Python 2 using pypy and pp: n = 15 in 3 minutes

Also just a simple brute force. Interesting to see, that I nearly get the same same speed as kuroi neko with C++. My code can reach n = 12 in about 5 minutes. And I only run it on one virtual core.

edit: Reduce search space by a factor of n

I noticed, that a cycled vector A* of A produces the same numbers as probabilities (same numbers) as the original vector A when I iterate over B. E.g. The vector (1, 1, 0, 1, 0, 0) has the same probabilities as each of the vectors (1, 0, 1, 0, 0, 1), (0, 1, 0, 0, 1, 1), (1, 0, 0, 1, 1, 0), (0, 0, 1, 1, 0, 1) and (0, 1, 1, 0, 1, 0) when choosing a random B. Therefore I don't have to iterate over each of these 6 vectors, but only about 1 and replace count[i] += 1 with count[i] += cycle_number.

This reduces the complexity from Theta(n) = 6^n to Theta(n) = 6^n / n. Therefore for n = 13 it's about 13 times as fast as my previous version. It calculates n = 13 in about 2 minutes 20 seconds. For n = 14 it is still a little bit too slow. It takes about 13 minutes.

edit 2: Multi-core-programming

Not really happy with the next improvement. I decided to also try to execute my program on multiple cores. On my 2 + 2 cores I now can calculate n = 14 in about 7 minutes. Only a factor of 2 improvement.

The code is available in this github repo: Link. The multi-core programming makes is a little bit ugly.

edit 3: Reducing search space for A vectors and B vectors

I noticed the same mirror-symmetry for the vectors A as kuroi neko did. Still not sure, why this works (and if it works for each n).

The reducing of the search space for B vectors is a bit cleverer. I replaced the generation of the vectors (itertools.product), with an own function. Basically, I start with an empty list and put it on a stack. Until the stack is empty, I remove a list, if it has not the same lenght as n, I generate 3 other lists (by appending -1, 0, 1) and pushing them onto the stack. I a list has the same length as n, I can evaluate the sums.

Now that I generate the vectors myself, I can filter them depending on if I can reach the sum = 0 or not. E.g. if my vector A is (1, 1, 1, 0, 0), and my vector B looks (1, 1, ?, ?, ?), I know, that I cannot fill the ? with values, so that A*B = 0. So I don't have to iterate over all those 6 vectors B of the form (1, 1, ?, ?, ?).

We can improve on this, if we ignore the values for 1. As noted in the question, for the values for i = 1 are the sequence A081671. There are many ways to calculate those. I choose the simple recurrence: a(n) = (4*(2*n-1)*a(n-1) - 12*(n-1)*a(n-2)) / n. Since we can calculate i = 1 in basically no time, we can filter more vectors for B. E.g. A = (0, 1, 0, 1, 1) and B = (1, -1, ?, ?, ?). We can ignore vectors, where the first ? = 1, because the A * cycled(B) > 0, for all these vectors. I hope you can follow. It's probably not the best example.

With this I can calculate n = 15 in 6 minutes.

edit 4:

Quickly implemented kuroi neko's great idea, which says, that B and -B produces the same results. Speedup x2. Implementation is only a quick hack, though. n = 15 in 3 minutes.

Code:

For the complete code visit Github. The following code is only a representation of the main features. I left out imports, multicore programming, printing the results, ...

count = [0] * n count[0] = oeis_A081671(n) #generating all important vector A visited = set(); todo = dict() for A in product((0, 1), repeat=n): if A not in visited: # generate all vectors, which have the same probability # mirrored and cycled vectors same_probability_set = set() for i in range(n): tmp = [A[(i+j) % n] for j in range(n)] same_probability_set.add(tuple(tmp)) same_probability_set.add(tuple(tmp[::-1])) visited.update(same_probability_set) todo[A] = len(same_probability_set) # for each vector A, create all possible vectors B stack = [] for A, cycled_count in dict_A.iteritems(): ones = [sum(A[i:]) for i in range(n)] + [0] # + [0], so that later ones[n] doesn't throw a exception stack.append(([0] * n, 0, 0, 0, False)) while stack: B, index, sum1, sum2, used_negative = stack.pop() if index < n: # fill vector B[index] in all possible ways, # so that it's still possible to reach 0. if used_negative: for v in (-1, 0, 1): sum1_new = sum1 + v * A[index] sum2_new = sum2 + v * A[index - 1 if index else n - 1] if abs(sum1_new) <= ones[index+1]: if abs(sum2_new) <= ones[index] - A[n-1]: C = B[:] C[index] = v stack.append((C, index + 1, sum1_new, sum2_new, True)) else: for v in (0, 1): sum1_new = sum1 + v * A[index] sum2_new = sum2 + v * A[index - 1 if index else n - 1] if abs(sum1_new) <= ones[index+1]: if abs(sum2_new) <= ones[index] - A[n-1]: C = B[:] C[index] = v stack.append((C, index + 1, sum1_new, sum2_new, v == 1)) else: # B is complete, calculate the sums count[1] += cycled_count # we know that the sum = 0 for i = 1 for i in range(2, n): sum_prod = 0 for j in range(n-i): sum_prod += A[j] * B[i+j] for j in range(i): sum_prod += A[n-i+j] * B[j] if sum_prod: break else: if used_negative: count[i] += 2*cycled_count else: count[i] += cycled_count 

Usage:

You have to install pypy (for Python 2!!!). The parallel python module isn't ported for Python 3. Then you have to install the parallel python module pp-1.6.4.zip. Extract it, cd into the folder and call pypy setup.py install.

Then you can call my program with

pypy you-do-the-math.py 15

It will automatically determine the number of cpu's. There may be some error messages after finishing the program, just ignore them. n = 16 should be possible on your machine.

Output:

Calculation for n = 15 took 2:50 minutes 1 83940771168 / 470184984576 17.85% 2 17379109692 / 470184984576 3.70% 3 3805906050 / 470184984576 0.81% 4 887959110 / 470184984576 0.19% 5 223260870 / 470184984576 0.05% 6 67664580 / 470184984576 0.01% 7 30019950 / 470184984576 0.01% 8 20720730 / 470184984576 0.00% 9 18352740 / 470184984576 0.00% 10 17730480 / 470184984576 0.00% 11 17566920 / 470184984576 0.00% 12 17521470 / 470184984576 0.00% 13 17510280 / 470184984576 0.00% 14 17507100 / 470184984576 0.00% 15 17506680 / 470184984576 0.00% 

Notes and ideas:

• I have a i7-4600m processor with 2 cores and 4 threads. It doesn't matter If I use 2 or 4 threads. The cpu-usage is 50% with 2 threads and 100% with 4 threads, but it still takes the same amount of time. I don't know why. I checked, that each thread only has to the the half amout of data, when there are 4 threads, checked the results, ...
• I use a lot of lists. Python isn't quite efficient in storing, I have to copy lots of lists, ... So I thought of using an integer instead. I could use the bits 00 (for 0) and 11 (for 1) in the vector A, and the bits 10 (for -1), 00 (for 0) and 01 (for 1) in the vector B. For the product of A and B, I would only have to calculate A & B and count the 01 and 10 blocks. Cycling can be done with shifting the vector and using masks, ... I actually implemented all this, you can find it in some of my older commits on Github. But it turned out, to be slower than with lists. I guess, pypy really optimizes list operations.

woolly bully - C++ - way too slow

Well since a better programmer took on the C++ implementation, I'm calling quits for this one.

#include <cstdlib> #include <cmath> #include <vector> #include <bitset> #include <future> #include <iostream> #include <iomanip> using namespace std; /* 6^^n events will be generated, so the absolute max that can be counted by a b bits integer is E(b*log(2)/log(6)), i.e. n=24 for a 64 bits counter To enumerate 3 possible values of a size n vector we need E(n*log(3)/log(2))+1 bits, i.e. 39 bits */ typedef unsigned long long Counter; // counts up to 6^^24 typedef unsigned long long Benumerator; // 39 bits typedef unsigned long Aenumerator; // 24 bits #define log2_over_log6 0.3869 #define A_LENGTH ((size_t)(8*sizeof(Counter)*log2_over_log6)) #define B_LENGTH (2*A_LENGTH) typedef bitset<B_LENGTH> vectorB; typedef vector<Counter> OccurenceCounters; // ----------------------------------------------------------------- // multithreading junk for CPUs detection and allocation // ----------------------------------------------------------------- int number_of_CPUs(void) { int res = thread::hardware_concurrency(); return res == 0 ? 8 : res; } #ifdef __linux__ #include <sched.h> void lock_on_CPU(int cpu) { cpu_set_t mask; CPU_ZERO(&mask); CPU_SET(cpu, &mask); sched_setaffinity(0, sizeof(mask), &mask); } #elif defined (_WIN32) #include <Windows.h> #define lock_on_CPU(cpu) SetThreadAffinityMask(GetCurrentThread(), 1 << cpu) #else // #warning is not really standard, so this might still cause compiler errors on some platforms. Sorry about that. #warning "Thread processor affinity settings not supported. Performances might be improved by providing a suitable alternative for your platform" #define lock_on_CPU(cpu) #endif // ----------------------------------------------------------------- // B values generator // ----------------------------------------------------------------- struct Bvalue { vectorB p1; vectorB m1; }; struct Bgenerator { int n; // A length Aenumerator stop; // computation limit Aenumerator zeroes; // current zeroes pattern Aenumerator plusminus; // current +1/-1 pattern Aenumerator pm_limit; // upper bound of +1/-1 pattern Bgenerator(int n, Aenumerator start=0, Aenumerator stop=0) : n(n), stop(stop) { // initialize generator so that first call to next() will generate first value zeroes = start - 1; plusminus = -1; pm_limit = 0; } // compute current B value Bvalue value(void) { Bvalue res; Aenumerator pm = plusminus; Aenumerator position = 1; int i_pm = 0; for (int i = 0; i != n; i++) { if (zeroes & position) { if (i_pm == 0) res.p1 |= position; // first non-zero value fixed to +1 else { if (pm & 1) res.m1 |= position; // next non-zero values else res.p1 |= position; pm >>= 1; } i_pm++; } position <<= 1; } res.p1 |= (res.p1 << n); // concatenate 2 Bpre instances res.m1 |= (res.m1 << n); return res; } // next value bool next(void) { if (++plusminus == pm_limit) { if (++zeroes == stop) return false; plusminus = 0; pm_limit = (1 << vectorB(zeroes).count()) >> 1; } return true; } // calibration: produces ranges that will yield the approximate same number of B values vector<Aenumerator> calibrate(int segments) { // setup generator for the whole B range zeroes = 0; stop = 1 << n; plusminus = -1; pm_limit = 0; // divide range into (nearly) equal chunks Aenumerator chunk_size = ((Aenumerator)pow (3,n)-1) / 2 / segments; // generate bounds for zeroes values vector<Aenumerator> res(segments + 1); int bound = 0; res[bound] = 1; Aenumerator count = 0; while (next()) if (++count % chunk_size == 0) res[++bound] = zeroes; res[bound] = stop; return res; } }; // ----------------------------------------------------------------- // equiprobable A values merging // ----------------------------------------------------------------- static char A_weight[1 << A_LENGTH]; struct Agroup { vectorB value; int count; Agroup(Aenumerator a = 0, int length = 0) : value(a), count(length) {} }; static vector<Agroup> A_groups; Aenumerator reverse(Aenumerator n) // this works on N-1 bits for a N bits word { Aenumerator res = 0; if (n != 0) // must have at least one bit set for the rest to work { // construct left-padded reverse value for (int i = 0; i != 8 * sizeof(n)-1; i++) { res |= (n & 1); res <<= 1; n >>= 1; } // shift right to elimitate trailing zeroes while (!(res & 1)) res >>= 1; } return res; } void generate_A_groups(int n) { static bitset<1 << A_LENGTH> lookup(0); Aenumerator limit_A = (Aenumerator)pow(2, n); Aenumerator overflow = 1 << n; for (char & w : A_weight) w = 0; // gather rotation cycles for (Aenumerator a = 0; a != limit_A; a++) { Aenumerator rotated = a; int cycle_length = 0; for (int i = 0; i != n; i++) { // check for new cycles if (!lookup[rotated]) { cycle_length++; lookup[rotated] = 1; } // rotate current value rotated <<= 1; if (rotated & overflow) rotated |= 1; rotated &= (overflow - 1); } // store new cycle if (cycle_length > 0) A_weight[a] = cycle_length; } // merge symetric groups for (Aenumerator a = 0; a != limit_A; a++) { // skip already grouped values if (A_weight[a] == 0) continue; // regroup a symetric pair Aenumerator r = reverse(a); if (r != a) { A_weight[a] += A_weight[r]; A_weight[r] = 0; } } // generate groups for (Aenumerator a = 0; a != limit_A; a++) { if (A_weight[a] != 0) A_groups.push_back(Agroup(a, A_weight[a])); } } // ----------------------------------------------------------------- // worker thread // ----------------------------------------------------------------- OccurenceCounters solve(int n, int index, Aenumerator Bstart, Aenumerator Bstop) { OccurenceCounters consecutive_zero_Z(n, 0); // counts occurences of the first i terms of Z being 0 // lock on assigned CPU lock_on_CPU(index); // enumerate B vectors Bgenerator Bgen(n, Bstart, Bstop); while (Bgen.next()) { // get next B value Bvalue B = Bgen.value(); // enumerate A vector groups for (const auto & group : A_groups) { // count consecutive occurences of inner product equal to zero vectorB sliding_A(group.value); for (int i = 0; i != n; i++) { if ((sliding_A & B.p1).count() != (sliding_A & B.m1).count()) break; consecutive_zero_Z[i] += group.count; sliding_A <<= 1; } } } return consecutive_zero_Z; } // ----------------------------------------------------------------- // main // ----------------------------------------------------------------- #define die(msg) { cout << msg << endl; exit (-1); } int main(int argc, char * argv[]) { int n = argc == 2 ? atoi(argv[1]) : 16; // arbitray value for debugging if (n < 1 || n > 24) die("vectors of lenght between 1 and 24 is all I can (try to) compute, guv"); auto begin = time(NULL); // one worker thread per CPU int num_workers = number_of_CPUs(); // regroup equiprobable A values generate_A_groups(n); // compute B generation ranges for proper load balancing vector<Aenumerator> ranges = Bgenerator(n).calibrate(num_workers); // set workers to work vector<future<OccurenceCounters>> workers(num_workers); for (int i = 0; i != num_workers; i++) { workers[i] = async( launch::async, // without this parameter, C++ will decide whether execution shall be sequential or asynchronous (isn't C++ fun?). solve, n, i, ranges[i], ranges[i+1]); } // collect results OccurenceCounters result(n + 1, 0); for (auto& worker : workers) { OccurenceCounters partial = worker.get(); for (size_t i = 0; i != partial.size(); i++) result[i] += partial[i]*2; // each result counts for a symetric B pair } for (Counter & res : result) res += (Counter)1 << n; // add null B vector contribution result[n] = result[n - 1]; // the last two probabilities are equal by construction auto duration = time(NULL) - begin; // output cout << "done in " << duration / 60 << ":" << setw(2) << setfill('0') << duration % 60 << setfill(' ') << " by " << num_workers << " worker thread" << ((num_workers > 1) ? "s" : "") << endl; Counter events = (Counter)pow(6, n); int width = (int)log10(events) + 2; cout.precision(5); for (int i = 0; i <= n; i++) cout << setw(2) << i << setw(width) << result[i] << " / " << events << " " << fixed << (float)result[i] / events << endl; return 0; } 

Building the executable

It's a standalone C++11 source that compiles without warnings and runs smoothly on:

• Win7 & MSVC2013
• Win7 & MinGW - g++4.7
• Ubuntu & g++4.8 (in a VirtualBox VM with 2 CPUs allocated)

If you compile with g++, use : g++ -O3 -pthread -std=c++11
forgetting about -pthread will produce a nice and friendly core dump.

Optimizations

1. The last Z term is equal to the first (Bpre x A in both cases), so the last two results are always equal, which dispenses of computing the last Z value.
   The gain is neglectible, but coding it costs nothing so you might as well use it.

2. As Jakube discovered, all cyclic values of a given A vector produce the same probabilities.
   You can compute these with a single instance of A and multiply the result by the number of its possible rotations. Rotation groups can easily be pre-computed in a neglectible amount of time, so this is a huge net speed gain.
   Since the number of permutations of an n length vector is n-1, the complexity drops from o(6ⁿ) to o(6ⁿ/(n-1)), basically going a step further for the same computation time.

3. It appears pairs of symetric patterns also generate the same probabilities. For instance, 100101 and 101001.
   I have no mathematical proof of that, but intuitively when presented with all possible B patterns, each symetric A value will be convoluted with the corresponding symetric B value for the same global outcome.
   This allows to regroup some more A vectors, for an approximative 30% reduction of A groups number.

4. WRONG For some semi-mysterious reason, all patterns with only one or two bits set produce the same result. This does not represent that many distinct groups, but still they can be merged at virtually no cost.

5. The vectors B and -B (B with all components multiplied by -1) produce the same probabilities.
   (for instance [ 1, 0,-1, 1] and [-1, 0, 1,-1]).
   Except for the null vector (all components equal to 0), B and -B form a pair of distinct vectors.
   This allows to cut the number of B values in half by considering only one of each pair and multiplying its contribution by 2, adding the known global contribution of null B to each probability only once.

How it works

The number of B values is huge (3ⁿ), so pre-computing them would require indecent amounts of memory, which would slow down computation and eventually exhaust available RAM.
Unfortunately, I could not find a simple way of enumerating the half set of optimized B values, so I resorted to coding a dedicated generator.

The mighty B generator was a lot of fun to code, though languages that support yield mechanisms would have allowed to program it in a much more elegant way.
What it does in a nutshell is consider the "skeleton" of a Bpre vector as a binary vector where 1s represent actual -1 or +1 values.
Among all these +1/-1 potential values, the first one is fixed to +1 (thus selecting one of the possible B/-B vector), and all remaining possible +1/-1 combinations are enumerated.
Lastly, a simple calibration system ensures each worker thread will process a range of values of approximately the same size.

A values are heavily filtered for regrouping in equiprobable chunks.
This is done in a pre-computing phase that brute-force examines all possible values.
This part has a neglectible O(2ⁿ) execution time and does not need to be optimized (the code being already unreadable enough as it is!).

To evaluate the inner product (which needs only be tested against zero), the -1 and 1 components of B are regrouped into binary vectors.
The inner product is null if (and only if) there is an equal number of +1s and -1s among the B values corresponding to non-zero A values.
This can be computed with simple masking and bit count operations, helped by std::bitset that will produce reasonably efficient bit count code without having to resort to ugly intrinsic instructions.

The work is equally divided among cores, with forced CPU affinity (every little bit helps, or so they say).

Example result

C:\Dev\PHP\_StackOverflow\C++\VectorCrunch>release\VectorCrunch.exe 16 done in 8:19 by 4 worker threads 0 487610895942 / 2821109907456 0.17284 1 97652126058 / 2821109907456 0.03461 2 20659337010 / 2821109907456 0.00732 3 4631534490 / 2821109907456 0.00164 4 1099762394 / 2821109907456 0.00039 5 302001914 / 2821109907456 0.00011 6 115084858 / 2821109907456 0.00004 7 70235786 / 2821109907456 0.00002 8 59121706 / 2821109907456 0.00002 9 56384426 / 2821109907456 0.00002 10 55686922 / 2821109907456 0.00002 11 55508202 / 2821109907456 0.00002 12 55461994 / 2821109907456 0.00002 13 55451146 / 2821109907456 0.00002 14 55449098 / 2821109907456 0.00002 15 55449002 / 2821109907456 0.00002 16 55449002 / 2821109907456 0.00002 

Performances

Multithreading should work perfectly on this, though only "true" cores will fully contribute to computation speed. My CPU has only 2 cores for 4 CPUs, and the gain over a single-threaded version is "only" about 3.5.

Compilers

An initial problem with multithreading led me to believe the GNU compilers were performing worse than Microsoft.

After more thorough testing, it appears g++ wins the day yet again, producing approximatively 30% faster code (the same ratio I noticed on two other computation-heavy projects).

Notably, the std::bitset library is implemented with dedicated bit count instructions by g++ 4.8, while MSVC 2013 uses only loops of conventional bit shifts.

As one could have expected, compiling in 32 or 64 bits makes no difference.

Further refinements

I noticed a few A groups producing the same probabilities after all the reduction operations, but I could not identify a pattern that would allow to regroup them.

Here are the pairs I noticed for n=11:

 10001011 and 10001101 100101011 and 100110101 100101111 and 100111101 100110111 and 100111011 101001011 and 101001101 101011011 and 101101011 101100111 and 110100111 1010110111 and 1010111011 1011011111 and 1011111011 1011101111 and 1011110111