chapter8 relu2deriv

[leeivan1007]

run relu2deriv() maybe not is effect,or it just demo for this case?

i think it has not any assist when it run for backpropagation,

any idea or suggest, thx

[leeivan1007]

i modify the code, isolate the relu in forward and back propagation
the performance look like better

for i in range(len(images)):
    layer_0 = images[i:i+1]
    layer_1 = np.dot(layer_0,weights_0_1) # modify
    layer_1_relu = relu(layer_1) # modify
    layer_2 = np.dot(layer_1_relu,weights_1_2)

    error += np.sum((labels[i:i+1] - layer_2) ** 2)
    correct_cnt += int(np.argmax(layer_2) == \
                                    np.argmax(labels[i:i+1]))

    layer_2_delta = (labels[i:i+1] - layer_2)
    layer_1_delta = layer_2_delta.dot(weights_1_2.T)\
                                * relu2deriv(layer_1)
    weights_1_2 += alpha * layer_1_relu.T.dot(layer_2_delta) # modify
    weights_0_1 += alpha * layer_0.T.dot(layer_1_delta)

before

after

[naruto678]

the output of reu2deriv(layer_1) will be equal to relu2deriv(layer_1_relu) in your code because whereever layer_1 > 0 the layer_1_relu >0 hence they produce the same boolean output.
The objective of layer_1_delta multiplying with relu is that we are making those delta's 0 whose inputs did not contribute to the delta at layer_2,which in this case occurs when the outputs of layer_1 <0 .So it should be immaterial whether you use layer_1 or layer_1_relu as input to the relu2deriv function .

Check this out by just adding this line:
`
assert(relu2deriv(layer_1)==relu2deriv(layer_1_relu)))

`

[maksabuzyarov]

I think there is an error in the relu2deriv function in chapter 8

relu2deriv = lambda x: x>=0 # returns 1 for input > 0, return 0 otherwise
relu = lambda x:(x>=0) * x # returns x if x > 0, return 0 otherwise

relu2deriv(1) # True = 1 - ok
relu(1) # 1 - ok

relu(0) # 0 - weight does not affect the output
relu2deriv(0) # True = 1 - but used in backpropagation

relu2deriv(-1) # False = 0 - ok
relu(-1) # 0 - ok

returns 1 for x > 0, return 0 otherwise

But in the function: x>=0

The correct code is shown in chapter 6: Backpropagation in Code

relu2deriv = lambda x: x>0 # returns 1 for input > 0, return 0 otherwise
relu = lambda x:(x>0) * x # returns x if x > 0, return 0 otherwise

relu2deriv(1) # True = 1 - ok
relu(1) # 1 - ok

relu(0) # 0 - ok
relu2deriv(0) # False = 0 - ok

relu2deriv(-1) # False = 0 - ok
relu(-1) # 0 - ok