+
+ for (i = 0; i < is->adj.npoints; i++) {
+ /*
+ * Now check to see if any currently empty direction must have
+ * at least one bridge in order to avoid forming an isolated
+ * subgraph. This differs from the check above in that it
+ * considers multiple target islands. For example:
+ *
+ * 2 2 4
+ * 1 3 2
+ * 3
+ * 4
+ *
+ * The example on the left can be handled by the above loop:
+ * it will observe that connecting the central 2 twice to the
+ * left would form an isolated subgraph, and hence it will
+ * restrict that 2 to at most one bridge in that direction.
+ * But the example on the right won't be handled by that loop,
+ * because the deduction requires us to imagine connecting the
+ * 3 to _both_ the 1 and 2 at once to form an isolated
+ * subgraph.
+ *
+ * This pass is necessary _as well_ as the above one, because
+ * neither can do the other's job. In the left one,
+ * restricting the direction which _would_ cause trouble can
+ * be done even if it's not yet clear which of the remaining
+ * directions has to have a compensatory bridge; whereas the
+ * pass below that can handle the right-hand example does need
+ * to know what direction to point the necessary bridge in.
+ *
+ * Neither pass can handle the most general case, in which we
+ * observe that an arbitrary subset of an island's neighbours
+ * would form an isolated subgraph with it if it connected
+ * maximally to them, and hence that at least one bridge must
+ * point to some neighbour outside that subset but we don't
+ * know which neighbour. To handle that, we'd have to have a
+ * richer data format for the solver, which could cope with
+ * recording the idea that at least one of two edges must have
+ * a bridge.
+ */
+ int got = 0;
+ int before[4];
+ int j;
+
+ spc = island_adjspace(is, 1, missing, i);
+ if (spc == 0) continue;
+
+ for (j = 0; j < is->adj.npoints; j++)
+ before[j] = GRIDCOUNT(is->state,
+ is->adj.points[j].x,
+ is->adj.points[j].y,
+ is->adj.points[j].dx ? G_LINEH : G_LINEV);
+ if (before[i] != 0) continue; /* this idea is pointless otherwise */
+
+ memcpy(ss->tmpdsf, ss->dsf, wh*sizeof(int));
+
+ for (j = 0; j < is->adj.npoints; j++) {
+ spc = island_adjspace(is, 1, missing, j);
+ if (spc == 0) continue;
+ if (j == i) continue;
+ solve_join(is, j, before[j] + spc, 0);
+ }
+ map_update_possibles(is->state);
+
+ if (solve_island_subgroup(is, -1, n))
+ got = 1;
+
+ for (j = 0; j < is->adj.npoints; j++)
+ solve_join(is, j, before[j], 0);
+ memcpy(ss->dsf, ss->tmpdsf, wh*sizeof(int));
+
+ if (got) {
+ debug(("island at (%d,%d) must connect in direction (%d,%d) to"
+ " avoid full subgroup.\n",
+ is->x, is->y, is->adj.points[i].dx, is->adj.points[i].dy));
+ solve_join(is, i, 1, 0);
+ didsth = 1;
+ }
+
+ map_update_possibles(is->state);
+ }
+