CODEWITHSUNDEEP
pythonnumpyneural-networkmlprelu
Luís Eduardo
Luís Eduardo

Reputation: 52

Simple ANN model converges with tanh(x) as the activation function, but it doesn't with leaky ReLu

I'm training a simple ANN model (MLP) using as the activation function tanh(x) and, after some interactions, it converges with error equal to 10^-5, here's my full code:

import numpy as np
import pandas as pd

# Base de dados a ser treinada
x = pd.DataFrame(
    [[1],
    [2],
    [3]],
    columns=['valores x'])

d = pd.DataFrame(
    [[5],
    [4],
    [3]],
    columns=['valores desejados'])

# Convertendo o dataframe em array e normalizando os valores desejados para ficar entre 0 e +1.
x = x.to_numpy()
d = d/(1.05*d.max())
d = d.to_numpy()


# Derivada de tanh(x) = sech²(x) = 1 - (tanh(x))²
def df(x):
    y = 1 - np.power(np.tanh(x), 2)
    return y

#def rede_mlp(n, x, d, net, k, precisao):

# Construindo a rede de duas camadas 
# net = número de neurônios na primeira camada
# n = taxa de aprendizagem
# precisao = precisão do erro quadrático médio
net=3
n = 0.1
precisao=0.00001
w1 = np.random.rand(net,len(x[0]))
w2 = np.random.rand(1,net)
E_M=1
epocas=0

while E_M>precisao:
    E_M=0
    errofinal=0
    for i in range(0,len(x)):

        # FOWARD
        i1 = np.matmul(w1, x[i].reshape(len(x[i]),1))
        y1 = np.tanh(i1)

        i2 = np.matmul(w2, y1)
        y2 = np.tanh(i2)

        # erro com o valor desejado
        erro = d[i].reshape(len(d[i]),1) - y2

        # BACKPROPAGATION
        delta_2 = erro*df(i2)
        w2 = w2 + n*(np.matmul(delta_2, y1.reshape(1, net)))

        delta_1 = (np.matmul(w2.T, delta_2))*df(i1)
        w1 = w1 + n*(np.matmul(delta_1, x[i].reshape(1, len(x[i]))))

        errofinal = errofinal + 0.5*erro**2

    E_M = errofinal/len(x)
    epocas+=1
    print(E_M)

After that, I tried to change the activation function to leaky ReLu, but it didn't converge. I have changed the learning rate n several times, but the error is still high. It's around 7.95, which is big for my data. Here's my try:

import numpy as np
import pandas as pd


# Base de dados a ser treinada
x = pd.DataFrame(
    [[1],
    [2],
    [3]],
    columns=['valores x'])

d = pd.DataFrame(
    [[5],
    [4],
    [3]],
    columns=['valores desejados'])

# Convertendo o dataframe em array e normalizando os valores desejados para ficar entre 0 e +1.
x = x.to_numpy()
d = d.to_numpy()


def df(x):
    x = np.array(x)
    x[x<=0] = 0.01
    x[x>0] = 1
    return x

def f(x):
    return(np.where(x > 0, x, x * 0.01))



#def rede_mlp(n, x, d, net, k, precisao):

# Construindo a rede de duas camadas 
# net = número de neurônios na primeira camada
# n = taxa de aprendizagem
# precisao = precisão do erro quadrático médio
net=3
n = 1e-4
precisao=0.0001
w1 = np.random.rand(net,len(x[0]))
w2 = np.random.rand(1,net)
E_M=20
epocas=0

while E_M>precisao:
    E_M=0
    errofinal=0
    for i in range(0,len(x)):

        # FOWARD
        i1 = np.matmul(w1, x[i].reshape(len(x[i]),1))
        y1 = f(i1)



        i2 = np.matmul(w2, y1)
        y2 = f(i2)


        # erro com o valor desejado
        erro = d[i].reshape(len(d[i]),1) - y2


        # BACKPROPAGATION
        delta_2 = erro*df(i2)
        w2 = w2 + n*(np.matmul(delta_2, y1.reshape(1, net)))


        delta_1 = (np.matmul(w2.T, delta_2))*df(i1)
        w1 = w1 + n*(np.matmul(delta_1, x[i].reshape(1, len(x[i]))))

        errofinal = errofinal + 0.5*erro**2

    #E_M = errofinal/len(x)
    E_M = errofinal
    epocas+=1
    print(E_M)

EDITED:

After some modifications, here's my ReLu code (but the error is still high ~7.77):

import numpy as np
import pandas as pd


# Base de dados a ser treinada
x = pd.DataFrame(
    [[1],
    [2],
    [3]],
    columns=['valores x'])

d = pd.DataFrame(
    [[5],
    [4],
    [3]],
    columns=['valores desejados'])

# Convertendo o dataframe em array e normalizando os valores desejados para ficar entre 0 e +1.
x = x.to_numpy()
d = d.to_numpy()


def df(x):
    return(np.where(x <= 0, 0.01, 1))

def f(x):
    return(np.where(x > 0, x, x * 0.01))


#def rede_mlp(n, x, d, net, k, precisao):

# Construindo a rede de duas camadas 
# net = número de neurônios na primeira camada
# n = taxa de aprendizagem
# precisao = precisão do erro quadrático médio
net=3
n = 1e-3
precisao=0.1
w1 = np.random.rand(net,len(x[0]))
w2 = np.random.rand(1,net)
E_M=20
epocas=0

while E_M>precisao:
    E_M=0
    errofinal=0
    for i in range(0,len(x)):

        # FOWARD
        i1 = np.matmul(w1, x[i].reshape(len(x[i]),1))
        y1 = f(i1)


        i2 = np.matmul(w2, y1)
        y2 = f(i2)


        # erro com o valor desejado
        erro = d[i].reshape(len(d[i]),1) - y2


        # BACKPROPAGATION
        delta_2 = erro*df(i2)
        delta_1 = (np.matmul(w2.T, delta_2))*df(i1)

        w2 = w2 + n*(np.matmul(delta_2, y1.reshape(1, net)))
        w1 = w1 + n*(np.matmul(delta_1, x[i].reshape(1, len(x[i]))))


        errofinal = errofinal + 0.5*erro**2

    #E_M = errofinal/len(x)
    E_M = errofinal
    epocas+=1
    print(E_M)

Upvotes: 0

Views: 211

Answers (1)

mcskinner
mcskinner

Reputation: 2748

You need to add a bias to the network.

The equation you are trying to model is y = 6 - x, which is trivial if you can use 6 as an intercept (bias), but I think actually impossible if you do not.

Many functions are much easier to represent once you add the bias, which is why including one is standard practice. This Q&A on the role of bias in NNs explains more thoroughly.

I modified your code to add the bias, as well as follow more typical naming conventions, and it converges for me.

net = 3
n = 1e-3
precisao = 0.0001 

w1 = np.random.rand(net, len(x[0])) 
bias1 = np.random.rand()

w2 = np.random.rand(1, net) 
bias2 = np.random.rand()

E_M = 20 
epocas = 0 

while E_M > precisao: 
    E_M = 0 
    errofinal = 0 
    for i in range(0,len(x)): 
        a0 = x[i].reshape(-1, 1) 
        targ = d[i].reshape(-1, 1) 

        z1 = np.matmul(w1, a0) + bias1
        a1 = f(z1) 

        z2 = np.matmul(w2, a1) + bias2
        a2 = f(z2) 

        erro = a2 - targ

        # BACKPROPAGATION 
        delta_2 = erro * df(z2) 
        delta_1 = np.matmul(w2.T, delta_2) * df(z1) 
        bias2 -= n * delta_2
        bias1 -= n * delta_1
        w2 -= n * np.matmul(delta_2, a1.T)
        w1 -= n * np.matmul(delta_1, a0.T)

        errofinal = errofinal + 0.5*erro**2 

    #E_M = errofinal/len(x) 
    E_M = errofinal 
    epocas += 1 
    if epocas % 1000 == 0:
        print(epocas, E_M)

I increased the learning rate so it would converge more quickly.

1000 [[0.14401507]]
2000 [[0.00028834]]

Earlier bug fix suggestion

You are setting the derivative always equal to 1.

def df(x):
    x = np.array(x)
    x[x<=0] = 0.01
    x[x>0] = 1
    return x

The line x[x<=0] = 0.01 sets all non-positive values to 1/100, a positive value. After that every value is positive, since the already-positive values go through unaffected and the negative-or-zero values just turned positive. So the next line x[x>0] = 1 sets all derivatives to 1.

Try this:

def df(x):
    return np.where(np.array(x) <= 0, 0.01, 1)

Upvotes: 2

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