/* OTHELLO -- The Game of Dramatic Reversals written by Bert Halstead modified for BDS C by Leor Zolman This program is a good example of: a) structured, heirarchical function organization b) arrays as formal parameters c) use of the "qsort" library function Object of the game is for two players to alternate placing their marker someplace on an 8 by 8 grid, so that at least one of the opponent's pieces becomes surrounded by the moving player's peices -- causing the flanked pieces to flip 'color' and belong to the moving player. After 60 moves have been played (or if no player has a legal move left), the player with the most of his own pieces on the board wins. The playing pieces are '*' and '@'. You may choose to play either '*' or '@' for the first game; thereafter, you and the computer will alternate going first for each game. Whoever goes first always plays `*'. You enter a move as a two digit number, each digit being from 1 to 8, first digit representing row and second representing column. For example: if playing '*', your first move might be '46', meaning 4th row down, 6th position across. As an alternative to entering a move, one of the following commands may be typed: g causes computer to play both sides until game is over a causes computer to print out an analysis of each of your possible moves. A letter from A to Z will appear at each of your legal move positions, where A is the machine's opinion of an excellant move and Z is a real loser. hn sets handicap. n is 1,2,3, or 4. If n is positive, gives n free pieces to the computer. If n is negative, gives YOU the free peices. f forfeit the current move. This happens automatically if you have no legal moves. q quit the current game. b prints out board again. s prints out the score, and tells who is winning. */ #define BLACK '*' #define WHITE '@' #define EMPTY '-' int handicap; char selfplay; /* true if computer playing with itself */ int h[4][2]; /* handicap position table */ char mine, his; /* who has black (*) and white (@) in current game */ char mefirst; /* true if computer goes first in current game */ main(argc,argv) int argc; char **argv; { char b[8][8]; int i; h[0][0] = h[0][1] = h[2][0] = h[3][1] = 0; h[1][0] = h[1][1] = h[2][1] = h[3][0] = 7; printf("\nWelcome to the BDS C OTHELLO program!\n"); printf("\nNote: `*' always goes first...Good luck!!!\n\n"); srand1("Do you want to go first? "); if (toupper(getchar()) == 'Y') mefirst = 0; else mefirst = 1; printf("\n\n"); do { clrbrd(b); prtbrd(b); i = game(b,4); mefirst = !mefirst; if (i==4) break; if (i=='Q') continue; printf("\n"); i = prtscr(b); if (i>0) printf(" You won by %d\n",i); else if (i<0) printf(" You lost by %d\n",-i); else printf(" A draw\n"); } while (ask("Another game? ")=='Y'); } game(b,n) char b[8][8]; int n; { char c; int ff; int i,j; handicap = 0; selfplay = ' '; ff=0; if (mefirst) { mine = BLACK; his = WHITE; printf("\nI go first:\n\n"); } else { mine = WHITE; his = BLACK; printf("\nYou go first:\n\n"); } while(1) { if (cntbrd(b,EMPTY)==0) return 'D'; if (cntbrd(b,EMPTY)==60 && mine == BLACK) goto Istart; if (chkmvs(b,his)==0) { printf(!mefirst ? "Forfeit" : " ...Forfeit\n"); ff |= 1; } else switch (c = getmov(&i,&j)) { case 'B': prtbrd(b); continue; case 'S': i= prtscr(b); if (i>0) printf(" You're winning\n"); else if (i<0)printf(" You're losing!\n"); else putchar('\n'); continue; case 'Q': case 4: return c; case 'H': if (n>abs(handicap)+4) printf("Illegal!\n"); else for (j=0; i!=0; j++) { b[h[j][0]][h[j][1]]= i>0?BLACK:WHITE; handicap += i>0 ? 1 : -1; ++n; i += i>0 ? -1 : 1; } prtbrd(b); continue; case 'A': analyze(b,his,mine,EMPTY); continue; case 'G': my_mov(b,his,mine,EMPTY,&i,&j); case 'M': if (chkmov(b,his,i,j)>0) { printf(!mefirst ? "%1d-%1d" : " ...%1d-%1d\n", i+1,j+1); putmov(b,his,i,j); } else { printf("Illegal!\n"); continue; } break; case 'F': if (n>abs(handicap)+4) { printf ("Illegal!\n"); continue; } else printf(!mefirst ? "Forfeit" : " ...Forfeit\n"); } Istart: if (cntbrd(b,EMPTY) == 0) return 'D'; if (chkmvs(b,mine)==0) { printf(!mefirst ? "...Forfeit\n": "Forfeit...\n"); ff |=2; } else { my_mov(b,mine,his,EMPTY,&i,&j); printf(!mefirst ? "...%1d-%1d\n" : "%1d-%1d...\n", i+1,j+1); putmov(b,mine,i,j); ++n; } if (ff==3 || n>64) return 'D'; if (!(ff & 1)) prtbrd(b); ff = 0; } } prtscr(b) char *b; { int i,j; printf("%1d-%1d",i = cntbrd(b,his), j=cntbrd(b,mine)); return i-j; } char getmov(i,j) int *i, *j; { char a,c; int n; char *p; char skipbl(); if (selfplay == 'G') { if (!kbhit()) return 'G'; selfplay = ' '; getchar(); } printf("Move: "); while(1) switch (c=skipbl()) { case '\n': printf("Move? "); continue; case 'G': if ((c = skipbl()) != '\n') goto flush; selfplay='G'; return 'G'; case 'B': case 'S': case 'Q': case 'F': case 'A': a=c; if (( c = skipbl()) != '\n') goto flush; return a; case 'H': if ((a=c=skipbl()) == EMPTY) c=getchar(); if (c<'1' || c>'4' || skipbl() !='\n') goto flush; *i = a==EMPTY? -(c-'0') : (c-'0'); return 'H'; case 4: return c; default: if (c<'1' || c>'8') goto flush; *i = c-'1'; c = skipbl(); if (c<'1' || c>'8') goto flush; *j = c- '1'; if ((c=skipbl()) == '\n') return 'M'; flush: while (c != '\n' && c != 4) c=getchar(); if (c==4) return c; printf ("Huh?? "); } } char ask(s) char *s; { char a,c; printf ("%s ",s); a=skipbl(); while (c != '\n' && c != 4) c= getchar(); return a; } char skipbl() { char c; while ((c = toupper(getchar())) == ' ' || c=='\t'); return c; } chkmvs(b,p) char b[8][8]; char p; { int i,j,k; k=0; for (i=0; i<8; i++) for (j=0; j<8; j++) k += chkmov(b,p,i,j); return k; } chkmov(b,p,x,y) char b[8][8],p; int x,y; { if (b[x][y] != EMPTY) return 0; return chkmv1(b,p,x,y,0,1) + chkmv1(b,p,x,y,1,0) + chkmv1(b,p,x,y,0,-1)+ chkmv1(b,p,x,y,-1,0)+ chkmv1(b,p,x,y,1,1) + chkmv1(b,p,x,y,1,-1)+ chkmv1(b,p,x,y,-1,1)+ chkmv1(b,p,x,y,-1,-1); } chkmv1(b,p,x,y,m,n) char b[8][8],p; int x,y,m,n; { int k; k=0; while ((x += m) >= 0 && x < 8 && (y += n) >= 0 && y<8) { if (b[x][y]==EMPTY) return 0; if (b[x][y]== p ) return k; if (x==0 || x==7 || y==0 || y==7) k += 10; else k++; } return 0; } notake(b,p,o,e,x,y) char b[8][8]; char p,o,e; int x,y; { return notak1(b,p,o,e,x,y,0,1)&& notak1(b,p,o,e,x,y,1,1)&& notak1(b,p,o,e,x,y,1,0)&& notak1(b,p,o,e,x,y,1,-1); } notak1(b,p,o,e,x,y,m,n) char b[8][8],p,o,e; int x,y,m,n; { int c1,c2; c1 = notak2(b,p,o,e,x,y,m,n); c2 = notak2(b,p,o,e,x,y,-m,-n); return !(c1==o && c2==e || c1==e && c2==o); } notak2(b,p,o,e,x,y,m,n) char b[8][8],p,o,e; int x,y,m,n; { x += m; y +=n; if (x>=0 && x<=7 && y>=0 && y<=7) while(b[x][y] == 0) { x += m; y+=n; if (x<0 || x>7 || y<0 || y>7 || b[x][y]==e) return o; } while (x>=0 && x<=7 && y>=0 && y<=7 && b[x][y]==p) { x +=m; y+=n; } if (x<0 || x>7 || y<0 || y>7) return p; return b[x][y]; } putmov(b,p,x,y) char b[8][8]; char p; int x,y; { int i,j; b[x][y] = p; for (i= -1; i<=1; i++) for (j= -1; j<=1; j++) { if ((i != 0 || j!=0)&&chkmv1(b,p,x,y,i,j)>0) putmv1(b,p,x,y,i,j); } } putmv1(b,p,x,y,m,n) char b[8][8]; char p; int x,y,m,n; { while ((x += m) >= 0 && x<8 && (y += n)>=0 && y<8) { if (b[x][y] == EMPTY || b[x][y] == p) return; b[x][y] = p; } } struct mt { int x; int y; int c; int s; }; my_mov(b,p,o,e,m,n) char b[8][8],p; int *m, *n; { struct mt t[64]; int i,k; int cmpmov(); k = fillmt(b,p,o,e,t); if (!k) return 0; qsort (&t, k, 8, &cmpmov); for (i=1; i=10) { s -= 4; oside |= 8; } } if (s< -oside) s= -oside; if (side>0) return s+side-7+10*ok; if (i==1 || i==6) {s--; side++;} if (j==1 || j==6) {s--; side++;} if (side>0) return s; if (i==2 || i==5) s++; if (j==2 || j==5) s++; return s; } cmpmov(a,b) struct mt *a, *b; { if ((*a).s > (*b).s) return -1; if ((*a).s < (*b).s) return 1; if ((*a).c > (*b).c) return -1; if ((*a).c < (*b).c) return 1; return 0; } clrbrd(b) char b[8][8]; { int i,j; for (i=0; i<8; i++) for (j=0; j<8; j++) b[i][j]= EMPTY; b[3][3] = b[4][4] = BLACK; b[3][4] = b[4][3] = WHITE; } prtbrd(b) char b[8][8]; { int i,j; printf(" 1 2 3 4 5 6 7 8\n"); for (i=0; i<8; i++) { printf("%2d",i+1); for (j=0; j<8; j++) { putchar(' '); putchar(b[i][j]); } putchar('\n'); } putchar('\n'); } cpybrd(a,b) char *a, *b; { int i; i=64; while (i--) *a++ = *b++; } cntbrd(b,p) char *b, p; { int i,j; i= 64; j=0; while (i--) if (*b++ == p) ++j; return (j); }