Time Complexity in code?

I have fun1 calls foo1() which takes O(n^6 + m^4). What do you think the time complexity of fun1? My guess is it would be O(2n^6 + m^4) !!

int fun1(int n, int G[MAX][MAX])
{
  int x, ans;
  if(n < 2)
     return 1;
  for(x = 0; x < n; x++){
    G[n][x] = G[x][n] = 1;
  }
  ans = foo1(n+1, G);
  return ans;
}

Also fun2 calls foo2() which takes O(n^3 + m^2). What do you think the time complexity of fun2? My guess is O(n^3 + m^2 + 2n^2) !!

 int fun2(int n, int G[MAX][MAX])
 {
   int x, y, i, j;
   int ans = y = 0;
   int arr[MAX][MAX] = {};
   for(i = 0; i < n; i++) {
     for(j = 0; j < n; j++)
       arr[i][j] = G[i][j];
   }

   if(n <= 2)
     return 0;
   for(x = 0; x < n; x++){
     if(arr[y][x] && arr[x][y]){
       arr[y][x] = arr[x][y] = 0;
       arr[n+1][x] = arr[x][n+1] = 1;
       arr[y][n] = arr[n][y] = 1;
       if(foo2(n+2, arr))
         ans = 1;
       arr[n][y] = arr[y][n] = 0;
       arr[n+1][x] = arr[x][n+1] = 0;
       arr[y][x] = arr[x][y] = 1;
       if(ans == 1)
         break;
     }    
   }
   return ans;
  }

Am I right?

For fun1(), I do not agree with you. In the body of the function there is a for loop that takes O(n) and then foo1(n+1, G), which takes O((n+1)⁶ + m⁴). Putting them together, we get:

O(n) + O((n+1)⁶ + m⁴) = O(n) + O(n⁶ + m⁴) = O(n⁶ + m⁴)

I am afraid I do not agree with your second guess either, since what you have there is two for loops, which made you guessed what you guessed.

However, please notice that the second for-loop calls foo2(n+2, arr) in its body. As a result, foo2() will be called n times!

Putting everything together, we have:

first for loop + second for loop = O(n) + O(n(n + 2)³ + nm²) = O(n) + O(n⁴ + nm²) = O(n⁴ + nm²)