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title:利用Python实现Strassen算法（2<=n,m）A,B为任意矩阵
strengths:
(1) A、B可以是任意矩阵不局限于阶数必须是2的n次方
(2)用Python语言编程实现，本程序使用的是Python3.5.2版本
@author: 向天笑
QQ:1500879657
执行结果
代码：
'''
Created on 2016年11月10日
@author: 向天笑
QQ:1500879657
title:利用Python实现Strassen算法（2<=n,m）A,B为任意矩阵
strengths:
(1) A、B可以是任意矩阵不局限于阶数必须是2的n次方
(2)用Python语言编程实现，本程序使用的是Python3.5.2版本
'''
import tkinter as tk
window=tk.Tk()
window.title("向天笑+软件工程+123456789")
window.geometry('300x200')
tk.Label(window, text='Strassen算法', font=('Arial', 20)).pack()
#函数
#分割矩阵
def Divide11(a,n):
k=int(n/2)
a11=[ [[0] for i in range(k)]for j in range(k)]
for i in range(0,k):
for j in range(0,k):
a11[i][j] = a[i][j]
return a11
def Divide12(a,n):
k=int(n/2)
a12=[ [[0] for i in range(k)]for i in range(k)]
for i in range(0,k):
for j in range(k,n):
a12[i][j-k] = a[i][j]
return a12
def Divide21(a,n):
k=int(n/2)
a21=[ [[0] for i in range(k)]for i in range(k)]
for i in range(k,n):
for j in range(0,k):
a21[i-k][j] = a[i][j]
return a21
def Divide22(a,n):
k=int(n/2)
a22=[ [[0] for i in range(k)]for i in range(k)]
for i in range(k,n):
for j in range(k,n):
a22[i-k][j-k] = a[i][j]
return a22
# 重组矩阵的方法
def Merge(a11,a12,a21,a22,n):
k=int(2*n)
a=[ [[0] for i in range(k)]for i in range(k)]
for i in range(0,n):
for j in range(0,n):
a[i][j] = a11[i][j]
a[i][j+n] = a12[i][j]
a[i+n][j] = a21[i][j]
a[i+n][j+n] = a22[i][j]
return a
#矩阵加法
def Plus(f, g, n):
h = [ [[0] for i in range(n)]for i in range(n)]
for i in range(0,n):
for j in range(0,n):
h[i][j]=f[i][j]+g[i][j]
return h
#矩阵减法
def Minus(f, g, n):
h = [ [[0] for i in range(n)]for i in range(n)]
for i in range(0,n):
for j in range(0,n):
h[i][j] = f[i][j] - g[i][j]
return h
#矩阵Strassen乘法方法
def Strassen(a,b,n):
k = n
if k == 2:
d = [[[0]for i in range(2)] for i in range(2)]
d[0][0] = a[0][0]*b[0][0] +a[0][1]*b[1][0]
d[0][1] = a[0][0]*b[0][1]+a[0][1]*b[1][1]
d[1][0] = a[1][0]*b[0][0]+a[1][1]*b[1][0]
d[1][1] = a[1][0]*b[0][1]+a[1][1]*b[1][1]
return d
else:
a11=Divide11(a,n)
a12=Divide12(a,n)
a21=Divide21(a,n)
a22=Divide22(a,n)
b11=Divide11(b,n)
b12=Divide12(b,n)
b21=Divide21(b,n)
b22=Divide22(b,n)
k = int(n / 2)
m1 = Strassen(a11, Minus(b12, b22, k), k)
m2 = Strassen(Plus(a11, a12, k), b22, k)
m3 = Strassen(Plus(a21, a22, k), b11, k)
m4 = Strassen(a22, Minus(b21, b11, k), k)
m5 = Strassen(Plus(a11, a22, k), Plus(b11, b22, k), k)
m6 = Strassen(Minus(a12, a22, k), Plus(b21, b22, k), k)
m7 = Strassen(Minus(a11, a21, k), Plus(b11, b12, k), k)
c11 = Plus(Minus(Plus(m5, m4, k), m2, k), m6, k)
c12 = Plus(m1, m2, k)
c21 = Plus(m3, m4, k)
c22 = Minus(Minus(Plus(m5, m1, k),m3,k),m7,k)
c = Merge(c11, c12, c21, c22, k)
return c
def Reshape(a,row,colum,n):
for i in range(0,row):
for j in range(colum,n):
a[i][j]=0
for i in range(row,n):
for j in range(0,n):
a[i][j]=0
return a
#主函数
def main1():
s=[]
print("请您输入矩阵的矩阵A的行数:")
s.append(int(input()))
print("请输入矩阵A的列数：")
s.append(int(input()))
print("请输入矩阵B的列数：")
s.append(int(input()))
print("您输入的矩阵A是"+str(s[0])+"*"+str(s[1])+"矩阵"+"矩阵B是"+str(s[1])+"*"+str(s[2])+"矩阵")
hangi=max(s)
hang=1
while(hang