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Given a string abcd how can I create a unique hashing method that will hash those 4 characters to match bcad or any other permutation of the letters abcd?
Currently I have this code
long hashString(string a) {
long hashed = 0;
for(int i = 0; i < a.length(); i++) {
hashed += a[i] * 7; // Timed by a prime to make the hash more unique?
}
return hashed;
}
Now this will not work becasue ad will hash with bc.
I know you can make it more unique by multiplying the position of the letter by the letter itself hashed += a[i] * i but then the string will not hash to its permutations.
Is it possible to create a hash that achieves this?
Edit
Some have suggested sorting the strings before you hash them. Which is a valid answer but the sorting would take O(nlog) time and I am looking for a hash function that runs in O(n) time.
I am looking to do this in O(1) memory.
Create an array of 26 integers, corresponding to letters a-z. Initialize it to 0. Scan the string from beginning to end, and increment the array element corresponding to the current letter. Note that up to this point the algorithm has O(n) time complexity and O(1) space complexity (since the array size is a constant).
Finally, hash the contents of the array using your favorite hash function.
The basic thing you can do is sort the strings before applying the hash function. So, to compute the hash of "adbc" or "dcba" you instead compute the hash of "abcd".
If you want to make sure that there are no collisions in your hash function, then the only way is to have the hash result be a string. There are many more strings than there are 32-bit (or 64-bit) integers so collisions are innevitable (collisions are unlikely with a good hash function though).
Easiest way to understand: sort the letters in the string, and then hash the resulting string.
Some variations on your original idea also work, like:
long hashString(string a) {
long hashed = 0;
for(int i = 0; i < a.length(); i++) {
long t = a[i] * 16777619;
hashed += t^(t>>8);
}
return hashed;
}
I suppose you need a hash such that two anagrams will hash to the same value. I'd suggest you sort them first and use any of the common hash function such as md5. I write the following code using Scala:
import java.security.MessageDigest
def hash(s: String) = {
MessageDigest.getInstance("MD5").digest(s.sorted.getBytes)
}
Note in scala:
scala> "hello".sorted
res0: String = ehllo
scala> "cinema".sorted
res1: String = aceimn
Synopsis: store a histogram of the letters in the hash value.
Step 1: compute a histogram of the letters (since a histogram uniquely identifies the letters in the string without regard to the order of the letters).
int histogram[26];
for ( int i = 0; i < a.length(); i++ )
histogram[a[i] - 'a']++;
Step 2: pack the histogram into the hash value. You have several options here. Which option to choose depends on what sort of limitations you can put on the strings.
If you knew that each letter would appear no more than 3 times, then it takes 2 bits to represent the count, so you could create a 52-bit hash that's guaranteed to be unique.
If you're willing to use a 128-bit hash, then you've got 5 bits for 24 letters, and 4 bits for 2 letters (e.g. q and z). The 128-bit hash allows each letter to appear 31 times (15 times for q and z).
But if you want a fixed sized hash, say 16-bit, then you need to pack the histogram into those 16 bits in a way that reduces collisions. The easiest way to do that is to create a 26 byte message (one byte for each entry in the histogram, allowing each letter to appear up to 255 times). Then take the 16-bit CRC of the message, using your favorite CRC generator.
I'm encoding binary data b1, b2, ... bn using an alphabet. But since the binary representations of the bs are more or less sequential, a simple mapping of bits to chars results in very similar strings. Example:
encode(b1) => "QXpB4IBsu0"
encode(b2) => "QXpB36Bsu0"
...
I'm looking for ways to make the output more "random", meaning more difficult to guess the input b when looking at the output string.
Some requirements:
For differenent bs, the output strings must be different. Encoding the same b multiple times does not necessarily have to result in the same output. As long as there are no collisions between the output strings of different input bs, everything is fine.
If it is of any importance: each b is around ~50-60 bits. The alphabet contains 64 characters.
The encoding function should not produce larger output strings than the ones you get by just using a simple mapping from the bits of bs to chars of the alphabet (given the values above, this means ~10 characters for each b). So just using a hash function like SHA is not an option.
Possible solutions for this problem don't need to be "cryptographically secure". If someone invests enough time and effort to reconstruct the binary data, then so be it. But the goal is to make it as difficult as possible. It maybe helps that a decode function is not needed anyway.
What I am doing at the moment:
take the next 4 bits from the binary data, let's say xxxx
prepend 2 random bits r to get rrxxxx
lookup the corresponding char in the alphabet: val char = alphabet[rrxxxx] and add it to the result (this works because the alphabet's size is 64)
continue with step 1
This appraoch adds some noise to the output string, however, the size of the string is increased by 50% due to the random bits. I could add more noise by adding more random bits (rrrxxx or even rrrrxx), but the output would get larger and larger. One of the requirements I mentioned above is not to increase the size of the output string. Currently I'm only using this approach because I have no better idea.
As an alternative procedure, I thought about shuffling the bits of an input b before applying the alphabet. But since it must be guaranteed that different bs result in different strings, the shuffle function should use some kind of determinism (maybe a secret number as an argument) instead of being completely random. I wasn't able to come up wih such a function.
I'm wondering if there is a better way, any hint is appreciated.
Basically, you need a reversible pseudo-random mapping from each possible 50-bit value to another 50-bit value. You can achieve this with a reversible Linear Congruential Generator (the kind used for some pseudo-random number generators).
When encoding, apply the LCG to your number in the forward direction, then encode with base64. If you need to decode, decode from base64, then apply the LCG in the opposite direction to get your original number back.
This answer contains some code for a reversible LCG. You'll need one with a period of 250. The constants used to define your LCG would be your secret numbers.
You want to use a multiplicative inverse. That will take the sequential key and transform it into a non-sequential number. There is a one-to-one relationship between the keys and their results. So no two numbers will create the same non-sequential key, and the process is reversible.
I have a small example, written in C#, that illustrates the process.
private void DoIt()
{
const long m = 101;
const long x = 387420489; // must be coprime to m
var multInv = MultiplicativeInverse(x, m);
var nums = new HashSet<long>();
for (long i = 0; i < 100; ++i)
{
var encoded = i*x%m;
var decoded = encoded*multInv%m;
Console.WriteLine("{0} => {1} => {2}", i, encoded, decoded);
if (!nums.Add(encoded))
{
Console.WriteLine("Duplicate");
}
}
}
private long MultiplicativeInverse(long x, long modulus)
{
return ExtendedEuclideanDivision(x, modulus).Item1%modulus;
}
private static Tuple<long, long> ExtendedEuclideanDivision(long a, long b)
{
if (a < 0)
{
var result = ExtendedEuclideanDivision(-a, b);
return Tuple.Create(-result.Item1, result.Item2);
}
if (b < 0)
{
var result = ExtendedEuclideanDivision(a, -b);
return Tuple.Create(result.Item1, -result.Item2);
}
if (b == 0)
{
return Tuple.Create(1L, 0L);
}
var q = a/b;
var r = a%b;
var rslt = ExtendedEuclideanDivision(b, r);
var s = rslt.Item1;
var t = rslt.Item2;
return Tuple.Create(t, s - q*t);
}
Code cribbed from the above-mentioned article, and supporting materials.
The idea, then, is to take your sequential number, compute the inverse, and then base-64 encode it. To reverse the process, base-64 decode the value you're given, run it through the reverse calculation, and you have the original number.
I would like to randomly iterate through a range. Each value will be visited only once and all values will eventually be visited. For example:
class Array
def shuffle
ret = dup
j = length
i = 0
while j > 1
r = i + rand(j)
ret[i], ret[r] = ret[r], ret[i]
i += 1
j -= 1
end
ret
end
end
(0..9).to_a.shuffle.each{|x| f(x)}
where f(x) is some function that operates on each value. A Fisher-Yates shuffle is used to efficiently provide random ordering.
My problem is that shuffle needs to operate on an array, which is not cool because I am working with astronomically large numbers. Ruby will quickly consume a large amount of RAM trying to create a monstrous array. Imagine replacing (0..9) with (0..99**99). This is also why the following code will not work:
tried = {} # store previous attempts
bigint = 99**99
bigint.times {
x = rand(bigint)
redo if tried[x]
tried[x] = true
f(x) # some function
}
This code is very naive and quickly runs out of memory as tried obtains more entries.
What sort of algorithm can accomplish what I am trying to do?
[Edit1]: Why do I want to do this? I'm trying to exhaust the search space of a hash algorithm for a N-length input string looking for partial collisions. Each number I generate is equivalent to a unique input string, entropy and all. Basically, I'm "counting" using a custom alphabet.
[Edit2]: This means that f(x) in the above examples is a method that generates a hash and compares it to a constant, target hash for partial collisions. I do not need to store the value of x after I call f(x) so memory should remain constant over time.
[Edit3/4/5/6]: Further clarification/fixes.
[Solution]: The following code is based on #bta's solution. For the sake of conciseness, next_prime is not shown. It produces acceptable randomness and only visits each number once. See the actual post for more details.
N = size_of_range
Q = ( 2 * N / (1 + Math.sqrt(5)) ).to_i.next_prime
START = rand(N)
x = START
nil until f( x = (x + Q) % N ) == START # assuming f(x) returns x
I just remembered a similar problem from a class I took years ago; that is, iterating (relatively) randomly through a set (completely exhausting it) given extremely tight memory constraints. If I'm remembering this correctly, our solution algorithm was something like this:
Define the range to be from 0 to
some number N
Generate a random starting point x[0] inside N
Generate an iterator Q less than N
Generate successive points x[n] by adding Q to
the previous point and wrapping around if needed. That
is, x[n+1] = (x[n] + Q) % N
Repeat until you generate a new point equal to the starting point.
The trick is to find an iterator that will let you traverse the entire range without generating the same value twice. If I'm remembering correctly, any relatively prime N and Q will work (the closer the number to the bounds of the range the less 'random' the input). In that case, a prime number that is not a factor of N should work. You can also swap bytes/nibbles in the resulting number to change the pattern with which the generated points "jump around" in N.
This algorithm only requires the starting point (x[0]), the current point (x[n]), the iterator value (Q), and the range limit (N) to be stored.
Perhaps someone else remembers this algorithm and can verify if I'm remembering it correctly?
As #Turtle answered, you problem doesn't have a solution. #KandadaBoggu and #bta solution gives you random numbers is some ranges which are or are not random. You get clusters of numbers.
But I don't know why you care about double occurence of the same number. If (0..99**99) is your range, then if you could generate 10^10 random numbers per second (if you have a 3 GHz processor and about 4 cores on which you generate one random number per CPU cycle - which is imposible, and ruby will even slow it down a lot), then it would take about 10^180 years to exhaust all the numbers. You have also probability about 10^-180 that two identical numbers will be generated during a whole year. Our universe has probably about 10^9 years, so if your computer could start calculation when the time began, then you would have probability about 10^-170 that two identical numbers were generated. In the other words - practicaly it is imposible and you don't have to care about it.
Even if you would use Jaguar (top 1 from www.top500.org supercomputers) with only this one task, you still need 10^174 years to get all numbers.
If you don't belive me, try
tried = {} # store previous attempts
bigint = 99**99
bigint.times {
x = rand(bigint)
puts "Oh, no!" if tried[x]
tried[x] = true
}
I'll buy you a beer if you will even once see "Oh, no!" on your screen during your life time :)
I could be wrong, but I don't think this is doable without storing some state. At the very least, you're going to need some state.
Even if you only use one bit per value (has this value been tried yes or no) then you will need X/8 bytes of memory to store the result (where X is the largest number). Assuming that you have 2GB of free memory, this would leave you with more than 16 million numbers.
Break the range in to manageable batches as shown below:
def range_walker range, batch_size = 100
size = (range.end - range.begin) + 1
n = size/batch_size
n.times do |i|
x = i * batch_size + range.begin
y = x + batch_size
(x...y).sort_by{rand}.each{|z| p z}
end
d = (range.end - size%batch_size + 1)
(d..range.end).sort_by{rand}.each{|z| p z }
end
You can further randomize solution by randomly choosing the batch for processing.
PS: This is a good problem for map-reduce. Each batch can be worked by independent nodes.
Reference:
Map-reduce in Ruby
you can randomly iterate an array with shuffle method
a = [1,2,3,4,5,6,7,8,9]
a.shuffle!
=> [5, 2, 8, 7, 3, 1, 6, 4, 9]
You want what's called a "full cycle iterator"...
Here is psudocode for the simplest version which is perfect for most uses...
function fullCycleStep(sample_size, last_value, random_seed = 31337, prime_number = 32452843) {
if last_value = null then last_value = random_seed % sample_size
return (last_value + prime_number) % sample_size
}
If you call this like so:
sample = 10
For i = 1 to sample
last_value = fullCycleStep(sample, last_value)
print last_value
next
It would generate random numbers, looping through all 10, never repeating If you change random_seed, which can be anything, or prime_number, which must be greater than, and not be evenly divisible by sample_size, you will get a new random order, but you will still never get a duplicate.
Database systems and other large-scale systems do this by writing the intermediate results of recursive sorts to a temp database file. That way, they can sort massive numbers of records while only keeping limited numbers of records in memory at any one time. This tends to be complicated in practice.
How "random" does your order have to be? If you don't need a specific input distribution, you could try a recursive scheme like this to minimize memory usage:
def gen_random_indices
# Assume your input range is (0..(10**3))
(0..3).sort_by{rand}.each do |a|
(0..3).sort_by{rand}.each do |b|
(0..3).sort_by{rand}.each do |c|
yield "#{a}#{b}#{c}".to_i
end
end
end
end
gen_random_indices do |idx|
run_test_with_index(idx)
end
Essentially, you are constructing the index by randomly generating one digit at a time. In the worst-case scenario, this will require enough memory to store 10 * (number of digits). You will encounter every number in the range (0..(10**3)) exactly once, but the order is only pseudo-random. That is, if the first loop sets a=1, then you will encounter all three-digit numbers of the form 1xx before you see the hundreds digit change.
The other downside is the need to manually construct the function to a specified depth. In your (0..(99**99)) case, this would likely be a problem (although I suppose you could write a script to generate the code for you). I'm sure there's probably a way to re-write this in a state-ful, recursive manner, but I can't think of it off the top of my head (ideas, anyone?).
[Edit]: Taking into account #klew and #Turtle's answers, the best I can hope for is batches of random (or close to random) numbers.
This is a recursive implementation of something similar to KandadaBoggu's solution. Basically, the search space (as a range) is partitioned into an array containing N equal-sized ranges. Each range is fed back in a random order as a new search space. This continues until the size of the range hits a lower bound. At this point the range is small enough to be converted into an array, shuffled, and checked.
Even though it is recursive, I haven't blown the stack yet. Instead, it errors out when attempting to partition a search space larger than about 10^19 keys. I has to do with the numbers being too large to convert to a long. It can probably be fixed:
# partition a range into an array of N equal-sized ranges
def partition(range, n)
ranges = []
first = range.first
last = range.last
length = last - first + 1
step = length / n # integer division
((first + step - 1)..last).step(step) { |i|
ranges << (first..i)
first = i + 1
}
# append any extra onto the last element
ranges[-1] = (ranges[-1].first)..last if last > step * ranges.length
ranges
end
I hope the code comments help shed some light on my original question.
pastebin: full source
Note: PW_LEN under # options can be changed to a lower number in order to get quicker results.
For a prohibitively large space, like
space = -10..1000000000000000000000
You can add this method to Range.
class Range
M127 = 170_141_183_460_469_231_731_687_303_715_884_105_727
def each_random(seed = 0)
return to_enum(__method__) { size } unless block_given?
unless first.kind_of? Integer
raise TypeError, "can't randomly iterate from #{first.class}"
end
sample_size = self.end - first + 1
sample_size -= 1 if exclude_end?
j = coprime sample_size
v = seed % sample_size
each do
v = (v + j) % sample_size
yield first + v
end
end
protected
def gcd(a,b)
b == 0 ? a : gcd(b, a % b)
end
def coprime(a, z = M127)
gcd(a, z) == 1 ? z : coprime(a, z + 1)
end
end
You could then
space.each_random { |i| puts i }
729815750697818944176
459631501395637888351
189447252093456832526
919263002791275776712
649078753489094720887
378894504186913665062
108710254884732609237
838526005582551553423
568341756280370497598
298157506978189441773
27973257676008385948
757789008373827330134
487604759071646274309
217420509769465218484
947236260467284162670
677052011165103106845
406867761862922051020
136683512560740995195
866499263258559939381
596315013956378883556
326130764654197827731
55946515352016771906
785762266049835716092
515578016747654660267
...
With a good amount of randomness so long as your space is a few orders smaller than M127.
Credit to #nick-steele and #bta for the approach.
This isn't really a Ruby-specific answer but I hope it's permitted. Andrew Kensler gives a C++ "permute()" function that does exactly this in his "Correlated Multi-Jittered Sampling" report.
As I understand it, the exact function he provides really only works if your "array" is up to size 2^27, but the general idea could be used for arrays of any size.
I'll do my best to sort of explain it. The first part is you need a hash that is reversible "for any power-of-two sized domain". Consider x = i + 1. No matter what x is, even if your integer overflows, you can determine what i was. More specifically, you can always determine the bottom n-bits of i from the bottom n-bits of x. Addition is a reversible hash operation, as is multiplication by an odd number, as is doing a bitwise xor by a constant. If you know a specific power-of-two domain, you can scramble bits in that domain. E.g. x ^= (x & 0xFF) >> 5) is valid for the 16-bit domain. You can specify that domain with a mask, e.g. mask = 0xFF, and your hash function becomes x = hash(i, mask). Of course you can add a "seed" value into that hash function to get different randomizations. Kensler lays out more valid operations in the paper.
So you have a reversible function x = hash(i, mask, seed). The problem is that if you hash your index, you might end up with a value that is larger than your array size, i.e. your "domain". You can't just modulo this or you'll get collisions.
The reversible hash is the key to using a technique called "cycle walking", introduced in "Ciphers with Arbitrary Finite Domains". Because the hash is reversible (i.e. 1-to-1), you can just repeatedly apply the same hash until your hashed value is smaller than your array! Because you're applying the same hash, and the mapping is one-to-one, whatever value you end up on will map back to exactly one index, so you don't have collisions. So your function could look something like this for 32-bit integers (pseudocode):
fun permute(i, length, seed) {
i = hash(i, 0xFFFF, seed)
while(i >= length): i = hash(i, 0xFFFF, seed)
return i
}
It could take a lot of hashes to get to your domain, so Kensler does a simple trick: he keeps the hash within the domain of the next power of two, which makes it require very few iterations (~2 on average), by masking out the unnecessary bits. The final algorithm looks like this:
fun next_pow_2(length) {
# This implementation is for clarity.
# See Kensler's paper for one way to do it fast.
p = 1
while (p < length): p *= 2
return p
}
permute(i, length, seed) {
mask = next_pow_2(length)-1
i = hash(i, mask, seed) & mask
while(i >= length): i = hash(i, mask, seed) & mask
return i
}
And that's it! Obviously the important thing here is choosing a good hash function, which Kensler provides in the paper but I wanted to break down the explanation. If you want to have different random permutations each time, you can add a "seed" value to the permute function which then gets passed to the hash function.
I have a string s and I want to search for the substring of length X that occurs most often in s. Overlapping substrings are allowed.
For example, if s="aoaoa" and X=3, the algorithm should find "aoa" (which appears 2 times in s).
Does an algorithm exist that does this in O(n) time?
You can do this using a rolling hash in O(n) time (assuming good hash distribution). A simple rolling hash would be the xor of the characters in the string, you can compute it incrementally from the previous substring hash using just 2 xors. (See the Wikipedia entry for better rolling hashes than xor.) Compute the hash of your n-x+1 substrings using the rolling hash in O(n) time. If there were no collisions, the answer is clear - if collisions happen, you'll need to do more work. My brain hurts trying to figure out if that can all be resolved in O(n) time.
Update:
Here's a randomized O(n) algorithm. You can find the top hash in O(n) time by scanning the hashtable (keeping it simple, assume no ties). Find one X-length string with that hash (keep a record in the hashtable, or just redo the rolling hash). Then use an O(n) string searching algorithm to find all occurrences of that string in s. If you find the same number of occurrences as you recorded in the hashtable, you're done.
If not, that means you have a hash collision. Pick a new random hash function and try again. If your hash function has log(n)+1 bits and is pairwise independent [Prob(h(s) == h(t)) < 1/2^{n+1} if s != t], then the probability that the most frequent x-length substring in s hash a collision with the <=n other length x substrings of s is at most 1/2. So if there is a collision, pick a new random hash function and retry, you will need only a constant number of tries before you succeed.
Now we only need a randomized pairwise independent rolling hash algorithm.
Update2:
Actually, you need 2log(n) bits of hash to avoid all (n choose 2) collisions because any collision may hide the right answer. Still doable, and it looks like hashing by general polynomial division should do the trick.
I don't see an easy way to do this in strictly O(n) time, unless X is fixed and can be considered a constant. If X is a parameter to the algorithm, then most simple ways of doing this will actually be O(n*X), as you will need to do comparison operations, string copies, hashes, etc., on a substring of length X at every iteration.
(I'm imagining, for a minute, that s is a multi-gigabyte string, and that X is some number over a million, and not seeing any simple ways of doing string comparison, or hashing substrings of length X, that are O(1), and not dependent on the size of X)
It might be possible to avoid string copies during scanning, by leaving everything in place, and to avoid re-hashing the entire substring -- perhaps by using an incremental hash algorithm where you can add a byte at a time, and remove the oldest byte -- but I don't know of any such algorithms that wouldn't result in huge numbers of collisions that would need to be filtered out with an expensive post-processing step.
Update
Keith Randall points out that this kind of hash is known as a rolling hash. It still remains, though, that you would have to store the starting string position for each match in your hash table, and then verify after scanning the string that all of your matches were true. You would need to sort the hashtable, which could contain n-X entries, based on the number of matches found for each hash key, and verify each result -- probably not doable in O(n).
It should be O(n*m) where m is the average length of a string in the list. For very small values of m then the algorithm will approach O(n)
Build a hashtable of counts for each string length
Iterate over your collection of strings, updating the hashtable accordingly, storing the current most prevelant number as an integer variable separate from the hashtable
done.
Naive solution in Python
from collections import defaultdict
from operator import itemgetter
def naive(s, X):
freq = defaultdict(int)
for i in range(len(s) - X + 1):
freq[s[i:i+X]] += 1
return max(freq.iteritems(), key=itemgetter(1))
print naive("aoaoa", 3)
# -> ('aoa', 2)
In plain English
Create mapping: substring of length X -> how many times it occurs in the s string
for i in range(len(s) - X + 1):
freq[s[i:i+X]] += 1
Find a pair in the mapping with the largest second item (frequency)
max(freq.iteritems(), key=itemgetter(1))
Here is a version I did in C. Hope that it helps.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char *string = NULL, *maxstring = NULL, *tmpstr = NULL, *tmpstr2 = NULL;
unsigned int n = 0, i = 0, j = 0, matchcount = 0, maxcount = 0;
string = "aoaoa";
n = 3;
for (i = 0; i <= (strlen(string) - n); i++) {
tmpstr = (char *)malloc(n + 1);
strncpy(tmpstr, string + i, n);
*(tmpstr + (n + 1)) = '\0';
for (j = 0; j <= (strlen(string) - n); j++) {
tmpstr2 = (char *)malloc(n + 1);
strncpy(tmpstr2, string + j, n);
*(tmpstr2 + (n + 1)) = '\0';
if (!strcmp(tmpstr, tmpstr2))
matchcount++;
}
if (matchcount > maxcount) {
maxstring = tmpstr;
maxcount = matchcount;
}
matchcount = 0;
}
printf("max string: \"%s\", count: %d\n", maxstring, maxcount);
free(tmpstr);
free(tmpstr2);
return 0;
}
You can build a tree of sub-strings. The idea is to organise your sub-strings like a telephone book. You then look up the sub-string and increase its count by one.
In your example above, the tree will have sections (nodes) starting with the letters: 'a' and 'o'. 'a' appears three times and 'o' appears twice. So those nodes will have a count of 3 and 2 respectively.
Next, under the 'a' node a sub-node of 'o' will appear corresponding to the sub-string 'ao'. This appears twice. Under the 'o' node 'a' also appears twice.
We carry on in this fashion until we reach the end of the string.
A representation of the tree for 'abac' might be (nodes on the same level are separated by a comma, sub-nodes are in brackets, counts appear after the colon).
a:2(b:1(a:1(c:1())),c:1()),b:1(a:1(c:1())),c:1()
If the tree is drawn out it will be a lot more obvious! What this all says for example is that the string 'aba' appears once, or the string 'a' appears twice etc. But, storage is greatly reduced and more importantly retrieval is greatly speeded up (compare this to keeping a list of sub-strings).
To find out which sub-string is most repeated, do a depth first search of the tree, every time a leaf node is reached, note the count, and keep a track of the highest one.
The running time is probably something like O(log(n)) not sure, but certainly better than O(n^2).
Python-3 Solution:
from collections import Counter
list = []
list.append([string[i: j] for i in range(len(string)) for j in range(i + 1, len(string) + 1) if len(string[i:j]) == K]) # Where K is length
# now find the most common value in this list
# you can do this natively, but I prefer using collections
most_frequent = Counter(list).most_common(1)[0][0]
print(most_freqent)
Here is the native way to get the most common (for those that are interested):
most_occurences = 0
current_most = ""
for i in list:
frequency = list.count(i)
if frequency > most_occurences:
most_occurences = frequency
current_most = list[i]
print(f"{current_most}, Occurences: {most_occurences}")
[Extract K length substrings (geeks for geeks)][1]
[1]: https://www.geeksforgeeks.org/python-extract-k-length-substrings/
LZW algorithm does this
This is exactly what Lempel-Ziv-Welch (LZW used in GIF image format) compression algorithm does. It finds prevalent repeated bytes and changes them for something short.
LZW on Wikipedia
There's no way to do this in O(n).
Feel free to downvote me if you can prove me wrong on this one, but I've got nothing.
Scroll down to see latest edit, I left all this text here just so that I don't invalidate the replies this question has received so far!
I have the following brain teaser I'd like to get a solution for, I have tried to solve this but since I'm not mathematically that much above average (that is, I think I'm very close to average) I can't seem wrap my head around this.
The problem: Given number x should be split to a serie of multipliers, where each multiplier <= y, y being a constant like 10 or 16 or whatever. In the serie (technically an array of integers) the last number should be added instead of multiplied to be able to convert the multipliers back to original number.
As an example, lets assume x=29 and y=10. In this case the expected array would be {10,2,9} meaning 10*2+9. However if y=5, it'd be {5,5,4} meaning 5*5+4 or if y=3, it'd be {3,3,3,2} which would then be 3*3*3+2.
I tried to solve this by doing something like this:
while x >= y, store y to multipliers, then x = x - y
when x < y, store x to multipliers
Obviously this didn't work, I also tried to store the "leftover" part separately and add that after everything else but that didn't work either. I believe my main problem is that I try to think this in a way too complex manner while the solution is blatantly obvious and simple.
To reiterate, these are the limits this algorithm should have:
has to work with 64bit longs
has to return an array of 32bit integers (...well, shorts are OK too)
while support for signed numbers (both + and -) would be nice, if it helps the task only unsigned numbers is a must
And while I'm doing this using Java, I'd rather take any possible code examples as pseudocode, I specifically do NOT want readily made answers, I just need a nudge (well, more of a strong kick) so that I can solve this at least partly myself. Thanks in advance.
Edit: Further clarification
To avoid some confusion, I think I should reword this a bit:
Every integer in the result array should be less or equal to y, including the last number.
Yes, the last number is just a magic number.
No, this is isn't modulus since then the second number would be larger than y in most cases.
Yes, there is multiple answers to most of the numbers available, however I'm looking for the one with least amount of math ops. As far as my logic goes, that means finding the maximum amount of as big multipliers as possible, for example x=1 000 000,y=100 is 100*100*100 even though 10*10*10*10*10*10 is equally correct answer math-wise.
I need to go through the given answers so far with some thought but if you have anything to add, please do! I do appreciate the interest you've already shown on this, thank you all for that.
Edit 2: More explanations + bounty
Okay, seems like what I was aiming for in here just can't be done the way I thought it could be. I was too ambiguous with my goal and after giving it a bit of a thought I decided to just tell you in its entirety what I'd want to do and see what you can come up with.
My goal originally was to come up with a specific method to pack 1..n large integers (aka longs) together so that their String representation is notably shorter than writing the actual number. Think multiples of ten, 10^6 and 1 000 000 are the same, however the representation's length in characters isn't.
For this I wanted to somehow combine the numbers since it is expected that the numbers are somewhat close to each other. I firsth thought that representing 100, 121, 282 as 100+21+161 could be the way to go but the saving in string length is neglible at best and really doesn't work that well if the numbers aren't very close to each other. Basically I wanted more than ~10%.
So I came up with the idea that what if I'd group the numbers by common property such as a multiplier and divide the rest of the number to individual components which I can then represent as a string. This is where this problem steps in, I thought that for example 1 000 000 and 100 000 can be expressed as 10^(5|6) but due to the context of my aimed usage this was a bit too flaky:
The context is Web. RESTful URL:s to be specific. That's why I mentioned of thinking of using 64 characters (web-safe alphanumberic non-reserved characters and then some) since then I could create seemingly random URLs which could be unpacked to a list of integers expressing a set of id numbers. At this point I thought of creating a base 64-like number system for expressing base 10/2 numbers but since I'm not a math genius I have no idea beyond this point how to do it.
The bounty
Now that I have written the whole story (sorry that it's a long one), I'm opening a bounty to this question. Everything regarding requirements for the preferred algorithm specified earlier is still valid. I also want to say that I'm already grateful for all the answers I've received so far, I enjoy being proven wrong if it's done in such a manner as you people have done.
The conclusion
Well, bounty is now given. I spread a few comments to responses mostly for future reference and myself, you can also check out my SO Uservoice suggestion about spreading bounty which is related to this question if you think we should be able to spread it among multiple answers.
Thank you all for taking time and answering!
Update
I couldn't resist trying to come up with my own solution for the first question even though it doesn't do compression. Here is a Python solution using a third party factorization algorithm called pyecm.
This solution is probably several magnitudes more efficient than Yevgeny's one. Computations take seconds instead of hours or maybe even weeks/years for reasonable values of y. For x = 2^32-1 and y = 256, it took 1.68 seconds on my core duo 1.2 ghz.
>>> import time
>>> def test():
... before = time.time()
... print factor(2**32-1, 256)
... print time.time()-before
...
>>> test()
[254, 232, 215, 113, 3, 15]
1.68499994278
>>> 254*232*215*113*3+15
4294967295L
And here is the code:
def factor(x, y):
# y should be smaller than x. If x=y then {y, 1, 0} is the best solution
assert(x > y)
best_output = []
# try all possible remainders from 0 to y
for remainder in xrange(y+1):
output = []
composite = x - remainder
factors = getFactors(composite)
# check if any factor is larger than y
bad_remainder = False
for n in factors.iterkeys():
if n > y:
bad_remainder = True
break
if bad_remainder: continue
# make the best factors
while True:
results = largestFactors(factors, y)
if results == None: break
output += [results[0]]
factors = results[1]
# store the best output
output = output + [remainder]
if len(best_output) == 0 or len(output) < len(best_output):
best_output = output
return best_output
# Heuristic
# The bigger the number the better. 8 is more compact than 2,2,2 etc...
# Find the most factors you can have below or equal to y
# output the number and unused factors that can be reinserted in this function
def largestFactors(factors, y):
assert(y > 1)
# iterate from y to 2 and see if the factors are present.
for i in xrange(y, 1, -1):
try_another_number = False
factors_below_y = getFactors(i)
for number, copies in factors_below_y.iteritems():
if number in factors:
if factors[number] < copies:
try_another_number = True
continue # not enough factors
else:
try_another_number = True
continue # a factor is not present
# Do we want to try another number, or was a solution found?
if try_another_number == True:
continue
else:
output = 1
for number, copies in factors_below_y.items():
remaining = factors[number] - copies
if remaining > 0:
factors[number] = remaining
else:
del factors[number]
output *= number ** copies
return (output, factors)
return None # failed
# Find prime factors. You can use any formula you want for this.
# I am using elliptic curve factorization from http://sourceforge.net/projects/pyecm
import pyecm, collections, copy
getFactors_cache = {}
def getFactors(n):
assert(n != 0)
# attempt to retrieve from cache. Returns a copy
try:
return copy.copy(getFactors_cache[n])
except KeyError:
pass
output = collections.defaultdict(int)
for factor in pyecm.factors(n, False, True, 10, 1):
output[factor] += 1
# cache result
getFactors_cache[n] = output
return copy.copy(output)
Answer to first question
You say you want compression of numbers, but from your examples, those sequences are longer than the undecomposed numbers. It is not possible to compress these numbers without more details to the system you left out (probability of sequences/is there a programmable client?). Could you elaborate more?
Here is a mathematical explanation as to why current answers to the first part of your problem will never solve your second problem. It has nothing to do with the knapsack problem.
This is Shannon's entropy algorithm. It tells you the theoretical minimum amount of bits you need to represent a sequence {X0, X1, X2, ..., Xn-1, Xn} where p(Xi) is the probability of seeing token Xi.
Let's say that X0 to Xn is the span of 0 to 4294967295 (the range of an integer). From what you have described, each number is as likely as another to appear. Therefore the probability of each element is 1/4294967296.
When we plug it into Shannon's algorithm, it will tell us what the minimum number of bits are required to represent the stream.
import math
def entropy():
num = 2**32
probability = 1./num
return -(num) * probability * math.log(probability, 2)
# the (num) * probability cancels out
The entropy unsurprisingly is 32. We require 32 bits to represent an integer where each number is equally likely. The only way to reduce this number, is to increase the probability of some numbers, and decrease the probability of others. You should explain the stream in more detail.
Answer to second question
The right way to do this is to use base64, when communicating with HTTP. Apparently Java does not have this in the standard library, but I found a link to a free implementation:
http://iharder.sourceforge.net/current/java/base64/
Here is the "pseudo-code" which works perfectly in Python and should not be difficult to convert to Java (my Java is rusty):
def longTo64(num):
mapping = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-_"
output = ""
# special case for 0
if num == 0:
return mapping[0]
while num != 0:
output = mapping[num % 64] + output
num /= 64
return output
If you have control over your web server and web client, and can parse the entire HTTP requests without problem, you can upgrade to base85. According to wikipedia, url encoding allows for up to 85 characters. Otherwise, you may need to remove a few characters from the mapping.
Here is another code example in Python
def longTo85(num):
mapping = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-_.~!*'();:#&=+$,/?%#[]"
output = ""
base = len(mapping)
# special case for 0
if num == 0:
return mapping[0]
while num != 0:
output = mapping[num % base] + output
num /= base
return output
And here is the inverse operation:
def stringToLong(string):
mapping = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-_.~!*'();:#&=+$,/?%#[]"
output = 0
base = len(mapping)
place = 0
# check each digit from the lowest place
for digit in reversed(string):
# find the number the mapping of symbol to number, then multiply by base^place
output += mapping.find(digit) * (base ** place)
place += 1
return output
Here is a graph of Shannon's algorithm in different bases.
As you can see, the higher the radix, the less symbols are needed to represent a number. At base64, ~11 symbols are required to represent a long. At base85, it becomes ~10 symbols.
Edit after final explanation:
I would think base64 is the best solution, since there are standard functions that deal with it, and variants of this idea don't give much improvement. This was answered with much more detail by others here.
Regarding the original question, although the code works, it is not guaranteed to run in any reasonable time, as was answered as well as commented on this question by LFSR Consulting.
Original Answer:
You mean something like this?
Edit - corrected after a comment.
shortest_output = {}
foreach (int R = 0; R <= X; R++) {
// iteration over possible remainders
// check if the rest of X can be decomposed into multipliers
newX = X - R;
output = {};
while (newX > Y) {
int i;
for (i = Y; i > 1; i--) {
if ( newX % i == 0) { // found a divider
output.append(i);
newX = newX /i;
break;
}
}
if (i == 1) { // no dividers <= Y
break;
}
}
if (newX != 1) {
// couldn't find dividers with no remainder
output.clear();
}
else {
output.append(R);
if (output.length() < shortest_output.length()) {
shortest_output = output;
}
}
}
It sounds as though you want to compress random data -- this is impossible for information theoretic reasons. (See http://www.faqs.org/faqs/compression-faq/part1/preamble.html question 9.) Use Base64 on the concatenated binary representations of your numbers and be done with it.
The problem you're attempting to solve (you're dealing with a subset of the problem, given you're restriction of y) is called Integer Factorization and it cannot be done efficiently given any known algorithm:
In number theory, integer factorization is the breaking down of a composite number into smaller non-trivial divisors, which when multiplied together equal the original integer.
This problem is what makes a number of cryptographic functions possible (namely RSA which uses 128 bit keys - long is half of that.) The wiki page contains some good resources that should move you in the right direction with your problem.
So, your brain teaser is indeed a brain teaser... and if you solve it efficiently we can elevate your math skills to above average!
Updated after the full story
Base64 is most likely your best option. If you want a custom solution you can try implementing a Base 65+ system. Just remember that just because 10000 can be written as "10^4" doesn't mean that everything can be written as 10^n where n is an integer. Different base systems are the simplest way to write numbers and the higher the base the less digits the number requires. Plus most framework libraries contain algorithms for Base64 encoding. (What language you are using?).
One way to further pack the urls is the one you mentioned but in Base64.
int[] IDs;
IDs.sort() // So IDs[i] is always smaller or equal to IDs[i-1].
string url = Base64Encode(IDs[0]);
for (int i = 1; i < IDs.length; i++) {
url += "," + Base64Encode(IDs[i-1] - IDs[i]);
}
Note that you require some separator as the initial ID can be arbitrarily large and the difference between two IDs CAN be more than 63 in which case one Base64 digit is not enough.
Updated
Just restating that the problem is unsolvable. For Y = 64 you can't write 87681 in multipliers + remainder where each of these is below 64. In other words, you cannot write any of the numbers 87617..87681 with multipliers that are below 64. Each of these numbers has an elementary term over 64. 87616 can be written in elementary terms below 64 but then you'd need those + 65 and so the remainder will be over 64.
So if this was just a brainteaser, it's unsolvable. Was there some practical purpose for this which could be achieved in some way other than using multiplication and a remainder?
And yes, this really should be a comment but I lost my ability to comment at some point. :p
I believe the solution which comes closest is Yevgeny's. It is also easy to extend Yevgeny's solution to remove the limit for the remainder in which case it would be able to find solution where multipliers are smaller than Y and remainder as small as possible, even if greater than Y.
Old answer:
If you limit that every number in the array must be below the y then there is no solution for this. Given large enough x and small enough y, you'll end up in an impossible situation. As an example with y of 2, x of 12 you'll get 2 * 2 * 2 + 4 as 2 * 2 * 2 * 2 would be 16. Even if you allow negative numbers with abs(n) below y that wouldn't work as you'd need 2 * 2 * 2 * 2 - 4 in the above example.
And I think the problem is NP-Complete even if you limit the problem to inputs which are known to have an answer where the last term is less than y. It sounds quite much like the [Knapsack problem][1]. Of course I could be wrong there.
Edit:
Without more accurate problem description it is hard to solve the problem, but one variant could work in the following way:
set current = x
Break current to its terms
If one of the terms is greater than y the current number cannot be described in terms greater than y. Reduce one from current and repeat from 2.
Current number can be expressed in terms less than y.
Calculate remainder
Combine as many of the terms as possible.
(Yevgeny Doctor has more conscise (and working) implementation of this so to prevent confusion I've skipped the implementation.)
OP Wrote:
My goal originally was to come up with
a specific method to pack 1..n large
integers (aka longs) together so that
their String representation is notably
shorter than writing the actual
number. Think multiples of ten, 10^6
and 1 000 000 are the same, however
the representation's length in
characters isn't.
I have been down that path before, and as fun as it was to learn all the math, to save you time I will just point you to: http://en.wikipedia.org/wiki/Kolmogorov_complexity
In a nutshell some strings can be easily compressed by changing your notation:
10^9 (4 characters) = 1000000000 (10 characters)
Others cannot:
7829203478 = some random number...
This is a great great simplification of the article I linked to above, so I recommend that you read it instead of taking my explanation at face value.
Edit:
If you are trying to make RESTful urls for some set of unique data, why wouldn't you use a hash, such as MD5? Then include the hash as part of the URL, then look up the data based on the hash. Or am I missing something obvious?
The original method you chose (a * b + c * d + e) would be very difficult to find optimal solutions for simply due to the large search space of possibilities. You could factorize the number but it's that "+ e" that complicates things since you need to factorize not just that number but quite a few immediately below it.
Two methods for compression spring immediately to mind, both of which give you a much-better-than-10% saving on space from the numeric representation.
A 64-bit number ranges from (unsigned):
0 to
18,446,744,073,709,551,616
or (signed):
-9,223,372,036,854,775,808 to
9,223,372,036,854,775,807
In both cases, you need to reduce the 20-characters taken (without commas) to something a little smaller.
The first is to simply BCD-ify the number the base64 encode it (actually a slightly modified base64 since "/" would not be kosher in a URL - you should use one of the acceptable characters such as "_").
Converting it to BCD will store two digits (or a sign and a digit) into one byte, giving you an immediate 50% reduction in space (10 bytes). Encoding it base 64 (which turns every 3 bytes into 4 base64 characters) will turn the first 9 bytes into 12 characters and that tenth byte into 2 characters, for a total of 14 characters - that's a 30% saving.
The only better method is to just base64 encode the binary representation. This is better because BCD has a small amount of wastage (each digit only needs about 3.32 bits to store [log210], but BCD uses 4).
Working on the binary representation, we only need to base64 encode the 64-bit number (8 bytes). That needs 8 characters for the first 6 bytes and 3 characters for the final 2 bytes. That's 11 characters of base64 for a saving of 45%.
If you wanted maximum compression, there are 73 characters available for URL encoding:
ABCDEFGHIJKLMNOPQRSTUVWXYZ
abcdefghijklmnopqrstuvwxyz
0123456789$-_.+!*'(),
so technically you could probably encode base-73 which, from rough calculations, would still take up 11 characters, but with more complex code which isn't worth it in my opinion.
Of course, that's the maximum compression due to the maximum values. At the other end of the scale (1-digit) this encoding actually results in more data (expansion rather than compression). You can see the improvements only start for numbers over 999, where 4 digits can be turned into 3 base64 characters:
Range (bytes) Chars Base64 chars Compression ratio
------------- ----- ------------ -----------------
< 10 (1) 1 2 -100%
< 100 (1) 2 2 0%
< 1000 (2) 3 3 0%
< 10^4 (2) 4 3 25%
< 10^5 (3) 5 4 20%
< 10^6 (3) 6 4 33%
< 10^7 (3) 7 4 42%
< 10^8 (4) 8 6 25%
< 10^9 (4) 9 6 33%
< 10^10 (5) 10 7 30%
< 10^11 (5) 11 7 36%
< 10^12 (5) 12 7 41%
< 10^13 (6) 13 8 38%
< 10^14 (6) 14 8 42%
< 10^15 (7) 15 10 33%
< 10^16 (7) 16 10 37%
< 10^17 (8) 17 11 35%
< 10^18 (8) 18 11 38%
< 10^19 (8) 19 11 42%
< 2^64 (8) 20 11 45%
Update: I didn't get everything, thus I rewrote the whole thing in a more Java-Style fashion. I didn't think of the prime number case that is bigger than the divisor. This is fixed now. I leave the original code in order to get the idea.
Update 2: I now handle the case of the big prime number in another fashion . This way a result is obtained either way.
public final class PrimeNumberException extends Exception {
private final long primeNumber;
public PrimeNumberException(long x) {
primeNumber = x;
}
public long getPrimeNumber() {
return primeNumber;
}
}
public static Long[] decompose(long x, long y) {
try {
final ArrayList<Long> operands = new ArrayList<Long>(1000);
final long rest = x % y;
// Extract the rest so the reminder is divisible by y
final long newX = x - rest;
// Go into recursion, actually it's a tail recursion
recDivide(newX, y, operands);
} catch (PrimeNumberException e) {
// return new Long[0];
// or do whatever you like, for example
operands.add(e.getPrimeNumber());
} finally {
// Add the reminder to the array
operands.add(rest);
return operands.toArray(new Long[operands.size()]);
}
}
// The recursive method
private static void recDivide(long x, long y, ArrayList<Long> operands)
throws PrimeNumberException {
while ((x > y) && (y != 1)) {
if (x % y == 0) {
final long rest = x / y;
// Since y is a divisor add it to the list of operands
operands.add(y);
if (rest <= y) {
// the rest is smaller than y, we're finished
operands.add(rest);
}
// go in recursion
x = rest;
} else {
// if the value x isn't divisible by y decrement y so you'll find a
// divisor eventually
if (--y == 1) {
throw new PrimeNumberException(x);
}
}
}
}
Original: Here some recursive code I came up with. I would have preferred to code it in some functional language but it was required in Java. I didn't bother converting the numbers to integer but that shouldn't be that hard (yes, I'm lazy ;)
public static Long[] decompose(long x, long y) {
final ArrayList<Long> operands = new ArrayList<Long>();
final long rest = x % y;
// Extract the rest so the reminder is divisible by y
final long newX = x - rest;
// Go into recursion, actually it's a tail recursion
recDivide(newX, y, operands);
// Add the reminder to the array
operands.add(rest);
return operands.toArray(new Long[operands.size()]);
}
// The recursive method
private static void recDivide(long newX, long y, ArrayList<Long> operands) {
long x = newX;
if (x % y == 0) {
final long rest = x / y;
// Since y is a divisor add it to the list of operands
operands.add(y);
if (rest <= y) {
// the rest is smaller than y, we're finished
operands.add(rest);
} else {
// the rest can still be divided, go one level deeper in recursion
recDivide(rest, y, operands);
}
} else {
// if the value x isn't divisible by y decrement y so you'll find a divisor
// eventually
recDivide(x, y-1, operands);
}
}
Are you married to using Java? Python has an entire package dedicated just for this exact purpose. It'll even sanitize the encoding for you to be URL-safe.
Native Python solution
The standard module I'm recommending is base64, which converts arbitrary stings of chars into sanitized base64 format. You can use it in conjunction with the pickle module, which handles conversion from lists of longs (actually arbitrary size) to a compressed string representation.
The following code should work on any vanilla installation of Python:
import base64
import pickle
# get some long list of numbers
a = (854183415,1270335149,228790978,1610119503,1785730631,2084495271,
1180819741,1200564070,1594464081,1312769708,491733762,243961400,
655643948,1950847733,492757139,1373886707,336679529,591953597,
2007045617,1653638786)
# this gets you the url-safe string
str64 = base64.urlsafe_b64encode(pickle.dumps(a,-1))
print str64
>>> gAIoSvfN6TJKrca3S0rCEqMNSk95-F9KRxZwakqn3z58Sh3hYUZKZiePR0pRlwlfSqxGP05KAkNPHUo4jooOSixVFCdK9ZJHdEqT4F4dSvPY41FKaVIRFEq9fkgjSvEVoXdKgoaQYnRxAC4=
# this unwinds it
a64 = pickle.loads(base64.urlsafe_b64decode(str64))
print a64
>>> (854183415, 1270335149, 228790978, 1610119503, 1785730631, 2084495271, 1180819741, 1200564070, 1594464081, 1312769708, 491733762, 243961400, 655643948, 1950847733, 492757139, 1373886707, 336679529, 591953597, 2007045617, 1653638786)
Hope that helps. Using Python is probably the closest you'll get from a 1-line solution.
Wrt the original algorithm request: Is there a limit on the size of the last number (beyond that it must be stored in a 32b int)?
(The original request is all I'm able to tackle lol.)
The one that produces the shortest list is:
bool negative=(n<1)?true:false;
int j=n%y;
if(n==0 || n==1)
{
list.append(n);
return;
}
while((long64)(n-j*y)>MAX_INT && y>1) //R has to be stored in int32
{
y--;
j=n%y;
}
if(y<=1)
fail //Number has no suitable candidate factors. This shouldn't happen
int i=0;
for(;i<j;i++)
{
list.append(y);
}
list.append(n-y*j);
if(negative)
list[0]*=-1;
return;
A little simplistic compared to most answers given so far but it achieves the desired functionality of the original post... It's a little dirty but hopefully useful :)
Isn't this modulus?
Let / be integer division (whole numbers) and % be modulo.
int result[3];
result[0] = y;
result[1] = x / y;
result[2] = x % y;
Just set x:=x/n where n is the largest number that is less both than x and y. When you end up with x<=y, this is your last number in the sequence.
Like in my comment above, I'm not sure I understand exactly the question. But assuming integers (n and a given y), this should work for the cases you stated:
multipliers[0] = n / y;
multipliers[1] = y;
addedNumber = n % y;