2D Arrays - sum diagonals, transpose, shift columns - c++

Question

I recently turned in homework which I had thought I had done correct but my teacher marked me for things that I honestly thought were right.
The first question: Fill up the following code for computing the sum of values on the principal and minor diagonal in a square matrix A[7][7]. Assume the matrix is defined
int sum = 0; // he marked this wrong. Why is initializing to 0 wrong?
for(int i = 0; i < 7; i++) {
for(int j = 0; j < 7; j++) { // also marked j < 7 wrong
sum = sum + A[i][i] + A[i][7-1-i]
}
return sum;
}
Question 2: Transpose a square matrix A[50][50] i.e., the first row becomes the first column and vice-versa, the second row becomes the second column and so on. Assume the matrix is defined.
void transposeSquare(){
for(int i=0;i<50;i++){
for(int j=0;j<50;j++){ //marked j=0 wrong
int temp = A[i][i];
A[i][j] = A[j][i];
A[j][i] = temp;
}
}
}
Why is j=0 in transposeSquare not right?
Why would sum be anything other than 0?
If i < 7 should j < 7 too if its a square matrix?
Thanks!

Answer 1

Your logic is wrong for couple of reasons.
By using a nested loop inside the outer loop, you are adding the numbers in the diagonals 7 times instead of just once.
When i is 3, the line
sum = sum + A[i][i] + A[i][7-1-i]
is equivalent to:
sum = sum + A[3][3] + A[3][7-1-3]
which is the same as:
sum = sum + A[3][3] + A[3][3]
Adding A[3][3] twice to the sum is not correct.
To fix the problems, use:
for(int i = 0; i < 7; i++){
sum = sum + A[i][i] + A[i][7-1-i];
}
sum -= A[3][3]; // Remove the element that was added twice.
You can use += to simplify the code a little bit.
for(int i = 0; i < 7; i++){
sum += A[i][i] + A[i][7-1-i];
}
sum -= A[3][3]; // Remove the element that was added twice.

Answer 2

Your code has small errors that's why your teacher has entitled your code as wrong.
For the first part,
int sum = 0;
int size = 7;
for(int i = 0; i < size; i++) {
sum += A[i][j] + A[i][sz-1-i]; // Error: Typo, it should be A[i][j] instead of A[i][i]
}
if(size%2 != 0)
sum -= A[size/2][size/2]; // Missed Case: Subtract the middle element since its repeated only when the size of array is an odd number
return sum; // Return the sum at the end!
For second Part You don't need to iterate the whole two-D array. Because if you do so it will copy upper triangle into lower triangle instead of calculating the Transpose. So traverse any triangle (Upper or Lower).
Like I have traversed Lower Triangle in this code.
void transposeSquare(){
for(int i=0;i<50;i++){
for(int j=0; j<i ;j++){ // Fix 1: Loop j to i-1
int temp = A[i][j]; // Fix 2: Typo it should be A[i][j] instead of A[i][i]
A[i][j] = A[j][i];
A[j][i] = temp;
}
}
}
Checkout Triangular Matrix for better explanation.