连续RBM:仅对负值输入数据表现不佳？

Question

我尝试将这个连续RBM的python实现移植到Matlab：http://imonad.com/rbm/restricted-boltzmann-machine/

我生成了一个(有噪声的)圆圈形状的二维训练数据，并训练了具有2个可见和8个隐藏层的rbm。为了测试实现，我向RBM提供了均匀分布的随机数据，并绘制了重建的数据(与上面链接中使用的过程相同)。

现在令人困惑的部分:当训练数据在(0,1)x(0,1)的范围内时，我得到了非常令人满意的结果，但是当训练数据在(-0.5，-0.5)x(-0.5，-0.5)或(-1,0)x(-1,0)的范围内时，RBM仅在圆圈的最右上角重建数据。我不明白是什么导致了这种情况，是不是我的实现中有一个我看不到的bug？

一些图，蓝点是训练数据，红点是重建数据。

以下是我实现的RBM: Training：

maxepoch = 300;
ksteps = 10;
sigma = 0.2;        % cd standard deviation
learnW = 0.5;       % learning rate W
learnA  = 0.5;      % learning rate A
nVis = 2;           % number of visible units
nHid = 8;           % number of hidden units
nDat = size(dat, 1);% number of training data points
cost = 0.00001;     % cost
moment = 0.9;      % momentum
W = randn(nVis+1, nHid+1) / 10; % weights
dW = randn(nVis+1, nHid+1) / 1000; % change of weights
sVis = zeros(1, nVis+1);    % state of visible neurons
sVis(1, end) = 1.0;         % bias
sVis0 = zeros(1, nVis+1);   % initial state of visible neurons
sVis0(1, end) = 1.0;        % bias
sHid = zeros(1, nHid+1);    % state of hidden neurons
sHid(1, end) = 1.0;         % bias
aVis  = 0.1*ones(1, nVis+1);% A visible
aHid  = ones(1, nHid+1);    % A hidden
err = zeros(1, maxepoch);
e = zeros(1, maxepoch);
for epoch = 1:maxepoch
    wPos = zeros(nVis+1, nHid+1);
    wNeg = zeros(nVis+1, nHid+1);
    aPos = zeros(1, nHid+1);
    aNeg = zeros(1, nHid+1);
    for point = 1:nDat
        sVis(1:nVis) = dat(point, :);
        sVis0(1:nVis) = sVis(1:nVis); % initial sVis
        % positive phase
        activHid;
        wPos = wPos + sVis' * sHid;
        aPos = aPos + sHid .* sHid;
        % negative phase
        activVis;
        activHid;
        for k = 1:ksteps
            activVis;
            activHid;
        end
        tmp = sVis' * sHid;
        wNeg = wNeg + tmp;
        aNeg = aNeg + sHid .* sHid;
        delta = sVis0(1:nVis) - sVis(1:nVis);
        err(epoch) = err(epoch) + sum(delta .* delta);
        e(epoch) = e(epoch) - sum(sum(W' * tmp));
    end
    dW = dW*moment + learnW * ((wPos - wNeg) / numel(dat)) - cost * W;
    W = W + dW;
    aHid = aHid + learnA * (aPos - aNeg) / (numel(dat) * (aHid .* aHid));
    % error
    err(epoch) = err(epoch) / (nVis * numel(dat));
    e(epoch) = e(epoch) / numel(dat);
    disp(['epoch: ' num2str(epoch) ' err: ' num2str(err(epoch)) ...
    ' ksteps: ' num2str(ksteps)]);
end
save(['rbm_' filename '.mat'], 'W', 'err', 'aVis', 'aHid');

activHid.m：

sHid = (sVis * W) + randn(1, nHid+1);
sHid = sigFun(aHid .* sHid, datRange);
sHid(end) = 1.; % bias

activVis.m：

sVis = (W * sHid')' + randn(1, nVis+1);
sVis = sigFun(aVis .* sVis, datRange);
sVis(end) = 1.; % bias

sigFun.m：

function [sig] = sigFun(X, datRange)
    a = ones(size(X)) * datRange(1);
    b = ones(size(X)) * (datRange(2) - datRange(1));
    c = ones(size(X)) + exp(-X);
    sig = a + (b ./ c);
end

重建：

nSamples = 2000;
ksteps = 10;
nVis = 2;
nHid = 8;
sVis = zeros(1, nVis+1);    % state of visible neurons
sVis(1, end) = 1.0;         % bias
sHid = zeros(1, nHid+1);    % state of hidden neurons
sHid(1, end) = 1.0;         % bias
input = rand(nSamples, 2);
output = zeros(nSamples, 2);
for sample = 1:nSamples
    sVis(1:nVis) = input(sample, :);
    for k = 1:ksteps
        activHid;
        activVis;
    end
    output(sample, :) = sVis(1:nVis);
end

Answer 1

RBM最初被设计为只处理二进制数据。但也可以处理0到1之间的数据。这是算法的一部分。Further reading

Answer 2

因为x和y的输入都在0-1的范围内，这就是为什么它们会留在那个区域。将输入更改为input = (rand(nSamples, 2)*2) -1;将导致从-1 1范围内采样的输入，因此红点将在圆周围更加分散。