- /* If we have atleastone set for this dline, infer
- * atmostone for each "opposite" dline (that is, each
- * dline without edges in common with this one).
- * Again, this test is only worth doing if both these
- * lines are UNKNOWN. For if one of these lines were YES,
- * the (yes == 1) test above would kick in instead. */
- if (is_atleastone(dlines, dline_index)) {
- int opp;
- for (opp = 0; opp < N; opp++) {
- int opp_dline_index;
- if (opp == j || opp == j+1 || opp == j-1)
- continue;
- if (j == 0 && opp == N-1)
- continue;
- if (j == N-1 && opp == 0)
- continue;
- opp_dline_index = dline_index_from_dot(g, d, opp);
- if (set_atmostone(dlines, opp_dline_index))
- diff = min(diff, DIFF_NORMAL);
- }
-
- if (yes == 0 && is_atmostone(dlines, dline_index)) {
- /* This dline has *exactly* one YES and there are no
- * other YESs. This allows more deductions. */
- if (unknown == 3) {
- /* Third unknown must be YES */
- for (opp = 0; opp < N; opp++) {
- int opp_index;
- if (opp == j || opp == k)
- continue;
- opp_index = d->edges[opp] - g->edges;
- if (state->lines[opp_index] == LINE_UNKNOWN) {
- solver_set_line(sstate, opp_index, LINE_YES);
- diff = min(diff, DIFF_EASY);
+ /* More advanced deduction that allows propagation along diagonal
+ * chains of faces connected by dots, for example: 3-2-...-2-3
+ * in square grids. */
+ if (sstate->diff >= DIFF_TRICKY) {
+ /* If we have atleastone set for this dline, infer
+ * atmostone for each "opposite" dline (that is, each
+ * dline without edges in common with this one).
+ * Again, this test is only worth doing if both these
+ * lines are UNKNOWN. For if one of these lines were YES,
+ * the (yes == 1) test above would kick in instead. */
+ if (is_atleastone(dlines, dline_index)) {
+ int opp;
+ for (opp = 0; opp < N; opp++) {
+ int opp_dline_index;
+ if (opp == j || opp == j+1 || opp == j-1)
+ continue;
+ if (j == 0 && opp == N-1)
+ continue;
+ if (j == N-1 && opp == 0)
+ continue;
+ opp_dline_index = dline_index_from_dot(g, d, opp);
+ if (set_atmostone(dlines, opp_dline_index))
+ diff = min(diff, DIFF_NORMAL);
+ }
+ if (yes == 0 && is_atmostone(dlines, dline_index)) {
+ /* This dline has *exactly* one YES and there are no
+ * other YESs. This allows more deductions. */
+ if (unknown == 3) {
+ /* Third unknown must be YES */
+ for (opp = 0; opp < N; opp++) {
+ int opp_index;
+ if (opp == j || opp == k)
+ continue;
+ opp_index = d->edges[opp] - g->edges;
+ if (state->lines[opp_index] == LINE_UNKNOWN) {
+ solver_set_line(sstate, opp_index,
+ LINE_YES);
+ diff = min(diff, DIFF_EASY);
+ }