HackerRank Linear Algebra Foundations – Linear Algebra Foundations #7 – The 1000th Power of a Matrix

Challenge: Linear Algebra Foundations #7 – The 1000th 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: 464718836

Limbaj: python3

Link challenge: https://www.hackerrank.com/challenges/linear-algebra-foundations-7-the-1000th-power-of-a-matrix/problem

Cerinta

You are provided a matrix $A$ =  

		   [-2     -9 ]                  
           [ 1      4 ]          
           
           
The 1000<sup>th</sup> power of $A$, i.e. $A$<sup>1000</sup> = 

			[A	B]
            [C	D]

In the text box below, enter the integers $A$, $B$, $C$ and $D$ each on a new line, respectively. Do not leave any leading or trailing spaces.

Cod sursa

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 and skip_indexes[r][c]:
                continue
            print(M[r][c])
            
#####
A = [[-2,-9],[1,4]]
B = A
EXP=1000
for _ in range(EXP-1):
    B = matrix_mul(B, A)
# print(B)
print_matrix(B)