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Rules of Toroidal Weaved Truncated Square Tiling Tapa:

  1. The grid wraps around horizontally and vertically and also is divided into several 1-cell and 3-cells tiles.
  2. Shade some tiles such that all shaded tiles form a single orthogonally connected region.
  3. In addition, for each intersection point, at least one of the three tiles remains unshaded (i.e. no 2x2 cells are shaded).
  4. Some tiles have clues and these tiles cannot be shaded.
  5. The clues represent the lengths of the blocks of consecutive shaded tiles (NOT cells!) surrounding the clues.

The "standard" Toroidal Weaved Truncated Square Tiling Tapa rules apply.

enter image description here

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  • 3
    $\begingroup$ Um, what are the '"standard" Toroidal Weaved Truncated Square Tiling Tapa rules'? $\endgroup$ Jul 2 at 12:57
  • 1
    $\begingroup$ @Pantuvarali yeah, the word "standard" here is a kind of joke, similar to my previous puzzle ;) $\endgroup$
    – athin
    Jul 2 at 14:22
  • 1
    $\begingroup$ @Pantuvarali Oh, in a more serious note, the rules are written just right above the puzzle :D $\endgroup$
    – athin
    Jul 2 at 14:54
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    $\begingroup$ @athin Rule 5. Are these two examples of one shaded correctly and the other one incorrectly? Upper correct, lower incorrect? (changed number) $\endgroup$ Jul 2 at 18:59
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    $\begingroup$ @Prim3numbah yes you're right, the upper one is a correct shading while the lower one is an incorrect shading. $\endgroup$
    – athin
    Jul 2 at 23:04
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23 hours and no takers? Here is my solution.

enter image description here

How to start:

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- If A is green then B and C are gray because of 33 and D is gray because of 3. But then BCD are gray which is forbidden. Therefore A is gray and so is E.
- If F is gray, since E is gray, G and H must be green but that is not possible because of 24. Therefore F must be green and 24 forces IJK to be gray.
- No 3 grays can joint at an intersection, so LMNO are green.
- L being green, with 33, forces P to be green and BCQQ to be gray.
- AB being gray, D must be green and because of 3, RS must be gray.
- Avoiding 3 grays forces TUV to be green.
- R and 2 force W to be green. N and 2 force X to be gray.
- If Y is gray, because of 2, Z must be gray also. But that is forbidden because of G. So Y is green.

enter image description here

And some more:

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- The 4 forces A to be gray.
- The 5 forces the B's to be gray, which in turn force the C's to be green.
- If one of DEFG is green, the corresponding 23 couldn't be satisfied. They must be gray.
- If H is gray, I muat also be because of the 5. But that would be 3 grays. So H must be green. And J must be gray because of the 5 again.
- K is now forced to green.
- The remaining options for 23 force LM to be gray, which in turn forces N's to green.
- 1111 has 2 ways, Shade the O's or the P's. But if you shade the O's, the P's are green and one O is surrounde by green P's and N. So that cannot be. The P's are shaded.
- E and the P below along with the 3's at the right forbid to shade the Q's. If a Q is gray it must extend up or down resulting in a junction of 3 grays.
- This forces R to be green and S to be gray to satisfy the 3.
- T must be green to avoid a triple gray. This forces the grays and the greens around the 4.
- This forces U to be green, V to be gray to satisfy the 24 and the W's to be green.
- The X's are gray because of the 3.
- This forces Y's to be green.
- Z must be green, forcing the 2 grays and 6 green around the 2.

enter image description here

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- 23 left forces A's to be gray and B to be green.
- 2 right forces C to be green.
- The placement of the 3 around 13 top left forces D to be gray and E to be green.
- F is now surrounded by green, it can only escape via G.
- Setting G forces the H's to be gray and the I's to be green to satisfy the 3's.
- Now J needs to escape via K. K is gray and the L's are green.
- If M is gray then it must extend to N, but 13 doesn't allow it. M is green and N is gray.
- O must be green to avoid a triple-gray.
- P is gray to satisfy the 2.
- Q can only connect via the R's. The R's are gray.
- This forces the gays and green around the 12 right of the R's.
- The remainng S's are forced green because of 2 grey neighbours.
- The T's are green because of a hint.

This completes the solution.

enter image description here

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  • $\begingroup$ @bobble Here is where I started, and I'm pretty sure this is the first step (since both tiles are 6 in total and close to each other). Red, meaning those can not be consecutively shaded. Which then leads to these shaded tiles, and so on... $\endgroup$ Jul 3 at 15:53
  • $\begingroup$ Very well-explained, nicely done! :D $\endgroup$
    – athin
    Jul 4 at 4:56
  • $\begingroup$ On the first step, why must D be shaded? Can't we also shade S to cover the 3? $\endgroup$
    – justhalf
    Jul 4 at 8:40
  • $\begingroup$ Assuming A is green, D, R and S must be shaded because of the 3. I mention only D because this one leads to a contradiction. $\endgroup$
    – Florian F
    Jul 4 at 9:07

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