calculate the sum of diagonals in a matrix - c++

Question

I need to calculate the sum of two diagonals in a matrix in C++, I already have a solution for that but I must be dumb because I cant understand what it is doing, so I would like to know if there is another version which I can understand. here is the code which does the job:
cout<<"Jepi rangun e matrices"<<endl; // pra bejme manipulim me matrice katrore ku rreshtat=kolonat
cin>>n;
cout<<"Tani jepi elementet e matrices"<<endl; // lexohet matrica
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
cin>>a[i][j];
}
d=0;
s=0; // ketu e keni kushtin si dhe mbledhjen per te dy diagonalet me dy variabla te ndryshme
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
{
if(i==j)
d=d+a[i][j];
if(j==n-i+1 || i==n-j+1)
s=s+a[i][j];
}
The part that is difficult to understand is
if(j==n-i+1 || i==n-j+1)
s=s+a[i][j];
Here is the entire code that I changed but it doesnt work for the secondary diagonal:
#include <iostream>
using namespace std;
int main()
{
int d=0,s=0; // ketu e keni kushtin si dhe mbledhjen per te dy diagonalet me dy variabla te ndryshme
int i,j,n;
int a[5][5];
cout<<"Jepi rangun e matrices"<<endl; // pra bejme manipulim me matrice katrore ku rreshtat=kolonat
cin>>n;
cout<<"Tani jepi elementet e matrices"<<endl; // lexohet matrica
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
cin>>a[i][j];
}
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(i==j)
d+=a[i][j]; //principal diagonal
if(i+j==n-1)
s+=a[i][j];//secondary diagonal
}
}
cout << d << endl;
cout << s << endl;
cin.get();
cin.get();
return 0;
}

Answer 1

It would be nice to have comments in English, but, the your code does (second loop):
browse all rows
browse all cells
if i == j (is in main diagonal):
increase one sum
if i == n - i + 1 (the other diagonal)
increase the second sum
The much nicer and much more effective code (using n, instead of n^2) would be:
for( int i = 0; i < n; i++){
d += a[i][i]; // main diagonal
s += a[i][n-i-1]; // second diagonal (you'll maybe need to update index)
}
This goes straight trough the diagonals (both at the one loop!) and doesn't go trough other items.
EDIT:
Main diagonal has coordinates {(1,1), (2,2), ..., (i,i)} (therefor i == j).
Secondary diagonal has coordinates (in matrix 3x3): {(1,3), (2,2),(3,1)} which in general is: {(1,n-1+1), (2, n-2+1), ... (i, n-i+1), .... (n,1)}. But in C, arrays are indexed from 0, not 1 so you won't need that +1 (probably).
All those items in secondary diagonal than has to fit condition: i == n - j + 1 (again due to C's indexing from 0 +1 changes to -1 (i=0,, n=3, j=2, j = n - i - 1)).
You can achieve all this in one loop (code above).

Answer 2

int diag1=0;
int diag2=0;
for (i=0;i<n;i++)
for (j=0;j<n;j++){
if(i==j) diag1+=a[i][j]; //principal diagonal
if(i+j==n-1) diag2+=a[i][j];//secondary diagonal
}
To understand this algorithm better you should paint a matrix on you notebook and number it's elements with their position in matrix,then apply the algorithm step by step.I'm 100% sure that you will understand

Answer 3

How about I try to explain this version? :D
There are 3 important parts of the code:
inputing the matrix
calculating major diagonal ( \ direction)
calculating minor diagonal ( / direction)
And here they are, explained:
// input elements
for(i=1;i<=n;i++) // from left to right
{
for(j=1;j<=n;j++) // from up to down
cin>>a[i][j]; // input element at (i,j) position
}
Here, d and s contain the inter-values of major and minor diagonal respectively. At the end of 2 loops, they will contain the results
for (i=1;i<=n;i++)
for (j=1;j<=n;j++)
{
if(i==j) // major diagonal - if coordinates are the same
d=d+a[i][j]; // e.g. (1,1), (2,2)
if(j==n-i+1 || i==n-j+1) // coordinates of the minor diagonal - check
s=s+a[i][j]; // e.g. n=3 (3,1) (2,2) ...
}
Hope this helps.
Note that this code starts matrix coordinates at 1 instead of 0, so you will actually need to allocate (n+1)x(n+1) space for the matrix:
double a[n+1][n+1];
before using it.
Also, the code you gave is not most effective. It has O(n^2) complexity, while the task can be done in O(n) like so:
// matrix coordinates now start from 0
for (int i=0; i < n; ++i){
d += a[i][i]; // major
s += a[i][n-1-i]; // minor
}

Answer 4

int num[5][5]={0}; //decleration
int i=0,j=0,sum=0;
for (int i=0;i<5;i++)
{
for (int j=0;j<5;j++)
{
cin>>num[i][j];
} //Taking Matrix input
}
cout<<endl<<"The Matrix is "<<endl;
for (int i=0;i<5;i++)
{
for (int j=0;j<5;j++)
{
cout<<num[i][j]<<" ";
}
cout<<endl; //Displaying the Matrix
}
cout<<endl<<"The sum of diagonals of the matrix is "<<endl;
if(i==j)
{
for (i=0;i<5;i++)
{
for (j=0;j<5;j++)
{
if (i==j) //This loop works where i and j will be equal
{
sum=sum+num[i][j];
}
}
}
cout<<sum;
}
else //Some times the user creates 4 x 3 matrix or so than diagonals not match so. . .
{
cout<<"The sum is not Possible";
}

Answer 5

you must use i + j == n + 1 instead of i + j == n - 1 for secondary diagonal i.e
for(i = 1; i <= n; i++)
{
for(j = 1; j <= n; j++)
{
if(i == j)
d += a[i][j]; //principal diagonal
if(i + j == n+1)
s += a[i][j];//secondary diagonal
}
}