[Question]
  [
This challenge is to write fast code that can perform a computationally difficult infinite sum.
**Input**
An `n` by `n` matrix `P` with integer entries that are smaller than `100` in absolute value. When testing I am happy to provide input to your code in any sensible format your code wants. The default will be one line per row of the matrix, space separated and provided on standard input.
`P` will be [positive definite](https://en.wikipedia.org/wiki/Positive-definite_matrix) which implies it will always be symmetric. Other than that you don't really need to know what positive definite means to answer the challenge. It does however mean that there will actually be an answer to the sum defined below.
You do however need to know what a [matrix-vector product](https://www.youtube.com/watch?v=Awcj447pYuk) is.
**Output**
Your code should compute the infinite sum:
> 
> [![enter image description here](https://i.stack.imgur.com/q4hqV.gif)](https://i.stack.imgur.com/q4hqV.gif)
> 
> 
> 
to within plus or minus 0.0001 of the correct answer. Here `Z` is the set of integers and so `Z^n` is all possible vectors with `n` integer elements and `e` is the [famous mathematical constant](https://en.wikipedia.org/wiki/E_(mathematical_constant)) that approximately equals 2.71828. Note that the value in the exponent is simply a number. See below for an explicit example.
**How does this relate to the Riemann Theta function?**
In the notation of this paper on [approximating the Riemann Theta function](http://page.math.tu-berlin.de/~bobenko/papers/2003_Bob_DHvHS.pdf) we are trying to compute [![enter image description here](https://i.stack.imgur.com/fegP6.png)](https://i.stack.imgur.com/fegP6.png). Our problem is a special case for at least two reasons.
* We set the initial parameter called `z` in the linked paper to 0.
* We create the matrix `P` in such a way that the minimum size of an eigenvalue is `1`. (See below for how the matrix is created.)
**Examples**
```
P = [[ 5.,  2.,  0.,  0.],
     [ 2.,  5.,  2., -2.],
     [ 0.,  2.,  5.,  0.],
     [ 0., -2.,  0.,  5.]]
Output: 1.07551411208
```
In more detail, let us see just one term in the sum for this P. Take for example just one term in the sum:
[![enter image description here](https://i.stack.imgur.com/pFOEs.gif)](https://i.stack.imgur.com/pFOEs.gif)
and `x^T P x = 30`. Notice that`e^(-30)` is about `10^(-14)` and so is unlikely to be important for getting the correct answer up to the given tolerance. Recall that the infinite sum will actually use every possible vector of length 4 where the elements are integers. I just picked one to give an explicit example.
```
P = [[ 5.,  2.,  2.,  2.],
     [ 2.,  5.,  4.,  4.],
     [ 2.,  4.,  5.,  4.],
     [ 2.,  4.,  4.,  5.]]
Output = 1.91841190706
P = [[ 6., -3.,  3., -3.,  3.],
     [-3.,  6., -5.,  5., -5.],
     [ 3., -5.,  6., -5.,  5.],
     [-3.,  5., -5.,  6., -5.],
     [ 3., -5.,  5., -5.,  6.]]
Output = 2.87091065342
P = [[6., -1., -3., 1., 3., -1., -3., 1., 3.],
     [-1., 6., -1., -5., 1., 5., -1., -5., 1.],
     [-3., -1., 6., 1., -5., -1., 5., 1., -5.],
     [1., -5., 1., 6., -1., -5., 1., 5., -1.],
     [3., 1., -5., -1., 6., 1., -5., -1., 5.],
     [-1., 5., -1., -5., 1., 6., -1., -5., 1.],
     [-3., -1., 5., 1., -5., -1., 6., 1., -5.],
     [1., -5., 1., 5., -1., -5., 1., 6., -1.],
     [3., 1., -5., -1., 5., 1., -5., -1., 6.]]
Output: 8.1443647932
P = [[ 7.,  2.,  0.,  0.,  6.,  2.,  0.,  0.,  6.],
     [ 2.,  7.,  0.,  0.,  2.,  6.,  0.,  0.,  2.],
     [ 0.,  0.,  7., -2.,  0.,  0.,  6., -2.,  0.],
     [ 0.,  0., -2.,  7.,  0.,  0., -2.,  6.,  0.],
     [ 6.,  2.,  0.,  0.,  7.,  2.,  0.,  0.,  6.],
     [ 2.,  6.,  0.,  0.,  2.,  7.,  0.,  0.,  2.],
     [ 0.,  0.,  6., -2.,  0.,  0.,  7., -2.,  0.],
     [ 0.,  0., -2.,  6.,  0.,  0., -2.,  7.,  0.],
     [ 6.,  2.,  0.,  0.,  6.,  2.,  0.,  0.,  7.]]
Output = 3.80639191181
```
**Score**
I will test your code on randomly chosen matrices P of increasing size. 
Your score is simply the largest `n` for which I get a correct answer in less than 30 seconds when averaged over 5 runs with randomly chosen matrices `P` of that size.
**What about a tie?**
If there is a tie, the winner will be the one whose code runs fastest averaged over 5 runs. In the event that those times are also equal, the winner is the first answer.
**How will the random input be created?**
1. Let M be a random m by n matrix with m<=n and entries which are -1 or 1. In Python/numpy `M = np.random.choice([0,1], size = (m,n))*2-1` . In practice I will set `m` to be about `n/2`.
2. Let P be the identity matrix + M^T M. In Python/numpy `P =np.identity(n)+np.dot(M.T,M)`. We are now guaranteed that `P` is positive definite and the entries are in a suitable range.
Note that this means that all eigenvalues of P are at least 1, making the problem potentially easier than the general problem of approximating the Riemann Theta function.
**Languages and libraries**
You can use any language or library you like. However, for the purposes of scoring I will run your code on my machine so please provide clear instructions for how to run it on Ubuntu.
**My Machine** The timings will be run on my machine. This is a standard Ubuntu install on an 8GB AMD FX-8350 Eight-Core Processor. This also means I need to be able to run your code.
---
**Leading answers**
* `n = 47` in **C++** by Ton Hospel
* `n = 8` in **Python** by Maltysen
  
[Answer]
# C++
No more naive approach. Only evaluate inside the ellipsoid.
Uses the armadillo, ntl, gsl and pthread libraries. Install using
```
apt-get install libarmadillo-dev libntl-dev libgsl-dev
```
Compile the program using something like:
```
g++ -Wall -std=c++11 -O3 -fno-math-errno -funsafe-math-optimizations -ffast-math -fno-signed-zeros -fno-trapping-math -fomit-frame-pointer -march=native -s infinity.cpp -larmadillo -lntl -lgsl -lpthread -o infinity
```
On some systems you may need to add `-lgslcblas` after `-lgsl`.
Run with the size of the matrix followed by the elements on STDIN:
```
./infinity < matrix.txt
```
`matrix.txt`:
```
4
5  2  0  0
2  5  2 -2
0  2  5  0
0 -2  0  5
```
Or to try a precision of 1e-5:
```
./infinity -p 1e-5 < matrix.txt
```
`infinity.cpp`:
```
// Based on http://arxiv.org/abs/nlin/0206009
#include <iostream>
#include <vector>
#include <stdexcept>
#include <cstdlib>
#include <cmath>
#include <string>
#include <thread>
#include <future>
#include <chrono>
using namespace std;
#include <getopt.h>
#include <armadillo>
using namespace arma;
#include <NTL/mat_ZZ.h>
#include <NTL/LLL.h>
using namespace NTL;
#include <gsl/gsl_sf_gamma.h>
#include <gsl/gsl_errno.h>
#include <gsl/gsl_roots.h>
double const EPSILON = 1e-4;       // default precision
double const GROW    = 2;          // By how much we grow the ellipsoid volume
double const UPSCALE = 1e9;        // lattice reduction, upscale real to integer
double const THREAD_SEC = 0.1;     // Use threads if need more time than this
double const RADIUS_MAX = 1e6;     // Maximum radius used in root finding
double const RADIUS_INTERVAL = 1e-6; // precision of target radius
int const ITER_MAX = 1000;         // Maximum iterations in root finding
unsigned long POINTS_MIN = 1000;   // Minimum points before getting fancy
struct Result {
    Result& operator+=(Result const& add) {
        sum     += add.sum;
        elapsed += add.elapsed;
        points  += add.points;
        return *this;
    }
    friend Result operator-(Result const& left, Result const& right) {
        return Result{left.sum - right.sum,
                left.elapsed - right.elapsed,
                left.points - right.points};
    }
    double sum, elapsed;
    unsigned long points;
};
struct Params {
    double half_rho, half_N, epsilon;
};
double fill_factor_error(double r, void *void_params) {
    auto params = static_cast<Params*>(void_params);
    r -= params->half_rho;
    return gsl_sf_gamma_inc(params->half_N, r*r) - params->epsilon;
}
// Calculate radius needed for target precision
double radius(int N, double rho, double lat_det, double epsilon) {
    Params params;
    params.half_rho = rho / 2.;
    params.half_N   = N   / 2.;
    params.epsilon = epsilon*lat_det*gsl_sf_gamma(params.half_N)/pow(M_PI, params.half_N);
    // Calculate minimum allowed radius
    auto r = sqrt(params.half_N)+params.half_rho;
    auto val = fill_factor_error(r, &params);
    cout << "Minimum R=" << r << " -> " << val << endl;
    if (val > 0) {
        // The minimum radius is not good enough. Work out a better one by
        // finding the root of a tricky function
        auto low  = r;
        auto high = RADIUS_MAX * 2 * params.half_rho;
        auto val = fill_factor_error(high, &params);
        if (val >= 0)
            throw(logic_error("huge RADIUS_MAX is still not big enough"));
        gsl_function F;
        F.function = fill_factor_error;
        F.params   = &params;
        auto T = gsl_root_fsolver_brent;
        auto s = gsl_root_fsolver_alloc (T);
        gsl_root_fsolver_set (s, &F, low, high);
        int status = GSL_CONTINUE;
        for (auto iter=1; status == GSL_CONTINUE && iter <= ITER_MAX; ++iter) {
            gsl_root_fsolver_iterate (s);
            low  = gsl_root_fsolver_x_lower (s);
            high = gsl_root_fsolver_x_upper (s);
            status = gsl_root_test_interval(low, high, 0, RADIUS_INTERVAL  * 2 * params.half_rho);
        }
        r = gsl_root_fsolver_root(s);
        gsl_root_fsolver_free(s);
        if (status == GSL_CONTINUE)
            throw(logic_error("Search for R did not converge"));
    }
    return r;
}
// Recursively walk down the ellipsoids in each dimension
void ellipsoid(int d, mat const& A, double const* InvD, mat& Accu,
               Result& result, double r2) {
    auto r = sqrt(r2);
    auto offset = Accu(d, d);
    // InvD[d] = 1/ A(d, d)
    auto from = ceil((-r-offset) * InvD[d]);
    auto to   = floor((r-offset) * InvD[d]);
    for (auto v = from; v <= to; ++v) {
        auto value  = v * A(d, d)+offset;
        auto residu = r2 - value*value;
        if (d == 0) {
            result.sum += exp(residu);
            ++result.points;
        } else {
            for (auto i=0; i<d; ++i) Accu(d-1, i) = Accu(d, i) + v * A(d, i);
            ellipsoid(d-1, A, InvD, Accu, result, residu);
        }
    }
}
// Specialised version of ellipsoid() that will only process points an octant
void ellipsoid(int d, mat const& A, double const* InvD, mat& Accu,
               Result& result, double r2, unsigned int octant) {
    auto r = sqrt(r2);
    auto offset = Accu(d, d);
    // InvD[d] = 1/ A(d, d)
    long from = ceil((-r-offset) * InvD[d]);
    long to   = floor((r-offset) * InvD[d]);
    auto points = to-from+1;
    auto base = from + points/2;
    if (points & 1) {
        auto value = base * A(d, d) + offset;
        auto residu = r2 - value * value;
        if (d == 0) {
            if ((octant & (octant - 1)) == 0) {
                result.sum += exp(residu);
                ++result.points;
            }
        } else {
            for (auto i=0; i<d; ++i) Accu(d-1, i) = Accu(d, i) + base * A(d, i);
            ellipsoid(d-1, A, InvD, Accu, result, residu, octant);
        }
        ++base;
    }
    if ((octant & 1) == 0) {
        to = from + points / 2 - 1;
        base = from;
    }
    octant /= 2;
    for (auto v = base; v <= to; ++v) {
        auto value = v * A(d,d)+offset;
        auto residu = r2 - value*value;
        if (d == 0) {
            if ((octant & (octant - 1)) == 0) {
                result.sum += exp(residu);
                ++result.points;
            }
        } else {
            for (auto i=0; i<d; ++i) Accu(d-1, i) = Accu(d, i) + v * A(d, i);
            if (octant == 1)
                ellipsoid(d-1, A, InvD, Accu, result, residu);
            else
                ellipsoid(d-1, A, InvD, Accu, result, residu, octant);
        }
    }
}
// Prepare call to ellipsoid()
Result sym_ellipsoid(int N, mat const& A, const vector<double>& InvD, double r,
                     unsigned int octant = 1) {
    auto start = chrono::steady_clock::now();
    auto r2 = r*r;
    mat Accu(N, N);
    Accu.row(N-1).zeros();
    Result result{0, 0, 0};
    // 2*octant+1 forces the points into the upper half plane, skipping 0
    // This way we use the lattice symmetry and calculate only half the points
    ellipsoid(N-1, A, &InvD[0], Accu, result, r2, 2*octant+1);
    // Compensate for the extra factor exp(r*r) we always add in ellipsoid()
    result.sum /= exp(r2);
    auto end = chrono::steady_clock::now();
    result.elapsed = chrono::duration<double>{end-start}.count();
    return result;
}
// Prepare multithreaded use of sym_ellipsoid(). Each thread gets 1 octant
Result sym_ellipsoid_t(int N, mat const& A, const vector<double>& InvD, double r, unsigned int nr_threads) {
    nr_threads = pow(2, ceil(log2(nr_threads)));
    vector<future<Result>> results;
    for (auto i=nr_threads+1; i<2*nr_threads; ++i)
        results.emplace_back(async(launch::async, sym_ellipsoid, N, ref(A), ref(InvD), r, i));
    auto result = sym_ellipsoid(N, A, InvD, r, nr_threads);
    for (auto i=0U; i<nr_threads-1; ++i) result += results[i].get();
    return result;
}
int main(int argc, char* const* argv) {
    cin.exceptions(ios::failbit | ios::badbit);
    cout.precision(12);
    double epsilon    = EPSILON; // Target absolute error
    bool inv_modular  = true;    // Use modular transform to get the best matrix
    bool lat_reduce   = true;    // Use lattice reduction to align the ellipsoid
    bool conservative = false;   // Use provable error bound instead of a guess
    bool eigen_values = false;   // Show eigenvalues
    int  threads_max  = thread::hardware_concurrency();
    int option_char;
    while ((option_char = getopt(argc, argv, "p:n:MRce")) != EOF)
        switch (option_char) {
            case 'p': epsilon      = atof(optarg); break;
            case 'n': threads_max  = atoi(optarg); break;
            case 'M': inv_modular  = false;        break;
            case 'R': lat_reduce   = false;        break;
            case 'c': conservative = true;         break;
            case 'e': eigen_values = true;         break;
            default:
              cerr << "usage: " << argv[0] << " [-p epsilon] [-n threads] [-M] [-R] [-e] [-c]" << endl;
              exit(EXIT_FAILURE);
        }
    if (optind < argc) {
        cerr << "Unexpected argument" << endl;
        exit(EXIT_FAILURE);
    }
    if (threads_max < 1) threads_max = 1;
    threads_max = pow(2, ceil(log2(threads_max)));
    cout << "Using up to " << threads_max << " threads" << endl;
    int N;
    cin >> N;
    mat P(N, N);
    for (auto& v: P) cin >> v;
    if (eigen_values) {
        vec eigval = eig_sym(P);
        cout << "Eigenvalues:\n" << eigval << endl;
    }
    // Decompose P = A * A.t()
    mat A = chol(P, "lower");
    // Calculate lattice determinant
    double lat_det = 1;
    for (auto i=0; i<N; ++i) {
        if (A(i,i) <= 0) throw(logic_error("Diagonal not Positive"));
        lat_det *= A(i,i);
    }
    cout << "Lattice determinant=" << lat_det << endl;
    auto factor = lat_det / pow(M_PI, N/2.0);
    if (inv_modular && factor < 1) {
        epsilon *= factor;
        cout << "Lattice determinant is small. Using inverse instead. Factor=" << factor << endl;
        P = M_PI * M_PI * inv(P);
        A = chol(P, "lower");
        // We could simple calculate the new lat_det as pow(M_PI,N)/lat_det
        lat_det = 1;
        for (auto i=0; i<N; ++i) {
            if (A(i,i) <= 0) throw(logic_error("Diagonal not Positive"));
            lat_det *= A(i,i);
        }
        cout << "New lattice determinant=" << lat_det << endl;
    } else
        factor = 1;
    // Prepare for lattice reduction.
    // Since the library works on integer lattices we will scale up our matrix
    double min = INFINITY;
    for (auto i=0; i<N; ++i) {
        for (auto j=0; j<N;++j)
            if (A(i,j) != 0 && abs(A(i,j) < min)) min = abs(A(i,j));
    }
    auto upscale = UPSCALE/min;
    mat_ZZ a;
    a.SetDims(N,N);
    for (auto i=0; i<N; ++i)
        for (auto j=0; j<N;++j) a[i][j] = to_ZZ(A(i,j)*upscale);
    // Finally do the actual lattice reduction
    mat_ZZ u;
    auto rank = G_BKZ_FP(a, u);
    if (rank != N) throw(logic_error("Matrix is singular"));
    mat U(N,N);
    for (auto i=0; i<N;++i)
        for (auto j=0; j<N;++j) U(i,j) = to_double(u[i][j]);
    // There should now be a short lattice vector at row 0
    ZZ sum = to_ZZ(0);
    for (auto j=0; j<N;++j) sum += a[0][j]*a[0][j];
    auto rho = sqrt(to_double(sum))/upscale;
    cout << "Rho=" << rho << " (integer square " <<
        rho*rho << " ~ " <<
        static_cast<int>(rho*rho+0.5) << ")" << endl;
    // Lattice reduction doesn't gain us anything conceptually.
    // The same number of points is evaluated for the same exponential values
    // However working through the ellipsoid dimensions from large lattice
    // base vectors to small makes ellipsoid() a *lot* faster
    if (lat_reduce) {
        mat B = U * A;
        P = B * B.t();
        A = chol(P, "lower");
        if (eigen_values) {
            vec eigval = eig_sym(P);
            cout << "New eigenvalues:\n" << eigval << endl;
        }
    }
    vector<double> InvD(N);;
    for (auto i=0; i<N; ++i) InvD[i] = 1 / A(i, i);
    // Calculate radius needed for target precision
    auto r = radius(N, rho, lat_det, epsilon);
    cout << "Safe R=" << r << endl;
    auto nr_threads = threads_max;
    Result result;
    if (conservative) {
        // Walk all points inside the ellipsoid with transformed radius r
        result = sym_ellipsoid_t(N, A, InvD, r, nr_threads);
    } else {
        // First grow the radius until we saw POINTS_MIN points or reach the
        // target radius
        double i = floor(N * log2(r/rho) / log2(GROW));
        if (i < 0) i = 0;
        auto R = r * pow(GROW, -i/N);
        cout << "Initial R=" << R << endl;
        result = sym_ellipsoid_t(N, A, InvD, R, nr_threads);
        nr_threads = result.elapsed < THREAD_SEC ? 1 : threads_max;
        auto max_new_points = result.points;
        while (--i >= 0 && result.points < POINTS_MIN) {
            R = r * pow(GROW, -i/N);
            auto change = result;
            result = sym_ellipsoid_t(N, A, InvD, R, nr_threads);
            nr_threads = result.elapsed < THREAD_SEC ? 1 : threads_max;
            change = result - change;
            if (change.points > max_new_points) max_new_points = change.points;
        }
        // Now we have enough points that it's worth bothering to use threads
        while (--i >= 0) {
            R = r * pow(GROW, -i/N);
            auto change = result;
            result = sym_ellipsoid_t(N, A, InvD, R, nr_threads);
            nr_threads = result.elapsed < THREAD_SEC ? 1 : threads_max;
            change = result - change;
            // This is probably too crude and might misestimate the error
            // I've never seen it fail though
            if (change.points > max_new_points) {
                max_new_points = change.points;
                if (change.sum < epsilon/2) break;
            }
        }
        cout << "Final R=" << R << endl;
    }
    // We calculated half the points and skipped 0.
    result.sum = 2*result.sum+1;
    // Modular transform factor
    result.sum /= factor;
    // Report result
    cout <<
        "Evaluated " << result.points << " points\n" <<
        "Sum = " << result.sum << endl;
}
```
[Answer]
# Python 3
**12 secs n=8 on my computer, ubuntu 4 core.**
Really naive, have no clue what I'm doing.
```
from itertools import product
from math import e
P = [[ 6., -3.,  3., -3.,  3.],
     [-3.,  6., -5.,  5., -5.],
     [ 3., -5.,  6., -5.,  5.],
     [-3.,  5., -5.,  6., -5.],
     [ 3., -5.,  5., -5.,  6.]]
N = 2
  
n = [1]
while e** -n[-1] > 0.0001:
    n = []
    for x in product(list(range(-N, N+1)), repeat = len(P)):
        n.append(sum(k[0] * k[1] for k in zip([sum(j[0] * j[1] for j in zip(i, x)) for i in P], x)))
    N += 1
print(sum(e** -i for i in n))
```
This will keep on increasing the range of `Z` that it uses until it gets a good enough answer. I wrote my own matrix multiplication, prolly should use numpy.
[Answer]
# Python, numpy
**one example program using `numpy` is as follows**.
Really naive, have no clue what I'm doing.
```
import numpy as np
from itertools import product
from math import e
from time import time
n=9
is_randomized=False
if is_randomized:
    m = n // 2
    M = np.random.choice([0, 1], size=(m, n)) * 2 - 1
    P = np.identity(n) + np.dot(M.T, M)
    print(P)
else:
    if n==4:
        P = [[5., 2., 2., 2.],
             [2., 5., 4., 4.],
             [2., 4., 5., 4.],
             [2., 4., 4., 5.]]
        P = np.array(P)
        print(P)
    elif n==5:
        P = np.array([[6., -3., 3., -3., 3.],
                      [-3., 6., -5., 5., -5.],
                      [3., -5., 6., -5., 5.],
                      [-3., 5., -5., 6., -5.],
                      [3., -5., 5., -5., 6.]])
        P = np.array(P)
        print(P)
    elif n==9:
        P = [[7., 2., 0., 0., 6., 2., 0., 0., 6.],
             [2., 7., 0., 0., 2., 6., 0., 0., 2.],
             [0., 0., 7., -2., 0., 0., 6., -2., 0.],
             [0., 0., -2., 7., 0., 0., -2., 6., 0.],
             [6., 2., 0., 0., 7., 2., 0., 0., 6.],
             [2., 6., 0., 0., 2., 7., 0., 0., 2.],
             [0., 0., 6., -2., 0., 0., 7., -2., 0.],
             [0., 0., -2., 6., 0., 0., -2., 7., 0.],
             [6., 2., 0., 0., 6., 2., 0., 0., 7.]]
        P = np.array(P)
        print(P)
# estimate how much N should be chosen?  based on matrix P or size n ?
if n==9:
    N = 2
elif n==5:
    N = 4
elif n==4:
    N = 5
else:
    N = 3
n_list = []
start_time=time()
for x in product(range(-N, N+1), repeat=len(P)):
    x = np.array(x)
    summand = np.dot(np.dot(x, P), x.T)
    n_list.append(summand)
result = sum(e**-i for i in n_list)
print(result)
print(f"duration: {time()-start_time}")
```
8 secs, n=9 on my computer.
]