Alternative Methods ...

[T.Jackson]

Algorithm to see how many variables are the same ...

I finished my final unit for Java last week. In the examination there was a question that asked for a method to be written to check to see how many elements in an array contained the same values. Given that there were only 3 elements in the array, the most obvious solution would be to resort to the humble 3-bit truth table and implement an IF statement containing ANDs / ORs. But I wanted to do things differently, I thought outside the square, this is what I came up with ...

public int getObjectsAtSameDistance()
{   
   int ObjectsSameDistance;

   // Scan thru array and see how many objs are at the same distance
   for (int j = 0, j < 2, j++)
   {
      for (int i = j + 1, i < 2, i++)
      {
         if (distances[j] == distances[i])
         {
            objectsSameDistance ++;
            break; 
         }
      }
   }

   if (objectsSameDistance > 0)
      objectsSameDistance ++;

   return objectsSameDistance;
}

The more traditional approach for something like that would be this ...

public int getObjectsAtSameDistance()
{
   int ObjectsSameDistance;

   if (distances[0] == distances[1] && distances[2] == distances[0])
   {
      objectsSameDistance = 3;
   }
   elseif (distances[0] == distances[1] || distances[1] == distances[2] || distances[0] == distances[2])
   {
      objectsSameDistance = 2;
   }
}

Doing things like can quickly become unmanagable when there's lots of elements in the array.
The truth table below supports the solution, whereas a = the first element in the array and c is the last ...

abc
000
001
010
011
100
101
110
111

I'm fully aware that this is a PBP forum and not Java, so I have ported the code to PBP below for those who wish to give it a whirl ..

Main:  
   objectsSameDistance var byte 

   // Scan thru array and see how many objs are at the same distance
   for j = 0 to 2
      for i = j + 1 to 2
         if distances[j] = distances[i] then
            objectsSameDistance = objectsSameDistance + 1
            exit for 
         end if
      next
   next

Best Regards,

Trent Jackson