难以实施Strassen';Python中的s算法
我不明白如何递归调用我的代码。以下是我目前的代码:难以实施Strassen';Python中的s算法,python,algorithm,recursion,runtime-error,strassen,Python,Algorithm,Recursion,Runtime Error,Strassen,我不明白如何递归调用我的代码。以下是我目前的代码: import numpy B = [[5,5,5,5,5,5,5,5],[6,6,6,6,6,6,6,6],[7,7,7,7,7,7,7,7],[8,8,8,8,8,8,8,8], [9,9,9,9,9,9,9,9], [10,10,10,10,10,10,10,10],[11,11,11,11,11,11,11,11], [12,12,12,12,12,12,12,12]] A = [[5,5,5,5,5,5,5,5],[6
import numpy
B = [[5,5,5,5,5,5,5,5],[6,6,6,6,6,6,6,6],[7,7,7,7,7,7,7,7],[8,8,8,8,8,8,8,8],
[9,9,9,9,9,9,9,9], [10,10,10,10,10,10,10,10],[11,11,11,11,11,11,11,11], [12,12,12,12,12,12,12,12]]
A = [[5,5,5,5,5,5,5,5],[6,6,6,6,6,6,6,6],[7,7,7,7,7,7,7,7],[8,8,8,8,8,8,8,8],
[1,1,1,1,1,1,1,1], [2,2,2,2,2,2,2,2],[3,3,3,3,3,3,3,3],[4,4,4,4,4,4,4,4]]
def main():
strassen(A,B)
def strassen(A, B):
A = numpy.asarray(A)
B = numpy.asarray(B)
lengthA = len(A)
lengthB = len(B)
if lengthA == 2:
print "will calculate"
else:
a, b = strassen(A[:lengthA//2, :lengthA//2], B[:lengthB//2, :lengthB//2])
lengthA = lengthA//2
lengthB = lengthB//2
print a
print b
return a, b
我正试图将a
减少到[[5,5],[6,6]]
和b
减少到[[5,5],[6,6]]
,但我得到了一个错误:
a, b = strassen(A[:lengthA//2, :lengthA//2], B[:lengthB//2, :lengthB//2])
TypeError: 'NoneType' object is not iterable.
a
和b
是在a和b的第二次整矩阵除法之后形成的第一个2x2矩阵。请有人给我解释一下。谢谢在递归终止条件中没有返回值。当我运行代码时,它会在给出错误之前打印“将计算”。之后会发生错误,因为最后一次调用(当lengthA==2
时)没有来自strassen函数的返回值