Reddit Programming Prompt

  /*

 (a / (b + c)) + (d / (e + f)) + (g / (h + i)) = 1

using only numbers 1 - 9, and only each number once!

When a solution is found, print the completed equation.

  */

   public class numProblem
   {
      static int tested, elim1, elim2, solutions,  recursions;

      public numProblem()
      {
      }

      public static void main(String[] args)
      {
      //loop from 1 to 9, running instances of solve
         for(int x=1; x<10; x++)
         {
            System.out.println("Root Parameter: "+x);
            solve(""+x);
         }

         int divides = elim1+ 2*(elim2)+3*(tested) ; //checks the number of divisions that occured


         System.out.println("Recursions Executed: "+recursions);//how many times the method "solve" recursed

         System.out.println("Combinations Tested: "+tested); //how many 9 number combinations were tested for the solution

         System.out.println("1st Round "+elim1);     //how many recursions occured after the first test

         System.out.println("2nd Round: "+elim2);   //how many recursions occured after the second test

         System.out.println("Number of float divisions "+divides);

         System.out.println("Number of Solutions: "+solutions);

      }

      public static void solve(String p) //p is the parameter that has been passed
      {
         float test=0; //variable used to check whether or not the current recurring pattern is possible
         boolean newclean = true; //determines whether or not the new integer being added to the pattern has been used previously in the pattern
         int prevlength = p.length(); //determine length

            //main solver
         if(prevlength == 8) //if contains length is 8 (check solutions) -- if correct, then print
         {
            for(int a = 1; a< 10;  a++) //checking 1 through 9
            {
               newclean = true;
               for(int b = 0; b< 8; b++) //checks to make sure a has not been previously used
               {
                  if( a == Integer.parseInt(""+p.charAt(b)) ) // if a has been used newclean is set to false, so the solution wont be tested. (having repeated numbers would make it an invalid solution)
                  {
                     newclean = false;
                  }
               }
               if(newclean)//tests to see the result of the combination of integers
               {
                  test =  ( Integer.parseInt(""+p.charAt(2))  /    (float)( Integer.parseInt(""+p.charAt(0)) + Integer.parseInt(""+p.charAt(1)) )   ) + (    Integer.parseInt(""+p.charAt(5))  / (float)( Integer.parseInt(""+p.charAt(3)) +Integer.parseInt(""+p.charAt(4)) )   )  + (    a  / (float)( Integer.parseInt(""+p.charAt(6)) +Integer.parseInt(""+p.charAt(7)) )   ) ;
                  tested++;
               }
               if( test==1  && newclean) // condition to fulfill prompt requirements, causes a print
               {
                  System.out.println("Solution: ( " +Integer.parseInt(""+p.charAt(2))+" / ("+Integer.parseInt(""+p.charAt(0))+" + "+Integer.parseInt(""+p.charAt(1))+") ) +( "+Integer.parseInt(""+p.charAt(5))+" / ("+Integer.parseInt(""+p.charAt(3))+" + "+Integer.parseInt(""+p.charAt(4))+") ) + ( " +a+" / ("+Integer.parseInt(""+p.charAt(6))+" + "+Integer.parseInt(""+p.charAt(7))+") ) = 1"   );
                  solutions++;
               }
            }
         }
         //Eliminator 2
         else if(prevlength == 5) //if length is 5 (loop going upwards, checking if total > 1 -- if not, then recurse)
         {
            for(int a = 1; a< 10;  a++) //checking 1 through 9
            {
               newclean = true;
               for(int b = 0; b< 5; b++) //checks to make sure a has not been previously used
               {
                  if( a == Integer.parseInt(""+p.charAt(b)) )  // if a has not been used it will be added on to the string and solved for
                  {
                     newclean = false;
                  }
               }
               if(newclean)
               {
                  test = ( Integer.parseInt(""+p.charAt(2))  /    (float)( Integer.parseInt(""+p.charAt(0)) + Integer.parseInt(""+p.charAt(1)) )  ) + (     a / (float)(Integer.parseInt(""+p.charAt(3)) +Integer.parseInt(""+p.charAt(4)))   ) ;
                  elim2++;
               }
               if(test>=1)
               {
                  break;
               }
               if(   test   < 1 && newclean) // condition to incur recursion, if this is not fulfilled, it is wasteful to calculate further possibilities
               {
                  solve(""+p+a);
                  recursions++;
               }
            }
         }
         //Eliminator 1
         else if(prevlength == 2) //if length is 2 (loop going upwards, checking if total > 1 -- if not, then recurse)
         {
            for(int a = 1; a< 10;  a++) //checking 1 through 9
            {
               newclean = true;
               for(int b = 0; b< 2; b++) //checks to make sure a has not been previously used
               {
                  if( a == Integer.parseInt(""+p.charAt(b)) ) // if a has not been used it will be added on to the string and solved for
                  {
                     newclean = false;
                  }
               }
               if(newclean)
               {
                  test = a  /    (float)( Integer.parseInt(""+p.charAt(0)) + Integer.parseInt(""+p.charAt(1)) ) ; //Tests the a, b, and c variables to see if they equal
                  elim1++;
               }
               if(test>=1)//breaks when the tested value exceeds
               {
                  break;
               }
               if( (test < 1.0) && newclean ) // condition to incur recursion, if this is not fulfilled, it is wasteful to calculate further possibilities
               {
                  solve(""+p+a);
                  recursions++;
               }
            }
         }
         //Main recursor for most checks
         else //else if length is anything, recurse with the addition of all unused values
         {
         //char current = '0'; - temp line, may not be needed
            for(int a = 1; a< 10;  a++) //checking 1 through 9
            {
               newclean = true;
               for(int b = 0; b< prevlength; b++) //checks to make sure a has not been previously used
               {
                  if(a == Integer.parseInt(""+p.charAt(b))) // if a has not been used it will be added on to the string and the string will be used as the new parameter
                  {
                     newclean=false;
                  }
               }
               if(newclean)
               {
                  solve(""+p+a);
                  recursions++;
               }
            }
         }
      }
   }