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  1. Prepare
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  3. Numpy
  4. Dot and Cross
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Dot and Cross

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356 Discussions

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  • SVVBathini_w
     + 0 comments

    For Python3 Platform

    import numpy

N = int(input())
A = numpy.array([input().split() for _ in range(N)], int)
B = numpy.array([input().split() for _ in range(N)], int)

print(numpy.matmul(A, B))

  • dilajevolves
     + 0 comments
    import numpy as np
N = int(input())
A = [list(map(int,input().split())) for _ in range(N)]
B = [list(map(int,input().split())) for _ in range(N)]
all_ind_sum = [[] for _ in range(N)]
for i in range(N):
    for j in range(N):
        ind_sum = 0
        for k in range(N):
            ind_sum += A[i][k] * B[k][j]
        all_ind_sum[i].append(ind_sum)
print(np.array(all_ind_sum))

  • lsloan
     + 0 comments

    My compact solution…

    import sys
import numpy as np

print([
    (n := int(input())),
   (ab := np.loadtxt(sys.stdin, int))[:n] @ ab[n:]
][1])

    It's a one-liner, broken here for clarity.

  • rohitambasta_ii1
     + 0 comments

    Using Built-ins:

    import numpy as np
N= int(input())

A = [list(map(int, input().split())) for i in range(N)]
B = [list(map(int, input().split())) for i in range(N)]

print(np.matmul(A, B))

    Without Using built-ins

    import numpy as np
N= int(input())

A = [list(map(int, input().split())) for i in range(N)]
B = [list(map(int, input().split())) for i in range(N)]



final_array =[]

for i in range(N):
    sum_total= []
    for j in range(N):
        sum =0
        for k in range(N):
            sum = sum + (A[i][k] * B[k][j])
        sum_total.append(sum)
     
    final_array.append(sum_total)
print(np.array(final_array))    
           
# For Your Understanding
# ------------------------------

# print(A)  #[['1', '2'], ['3', '4']]
# print(B)  #[['1', '2'], ['3', '4']]  

        #    i, j, k= 0, 0, 0   1 * 1 = 1
        #    i, j, k= 0, 0, 1   2 * 3 = 6
        #    i, j, k= 0, 1, 0   1 * 2 = 2
        #    i, j, k= 0, 1, 1   2 * 4 = 8
        #    i, j, k= 1, 0, 0   3 * 1 = 3
        #    i, j, k= 1, 0, 1   4 * 3 = 12
        #    i, j, k= 1, 1, 0   3 * 2 = 6
        #    i, j, k= 1, 1, 1   4 * 4 = 16
    

# (i[0][0]*j[0][0]) +  (i[0][1]*j[1][0])      (i[0][0]*j[0][1]) + (i[0][1]*j[1][1])   .............


# Ex- Matrix A & B
# [[1 2]       [[1 2]
#  [3 4]]       [3 4]]

# Product of A & B
# (1*1) + (2*3)         (1*2) + (2*4)
# (3*1) + (4*3)         (3*2) + (4*4)

# Result of above Matrix Multiplication
# 7                      10
# 15                     22

  • yashdeora98294
     + 0 comments

    Here is HackerRank Dot and Cross in Python solution - https://programmingoneonone.com/hackerrank-dot-and-cross-problem-solution-in-python.html