Data type:
Integer
Explanation:
Find the summation of each row and column individually. If All summation is Even then output: OK
or, Find the corrupted row and column. Then output: Change bit (row,column)
Now,
a) if more than one row r corrupted or
more than one column r corrupted then output: Corrupt
b) if one row and one column corrupted but there is no common bit then output: Corrupt.
It’s an Ad-hoc problem so u can use any other algorithm while it’s produce correct output.
But remember, Parity property can be established iff, only one row and only column corrupted and these two have a common bit.
Critical input:
4
1 1 1 1
1 1 1 1
1 1 1 1
0 0 0 0
4
1 1 1 1
0 0 0 0
1 1 1 1
0 0 1 0
4
1 0 0 0
0 0 0 0
0 0 1 1
0 0 1 1
0
Critical output:
Corrupt
Change bit (4,3)
Change bit (1,1)
