[Question]
  [
The challenge is to write the fastest code possible for computing the [permanent of a matrix](https://en.wikipedia.org/wiki/Permanent_(mathematics)). 
The permanent of an `n`-by-`n` matrix `A` = (`a``i,j`) is defined as
[![enter image description here](https://i.stack.imgur.com/GOz7G.png)](https://i.stack.imgur.com/GOz7G.png)
Here `S_n` represents the set of all permutations of `[1, n]`.
As an example (from the wiki):
[![enter image description here](https://i.stack.imgur.com/Qr1OJ.png)](https://i.stack.imgur.com/Qr1OJ.png)
In this question matrices are all square and will only have the values `-1` and `1` in them. 
**Examples**
Input:
```
[[ 1 -1 -1  1]
 [-1 -1 -1  1]
 [-1  1 -1  1]
 [ 1 -1 -1  1]]
```
Permanent:
```
-4
```
Input:
```
[[-1 -1 -1 -1]
 [-1  1 -1 -1]
 [ 1 -1 -1 -1]
 [ 1 -1  1 -1]]
```
Permanent:
```
0
```
Input:
```
[[ 1 -1  1 -1 -1 -1 -1 -1]
 [-1 -1  1  1 -1  1  1 -1]
 [ 1 -1 -1 -1 -1  1  1  1]
 [-1 -1 -1  1 -1  1  1  1]
 [ 1 -1 -1  1  1  1  1 -1]
 [-1  1 -1  1 -1  1  1 -1]
 [ 1 -1  1 -1  1 -1  1 -1]
 [-1 -1  1 -1  1  1  1  1]]
```
Permanent:
```
192
```
Input:
```
[[1, -1, 1, -1, -1, 1, 1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, 1, 1, -1],
 [1, -1, 1, 1, 1, 1, 1, -1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1],
 [-1, -1, 1, 1, 1, -1, -1, -1, -1, 1, -1, 1, 1, 1, -1, -1, -1, 1, -1, -1],
 [-1, -1, -1, 1, 1, -1, 1, 1, 1, 1, 1, 1, -1, -1, -1, -1, -1, -1, 1, -1],
 [-1, 1, 1, 1, -1, 1, 1, 1, -1, -1, -1, 1, -1, 1, -1, 1, 1, 1, 1, 1],
 [1, -1, 1, 1, -1, -1, 1, -1, 1, 1, 1, 1, -1, 1, 1, -1, 1, -1, -1, -1],
 [1, -1, -1, 1, -1, -1, -1, 1, -1, 1, 1, 1, 1, -1, -1, -1, 1, 1, 1, -1],
 [1, -1, -1, 1, -1, 1, 1, -1, 1, 1, 1, -1, 1, -1, 1, 1, 1, -1, 1, 1],
 [1, -1, -1, -1, -1, -1, 1, 1, 1, -1, -1, -1, -1, -1, 1, 1, -1, 1, 1, -1],
 [-1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, 1, 1, 1, 1, 1, -1, 1, 1],
 [-1, -1, -1, -1, -1, -1, -1, 1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1],
 [1, 1, -1, -1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -1, 1, 1, 1, 1, 1, 1],
 [-1, 1, 1, -1, -1, -1, -1, -1, 1, 1, 1, 1, -1, -1, -1, -1, -1, 1, -1, 1],
 [1, 1, -1, -1, -1, 1, -1, 1, -1, -1, -1, -1, 1, -1, 1, 1, -1, 1, -1, 1],
 [1, 1, 1, 1, 1, -1, -1, -1, 1, 1, 1, -1, 1, -1, 1, 1, 1, -1, 1, 1],
 [1, -1, -1, 1, -1, -1, -1, -1, 1, -1, -1, 1, 1, -1, 1, -1, -1, -1, -1, -1],
 [-1, 1, 1, 1, -1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1],
 [1, 1, -1, -1, 1, 1, -1, 1, 1, -1, 1, 1, 1, -1, 1, 1, -1, 1, -1, 1],
 [1, 1, 1, -1, -1, -1, 1, -1, -1, 1, 1, -1, -1, -1, 1, -1, -1, -1, -1, 1],
 [-1, 1, 1, 1, -1, -1, -1, -1, -1, -1, -1, 1, 1, -1, 1, 1, -1, 1, -1, -1]]
```
Permanent:
```
1021509632
```
**The task**
You should write code that, given an `n` by `n` matrix, outputs its permanent.
As I will need to test your code it would be helpful if you could give a simple way for me to give a matrix as input to your code, for example by reading from standard in.
Be warned that the permanent can be large (the all 1s matrix is the extreme case).
**Scores and ties** 
I will test your code on random +-1 matrices of increasing size and stop the first time your code takes more than 1 minute on my computer. The scoring matrices will be consistent for all submissions in order to ensure fairness.
If two people get the same score then the winner is the one which is fastest for that value of `n`. If those are within 1 second of each other then it is the one posted first.
**Languages and libraries**
You can use any available language and libraries you like but no pre-existing function to compute the permanent. Where feasible, it would be good to be able to run your code so please include a full explanation for how to run/compile your code in Linux if at all possible.`
**Reference implementations**
There is already a [codegolf question](https://codegolf.stackexchange.com/questions/96825/codegolf-the-permanent) question with lots of code in different languages for computing the permanent for small matrices. [Mathematica](https://reference.wolfram.com/language/ref/Permanent.html) and [Maple](https://www.maplesoft.com/support/help/Maple/view.aspx?path=LinearAlgebra%2FPermanent) also both have permanent implementations if you can access those. 
**My Machine** The timings will be run on my 64-bit machine. This is a standard ubuntu install with 8GB RAM, AMD FX-8350 Eight-Core Processor and Radeon HD 4250. This also means I need to be able to run your code.
**Low level information about my machine**
`cat /proc/cpuinfo/|grep flags` gives 
> 
> flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov
>  pat pse36 clflush mmx fxsr sse sse2 ht syscall nx mmxext fxsr\_opt
>  pdpe1gb rdtscp lm constant\_tsc rep\_good nopl nonstop\_tsc extd\_apicid
>  aperfmperf pni pclmulqdq monitor ssse3 fma cx16 sse4\_1 sse4\_2 popcnt
>  aes xsave avx f16c lahf\_lm cmp\_legacy svm extapic cr8\_legacy abm sse4a
>  misalignsse 3dnowprefetch osvw ibs xop skinit wdt lwp fma4 tce
>  nodeid\_msr tbm topoext perfctr\_core perfctr\_nb cpb hw\_pstate vmmcall
>  bmi1 arat npt lbrv svm\_lock nrip\_save tsc\_scale vmcb\_clean flushbyasid
>  decodeassists pausefilter pfthreshold
> 
> 
> 
I will ask a closely related follow up multi-language question that doesn't suffer from the big Int problem so lovers of [Scala](https://en.wikipedia.org/wiki/Scala_(programming_language)), [Nim](https://en.wikipedia.org/wiki/Nim_(programming_language)), [Julia](https://en.wikipedia.org/wiki/Julia_(programming_language)), [Rust](https://en.wikipedia.org/wiki/Rust_(programming_language)), [Bash](https://en.wikipedia.org/wiki/Bash_(Unix_shell)) can show off their languages too.
**Leader board**
* n = 33 (45 seconds. 64 seconds for n = 34). **Ton Hospel** in *C++* with g++ 5.4.0.
* n = 32 (32 seconds). **Dennis** in **C** with gcc 5.4.0 using Ton Hospel's gcc flags.
* n = 31 (54 seconds). **Christian Sievers** in **Haskell**
* n = 31 (60 seconds). **primo** in **rpython**
* n = 30 (26 seconds). **ezrast** in **Rust**
* n = 28 (49 seconds). **xnor** with **Python + pypy 5.4.1**
* n = 22 (25 seconds). **Shebang** with **Python + pypy 5.4.1**
*Note*. In practice the timings for Dennis and Ton Hospel's vary a lot for mysterious reasons. For example they seem to be faster after I have loaded a web browser! The timings quoted are the fastest in all the tests I have done.
  
[Answer]
# gcc C++ n ≈ 36 (57 seconds on my system)
Uses Glynn formula with a Gray code for updates if all column sums are even, otherwise uses Ryser's method. Threaded and vectorized. Optimized for AVX, so don't expect much on older processors. Don't bother with `n>=35` for a matrix with only +1's even if your system is fast enough since the signed 128 bit accumulator will overflow. For random matrices you will probably not hit the overflow. For `n>=37` the internal multipliers will start to overflow for an all `1/-1` matrix. So only use this program for `n<=36`.
Just give the matrix elements on STDIN separated by any kind of whitespace
```
permanent
1 2
3 4
^D
```
`permanent.cpp`:
```
/*
  Compile using something like:
    g++ -Wall -O3 -march=native -fstrict-aliasing -std=c++11 -pthread -s permanent.cpp -o permanent
*/
#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <cstdint>
#include <climits>
#include <array>
#include <vector>
#include <thread>
#include <future>
#include <ctgmath>
#include <immintrin.h>
using namespace std;
bool const DEBUG = false;
int const CACHE = 64;
using Index  = int_fast32_t;
Index glynn;
// Number of elements in our vectors
Index const POW   = 3;
Index const ELEMS = 1 << POW;
// Over how many floats we distribute each row
Index const WIDTH = 9;
// Number of bits in the fraction part of a floating point number
int const FLOAT_MANTISSA = 23;
// Type to use for the first add/multiply phase
using Sum  = float;
using SumN = __restrict__ Sum __attribute__((vector_size(ELEMS*sizeof(Sum))));
// Type to convert to between the first and second phase
using ProdN = __restrict__ int32_t __attribute__((vector_size(ELEMS*sizeof(int32_t))));
// Type to use for the third and last multiply phase.
// Also used for the final accumulator
using Value = __int128;
using UValue = unsigned __int128;
// Wrap Value so C++ doesn't really see it and we can put it in vectors etc.
// Needed since C++ doesn't fully support __int128
struct Number {
    Number& operator+=(Number const& right) {
        value += right.value;
        return *this;
    }
    // Output the value
    void print(ostream& os, bool dbl = false) const;
    friend ostream& operator<<(ostream& os, Number const& number) {
        number.print(os);
        return os;
    }
    Value value;
};
using ms = chrono::milliseconds;
auto nr_threads = thread::hardware_concurrency();
vector<Sum> input;
// Allocate cache aligned datastructures
template<typename T>
T* alloc(size_t n) {
    T* mem = static_cast<T*>(aligned_alloc(CACHE, sizeof(T) * n));
    if (mem == nullptr) throw(bad_alloc());
    return mem;
}
// Work assigned to thread k of nr_threads threads
Number permanent_part(Index n, Index k, SumN** more) {
    uint64_t loops = (UINT64_C(1) << n) / nr_threads;
    if (glynn) loops /= 2;
    Index l = loops < ELEMS ? loops : ELEMS;
    loops /= l;
    auto from = loops * k;
    auto to   = loops * (k+1);
    if (DEBUG) cout << "From=" << from << "\n";
    uint64_t old_gray = from ^ from/2;
    uint64_t bit = 1;
    bool bits = (to-from) & 1;
    Index nn = (n+WIDTH-1)/WIDTH;
    Index ww = nn * WIDTH;
    auto column = alloc<SumN>(ww);
    for (Index i=0; i<n; ++i)
        for (Index j=0; j<ELEMS; ++j) column[i][j] = 0;
    for (Index i=n; i<ww; ++i)
        for (Index j=0; j<ELEMS; ++j) column[i][j] = 1;
    Index b;
    if (glynn) {
        b = n > POW+1 ? n - POW - 1: 0;
        auto c = n-1-b;
        for (Index k=0; k<l; k++) {
            Index gray = k ^ k/2;
            for (Index j=0; j< c; ++j)
                if (gray & 1 << j)
                    for (Index i=0; i<n; ++i)
                        column[i][k] -= input[(b+j)*n+i];
                else
                    for (Index i=0; i<n; ++i)
                        column[i][k] += input[(b+j)*n+i];
        }
        for (Index i=0; i<n; ++i)
            for (Index k=0; k<l; k++)
                column[i][k] += input[n*(n-1)+i];
        for (Index k=1; k<l; k+=2)
            column[0][k] = -column[0][k];
        for (Index i=0; i<b; ++i, bit <<= 1) {
            if (old_gray & bit) {
                bits = bits ^ 1;
                for (Index j=0; j<ww; ++j)
                    column[j] -= more[i][j];
            } else {
                for (Index j=0; j<ww; ++j)
                    column[j] += more[i][j];
            }
        }
        for (Index i=0; i<n; ++i)
            for (Index k=0; k<l; k++)
                column[i][k] /= 2;
    } else {
        b = n > POW ? n - POW : 0;
        auto c = n-b;
        for (Index k=0; k<l; k++) {
            Index gray = k ^ k/2;
            for (Index j=0; j<c; ++j)
                if (gray & 1 << j)
                    for (Index i=0; i<n; ++i)
                        column[i][k] -= input[(b+j)*n+i];
        }
        for (Index k=1; k<l; k+=2)
            column[0][k] = -column[0][k];
        for (Index i=0; i<b; ++i, bit <<= 1) {
            if (old_gray & bit) {
                bits = bits ^ 1;
                for (Index j=0; j<ww; ++j)
                    column[j] -= more[i][j];
            }
        }
    }
    if (DEBUG) {
        for (Index i=0; i<ww; ++i) {
            cout << "Column[" << i << "]=";
            for (Index j=0; j<ELEMS; ++j) cout << " " << column[i][j];
            cout << "\n";
        }
    }
    --more;
    old_gray = (from ^ from/2) | UINT64_C(1) << b;
    Value total = 0;
    SumN accu[WIDTH];
    for (auto p=from; p<to; ++p) {
        uint64_t new_gray = p ^ p/2;
        uint64_t bit = old_gray ^ new_gray;
        Index i = __builtin_ffsl(bit);
        auto diff = more[i];
        auto c = column;
        if (new_gray > old_gray) {
            // Phase 1 add/multiply.
            // Uses floats until just before loss of precision
            for (Index i=0; i<WIDTH; ++i) accu[i] = *c++ -= *diff++;
            for (Index j=1; j < nn; ++j)
                for (Index i=0; i<WIDTH; ++i) accu[i] *= *c++ -= *diff++;
        } else {
            // Phase 1 add/multiply.
            // Uses floats until just before loss of precision
            for (Index i=0; i<WIDTH; ++i) accu[i] = *c++ += *diff++;
            for (Index j=1; j < nn; ++j)
                for (Index i=0; i<WIDTH; ++i) accu[i] *= *c++ += *diff++;
        }
        if (DEBUG) {
            cout << "p=" << p << "\n";
            for (Index i=0; i<ww; ++i) {
                cout << "Column[" << i << "]=";
                for (Index j=0; j<ELEMS; ++j) cout << " " << column[i][j];
                cout << "\n";
            }
        }
        // Convert floats to int32_t
        ProdN prod32[WIDTH] __attribute__((aligned (32)));
        for (Index i=0; i<WIDTH; ++i)
            // Unfortunately gcc doesn't recognize the static_cast<int32_t>
            // as a vector pattern, so force it with an intrinsic
#ifdef __AVX__
            //prod32[i] = static_cast<ProdN>(accu[i]);
            reinterpret_cast<__m256i&>(prod32[i]) = _mm256_cvttps_epi32(accu[i]);
#else   // __AVX__
            for (Index j=0; j<ELEMS; ++j)
                prod32[i][j] = static_cast<int32_t>(accu[i][j]);
#endif  // __AVX__
        // Phase 2 multiply. Uses int64_t until just before overflow
        int64_t prod64[3][ELEMS];
        for (Index i=0; i<3; ++i) {
            for (Index j=0; j<ELEMS; ++j)
                prod64[i][j] = static_cast<int64_t>(prod32[i][j]) * prod32[i+3][j] * prod32[i+6][j];
        }
        // Phase 3 multiply. Collect into __int128. For large matrices this will
        // actually overflow but that's ok as long as all 128 low bits are
        // correct. Terms will cancel and the final sum can fit into 128 bits
        // (This will start to fail at n=35 for the all 1 matrix)
        // Strictly speaking this needs the -fwrapv gcc option
        for (Index j=0; j<ELEMS; ++j) {
            auto value = static_cast<Value>(prod64[0][j]) * prod64[1][j] * prod64[2][j];
            if (DEBUG) cout << "value[" << j << "]=" << static_cast<double>(value) << "\n";
            total += value;
        }
        total = -total;
        old_gray = new_gray;
    }
    return bits ? Number{-total} : Number{total};
}
// Prepare datastructures, Assign work to threads
Number permanent(Index n) {
    Index nn = (n+WIDTH-1)/WIDTH;
    Index ww = nn*WIDTH;
    Index rows  = n > (POW+glynn) ? n-POW-glynn : 0;
    auto data = alloc<SumN>(ww*(rows+1));
    auto pointers = alloc<SumN *>(rows+1);
    auto more = &pointers[0];
    for (Index i=0; i<rows; ++i)
        more[i] = &data[ww*i];
    more[rows] = &data[ww*rows];
    for (Index j=0; j<ww; ++j)
        for (Index i=0; i<ELEMS; ++i)
            more[rows][j][i] = 0;
    Index loops = n >= POW+glynn ? ELEMS : 1 << (n-glynn);
    auto a = &input[0];
    for (Index r=0; r<rows; ++r) {
        for (Index j=0; j<n; ++j) {
            for (Index i=0; i<loops; ++i)
                more[r][j][i] = j == 0 && i %2 ? -*a : *a;
            for (Index i=loops; i<ELEMS; ++i)
                more[r][j][i] = 0;
            ++a;
        }
        for (Index j=n; j<ww; ++j)
            for (Index i=0; i<ELEMS; ++i)
                more[r][j][i] = 0;
    }
    if (DEBUG)
        for (Index r=0; r<=rows; ++r)
            for (Index j=0; j<ww; ++j) {
                cout << "more[" << r << "][" << j << "]=";
                for (Index i=0; i<ELEMS; ++i)
                    cout << " " << more[r][j][i];
                cout << "\n";
            }
    // Send work to threads...
    vector<future<Number>> results;
    for (auto i=1U; i < nr_threads; ++i)
        results.emplace_back(async(DEBUG ? launch::deferred: launch::async, permanent_part, n, i, more));
    // And collect results
    auto r = permanent_part(n, 0, more);
    for (auto& result: results)
        r += result.get();
    free(data);
    free(pointers);
    // For glynn we should double the result, but we will only do this during
    // the final print. This allows n=34 for an all 1 matrix to work
    // if (glynn) r *= 2;
    return r;
}
// Print 128 bit number
void Number::print(ostream& os, bool dbl) const {
    const UValue BILLION = 1000000000;
    UValue val;
    if (value < 0) {
        os << "-";
        val = -value;
    } else
        val = value;
    if (dbl) val *= 2;
    uint32_t output[5];
    for (int i=0; i<5; ++i) {
        output[i] = val % BILLION;
        val /= BILLION;
    }
    bool print = false;
    for (int i=4; i>=0; --i) {
        if (print) {
            os << setfill('0') << setw(9) << output[i];
        } else if (output[i] || i == 0) {
            print = true;
            os << output[i];
        }
    }
}
// Read matrix, check for sanity
void my_main() {
    Sum a;
    while (cin >> a)
        input.push_back(a);
    size_t n = sqrt(input.size());
    if (input.size() != n*n)
        throw(logic_error("Read " + to_string(input.size()) +
                          " elements which does not make a square matrix"));
    vector<double> columns_pos(n, 0);
    vector<double> columns_neg(n, 0);
    Sum *p = &input[0];
    for (size_t i=0; i<n; ++i)
        for (size_t j=0; j<n; ++j, ++p) {
            if (*p >= 0) columns_pos[j] += *p;
            else         columns_neg[j] -= *p;
        }
    std::array<double,WIDTH> prod;
    prod.fill(1);
    int32_t odd = 0;
    for (size_t j=0; j<n; ++j) {
        prod[j%WIDTH] *= max(columns_pos[j], columns_neg[j]);
        auto sum = static_cast<int32_t>(columns_pos[j] - columns_neg[j]);
        odd |= sum;
    }
    glynn = (odd & 1) ^ 1;
    for (Index i=0; i<WIDTH; ++i)
        // A float has an implicit 1. in front of the fraction so it can
        // represent 1 bit more than the mantissa size. And 1 << (mantissa+1)
        // itself is in fact representable
        if (prod[i] && log2(prod[i]) > FLOAT_MANTISSA+1)
            throw(range_error("Values in matrix are too large. A subproduct reaches " + to_string(prod[i]) + " which doesn't fit in a float without loss of precision"));
    for (Index i=0; i<3; ++i) {
        auto prod3 = prod[i] * prod[i+3] * prod[i+6];
        if (log2(prod3) >= CHAR_BIT*sizeof(int64_t)-1)
            throw(range_error("Values in matrix are too large. A subproduct reaches " + to_string(prod3) + " which doesn't fit in an int64"));
    }
    nr_threads = pow(2, ceil(log2(static_cast<float>(nr_threads))));
    uint64_t loops = UINT64_C(1) << n;
    if (glynn) loops /= 2;
    if (nr_threads * ELEMS > loops)
        nr_threads = max(loops / ELEMS, UINT64_C(1));
    // if (DEBUG) nr_threads = 1;
    cout << n << " x " << n << " matrix, method " << (glynn ? "Glynn" : "Ryser") << ", " << nr_threads << " threads" << endl;
    // Go for the actual calculation
    auto start = chrono::steady_clock::now();
    auto perm = permanent(n);
    auto end = chrono::steady_clock::now();
    auto elapsed = chrono::duration_cast<ms>(end-start).count();
    cout << "Permanent=";
    perm.print(cout, glynn);
    cout << " (" << elapsed / 1000. << " s)" << endl;
}
// Wrapper to print any exceptions
int main() {
    try {
        my_main();
    } catch(exception& e) {
        cerr << "Error: " << e.what() << endl;
        exit(EXIT_FAILURE);
    }
    exit(EXIT_SUCCESS);
}
```
[Answer]
# C99, n ≈ 33 (35 seconds)
```
#include <stdint.h>
#include <stdio.h>
#define CHUNK_SIZE 12
#define NUM_THREADS 8
#define popcnt __builtin_popcountll
#define BILLION (1000 * 1000 * 1000)
#define UPDATE_ROW_PPROD() \
    update_row_pprod(row_pprod, row, rows, row_sums, mask, mask_popcnt)
typedef __int128 int128_t;
static inline int64_t update_row_pprod
(
    int64_t* row_pprod, int64_t row, int64_t* rows,
    int64_t* row_sums, int64_t mask, int64_t mask_popcnt
)
{
    int64_t temp = 2 * popcnt(rows[row] & mask) - mask_popcnt;
    row_pprod[0] *= temp;
    temp -= 1;
    row_pprod[1] *= temp;
    temp -= row_sums[row];
    row_pprod[2] *= temp;
    temp += 1;
    row_pprod[3] *= temp;
    return row + 1;
}
int main(int argc, char* argv[])
{
    int64_t size = argc - 1, rows[argc - 1];
    int64_t row_sums[argc - 1];
    int128_t permanent = 0, sign = size & 1 ? -1 : 1;
    if (argc == 2)
    {
        printf("%d\n", argv[1][0] == '-' ? -1 : 1);
        return 0;
    }
    for (int64_t row = 0; row < size; row++)
    {
        char positive = argv[row + 1][0] == '+' ? '-' : '+';
        sign *= ',' - positive;
        rows[row] = row_sums[row] = 0;
        for (char* p = &argv[row + 1][1]; *p; p++)
        {
            rows[row] <<= 1;
            rows[row] |= *p == positive;
            row_sums[row] += *p == positive;
        }
        row_sums[row] = 2 * row_sums[row] - size;
    }
    #pragma omp parallel for reduction(+:permanent) num_threads(NUM_THREADS)
    for (int64_t mask = 1; mask < 1LL << (size - 1); mask += 2)
    {
        int64_t mask_popcnt = popcnt(mask);
        int64_t row = 0;
        int128_t row_prod = 1 - 2 * (mask_popcnt & 1);
        int128_t row_prod_high = -row_prod;
        int128_t row_prod_inv = row_prod;
        int128_t row_prod_inv_high = -row_prod;
        for (int64_t chunk = 0; chunk < size / CHUNK_SIZE; chunk++)
        {
            int64_t row_pprod[4] = {1, 1, 1, 1};
            for (int64_t i = 0; i < CHUNK_SIZE; i++)
                row = UPDATE_ROW_PPROD();
            row_prod *= row_pprod[0], row_prod_high *= row_pprod[1];
            row_prod_inv *= row_pprod[3], row_prod_inv_high *= row_pprod[2];
        }
        int64_t row_pprod[4] = {1, 1, 1, 1};
        while (row < size)
            row = UPDATE_ROW_PPROD();
        row_prod *= row_pprod[0], row_prod_high *= row_pprod[1];
        row_prod_inv *= row_pprod[3], row_prod_inv_high *= row_pprod[2];
        permanent += row_prod + row_prod_high + row_prod_inv + row_prod_inv_high;
    }
    permanent *= sign;
    if (permanent < 0)
        printf("-"), permanent *= -1;
    int32_t output[5], print = 0;
    output[0] = permanent % BILLION, permanent /= BILLION;
    output[1] = permanent % BILLION, permanent /= BILLION;
    output[2] = permanent % BILLION, permanent /= BILLION;
    output[3] = permanent % BILLION, permanent /= BILLION;
    output[4] = permanent % BILLION;
    if (output[4])
        printf("%u", output[4]), print = 1;
    if (print)
        printf("%09u", output[3]);
    else if (output[3])
        printf("%u", output[3]), print = 1;
    if (print)
        printf("%09u", output[2]);
    else if (output[2])
        printf("%u", output[2]), print = 1;
    if (print)
        printf("%09u", output[1]);
    else if (output[1])
        printf("%u", output[1]), print = 1;
    if (print)
        printf("%09u\n", output[0]);
    else
        printf("%u\n", output[0]);
}
```
Input is currently a bit cumbersome; it is taken with rows as command line arguments, where each entry is represented by its sign, i.e., **+** indicates a **1** and **-** indicates a **-1**.
### Test run
```
$ gcc -Wall -std=c99 -march=native -Ofast -fopenmp -fwrapv -o permanent permanent.c
$ ./permanent +--+ ---+ -+-+ +--+
-4
$ ./permanent ---- -+-- +--- +-+-
0
$ ./permanent +-+----- --++-++- +----+++ ---+-+++ +--++++- -+-+-++- +-+-+-+- --+-++++
192
$ ./permanent +-+--+++----++++-++- +-+++++-+--+++--+++- --+++----+-+++---+-- ---++-++++++------+- -+++-+++---+-+-+++++ +-++--+-++++-++-+--- +--+---+-++++---+++- +--+-++-+++-+-+++-++ +-----+++-----++-++- --+-+-++-+-++++++-++ -------+----++++---- ++---++--+-++-++++++ -++-----++++-----+-+ ++---+-+----+-++-+-+ +++++---+++-+-+++-++ +--+----+--++-+----- -+++-++--+++--++--++ ++--++-++-+++-++-+-+ +++---+--++---+----+ -+++-------++-++-+--
1021509632
$ time ./permanent +++++++++++++++++++++++++++++++{,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,}     # 31
8222838654177922817725562880000000
real    0m8.365s
user    1m6.504s
sys     0m0.000s
$ time ./permanent ++++++++++++++++++++++++++++++++{,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,}   # 32
263130836933693530167218012160000000
real    0m17.013s
user    2m15.226s
sys     0m0.001s
$ time ./permanent +++++++++++++++++++++++++++++++++{,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,} # 33
8683317618811886495518194401280000000
real    0m34.592s
user    4m35.354s
sys     0m0.001s
```
[Answer]
## Python 2, n ≈ 28
```
from operator import mul
def fast_glynn_perm(M):
    row_comb = [sum(c) for c in zip(*M)]
    n=len(M)
    total = 0
    old_grey = 0 
    sign = +1
    binary_power_dict = {2**i:i for i in range(n)}
    num_loops = 2**(n-1)
    for bin_index in xrange(1, num_loops + 1):  
        total += sign * reduce(mul, row_comb)
        new_grey = bin_index^(bin_index/2)
        grey_diff = old_grey ^ new_grey
        grey_diff_index = binary_power_dict[grey_diff]
        new_vector = M[grey_diff_index]
        direction = 2 * cmp(old_grey,new_grey)      
        for i in range(n):
            row_comb[i] += new_vector[i] * direction
        sign = -sign
        old_grey = new_grey
    return total/num_loops
```
Uses the [Glynn formula](https://en.wikipedia.org/wiki/Computing_the_permanent#CITEREFBalasubramanian1980) with a [Gray code](https://en.wikipedia.org/wiki/Gray_code) for updates. Runs up to `n=23` in a minute on my machine. One can surely do better implementing this in a faster language and with better data structures. This doesn't use that the matrix is ±1-valued.
A Ryser formula implementation is very similar, summing over all 0/1 vectors of coefficients rather than ±1-vectors. It takes about twice as long as Glynn's formula because adds over all 2^n such vectors, whereas Glynn's halves that using symmetry to only those starting with `+1`.
```
from operator import mul
def fast_ryser_perm(M):
    n=len(M)
    row_comb = [0] * n
    total = 0
    old_grey = 0 
    sign = +1
    binary_power_dict = {2**i:i for i in range(n)}
    num_loops = 2**n
    for bin_index in range(1, num_loops) + [0]: 
        total += sign * reduce(mul, row_comb)
        new_grey = bin_index^(bin_index/2)
        grey_diff = old_grey ^ new_grey
        grey_diff_index = binary_power_dict[grey_diff]
        new_vector = M[grey_diff_index]
        direction = cmp(old_grey, new_grey)
        for i in range(n):
            row_comb[i] += new_vector[i] * direction
        sign = -sign
        old_grey = new_grey
    return total * (-1)**n
```
[Answer]
## Haskell, n=31 (54s)
With a lot of invaluable contributions by @Angs: use `Vector`, use short circuit products, look at odd n.
```
import Control.Parallel.Strategies
import qualified Data.Vector.Unboxed as V
import Data.Int
type Row = V.Vector Int8
x :: Row -> [Row] -> Integer -> Int -> Integer
x p (v:vs) m c = let c' = c - 1
                     r = if c>0 then parTuple2 rseq rseq else r0
                     (a,b) = ( x p                  vs m    c' ,
                               x (V.zipWith(-) p v) vs (-m) c' )
                             `using` r
                 in a+b
x p _      m _ = prod m p
prod :: Integer -> Row -> Integer
prod m p = if 0 `V.elem` p then 0 
                           else V.foldl' (\a b->a*fromIntegral b) m p
p, pt :: [Row] -> Integer
p (v:vs) = x (foldl (V.zipWith (+)) v vs) (map (V.map (2*)) vs) 1 11
           `div` 2^(length vs)
p [] = 1 -- handle 0x0 matrices too  :-)
pt (v:vs) | even (length vs) = p ((V.map (2*) v) : vs ) `div` 2
pt mat                       = p mat
main = getContents >>= print . pt . map V.fromList . read
```
My first attempts at parallelism in Haskell. You can see a lot of optimization steps through the revision history. Amazingly, it were mostly very small changes. The code is based on the formula in the section "Balasubramanian-Bax/Franklin-Glynn formula" in the Wikipedia article on [computing the permanent](https://en.wikipedia.org/wiki/Computing_the_permanent).
`p` computes the permanent. It is called via `pt` which transforms the matrix in a way that is always valid, but especially useful for the matrices that we get here.
Compile with `ghc -O2 -threaded -fllvm -feager-blackholing -o <name> <name>.hs`. To run with parallelisation, give it runtime parameters like this: `./<name> +RTS -N`. Input is from stdin with nested comma separated lists in brackets like `[[1,2],[3,4]]` as in the last example (newlines allowed everywhere). 
[Answer]
# Python 2, n = 22 [Reference]
This is the 'reference' implementation I shared with Lembik yesterday, it misses making it to `n=23` by a few seconds on his machine, on my machine it does it in about 52 seconds. To achieve these speeds you need to run this through PyPy.
The first function calculates the permanent similar to how the determinant could be calculated, by going over each submatrix until you are left with a 2x2 which you can apply the basic rule to. It is *incredibly slow*.
The second function is the one implementing the Ryser function (the second equation listed in Wikipedia). The set `S` is essentially the powerset of the numbers `{1,...,n}` (variable `s_list` in the code).
```
from random import *
from time import time
from itertools import*
def perm(a): # naive method, recurses over submatrices, slow 
    if len(a) == 1:
        return a[0][0]
    elif len(a) == 2:
        return a[0][0]*a[1][1]+a[1][0]*a[0][1]
    else:
        tsum = 0
        for i in xrange(len(a)):
            transposed = [zip(*a)[j] for j in xrange(len(a)) if j != i]
            tsum += a[0][i] * perm(zip(*transposed)[1:])
        return tsum
def perm_ryser(a): # Ryser's formula, using matrix entries
    maxn = len(a)
    n_list = range(1,maxn+1)
    s_list = chain.from_iterable(combinations(n_list,i) for i in range(maxn+1))
    total = 0
    for st in s_list:
        stotal = (-1)**len(st)
        for i in xrange(maxn):
            stotal *= sum(a[i][j-1] for j in st)
        total += stotal
    return total*((-1)**maxn)
def genmatrix(d):
    mat = []
    for x in xrange(d):
        row = []
        for y in xrange(d):
            row.append([-1,1][randrange(0,2)])
        mat.append(row)
    return mat
def main():
    for i in xrange(1,24):
        k = genmatrix(i)
        print 'Matrix: (%dx%d)'%(i,i)
        print '\n'.join('['+', '.join(`j`.rjust(2) for j in a)+']' for a in k)
        print 'Permanent:',
        t = time()
        p = perm_ryser(k)
        print p,'(took',time()-t,'seconds)'
if __name__ == '__main__':
    main()
```
[Answer]
# Rust + extprim
This straightforward Ryser with Gray code implementation takes about ~~65~~ 90 seconds to run n=31 on my laptop. ~~I imagine your machine will get there in well under 60s.~~ I'm using extprim 1.1.1 for `i128`.
I have never used Rust and have no idea what I'm doing. No compiler options other than whatever `cargo build --release` does. Comments/suggestions/optimizations are appreciated.
Invocation is identical to Dennis' program.
```
use std::env;
use std::thread;
use std::sync::Arc;
use std::sync::mpsc;
extern crate extprim;
use extprim::i128::i128;
static THREADS : i64 = 8; // keep this a power of 2.
fn main() {
  // Read command line args for the matrix, specified like
  // "++- --- -+-" for [[1, 1, -1], [-1, -1, -1], [-1, 1, -1]].
  let mut args = env::args();
  args.next();
  let mat : Arc<Vec<Vec<i64>>> = Arc::new(args.map( |ss|
    ss.trim().bytes().map( |cc| if cc == b'+' {1} else {-1}).collect()
  ).collect());
  // Figure how many iterations each thread has to do.
  let size = 2i64.pow(mat.len() as u32);
  let slice_size = size / THREADS; // Assumes divisibility.
  let mut accumulator : i128;
  if slice_size >= 4 { // permanent() requires 4 divides slice_size
    let (tx, rx) = mpsc::channel();
    // Launch threads.
    for ii in 0..THREADS {
      let mat = mat.clone();
      let tx = tx.clone();
      thread::spawn(move ||
        tx.send(permanent(&mat, ii * slice_size, (ii+1) * slice_size))
      );
    }
    // Accumulate results.
    accumulator = extprim::i128::ZERO;
    for _ in 0..THREADS {
      accumulator += rx.recv().unwrap();
    }
  }
  else { // Small matrix, don't bother threading.
    accumulator = permanent(&mat, 0, size);
  }
  println!("{}", accumulator);
}
fn permanent(mat: &Vec<Vec<i64>>, start: i64, end: i64) -> i128 {
  let size = mat.len();
  let sentinel = std::i64::MAX / size as i64;
  let mut bits : Vec<bool> = Vec::with_capacity(size);
  let mut sums : Vec<i64> = Vec::with_capacity(size);
  // Initialize gray code bits.
  let gray_number = start ^ (start / 2);
  for row in 0..size {
    bits.push((gray_number >> row) % 2 == 1);
    sums.push(0);
  }
  // Initialize column sums
  for row in 0..size {
    if bits[row] {
      for column in 0..size {
        sums[column] += mat[row][column];
      }
    }
  }
  // Do first two iterations with initial sums
  let mut total = product(&sums, sentinel);
  for column in 0..size {
    sums[column] += mat[0][column];
  }
  bits[0] = true;
  total -= product(&sums, sentinel);
  // Do rest of iterations updating gray code bits incrementally
  let mut gray_bit : usize;
  let mut idx = start + 2;
  while idx < end {
    gray_bit = idx.trailing_zeros() as usize;
    if bits[gray_bit] {
      for column in 0..size {
        sums[column] -= mat[gray_bit][column];
      }
      bits[gray_bit] = false;
    }
    else {
      for column in 0..size {
        sums[column] += mat[gray_bit][column];
      }
      bits[gray_bit] = true;
    }
    total += product(&sums, sentinel);
    if bits[0] {
      for column in 0..size {
        sums[column] -= mat[0][column];
      }
      bits[0] = false;
    }
    else {
      for column in 0..size {
        sums[column] += mat[0][column];
      }
      bits[0] = true;
    }
    total -= product(&sums, sentinel);
    idx += 2;
  }
  return if size % 2 == 0 {total} else {-total};
}
#[inline]
fn product(sums : &Vec<i64>, sentinel : i64) -> i128 {
  let mut ret : Option<i128> = None;
  let mut tally = sums[0];
  for ii in 1..sums.len() {
    if tally.abs() >= sentinel {
      ret = Some(ret.map_or(i128::new(tally), |n| n * i128::new(tally)));
      tally = sums[ii];
    }
    else {
      tally *= sums[ii];
    }
  }
  if ret.is_none() {
    return i128::new(tally);
  }
  return ret.unwrap() * i128::new(tally);
}
```
[Answer]
# Mathematica, n ≈ 20
```
p[m_] := Last[Fold[Take[ListConvolve[##, {1, -1}, 0], 2^Length[m]]&,
  Table[If[IntegerQ[Log2[k]], m[[j, Log2[k] + 1]], 0], {j, n}, {k, 0, 2^Length[m] - 1}]]]
```
Using the `Timing` command, a 20x20 matrix requires about 48 seconds on my system. This is not exactly as efficient as the other since it relies on the fact that the permanent can be found as the coefficient of the product of polymomials from each row of the matrix. Efficient polynomial multiplication is performed by creating the coefficient lists and performing convolution using `ListConvolve`. This requires about *O*(2*n* *n*2) time assuming convolution is performed using a Fast Fourier transform or similar which requires *O*(*n* log *n*) time.
[Answer]
# RPython 5.4.1, n ≈ 32 (37 seconds)
```
from rpython.rlib.rtime import time
from rpython.rlib.rarithmetic import r_int, r_uint
from rpython.rlib.rrandom import Random
from rpython.rlib.rposix import pipe, close, read, write, fork, waitpid
from rpython.rlib.rbigint import rbigint
from math import log, ceil
from struct import pack
bitsize = len(pack('l', 1)) * 8 - 1
bitcounts = bytearray([0])
for i in range(16):
  b = bytearray([j+1 for j in bitcounts])
  bitcounts += b
def bitcount(n):
  bits = 0
  while n:
    bits += bitcounts[n & 65535]
    n >>= 16
  return bits
def main(argv):
  if len(argv) < 2:
    write(2, 'Usage: %s NUM_THREADS [N]'%argv[0])
    return 1
  threads = int(argv[1])
  if len(argv) > 2:
    n = int(argv[2])
    rnd = Random(r_uint(time()*1000))
    m = []
    for i in range(n):
      row = []
      for j in range(n):
        row.append(1 - r_int(rnd.genrand32() & 2))
      m.append(row)
  else:
    m = []
    strm = ""
    while True:
      buf = read(0, 4096)
      if len(buf) == 0:
        break
      strm += buf
    rows = strm.split("\n")
    for row in rows:
      r = []
      for val in row.split(' '):
        r.append(int(val))
      m.append(r)
    n = len(m)
  a = []
  for row in m:
    val = 0
    for v in row:
      val = (val << 1) | -(v >> 1)
    a.append(val)
  batches = int(ceil(n * log(n) / (bitsize * log(2))))
  pids = []
  handles = []
  total = rbigint.fromint(0)
  for i in range(threads):
    r, w = pipe()
    pid = fork()
    if pid:
      close(w)
      pids.append(pid)
      handles.append(r)
    else:
      close(r)
      total = run(n, a, i, threads, batches)
      write(w, total.str())
      close(w)
      return 0
  for pid in pids:
    waitpid(pid, 0)
  for handle in handles:
    strval = read(handle, 256)
    total = total.add(rbigint.fromdecimalstr(strval))
    close(handle)
  print total.rshift(n-1).str()
  return 0
def run(n, a, mynum, threads, batches):
  start = (1 << n-1) * mynum / threads
  end = (1 << n-1) * (mynum+1) / threads
  dtotal = rbigint.fromint(0)
  for delta in range(start, end):
    pdelta = rbigint.fromint(1 - ((bitcount(delta) & 1) << 1))
    for i in range(batches):
      pbatch = 1
      for j in range(i, n, batches):
        pbatch *= n - (bitcount(delta ^ a[j]) << 1)
      pdelta = pdelta.int_mul(pbatch)
    dtotal = dtotal.add(pdelta)
  return dtotal
def target(*args):
  return main
```
To compile, [download](http://pypy.org/download.html) the most recent PyPy source, and execute the following:
```
pypy /path/to/pypy-src/rpython/bin/rpython matrix-permanent.py
```
The resulting executable will be named `matrix-permanent-c` or similiar in the current working directory.
As of PyPy 5.0, RPython's threading primitives are a lot less primitive than they used to be. Newly spawned threads require the GIL, which is more or less useless for parallel computations. I've used `fork` instead, so it may not work as expected on Windows, ~~although I haven't tested~~ fails to compile (`unresolved external symbol _fork`).
The executable accepts up to two command line parameters. The first is the number of threads, the second optional parameter is `n`. If it is provided, a random matrix will be generated, otherwise it will be read from stdin. Each row must be newline separated (without a trailing newline), and each value space separated. The third example input would be given as:
```
1 -1 1 -1 -1 1 1 1 -1 -1 -1 -1 1 1 1 1 -1 1 1 -1
1 -1 1 1 1 1 1 -1 1 -1 -1 1 1 1 -1 -1 1 1 1 -1
-1 -1 1 1 1 -1 -1 -1 -1 1 -1 1 1 1 -1 -1 -1 1 -1 -1
-1 -1 -1 1 1 -1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1
-1 1 1 1 -1 1 1 1 -1 -1 -1 1 -1 1 -1 1 1 1 1 1
1 -1 1 1 -1 -1 1 -1 1 1 1 1 -1 1 1 -1 1 -1 -1 -1
1 -1 -1 1 -1 -1 -1 1 -1 1 1 1 1 -1 -1 -1 1 1 1 -1
1 -1 -1 1 -1 1 1 -1 1 1 1 -1 1 -1 1 1 1 -1 1 1
1 -1 -1 -1 -1 -1 1 1 1 -1 -1 -1 -1 -1 1 1 -1 1 1 -1
-1 -1 1 -1 1 -1 1 1 -1 1 -1 1 1 1 1 1 1 -1 1 1
-1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1
1 1 -1 -1 -1 1 1 -1 -1 1 -1 1 1 -1 1 1 1 1 1 1
-1 1 1 -1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1 -1 1 -1 1
1 1 -1 -1 -1 1 -1 1 -1 -1 -1 -1 1 -1 1 1 -1 1 -1 1
1 1 1 1 1 -1 -1 -1 1 1 1 -1 1 -1 1 1 1 -1 1 1
1 -1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 -1 -1 -1 -1 -1
-1 1 1 1 -1 1 1 -1 -1 1 1 1 -1 -1 1 1 -1 -1 1 1
1 1 -1 -1 1 1 -1 1 1 -1 1 1 1 -1 1 1 -1 1 -1 1
1 1 1 -1 -1 -1 1 -1 -1 1 1 -1 -1 -1 1 -1 -1 -1 -1 1
-1 1 1 1 -1 -1 -1 -1 -1 -1 -1 1 1 -1 1 1 -1 1 -1 -1
```
---
**Sample Usage**
```
$ time ./matrix-permanent-c 8 30
8395059644858368
real    0m8.582s
user    1m8.656s
sys     0m0.000s
```
---
**Method**
I've used the [Balasubramanian-Bax/Franklin-Glynn formula](https://en.wikipedia.org/wiki/Computing_the_permanent#Balasubramanian-Bax.2FFranklin-Glynn_formula), with a runtime complexity of *O(2nn)*. However, instead of iterating the *δ* in grey code order, I've instead replaced vector-row multiplication with a single xor operation (mapping (1, -1) → (0, 1)). The vector sum can likewise be found in a single operation, by taking n minus twice the popcount.
[Answer]
## Racket 84 bytes
Following simple function works for smaller matrices but hangs on my machine for larger matrices: 
```
(for/sum((p(permutations(range(length l)))))(for/product((k l)(c p))(list-ref k c)))
```
Ungolfed: 
```
(define (f ll) 
  (for/sum ((p (permutations (range (length ll))))) 
    (for/product ((l ll)(c p)) 
      (list-ref l c))))
```
The code can easily be modified for unequal number of rows and columns.
Testing: 
```
(f '[[ 1 -1 -1  1]
     [-1 -1 -1  1]
     [-1  1 -1  1]
     [ 1 -1 -1  1]])
(f '[[ 1 -1  1 -1 -1 -1 -1 -1]
 [-1 -1  1  1 -1  1  1 -1]
 [ 1 -1 -1 -1 -1  1  1  1]
 [-1 -1 -1  1 -1  1  1  1]
 [ 1 -1 -1  1  1  1  1 -1]
 [-1  1 -1  1 -1  1  1 -1]
 [ 1 -1  1 -1  1 -1  1 -1]
 [-1 -1  1 -1  1  1  1  1]])
```
Output: 
```
-4
192
```
As I mentioned above, it hangs on testing following: 
```
(f '[[1 -1 1 -1 -1 1 1 1 -1 -1 -1 -1 1 1 1 1 -1 1 1 -1]
 [1 -1 1 1 1 1 1 -1 1 -1 -1 1 1 1 -1 -1 1 1 1 -1]
 [-1 -1 1 1 1 -1 -1 -1 -1 1 -1 1 1 1 -1 -1 -1 1 -1 -1]
 [-1 -1 -1 1 1 -1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 1 -1]
 [-1 1 1 1 -1 1 1 1 -1 -1 -1 1 -1 1 -1 1 1 1 1 1]
 [1 -1 1 1 -1 -1 1 -1 1 1 1 1 -1 1 1 -1 1 -1 -1 -1]
 [1 -1 -1 1 -1 -1 -1 1 -1 1 1 1 1 -1 -1 -1 1 1 1 -1]
 [1 -1 -1 1 -1 1 1 -1 1 1 1 -1 1 -1 1 1 1 -1 1 1]
 [1 -1 -1 -1 -1 -1 1 1 1 -1 -1 -1 -1 -1 1 1 -1 1 1 -1]
 [-1 -1 1 -1 1 -1 1 1 -1 1 -1 1 1 1 1 1 1 -1 1 1]
 [-1 -1 -1 -1 -1 -1 -1 1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1]
 [1 1 -1 -1 -1 1 1 -1 -1 1 -1 1 1 -1 1 1 1 1 1 1]
 [-1 1 1 -1 -1 -1 -1 -1 1 1 1 1 -1 -1 -1 -1 -1 1 -1 1]
 [1 1 -1 -1 -1 1 -1 1 -1 -1 -1 -1 1 -1 1 1 -1 1 -1 1]
 [1 1 1 1 1 -1 -1 -1 1 1 1 -1 1 -1 1 1 1 -1 1 1]
 [1 -1 -1 1 -1 -1 -1 -1 1 -1 -1 1 1 -1 1 -1 -1 -1 -1 -1]
 [-1 1 1 1 -1 1 1 -1 -1 1 1 1 -1 -1 1 1 -1 -1 1 1]
 [1 1 -1 -1 1 1 -1 1 1 -1 1 1 1 -1 1 1 -1 1 -1 1]
 [1 1 1 -1 -1 -1 1 -1 -1 1 1 -1 -1 -1 1 -1 -1 -1 -1 1]
 [-1 1 1 1 -1 -1 -1 -1 -1 -1 -1 1 1 -1 1 1 -1 1 -1 -1]])
```
]