Puzzle Description:

i was studing for my french test, i know some people dont like words in the puzzle, but oh well.

#1: Gator (Gator) on Oct 13, 2010 [HINT]

It's not hard to look at the image and figure out what needs to be happen, but I'm having trouble getting at the solution logically. The edge/color logic gets complex in a lot of areas, but I have been able to progress somewhat. I can also look about 4 moves ahead to see some contradictions, but this is considered guessing.

I'm experimenting with using my solving program to generate sensible hints for human solvers, and decided to use this as a test case. Deriving hints is still largely a manual process, but if this works, maybe I can automate it.

Here are some hints:

1. Plain old color logic allows us to mark a lot of cells in the top row and the two side columns and a few on the bottom row too.
2. Use color logic to consider where the black 2's in row three can be.
3. The red 4 in the second to last row can't be columns 6 or 7, and those cells also can't be black, so they must be white.
4. The first three columns of rows 22, 23 and 24 must be white.
5. There are some cells in column 6 that can't be red. This limits the placement of the black 6 and 5 enough that we can conclude that r7c6 and r8c6 must be black.
6. Speaking of column 6, r27c6 must be white. Ditto r27c7.
7. By now it is plain that the first ten columns of the bottom row are white.
8. You can dot a lot of cells in column 13.
9. r27c22 cannot be red. Because of this the black 5 and 6 in column 22 cannot both fit below row 13. So the cells in row 7 through 9 must be black.
10. r7c2 can't be black, so r7c7 must be black.
11. In column 6, rows 22 and 24 can't be black, so rows 17 and 18 must be black.
12. In column 13, we can now dot rows 17 and 18.

So now we are in the following position:

Now pbnsolve gets a bit tricky. It says r1c2 can't be black. If it was, then r1c3 could obviously not be black. Also r3c3 could not be black, because r3c1 would be black. If neither r1c3 nor r3c3 can be black, then column 3 is unsolvable.

Well,that's true enough, but I think that's four levels of look-ahead:

  r1c2 black -> r1c1 not black and r1c3 not black
  r1c1 black -> r3c1 black
  r3c1 black -> r3c3 not black
  r1c3 not black and r3c3 not black -> column 3 unsolvable

I think that's a bit too much to ask of a human solver. I'm afraid I'm going to have to limit pbnsolve's lookahead more strictly if it is going to help me identify human solvable puzzles.

OK, so I tightened its screws a bit, and asked it again what it proposed for the snapshot above.

It says r3c2 may not be red. Because:

   r3c2 red -> r3c3 red
   r3c2 red -> r1c2 black
   r1c2 black -> r1c3 white
   r3c3 red and r1c3 white -> column 3 unsolvable

Well, this isn't too bad to see once it's pointed out to you, though I wouldn't have spotted it in a month. I think it's still overestimating how much lookahead humans can do, and I'm going to have to take it back to the drawing board to get finer controls on this. It should be erring on the side of conservatism when it calls a puzzle logically solvable, leaving the close calls to humans.

By the way, with the more liberal lookahead settings I used in the previous response, it could solve this puzzle, but when I tightened them down for the second test, it was no longer able to solve the puzzle. It got a bit further than my snapshot, but not to the end.

Well, it's good to know that it's having as much trouble as I am solving this.

I got a lot further along that your screen shot. Would it help you to know some of what I did, and then figure out how to make the program do the same thing?

Not particularly. The computer had actually found several ways forward from the state I showed. My goal was for it to be able to pick from the many possibilities the ones that were easiest for humans to understand. This would make it useful for (1) testing puzzles for solvability, and (2) hint generation.

Clearly the criteria it had for selecting ways forward were too loose, allow it to take steps that were a bit too complex for my tastes. I need to design better criteria. I think I am going to have to make to backprune the contradictions it finds to reduce them to the simplest possible proof of the contradiction, then evaluate the "size" of the resulting proof. That won't be trivial to do.

The alternative is to build specific modules for specific kinds of reasoning - edge logic, etc.

I really enjoyed this puzzle!

I believe this puzzle is logically solvable. Here is the way I solved it. There may be a more efficient way, and some of these steps may be unnecessary.

I use the option for “Unknown cells are grey, White cells are white.” If you use dots, please make the substitution in the logic below. The steps are numbered to help keep track of so many steps.

1) Color logic in column 30 makes R9-14 red.

2) Color logic in row 1 makes C4-12 white and C18-27 white.

3) Color logic in row 30 makes C1-5 white.

4) Color logic in columns 1 makes R8-14 red, R22-24 white, and R2 black.

5) Line logic now makes R2C2 black and R13C2-29 white.

6) Color logic on the red 11 clue in column 1 requires the 11 clue to fit within rows 4 to 18. Because the clues on either side of the red 11 are black, any spaces in the R4-18 that are not part of the red 11 must be white. Edge logic on that same clue in that range makes R4C1 and R18C1 white. Line logic then makes R7C1 and R15C1 red.

7) Color logic on the red 11 clue in column 30 requires the 11 clue to fit within rows 4 to 19. Because the clues on either side of the red 11 are black, any spaces in the R4-19 that are not part of the red 11 must be white. Edge logic on that same clue in that range makes R4C30 and R18-19C30 white. Line logic then makes R7-8C30 and R15C30 red.

8) Color logic makes R24C2-3 white. Color logic then makes R23C2-3 white. Color logic then makes R22C2-3 white.

9) Edge logic on the red 2 clue in R30C6-7 leaves no place to fit the red 4 clue in row 29. Therefore R30C6-7 are white. The same logic makes R29C6-7 white.

10) Color logic makes R30C8-10 white.

11) Color logic in row 2 means that the two red 4 clues must fit within columns 3-12 and 18-27. Edge logic can then be used to make R2C10-12 and R2C18-21 white. This reduces the locations of the red 4 clues so that color logic makes R2C6 and R2C24-25 red.

12) Line logic now fills in quite a few squares, bringing the puzzle to 14%.

13) Color logic makes R2C29 black. Line logic then makes R3C29 black.

14) Color logic also makes R3C13-14 and R3C16-17 black, with R3C15 being white between the two clues.

15) Line logic now fills in the rest of the black clues in rows 4 through 6, bringing the puzzle to 16%.

16) Edge logic on the leftmost 4 clue in row 2 makes R2C3 white. Line logic then makes R2C7 red.

17) Color logic makes the follow squares white: R3-6C6, R3-6C7, R3-6C24, R3-6C25.

18) Color logic in column 2 limits the spaces that can be red to rows 4-6 and 16-18. Since there are two red 2 clues in this column, R5C2 and R17C2 must be red.

19) Line logic makes R7-15C2 white.

20) As long as you start at the top in square R7C3, color logic makes R7-15C3 white.

21) Color logic in column 3 limits the spaces that can be red to rows 3-6 and 16-18. This means the lower red 2 clue must be in the lower set, making R17C3 red.

22) Line logic now brings the puzzle to 21%. Don’t miss the white square in R3C10, which the hint button ignores.

23) Color logic makes the following squares white: R4-6C8, R5-6C9, R6C4.

24) Line logic fills in a few more squares, bringing the puzzle to 23%.

25) Examining column 2, color logic makes R20C2 black. Line logic then makes R20C3 black.

26) Examining column 6, color logic makes R7-12C6 and R17-18C6 black.

27) Line logic fills in a few more squares, bringing the puzzle to 25%. Don’t miss the white squares in R9C7, R17C5, and R17C7, which the hint button ignores.

28) Color logic makes R9C8 white. Continuing down the column, R10-15C8 are also white.

29) Line logic moves the puzzle along to 29% complete.

30) Now, because nearly every red clue left in the puzzle is a 2, whether vertical or horizontal, it is possible to use color logic to finish all the red. I found it easiest to work counterclockwise around the puzzle.

31) After completing all the red, line logic solves the rest of the puzzle.

Happy Solving!

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