这种多层感知器反向传播实现有什么问题？

Question

我正在尝试编写一个简单的多层感知器来解决纯 numpy 中的 XOR 问题。具有 2 个神经元的单隐藏层，sigmoid 激活。

我一直在尝试直接翻译算法上的Wikipedia article，以及查看无数其他资源，但我无法让反向传播正常工作。我在 SO 上发帖是因为我认为这是滥用/误解某些 numpy 操作的问题，而不是算法本身，但可能不是。

这是完整的、可运行的程序：

import numpy as np

X = np.array([[0, 0], [0, 1], [1, 0], [1, 1]])    
y = np.array([0, 1, 1, 0])

def activation(x): # sigmoid
    return 1 / (1 + np.exp(-x))

def activation_d(x): # sigmoid derivative
    s = activation(x)
    return s * (1 - s)

def cost(y1, y2):
    return (np.linalg.norm(y1 - y2) ** 2) / 2

def mlp_train(X, y, n_h, learning_rate=1e-2, max_iterations=10000):
    n_i = 1 if len(X.shape) == 1 else len(X[0]) # input neurons count
    n_o = 1 if len(y.shape) == 1 else len(y[0]) # output neurons count
    h_layer = [np.random.rand(n_i) for i in range(n_h)]
    o_layer = [np.random.rand(n_h) for i in range(n_o)]

    for iteration in range(max_iterations):
        if (iteration % 2000 == 0): print('iteration', iteration)
        for j in range(len(X)):
            x = X[j]
            h = [activation(np.dot(x, n)) for n in h_layer]
            o = [activation(np.dot(h, n)) for n in o_layer]
            o = np.array(o)
            c = cost(o, np.array(y[j]))
            a_d = activation_d(x)
            o_grad = c * a_d
            o_delta = learning_rate * o_grad * h
            o_layer += o_delta
            h_grad = a_d * np.dot(o_delta, o_layer.T)
            h_delta = learning_rate * h_grad * x
            h_layer += h_delta
            if (iteration % 2000 == 0): print(x, '->', o, 'cost', c)

mlp_train(X, y, n_h=2)

成本并没有被最小化，所有的输出都收敛到 0：

iteration 0
[0 0] -> [0.70755] cost 0.25031025599858575
[0 1] -> [0.74962] cost 0.031344966778714914
[1 0] -> [0.7546] cost 0.030109871312169207
[1 1] -> [0.78708] cost 0.30974627646512554
iteration 2000
[0 0] -> [0.9568] cost 0.45773097730807827
[0 1] -> [0.97965] cost 0.00020711262741391742
[1 0] -> [0.98117] cost 0.00017728427582410072
[1 1] -> [0.99024] cost 0.4902891867523237
iteration 4000
[0 0] -> [0.99691] cost 0.49691698274069973
[0 1] -> [0.99932] cost 2.28713104303751e-07
[1 0] -> [0.99941] cost 1.7196664273121246e-07
[1 1] -> [0.99984] cost 0.4998383598602833
iteration 6000
[0 0] -> [0.9998] cost 0.49980132025306195
[0 1] -> [0.99998] cost 1.587647769449268e-10
[1 0] -> [0.99999] cost 1.0506374041454625e-10
[1 1] -> [1.] cost 0.49999803556631833
iteration 8000
[0 0] -> [0.99999] cost 0.49998771962960453
[0 1] -> [1.] cost 6.768208468242089e-14
[1 0] -> [1.] cost 3.953230074403547e-14
[1 1] -> [1.] cost 0.49999998306930477

Answer 1

这是一个重新实现但重写的优化，看来你没有掌握backpropagation。 Michael Nielsen 的Neural Networks and Deep Learning 详细讨论了反向传播。

import numpy as np

X = np.array([[0, 0], [0, 1], [1, 0], [1, 1]])
y = np.array([0, 1, 1, 0])


def activation(x):  # sigmoid
    return 1 / (1 + np.exp(-x))


def activation_d(x):  # sigmoid derivative
    s = activation(x)
    return s * (1 - s)


def cost(y1, y2):
    return (np.linalg.norm(y1 - y2) ** 2) / 2

def cost_d(y1, y2):
    """ Compute MSE loss, gradient

    Args:
        y1: prediction
        y2: target

    Returns:
        grads: same shape with y1 / y2
    """
    # d_sigmoid = sigmoid * (1 - sigmoid)
    return np.array(y1 - y2) * (y1 * (1 - y1))


def mlp_train(X, y, n_h, learning_rate=1e-2, max_iterations=10000):
    n_i = 1 if len(X.shape) == 1 else len(X[0])  # input neurons count
    n_o = 1 if len(y.shape) == 1 else len(y[0])  # output neurons count
    # add bias
    hw = np.random.randn(n_i, n_h)
    hb = np.zeros((1, n_h))
    ow = np.random.randn(n_h, n_o)
    ob = np.zeros((1, n_o))
    for iteration in range(max_iterations):
        if iteration % 2000 == 0:
            print('iteration', iteration)
        for x_i, y_i in zip(X, y):
            # forwardprop
            x_i = x_i[np.newaxis, :]
            hz = np.dot(x_i, hw) + hb   # (1, n_h)
            ho = activation(hz)

            oz = np.dot(ho, ow) + ob    # (1, n_o)
            oo = activation(oz)

            # cost
            c = cost(oo, y_i)

            # backwardprop
            grad_oz = cost_d(oo, y_i)   # (1, n_o)
            grad_ob = grad_oz
            grad_ow = np.dot(ho.T, grad_oz) # (n_h, n_o)
            # update
            ow -= learning_rate * grad_ow
            ob -= learning_rate * grad_ob

            grad_h = np.dot(grad_oz, ow.T)  # (1, n_h)
            grad_hz = grad_h * (ho * (1 - ho))
            grad_hb = grad_hz       # (1, n_h)
            grad_hw = np.dot(x_i.T, grad_hz)    # (n_i, n_h)
            # update
            hw -= learning_rate * grad_hw
            hb -= learning_rate * grad_hb

            if iteration % 2000 == 0:
                print(x_i, '->', oo, 'cost', c)


mlp_train(X, y, n_h=2, max_iterations=int(1e5))

输出：

...
[[1 0]] -> [[0.94364581]] cost 0.0015878973434031022
[[1 1]] -> [[0.04349045]] cost 0.0009457095065652823
iteration 96000
[[0 0]] -> [[0.04870092]] cost 0.0011858898326805463
[[0 1]] -> [[0.95518092]] cost 0.0010043748998508786
[[1 0]] -> [[0.94458789]] cost 0.001535251186790804
[[1 1]] -> [[0.04277648]] cost 0.0009149137866793687
iteration 98000
[[0 0]] -> [[0.04791496]] cost 0.0011479218121198723
[[0 1]] -> [[0.95588406]] cost 0.0009731082050768009
[[1 0]] -> [[0.94548701]] cost 0.0014858330062528543
[[1 1]] -> [[0.04209458]] cost 0.0008859767334115659