Here is a standard Yin-Yang puzzle.

Rules of Yin-Yang:

  1. Fill each empty cell with either a black circle or a white circle.
  2. All white circles should be orthogonally connected, so do all black circles.
  3. There may not be any 2x2 cell region consisting of the same circle color.

enter image description here

  • 1
    $\begingroup$ I don't undestand your second point could you be clearer. $\endgroup$ – Prince Deepthinker Nov 17 '20 at 9:56
  • $\begingroup$ Like are you saying they should all touch but do diagonals count? $\endgroup$ – Prince Deepthinker Nov 17 '20 at 9:57
  • $\begingroup$ @PrinceDeepthinker ah yes, so all white circles should be connected orthogonally (not counting diagonal); all black circles should be connected orthogonally too $\endgroup$ – athin Nov 17 '20 at 10:03
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    $\begingroup$ I first have to apologize.. As an entry of this genre in this site, I shouldn't set the puzzle this hard haha.. >< Here are two tips (which are believed to be sufficient to solve most Yin-Yang puzzles) for beginners: Edge Connection and Opposite Pairs :D $\endgroup$ – athin Nov 17 '20 at 11:08

I knew that there was already an answer; I'm posting this to show a cleaner path to the solution.

First of all,

Basic deductions give this:

enter image description here

Then we apply the boundary logic:

The white region at lower right cannot extend to R10C4 or R1C9, so we get a giant black boundary on the left side.

enter image description hereenter image description here

Then there's a highway of basic deductions.

One crucial "basic deduction" is that a 2x2 checkerboard is not allowed. If we have such a pattern, trying to connect one pair of same color will divide the other. Using that, and avoiding 2x2 monochrome blocks and isolated islands, we can get this far:

enter image description hereenter image description here

We can extend the white boundary a bit

because the black can't extend to R4C10, white can extend up to there.

enter image description hereenter image description here

Finally, spotting a small contradiction finishes the game:

If R6C8 is white, R5C8 can't be white (2x2 block) nor black (2x2 checkerboard). So R6C8 is black. The rest is just some more basic deductions.

enter image description hereenter image description here

  • 1
    $\begingroup$ Yeah, +1 for spotting the contradiction in R6C8. I didn't see that. $\endgroup$ – Alaiko Nov 17 '20 at 14:06
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    $\begingroup$ I'm accepting this one as this is much cleaner to avoid some trial-and-errors (and indeed this is the intended path -- the extension of the 2 basic strategies yield the 2x3 pattern as shown in R5R6-C7C9). Nicely done for both of you nevertheless! :) $\endgroup$ – athin Nov 18 '20 at 1:12

First of all,

We can try placing a white circle in R5C1, but using the Edge Connection rule to try and connect it from the bottom right white circles either clockwise or anti-clockwise, we will run into 2 x 2 white circles somwehere. Therefore, R5C1 must be a black circle. Then, we try to place a white circle in R1C9, but that results in 2 x 2 white circles as well. Therefore, R1C9 is a black circle as well. Now, we have two black circles on the edges, we can use Edge Connection to connect them. This gets us to:


From here, we can simply use the Opposite Pairs rule and avoid making any 2 x 2 black or white circles. This leads to a series of chain deductions resulting in:


Now, we turn our attention to the R10C3 cell. If that cell was a white circle, then Edge Connection rule will either create a 2 x 2 white square or a solo black circle in R9C4. Therefore, that cell must be a black circle and using Edge Connection, we get


Then, again applying opposite pair rule and avoiding 2 x 2 squares, we can use chain deductions to get to:


Now, the mass of black circles in the middle is trapped and it needs a way out. So, using connectivity rules, we get to:

enter image description here

Note that R4C10 cannot be a black circle, as it will then need to extend upwards and make a 2 x 2 black square. Therefore, that must be white circle and we then use Edge Connection to get:


Now, trying to place a white circle in R10C5 will lead to a contradiction eventually.


Therefore, that cell must be a black circle and using Opposite Pair rule, we get


Now, trying to place a white circle in R5C8 eventually led to a contradiction as shown here. We cannot place either a white circle or black circle in R9C7.


So, R5C8 must be a black circle instead. Carefully using Opposite Pair rule and avoiding 2 x 2 squares eventually leads to the solution:



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