Soluție HackerRank pentru Linear Algebra Foundations #5 - The 100th Power of a Matrix, subdomeniul Linear Algebra Foundations, în Python 3. Include…
- Problemă: Linear Algebra Foundations #5 - The 100th Power of a Matrix
- Domeniu: Linear Algebra Foundations
- Limbaj: Python 3
Challenge: Linear Algebra Foundations #5 – The 100th Power of a Matrix
Subdomeniu: Linear Algebra Foundations (linear-algebra-foundations)
Scor cont: 5.0 / 5
Submission status: Accepted
Submission score: 1.0
Submission ID: 464718707
Limbaj: python3
Link challenge: https://www.hackerrank.com/challenges/linear-algebra-foundations-5-the-100th-power-of-a-matrix/problem
Cerință
Given the following matrix A:
[1 1 0 ]
A = [0 1 0 ]
[0 0 1 ]
We compute that
A<sup>100</sup> =
[A B 0]
[0 C 0]
[0 D E]
In the text box below, enter the values of the integers A, B, C, D, E each in a new line. Do not leave any extra leading or trailing spaces.
Cod sursă
def matrix_mul(A, B):
ROWS_A, COLS_A = len(A), len(A[0])
ROWS_B, COLS_B = len(B), len(B[0])
assert COLS_A == ROWS_B
COMMON_D = ROWS_B
# zeros for Rows-A x Cols-B
M_TIMES = []
for r in range(ROWS_A):
M_TIMES.append( [0] * COLS_B )
# Loop over Rows-A, then Cols-B, then the common dimension
for rA in range(ROWS_A):
for cB in range(COLS_B):
for t in range(COMMON_D):
M_TIMES[rA][cB] += A[rA][t]*B[t][cB]
return M_TIMES
def print_matrix(M, skip_indexes=[]):
ROWS, COLS = len(M), len(M[0])
for r in range(ROWS):
for c in range(COLS):
if skip_indexes[r][c]:
continue
print(M[r][c])
#####
A = [[1,1,0],[0,1,0],[0,0,1]]
B = A
EXP=100
for _ in range(EXP-1):
B = matrix_mul(B, A)
# print(B)
print_matrix(B, skip_indexes=[[False, False,True],[True, False, True],[True, False, False]])