# Yin-Yang: All-White

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.

• I don't undestand your second point could you be clearer.
– PDT
Commented Nov 17, 2020 at 9:56
• Like are you saying they should all touch but do diagonals count?
– PDT
Commented Nov 17, 2020 at 9:57
• @PrinceDeepthinker ah yes, so all white circles should be connected orthogonally (not counting diagonal); all black circles should be connected orthogonally too Commented Nov 17, 2020 at 10:03
• 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 Commented Nov 17, 2020 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:

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.

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:

We can extend the white boundary a bit

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

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.

• Yeah, +1 for spotting the contradiction in R6C8. I didn't see that. Commented Nov 17, 2020 at 14:06
• 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! :) Commented Nov 18, 2020 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:

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: