first 0} {$first < [string length $input]-2} {incr first} {
if {[string index $input $first] != 1} {
continue
}
set start [expr {$first + 1}]
set end [expr {1+ $first + (([string length $input] - $first) /2)}]
for {set second $start} {$second < $end} {incr second} {
if {[string index $input $second] != 1} {
continue
}
set last [expr {($second - $first) + $second}]
if {[string index $input $last] == 1} {
return [list $first $second $last]
}
}
}
return {}
}
get-triplet 10101 ;# 0 2 4
get-triplet 10111 ;# 0 2 4
get-triplet 11100000 ;# 0 1 2
get-triplet 0100100100 ;# 1 4 7
I think I have found a way of solving the problem, but I can't construct a formal proof. The solution I made is written in Java, and it uses a counter 'n' to count how many list/array accesses it does. So n should be less than or equal to stringLength*log(stringLength) if it is correct. I tried it for the numbers 0 to 2^22, and it works.
It starts by iterating over the input string and making a list of all the indexes which hold a one. This is just O(n).
Then from the list of indexes it picks a firstIndex, and a secondIndex which is greater than the first. These two indexes must hold ones, because they are in the list of indexes. From there the thirdIndex can be calculated. If the inputString[thirdIndex] is a 1 then it halts.
public static int testString(String input){
//n is the number of array/list accesses in the algorithm
int n=0;
//Put the indices of all the ones into a list, O(n)
ArrayList<Integer> ones = new ArrayList<Integer>();
for(int i=0;i<input.length();i++){
if(input.charAt(i)=='1'){
ones.add(i);
}
}
//If less than three ones in list, just stop
if(ones.size()<3){
return n;
}
int firstIndex, secondIndex, thirdIndex;
for(int x=0;x<ones.size()-2;x++){
n++;
firstIndex = ones.get(x);
for(int y=x+1; y<ones.size()-1; y++){
n++;
secondIndex = ones.get(y);
thirdIndex = secondIndex*2 - firstIndex;
if(thirdIndex >= input.length()){
break;
}
n++;
if(input.charAt(thirdIndex) == '1'){
//This case is satisfied if it has found three evenly spaced ones
//System.out.println("This one => " + input);
return n;
}
}
}
return n;
}
additional note: the counter n is not incremented when it iterates over the input string to construct the list of indexes. This operation is O(n), so it won't have an effect on the algorithm complexity anyway.
One inroad into the problem is to think of factors and shifting.
With shifting, you compare the string of ones and zeroes with a shifted version of itself. You then take matching ones. Take this example shifted by two:
1010101010
1010101010
------------
001010101000
The resulting 1's (bitwise ANDed), must represent all those 1's which are evenly spaced by two. The same example shifted by three:
1010101010
1010101010
-------------
0000000000000
In this case there are no 1's which are evenly spaced three apart.
So what does this tell you? Well that you only need to test shifts which are prime numbers. For example say you have two 1's which are six apart. You would only have to test 'two' shifts and 'three' shifts (since these divide six). For example:
10000010
10000010 (Shift by two)
10000010
10000010 (We have a match)
10000010
10000010 (Shift by three)
10000010 (We have a match)
So the only shifts you ever need to check are 2,3,5,7,11,13 etc. Up to the prime closest to the square root of size of the string of digits.
Nearly solved?
I think I am closer to a solution. Basically:
Scan the string for 1's. For each 1 note it's remainder after taking a modulus of its position. The modulus ranges from 1 to half the size of the string. This is because the largest possible separation size is half the string. This is done in O(n^2). BUT. Only prime moduli need be checked so O(n^2/log(n))
Sort the list of modulus/remainders in order largest modulus first, this can be done in O(n*log(n)) time.
Look for three consecutive moduli/remainders which are the same.
Somehow retrieve the position of the ones!
I think the biggest clue to the answer, is that the fastest sort algorithms, are O(n*log(n)).
WRONG
Step 1 is wrong as pointed out by a colleague. If we have 1's at position 2,12 and 102. Then taking a modulus of 10, they would all have the same remainders, and yet are not equally spaced apart! Sorry.
Here are some thoughts that, despite my best efforts, will not seem to wrap themselves up in a bow. Still, they might be a useful starting point for someone's analysis.
Consider the proposed solution as follows, which is the approach that several folks have suggested, including myself in a prior version of this answer. :)
Trim leading and trailing zeroes.
Scan the string looking for 1's.
When a 1 is found:
Assume that it is the middle 1 of the solution.
For each prior 1, use its saved position to compute the anticipated position of the final 1.
If the computed position is after the end of the string it cannot be part of the solution, so drop the position from the list of candidates.
Check the solution.
If the solution was not found, add the current 1 to the list of candidates.
Repeat until no more 1's are found.
Now consider input strings strings like the following, which will not have a solution:
101
101001
1010010001
101001000100001
101001000100001000001
In general, this is the concatenation of k strings of the form j 0's followed by a 1 for j from zero to k-1.
k=2 101
k=3 101001
k=4 1010010001
k=5 101001000100001
k=6 101001000100001000001
Note that the lengths of the substrings are 1, 2, 3, etc. So, problem size n has substrings of lengths 1 to k such that n = k(k+1)/2.
k=2 n= 3 101
k=3 n= 6 101001
k=4 n=10 1010010001
k=5 n=15 101001000100001
k=6 n=21 101001000100001000001
Note that k also tracks the number of 1's that we have to consider. Remember that every time we see a 1, we need to consider all the 1's seen so far. So when we see the second 1, we only consider the first, when we see the third 1, we reconsider the first two, when we see the fourth 1, we need to reconsider the first three, and so on. By the end of the algorithm, we've considered k(k-1)/2 pairs of 1's. Call that p.
k=2 n= 3 p= 1 101
k=3 n= 6 p= 3 101001
k=4 n=10 p= 6 1010010001
k=5 n=15 p=10 101001000100001
k=6 n=21 p=15 101001000100001000001
The relationship between n and p is that n = p + k.
The process of going through the string takes O(n) time. Each time a 1 is encountered, a maximum of (k-1) comparisons are done. Since n = k(k+1)/2, n > k**2, so sqrt(n) > k. This gives us O(n sqrt(n)) or O(n**3/2). Note however that may not be a really tight bound, because the number of comparisons goes from 1 to a maximum of k, it isn't k the whole time. But I'm not sure how to account for that in the math.
It still isn't O(n log(n)). Also, I can't prove those inputs are the worst cases, although I suspect they are. I think a denser packing of 1's to the front results in an even sparser packing at the end.
Since someone may still find it useful, here's my code for that solution in Perl:
#!/usr/bin/perl
# read input as first argument
my $s = $ARGV[0];
# validate the input
$s =~ /^[01]+$/ or die "invalid input string\n";
# strip leading and trailing 0's
$s =~ s/^0+//;
$s =~ s/0+$//;
# prime the position list with the first '1' at position 0
my #p = (0);
# start at position 1, which is the second character
my $i = 1;
print "the string is $s\n\n";
while ($i < length($s)) {
if (substr($s, $i, 1) eq '1') {
print "found '1' at position $i\n";
my #t = ();
# assuming this is the middle '1', go through the positions
# of all the prior '1's and check whether there's another '1'
# in the correct position after this '1' to make a solution
while (scalar #p) {
# $p is the position of the prior '1'
my $p = shift #p;
# $j is the corresponding position for the following '1'
my $j = 2 * $i - $p;
# if $j is off the end of the string then we don't need to
# check $p anymore
next if ($j >= length($s));
print "checking positions $p, $i, $j\n";
if (substr($s, $j, 1) eq '1') {
print "\nsolution found at positions $p, $i, $j\n";
exit 0;
}
# if $j isn't off the end of the string, keep $p for next time
push #t, $p;
}
#p = #t;
# add this '1' to the list of '1' positions
push #p, $i;
}
$i++;
}
print "\nno solution found\n";
While scanning 1s, add their positions to a List. When adding the second and successive 1s, compare them to each position in the list so far. Spacing equals currentOne (center) - previousOne (left). The right-side bit is currentOne + spacing. If it's 1, the end.
The list of ones grows inversely with the space between them. Simply stated, if you've got a lot of 0s between the 1s (as in a worst case), your list of known 1s will grow quite slowly.
using System;
using System.Collections.Generic;
namespace spacedOnes
{
class Program
{
static int[] _bits = new int[8] {128, 64, 32, 16, 8, 4, 2, 1};
static void Main(string[] args)
{
var bytes = new byte[4];
var r = new Random();
r.NextBytes(bytes);
foreach (var b in bytes) {
Console.Write(getByteString(b));
}
Console.WriteLine();
var bitCount = bytes.Length * 8;
var done = false;
var onePositions = new List<int>();
for (var i = 0; i < bitCount; i++)
{
if (isOne(bytes, i)) {
if (onePositions.Count > 0) {
foreach (var knownOne in onePositions) {
var spacing = i - knownOne;
var k = i + spacing;
if (k < bitCount && isOne(bytes, k)) {
Console.WriteLine("^".PadLeft(knownOne + 1) + "^".PadLeft(spacing) + "^".PadLeft(spacing));
done = true;
break;
}
}
}
if (done) {
break;
}
onePositions.Add(i);
}
}
Console.ReadKey();
}
static String getByteString(byte b) {
var s = new char[8];
for (var i=0; i<s.Length; i++) {
s[i] = ((b & _bits[i]) > 0 ? '1' : '0');
}
return new String(s);
}
static bool isOne(byte[] bytes, int i)
{
var byteIndex = i / 8;
var bitIndex = i % 8;
return (bytes[byteIndex] & _bits[bitIndex]) > 0;
}
}
}
I thought I'd add one comment before posting the 22nd naive solution to the problem. For the naive solution, we don't need to show that the number of 1's in the string is at most O(log(n)), but rather that it is at most O(sqrt(n*log(n)).
Solver:
def solve(Str):
indexes=[]
#O(n) setup
for i in range(len(Str)):
if Str[i]=='1':
indexes.append(i)
#O((number of 1's)^2) processing
for i in range(len(indexes)):
for j in range(i+1, len(indexes)):
indexDiff = indexes[j] - indexes[i]
k=indexes[j] + indexDiff
if k<len(Str) and Str[k]=='1':
return True
return False
It's basically a fair bit similar to flybywire's idea and implementation, though looking ahead instead of back.
Greedy String Builder:
#assumes final char hasn't been added, and would be a 1
def lastCharMakesSolvable(Str):
endIndex=len(Str)
j=endIndex-1
while j-(endIndex-j) >= 0:
k=j-(endIndex-j)
if k >= 0 and Str[k]=='1' and Str[j]=='1':
return True
j=j-1
return False
def expandString(StartString=''):
if lastCharMakesSolvable(StartString):
return StartString + '0'
return StartString + '1'
n=1
BaseStr=""
lastCount=0
while n<1000000:
BaseStr=expandString(BaseStr)
count=BaseStr.count('1')
if count != lastCount:
print(len(BaseStr), count)
lastCount=count
n=n+1
(In my defense, I'm still in the 'learn python' stage of understanding)
Also, potentially useful output from the greedy building of strings, there's a rather consistent jump after hitting a power of 2 in the number of 1's... which I was not willing to wait around to witness hitting 2096.
strlength # of 1's
1 1
2 2
4 3
5 4
10 5
14 8
28 9
41 16
82 17
122 32
244 33
365 64
730 65
1094 128
2188 129
3281 256
6562 257
9842 512
19684 513
29525 1024
I'll try to present a mathematical approach. This is more a beginning than an end, so any help, comment, or even contradiction - will be deeply appreciated. However, if this approach is proven - the algorithm is a straight-forward search in the string.
Given a fixed number of spaces k and a string S, the search for a k-spaced-triplet takes O(n) - We simply test for every 0<=i<=(n-2k) if S[i]==S[i+k]==S[i+2k]. The test takes O(1) and we do it n-k times where k is a constant, so it takes O(n-k)=O(n).
Let us assume that there is an Inverse Proportion between the number of 1's and the maximum spaces we need to search for. That is, If there are many 1's, there must be a triplet and it must be quite dense; If there are only few 1's, The triplet (if any) can be quite sparse. In other words, I can prove that if I have enough 1's, such triplet must exist - and the more 1's I have, a more dense triplet must be found. This can be explained by the Pigeonhole principle - Hope to elaborate on this later.
Say have an upper bound k on the possible number of spaces I have to look for. Now, for each 1 located in S[i] we need to check for 1 in S[i-1] and S[i+1], S[i-2] and S[i+2], ... S[i-k] and S[i+k]. This takes O((k^2-k)/2)=O(k^2) for each 1 in S - due to Gauss' Series Summation Formula. Note that this differs from section 1 - I'm having k as an upper bound for the number of spaces, not as a constant space.
We need to prove O(n*log(n)). That is, we need to show that k*(number of 1's) is proportional to log(n).
If we can do that, the algorithm is trivial - for each 1 in S whose index is i, simply look for 1's from each side up to distance k. If two were found in the same distance, return i and k. Again, the tricky part would be finding k and proving the correctness.
I would really appreciate your comments here - I have been trying to find the relation between k and the number of 1's on my whiteboard, so far without success.
Assumption:
Just wrong, talking about log(n) number of upper limit of ones
EDIT:
Now I found that using Cantor numbers (if correct), density on set is (2/3)^Log_3(n) (what a weird function) and I agree, log(n)/n density is to strong.
If this is upper limit, there is algorhitm who solves this problem in at least O(n*(3/2)^(log(n)/log(3))) time complexity and O((3/2)^(log(n)/log(3))) space complexity. (check Justice's answer for algorhitm)
This is still by far better than O(n^2)
This function ((3/2)^(log(n)/log(3))) really looks like n*log(n) on first sight.
How did I get this formula?
Applaying Cantors number on string.
Supose that length of string is 3^p == n
At each step in generation of Cantor string you keep 2/3 of prevous number of ones. Apply this p times.
That mean (n * ((2/3)^p)) -> (((3^p)) * ((2/3)^p)) remaining ones and after simplification 2^p.
This mean 2^p ones in 3^p string -> (3/2)^p ones . Substitute p=log(n)/log(3) and get
((3/2)^(log(n)/log(3)))
How about a simple O(n) solution, with O(n^2) space? (Uses the assumption that all bitwise operators work in O(1).)
The algorithm basically works in four stages:
Stage 1: For each bit in your original number, find out how far away the ones are, but consider only one direction. (I considered all the bits in the direction of the least significant bit.)
Stage 2: Reverse the order of the bits in the input;
Stage 3: Re-run step 1 on the reversed input.
Stage 4: Compare the results from Stage 1 and Stage 3. If any bits are equally spaced above AND below we must have a hit.
Keep in mind that no step in the above algorithm takes longer than O(n). ^_^
As an added benefit, this algorithm will find ALL equally spaced ones from EVERY number. So for example if you get a result of "0x0005" then there are equally spaced ones at BOTH 1 and 3 units away
I didn't really try optimizing the code below, but it is compilable C# code that seems to work.
using System;
namespace ThreeNumbers
{
class Program
{
const int uint32Length = 32;
static void Main(string[] args)
{
Console.Write("Please enter your integer: ");
uint input = UInt32.Parse(Console.ReadLine());
uint[] distancesLower = Distances(input);
uint[] distancesHigher = Distances(Reverse(input));
PrintHits(input, distancesLower, distancesHigher);
}
/// <summary>
/// Returns an array showing how far the ones away from each bit in the input. Only
/// considers ones at lower signifcant bits. Index 0 represents the least significant bit
/// in the input. Index 1 represents the second least significant bit in the input and so
/// on. If a one is 3 away from the bit in question, then the third least significant bit
/// of the value will be sit.
///
/// As programed this algorithm needs: O(n) time, and O(n*log(n)) space.
/// (Where n is the number of bits in the input.)
/// </summary>
public static uint[] Distances(uint input)
{
uint[] distanceToOnes = new uint[uint32Length];
uint result = 0;
//Sets how far each bit is from other ones. Going in the direction of LSB to MSB
for (uint bitIndex = 1, arrayIndex = 0; bitIndex != 0; bitIndex <<= 1, ++arrayIndex)
{
distanceToOnes[arrayIndex] = result;
result <<= 1;
if ((input & bitIndex) != 0)
{
result |= 1;
}
}
return distanceToOnes;
}
/// <summary>
/// Reverses the bits in the input.
///
/// As programmed this algorithm needs O(n) time and O(n) space.
/// (Where n is the number of bits in the input.)
/// </summary>
/// <param name="input"></param>
/// <returns></returns>
public static uint Reverse(uint input)
{
uint reversedInput = 0;
for (uint bitIndex = 1; bitIndex != 0; bitIndex <<= 1)
{
reversedInput <<= 1;
reversedInput |= (uint)((input & bitIndex) != 0 ? 1 : 0);
}
return reversedInput;
}
/// <summary>
/// Goes through each bit in the input, to check if there are any bits equally far away in
/// the distancesLower and distancesHigher
/// </summary>
public static void PrintHits(uint input, uint[] distancesLower, uint[] distancesHigher)
{
const int offset = uint32Length - 1;
for (uint bitIndex = 1, arrayIndex = 0; bitIndex != 0; bitIndex <<= 1, ++arrayIndex)
{
//hits checks if any bits are equally spaced away from our current value
bool isBitSet = (input & bitIndex) != 0;
uint hits = distancesLower[arrayIndex] & distancesHigher[offset - arrayIndex];
if (isBitSet && (hits != 0))
{
Console.WriteLine(String.Format("The {0}-th LSB has hits 0x{1:x4} away", arrayIndex + 1, hits));
}
}
}
}
}
Someone will probably comment that for any sufficiently large number, bitwise operations cannot be done in O(1). You'd be right. However, I'd conjecture that every solution that uses addition, subtraction, multiplication, or division (which cannot be done by shifting) would also have that problem.
Below is a solution. There could be some little mistakes here and there, but the idea is sound.
Edit: It's not n * log(n)
PSEUDO CODE:
foreach character in the string
if the character equals 1 {
if length cache > 0 { //we can skip the first one
foreach location in the cache { //last in first out kind of order
if ((currentlocation + (currentlocation - location)) < length string)
if (string[(currentlocation + (currentlocation - location))] equals 1)
return found evenly spaced string
else
break;
}
}
remember the location of this character in a some sort of cache.
}
return didn't find evenly spaced string
C# code:
public static Boolean FindThreeEvenlySpacedOnes(String str) {
List<int> cache = new List<int>();
for (var x = 0; x < str.Length; x++) {
if (str[x] == '1') {
if (cache.Count > 0) {
for (var i = cache.Count - 1; i > 0; i--) {
if ((x + (x - cache[i])) >= str.Length)
break;
if (str[(x + (x - cache[i]))] == '1')
return true;
}
}
cache.Add(x);
}
}
return false;
}
How it works:
iteration 1:
x
|
101101001
// the location of this 1 is stored in the cache
iteration 2:
x
|
101101001
iteration 3:
a x b
| | |
101101001
//we retrieve location a out of the cache and then based on a
//we calculate b and check if te string contains a 1 on location b
//and of course we store x in the cache because it's a 1
iteration 4:
axb
|||
101101001
a x b
| | |
101101001
iteration 5:
x
|
101101001
iteration 6:
a x b
| | |
101101001
a x b
| | |
101101001
//return found evenly spaced string
Obviously we need to at least check bunches of triplets at the same time, so we need to compress the checks somehow. I have a candidate algorithm, but analyzing the time complexity is beyond my ability*time threshold.
Build a tree where each node has three children and each node contains the total number of 1's at its leaves. Build a linked list over the 1's, as well. Assign each node an allowed cost proportional to the range it covers. As long as the time we spend at each node is within budget, we'll have an O(n lg n) algorithm.
--
Start at the root. If the square of the total number of 1's below it is less than its allowed cost, apply the naive algorithm. Otherwise recurse on its children.
Now we have either returned within budget, or we know that there are no valid triplets entirely contained within one of the children. Therefore we must check the inter-node triplets.
Now things get incredibly messy. We essentially want to recurse on the potential sets of children while limiting the range. As soon as the range is constrained enough that the naive algorithm will run under budget, you do it. Enjoy implementing this, because I guarantee it will be tedious. There's like a dozen cases.
--
The reason I think that algorithm will work is because the sequences without valid triplets appear to go alternate between bunches of 1's and lots of 0's. It effectively splits the nearby search space, and the tree emulates that splitting.
The run time of the algorithm is not obvious, at all. It relies on the non-trivial properties of the sequence. If the 1's are really sparse then the naive algorithm will work under budget. If the 1's are dense, then a match should be found right away. But if the density is 'just right' (eg. near ~n^0.63, which you can achieve by setting all bits at positions with no '2' digit in base 3), I don't know if it will work. You would have to prove that the splitting effect is strong enough.
No theoretical answer here, but I wrote a quick Java program to explore the running-time behavior as a function of k and n, where n is the total bit length and k is the number of 1's. I'm with a few of the answerers who are saying that the "regular" algorithm that checks all the pairs of bit positions and looks for the 3rd bit, even though it would require O(k^2) in the worst case, in reality because the worst-case needs sparse bitstrings, is O(n ln n).
Anyway here's the program, below. It's a Monte-Carlo style program which runs a large number of trials NTRIALS for constant n, and randomly generates bitsets for a range of k-values using Bernoulli processes with ones-density constrained between limits that can be specified, and records the running time of finding or failing to find a triplet of evenly spaced ones, time measured in steps NOT in CPU time. I ran it for n=64, 256, 1024, 4096, 16384* (still running), first a test run with 500000 trials to see which k-values take the longest running time, then another test with 5000000 trials with narrowed ones-density focus to see what those values look like. The longest running times do happen with very sparse density (e.g. for n=4096 the running time peaks are in the k=16-64 range, with a gentle peak for mean runtime at 4212 steps # k=31, max runtime peaked at 5101 steps # k=58). It looks like it would take extremely large values of N for the worst-case O(k^2) step to become larger than the O(n) step where you scan the bitstring to find the 1's position indices.
package com.example.math;
import java.io.PrintStream;
import java.util.BitSet;
import java.util.Random;
public class EvenlySpacedOnesTest {
static public class StatisticalSummary
{
private int n=0;
private double min=Double.POSITIVE_INFINITY;
private double max=Double.NEGATIVE_INFINITY;
private double mean=0;
private double S=0;
public StatisticalSummary() {}
public void add(double x) {
min = Math.min(min, x);
max = Math.max(max, x);
++n;
double newMean = mean + (x-mean)/n;
S += (x-newMean)*(x-mean);
// this algorithm for mean,std dev based on Knuth TAOCP vol 2
mean = newMean;
}
public double getMax() { return (n>0)?max:Double.NaN; }
public double getMin() { return (n>0)?min:Double.NaN; }
public int getCount() { return n; }
public double getMean() { return (n>0)?mean:Double.NaN; }
public double getStdDev() { return (n>0)?Math.sqrt(S/n):Double.NaN; }
// some may quibble and use n-1 for sample std dev vs population std dev
public static void printOut(PrintStream ps, StatisticalSummary[] statistics) {
for (int i = 0; i < statistics.length; ++i)
{
StatisticalSummary summary = statistics[i];
ps.printf("%d\t%d\t%.0f\t%.0f\t%.5f\t%.5f\n",
i,
summary.getCount(),
summary.getMin(),
summary.getMax(),
summary.getMean(),
summary.getStdDev());
}
}
}
public interface RandomBernoulliProcess // see http://en.wikipedia.org/wiki/Bernoulli_process
{
public void setProbability(double d);
public boolean getNextBoolean();
}
static public class Bernoulli implements RandomBernoulliProcess
{
final private Random r = new Random();
private double p = 0.5;
public boolean getNextBoolean() { return r.nextDouble() < p; }
public void setProbability(double d) { p = d; }
}
static public class TestResult {
final public int k;
final public int nsteps;
public TestResult(int k, int nsteps) { this.k=k; this.nsteps=nsteps; }
}
////////////
final private int n;
final private int ntrials;
final private double pmin;
final private double pmax;
final private Random random = new Random();
final private Bernoulli bernoulli = new Bernoulli();
final private BitSet bits;
public EvenlySpacedOnesTest(int n, int ntrials, double pmin, double pmax) {
this.n=n; this.ntrials=ntrials; this.pmin=pmin; this.pmax=pmax;
this.bits = new BitSet(n);
}
/*
* generate random bit string
*/
private int generateBits()
{
int k = 0; // # of 1's
for (int i = 0; i < n; ++i)
{
boolean b = bernoulli.getNextBoolean();
this.bits.set(i, b);
if (b) ++k;
}
return k;
}
private int findEvenlySpacedOnes(int k, int[] pos)
{
int[] bitPosition = new int[k];
for (int i = 0, j = 0; i < n; ++i)
{
if (this.bits.get(i))
{
bitPosition[j++] = i;
}
}
int nsteps = n; // first, it takes N operations to find the bit positions.
boolean found = false;
if (k >= 3) // don't bother doing anything if there are less than 3 ones. :(
{
int lastBitSetPosition = bitPosition[k-1];
for (int j1 = 0; !found && j1 < k; ++j1)
{
pos[0] = bitPosition[j1];
for (int j2 = j1+1; !found && j2 < k; ++j2)
{
pos[1] = bitPosition[j2];
++nsteps;
pos[2] = 2*pos[1]-pos[0];
// calculate 3rd bit index that might be set;
// the other two indices point to bits that are set
if (pos[2] > lastBitSetPosition)
break;
// loop inner loop until we go out of bounds
found = this.bits.get(pos[2]);
// we're done if we find a third 1!
}
}
}
if (!found)
pos[0]=-1;
return nsteps;
}
/*
* run an algorithm that finds evenly spaced ones and returns # of steps.
*/
public TestResult run()
{
bernoulli.setProbability(pmin + (pmax-pmin)*random.nextDouble());
// probability of bernoulli process is randomly distributed between pmin and pmax
// generate bit string.
int k = generateBits();
int[] pos = new int[3];
int nsteps = findEvenlySpacedOnes(k, pos);
return new TestResult(k, nsteps);
}
public static void main(String[] args)
{
int n;
int ntrials;
double pmin = 0, pmax = 1;
try {
n = Integer.parseInt(args[0]);
ntrials = Integer.parseInt(args[1]);
if (args.length >= 3)
pmin = Double.parseDouble(args[2]);
if (args.length >= 4)
pmax = Double.parseDouble(args[3]);
}
catch (Exception e)
{
System.out.println("usage: EvenlySpacedOnesTest N NTRIALS [pmin [pmax]]");
System.exit(0);
return; // make the compiler happy
}
final StatisticalSummary[] statistics;
statistics=new StatisticalSummary[n+1];
for (int i = 0; i <= n; ++i)
{
statistics[i] = new StatisticalSummary();
}
EvenlySpacedOnesTest test = new EvenlySpacedOnesTest(n, ntrials, pmin, pmax);
int printInterval=100000;
int nextPrint = printInterval;
for (int i = 0; i < ntrials; ++i)
{
TestResult result = test.run();
statistics[result.k].add(result.nsteps);
if (i == nextPrint)
{
System.err.println(i);
nextPrint += printInterval;
}
}
StatisticalSummary.printOut(System.out, statistics);
}
}
# <algorithm>
def contains_evenly_spaced?(input)
return false if input.size < 3
one_indices = []
input.each_with_index do |digit, index|
next if digit == 0
one_indices << index
end
return false if one_indices.size < 3
previous_indexes = []
one_indices.each do |index|
if !previous_indexes.empty?
previous_indexes.each do |previous_index|
multiple = index - previous_index
success_index = index + multiple
return true if input[success_index] == 1
end
end
previous_indexes << index
end
return false
end
# </algorithm>
def parse_input(input)
input.chars.map { |c| c.to_i }
end
I'm having trouble with the worst-case scenarios with millions of digits. Fuzzing from /dev/urandom essentially gives you O(n), but I know the worst case is worse than that. I just can't tell how much worse. For small n, it's trivial to find inputs at around 3*n*log(n), but it's surprisingly hard to differentiate those from some other order of growth for this particular problem.
Can anyone who was working on worst-case inputs generate a string with length greater than say, one hundred thousand?
An adaptation of the Rabin-Karp algorithm could be possible for you.
Its complexity is 0(n) so it could help you.
Take a look http://en.wikipedia.org/wiki/Rabin-Karp_string_search_algorithm
Could this be a solution? I', not sure if it's O(nlogn) but in my opinion it's better than O(n²) because the the only way not to find a triple would be a prime number distribution.
There's room for improvement, the second found 1 could be the next first 1. Also no error checking.
#include <iostream>
#include <string>
int findIt(std::string toCheck) {
for (int i=0; i<toCheck.length(); i++) {
if (toCheck[i]=='1') {
std::cout << i << ": " << toCheck[i];
for (int j = i+1; j<toCheck.length(); j++) {
if (toCheck[j]=='1' && toCheck[(i+2*(j-i))] == '1') {
std::cout << ", " << j << ":" << toCheck[j] << ", " << (i+2*(j-i)) << ":" << toCheck[(i+2*(j-i))] << " found" << std::endl;
return 0;
}
}
}
}
return -1;
}
int main (int agrc, char* args[]) {
std::string toCheck("1001011");
findIt(toCheck);
std::cin.get();
return 0;
}
I think this algorithm has O(n log n) complexity (C++, DevStudio 2k5). Now, I don't know the details of how to analyse an algorithm to determine its complexity, so I have added some metric gathering information to the code. The code counts the number of tests done on the sequence of 1's and 0's for any given input (hopefully, I've not made a balls of the algorithm). We can compare the actual number of tests against the O value and see if there's a correlation.
#include <iostream>
using namespace std;
bool HasEvenBits (string &sequence, int &num_compares)
{
bool
has_even_bits = false;
num_compares = 0;
for (unsigned i = 1 ; i <= (sequence.length () - 1) / 2 ; ++i)
{
for (unsigned j = 0 ; j < sequence.length () - 2 * i ; ++j)
{
++num_compares;
if (sequence [j] == '1' && sequence [j + i] == '1' && sequence [j + i * 2] == '1')
{
has_even_bits = true;
// we could 'break' here, but I want to know the worst case scenario so keep going to the end
}
}
}
return has_even_bits;
}
int main ()
{
int
count;
string
input = "111";
for (int i = 3 ; i < 32 ; ++i)
{
HasEvenBits (input, count);
cout << i << ", " << count << endl;
input += "0";
}
}
This program outputs the number of tests for each string length up to 32 characters. Here's the results:
n Tests n log (n)
=====================
3 1 1.43
4 2 2.41
5 4 3.49
6 6 4.67
7 9 5.92
8 12 7.22
9 16 8.59
10 20 10.00
11 25 11.46
12 30 12.95
13 36 14.48
14 42 16.05
15 49 17.64
16 56 19.27
17 64 20.92
18 72 22.59
19 81 24.30
20 90 26.02
21 100 27.77
22 110 29.53
23 121 31.32
24 132 33.13
25 144 34.95
26 156 36.79
27 169 38.65
28 182 40.52
29 196 42.41
30 210 44.31
31 225 46.23
I've added the 'n log n' values as well. Plot these using your graphing tool of choice to see a correlation between the two results. Does this analysis extend to all values of n? I don't know.
I am trying to test the likelihood that a particular clustering of data has occurred by chance. A robust way to do this is Monte Carlo simulation, in which the associations between data and groups are randomly reassigned a large number of times (e.g. 10,000), and a metric of clustering is used to compare the actual data with the simulations to determine a p value.
I've got most of this working, with pointers mapping the grouping to the data elements, so I plan to randomly reassign pointers to data. THE QUESTION: what is a fast way to sample without replacement, so that every pointer is randomly reassigned in the replicate data sets?
For example (these data are just a simplified example):
Data (n=12 values) - Group A: 0.1, 0.2, 0.4 / Group B: 0.5, 0.6, 0.8 / Group C: 0.4, 0.5 / Group D: 0.2, 0.2, 0.3, 0.5
For each replicate data set, I would have the same cluster sizes (A=3, B=3, C=2, D=4) and data values, but would reassign the values to the clusters.
To do this, I could generate random numbers in the range 1-12, assign the first element of group A, then generate random numbers in the range 1-11 and assign the second element in group A, and so on. The pointer reassignment is fast, and I will have pre-allocated all data structures, but the sampling without replacement seems like a problem that might have been solved many times before.
Logic or pseudocode preferred.
Here's some code for sampling without replacement based on Algorithm 3.4.2S of Knuth's book Seminumeric Algorithms.
void SampleWithoutReplacement
(
int populationSize, // size of set sampling from
int sampleSize, // size of each sample
vector<int> & samples // output, zero-offset indicies to selected items
)
{
// Use Knuth's variable names
int& n = sampleSize;
int& N = populationSize;
int t = 0; // total input records dealt with
int m = 0; // number of items selected so far
double u;
while (m < n)
{
u = GetUniform(); // call a uniform(0,1) random number generator
if ( (N - t)*u >= n - m )
{
t++;
}
else
{
samples[m] = t;
t++; m++;
}
}
}
There is a more efficient but more complex method by Jeffrey Scott Vitter in "An Efficient Algorithm for Sequential Random Sampling," ACM Transactions on Mathematical Software, 13(1), March 1987, 58-67.
A C++ working code based on the answer by John D. Cook.
#include <random>
#include <vector>
// John D. Cook, https://stackoverflow.com/a/311716/15485
void SampleWithoutReplacement
(
int populationSize, // size of set sampling from
int sampleSize, // size of each sample
std::vector<int> & samples // output, zero-offset indicies to selected items
)
{
// Use Knuth's variable names
int& n = sampleSize;
int& N = populationSize;
int t = 0; // total input records dealt with
int m = 0; // number of items selected so far
std::default_random_engine re;
std::uniform_real_distribution<double> dist(0,1);
while (m < n)
{
double u = dist(re); // call a uniform(0,1) random number generator
if ( (N - t)*u >= n - m )
{
t++;
}
else
{
samples[m] = t;
t++; m++;
}
}
}
#include <iostream>
int main(int,char**)
{
const size_t sz = 10;
std::vector< int > samples(sz);
SampleWithoutReplacement(10*sz,sz,samples);
for (size_t i = 0; i < sz; i++ ) {
std::cout << samples[i] << "\t";
}
return 0;
}
See my answer to this question Unique (non-repeating) random numbers in O(1)?. The same logic should accomplish what you are looking to do.
Inspired by #John D. Cook's answer, I wrote an implementation in Nim. At first I had difficulties understanding how it works, so I commented extensively also including an example. Maybe it helps to understand the idea. Also, I have changed the variable names slightly.
iterator uniqueRandomValuesBelow*(N, M: int) =
## Returns a total of M unique random values i with 0 <= i < N
## These indices can be used to construct e.g. a random sample without replacement
assert(M <= N)
var t = 0 # total input records dealt with
var m = 0 # number of items selected so far
while (m < M):
let u = random(1.0) # call a uniform(0,1) random number generator
# meaning of the following terms:
# (N - t) is the total number of remaining draws left (initially just N)
# (M - m) is the number how many of these remaining draw must be positive (initially just M)
# => Probability for next draw = (M-m) / (N-t)
# i.e.: (required positive draws left) / (total draw left)
#
# This is implemented by the inequality expression below:
# - the larger (M-m), the larger the probability of a positive draw
# - for (N-t) == (M-m), the term on the left is always smaller => we will draw 100%
# - for (N-t) >> (M-m), we must get a very small u
#
# example: (N-t) = 7, (M-m) = 5
# => we draw the next with prob 5/7
# lets assume the draw fails
# => t += 1 => (N-t) = 6
# => we draw the next with prob 5/6
# lets assume the draw succeeds
# => t += 1, m += 1 => (N-t) = 5, (M-m) = 4
# => we draw the next with prob 4/5
# lets assume the draw fails
# => t += 1 => (N-t) = 4
# => we draw the next with prob 4/4, i.e.,
# we will draw with certainty from now on
# (in the next steps we get prob 3/3, 2/2, ...)
if (N - t)*u >= (M - m).toFloat: # this is essentially a draw with P = (M-m) / (N-t)
# no draw -- happens mainly for (N-t) >> (M-m) and/or high u
t += 1
else:
# draw t -- happens when (M-m) gets large and/or low u
yield t # this is where we output an index, can be used to sample
t += 1
m += 1
# example use
for i in uniqueRandomValuesBelow(100, 5):
echo i
When the population size is much greater than the sample size, the above algorithms become inefficient, since they have complexity O(n), n being the population size.
When I was a student I wrote some algorithms for uniform sampling without replacement, which have average complexity O(s log s), where s is the sample size. Here is the code for the binary tree algorithm, with average complexity O(s log s), in R:
# The Tree growing algorithm for uniform sampling without replacement
# by Pavel Ruzankin
quicksample = function (n,size)
# n - the number of items to choose from
# size - the sample size
{
s=as.integer(size)
if (s>n) {
stop("Sample size is greater than the number of items to choose from")
}
# upv=integer(s) #level up edge is pointing to
leftv=integer(s) #left edge is poiting to; must be filled with zeros
rightv=integer(s) #right edge is pointig to; must be filled with zeros
samp=integer(s) #the sample
ordn=integer(s) #relative ordinal number
ordn[1L]=1L #initial value for the root vertex
samp[1L]=sample(n,1L)
if (s > 1L) for (j in 2L:s) {
curn=sample(n-j+1L,1L) #current number sampled
curordn=0L #currend ordinal number
v=1L #current vertice
from=1L #how have come here: 0 - by left edge, 1 - by right edge
repeat {
curordn=curordn+ordn[v]
if (curn+curordn>samp[v]) { #going down by the right edge
if (from == 0L) {
ordn[v]=ordn[v]-1L
}
if (rightv[v]!=0L) {
v=rightv[v]
from=1L
} else { #creating a new vertex
samp[j]=curn+curordn
ordn[j]=1L
# upv[j]=v
rightv[v]=j
break
}
} else { #going down by the left edge
if (from==1L) {
ordn[v]=ordn[v]+1L
}
if (leftv[v]!=0L) {
v=leftv[v]
from=0L
} else { #creating a new vertex
samp[j]=curn+curordn-1L
ordn[j]=-1L
# upv[j]=v
leftv[v]=j
break
}
}
}
}
return(samp)
}
The complexity of this algorithm is discussed in:
Rouzankin, P. S.; Voytishek, A. V. On the cost of algorithms for random selection. Monte Carlo Methods Appl. 5 (1999), no. 1, 39-54.
http://dx.doi.org/10.1515/mcma.1999.5.1.39
If you find the algorithm useful, please make a reference.
See also:
P. Gupta, G. P. Bhattacharjee. (1984) An efficient algorithm for random sampling without replacement. International Journal of Computer Mathematics 16:4, pages 201-209.
DOI: 10.1080/00207168408803438
Teuhola, J. and Nevalainen, O. 1982. Two efficient algorithms for random sampling without replacement. /IJCM/, 11(2): 127–140.
DOI: 10.1080/00207168208803304
In the last paper the authors use hash tables and claim that their algorithms have O(s) complexity. There is one more fast hash table algorithm, which will soon be implemented in pqR (pretty quick R):
https://stat.ethz.ch/pipermail/r-devel/2017-October/075012.html
I wrote a survey of algorithms for sampling without replacement. I may be biased but I recommend my own algorithm, implemented in C++ below, as providing the best performance for many k, n values and acceptable performance for others. randbelow(i) is assumed to return a fairly chosen random non-negative integer less than i.
void cardchoose(uint32_t n, uint32_t k, uint32_t* result) {
auto t = n - k + 1;
for (uint32_t i = 0; i < k; i++) {
uint32_t r = randbelow(t + i);
if (r < t) {
result[i] = r;
} else {
result[i] = result[r - t];
}
}
std::sort(result, result + k);
for (uint32_t i = 0; i < k; i++) {
result[i] += i;
}
}
Another algorithm for sampling without replacement is described here.
It is similar to the one described by John D. Cook in his answer and also from Knuth, but it has different hypothesis: The population size is unknown, but the sample can fit in memory. This one is called "Knuth's algorithm S".
Quoting the rosettacode article:
Select the first n items as the sample as they become available;
For the i-th item where i > n, have a random chance of n/i of keeping it. If failing this chance, the sample remains the same. If
not, have it randomly (1/n) replace one of the previously selected n
items of the sample.
Repeat #2 for any subsequent items.