codewars（python）练习笔记二十一：Matrix Determinant
题目
Write a function that accepts a square matrix (n x n 2D array) and returns the determinant of the matrix.
How to take the determinant of a matrix -- it is simplest to start with the smallest cases: A 1x1 matrix |a| has determinant a. A 2x2 matrix [[a, b], [c, d]] or
|a b|
|c d|
has determinant ad - bc.
The determinant of an n x n sized matrix is calculated by reducing the problem to the calculation of the determinants of n n-1 x n-1 matrices. For the 3x3 case, [[a, b, c], [d, e, f], [g, h, i]] or
|a b c|
|d e f|
|g h i|
the determinant is: a * det(a_minor) - b * det(b_minor) + c * det(c_minor) where det(a_minor) refers to taking the determinant of the 2x2 matrix created by crossing out the row and column in which the element a occurs, or
|e f|
|h i|
Note the alternation of signs.
The determinant of larger matrices are calculated analogously, e.g. if M is a 4x4 matrix with first row [a, b, c, d], det(M) = a * det(a_minor) - b * det(b_minor) + c * det(c_minor) - d * det(d_minor)
Sample Tests:
m1 = [ [1, 3], [2,5]]
m2 = [ [2,5,3], [1,-2,-1], [1, 3, 4]]
Test.assert_equals(determinant([[1]]), 1, "Determinant of a 1 x 1 matrix yields the value of the one element")
Test.assert_equals(determinant(m1), -1, "Should return 1 * 5 - 3 * 2, i.e., -1 ")
Test.expect(determinant(m2) == -20)
题目大意：
求出给定矩阵的行列式。
写出一个determinant(A)函数。输入项A为n行n列的矩阵|A|(n>=1)。
若A是一个1X1的矩阵，则determinant(A) 返回 A[0][0];
若A是一个2X2的矩阵,如：
|a b|
|c d|
则determinant(A) 返回 ad - bc;
若A是一个nXn的矩阵,如：
|a b c|
|d e f|
|g h i|
则determinant(A)返回 a * det(a_minor) - b * det(b_minor) + c * det(c_minor)，(注意加减符号的交替)
det(a_minor)是指通过划掉元素a所在的行和列而创建的2x2矩阵的矩阵，如
|e f|
|h i|
若A是更大的矩阵比如5行5列，则determinant(A)返回 a * det(a_minor) - b * det(b_minor) + c * det(c_minor) - d * det(d_minor) + e * det(e_minor)。
依次类推。
我的解法：
#!/usr/bin/python
def determinant(matrix):
if len(matrix) < 2:
return matrix[0][0]
while len(matrix) >= 2:
return sum([((y%2)*2-1)*(matrix[0][y])*determinant([[matrix[i][j] for j in range(0,len(matrix[i])) if j != y] for i in range(1,len(matrix))]) for y in range(0,len(matrix[0]))])
我的思路：
先是常规思路，递归求值,将找出matrix[x][y] 的 minor 提出为函数 minor(matrix,x,y)，以优化算法可读性。
下面是优化前的第一版能跑通的函数：
#!/usr/bin/python
def minor(matrix,x,y):
arr = []
for i in range(1,len(matrix)):
if i != x:
arr_y = []
for j in range(0,len(matrix[i])):
if j != y:arr_y.append(matrix[i][j])
arr.append(arr_y)
return arr
def determinant(matrix):
if len(matrix) < 2:
return matrix[0][0]
while len(matrix) >= 2:
if len(matrix) == 2:
return matrix[0][0]*matrix[1][1] - matrix[0][1]*matrix[1][0]
else:
temp = 0
for y in range(0,len(matrix[0])):
temp += (1 if y%2 == 0 else -1)*(matrix[0][y])*determinant(minor(matrix,0,y))
return temp
print determinant([ [2,5,3], [1,-2,-1], [1, 3, 4]])
1.优化函数 minor(matrix,x,y)
因为函数 minor(matrix,x,y)三个参数中，x一直为0，可以去掉
#!/usr/bin/python
def minor(matrix,y):
arr = []
for i in range(1,len(matrix)):
arr_y = []
for j in range(0,len(matrix[i])):
if j != y:arr_y.append(matrix[i][j])
arr.append(arr_y)
return arr
def determinant(matrix):
if len(matrix) < 2:
return matrix[0][0]
while len(matrix) >= 2:
if len(matrix) == 2:
return matrix[0][0]*matrix[1][1] - matrix[0][1]*matrix[1][0]
else:
temp = 0
for y in range(0,len(matrix[0])):
temp += (1 if y%2 == 0 else -1)*(matrix[0][y])*determinant(minor(matrix,y))
return temp
print determinant([ [2,5,3], [1,-2,-1], [1, 3, 4]])
2.继续优化minor()函数：
通过嵌套，将函数缩短为一行
#!/usr/bin/python
def minor(matrix,y):
return [[matrix[i][j] for j in range(0,len(matrix[i])) if j != y] for i in range(1,len(matrix))]
def determinant(matrix):
if len(matrix) < 2:
return matrix[0][0]
while len(matrix) >= 2:
if len(matrix) == 2:
return matrix[0][0]*matrix[1][1] - matrix[0][1]*matrix[1][0]
else:
temp = 0
for y in range(0,len(matrix[0])):
temp += (1 if y%2 == 0 else -1)*(matrix[0][y])*determinant(minor(matrix,y))
return temp
print determinant([ [2,5,3], [1,-2,-1], [1, 3, 4]])
3.优化递归的语法：
#!/usr/bin/python
def minor(matrix,y):
return [[matrix[i][j] for j in range(0,len(matrix[i])) if j != y] for i in range(1,len(matrix))]
def determinant(matrix):
if len(matrix) < 2:
return matrix[0][0]
while len(matrix) > 2:
if len(matrix) == 2:
return matrix[0][0]*matrix[1][1] - matrix[0][1]*matrix[1][0]
else:
return sum([(1 if y%2 == 0 else -1)*(matrix[0][y])*determinant(minor(matrix,y)) for y in range(0,len(matrix[0]))])
print determinant([ [2,5,3], [1,-2,-1], [1, 3, 4]])
4.调整判断矩阵的行数的顺序：
#!/usr/bin/python
def minor(matrix,y):
return [[matrix[i][j] for j in range(0,len(matrix[i])) if j != y] for i in range(1,len(matrix))]
def determinant(matrix):
if len(matrix) < 2:
return matrix[0][0]
elif len(matrix) == 2:
return matrix[0][0]*matrix[1][1] - matrix[0][1]*matrix[1][0]
while len(matrix) > 2:
minor(matrix,y) = [[matrix[i][j] for j in range(0,len(matrix[i])) if j != y] for i in range(1,len(matrix))]
return sum([(1 if y%2 == 0 else -1)*(matrix[0][y])*determinant(minor(matrix,y)) for y in range(0,len(matrix[0]))])
print determinant([ [2,5,3], [1,-2,-1], [1, 3, 4]])
5.将minor()函数嵌套进determinant()。
#!/usr/bin/python
def determinant(matrix):
if len(matrix) < 2:
return matrix[0][0]
elif len(matrix) == 2:
return matrix[0][0]*matrix[1][1] - matrix[0][1]*matrix[1][0]
def minor(matrix,y):return [[matrix[i][j] for j in range(0,len(matrix[i])) if j != y] for i in range(1,len(matrix))]
while len(matrix) > 2:
return sum([(1 if y%2 == 0 else -1)*(matrix[0][y])*determinant(minor(matrix,y)) for y in range(0,len(matrix[0]))])
print determinant([ [2,5,3], [1,-2,-1], [1, 3, 4]])
6.可以直接将 minor() 嵌套进语句：
#!/usr/bin/python
def determinant(matrix):
if len(matrix) < 2:
return matrix[0][0]
elif len(matrix) == 2:
return matrix[0][0]*matrix[1][1] - matrix[0][1]*matrix[1][0]
while len(matrix) > 2:
return sum([(1 if y%2 == 0 else -1)*(matrix[0][y])*determinant([[matrix[i][j] for j in range(0,len(matrix[i])) if j != y] for i in range(1,len(matrix))]) for y in range(0,len(matrix[0]))])
print determinant([ [2,5,3], [1,-2,-1], [1, 3, 4]])
代码效果：
7.可以直接 elif len(matrix) == 2这一步去掉：
#!/usr/bin/python
def determinant(matrix):
if len(matrix) < 2: return matrix[0][0]
while len(matrix) >= 2:
return sum([(1 if y%2 == 0 else -1)*(matrix[0][y])*determinant([[matrix[i][j] for j in range(0,len(matrix[i])) if j != y] for i in range(1,len(matrix))]) for y in range(0,len(matrix[0]))])
print determinant([ [2,5,3], [1,-2,-1], [1, 3, 4]])
8.(1 if y%2 == 0 else -1) => (y%2)*2-1
#!/usr/bin/python
def determinant(matrix):
if len(matrix) < 2:
return matrix[0][0]
while len(matrix) >= 2:
return sum([((y%2)*2-1)*(matrix[0][y])*determinant([[matrix[i][j] for j in range(0,len(matrix[i])) if j != y] for i in range(1,len(matrix))]) for y in range(0,len(matrix[0]))])
这个是我最为满意的一次编写算法过程。
其他解法：
解法一：
def determinant(m):
a = 0
if len(m) == 1:
a = m[0][0]
else:
for n in xrange(len(m)):
if (n + 1) % 2 == 0:
a -= m[0][n] * determinant([o[:n] + o[n+1:] for o in m[1:]])
else:
a += m[0][n] * determinant([o[:n] + o[n+1:] for o in m[1:]])
return a
解法二：
import numpy as np
def determinant(a):
return round(np.linalg.det(np.matrix(a)))
解法三：
def determinant(matrix):
return reduce(lambda r, i:r+(-1)**i*matrix[0][i]*determinant([m[:i]+m[i+1:] for m in matrix[1:]]),range(len(matrix[0])),0) if len(matrix) != 1 else matrix[0][0]
解法四：
from numpy.linalg import det
def determinant(matrix):
return round(det(matrix))
解法五：
def determinant(m):
return m[0][0] if len(m) == 1 else sum([n*determinant(map(lambda x: x[:i]+x[i+1:],m[1:]))*(-1)**i for i,n in enumerate(m[0])])
解法六：
def determinant(matrix):
det = 0
l=len(matrix)
if(l==1):
return matrix[0][0]
for i in range(l):
minor = [matrix[r][1:l] for r in range(l) if r!=i]
det+=(1-i%2*2)*matrix[i][0]*determinant(minor)
return det
解法七：
determinant = lambda m: m[0][0] if len(m) == 1 else sum([(1 if i % 2 == 0 else -1) * m[0][i] * determinant([[r[j] for j in range(len(r)) if j != i] for r in m[1:]]) for i in range(len(m))])
看了codewars 上大神的解法，的的确确有比我更好的解法。
想起来那句经典的话，你考98分是实力只有这么多人家考100分是试卷只有这么多分。