How to speed up multiple inner products in python

Question

I have some simple code that does the following.

It iterates over all possible length n lists F with +-1 entries. For each one, it iterates over all possible length 2n lists S with +-1 entries, where the first half of $S$ is simply a copy of the second half. The code computes the inner product of F with each sublist of S of length n. For each F, S it counts the inner products that are zero until the first non-zero inner product.

Here is the code.

#!/usr/bin/python

from __future__ import division
import itertools
import operator
import math

n=14
m=n+1
def innerproduct(A, B):
    assert (len(A) == len(B))
    s = 0 
    for k in xrange(0,n):
        s+=A[k]*B[k]
    return s

leadingzerocounts = [0]*m
for S in itertools.product([-1,1], repeat = n):
    S1 = S + S
    for F in itertools.product([-1,1], repeat = n):
        i = 0
        while (i<m):
            ip = innerproduct(F, S1[i:i+n])
            if (ip == 0):
                leadingzerocounts[i] +=1
                i+=1
            else:
                break

print leadingzerocounts

The correct output for n=14 is

[56229888, 23557248, 9903104, 4160640, 1758240, 755392, 344800, 172320, 101312, 75776, 65696, 61216, 59200, 59200, 59200]

Using pypy, this takes 1 min 18 seconds for n = 14. Unfortunately, I would really like to run it for 16,18,20,22,24,26. I don't mind using numba or cython but I would like to stay close to python if at all possible.

Any help speeding this up is very much appreciated.

---

I'll keep a record of the fastest solutions here. (Please let me know if I miss an updated answer.)

• n = 22 at 9m35.081s by Eisenstat (C)
• n = 18 at 1m16.344s by Eisenstat (pypy)
• n = 18 at 2m54.998s by Tupteq (pypy)
• n = 14 at 26s by Neil (numpy)
• n - 14 at 11m59.192s by kslote1 (pypy)

Answer 1

This new code gets another order of magnitude speedup by taking advantage of the cyclic symmetry of the problem. This Python version enumerates necklaces with Duval's algorithm; the C version uses brute force. Both incorporate the speedups described below. On my machine, the C version solves n = 20 in 100 seconds! A back-of-the-envelope calculation suggests that, if you were to let it run for a week on a single core, it could do n = 26, and, as noted below, it's amenable to parallelism.

import itertools


def necklaces_with_multiplicity(n):
    assert isinstance(n, int)
    assert n > 0
    w = [1] * n
    i = 1
    while True:
        if n % i == 0:
            s = sum(w)
            if s > 0:
                yield (tuple(w), i * 2)
            elif s == 0:
                yield (tuple(w), i)
        i = n - 1
        while w[i] == -1:
            if i == 0:
                return
            i -= 1
        w[i] = -1
        i += 1
        for j in range(n - i):
            w[i + j] = w[j]


def leading_zero_counts(n):
    assert isinstance(n, int)
    assert n > 0
    assert n % 2 == 0
    counts = [0] * n
    necklaces = list(necklaces_with_multiplicity(n))
    for combo in itertools.combinations(range(n - 1), n // 2):
        for v, multiplicity in necklaces:
            w = list(v)
            for j in combo:
                w[j] *= -1
            for i in range(n):
                counts[i] += multiplicity * 2
                product = 0
                for j in range(n):
                    product += v[j - (i + 1)] * w[j]
                if product != 0:
                    break
    return counts


if __name__ == '__main__':
    print(leading_zero_counts(12))

C version:

#include <stdio.h>

enum {
  N = 14
};

struct Necklace {
  unsigned int v;
  int multiplicity;
};

static struct Necklace g_necklace[1 << (N - 1)];
static int g_necklace_count;

static void initialize_necklace(void) {
  g_necklace_count = 0;
  for (unsigned int v = 0; v < (1U << (N - 1)); v++) {
    int multiplicity;
    unsigned int w = v;
    for (multiplicity = 2; multiplicity < 2 * N; multiplicity += 2) {
      w = ((w & 1) << (N - 1)) | (w >> 1);
      unsigned int x = w ^ ((1U << N) - 1);
      if (w < v || x < v) goto nope;
      if (w == v || x == v) break;
    }
    g_necklace[g_necklace_count].v = v;
    g_necklace[g_necklace_count].multiplicity = multiplicity;
    g_necklace_count++;
   nope:
    ;
  }
}

int main(void) {
  initialize_necklace();
  long long leading_zero_count[N + 1];
  for (int i = 0; i < N + 1; i++) leading_zero_count[i] = 0;
  for (unsigned int v_xor_w = 0; v_xor_w < (1U << (N - 1)); v_xor_w++) {
    if (__builtin_popcount(v_xor_w) != N / 2) continue;
    for (int k = 0; k < g_necklace_count; k++) {
      unsigned int v = g_necklace[k].v;
      unsigned int w = v ^ v_xor_w;
      for (int i = 0; i < N + 1; i++) {
        leading_zero_count[i] += g_necklace[k].multiplicity;
        w = ((w & 1) << (N - 1)) | (w >> 1);
        if (__builtin_popcount(v ^ w) != N / 2) break;
      }
    }
  }
  for (int i = 0; i < N + 1; i++) {
    printf(" %lld", 2 * leading_zero_count[i]);
  }
  putchar('\n');
  return 0;
}

---

You can get a bit of speedup by exploiting the sign symmetry (4x) and by iterating over only those vectors that pass the first inner product test (asymptotically, O(sqrt(n))x).

import itertools


n = 10
m = n + 1


def innerproduct(A, B):
    s = 0
    for k in range(n):
        s += A[k] * B[k]
    return s


leadingzerocounts = [0] * m
for S in itertools.product([-1, 1], repeat=n - 1):
    S1 = S + (1,)
    S1S1 = S1 * 2
    for C in itertools.combinations(range(n - 1), n // 2):
        F = list(S1)
        for i in C:
            F[i] *= -1
        leadingzerocounts[0] += 4
        for i in range(1, m):
            if innerproduct(F, S1S1[i:i + n]):
                break
            leadingzerocounts[i] += 4
print(leadingzerocounts)

C version, to get a feel for how much performance we're losing to PyPy (16 for PyPy is roughly equivalent to 18 for C):

#include <stdio.h>

enum {
  HALFN = 9,
  N = 2 * HALFN
};

int main(void) {
  long long lzc[N + 1];
  for (int i = 0; i < N + 1; i++) lzc[i] = 0;
  unsigned int xor = 1 << (N - 1);
  while (xor-- > 0) {
    if (__builtin_popcount(xor) != HALFN) continue;
    unsigned int s = 1 << (N - 1);
    while (s-- > 0) {
      lzc[0]++;
      unsigned int f = xor ^ s;
      for (int i = 1; i < N + 1; i++) {
        f = ((f & 1) << (N - 1)) | (f >> 1);
        if (__builtin_popcount(f ^ s) != HALFN) break;
        lzc[i]++;
      }
    }
  }
  for (int i = 0; i < N + 1; i++) printf(" %lld", 4 * lzc[i]);
  putchar('\n');
  return 0;
}

This algorithm is embarrassingly parallel because it's just accumulating over all values of xor. With the C version, a back-of-the-envelope calculation suggests that a few thousand hours of CPU time would suffice to calculate n = 26, which works out to a couple hundred dollars at current rates on EC2. There are undoubtedly some optimizations to be made (e.g., vectorization), but for a one-off like this I'm not sure how much more programmer effort is worthwhile.

Answer 2

One very simple speed up of a factor of n is to change this code:

def innerproduct(A, B):
    assert (len(A) == len(B))
    for j in xrange(len(A)):
        s = 0 
        for k in xrange(0,n):
            s+=A[k]*B[k]
    return s

to

def innerproduct(A, B):
    assert (len(A) == len(B))
    s = 0 
    for k in xrange(0,n):
        s+=A[k]*B[k]
    return s

(I don't know why you have the loop over j, but it just does the same calculation each time so is unnecessary.)