15
$\begingroup$

Yesterday's post and Stiv's answer provided inspiration for a new puzzle. What I envision will be a big effort, so I spent part of yesterday creating a study, a sample part of the bigger puzzle to get some practice. It's not too hard, but I think this small puzzle came out pretty well, so I figured I'd share.

This puzzle contains both a Nurikabe and a Slitherlink. In the grid below, the boxed numbers are the Nurikabe clues. The other numbers are Slitherlink clues, but with the following caveat: if a Slitherlink clue is not shaded by the Nurikabe, then it is an accurate clue. If a Slitherlink clue is shaded by the Nurikabe, the Slitherlink clue is wrong...the Slitherlink path will use a different number of sides of that square. The squares with Nurikabe clues provide no information about the Slitherlink.

Accepted solution will solve both the Nurikabe and Slitherlink grids, and will contain at least some indications of the logic path. There is a unique solution, and it is obtainable purely logically, but you will have to work both parts of the puzzle together. As the Joker says...here...we...go:

Full Grid

Solver Note: As I was solving this puzzle, I found it easier to solve the Nurikabe and Slitherlink in different grids, so for your solving pleasure:

Nurikabe

Slitherlink

Improve this question
$\endgroup$
5
  • $\begingroup$ Just to check something (and it's probably just me doing something wrong) but in the Nurikabe solution is the shaded area all connected like normal, and is the bottom right corner uniquely resolvable? I keep getting an ambiguity for the '3'... $\endgroup$
    – Stiv
    Commented Sep 14, 2020 at 12:41
  • $\begingroup$ Of course, the other possibility is that we cannot solve the Nurikabe on its own before solving the slitherlink, and the solution of one depends on the simultaneous co-solving of the other. If that's the case, maybe you ought to be explicit about that in the text? Just a suggestion, thanks :) $\endgroup$
    – Stiv
    Commented Sep 14, 2020 at 12:43
  • $\begingroup$ @Stiv...I was trying to be coy about it, but your second comment is correct...you need to co-solve. $\endgroup$
    – Jeremy Dover
    Commented Sep 14, 2020 at 12:51
  • $\begingroup$ Aha, well if you wish it to be unstated then you are at liberty to leave it as it is! I'm sure many others will hit the same moment of realisation as me, so I'll leave my comments for their confirmation... I imagine a grid-deduction pro will pick this one off in the next few hours while I am otherwise engaged, so I look forward to seeing it solved if it's done and dusted by the time I get back to it! $\endgroup$
    – Stiv
    Commented Sep 14, 2020 at 12:55
  • 1
    $\begingroup$ @Stiv No worries! I went ahead and threw in a comment about needing to solve both together. $\endgroup$
    – Jeremy Dover
    Commented Sep 14, 2020 at 12:59

1 Answer 1

Reset to default
14
$\begingroup$

With just Nurikabe logic, we can get this far:

enter image description here

The cell in R3C4 gives us another step: if it were unshaded, the shaded cell above it would have no way to connect to the rest of the shaded cells without breaking the 4 or 5.

Similar logic can be applied to finish off the 4 region, then the 2 region:
enter image description here

Now it's time to switch to the Slitherlink:

The 1-3-1 can only be resolved in one way.
enter image description here

Then we have to make sure the nearby 2s stay false:

enter image description here

Now, notice that the bottom left 2 cannot be satisfied. So it must be shaded in the Nurikabe.

Switching back to the Nurikabe,

with that extra shaded cell we can get this far:
enter image description here

And this gives us more information for the Slitherlink:

enter image description here

Lots of clues in the upper left - we can get enough to determine that we can't satisfy that 0 clue.

At the same time, if that lowest endpoint goes upwards, then we must satisfy one of the two false 2s. So the lowest endpoint goes right...

enter image description here

...and if we try to make the first 3 true, we end up with a contradiction.
enter image description here

So that first 3 must be false, meaning R7C6 must be shaded.

And now we have enough to finish off both halves of the puzzle:

The Nurikabe has had its last two shaded cells determined.
enter image description here

And now that all the clues are determined, the other two corners of the Slitherlink can be finished off without any tricky deductions.
enter image description here

Improve this answer
$\endgroup$
1
  • 3
    $\begingroup$ A perfect solution! Well-explained in an excellent presentation. Hope you enjoyed! $\endgroup$
    – Jeremy Dover
    Commented Sep 14, 2020 at 13:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.