i need to write a code that solve this issue... if the number of circulating was 4 then.... if the matrix first was : 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 Then after 4 rotation to the right the matrix will be: 2 1 1 1 3 2 3 1 4 2 3 2 4 4 4 3 the input could be any number of circulating... thank Tomas

Hope this is what u asked... Code: /* *author SHIBIN K.REENY(skr1986.wordpress.com) */ #include <stdio.h> main() { int a[4][4], i, j, k, e1, e2, e3, e4, rotate; for(i = 0; i < 4; i++)for(j = 0; j < 4; j++) scanf("%d", &a[i][j]); printf("Enter the rotation number : "); scanf("%d", &rotate); printf("Orginal matrix :-\n"); for(i = 0; i < 4; i++){ // displaying the original matrix for(j = 0; j < 4; j++) printf("%d ", a[i][j]); printf("\n"); } printf("\n"); k = 0; while(k < rotate){ // saving the edges... e1 = a[0][0]; e2 = a[0][3]; e3 = a[3][3]; e4 = a[3][0]; for(i = 3; i > 0; i--) // top edge moving by one position a[0][i] = a[0][i - 1]; for(i = 3; i > 1; i--) // rigth edge moving by one position a[i][3] = a[i - 1][3]; a[1][3] = e2; for(i = 0; i < 2; i++) // bottom edge moving by one position a[3][i] = a[3][i + 1]; a[3][2] = e3; for(i = 0; i < 2; i++) // left edge moving by one position a[i][0] = a[i + 1][0]; a[2][0] = e4; k++; } printf("NEW Matrix :-\n"); for(i = 0; i < 4; i++){ // final output after rotation for(j = 0; j < 4; j++) printf("%d ", a[i][j]); printf("\n"); } return 1; }