Basic concepts
Image you had to multiply two small matrices in Python. You could just use the definition of a matrix product:
\(A, B \in \mathbb{R}^{n \times n}\): \(C = A \cdot B, C \in \mathbb{R}^{n \times n}\) where the components of C are definied by \(c_{i,j} = \sum_{k=1}^n a_{i,k} \cdot b_{k, j}\)
Note that this means:
\(
You might also have heard of Pythons overloaded multiplication:
print([0]*4)
print([[0]*4]*4)
print("abc"*4)
Output:
[0, 0, 0, 0]
[[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
abcabcabcabc
Question
What do you think does the following piece of Python-Code print?
#!/usr/bin/python
# -*- coding: utf-8 -*-
def standardMatrixProduct(A, B):
n = len(A)
C = [[0]*n]*n
for i in xrange(n):
for j in xrange(n):
for k in xrange(n):
C[i][j] += A[i][k] * B[k][j]
return C
A = [[1,2], [3,4]]
B = [[5,6], [7,8]]
print standardMatrixProduct(A, B)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answer
[[32, 32], [32, 32]]
Python creates only one list and makes pointers to it!
So this is one that works:
def standardMatrixProduct(A, B):
n = len(A)
C = [[0 for i in xrange(n)] for j in xrange(n)]
for i in xrange(n):
for j in xrange(n):
for k in xrange(n):
print C
C[i][j] += A[i][k] * B[k][j]
return C