For what cases will this logic of this code fail? - c

Question

In a coding site I came across a question which seem to be very easy. But when I implement a solution it gives me wrong answers for the hidden test cases. I tried to find for what cases it failed but I just couldn't get any.
The question goes as follows
Construct an N×M matrix with entries having positive integers such that
If M>1, ith row is strictly increasing from left to right with a fixed common difference di
for all 1≤i≤N.
If N>1, jth column is strictly increasing from top to bottom with a fixed common difference cj
for all 1≤j≤M.
All the common differences should be distinct.
In the case of multiple answers, find an arrangement that minimizes the maximum element of the
matrix.
The code with the logic
#include<stdio.h>
#include<stdlib.h>
int main(){
int t=0,n=0,m=0,d=0;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
if(n>=m){
d=2;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
printf("%d ",(i+1) + j*d);
}
d+=2;
printf("\n");
}
}
else{
d=1;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
printf("%d ",(2*i+1) + j*d);
}
d+=2;
printf("\n");
}
}
}
return 0;
}
Sample input
3
3 3
1 2
1 1
Sample Output
1 2 3
3 6 9
5 10 15
1 2
1
Simple logic I tried was, check if the rows are more or the columns, whichever is more increment that by even number C.D (from 2,4,6....) and the other in odd number C.D(from 1,3,5...)
For example: for 2X3
The matrix will be
1 3
2 6
3 9
This logic seems to be correct and while discussing with my friends even they say its right. But I get wrong answers while submitting to the hidden cases.
What logic error might have I done? Could you give me a example for which this code will fail?

Answer 1

The matrix shown in the question for the 2×3 case is:
1 3
2 6
3 9
However, a better solution is:
1 4
2 6
3 8
We see this satisfies the constraints:
Each row and each column is strictly increasing.
Each row has a fixed difference: 3, 4, and 5.
Each column has a fixed difference: 1 and 2.
All those differences are distinct.
A general solution follows.
Let the matrix be Ai, j. It is clear A0, 0 should be 1. Let A0, 1, A1, 0, and A1, 1 be 1+a, 1+b, and 1+c, respectively. Then simple algebra and induction shows that Ai, j is 1 + (1−i)ja + i(1−j)b + ijc.
The common difference in column j is b+j(c−a−b), and the common difference is row i is a+i(c−a−b). These form two arithmetic series with the same step size, c−a−b, so they do not intersect only if they either interleave (for example, c−a−b could be 2 and one of a or b can be odd while the other is even) or one series is entirely less than the other (for example, c−a−b could be 1). We will consider these cases below.
Let I and J be the maximum subscripts in the array, m−1 and n−1, respectively. (These are introduced for brevity, as m−1 and n−1 appear repeatedly below.) The maximum element in the array, AI,J, is 1 + (1−I)Ja + I(1−J)b + IJc = 1 + IJ(c−a−b) + Ja + Ib. Observe the derivative with respect to c is positive, so, once a and b are chosen, the minimum value for AI,J is obtained by making c as small as the constraints allow, hence either c = a+b+1 (for the case when one series is entirely less than the other) or c = a+b+2 (for the case when the series interleave). For the former, AI,J = 1 + IJ + Ja + Ib. For the latter, AI,J = 1 + 2IJ + Ja + Ib. This reveals we also wish to minimize a and b subject to the constraints.
The smallest a and b can be is 1 and 2 (in either order). So in the interleaving case, we have either AI,J = 1 + 2IJ + J + 2I or AI,J = 1 + 2IJ + 2J + I. To obtain the lesser of these, we choose a = 1, b = 2 if J < I and a = 2, b = 1 otherwise. Combinging these, the maximum is 1 + 2IJ + I + J + min(I, J). Restoring m and n from I and J and doing a little algebra gives us a maximum of 2mn − max(m, n).
For the non-interleaving case, we could have a = 1, the largest row difference (generally a+i(c−a−b)) of 1+I•1 = 1+I, so b = 2+I to start the column differences (generally b+j(c−a−b)) with b + 0(c−a−b) = 2+I. Conversely, we could have b = 1 and a = 2+J. Then we have AI,J = 1+ IJ + J + I(2+I) or 1 + IJ + J(2+J) + I, or, equivalently, mn+m2−m or mn+n2−n. The former is less when m < n and vice-versa.
Note than when m ≤ n, the interleaved case gives us a maximum of 2mn − n = mn + m2 + (m−1)(n−m) − m. Since (m−1)(n−m) is non-negative, that is at least as large as mn+m2−m, which is the maximum for the non-interleaved case. The analogous result holds for n ≤ m. Therefore, the interleaved case never has any advantage. The minimum is always obtained with the non-interleaved case with either a = 1, b = 2+I = m+1 or a = 2+J = n+1, b = 1, according to which of m and n is smaller.