我正在尝试使用 FFT 加速神经模拟器的计算。
方程是:
(1) \sum(j=1 to N) (w(i - j) * s_NMDA[j])
其中 s_NMDA 是长度为 N 的向量,w 定义为:
(2) w(j) = tanh[1/(2 * sigma * p)] * exp(-abs(j) / (sigma * p)]
其中 sigma 和 p 是常数。
(有没有更好的方法在stackoverflow上渲染方程?)
必须对 N 个神经元进行计算。由于 (1) 仅取决于绝对距离 abs(i - j),因此应该可以使用 FFT(卷积定理)进行计算。
我尝试使用 FFTW 来实现这一点,但结果与预期结果不匹配。我以前从未使用过 FFTW,现在我不确定我的实现是否不正确,我对卷积定理的假设是否错误。
void f_I_NMDA_FFT(
const double **states, // states[i][6] == s_NMDA[i]
const unsigned int numNeurons)
{
fftw_complex *distances, *sNMDAs, *convolution;
fftw_complex *distances_f, *sNMDAs_f, *convolution_f;
fftw_plan p, pinv;
const double scale = 1./numNeurons;
distances = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
sNMDAs = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
convolution = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
distances_f = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
sNMDAs_f = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
convolution_f = (fftw_complex *)fftw_malloc(sizeof(fftw_complex) * numNeurons);
// fill input array for distances
for (unsigned int i = 0; i < numNeurons; ++i)
{
distances[i][0] = w(i);
distances[i][1] = 0;
}
// fill input array for sNMDAs
for (unsigned int i = 0; i < numNeurons; ++i)
{
sNMDAs[i][0] = states[i][6];
sNMDAs[i][1] = 0;
}
p = fftw_plan_dft_1d(numNeurons,
distances,
distances_f,
FFTW_FORWARD,
FFTW_ESTIMATE);
fftw_execute(p);
p = fftw_plan_dft_1d(numNeurons,
sNMDAs,
sNMDAs_f,
FFTW_FORWARD,
FFTW_ESTIMATE);
fftw_execute(p);
// convolution in frequency domain
for(unsigned int i = 0; i < numNeurons; ++i)
{
convolution_f[i][0] = (distances_f[i][0] * sNMDAs_f[i][0]
- distances_f[i][1] * sNMDAs_f[i][1]) * scale;
convolution_f[i][1] = (distances_f[i][0] * sNMDAs_f[i][1]
- distances_f[i][1] * sNMDAs_f[i][0]) * scale;
}
pinv = fftw_plan_dft_1d(numNeurons,
convolution_f,
convolution,
FFTW_FORWARD,
FFTW_ESTIMATE);
fftw_execute(pinv);
// compute and compare with expected result
for (unsigned int i = 0; i < numNeurons; ++i)
{
double expected = 0;
for (int j = 0; j < numNeurons; ++j)
{
expected += w(i - j) * states[j][6];
}
printf("i=%d, FFT: r%f, i%f : Expected: %f\n", i, convolution[i][0], convolution[i][1], expected);
}
fftw_destroy_plan(p);
fftw_destroy_plan(pinv);
fftw_free(distances), fftw_free(sNMDAs), fftw_free(convolution);
fftw_free(distances_f), fftw_free(sNMDAs_f), fftw_free(convolution_f);
这是 20 个神经元的示例输出:
i=0, FFT: r0.042309, i0.000000 : Expected: 0.041504
i=1, FFT: r0.042389, i0.000000 : Expected: 0.042639
i=2, FFT: r0.042466, i0.000000 : Expected: 0.043633
i=3, FFT: r0.042543, i0.000000 : Expected: 0.044487
i=4, FFT: r0.041940, i0.000000 : Expected: 0.045203
i=5, FFT: r0.041334, i0.000000 : Expected: 0.045963
i=6, FFT: r0.041405, i0.000000 : Expected: 0.046585
i=7, FFT: r0.041472, i0.000000 : Expected: 0.047070
i=8, FFT: r0.041537, i0.000000 : Expected: 0.047419
i=9, FFT: r0.041600, i0.000000 : Expected: 0.047631
i=10, FFT: r0.041660, i0.000000 : Expected: 0.047708
i=11, FFT: r0.041717, i0.000000 : Expected: 0.047649
i=12, FFT: r0.041773, i0.000000 : Expected: 0.047454
i=13, FFT: r0.041826, i0.000000 : Expected: 0.047123
i=14, FFT: r0.041877, i0.000000 : Expected: 0.046656
i=15, FFT: r0.041926, i0.000000 : Expected: 0.046052
i=16, FFT: r0.041294, i0.000000 : Expected: 0.045310
i=17, FFT: r0.042059, i0.000000 : Expected: 0.044430
i=18, FFT: r0.042144, i0.000000 : Expected: 0.043412
i=19, FFT: r0.042228, i0.000000 : Expected: 0.042253
结果似乎几乎是正确的,但误差随着神经元数量的增加而增加。此外,对于非常低或非常高的位置 (i),结果似乎更准确。这里发生了什么?
更新:正如 Oli Charlesworth 所建议的,我在 octave 中实现了算法,看看它是实现还是数学问题:
input = [0.186775; 0.186775; 0.186775; 0.186775; 0.186775; 0; 0.186775; 0.186775; 0.186775; 0.186775];
function ret = _w(i)
ret = tanh(1 / (2* 1 * 32)) * exp(-abs(i) / (1 * 32));
end
for i = linspace(1, 10, 10)
expected = 0;
for j = linspace(1, 10, 10)
expected += _w(i-j) * input(j);
end
expected
end
distances = _w(transpose(linspace(0, 9, 10)));
input_f = fft(input);
distances_f = fft(distances);
convolution_f = input_f .* distances_f;
convolution = ifft(convolution_f)
结果:
expected = 0.022959
expected = 0.023506
expected = 0.023893
expected = 0.024121
expected = 0.024190
expected = 0.024100
expected = 0.024034
expected = 0.023808
expected = 0.023424
expected = 0.022880
convolution =
0.022959
0.023036
0.023111
0.023183
0.023253
0.022537
0.022627
0.022714
0.022798
0.022880
结果非常相似。因此,我对卷积定理/FFT的理解一定有问题。