感知机perceptron 与或非异或问题的编程实现

含义

感知机是二类分类的线性分类模型，其输入为实例的特征向量，输出为实例的类别，取+1和-1二值。
f(x)=sign(w*x+b)

与门

与门仅在两个输入均为1时输出1，其他时候则输出0:

# coding: utf-8
import numpy as np
import matplotlib.pyplot as plt


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

if __name__ == '__main__':
    for xs in [(0, 0), (1, 0), (0, 1), (1, 1)]:
        y = AND(xs[0], xs[1])
        print(str(xs) + " -> " + str(y))

输入结果：
(0, 0) -> 0
(1, 0) -> 0
(0, 1) -> 0
(1, 1) -> 1

或门

有1则输出为1，否则为0

# coding: utf-8
import numpy as np


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

if __name__ == '__main__':
    for xs in [(0, 0), (1, 0), (0, 1), (1, 1)]:
        y = NAND(xs[0], xs[1])
        print(str(xs) + " -> " + str(y))

输出结果：
(0, 0) -> 0
(1, 0) -> 1
(0, 1) -> 1
(1, 1) -> 1

非门

x1和x2同时为1时输出0，其他时候则输出1

# coding: utf-8
import numpy as np


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

if __name__ == '__main__':
    for xs in [(0, 0), (1, 0), (0, 1), (1, 1)]:
        y = OR(xs[0], xs[1])
        print(str(xs) + " -> " + str(y))

输出结果：
(0, 0) -> 1
(1, 0) -> 1
(0, 1) -> 1
(1, 1) -> 0

异或门

相同为0，相异为1 用前面介绍的感知机模型无法实现异或门，多层感知机实现如下：

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

if __name__ == '__main__':
    for xs in [(0, 0), (1, 0), (0, 1), (1, 1)]:
        y = XOR(xs[0], xs[1])
        print(str(xs) + " -> " + str(y))

输出结果
(0, 0) -> 0
(1, 0) -> 1
(0, 1) -> 1
(1, 1) -> 0