Chap2-Perceptron(感知机)

感知机原理：

\[y = \begin{cases}0 & (w_1x_1 + w_2x_2 \leq \theta) \\ 1 & (w_1x_1 + w_2x_2 > \theta)\end{cases}\]

其中\(w_1,w_2\)为权重，\(x_1,x_2\)为输入信号，\(y\)为输出信号，\(\theta\)为阈值.

传入信号总和大于阈值时称为激活，输出1；否则输出0.

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感知机可以表示AND, NAND, OR.

def AND(x1, x2):
    w1, w2, theta = 0.5, 0.5, 0.7
    tmp = w1*x1 + w2*x2
    if tmp <= theta:
        return 0
    else:
        return 1

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导入权重与偏置：

\(y = \begin{cases}0 & (w_1x_1 + w_2x_2 + b \leq 0) \\ 1 & (w_1x_1 + w_2x_2 + b > 0)\end{cases},\quad b = -\theta\)

import numpy as np
x = np.array([0,1])
w = np.array([0.5, 0.5])
b = -0.7

mul = x*w               # = np.array([0., 0.5])
prod = np.sum(mul)       # = 0.5
res = np.sum(w*x) + b    # = -0.2

于是AND的实现：

def AND(x1, x2):
    x = np.array([x1, x2])
    w = np.array([0.5, 0.5])
    b = -0.7
    return 0 if (np.sum(w*x) + b <= 0) else 1

NAND:

def NAND(x1, x2):
    x = np.array([x1, x2])
    w = np.array([-0.5, -0.5])
    b = 0.7
    return 0 if (np.sum(w*x) + b <= 0) else 1

OR:

def OR(x1, x2):
    x = np.array([x1, x2])
    w = np.array([0.5, 0.5])
    b = -0.2
    return 0 if (np.sum(w*x) + b <= 0) else 1

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局限性：异或门无法实现 ---> 多层感知机

def XOR(x1, x2):
    s1 = NAND(x1, x2)
    s2 = OR(x1, x2)
    y = AND(s1, s2)
    return y