According to the following code which pattern will be generated on the console?
# init matrix
for i in range(6):
row = []
for j in range(6):
if(i%2 ==0 or j%2 == 0):
row.append(0)
else:
row.append(1)
matrix.append(row)
for i in range(6):
for j in range(6):
print(matrix[i][j], " ",end="")
print("\n")
a)
0 0 0 0 0 0
0 1 1 1 0 1
0 1 1 1 0 1
0 1 1 1 0 1
0 0 0 0 0 0
0 1 1 1 0 1
b)
0 0 0 0 0 0
0 1 0 1 0 1
0 0 0 0 0 0
0 1 0 1 0 1
0 0 0 0 0 0
0 1 0 1 0 1
c)
0 0 0 0 0 0
0 1 1 0 1 1
0 1 1 0 1 1
0 0 0 0 0 0
0 1 1 0 1 1
0 1 1 0 1 1
d)
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
The correct answer is option (a):
0 0 0 0 0 0
0 1 1 1 0 1
0 1 1 1 0 1
0 1 1 1 0 1
0 0 0 0 0 0
0 1 1 1 0 1
To determine the pattern that will be generated by the code, let's break down the code:
1. The code starts by initializing an empty matrix variable.
2. Then, it enters a nested loop with the outer loop iterating over the range of 6.
3. Inside the outer loop, an empty row list is initialized.
4. The inner loop iterates over the range of 6 as well.
5. In the inner loop, an if condition checks if either the current row index (i) or the current column index (j) is divisible by 2 with no remainder.
6. If either i or j is divisible by 2, the if condition evaluates to True and appends a 0 to the row list.
7. If neither i nor j is divisible by 2, the if condition evaluates to False and appends a 1 to the row list.
8. After the inner loop completes, the row list is appended to the matrix variable.
9. This process is repeated for all values of i in the outer loop.
10. Finally, another set of nested loops is used to print the elements of the matrix to the console, row by row, with a space separator.
Based on this code logic, the pattern that will be generated is option (a) because the code initializes the matrix with 0s and 1s based on the conditions mentioned above.