4
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I'm working on another idea for a combination logic puzzle, and as before I've been gearing up by creating some studies. This first one came out decently, I think, so I figured I'd share it. This puzzle contains a Nonogram and a Slitherlink. The connection between the two is simple: for each clue in the Slitherlink, if the corresponding square in the Nonogram is shaded, the Slitherlink clue should be increased by 1. Unclued squares in the Slitherlink are unclued, regardless of the Nonogram. Here's the grid:

Grid

The different colors on the numbers in the grid are simply to differentiate the clue types, and do not have any bearing on the puzzle. You will need to solve both puzzles simultaneously; there is a unique solution for the pair of puzzles as connected. Hope you enjoy!

Solver Notes

Text versions of the puzzles:

       2      
       1   1 1   1 2
       1 1 2 1 2 1 1
      ---------------
    2 | | | | | | | |
      ---------------
1 1 1 | | | | | | | |
      ---------------
  1 1 | | | | | | | |
      ---------------
    1 | | | | | | | |
      ---------------
  1 1 | | | | | | | |
      ---------------
  1 2 | | | | | | | |
      ---------------
  1 3 | | | | | | | |
      ---------------
• • • • • • • •
 0 1       2 1 
• • • • • • • •
 0   1     0 1 
• • • • • • • •
 2   1 0    
• • • • • • • •
 2 1         2 
• • • • • • • •
 0     2     1 
• • • • • • • •
     1 0   0 2 
• • • • • • • •
 2   2   2 1 2 
• • • • • • • •
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3
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Here is the solution with the "true" slitherlink clues and happy stars representing shaded cells in the nonogram puzzle.

Solution is enter image description here

To start the puzzle observe that

r7c5 in the nonogram must be a happy star. If r6c4 is also happy star then we readily deduce a contradiction in the lower right corner.

After that, the rest of the solve is relatively straightforward.

| improve this answer | |
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  • $\begingroup$ That is the correct answer! Hope you enjoyed this small puzzle :-) $\endgroup$ – Jeremy Dover Sep 19 at 2:50
  • $\begingroup$ happy stars OHH I GET IT NOW $\endgroup$ – matt Sep 28 at 17:46

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