为什么维基百科的感知器正确地将XOR分开?

时间:2014-01-15 16:57:25

标签: python neural-network perceptron

我理解感知器只能在线性可分离的集合上正常工作,例如NAND,AND,OR函数的输出。我一直在阅读Wikipedia's entry on the perceptron,并开始使用它的代码。

XOR是单层感知器失败的情况,因为它不是线性可分离的集合。

#xor
print ("xor")
t_s           = [((1, 1, 1), 0), ((1, 0, 1), 1), ((1, 1, 0), 1), ((1, 1, 1), 0)] 


threshold     = 0.5
learning_rate = 0.1
w             = [0, 0, 0]

def dot_product(values, weights):
    return sum(value * weight for value, weight in zip(values, weights))

def train_perceptron(threshold, learning_rate, weights, training_set):
    while True:
        #print('-' * 60)
        error_count = 0

        for input_vector, desired_output in training_set:
            #print(weights)
            result = dot_product(input_vector, weights) > threshold
            error  = desired_output - result

            if error != 0:
                error_count += 1
                for index, value in enumerate(input_vector):
                    weights[index] += learning_rate * error * value

        if error_count == 0: #iterate till there's no error 
            break
    return training_set

t_s = train_perceptron(threshold, learning_rate, w, t_s)

t_s = [(a[1:], b) for a, b in t_s]

for a, b in t_s:
    print "input: " + str(a) + ", output: " + str(b)

The output for this Ideone run is correct for XOR。怎么会?

xor
input: (1, 1), output: 0
input: (0, 1), output: 1
input: (1, 0), output: 1
input: (1, 1), output: 0

3 个答案:

答案 0 :(得分:6)

您将t_s输入train_perceptron并返回,而不进行修改。然后输出它。当然这很有效......

t_s = train_perceptron(threshold, learning_rate, w, t_s)

这根本不会改变t_strain_perceptron不会修改training_set。但是返回它:return training_set

然后在这里输出:

t_s = [(a[1:], b) for a, b in t_s]

for a, b in t_s:
    print "input: " + str(a) + ", output: " + str(b)

答案 1 :(得分:1)

尝试更改训练集:

t_s = [((1, 1, 1), 0), ((1, 0, 1), 1), ((1, 1, 0), 1), ((0, 0, 0), 0)]

答案 2 :(得分:0)

如果我的记忆是正确的,以破坏感知器的非线性问题,你需要至少一个隐藏层,对该层中的神经元进行非线性激活。