multithreading - 如何使用 Grand Central Dispatch 并行化数独求解器？

Question

作为编程练习，我刚刚完成了使用回溯算法的数独求解器的编写(参见 Wikipedia 以获取用 C 编写的简单示例)。

为了更进一步，我想使用 Snow Leopard 的 GCD 来并行化它，以便它在我机器的所有内核上运行。有人能给我一些关于我应该如何去做以及我应该进行哪些代码更改的指示吗？谢谢!

马特

Answer 1

如果您最终使用它，请告诉我。它是工厂运行的 ANSI C，所以应该在所有东西上运行。用法见其他帖子。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

short sudoku[9][9];
unsigned long long cubeSolutions=0;
void* cubeValues[10];
const unsigned char oneLookup[64] = {0x8b, 0x80, 0, 0x80, 0, 0, 0, 0x80, 0, 0,0,0,0,0,0, 0x80, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0x80,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0};

int ifOne(int val) {
  if ( oneLookup[(val-1) >> 3] & (1 << ((val-1) & 0x7))  )
    return val;
  return 0;
}


void init_sudoku() {
  int i,j;
  for (i=0; i<9; i++)
    for (j=0; j<9; j++)
      sudoku[i][j]=0x1ff;
}

void set_sudoku( char* initialValues) {
  int i;
  if ( strlen (initialValues) !=  81 ) {
    printf("Error: inputString should have length=81, length is %2.2d\n", strlen(initialValues) );
    exit (-12);
  }
  for (i=0; i < 81; i++)
    if ((initialValues[i] > 0x30) && (initialValues[i] <= 0x3a))
      sudoku[i/9][i%9] = 1 << (initialValues[i] - 0x31) ;
}

void print_sudoku ( int style ) {
  int i, j, k;
  for (i=0; i < 9; i++) {
    for (j=0; j < 9; j++) {
      if ( ifOne(sudoku[i][j]) || !style) {
        for (k=0; k < 9; k++)
          if (sudoku[i][j] & 1<<k)
            printf("%d", k+1);
      } else
        printf("*");
      if ( !((j+1)%3) )
        printf("\t");
      else
        printf(",");
    }
    printf("\n");
    if (!((i+1) % 3) )
      printf("\n");
  }
}

void print_HTML_sudoku () {
  int i, j, k, l, m;
  printf("<TABLE>\n");
  for (i=0; i<3; i++) {
    printf("  <TR>\n");
    for (j=0; j<3; j++) {
      printf("    <TD><TABLE>\n");
      for (l=0; l<3; l++) { printf("      <TR>"); for (m=0; m<3; m++) { printf("<TD>"); for (k=0; k < 9; k++)  { if (sudoku[i*3+l][j*3+m] & 1<<k)
            printf("%d", k+1);
          }
          printf("</TD>");
        }
        printf("</TR>\n");
      }
    printf("    </TABLE></TD>\n");
    }
    printf("  </TR>\n");
  }
  printf("</TABLE>");
}



int doRow () {
  int count=0, new_value, row_value, i, j;
  for (i=0; i<9; i++) {
    row_value=0x1ff;
    for (j=0; j<9; j++)
      row_value&=~ifOne(sudoku[i][j]);
    for (j=0; j<9; j++) {
      new_value=sudoku[i][j] & row_value;
      if (new_value && (new_value != sudoku[i][j]) ) {
        count++;
        sudoku[i][j] = new_value;
      }
    }
  }
  return count;
}

int doCol () {
  int count=0, new_value, col_value, i, j;
  for (i=0; i<9; i++) {
    col_value=0x1ff;
    for (j=0; j<9; j++)
      col_value&=~ifOne(sudoku[j][i]);
    for (j=0; j<9; j++) {
      new_value=sudoku[j][i] & col_value;
      if (new_value && (new_value != sudoku[j][i]) ) {
        count++;
        sudoku[j][i] = new_value;
      }
    }
  }
  return count;
}

int doCube () {
  int count=0, new_value, cube_value, i, j, l, m;
  for (i=0; i<3; i++)
    for (j=0; j<3; j++) {
      cube_value=0x1ff;
      for (l=0; l<3; l++)
        for (m=0; m<3; m++)
          cube_value&=~ifOne(sudoku[i*3+l][j*3+m]);
      for (l=0; l<3; l++)
        for (m=0; m<3; m++) {
          new_value=sudoku[i*3+l][j*3+m] & cube_value;
          if (new_value && (new_value != sudoku[i*3+l][j*3+m]) ) {
            count++;
            sudoku[i*3+l][j*3+m] = new_value;
          }
        }
    }
  return count;
}

#define FALSE -1
#define TRUE 1
#define INCOMPLETE 0

int validCube () {
  int i, j, l, m, r, c;
  int pigeon;
  int solved=TRUE;

  //check horizontal
  for (i=0; i<9; i++) {
    pigeon=0;
    for (j=0; j<9; j++)
      if (ifOne(sudoku[i][j])) {
        if (pigeon & sudoku[i][j]) return FALSE;
        pigeon |= sudoku[i][j];
      } else {
        solved=INCOMPLETE;
      }
  }

  //check vertical
  for (i=0; i<9; i++) {
    pigeon=0;
    for (j=0; j<9; j++)
      if (ifOne(sudoku[j][i])) {
        if (pigeon & sudoku[j][i]) return FALSE;
        pigeon |= sudoku[j][i];
      }
      else {
        solved=INCOMPLETE;
      }
  }

  //check cube
  for (i=0; i<3; i++)
    for (j=0; j<3; j++) {
      pigeon=0;
      r=j*3; c=i*3;
      for (l=0; l<3; l++)
        for (m=0; m<3; m++)
        if (ifOne(sudoku[r+l][c+m])) {
          if (pigeon & sudoku[r+l][c+m]) return FALSE;
          pigeon |= sudoku[r+l][c+m];
        }
        else {
          solved=INCOMPLETE;
        }
    }

  return solved;
}

int solveSudoku(int position ) {
  int status, i, k;
  short oldCube[9][9];

  for (i=position; i < 81; i++) {

    while ( doCube() + doRow() + doCol() );

    status = validCube() ;
    if ((status == TRUE) || (status == FALSE))
      return status;


    if ((status == INCOMPLETE) && !ifOne(sudoku[i/9][i%9]) ) {
      memcpy( &oldCube, &sudoku, sizeof(short) * 81) ;
      for (k=0; k < 9; k++) {
        if ( sudoku[i/9][i%9] & (1<<k) ) {
          sudoku[i/9][i%9] = 1 << k ;
          if (solveSudoku(i+1) == TRUE ) {

            /* return TRUE; */
            /* Or look for entire set of solutions */

            if (cubeSolutions < 10) {
              cubeValues[cubeSolutions] = malloc ( sizeof(short) * 81 ) ;
              memcpy( cubeValues[cubeSolutions], &sudoku, sizeof(short) * 81) ;
            }

            cubeSolutions++;
            if ((cubeSolutions & 0x3ffff) == 0x3ffff ) {
              printf ("cubeSolutions = %llx\n", cubeSolutions+1 );
            }

            //if ( cubeSolutions > 10 ) 
            //    return TRUE;

          }

          memcpy( &sudoku, &oldCube, sizeof(short) * 81) ;
        }
        if (k==8)
          return FALSE;
      }

    }
  }

  return FALSE;
}


int main ( int argc, char** argv)  {
  int i;
  if (argc != 2) {
    printf("Error: number of arguments on command line is incorrect\n");
    exit (-12);
  }

  init_sudoku();
  set_sudoku(argv[1]);

  printf("[----------------------- Input  Data ------------------------]\n\n");
  print_sudoku(1);

  solveSudoku(0);
  if ((validCube()==1) && !cubeSolutions)  {
    // If sudoku is effectively already solved, cubeSolutions will not be set
    printf ("\n  This is a trivial sudoku. \n\n");
    print_sudoku(1);
  }


  if (!cubeSolutions && validCube()!=1)
    printf("Not Solvable\n");
  if (cubeSolutions > 1) {
    if (cubeSolutions >= 10)
      printf("10+ Solutions, returning first 10 (%lld) [%llx] \n", cubeSolutions, cubeSolutions);
    else
      printf("%llx Solutions. \n", cubeSolutions);
  }

  for (i=0; (i < cubeSolutions) && (i < 10); i++) {
    memcpy ( &sudoku, cubeValues[i], sizeof(short) * 81 );
    printf("[----------------------- Solution %2.2d ------------------------]\n\n", i+1);
    print_sudoku(0);
    //print_HTML_sudoku();
  }
  return 0;
}