I have been battling to put up an algorithm to solve this problem.
Let say i have a set of number {1, 2, 5} and each element of the this set as unlimited supply, and i given another number 6, then ask to determine the number of ways you can sum the elements to get the number 6. For illustration purpose i do this
1 + 1 + 1 + 1 + 1 + 1 = 6
1 + 1 + 2 + 2 = 6
2 + 2 + 2 = 6
1 + 5 = 6
1 + 1 + 1 + 1 + 2 = 6
So in this case the program will output 5 as the number of ways. Again let say you are to find the sum for 4,
1 + 1 + 1 + 1 = 4
2 + 2 = 4
1 + 1 + 2 = 4
In this case the algorithm will output 3 as the number of way
This is similar to sum of subsets problem . I am sure you have to use branch and bound method or backtracking method.
1)Create a state space tree which consist of all possible cases.
0
/ | \
1 2 5
/ | \
1 2 5 ........
2) Continue the process until the sum of nodes in depth first manner is greater or equal to your desired number.
3) Count the no. of full branches that satisfy your condition.
The python implementation of similar problem can be found here.
This is a good problem to use recursion and dynamic programming techniques. Here is an implementation in Python using the top-down approach (memoization) to avoid doing the same calculation multiple times:
# Remember answers for subsets
cache = {}
# Return the ways to get the desired sum from combinations of the given numbers
def possible_sums(numbers, desired_sum):
# See if we have already calculated this possibility
key = (tuple(set(numbers)), desired_sum)
if key in cache:
return cache[key]
answers = {}
for n in numbers:
if desired_sum % n == 0:
# The sum is a multiple of the number
answers[tuple([n] * (desired_sum / n))] = True
if n < desired_sum:
for a in possible_sums(numbers, desired_sum - n):
answers[tuple([n] + a)] = True
cache[key] = [list(k) for k in answers.iterkeys()]
return cache[key]
# Return only distinct combinations of sums, ignoring order
def unique_possible_sums(numbers, desired_sum):
answers = {}
for s in possible_sums(numbers, desired_sum):
answers[tuple(sorted(s))] = True
return [list(k) for k in answers.iterkeys()]
for s in unique_possible_sums([1, 2, 5], 6):
print '6: ' + repr(s)
for s in unique_possible_sums([1, 2, 5], 4):
print '4: ' + repr(s)
For smaller target number(~1000000) and 1000{supply} n try this:
The supply of numbers you have
supply {a,b,c....}
The target you need
steps[n]
1 way to get to 0 use nothing
steps[0]=1
Scan till target number
for i from 1 to n:
for each supply x:
if i - x >=0
steps[i] += steps[i-x]
Steps at n will contain the number of ways
steps[n]
Visualization of the above:
supply {1, 2, 5} , target 6
i = 1, x=1 and steps required is 1
i = 2, x=1 and steps required is 1
i = 2, x=2 and steps required is 2
i = 3, x=1 and steps required is 2
i = 3, x=2 and steps required is 3
i = 4, x=1 and steps required is 3
i = 4, x=2 and steps required is 5
i = 5, x=1 and steps required is 5
i = 5, x=2 and steps required is 8
i = 5, x=5 and steps required is 9
i = 6, x=1 and steps required is 9
i = 6, x=2 and steps required is 14
i = 6, x=5 and steps required is 15
Some Java Code
private int test(int targetSize, int supply[]){
int target[] = new int[targetSize+1];
target[0]=1;
for(int i=0;i<=targetSize;i++){
for(int x:supply){
if(i-x >= 0){
target[i]+=target[i-x];
}
}
}
return target[targetSize];
}
#Test
public void test(){
System.err.println(test(12, new int[]{1,2,3,4,5,6}));
}
I need to write a program to input a number and output its factorial's prime factorization in the form:
4!=(2^3)*(3^1)
5!=(2^3)*(3^1)*(5^1)
The problem is I still can't figure out how to get that result.
Apparently each first number in brackets is for the ascending prime numbers up until the actual factorial. The second number in brackets is the amount of times the number occurs in the factorial.
What I can't figure out is for example in 5!=(2^3)*(3^1)*(5^1), how does 2 only occur 3 times, 3 only 1 time and 5 only one time in 120 (5!=120).
I have now solved this thanks to the helpful people who commented but I'm now having trouble trying to figure out how could I take a number and get the factorial in this format without actually calculating the factorial.
Every number can be represented by a unique (up to re-ordering) multiplication of prime numbers, called the prime factorization of the number, as you are finding the prime factors that can uniquely create that number.
2^3=8
3^1=3
5^1=5
and 8*3*5=120
But this also means that: (2^3)*(3^1)*(5^1) = 120
It's not saying that 2 occurs 3 times as a digit in the number 120, which it obviously does not, but rather to multiply 2 by 2 by 2, for a total of 3 twos. Likewise for the 3 and 5, which occur once in the prime factorization of 120. The expression which you mention is showing you this unique prime factorization of the number 120. This is one way of getting the prime factorization of a number in Python:
def pf(number):
factors=[]
d=2
while(number>1):
while(number%d==0):
factors.append(d)
number=number/d
d+=1
return factors
Running it you get:
>>> pf(120)
[2, 2, 2, 3, 5]
Which multiplied together give you 120, as explained above. Here's a little diagram to illustrate this more clearly:
Consider e.g. 33!. It's a product of:
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33
the factors are:
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2
2 2 2 2
2 2
2
3 3 3 3 3 3 3 3 3 3 3
3 3 3
3
5 5 5 5 5 5
5
7 7 7 7
11 11 11
13 13
17
19
23
29 31
Do you see the pattern?
33! = 2^( 33 div 2 + 33 div 4 + 33 div 8 + 33 div 16 + 33 div 32) *
3^( 33 div 3 + 33 div 9 + 33 div 27) *
5^( 33 div 5 + 33 div 25) *
----
7^( 33 div 7) * 11^( 33 div 11) * 13^( 33 div 13) *
----
17 * 19 * 23 * 29 * 31
Thus, to find prime factorization of n! without doing any multiplications or factorizations, we just need to have the ordered list of primes not greater than n, which we process (with a repeated integer division and a possible summation) in three stages - primes that are smaller or equal to the square root of n; such that are smaller or equal to n/2; and the rest.
Actually with lazy evaluation it's even simpler than that. Assuming primes is already implemented returning a stream of prime numbers in order, in Haskell, factorial factorization is found as
ff n = [(p, sum . takeWhile (> 0) . tail . iterate (`div` p) $ n)
| p <- takeWhile (<= n) primes]
-- Prelude> ff 33
-- [(2,31),(3,15),(5,7),(7,4),(11,3),(13,2),(17,1),(19,1),(23,1),(29,1),(31,1)]
because 33 div 4 is (33 div 2) div 2, etc..
2^3 is another way of writing 23, or two to the third power. (2^3)(3^1)(5^1) = 23 × 3 × 5 = 120.
(2^3)(3^1)(5^1) is just the prime factorization of 120 expressed in plain ASCII text rather than with pretty mathematical formatting. Your assignment requires output in this form simply because it's easier for you to output than it would be for you to figure out how to output formatted equations (and probably because it's easier to process for grading).
The conventions used here for expressing equations in plain text are standard enough that you can directly type this text into google.com or wolframalpha.com and it will calculate the result as 120 for you: (2^3)(3^1)(5^1) on wolframalpha.com / (2^3)(3^1)(5^1) on google.com
WolframAlpha can also compute prime factorizations, which you can use to get test results to compare your program with. For example: prime factorization of 1000!
A naïve solution that actually calculates the factorial will only handle numbers up to 12 (if using 32 bit ints). This is because 13! is ~6.2 billion, larger than the largest number that can be represented in a 32 bit int.
However it's possible to handle much larger inputs if you avoid calculating the factorial first. I'm not going to tell you exactly how to do that because either figuring it out is part of your assignment or you can ask your prof/TAs. But below are some hints.
ab × ac = ab+c
equation (a) 10 = 21 × 51
equation (b) 15 = 31 × 51
10 × 15 = ? Answer using the right hand sides of equations (a) and (b), not with the number 150.
10 × 15 = (21 × 51) × (31 × 51) = 21 × 31 × (51 × 51) = 21 × 31 × 52
As you can see, computing the prime factorization of 10 × 15 can be done without multiplying 10 by 15; You can instead compute the prime factorization of the individual terms and then combine those factorizations.
If you write out the factorial 5!:
1 * 2 * 3 * 4 * 5,
you will notice that there is one non-prime number: 4. 4 can be written as 2 * 2 or 2^2, which is where the extra 2s come from.
Add up all of the occurrences (exponential forms are in parentheses; add exponents for like bases):
2 (2^1) * 3 (3^1) * 4 (2^2) * 5 (5^1), you get the proper answer.
You can use O(n/2 log log n) algorithm using only sums (no need precalc primes).
This is a sieve using relation
f = a * b ~> f^k = a^k * b^k
then, we reduce all initial factors 1 * 2 * 3 * ... * n moving k from big numbers to small numbers.
Using Sieve of Atkin the Will Ness algorithm could be better for very big n if not, I think it could be better
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char **argv) {
int n = atoi(argv[1]);
int *p = (int *) malloc(sizeof(int) * (n + 1));
int i, j, d;
for(i = 0; i <= n; i++)
p[i] = 1;
for(i = n >> 1; i > 1; i--)
if(p[i]) {
for(j = i + i, d = 2; j <= n; j += i, d++) {
if(p[j]) {
p[i] += p[j];
p[d] += p[j];
p[j] = 0;
}
}
}
printf("1");
for(i = 2; i <= n; i++)
if(p[i])
printf(" * %i^%i", i, p[i]);
printf("\n");
return 0;
}
I was wondering if there is an algorithm that checks wether a given number is factorable into a set of prime numbers and if no give out the nearest number.
The problem comes always up when I use the FFT.
Thanks a lot for your help guys.
In general this looks like a hard problem, particularly finding the next largest integer that factors into your set of primes. However, if your set of primes isn't too big, one approach would be to turn this into an integer optimization problem by taking the logs. Here is how to find the smallest number > n that factors into a set of primes p_1...p_k
choose integers x_1,...,x_k to minimize (x_1 log p_1 + ... + x_k log p_k - log n)
Subject to:
x_1 log p_1 + ... + x_k log p_k >= log n
x_i >= 0 for all i
The x_i will give you the exponents for the primes. Here is an implementation in R using lpSolve:
minfact<-function(x,p){
sol<-lp("min",log(p),t(log(p)),">=",log(x),all.int=T)
prod(p^sol$solution)
}
> p<-c(2,3,13,31)
> x<-124363183
> y<-minfact(x,p)
> y
[1] 124730112
> factorize(y)
Big Integer ('bigz') object of length 13:
[1] 2 2 2 2 2 2 2 2 3 13 13 31 31
> y-x
[1] 366929
>
Using big integers, this works pretty well even for large numbers:
> p<-c(2,3,13,31,53,79)
> x<-as.bigz("1243631831278461278641361")
> y<-minfact(x,p)
y
>
Big Integer ('bigz') :
[1] 1243634072805560436129792
> factorize(y)
Big Integer ('bigz') object of length 45:
[1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
[26] 2 2 2 2 2 2 2 2 3 3 3 3 13 31 31 31 31 53 53 53
>
Your question is about well-known factorization problem - which could not be resolved with 'fast' (polynomial) time. Lenstra's elliptic algorithm is the most efficient (known) way in common case, but it requires strong knowledge of numbers theory - and it's also sub-exponential (but not polynomial).
Other algorithms are listed in the page by first link in my post, but such things as direct try (brute force) are much more slower, of cause.
Please, note, that under "could not be resolved with polynomial time" - I mean that there's no way to do this now - but not that such way does not exist (at least now, number theory can not provide such solution for this problem)
Here is a brute force method in C++. It returns the factorization of the nearest factorable number. If N has two equidistant factorable neighbours, it returns the smallest one.
GCC 4.7.3: g++ -Wall -Wextra -std=c++0x factorable-neighbour.cpp
#include <iostream>
#include <vector>
using ints = std::vector<int>;
ints factor(int n, const ints& primes) {
ints f(primes.size(), 0);
for (int i = 0; i < primes.size(); ++i) {
while (0< n && !(n % primes[i])) {
n /= primes[i];
++f[i]; } }
// append the "remainder"
f.push_back(n);
return f;
}
ints closest_factorable(int n, const ints& primes) {
int d = 0;
ints r;
while (true) {
r = factor(n + d, primes);
if (r[r.size() - 1] == 1) { break; }
++d;
r = factor(n - d, primes);
if (r[r.size() - 1] == 1) { break; }
}
r.pop_back();
return r; }
int main() {
for (int i = 0; i < 30; ++i) {
for (const auto& f : closest_factorable(i, {2, 3, 5, 7, 11})) {
std::cout << f << " "; }
std::cout << "\n"; }
}
I suppose that you have a (small) set of prime numbers S and an integer n and you want to know is n factors only using number in S. The easiest way seems to be the following:
P <- product of s in S
while P != 1 do
P <- GCD(P, n)
n <- n/P
return n == 1
You compute the GCD using Euclid's algorithm.
The idea is the following: Suppose that S = {p1, p2, ... ,pk}. You can write n uniquely as
n = p1^n1 p2^n2 ... pk^nk * R
where R is coprime wrt the pi. You want to know whether R=1.
Then
GCD(n, P) = prod ( pi such that ni <> 0 ).
Therefore n/p decrease all those non zeros ni by 1 so that they eventually become 0. At the end only R remains.
For example: S = {2,3,5}, n = 5600 = 2^5*5^2*7. Then P = 2*3*5 = 30. One gets GCD(n, p)=10=2*5. And therefore n/GCD(n,p) = 560 = 2^4*5*7.
You are now back to the same problem: You want to know if 560 can be factored using S = {2,5} hence the loop. So the next steps are
GCD(560, 10) = 10. 560/GCD = 56 = 2^3 * 7.
GCD(56, 10) = 2. 56/2 = 28 = 2^2 * 7
GCD(28, 2) = 2. 28/2 = 14 = 2 * 7
GCD(14, 2) = 2. 14/2 = 7
GCD(7, 2) = 1 so that R = 7 ! Your answer if FALSE.
kissfft has a function
int kiss_fft_next_fast_size(int n)
that returns the next largest N that is an aggregate of 2,3,5.
Also related is a kf_factor function that factorizes a number n, pulling out the "nice" FFT primes first (e.g. 4's are pulled out before 2's)
Numbers whose only prime factors are 2, 3, or 5 are called ugly numbers.
Example:
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, ...
1 can be considered as 2^0.
I am working on finding nth ugly number. Note that these numbers are extremely sparsely distributed as n gets large.
I wrote a trivial program that computes if a given number is ugly or not. For n > 500 - it became super slow. I tried using memoization - observation: ugly_number * 2, ugly_number * 3, ugly_number * 5 are all ugly. Even with that it is slow. I tried using some properties of log - since that will reduce this problem from multiplication to addition - but, not much luck yet. Thought of sharing this with you all. Any interesting ideas?
Using a concept similar to Sieve of Eratosthenes (thanks Anon)
for (int i(2), uglyCount(0); ; i++) {
if (i % 2 == 0)
continue;
if (i % 3 == 0)
continue;
if (i % 5 == 0)
continue;
uglyCount++;
if (uglyCount == n - 1)
break;
}
i is the nth ugly number.
Even this is pretty slow. I am trying to find the 1500th ugly number.
A simple fast solution in Java. Uses approach described by Anon..
Here TreeSet is just a container capable of returning smallest element in it. (No duplicates stored.)
int n = 20;
SortedSet<Long> next = new TreeSet<Long>();
next.add((long) 1);
long cur = 0;
for (int i = 0; i < n; ++i) {
cur = next.first();
System.out.println("number " + (i + 1) + ": " + cur);
next.add(cur * 2);
next.add(cur * 3);
next.add(cur * 5);
next.remove(cur);
}
Since 1000th ugly number is 51200000, storing them in bool[] isn't really an option.
edit
As a recreation from work (debugging stupid Hibernate), here's completely linear solution. Thanks to marcog for idea!
int n = 1000;
int last2 = 0;
int last3 = 0;
int last5 = 0;
long[] result = new long[n];
result[0] = 1;
for (int i = 1; i < n; ++i) {
long prev = result[i - 1];
while (result[last2] * 2 <= prev) {
++last2;
}
while (result[last3] * 3 <= prev) {
++last3;
}
while (result[last5] * 5 <= prev) {
++last5;
}
long candidate1 = result[last2] * 2;
long candidate2 = result[last3] * 3;
long candidate3 = result[last5] * 5;
result[i] = Math.min(candidate1, Math.min(candidate2, candidate3));
}
System.out.println(result[n - 1]);
The idea is that to calculate a[i], we can use a[j]*2 for some j < i. But we also need to make sure that 1) a[j]*2 > a[i - 1] and 2) j is smallest possible.
Then, a[i] = min(a[j]*2, a[k]*3, a[t]*5).
I am working on finding nth ugly number. Note that these numbers are extremely sparsely distributed as n gets large.
I wrote a trivial program that computes if a given number is ugly or not.
This looks like the wrong approach for the problem you're trying to solve - it's a bit of a shlemiel algorithm.
Are you familiar with the Sieve of Eratosthenes algorithm for finding primes? Something similar (exploiting the knowledge that every ugly number is 2, 3 or 5 times another ugly number) would probably work better for solving this.
With the comparison to the Sieve I don't mean "keep an array of bools and eliminate possibilities as you go up". I am more referring to the general method of generating solutions based on previous results. Where the Sieve gets a number and then removes all multiples of it from the candidate set, a good algorithm for this problem would start with an empty set and then add the correct multiples of each ugly number to that.
My answer refers to the correct answer given by Nikita Rybak.
So that one could see a transition from the idea of the first approach to that of the second.
from collections import deque
def hamming():
h=1;next2,next3,next5=deque([]),deque([]),deque([])
while True:
yield h
next2.append(2*h)
next3.append(3*h)
next5.append(5*h)
h=min(next2[0],next3[0],next5[0])
if h == next2[0]: next2.popleft()
if h == next3[0]: next3.popleft()
if h == next5[0]: next5.popleft()
What's changed from Nikita Rybak's 1st approach is that, instead of adding next candidates into single data structure, i.e. Tree set, one can add each of them separately into 3 FIFO lists. This way, each list will be kept sorted all the time, and the next least candidate must always be at the head of one ore more of these lists.
If we eliminate the use of the three lists above, we arrive at the second implementation in Nikita Rybak' answer. This is done by evaluating those candidates (to be contained in three lists) only when needed, so that there is no need to store them.
Simply put:
In the first approach, we put every new candidate into single data structure, and that's bad because too many things get mixed up unwisely. This poor strategy inevitably entails O(log(tree size)) time complexity every time we make a query to the structure. By putting them into separate queues, however, you will see that each query takes only O(1) and that's why the overall performance reduces to O(n)!!! This is because each of the three lists is already sorted, by itself.
I believe you can solve this problem in sub-linear time, probably O(n^{2/3}).
To give you the idea, if you simplify the problem to allow factors of just 2 and 3, you can achieve O(n^{1/2}) time starting by searching for the smallest power of two that is at least as large as the nth ugly number, and then generating a list of O(n^{1/2}) candidates. This code should give you an idea how to do it. It relies on the fact that the nth number containing only powers of 2 and 3 has a prime factorization whose sum of exponents is O(n^{1/2}).
def foo(n):
p2 = 1 # current power of 2
p3 = 1 # current power of 3
e3 = 0 # exponent of current power of 3
t = 1 # number less than or equal to the current power of 2
while t < n:
p2 *= 2
if p3 * 3 < p2:
p3 *= 3
e3 += 1
t += 1 + e3
candidates = [p2]
c = p2
for i in range(e3):
c /= 2
c *= 3
if c > p2: c /= 2
candidates.append(c)
return sorted(candidates)[n - (t - len(candidates))]
The same idea should work for three allowed factors, but the code gets more complex. The sum of the powers of the factorization drops to O(n^{1/3}), but you need to consider more candidates, O(n^{2/3}) to be more precise.
A lot of good answers here, but I was having trouble understanding those, specifically how any of these answers, including the accepted one, maintained the axiom 2 in Dijkstra's original paper:
Axiom 2. If x is in the sequence, so is 2 * x, 3 * x, and 5 * x.
After some whiteboarding, it became clear that the axiom 2 is not an invariant at each iteration of the algorithm, but actually the goal of the algorithm itself. At each iteration, we try to restore the condition in axiom 2. If last is the last value in the result sequence S, axiom 2 can simply be rephrased as:
For some x in S, the next value in S is the minimum of 2x,
3x, and 5x, that is greater than last. Let's call this axiom 2'.
Thus, if we can find x, we can compute the minimum of 2x, 3x, and 5x in constant time, and add it to S.
But how do we find x? One approach is, we don't; instead, whenever we add a new element e to S, we compute 2e, 3e, and 5e, and add them to a minimum priority queue. Since this operations guarantees e is in S, simply extracting the top element of the PQ satisfies axiom 2'.
This approach works, but the problem is that we generate a bunch of numbers we may not end up using. See this answer for an example; if the user wants the 5th element in S (5), the PQ at that moment holds 6 6 8 9 10 10 12 15 15 20 25. Can we not waste this space?
Turns out, we can do better. Instead of storing all these numbers, we simply maintain three counters for each of the multiples, namely, 2i, 3j, and 5k. These are candidates for the next number in S. When we pick one of them, we increment only the corresponding counter, and not the other two. By doing so, we are not eagerly generating all the multiples, thus solving the space problem with the first approach.
Let's see a dry run for n = 8, i.e. the number 9. We start with 1, as stated by axiom 1 in Dijkstra's paper.
+---------+---+---+---+----+----+----+-------------------+
| # | i | j | k | 2i | 3j | 5k | S |
+---------+---+---+---+----+----+----+-------------------+
| initial | 1 | 1 | 1 | 2 | 3 | 5 | {1} |
+---------+---+---+---+----+----+----+-------------------+
| 1 | 1 | 1 | 1 | 2 | 3 | 5 | {1,2} |
+---------+---+---+---+----+----+----+-------------------+
| 2 | 2 | 1 | 1 | 4 | 3 | 5 | {1,2,3} |
+---------+---+---+---+----+----+----+-------------------+
| 3 | 2 | 2 | 1 | 4 | 6 | 5 | {1,2,3,4} |
+---------+---+---+---+----+----+----+-------------------+
| 4 | 3 | 2 | 1 | 6 | 6 | 5 | {1,2,3,4,5} |
+---------+---+---+---+----+----+----+-------------------+
| 5 | 3 | 2 | 2 | 6 | 6 | 10 | {1,2,3,4,5,6} |
+---------+---+---+---+----+----+----+-------------------+
| 6 | 4 | 2 | 2 | 8 | 6 | 10 | {1,2,3,4,5,6} |
+---------+---+---+---+----+----+----+-------------------+
| 7 | 4 | 3 | 2 | 8 | 9 | 10 | {1,2,3,4,5,6,8} |
+---------+---+---+---+----+----+----+-------------------+
| 8 | 5 | 3 | 2 | 10 | 9 | 10 | {1,2,3,4,5,6,8,9} |
+---------+---+---+---+----+----+----+-------------------+
Notice that S didn't grow at iteration 6, because the minimum candidate 6 had already been added previously. To avoid this problem of having to remember all of the previous elements, we amend our algorithm to increment all the counters whenever the corresponding multiples are equal to the minimum candidate. That brings us to the following Scala implementation.
def hamming(n: Int): Seq[BigInt] = {
#tailrec
def next(x: Int, factor: Int, xs: IndexedSeq[BigInt]): Int = {
val leq = factor * xs(x) <= xs.last
if (leq) next(x + 1, factor, xs)
else x
}
#tailrec
def loop(i: Int, j: Int, k: Int, xs: IndexedSeq[BigInt]): IndexedSeq[BigInt] = {
if (xs.size < n) {
val a = next(i, 2, xs)
val b = next(j, 3, xs)
val c = next(k, 5, xs)
val m = Seq(2 * xs(a), 3 * xs(b), 5 * xs(c)).min
val x = a + (if (2 * xs(a) == m) 1 else 0)
val y = b + (if (3 * xs(b) == m) 1 else 0)
val z = c + (if (5 * xs(c) == m) 1 else 0)
loop(x, y, z, xs :+ m)
} else xs
}
loop(0, 0, 0, IndexedSeq(BigInt(1)))
}
Basicly the search could be made O(n):
Consider that you keep a partial history of ugly numbers. Now, at each step you have to find the next one. It should be equal to a number from the history multiplied by 2, 3 or 5. Chose the smallest of them, add it to history, and drop some numbers from it so that the smallest from the list multiplied by 5 would be larger than the largest.
It will be fast, because the search of the next number will be simple:
min(largest * 2, smallest * 5, one from the middle * 3),
that is larger than the largest number in the list. If they are scarse, the list will always contain few numbers, so the search of the number that have to be multiplied by 3 will be fast.
Here is a correct solution in ML. The function ugly() will return a stream (lazy list) of hamming numbers. The function nth can be used on this stream.
This uses the Sieve method, the next elements are only calculated when needed.
datatype stream = Item of int * (unit->stream);
fun cons (x,xs) = Item(x, xs);
fun head (Item(i,xf)) = i;
fun tail (Item(i,xf)) = xf();
fun maps f xs = cons(f (head xs), fn()=> maps f (tail xs));
fun nth(s,1)=head(s)
| nth(s,n)=nth(tail(s),n-1);
fun merge(xs,ys)=if (head xs=head ys) then
cons(head xs,fn()=>merge(tail xs,tail ys))
else if (head xs<head ys) then
cons(head xs,fn()=>merge(tail xs,ys))
else
cons(head ys,fn()=>merge(xs,tail ys));
fun double n=n*2;
fun triple n=n*3;
fun ij()=
cons(1,fn()=>
merge(maps double (ij()),maps triple (ij())));
fun quint n=n*5;
fun ugly()=
cons(1,fn()=>
merge((tail (ij())),maps quint (ugly())));
This was first year CS work :-)
To find the n-th ugly number in O (n^(2/3)), jonderry's algorithm will work just fine. Note that the numbers involved are huge so any algorithm trying to check whether a number is ugly or not has no chance.
Finding all of the n smallest ugly numbers in ascending order is done easily by using a priority queue in O (n log n) time and O (n) space: Create a priority queue of numbers with the smallest numbers first, initially including just the number 1. Then repeat n times: Remove the smallest number x from the priority queue. If x hasn't been removed before, then x is the next larger ugly number, and we add 2x, 3x and 5x to the priority queue. (If anyone doesn't know the term priority queue, it's like the heap in the heapsort algorithm). Here's the start of the algorithm:
1 -> 2 3 5
1 2 -> 3 4 5 6 10
1 2 3 -> 4 5 6 6 9 10 15
1 2 3 4 -> 5 6 6 8 9 10 12 15 20
1 2 3 4 5 -> 6 6 8 9 10 10 12 15 15 20 25
1 2 3 4 5 6 -> 6 8 9 10 10 12 12 15 15 18 20 25 30
1 2 3 4 5 6 -> 8 9 10 10 12 12 15 15 18 20 25 30
1 2 3 4 5 6 8 -> 9 10 10 12 12 15 15 16 18 20 24 25 30 40
Proof of execution time: We extract an ugly number from the queue n times. We initially have one element in the queue, and after extracting an ugly number we add three elements, increasing the number by 2. So after n ugly numbers are found we have at most 2n + 1 elements in the queue. Extracting an element can be done in logarithmic time. We extract more numbers than just the ugly numbers but at most n ugly numbers plus 2n - 1 other numbers (those that could have been in the sieve after n-1 steps). So the total time is less than 3n item removals in logarithmic time = O (n log n), and the total space is at most 2n + 1 elements = O (n).
I guess we can use Dynamic Programming (DP) and compute nth Ugly Number. Complete explanation can be found at http://www.geeksforgeeks.org/ugly-numbers/
#include <iostream>
#define MAX 1000
using namespace std;
// Find Minimum among three numbers
long int min(long int x, long int y, long int z) {
if(x<=y) {
if(x<=z) {
return x;
} else {
return z;
}
} else {
if(y<=z) {
return y;
} else {
return z;
}
}
}
// Actual Method that computes all Ugly Numbers till the required range
long int uglyNumber(int count) {
long int arr[MAX], val;
// index of last multiple of 2 --> i2
// index of last multiple of 3 --> i3
// index of last multiple of 5 --> i5
int i2, i3, i5, lastIndex;
arr[0] = 1;
i2 = i3 = i5 = 0;
lastIndex = 1;
while(lastIndex<=count-1) {
val = min(2*arr[i2], 3*arr[i3], 5*arr[i5]);
arr[lastIndex] = val;
lastIndex++;
if(val == 2*arr[i2]) {
i2++;
}
if(val == 3*arr[i3]) {
i3++;
}
if(val == 5*arr[i5]) {
i5++;
}
}
return arr[lastIndex-1];
}
// Starting point of program
int main() {
long int num;
int count;
cout<<"Which Ugly Number : ";
cin>>count;
num = uglyNumber(count);
cout<<endl<<num;
return 0;
}
We can see that its quite fast, just change the value of MAX to compute higher Ugly Number
Using 3 generators in parallel and selecting the smallest at each iteration, here is a C program to compute all ugly numbers below 2128 in less than 1 second:
#include <limits.h>
#include <stdio.h>
#if 0
typedef unsigned long long ugly_t;
#define UGLY_MAX (~(ugly_t)0)
#else
typedef __uint128_t ugly_t;
#define UGLY_MAX (~(ugly_t)0)
#endif
int print_ugly(int i, ugly_t u) {
char buf[64], *p = buf + sizeof(buf);
*--p = '\0';
do { *--p = '0' + u % 10; } while ((u /= 10) != 0);
return printf("%d: %s\n", i, p);
}
int main() {
int i = 0, n2 = 0, n3 = 0, n5 = 0;
ugly_t u, ug2 = 1, ug3 = 1, ug5 = 1;
#define UGLY_COUNT 110000
ugly_t ugly[UGLY_COUNT];
while (i < UGLY_COUNT) {
u = ug2;
if (u > ug3) u = ug3;
if (u > ug5) u = ug5;
if (u == UGLY_MAX)
break;
ugly[i++] = u;
print_ugly(i, u);
if (u == ug2) {
if (ugly[n2] <= UGLY_MAX / 2)
ug2 = 2 * ugly[n2++];
else
ug2 = UGLY_MAX;
}
if (u == ug3) {
if (ugly[n3] <= UGLY_MAX / 3)
ug3 = 3 * ugly[n3++];
else
ug3 = UGLY_MAX;
}
if (u == ug5) {
if (ugly[n5] <= UGLY_MAX / 5)
ug5 = 5 * ugly[n5++];
else
ug5 = UGLY_MAX;
}
}
return 0;
}
Here are the last 10 lines of output:
100517: 338915443777200000000000000000000000000
100518: 339129266201729628114355465608000000000
100519: 339186548067800934969350553600000000000
100520: 339298130282929870605468750000000000000
100521: 339467078447341918945312500000000000000
100522: 339569540691046437734055936000000000000
100523: 339738624000000000000000000000000000000
100524: 339952965770562084651663360000000000000
100525: 340010386766614455386112000000000000000
100526: 340122240000000000000000000000000000000
Here is a version in Javascript usable with QuickJS:
import * as std from "std";
function main() {
var i = 0, n2 = 0, n3 = 0, n5 = 0;
var u, ug2 = 1n, ug3 = 1n, ug5 = 1n;
var ugly = [];
for (;;) {
u = ug2;
if (u > ug3) u = ug3;
if (u > ug5) u = ug5;
ugly[i++] = u;
std.printf("%d: %s\n", i, String(u));
if (u >= 0x100000000000000000000000000000000n)
break;
if (u == ug2)
ug2 = 2n * ugly[n2++];
if (u == ug3)
ug3 = 3n * ugly[n3++];
if (u == ug5)
ug5 = 5n * ugly[n5++];
}
return 0;
}
main();
here is my code , the idea is to divide the number by 2 (till it gives remainder 0) then 3 and 5 . If at last the number becomes one it's a ugly number.
you can count and even print all ugly numbers till n.
int count = 0;
for (int i = 2; i <= n; i++) {
int temp = i;
while (temp % 2 == 0) temp=temp / 2;
while (temp % 3 == 0) temp=temp / 3;
while (temp % 5 == 0) temp=temp / 5;
if (temp == 1) {
cout << i << endl;
count++;
}
}
This problem can be done in O(1).
If we remove 1 and look at numbers between 2 through 30, we will notice that there are 22 numbers.
Now, for any number x in the 22 numbers above, there will be a number x + 30 in between 31 and 60 that is also ugly. Thus, we can find at least 22 numbers between 31 and 60. Now for every ugly number between 31 and 60, we can write it as s + 30. So s will be ugly too, since s + 30 is divisible by 2, 3, or 5. Thus, there will be exactly 22 numbers between 31 and 60. This logic can be repeated for every block of 30 numbers after that.
Thus, there will be 23 numbers in the first 30 numbers, and 22 for every 30 after that. That is, first 23 uglies will occur between 1 and 30, 45 uglies will occur between 1 and 60, 67 uglies will occur between 1 and 30 etc.
Now, if I am given n, say 137, I can see that 137/22 = 6.22. The answer will lie between 6*30 and 7*30 or between 180 and 210. By 180, I will have 6*22 + 1 = 133rd ugly number at 180. I will have 154th ugly number at 210. So I am looking for 4th ugly number (since 137 = 133 + 4)in the interval [2, 30], which is 5. The 137th ugly number is then 180 + 5 = 185.
Another example: if I want the 1500th ugly number, I count 1500/22 = 68 blocks. Thus, I will have 22*68 + 1 = 1497th ugly at 30*68 = 2040. The next three uglies in the [2, 30] block are 2, 3, and 4. So our required ugly is at 2040 + 4 = 2044.
The point it that I can simply build a list of ugly numbers between [2, 30] and simply find the answer by doing look ups in O(1).
Here is another O(n) approach (Python solution) based on the idea of merging three sorted lists. The challenge is to find the next ugly number in increasing order. For example, we know the first seven ugly numbers are [1,2,3,4,5,6,8]. The ugly numbers are actually from the following three lists:
list 1: 1*2, 2*2, 3*2, 4*2, 5*2, 6*2, 8*2 ... ( multiply each ugly number by 2 )
list 2: 1*3, 2*3, 3*3, 4*3, 5*3, 6*3, 8*3 ... ( multiply each ugly number by 3 )
list 3: 1*5, 2*5, 3*5, 4*5, 5*5, 6*5, 8*5 ... ( multiply each ugly number by 5 )
So the nth ugly number is the nth number of the list merged from the three lists above:
1, 1*2, 1*3, 2*2, 1*5, 2*3 ...
def nthuglynumber(n):
p2, p3, p5 = 0,0,0
uglynumber = [1]
while len(uglynumber) < n:
ugly2, ugly3, ugly5 = uglynumber[p2]*2, uglynumber[p3]*3, uglynumber[p5]*5
next = min(ugly2, ugly3, ugly5)
if next == ugly2: p2 += 1 # multiply each number
if next == ugly3: p3 += 1 # only once by each
if next == ugly5: p5 += 1 # of the three factors
uglynumber += [next]
return uglynumber[-1]
STEP I: computing three next possible ugly numbers from the three lists
ugly2, ugly3, ugly5 = uglynumber[p2]*2, uglynumber[p3]*3, uglynumber[p5]*5
STEP II, find the one next ugly number as the smallest of the three above:
next = min(ugly2, ugly3, ugly5)
STEP III: moving the pointer forward if its ugly number was the next ugly number
if next == ugly2: p2+=1
if next == ugly3: p3+=1
if next == ugly5: p5+=1
note: not using if with elif nor else
STEP IV: adding the next ugly number into the merged list uglynumber
uglynumber += [next]