Phil Bordelon points out that the Unequal difficulty settings

documentation is a bit odd, and also offers a signedness fix in
latin.c.

[originally from svn r7112]

diff --git a/latin.c b/latin.c
index fa981f296829c9de29d80fe79c8a3fa39dd7b866..4f6e1f37a00af6686f17409eb5c17a08475d0f13 100644 (file)
--- a/latin.c
+++ b/latin.c
@@ -947,7 +947,7 @@ void latin_solver_debug(unsigned char *cube, int o)
 #ifdef STANDALONE_SOLVER
     if (solver_show_working) {
         struct latin_solver ls, *solver = &ls;
-        char *dbg;
+        unsigned char *dbg;
         int x, y, i, c = 0;
 
         ls.cube = cube; ls.o = o; /* for cube() to work */

diff --git a/puzzles.but b/puzzles.but
index 38300035c155823717a97b07a14a4be278c18ef8..762baff279145538a293d019229fe8d4e83c148c 100644 (file)
--- a/puzzles.but
+++ b/puzzles.but
@@ -2136,10 +2136,11 @@ These parameters are available from the \q{Custom...} option on the
 
 \dt \e{Difficulty}
 
-\dd Controls the difficulty of the generated puzzle. At Trivial level,
-there are no greater-than signs (the puzzle is to solve the Latin
-square only); at Tricky level, some recursion may be required (but the
-solutions should always be unique).
+\dd Controls the difficulty of the generated puzzle. At Trivial
+level, there are no greater-than signs (the puzzle is to solve the
+Latin square only); at Recursive level backtracking will be required
+(but the solution should still be unique); the levels in between
+require increasingly complex reasoning to avoid having to backtrack.
 
 
 \A{licence} \I{MIT licence}\ii{Licence}