*what is the fastest way to enumerate all possible combinations of 4 items out of a possible 64?*Actually, it doesn't have to be 4 of 64, it could be 7 of 48, or 2 specific aces from 52 cards or 3 bits of 10. The field of combinatorics can tell us

*how many*there are. This is expressed as

**{ n choose r }**and has the formula

*factorial(n) / (factorial(r) * factorial(n - r))*. In the case of

*n = 64*and

*r = 4*it yields 635,376 different combinations.

So, the task is to enumerate each of these unique combinations exactly one time until all 635,376 have been generated. This is one of those problems where the right approach makes all the difference.

**The Naive Approach**

Let's look, for a moment, at the most basic brute force approach: iterate through every value that can be contained in 64 bits and check if it has four bits. How long would this take? On my machine it takes 1 second to check 250 million numbers. Pretty fast, right? At this rate, however, it will take

**2,340 years**to check all 2^64 numbers! Clearly not a usable solution.

**A Better Approach**

Poking around the internet I found a function that takes a number and returns the next highest number with the same number of bits. I adapt it a bit and bam!

int enumerate_combinations(int n, int r, unsigned __int64 *v)

{

unsigned __int64 y, r, x;

unsigned count = (unsigned)math::choose(n, r);

v[0] = ((unsigned __int64)1 << r) - 1;

for (unsigned i = 1; i < count; i++)

{

x = v[i - 1];

y = x & -(__int64)x;

r = x + y;

v[i] = r | (((x ^ r) >> 2) / y);

}

return count;

}

This variation turns in

**32 milliseconds**for complete enumeration. Not a bad improvement over 2,340 years, eh?

**The Best Approach**

I actually developed the following function before the above. However, I figured (from looking at the code) that the above would blow this one away. How wrong I was. One problem with the above approach is that it has a nasty little division. Another thing is that this approach takes advantage of certain special cases like when r == 1 and n == r. The following approach is based on my initial recursive approach, but I removed the recursion so that I could rewrite it as an iterator. Removing the recursion did not seem to have a significant impact on performance. Anyhow the following runs to completion in just

**3 milliseconds**, over 10 times faster than the above version.

template <typename T = unsigned __int64>

int enumerate_combinations(int n, int r, T *v)

{

struct { int n, r; T h; } s[sizeof(T) * 8] = { { n, r, 0 } }, q;

int si = 1, i = 0;

T one = 1;

while (si)

{

q = s[--si];

tail:

if (q.r != 0)

{

one = 1;

if (q.r == 1)

{

for (int j = 0; j < q.n; j++)

{

v[i++] = q.h | one;

one <<= 1;

}

}

else if (q.r == q.n)

{

v[i++] = q.h | (one << q.n) - 1;

}

else

{

--q.n; s[si++] = q; q.r--;

q.h |= one << q.n;

goto tail;

}

}

}

return i;

}

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