Maximum PC

/* an example of basic Hill Climbing, constrained by a step maximum
Adak, June 16, 2010
status: OK
*/

#include <stdio.h>
#include <conio.h>
#include <dos.h>

#define ROWS 8

struct point {
  int row;
  int col;
  int value;
}p1;

void info(void);

int main() {
  int i, j, step, max, last_value, new_value;
  int x, y, x1, y1;
  int hill[ROWS][ROWS] = {      //correct local maxima is at hill[0][7]
  { 0,  1,  0,  0,  0,  0,  0, 17 },
  { 3,  1,  2,  0,  0,  0, 16,  0 },
  { 3,  3,  0,  8, 20, 15,  0,  0 },
  { 0,  4,  5,  9,  0,  8, 14,  0 },
  { 0,  0,  6,  7,  9, 13,  8,  0 },
  { 0,  0,  8,  0, 12,  8,  8,  0 },
  { 0,  9,  8, 11,  0,  0,  0,  0 },
  { 0,  0, 10,  0,  0,  0,  0,  0 }
  };
  struct point p2;
  step=1;
  printf("\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n");
  info();
  p2 = p1;
  gotoxy(1,5);
  textcolor(14);  //affects only cprintf() "direct video" output
  printf("\n    0   1   2   3   4   5   6   7");
  printf("\n    =============================\n");

  for(i=0;i<ROWS;i++) {
    printf("%d: ", i);
    for(j=0;j<ROWS;j++) {
      if(i==0 && j==0) {
        printf("%2d",hill[i][j]);
        cprintf("* ");
      }
      else
        printf("%2d  ",hill[i][j]);
    }
    putchar('\n');
  }
  printf("\n From row 0, column 0, climb to:");
  do {
    p1.value = hill[p1.row][p1.col];
    last_value = p1.value;
    max = p1.value + step;
    for(i=0; i<8; i++) {
      if(i==0 && p1.row > 0) {                    //12 o'clock
        p2.row = p1.row -1;
      }
      else if(i==1 && p1.row > 0) {               //1:30 o'clock
        p2.row = p1.row -1;
        p2.col = p1.col +1; 
      }
      else if(i==2 && p1.col < ROWS -1)           //3 o'clock
        p2.col = p1.col +1;
      else if(i==3 && p1.row < ROWS -1 && p1.col < ROWS -1) { //4:30 o'clock
        p2.row = p1.row +1;
        p2.col = p1.col +1;
      }
      else if(i==4 && p1.row < ROWS -1)           //6 o'clock
        p2.row = p1.row +1;
      else if(i==5 && p1.row < ROWS -1 && p1.col > 0) {  //7:30 o'clock
        p2.row = p1.row +1;
        p2.col = p1.col -1;
      }
      else if(i==6 && p1.col > 0)                  //9 o'clock
        p2.col = p1.col -1;
      else if(i==7 && p1.row > 0 && p1.col > 0) {  //10:30 o'clock
        p2.row = p1.row -1;
        p2.col = p1.col -1;
      }
      p2.value = hill[p2.row][p2.col];
      if(p2.value > p1.value && p2.value <= max) {
        p1 = p2;
        break;
      }   
      else
        p2 = p1;
    }
    new_value = p1.value;
    if(new_value > last_value) {
      x1 = wherex();
      y1 = wherey();
      x=6+(p1.col*4);
      y=8+p1.row;
      gotoxy(x,y);
      cprintf("*");
      gotoxy(x1,y1);
      if(new_value % 2)
        putchar('\n');
      printf("  >> row: %d, col: %d, value: %2d", p1.row,p1.col,p1.value);
    }
    sleep(1); //pause for 1 second
  }while(new_value > last_value);
  printf("\n\n\t\t\t     press enter when ready");

  (void) getchar();
  return 0;
}
void info(void) {
  int i;
  printf("  An example of Hill Climbing, in an array. From the start at\n");
  printf("  the top left row and column, each step must be:\n\n");
  printf("    * to a higher value, and \n    * no more than the current value + 1\n");
  printf("\n  This algorithm finds the local maximal value, which may or may not\n\
  be the greatest value in the array.");
  printf("  Here, it misses the highest\n  value of 20, near the center of the array.\n");
  printf("\n\n\n\n\n\n\n\n\n\n\n\t\t\t     press enter when ready\n");
  (void) getchar();
  for(i=0;i<4;i++)
    printf("\n\n\n\n\n\n\n\n\n\n\n\n");
}

a..bc
de..f
.g.hi
.jkl.

start(a).
goal(c).
goal(f).
goal(i).

connected(a,d).
connected(d,e).
connected(e,g).
connected(g,j).
connected(j,k).
connected(k,l).
connected(l,h).
connected(h,i).
connected(i,f).
connected(f,c).
connected(c,b).

path(A,B):-connected(A,B).
path(A,B):-connected(A,C),path(C,B).

[1] 1 ?- [maze],
[maze].
% maze compiled 0.00 sec, 1,764 bytes
true.

[1] 2 ?- start(X).
X = a.

[1] 3 ?- goal(X).
X = c ;
X = f ;
X = i.

[1] 4 ?- path(a,c).
true .

[1] 5 ?- path(a,f).
true .

[1] 6 ?- path(a,i).
true .