Given n, print the checkboard pattern for a n x n matrix
Checkboard Pattern for n = 8:
It consists of n * n squares of alternating 0 for white and 1 for black.
We can do the same using nested for loops and some if conditions, but using Python’s numpy library, we can import a 2-D matrix and get the checkboard pattern using slicing.
W2’ll be using following python function to print pattern :
x = np.zeros((n, n), dtype=int)
Using this function, we initialize a 2-D matrix with 0’s at all index using numpy
- x[1::2, ::2] = 1 : Slice from 1st index row till 1+2+2… and fill all columns with 1 starting from 0th to 0+2+2… and so on.
- x[::2, 1::2] = 1 : Slice from 0th row till 0+2+2… and fill all columns with 1 starting from 1 to 1+2+2+…..
Function of np.zeros((n, n), dtype=int) : Often, the elements of an array are originally unknown, but its size is known. Hence, NumPy offers several functions to create arrays with initial placeholder content. These minimize the necessity of growing arrays, an expensive operation. Using the dtype parameter initializes all the values with int data-type.
For example: np.zeros, np.ones etc.
# Python program to print nXn # checkerboard pattern using numpy import numpy as np # function to print Checkerboard pattern def printcheckboard(n): print ( "Checkerboard pattern:" ) # create a n * n matrix x = np.zeros((n, n), dtype = int ) # fill with 1 the alternate rows and columns x[ 1 :: 2 , :: 2 ] = 1 x[:: 2 , 1 :: 2 ] = 1 # print the pattern for i in range (n): for j in range (n): print (x[i][j], end = " " ) print () # driver code n = 8 printcheckboard(n) |
Output:
Checkerboard pattern: 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 1 0
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