2
$\begingroup$

I am stuck on a Tapa puzzle. I have tried solving it a number of times and keep running into the same problems. I am starting to wonder if it is solvable at all. As far as the puzzle itself goes, best I can understand, both the black squares and the clues mirror across the dotted line. (For the sake of completeness: Normal Tapa rules are that you need to make a continuous wall of black squares. The numerical clues indicate how many black squares touch that clue cell, and those black cells must be consecutive. If there are two numbers in a clue cell, the corresponding black blocks must be separated by a white space.)

I find a unique option for the location of the "1" around the clue on the bottom left and then work from there and wind up with problems trying to get everything to connect, especially around the clue in the third column.

I would appreciate some hints/indications of where I went astray if I am wrong, or some confirmation that the puzzle is at fault if indeed it is. Thank you all so much! Here is where I pulled the puzzle from: http://oapc.wpc2009.org/archive.php?id=56

The Puzzle

$\endgroup$
  • $\begingroup$ If I remember correctly, the solution doesn't have to be mirrored? But in this case it will be, since all the clues are the same when mirrored. $\endgroup$ – Deusovi Sep 26 at 18:05
  • $\begingroup$ What do the arrows mean? $\endgroup$ – Dr Xorile Sep 26 at 18:08
  • $\begingroup$ Hm, maybe I'm wrong about that, actually. I think it is the case that it has to be mirrored. $\endgroup$ – Deusovi Sep 26 at 18:08
  • $\begingroup$ @DrXorile Those are for the answer key for the competition - they don't have any bearing on the puzzle itself. $\endgroup$ – Deusovi Sep 26 at 18:08
4
$\begingroup$

First of all, the 'real' puzzle:

enter image description here

Now, we start in

the bottom right corner. There must be at least one shaded cell in the bottom right 2x2 - if it's part of the 1, then it cannot escape. So it must be part of the 4. Then, R6C6 cannot be the 1, because it could not escape as well.

enter image description here

Time for some case-bashing! (This may be doable more cleanly, but this works and it's not too difficult.)

Either way you try to extend this will not allow connectivity.

enter image description here

Ditto here: the single cell from the central (1,4) clues cannot escape its position, no matter which way you place it.

enter image description here

And here too: once again, the lone shaded cell cannot escape.
enter image description here

And in this one, the 4-cell paths cannot escape.

enter image description here

So this gives two shaded cells:

R4C5 and R4C6 must both be part of the 4-group in the central clues.
enter image description here

Now, the bottom right group must connect to the middle groups somewhere in R6-8 C3-4. If it connects on the left side (that is, R7C4 is unshaded), then we get a problem: no matter how we place the 1s from the clues, they can't be connected to the rest. So R7C4 is shaded.

With that shaded cell,

enter image description here
the 4 from the outer clues must connect to the left/top wall. From the determined shaded cell, it must go all the way in one direction. If it goes all the way around the outside (first three cells of row/column 8) we can't get connectivity on the inner clue's single shaded cell. So it must go through the inside (row/column 6).

And this gives a unique solution:

enter image description here

$\endgroup$
  • $\begingroup$ Ah. You just beat me to it. I had just got the same answer! $\endgroup$ – Dr Xorile Sep 26 at 18:30
  • $\begingroup$ Oh my thank you. I have tried this a good 7 or 8 times. I am not 100% sure how I pulled this off, but I convinced myself the connection between the two "halves" had to be between the clues in C3 and C5... even though I had a connection between the halves in the proper place all along, in the bottom right corner. Thank you! $\endgroup$ – Jobo Sep 26 at 18:34
1
$\begingroup$

Devious @Deusovi has beaten me to the punch. However, I thought I'd write mine up anyway, just for the sake of a slightly different logic stream:

tapa solved

Here's the logic:

Firstly, the 0's are obvious. And the 1s in the bottom right cannot be inside or they would be blocked. That leaves two options for the 1 (W and N), and so you can fill in a bunch of the bottom right 4s:
tapa002
Next, if the 1 is as shown below, you'll get yourself isolated again:
tapa003
So the 1 must be as shown and we can finish off the bottom right and do the obvious extensions:
tapa004
Next, there's only 1 place for the 1 in the top-right/bottom-left, as shown. If should be easy to convince yourself that it can't be anywhere else because you couldn't get around to it:
tapa005
Next, we extend the black wall in the only route that it has, which also forces the top-right 4 to extend its possible locations:
tapa006
Now consider the central 1/4 and where the 1 is going to be. Looking at the original rather than the reflection, it can't be the SE corner because that would isolate it. It can't be the SW corner, because you'd need at least 5 blocks to get around to it. So SW and SE must both be part of the 4, which leaves only one option:
tapa007

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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