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This is part 68 of the puzzle series Around the World in Many Days. Each part is solvable on its own.

Dear Puzzling,

The top grid here is a Yin-yang puzzle. Shade some cells in the grid so that shaded and unshaded cells both form a single orthogonally connected area and no 2x2 square is fully shaded or unshaded. Cells containing a black circle must be shaded, and cells containing a white circle must be unshaded. Once you’ve solved the Yin-yang, use some of the given 24 pieces from the bottom grid to find out the final answer.

Today I have visited an archaeological site featuring the remains of a very ancient civilisation. Can you guess where I am?

Love, Gladys.

Empty Yin-yang grid
Solve on Penpa+
Grid with jigsaw pieces

Gladys will return in Nice.

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    $\begingroup$ wait why are you telling us where gladys is going to be next time? isn't that the whole point of the puzzles? $\endgroup$
    – juicifer
    Commented 2 days ago
  • $\begingroup$ Jafe is telling us the name of the next puzzle, not the answer ;) $\endgroup$
    – Jujustum
    Commented 2 days ago
  • $\begingroup$ The rules of ying-yang are not complete. No 2x2 square is colored like a checkerboard. $\endgroup$
    – Florian F
    Commented yesterday
  • $\begingroup$ @FlorianF Doesn't that rwquirement follow from the stated rules? I can't come up with an example grid with a 2x2 checkerboard that would meet the rules stated here. $\endgroup$
    – Jafe
    Commented yesterday
  • $\begingroup$ Of course you are right. You cannot have a 2x2 checkerboard pattern and still have both regions orthogonally connected. So no more objection. $\endgroup$
    – Florian F
    Commented 15 hours ago

1 Answer 1

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Gladys is visiting

Mohenjo-daro

Solution

Step-by-step

A basic deduction of Yin-yang is that any two cells of same colour on the border must be connected through the border. Since at least one of R4C9, R5C9 and one of R8C6, R8C7 must be black while one of R6C9, R7C9 must be white, we know that the black region must cover clockwise the border from R8C6 to R6C9


Next, we can colour some cells just based on the connectivity and 2x2 rules
enter image description here

Another basic deduction of Yin-yang is that in any 2x2 region the two colours cannot "cross" each other. This gives us R3C4 and R6C7


And the rest is filled up with just those basic deductions

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  • $\begingroup$ Couldn’t we also place the other N piece on R4C3? $\endgroup$
    – Jujustum
    Commented 2 days ago
  • $\begingroup$ @Jujustum Argh, you're right. I actually "fixed" that at the last minute by changing one of the N pieces slightly but looks like I somehow edited the wrong one. D'oh! $\endgroup$
    – Jafe
    Commented 2 days ago
  • $\begingroup$ I've edited the question to remove that ambiguity (shaded one more cell in piece 16) and edited your answer to correspond to the edited question. Sorry about that! I should have done a full double-check again after making a last-minute edit. $\endgroup$
    – Jafe
    Commented 2 days ago
  • $\begingroup$ @Jafe No worries, thanks for the nice puzzle as always! I wouldn't have even noticed the ambiguity so good catch from Jujustum! $\endgroup$
    – user39583
    Commented 2 days ago

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