enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF2, SYMM_REF2D, SYMM_REF4,
SYMM_REF4D, SYMM_REF8 };
-enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT,
- DIFF_SET, DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
+enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, DIFF_SET, DIFF_EXTREME,
+ DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
enum {
COL_BACKGROUND,
{ "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE } },
{ "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT } },
{ "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET } },
+ { "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME } },
{ "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE } },
#ifndef SLOW_SYSTEM
{ "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } },
string++, ret->diff = DIFF_INTERSECT;
else if (*string == 'a') /* advanced */
string++, ret->diff = DIFF_SET;
+ else if (*string == 'e') /* extreme */
+ string++, ret->diff = DIFF_EXTREME;
else if (*string == 'u') /* unreasonable */
string++, ret->diff = DIFF_RECURSIVE;
} else
case DIFF_SIMPLE: strcat(str, "db"); break;
case DIFF_INTERSECT: strcat(str, "di"); break;
case DIFF_SET: strcat(str, "da"); break;
+ case DIFF_EXTREME: strcat(str, "de"); break;
case DIFF_RECURSIVE: strcat(str, "du"); break;
}
}
ret[3].name = "Difficulty";
ret[3].type = C_CHOICES;
- ret[3].sval = ":Trivial:Basic:Intermediate:Advanced:Unreasonable";
+ ret[3].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable";
ret[3].ival = params->diff;
ret[4].name = NULL;
/* ----------------------------------------------------------------------
* Solver.
*
- * This solver is used for several purposes:
- * + to generate filled grids as the basis for new puzzles (by
- * supplying no clue squares at all)
+ * This solver is used for two purposes:
* + to check solubility of a grid as we gradually remove numbers
* from it
* + to solve an externally generated puzzle when the user selects
* the numbers' possible positions (or the spaces' possible
* contents).
*
+ * - Mutual neighbour elimination: find two squares A,B and a
+ * number N in the possible set of A, such that putting N in A
+ * would rule out enough possibilities from the mutual
+ * neighbours of A and B that there would be no possibilities
+ * left for B. Thereby rule out N in A.
+ * + The simplest case of this is if B has two possibilities
+ * (wlog {1,2}), and there are two mutual neighbours of A and
+ * B which have possibilities {1,3} and {2,3}. Thus, if A
+ * were to be 3, then those neighbours would contain 1 and 2,
+ * and hence there would be nothing left which could go in B.
+ * + There can be more complex cases of it too: if A and B are
+ * in the same column of large blocks, then they can have
+ * more than two mutual neighbours, some of which can also be
+ * neighbours of one another. Suppose, for example, that B
+ * has possibilities {1,2,3}; there's one square P in the
+ * same column as B and the same block as A, with
+ * possibilities {1,4}; and there are _two_ squares Q,R in
+ * the same column as A and the same block as B with
+ * possibilities {2,3,4}. Then if A contained 4, P would
+ * contain 1, and Q and R would have to contain 2 and 3 in
+ * _some_ order; therefore, once again, B would have no
+ * remaining possibilities.
+ *
* - Recursion. If all else fails, we pick one of the currently
* most constrained empty squares and take a random guess at its
* contents, then continue solving on that basis and see if we
struct solver_scratch {
unsigned char *grid, *rowidx, *colidx, *set;
+ int *neighbours, *bfsqueue;
+#ifdef STANDALONE_SOLVER
+ int *bfsprev;
+#endif
};
static int solver_set(struct solver_usage *usage,
return 0;
}
+/*
+ * Try to find a number in the possible set of (x1,y1) which can be
+ * ruled out because it would leave no possibilities for (x2,y2).
+ */
+static int solver_mne(struct solver_usage *usage,
+ struct solver_scratch *scratch,
+ int x1, int y1, int x2, int y2)
+{
+ int c = usage->c, r = usage->r, cr = c*r;
+ int *nb[2];
+ unsigned char *set = scratch->set;
+ unsigned char *numbers = scratch->rowidx;
+ unsigned char *numbersleft = scratch->colidx;
+ int nnb, count;
+ int i, j, n, nbi;
+
+ nb[0] = scratch->neighbours;
+ nb[1] = scratch->neighbours + cr;
+
+ /*
+ * First, work out the mutual neighbour squares of the two. We
+ * can assert that they're not actually in the same block,
+ * which leaves two possibilities: they're in different block
+ * rows _and_ different block columns (thus their mutual
+ * neighbours are precisely the other two corners of the
+ * rectangle), or they're in the same row (WLOG) and different
+ * columns, in which case their mutual neighbours are the
+ * column of each block aligned with the other square.
+ *
+ * We divide the mutual neighbours into two separate subsets
+ * nb[0] and nb[1]; squares in the same subset are not only
+ * adjacent to both our key squares, but are also always
+ * adjacent to one another.
+ */
+ if (x1 / r != x2 / r && y1 % r != y2 % r) {
+ /* Corners of the rectangle. */
+ nnb = 1;
+ nb[0][0] = cubepos(x2, y1, 1);
+ nb[1][0] = cubepos(x1, y2, 1);
+ } else if (x1 / r != x2 / r) {
+ /* Same row of blocks; different blocks within that row. */
+ int x1b = x1 - (x1 % r);
+ int x2b = x2 - (x2 % r);
+
+ nnb = r;
+ for (i = 0; i < r; i++) {
+ nb[0][i] = cubepos(x2b+i, y1, 1);
+ nb[1][i] = cubepos(x1b+i, y2, 1);
+ }
+ } else {
+ /* Same column of blocks; different blocks within that column. */
+ int y1b = y1 % r;
+ int y2b = y2 % r;
+
+ assert(y1 % r != y2 % r);
+
+ nnb = c;
+ for (i = 0; i < c; i++) {
+ nb[0][i] = cubepos(x2, y1b+i*r, 1);
+ nb[1][i] = cubepos(x1, y2b+i*r, 1);
+ }
+ }
+
+ /*
+ * Right. Now loop over each possible number.
+ */
+ for (n = 1; n <= cr; n++) {
+ if (!cube(x1, y1, n))
+ continue;
+ for (j = 0; j < cr; j++)
+ numbersleft[j] = cube(x2, y2, j+1);
+
+ /*
+ * Go over every possible subset of each neighbour list,
+ * and see if its union of possible numbers minus n has the
+ * same size as the subset. If so, add the numbers in that
+ * subset to the set of things which would be ruled out
+ * from (x2,y2) if n were placed at (x1,y1).
+ */
+ memset(set, 0, nnb);
+ count = 0;
+ while (1) {
+ /*
+ * Binary increment: change the rightmost 0 to a 1, and
+ * change all 1s to the right of it to 0s.
+ */
+ i = nnb;
+ while (i > 0 && set[i-1])
+ set[--i] = 0, count--;
+ if (i > 0)
+ set[--i] = 1, count++;
+ else
+ break; /* done */
+
+ /*
+ * Examine this subset of each neighbour set.
+ */
+ for (nbi = 0; nbi < 2; nbi++) {
+ int *nbs = nb[nbi];
+
+ memset(numbers, 0, cr);
+
+ for (i = 0; i < nnb; i++)
+ if (set[i])
+ for (j = 0; j < cr; j++)
+ if (j != n-1 && usage->cube[nbs[i] + j])
+ numbers[j] = 1;
+
+ for (i = j = 0; j < cr; j++)
+ i += numbers[j];
+
+ if (i == count) {
+ /*
+ * Got one. This subset of nbs, in the absence
+ * of n, would definitely contain all the
+ * numbers listed in `numbers'. Rule them out
+ * of `numbersleft'.
+ */
+ for (j = 0; j < cr; j++)
+ if (numbers[j])
+ numbersleft[j] = 0;
+ }
+ }
+ }
+
+ /*
+ * If we've got nothing left in `numbersleft', we have a
+ * successful mutual neighbour elimination.
+ */
+ for (j = 0; j < cr; j++)
+ if (numbersleft[j])
+ break;
+
+ if (j == cr) {
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ printf("%*smutual neighbour elimination, (%d,%d) vs (%d,%d):\n",
+ solver_recurse_depth*4, "",
+ 1+x1, 1+YUNTRANS(y1), 1+x2, 1+YUNTRANS(y2));
+ printf("%*s ruling out %d at (%d,%d)\n",
+ solver_recurse_depth*4, "",
+ n, 1+x1, 1+YUNTRANS(y1));
+ }
+#endif
+ cube(x1, y1, n) = FALSE;
+ return +1;
+ }
+ }
+
+ return 0; /* nothing found */
+}
+
+/*
+ * Look for forcing chains. A forcing chain is a path of
+ * pairwise-exclusive squares (i.e. each pair of adjacent squares
+ * in the path are in the same row, column or block) with the
+ * following properties:
+ *
+ * (a) Each square on the path has precisely two possible numbers.
+ *
+ * (b) Each pair of squares which are adjacent on the path share
+ * at least one possible number in common.
+ *
+ * (c) Each square in the middle of the path shares _both_ of its
+ * numbers with at least one of its neighbours (not the same
+ * one with both neighbours).
+ *
+ * These together imply that at least one of the possible number
+ * choices at one end of the path forces _all_ the rest of the
+ * numbers along the path. In order to make real use of this, we
+ * need further properties:
+ *
+ * (c) Ruling out some number N from the square at one end
+ * of the path forces the square at the other end to
+ * take number N.
+ *
+ * (d) The two end squares are both in line with some third
+ * square.
+ *
+ * (e) That third square currently has N as a possibility.
+ *
+ * If we can find all of that lot, we can deduce that at least one
+ * of the two ends of the forcing chain has number N, and that
+ * therefore the mutually adjacent third square does not.
+ *
+ * To find forcing chains, we're going to start a bfs at each
+ * suitable square, once for each of its two possible numbers.
+ */
+static int solver_forcing(struct solver_usage *usage,
+ struct solver_scratch *scratch)
+{
+ int c = usage->c, r = usage->r, cr = c*r;
+ int *bfsqueue = scratch->bfsqueue;
+#ifdef STANDALONE_SOLVER
+ int *bfsprev = scratch->bfsprev;
+#endif
+ unsigned char *number = scratch->grid;
+ int *neighbours = scratch->neighbours;
+ int x, y;
+
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++) {
+ int count, t, n;
+
+ /*
+ * If this square doesn't have exactly two candidate
+ * numbers, don't try it.
+ *
+ * In this loop we also sum the candidate numbers,
+ * which is a nasty hack to allow us to quickly find
+ * `the other one' (since we will shortly know there
+ * are exactly two).
+ */
+ for (count = t = 0, n = 1; n <= cr; n++)
+ if (cube(x, y, n))
+ count++, t += n;
+ if (count != 2)
+ continue;
+
+ /*
+ * Now attempt a bfs for each candidate.
+ */
+ for (n = 1; n <= cr; n++)
+ if (cube(x, y, n)) {
+ int orign, currn, head, tail;
+
+ /*
+ * Begin a bfs.
+ */
+ orign = n;
+
+ memset(number, cr+1, cr*cr);
+ head = tail = 0;
+ bfsqueue[tail++] = y*cr+x;
+#ifdef STANDALONE_SOLVER
+ bfsprev[y*cr+x] = -1;
+#endif
+ number[y*cr+x] = t - n;
+
+ while (head < tail) {
+ int xx, yy, nneighbours, xt, yt, xblk, i;
+
+ xx = bfsqueue[head++];
+ yy = xx / cr;
+ xx %= cr;
+
+ currn = number[yy*cr+xx];
+
+ /*
+ * Find neighbours of yy,xx.
+ */
+ nneighbours = 0;
+ for (yt = 0; yt < cr; yt++)
+ neighbours[nneighbours++] = yt*cr+xx;
+ for (xt = 0; xt < cr; xt++)
+ neighbours[nneighbours++] = yy*cr+xt;
+ xblk = xx - (xx % r);
+ for (yt = yy % r; yt < cr; yt += r)
+ for (xt = xblk; xt < xblk+r; xt++)
+ neighbours[nneighbours++] = yt*cr+xt;
+
+ /*
+ * Try visiting each of those neighbours.
+ */
+ for (i = 0; i < nneighbours; i++) {
+ int cc, tt, nn;
+
+ xt = neighbours[i] % cr;
+ yt = neighbours[i] / cr;
+
+ /*
+ * We need this square to not be
+ * already visited, and to include
+ * currn as a possible number.
+ */
+ if (number[yt*cr+xt] <= cr)
+ continue;
+ if (!cube(xt, yt, currn))
+ continue;
+
+ /*
+ * Don't visit _this_ square a second
+ * time!
+ */
+ if (xt == xx && yt == yy)
+ continue;
+
+ /*
+ * To continue with the bfs, we need
+ * this square to have exactly two
+ * possible numbers.
+ */
+ for (cc = tt = 0, nn = 1; nn <= cr; nn++)
+ if (cube(xt, yt, nn))
+ cc++, tt += nn;
+ if (cc == 2) {
+ bfsqueue[tail++] = yt*cr+xt;
+#ifdef STANDALONE_SOLVER
+ bfsprev[yt*cr+xt] = yy*cr+xx;
+#endif
+ number[yt*cr+xt] = tt - currn;
+ }
+
+ /*
+ * One other possibility is that this
+ * might be the square in which we can
+ * make a real deduction: if it's
+ * adjacent to x,y, and currn is equal
+ * to the original number we ruled out.
+ */
+ if (currn == orign &&
+ (xt == x || yt == y ||
+ (xt / r == x / r && yt % r == y % r))) {
+#ifdef STANDALONE_SOLVER
+ if (solver_show_working) {
+ char *sep = "";
+ int xl, yl;
+ printf("%*sforcing chain, %d at ends of ",
+ solver_recurse_depth*4, "", orign);
+ xl = xx;
+ yl = yy;
+ while (1) {
+ printf("%s(%d,%d)", sep, 1+xl,
+ 1+YUNTRANS(yl));
+ xl = bfsprev[yl*cr+xl];
+ if (xl < 0)
+ break;
+ yl = xl / cr;
+ xl %= cr;
+ sep = "-";
+ }
+ printf("\n%*s ruling out %d at (%d,%d)\n",
+ solver_recurse_depth*4, "",
+ orign, 1+xt, 1+YUNTRANS(yt));
+ }
+#endif
+ cube(xt, yt, orign) = FALSE;
+ return 1;
+ }
+ }
+ }
+ }
+ }
+
+ return 0;
+}
+
static struct solver_scratch *solver_new_scratch(struct solver_usage *usage)
{
struct solver_scratch *scratch = snew(struct solver_scratch);
scratch->rowidx = snewn(cr, unsigned char);
scratch->colidx = snewn(cr, unsigned char);
scratch->set = snewn(cr, unsigned char);
+ scratch->neighbours = snewn(3*cr, int);
+ scratch->bfsqueue = snewn(cr*cr, int);
+#ifdef STANDALONE_SOLVER
+ scratch->bfsprev = snewn(cr*cr, int);
+#endif
return scratch;
}
static void solver_free_scratch(struct solver_scratch *scratch)
{
+#ifdef STANDALONE_SOLVER
+ sfree(scratch->bfsprev);
+#endif
+ sfree(scratch->bfsqueue);
+ sfree(scratch->neighbours);
sfree(scratch->set);
sfree(scratch->colidx);
sfree(scratch->rowidx);
struct solver_usage *usage;
struct solver_scratch *scratch;
int cr = c*r;
- int x, y, n, ret;
+ int x, y, x2, y2, n, ret;
int diff = DIFF_BLOCK;
/*
}
/*
+ * Row-vs-column set elimination on a single number.
+ */
+ for (n = 1; n <= cr; n++) {
+ ret = solver_set(usage, scratch, cubepos(0,0,n), cr*cr, cr
+#ifdef STANDALONE_SOLVER
+ , "positional set elimination, number %d", n
+#endif
+ );
+ if (ret < 0) {
+ diff = DIFF_IMPOSSIBLE;
+ goto got_result;
+ } else if (ret > 0) {
+ diff = max(diff, DIFF_EXTREME);
+ goto cont;
+ }
+ }
+
+ /*
+ * Mutual neighbour elimination.
+ */
+ for (y = 0; y+1 < cr; y++) {
+ for (x = 0; x+1 < cr; x++) {
+ for (y2 = y+1; y2 < cr; y2++) {
+ for (x2 = x+1; x2 < cr; x2++) {
+ /*
+ * Can't do mutual neighbour elimination
+ * between elements of the same actual
+ * block.
+ */
+ if (x/r == x2/r && y%r == y2%r)
+ continue;
+
+ /*
+ * Otherwise, try (x,y) vs (x2,y2) in both
+ * directions, and likewise (x2,y) vs
+ * (x,y2).
+ */
+ if (!usage->grid[YUNTRANS(y)*cr+x] &&
+ !usage->grid[YUNTRANS(y2)*cr+x2] &&
+ (solver_mne(usage, scratch, x, y, x2, y2) ||
+ solver_mne(usage, scratch, x2, y2, x, y))) {
+ diff = max(diff, DIFF_EXTREME);
+ goto cont;
+ }
+ if (!usage->grid[YUNTRANS(y)*cr+x2] &&
+ !usage->grid[YUNTRANS(y2)*cr+x] &&
+ (solver_mne(usage, scratch, x2, y, x, y2) ||
+ solver_mne(usage, scratch, x, y2, x2, y))) {
+ diff = max(diff, DIFF_EXTREME);
+ goto cont;
+ }
+ }
+ }
+ }
+ }
+
+ /*
+ * Forcing chains.
+ */
+ if (solver_forcing(usage, scratch)) {
+ diff = max(diff, DIFF_EXTREME);
+ goto cont;
+ }
+
+ /*
* If we reach here, we have made no deductions in this
* iteration, so the algorithm terminates.
*/
int nlocs;
char *desc;
int coords[16], ncoords;
- int *symmclasses, nsymmclasses;
- int maxdiff, recursing;
+ int maxdiff;
+ int x, y, i, j;
/*
* Adjust the maximum difficulty level to be consistent with
grid2 = snewn(area, digit);
/*
- * Find the set of equivalence classes of squares permitted
- * by the selected symmetry. We do this by enumerating all
- * the grid squares which have no symmetric companion
- * sorting lower than themselves.
- */
- nsymmclasses = 0;
- symmclasses = snewn(cr * cr, int);
- {
- int x, y;
-
- for (y = 0; y < cr; y++)
- for (x = 0; x < cr; x++) {
- int i = y*cr+x;
- int j;
-
- ncoords = symmetries(params, x, y, coords, params->symm);
- for (j = 0; j < ncoords; j++)
- if (coords[2*j+1]*cr+coords[2*j] < i)
- break;
- if (j == ncoords)
- symmclasses[nsymmclasses++] = i;
- }
- }
-
- /*
* Loop until we get a grid of the required difficulty. This is
* nasty, but it seems to be unpleasantly hard to generate
* difficult grids otherwise.
* Now we have a solved grid, start removing things from it
* while preserving solubility.
*/
- recursing = FALSE;
- while (1) {
- int x, y, i, j;
- /*
- * Iterate over the grid and enumerate all the filled
- * squares we could empty.
- */
- nlocs = 0;
+ /*
+ * Find the set of equivalence classes of squares permitted
+ * by the selected symmetry. We do this by enumerating all
+ * the grid squares which have no symmetric companion
+ * sorting lower than themselves.
+ */
+ nlocs = 0;
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++) {
+ int i = y*cr+x;
+ int j;
- for (i = 0; i < nsymmclasses; i++) {
- x = symmclasses[i] % cr;
- y = symmclasses[i] / cr;
- if (grid[y*cr+x]) {
+ ncoords = symmetries(params, x, y, coords, params->symm);
+ for (j = 0; j < ncoords; j++)
+ if (coords[2*j+1]*cr+coords[2*j] < i)
+ break;
+ if (j == ncoords) {
locs[nlocs].x = x;
locs[nlocs].y = y;
nlocs++;
}
}
- /*
- * Now shuffle that list.
- */
- shuffle(locs, nlocs, sizeof(*locs), rs);
-
- /*
- * Now loop over the shuffled list and, for each element,
- * see whether removing that element (and its reflections)
- * from the grid will still leave the grid soluble by
- * solver.
- */
- for (i = 0; i < nlocs; i++) {
- int ret;
+ /*
+ * Now shuffle that list.
+ */
+ shuffle(locs, nlocs, sizeof(*locs), rs);
- x = locs[i].x;
- y = locs[i].y;
+ /*
+ * Now loop over the shuffled list and, for each element,
+ * see whether removing that element (and its reflections)
+ * from the grid will still leave the grid soluble.
+ */
+ for (i = 0; i < nlocs; i++) {
+ int ret;
- memcpy(grid2, grid, area);
- ncoords = symmetries(params, x, y, coords, params->symm);
- for (j = 0; j < ncoords; j++)
- grid2[coords[2*j+1]*cr+coords[2*j]] = 0;
+ x = locs[i].x;
+ y = locs[i].y;
- ret = solver(c, r, grid2, maxdiff);
- if (ret != DIFF_IMPOSSIBLE && ret != DIFF_AMBIGUOUS) {
- for (j = 0; j < ncoords; j++)
- grid[coords[2*j+1]*cr+coords[2*j]] = 0;
- break;
- }
- }
+ memcpy(grid2, grid, area);
+ ncoords = symmetries(params, x, y, coords, params->symm);
+ for (j = 0; j < ncoords; j++)
+ grid2[coords[2*j+1]*cr+coords[2*j]] = 0;
- if (i == nlocs) {
- /*
- * There was nothing we could remove without
- * destroying solvability. Give up.
- */
- break;
+ ret = solver(c, r, grid2, maxdiff);
+ if (ret <= maxdiff) {
+ for (j = 0; j < ncoords; j++)
+ grid[coords[2*j+1]*cr+coords[2*j]] = 0;
}
}
sfree(grid2);
sfree(locs);
- sfree(symmclasses);
-
/*
* Now we have the grid as it will be presented to the user.
* Encode it in a game desc.
} else if (n == '_') {
/* do nothing */;
} else if (n > '0' && n <= '9') {
+ int val = atoi(desc-1);
+ if (val < 1 || val > params->c * params->r)
+ return "Out-of-range number in game description";
squares++;
while (*desc >= '0' && *desc <= '9')
desc++;
return NULL;
}
-static game_state *new_game(midend_data *me, game_params *params, char *desc)
+static game_state *new_game(midend *me, game_params *params, char *desc)
{
game_state *state = snew(game_state);
int c = params->c, r = params->r, cr = c*r, area = cr * cr;
for (x = 0; x < cr; x++) {
int ch = grid[y * cr + x];
if (ch == 0)
- ch = ' ';
+ ch = '.';
else if (ch <= 9)
ch = '0' + ch;
else
((button >= '1' && button <= '9' && button - '0' <= cr) ||
(button >= 'a' && button <= 'z' && button - 'a' + 10 <= cr) ||
(button >= 'A' && button <= 'Z' && button - 'A' + 10 <= cr) ||
- button == ' ')) {
+ button == ' ' || button == '\010' || button == '\177')) {
int n = button - '0';
if (button >= 'A' && button <= 'Z')
n = button - 'A' + 10;
if (button >= 'a' && button <= 'z')
n = button - 'a' + 10;
- if (button == ' ')
+ if (button == ' ' || button == '\010' || button == '\177')
n = 0;
/*
*y = SIZE(params->c * params->r);
}
-static void game_set_size(game_drawstate *ds, game_params *params,
- int tilesize)
+static void game_set_size(drawing *dr, game_drawstate *ds,
+ game_params *params, int tilesize)
{
ds->tilesize = tilesize;
}
-static float *game_colours(frontend *fe, game_state *state, int *ncolours)
+static float *game_colours(frontend *fe, int *ncolours)
{
float *ret = snewn(3 * NCOLOURS, float);
return ret;
}
-static game_drawstate *game_new_drawstate(game_state *state)
+static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
{
struct game_drawstate *ds = snew(struct game_drawstate);
int c = state->c, r = state->r, cr = c*r;
return ds;
}
-static void game_free_drawstate(game_drawstate *ds)
+static void game_free_drawstate(drawing *dr, game_drawstate *ds)
{
sfree(ds->hl);
sfree(ds->pencil);
sfree(ds);
}
-static void draw_number(frontend *fe, game_drawstate *ds, game_state *state,
+static void draw_number(drawing *dr, game_drawstate *ds, game_state *state,
int x, int y, int hl)
{
int c = state->c, r = state->r, cr = c*r;
if ((y+1) % c)
ch++;
- clip(fe, cx, cy, cw, ch);
+ clip(dr, cx, cy, cw, ch);
/* background needs erasing */
- draw_rect(fe, cx, cy, cw, ch, (hl & 15) == 1 ? COL_HIGHLIGHT : COL_BACKGROUND);
+ draw_rect(dr, cx, cy, cw, ch, (hl & 15) == 1 ? COL_HIGHLIGHT : COL_BACKGROUND);
/* pencil-mode highlight */
if ((hl & 15) == 2) {
coords[3] = cy;
coords[4] = cx;
coords[5] = cy+ch/2;
- draw_polygon(fe, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT);
+ draw_polygon(dr, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT);
}
/* new number needs drawing? */
str[0] = state->grid[y*cr+x] + '0';
if (str[0] > '9')
str[0] += 'a' - ('9'+1);
- draw_text(fe, tx + TILE_SIZE/2, ty + TILE_SIZE/2,
+ draw_text(dr, tx + TILE_SIZE/2, ty + TILE_SIZE/2,
FONT_VARIABLE, TILE_SIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE,
state->immutable[y*cr+x] ? COL_CLUE : (hl & 16) ? COL_ERROR : COL_USER, str);
} else {
str[0] = i + '1';
if (str[0] > '9')
str[0] += 'a' - ('9'+1);
- draw_text(fe, tx + (4*dx+3) * TILE_SIZE / (4*pw+2),
+ draw_text(dr, tx + (4*dx+3) * TILE_SIZE / (4*pw+2),
ty + (4*dy+3) * TILE_SIZE / (4*ph+2),
FONT_VARIABLE, fontsize,
ALIGN_VCENTRE | ALIGN_HCENTRE, COL_PENCIL, str);
}
}
- unclip(fe);
+ unclip(dr);
- draw_update(fe, cx, cy, cw, ch);
+ draw_update(dr, cx, cy, cw, ch);
ds->grid[y*cr+x] = state->grid[y*cr+x];
memcpy(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr);
ds->hl[y*cr+x] = hl;
}
-static void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate,
+static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
game_state *state, int dir, game_ui *ui,
float animtime, float flashtime)
{
* all games should start by drawing a big
* background-colour rectangle covering the whole window.
*/
- draw_rect(fe, 0, 0, SIZE(cr), SIZE(cr), COL_BACKGROUND);
+ draw_rect(dr, 0, 0, SIZE(cr), SIZE(cr), COL_BACKGROUND);
/*
* Draw the grid.
*/
for (x = 0; x <= cr; x++) {
int thick = (x % r ? 0 : 1);
- draw_rect(fe, BORDER + x*TILE_SIZE - thick, BORDER-1,
+ draw_rect(dr, BORDER + x*TILE_SIZE - thick, BORDER-1,
1+2*thick, cr*TILE_SIZE+3, COL_GRID);
}
for (y = 0; y <= cr; y++) {
int thick = (y % c ? 0 : 1);
- draw_rect(fe, BORDER-1, BORDER + y*TILE_SIZE - thick,
+ draw_rect(dr, BORDER-1, BORDER + y*TILE_SIZE - thick,
cr*TILE_SIZE+3, 1+2*thick, COL_GRID);
}
}
(ds->entered_items[((x/r)+(y/c)*c)*cr+d-1] & 32)))
highlight |= 16;
- draw_number(fe, ds, state, x, y, highlight);
+ draw_number(dr, ds, state, x, y, highlight);
}
}
* Update the _entire_ grid if necessary.
*/
if (!ds->started) {
- draw_update(fe, 0, 0, SIZE(cr), SIZE(cr));
+ draw_update(dr, 0, 0, SIZE(cr), SIZE(cr));
ds->started = TRUE;
}
}
return 0.0F;
}
-static int game_wants_statusbar(void)
+static int game_timing_state(game_state *state, game_ui *ui)
{
- return FALSE;
+ return TRUE;
}
-static int game_timing_state(game_state *state, game_ui *ui)
+static void game_print_size(game_params *params, float *x, float *y)
{
- return TRUE;
+ int pw, ph;
+
+ /*
+ * I'll use 9mm squares by default. They should be quite big
+ * for this game, because players will want to jot down no end
+ * of pencil marks in the squares.
+ */
+ game_compute_size(params, 900, &pw, &ph);
+ *x = pw / 100.0;
+ *y = ph / 100.0;
+}
+
+static void game_print(drawing *dr, game_state *state, int tilesize)
+{
+ int c = state->c, r = state->r, cr = c*r;
+ int ink = print_mono_colour(dr, 0);
+ int x, y;
+
+ /* Ick: fake up `ds->tilesize' for macro expansion purposes */
+ game_drawstate ads, *ds = &ads;
+ game_set_size(dr, ds, NULL, tilesize);
+
+ /*
+ * Border.
+ */
+ print_line_width(dr, 3 * TILE_SIZE / 40);
+ draw_rect_outline(dr, BORDER, BORDER, cr*TILE_SIZE, cr*TILE_SIZE, ink);
+
+ /*
+ * Grid.
+ */
+ for (x = 1; x < cr; x++) {
+ print_line_width(dr, (x % r ? 1 : 3) * TILE_SIZE / 40);
+ draw_line(dr, BORDER+x*TILE_SIZE, BORDER,
+ BORDER+x*TILE_SIZE, BORDER+cr*TILE_SIZE, ink);
+ }
+ for (y = 1; y < cr; y++) {
+ print_line_width(dr, (y % c ? 1 : 3) * TILE_SIZE / 40);
+ draw_line(dr, BORDER, BORDER+y*TILE_SIZE,
+ BORDER+cr*TILE_SIZE, BORDER+y*TILE_SIZE, ink);
+ }
+
+ /*
+ * Numbers.
+ */
+ for (y = 0; y < cr; y++)
+ for (x = 0; x < cr; x++)
+ if (state->grid[y*cr+x]) {
+ char str[2];
+ str[1] = '\0';
+ str[0] = state->grid[y*cr+x] + '0';
+ if (str[0] > '9')
+ str[0] += 'a' - ('9'+1);
+ draw_text(dr, BORDER + x*TILE_SIZE + TILE_SIZE/2,
+ BORDER + y*TILE_SIZE + TILE_SIZE/2,
+ FONT_VARIABLE, TILE_SIZE/2,
+ ALIGN_VCENTRE | ALIGN_HCENTRE, ink, str);
+ }
}
#ifdef COMBINED
#endif
const struct game thegame = {
- "Solo", "games.solo",
+ "Solo", "games.solo", "solo",
default_params,
game_fetch_preset,
decode_params,
game_redraw,
game_anim_length,
game_flash_length,
- game_wants_statusbar,
+ TRUE, FALSE, game_print_size, game_print,
+ FALSE, /* wants_statusbar */
FALSE, game_timing_state,
- 0, /* mouse_priorities */
+ 0, /* flags */
};
#ifdef STANDALONE_SOLVER
-/*
- * gcc -DSTANDALONE_SOLVER -o solosolver solo.c malloc.c
- */
-
-void frontend_default_colour(frontend *fe, float *output) {}
-void draw_text(frontend *fe, int x, int y, int fonttype, int fontsize,
- int align, int colour, char *text) {}
-void draw_rect(frontend *fe, int x, int y, int w, int h, int colour) {}
-void draw_line(frontend *fe, int x1, int y1, int x2, int y2, int colour) {}
-void draw_polygon(frontend *fe, int *coords, int npoints,
- int fillcolour, int outlinecolour) {}
-void clip(frontend *fe, int x, int y, int w, int h) {}
-void unclip(frontend *fe) {}
-void start_draw(frontend *fe) {}
-void draw_update(frontend *fe, int x, int y, int w, int h) {}
-void end_draw(frontend *fe) {}
-unsigned long random_bits(random_state *state, int bits)
-{ assert(!"Shouldn't get randomness"); return 0; }
-unsigned long random_upto(random_state *state, unsigned long limit)
-{ assert(!"Shouldn't get randomness"); return 0; }
-void shuffle(void *array, int nelts, int eltsize, random_state *rs)
-{ assert(!"Shouldn't get randomness"); }
-
-void fatal(char *fmt, ...)
-{
- va_list ap;
-
- fprintf(stderr, "fatal error: ");
-
- va_start(ap, fmt);
- vfprintf(stderr, fmt, ap);
- va_end(ap);
-
- fprintf(stderr, "\n");
- exit(1);
-}
-
int main(int argc, char **argv)
{
game_params *p;
ret==DIFF_SIMPLE ? "Basic (row/column/number elimination required)":
ret==DIFF_INTERSECT ? "Intermediate (intersectional analysis required)":
ret==DIFF_SET ? "Advanced (set elimination required)":
+ ret==DIFF_EXTREME ? "Extreme (complex non-recursive techniques required)":
ret==DIFF_RECURSIVE ? "Unreasonable (guesswork and backtracking required)":
ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)":
ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)":
~mdw / sgt / puzzles / blobdiff