Thread: Simple 2D rubiks cube algorithm.

I'm sorry for not responding earlier. I've a busy weekend so I won't be able to code anything 'til mon-tue. Anyway, I asked my lecturer and he was very vague regarding the solution, but he said something like:

Start with a tree and test each depth and go deeper:

. A....B....C
. /|\ /|\ /|\
.ABCABCABC

etc. (hard to write up but you get the idea).

Doesn't really add much to our discussion though.

I might have a solution in mind that I'll try to code up soon. I'll also take a look at your code.

As for admin moving this thread to the game programming forum, I've no objections if that would keep it on the first page for longer.

An update that actually runs.

Code:

/* Will solve via brute force, the 8 tile 2D "rubik's cube" puzzle, when finished
Adak, Nov. 2008 

Status: just started 7/11/08
*/

#include <stdio.h>

void copyB(int);
int move1(int);
int move2(int);
int move3(int);

void showB(int);
int isSolved(int);

int b[2][4][20];    //b is shorthand for the board
int t[1][4];        //t is a temporary holding row
int line[20];       //the line of moves that are played

int main(void)  {
   int i, r, row, c, col, move, deep, depth, solved, reset, gar;
// test right shift    
/*   b[0][0][0] = 2; b[0][1][0] = 3; b[0][2][0] = 4; b[0][3][0] = 1;
   b[1][0][0] = 7; b[1][1][0] = 6; b[1][2][0] = 5; b[1][3][0] = 8;
*/
// test row swap
/*   b[0][0][0] = 8; b[0][1][0] = 7; b[0][2][0] = 6; b[0][3][0] = 5;
   b[1][0][0] = 1; b[1][1][0] = 2; b[1][2][0] = 3; b[1][3][0] = 4;
*/
// test Rotate 4 inner tiles
/*   b[0][0][0] = 1; b[0][1][0] = 3; b[0][2][0] = 6; b[0][3][0] = 4;
   b[1][0][0] = 8; b[1][1][0] = 2; b[1][2][0] = 7; b[1][3][0] = 5;
*/
// test depth of 2, solution 1 + 2
/*   b[0][0][0] = 7; b[0][1][0] = 6; b[0][2][0] = 5; b[0][3][0] = 8;
   b[1][0][0] = 2; b[1][1][0] = 3; b[1][2][0] = 4; b[1][3][0] = 1;
*/
//test depth of 2, solution 2 + 3  no depth=2 solution found for 1+3
   b[0][0][0] = 8; b[0][1][0] = 2; b[0][2][0] = 7; b[0][3][0] = 5;
   b[1][0][0] = 1; b[1][1][0] = 3; b[1][2][0] = 6; b[1][3][0] = 4;

//test depth of , solution 
/*   b[0][0][0] = ; b[0][1][0] = 2; b[0][2][0] = 1; b[0][3][0] = 4;
   b[1][0][0] = 8; b[1][1][0] = 7; b[1][2][0] = 6; b[1][3][0] = 5;
*/
   printf("\n Starting position:\n");
   showB(0);
   //getch();
   depth = deep = solved = reset = 0;

   do  {
      move = 1; depth = 2;

        
      while(1)  {   // this is the main loop, for now
         ++deep;
         if(move == 1)
            solved = move1(deep);
         else if(move == 2)
            solved = move2(deep);
         else
            solved = move3(deep);

         if(solved) 
            break;

         if(reset)  {    //depth was reset, move needs to reset to 1 again
            move = 1;
            reset = 0;
            continue;
         }


         if(deep >= depth)  {
            line[deep-1] = 0;
            deep = depth - 1;
            if(++move > 3) { 
               deep -= 1;
               move = line[deep] + 1;
               if(move > 1)
                  reset = 1;
            }
         }
      }

   }while(!solved);

   printf("\n\n\t Press Enter to Quit\n ");
   gar = getchar(); gar++;
   return 0;
}
void copyB(int d)  {  //copy the old board position, into the new depth
   int r, c, temp;    
   
   for(r = 0; r < 2; r++) 
      for(c = 0; c < 4; c++)
         b[r][c][d] = b[r][c][d-1];
}
int move1(int d)  {  //shifts all numbers one column to the right &  wraps
   int r, c, solved, temp;
   solved = 0;
   printf("\n Right shift:");
   copyB(d);
   //showB(d);
   //++d;

   for(r = 0; r < 2; r++)  {
      temp = b[r][3][d];
      for(c = 2; c >-1 ; c--)  
         b[r][c+1][d] = b[r][c][d];
      b[r][0][d] = temp;
   }
   
   showB(d);
   line[d-1] = 1;
   solved = isSolved(d);
   
   return solved;
}
int move2(int d)  { //swaps rows
   int r, c, solved;
   solved = 0;
   printf("\n Swap rows:");
   copyB(d);
   //showB(d);
   
   for(c = 0; c < 4; c++)  {
      t[0][c] = b[0][c][d];
      b[0][c][d] = b[1][c][d];
      b[1][c][d] = t[0][c];
   }

   showB(d);
   line[d-1] = 2;
   solved = isSolved(d);
   
   return solved;
}
int move3(int d) { //turns the 4 inner tiles, clockwise
   int r, c, solved, temp;
   solved = 0;
   printf("\n Rotate 4 inner tiles:");
   copyB(d);
   //showB(d);  

   temp = b[0][1][d];
   b[0][1][d] = b[1][1][d];
   b[1][1][d] = b[1][2][d];
   b[1][2][d] = b[0][2][d];
   b[0][2][d] = temp;

   showB(d);
   line[d-1] = 3;
   solved = isSolved(d);
   
   return solved;
}         
void showB(int d)   {
   int r, c, tab;
   printf("\n");
   for(tab = 0; tab < d; tab++)
      putchar('\t');
   for(r = 0; r < 2; r++) {
      for(c = 0; c < 4; c++)  
         printf(" %d", b[r][c][d]);
      putchar('\n');
      for(tab = 0; tab < d; tab++)
         putchar('\t');
   }
}
int isSolved(int d)  {
   int i, gar;
   int done = 0;
   if(b[0][0][d]==1 && b[0][1][d]==2 && b[0][2][d]==3 && b[0][3][d]==4)
      if(b[1][0][d]==8 && b[1][1][d]==7 && b[1][2][d]==6 && b[1][3][d]==5)  {
         done = 1;
         printf("\n\n\t\t It's solved!");
         showB(d);
         printf("\t Line: ");
         for(i = 0; i < d; i++)
           printf(" %d", line[i]);

         //printf("\n\t Press Enter to Continue");
         //gar = getchar(); ++gar;
      }
      
   return done;
}

This shows a very basic graphic of the various depths and the moves being made, to a
depth of 2 ply. The leftmost board being the starting board, and is at depth 0. The next board column to the right is depth 1. Going right one more board width, the boards for depth = 2 are shown (if the search went that far), etc.

So it's the search tree, with the root on the far left, and the leaves to the right side.

Code:

Starting 
board:
nnnn
nnnn
Description of move 1 at depth 1:
         nnnn
         nnnn
Description of move 1 at depth 2:
                  nnnn
                  nnnn
Description of move 2 at depth 2:
                  nnnn
                  nnnn
Description of move 3 at depth 2:
                  nnnn
                  nnnn
Description of move 2 at depth 1:
         nnnn
         nnnn

etc.

I'm a hobby programmer, so you may very well find something you like better. I'd suggest this as a good starting off point or food for thought, rather than a hard and fast "this is the way it should be done" kind of thing.

There are several ways to search through a tree, this is just one that felt right (DFS), in this instance, to me.

The above program handles 2 ply searches, but not 3 or more. The part after

Code:

if(solved)
   break;

needs to be adjusted to handle multiple "back ups" in ply. After moves of 1, 3, 3, you need to back up 2 ply, for instance, not just one. That makes your next move 2.

The program prints the board array now, but even more helpful is getting it to print the line of moves as it goes through it's search.

If you change the lines array to print < deep right after the if(solved) break, line of code, then you can get a better reference for debugging the DFS.

The moves should be:
1
11
111
112
113
12
121
122
123
13
131
132
133
2
21
211
212
213
22
221
222
223
23
231
232
233

Which will solve this position:
b[0][0][0] = 8; b[0][1][0] = 3; b[0][2][0] = 2; b[0][3][0] = 5;
b[1][0][0] = 1; b[1][1][0] = 6; b[1][2][0] = 7; b[1][3][0] = 4;

Since this is a student's project, I'm going to refrain from posting any more code on this.

Original starting position of:

2684
1375

the first solution found for it was: 3 3 1 3 1 1 1

Here's a laugh for you.

I thought I'd try an easy puzzle position, something with just one pair of numbers in the wrong position - should be a good way to get started with a new puzzle, right?


2134
8765

Wrong!

I just couldn't figure the dang thing out, but now my program has. The shortest answer found was 18 ply's deep:


depth is 18 ply, board is (2134/8765), solution is 231231313131231312


I still haven't figured out how to solve a lot of these positions by hand. LOL