Presenter of peg-solitaire solution from 2⁽n-3) length solution file

sol2.c

/* gcc -O6 -Wall -pedantic -Wextra sol2.c -Wno-long-long -o sol2 */
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <inttypes.h>
#include <unistd.h>


char B[9][9]={
#if 0
".........",  /* English */
"...ooo...",
"...ooo...",
".ooooooo.",
".oooxooo.",
".ooooooo.",
"...ooo...",
"...ooo...",
"........."
#elif 0
".........",  /* French */
"...ooo...",
"..ooooo..",
".oooxooo.",
".ooooooo.",
".ooooooo.",
"..ooooo..",
"...ooo...",
"........."
#else
"...ooo...",  /* 3-3-2-2 */
"...ooo...",
"...ooo...",
".oooooooo",
".oooxoooo",
".oooooooo",
"...ooo...",
"...ooo...",
"........."
#endif
};

int N[9][9];

int m,cnt;
uint64_t M[7*9*2*2];
uint64_t D[7*9*2*2];
uint64_t P[7*9*2*2];

void draw(uint64_t u, int p)
{
  int i,j;
  printf("\x1b[2J");
  for(i=0; i<9; ++i)
  {
    for(j=0; j<9; ++j)
      printf("%s", N[i][j]==255 ? "." : (u & (1LL<<N[i][j])) ? 
                   ((N[i][j]==p) ? "\x1b[7mo\x1b[0m" : "o") : " ");
    printf("\n");
  }
  usleep(p==255 ? 900000 : 600000);
}

uint8_t fgtc(FILE *src, uint64_t u)
{
  int c;
  rewind(src);
  assert(0 == fseek(src, u, SEEK_SET));
  c = fgetc(src);
  assert(c != EOF);
  return c;
}

int main(int argc, char *argv[])
{
  uint64_t u;
  FILE *src;

  int i,d;
  uint64_t j,n;

  assert(argc==2);
  sscanf(argv[1],"%"SCNx64,&u);

  m=0;
  for(i=0; i<9; ++i)
  {
    for(j=0; j<9; ++j)
    {
      N[i][j] = (B[i][j]!='.') ? cnt++ : 255;

      if (i>1)
        if (N[i][j]!=255 && N[i-1][j]!=255 && N[i-2][j]!=255)
        {
          M[m]   = (1LL<<N[i-2][j]);
          P[m]   = N[i][j];
          D[m++] = (1LL<<N[i][j]) | (1LL<<N[i-1][j]);
          M[m]   = (1LL<<N[i][j]);
          P[m]   = N[i-2][j];
          D[m++] = (1LL<<N[i-1][j]) | (1LL<<N[i-2][j]);
        }  

      if (j>1)
        if (N[i][j]!=255 && N[i][j-1]!=255 && N[i][j-2]!=255)
        {
          M[m]   = (1LL<<N[i][j-2]);
          P[m]   = N[i][j];
          D[m++] = (1LL<<N[i][j]) | (1LL<<N[i][j-1]);
          M[m]   = (1LL<<N[i][j]);
          P[m]   = N[i][j-2];
          D[m++] = (1LL<<N[i][j-1]) | (1LL<<N[i][j-2]);
        }  
    }
  }

  assert( (src=fopen("3-3-2-2","rb")) );

  i = fgtc(src, u>>3);
  assert(i & 1<<(u%8));

  for(;;)
  {
/*    printf("%"PRIx64"\n",u); */
    draw(u, 255);

    n=0;
    for(d=0; d<m; ++d)
    {
      if (!(u & M[d]) && ((u & D[d]) == D[d]))
      {
        n = u^(M[d]|D[d]);
        
        i = fgtc(src, n>>3);
        if (i & 1<<(n%8))
          break;
      }
    }

    if (n==0)
      break;

    draw(u, P[d]);

    u=n;
  }

  return 0;
}