Determine if a relation is symmetric

[trevormoon]

Hello all I am having trouble writing a function that will determine if a relation is symmetric. Any tips as to what I am doing wrong would be greatly appreciated. The arr arrays are the relations, arm arrays are the unique number sin each relation.

 int arr[]  =  {1,1,2,2,3,3,4,4,5,5};
int arm[]  =  {1,2,3,4,5};
int arr2[] =  {1, 1, 2, 2, 3, 3, 4, 4, 4, 7, 7, 4, 7, 7};
int arm2[] =  {1,2,3,4,7};
int arr3[] =  {1, 1, 2, 4, 3, 9, 4, 16, 5, 25, 6,36, 7, 49, 8, 64};
int arm3[] =  {1,2,3,4,9,16,5,25,6,36,7,49,8,64};

int n   = sizeof(arm)/sizeof(int);
int nn  = sizeof(arr)/sizeof(int);
int n2  = sizeof(arm2)/sizeof(int);
int nn2 = sizeof(arr2)/sizeof(int);
int n3  = sizeof(arm3)/sizeof(int);
int nn3 = sizeof(arr3)/sizeof(int);


bool pair_exist(int left, int right, int b[], int nn)
{
   for (int i=0; i<nn; i+=2)
   {
      if (left==b[i] && right==b[i+1])  return true;
   }

   return false;
}


  bool symmetric(int arr[],int nn)
{
   for(int i=0;i<nn;i+=2)
    {
     int e = arr[i];
     int f = arr[i+1];
       cout<<"f " <<f<<endl;
       cout<<"e " <<e<<endl;

      if(!pair_exist(e,f,arr,nn))
          return false;
    }
return true;
}

[threeright]

You can't sizeof an array in c++... if you try, it will jus return the size of the pointer and not the array.

If you want to use size of, use vectors

[Danny Toledo]

Any tips as to what I am doing wrong would be greatly appreciated.

Could you expand on that (what is happening and what is expected to happen)?

It appears that you go out of bounds here:
if (left==b[i] && right==b[i+1]) return true;
and here:
int f = arr[i+1];
When i = nn-1

You can't sizeof an array in c++... if you try, it will jus return the size of the pointer and not the array.

Nope. If you use sizeof on a pointer, it will return the size of a pointer:

http://ideone.com/P51Osj