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This morning, I made three small and easy puzzles to be posted today. However, when I come back from my lunch, all of my puzzles are corrupted! :(

  • The first one is a standard Curve Data puzzle. Somehow, all clue shapes are "blurred" and replaced by '#' symbols.
  • The second one is a standard Yin-Yang puzzle. Both black and white circles given in the grid are all replaced by '#' symbols too.
  • The third one is a standard Tapa puzzle. I believe I wrote all clues with positive integers, but, you guess it, all of them are now replaced by '#'!

One thing for sure is that all of the puzzles I made have a unique solution. Can you somehow recover the puzzles and solve them? You may first ignore the red letters when recovering and solving the puzzles. After that, use the red letters to answer the following question:

What should I do to avoid this from happening again?

enter image description here enter image description here enter image description here

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  • $\begingroup$ I'm not sure, but are the pictures in the wrong order? $\endgroup$
    – Jerry Dean
    Oct 7 '21 at 10:02
  • $\begingroup$ @JerryDean the pictures are in the correct order (i.e. Curve Data is on a 7x7 grid and the Tapa has several multiple digits as clues) $\endgroup$
    – athin
    Oct 7 '21 at 10:41
  • $\begingroup$ Can a number in a Tapa grid be 0? The third board can't seem to be solved otherwise. $\endgroup$
    – Nautilus
    Oct 7 '21 at 13:48
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    $\begingroup$ @Nautilus I don't understand what exactly is going on in your image. Are you assuming that all shaded cells are touching at least one clue? $\endgroup$
    – Deusovi
    Oct 7 '21 at 20:23
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    $\begingroup$ @Bubbler Correct! A dot or an "empty zero-length" segment is not a valid clue for Curve Data. $\endgroup$
    – athin
    Oct 7 '21 at 22:34
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Partial answer - Curve Data and Tapa solved, stuck on Yin-Yang

Curve Data

Note that

a Curve Data clue cannot be a single cell. So we can get a significant amount of the puzzle solved with just that:
curve data, partial progress

Now look at the section

on top, with the single unassigned cell. If it goes to the left or right, then we have an ambiguity -- the clues would both just be horizontal lines, and then either solution would work. So instead, the third region must use that cell, to prevent this.

Repeating this logic leads to the solution:
solved curve data

Yin-Yang (unsolved)

Some important lemmas for Yin-Yang that it is good to know:

There can be no 2×2 checkerboard pattern anywhere in the grid in a solved Yin-Yang. This would make it so that you couldn't connect both opposite sides.

The border must have at most one "run" of each color. If it had multiple, you wouldn't be able to connect them both.

So, let's think about this puzzle in a different way.

As hinted by the letters, we're drawing the border between the two regions to extract. (This makes sense, because we could flip black and white in any solution.) So what if we rephrase it in terms of this border?

The rules now become: "Draw a line separating all the black circles from all the white circles. Your line should either start and end on an exterior vertex of the grid, or form a complete loop. Every interior vertex must be visited."

And now we can start getting somewhere!

First of all, our path can't start or end in a corner -- if it did, it wouldn't be unique, because we could flip it to start/end from the other nearby point.

yinyang stage 1

But there's a more interesting deduction involving parity.
We have to visit 55 interior vertices total. That's an odd number.

yinyang parity labels
Our path alternates black and white vertices. There are more black than white vertices (specifically, exactly one more), and therefore our path must start and end on a black vertex. (And so it can't be a loop.)

yinyang stage 2, with parity deductions made

...and this is where I'm stuck.

Tapa

To start,

the ## clue on top tells us that R2C2 must be shaded. It needs to escape downwards in order to not break the ## clue.
tapa stage 1

Now, the bottom left can only be made unique if the # clue on the left side forces it. That means # must be 5, and to make the top left unique, ## must be 2,2.

What about the right side? Well, we don't want the region to be able to reach the lower right, because that is necessarily ambiguous. This means that (1) F=G, and (2) E<5.
tapa stage 2

If we don't use R2C7, we'll use all three cells in R1-3C8. But then there's an ambiguity with the one cell seen by F and G. So that's not possible; we must use R2C7. Then, to stop an ambiguity, we must block off R2C8 by creating a near-2×2; this means R3C8 must be used.

tapa stage 3

To prevent the solver from being able to "twist" E's cells leftwards, we need to shade R3C5.

Now clues A-E have been determined. A solver can logically get to here with that knowledge (and F=G):
tapa stage 4

To make this unique, we need to set F and G to 2.

tapa solved

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    $\begingroup$ I was really hoping to finish the Yin-Yang, but I've been bashing at it for a while now and haven't found any concrete progress. Not sure what I'm missing. $\endgroup$
    – Deusovi
    Oct 8 '21 at 0:38
  • $\begingroup$ Greatly explained as always! (and welp sorry didn't respond here fast.) However, I'm going to accept this answer even if it's partial as it has the most substantial value to get the final answer; including the argument for Yin-Yang. :) $\endgroup$
    – athin
    Oct 10 '21 at 0:06
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solving yin-yang:

To force as much as possible, color b1 and b5 black, and c1 white. This forces the the entire border (taking Deuvosi's parity argument into account)

If we now make e2 black, we can make e2-e5 g2-g5 and i2-i5 black - there is a solution on the left side for all 8 possibilities of the given circles while on the right the solution is not unique (even if all circle there are black there is a choice of j5 or k4). Conclusion: e2 must be white

Since making f3,h3 and j3 black was the most restrictive in the previous step; lets try that first: j2 white leads to a contradiction; j2 black forces almost everything trivially.

Connecting the last parts give the solution (though I'm not sure what to do with the letter part)
enter image description here

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  • $\begingroup$ Once the rot13(vaghvgvba ba gur obeqre unf orra qbar), and before rot13(frggvat hc gur guveq ebj), you can focus first on the other circles -- rot13(ba gur svefg 4 pbyhzaf). After that, the rest will flow and be resolved to the right (and can even be extended infinitely!) $\endgroup$
    – athin
    Oct 9 '21 at 23:59
  • $\begingroup$ That b probably could not be black caught my eye, after I did not see easy progress on the left. It is certainly possible that there are easier ways to make progress. $\endgroup$
    – Retudin
    Oct 10 '21 at 8:27
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The unique solutions to the 3 puzzles have all been found by @Deusovi (Curve Data & Tapa) and @Retudin (Yin-Yang) - here's how to extract the final message... (See their answers for explanations of the puzzles' individual solutions...)

1. Tapa:

The only 'obvious' way to differentiate a message from the final grid as solved by @Deusovi is to note that every region is either a two-space domino or a three-space 'corner shape'.

Focusing on the three corner-shaped pieces, we can read a word by focusing on the central tile of each and reading down the grid: ADD.

2. Yin-Yang

Overlaying @Retudin's solution over the original letter grid, there are six letters which fall on the boundary between the two colours. If we read these off from top to bottom, left to right, they spell COMMIT.

3. Tapa

Reading the letters contained in unshaded spaces (again top to bottom, left to right) spells THEN PUSH.

Extracting the message

Overall then, the answer to "What should I do to avoid this from happening again?" is:

ADD COMMIT THEN PUSH. In other words, (i.e. in computing speak) you need to commit your changes (add them to the local repository) and then push them (to the remote repository), meaning that if disaster strikes again you should be able to recover past versions safely...

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  • $\begingroup$ The final answer is correct, well done! And yes they are the reference for rot13(n trrx fghss bs Tvg) :) $\endgroup$
    – athin
    Oct 10 '21 at 0:01

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