Partial answer - Curve Data and Tapa solved, stuck on Yin-Yang
a Curve Data clue cannot be a single cell. So we can get a significant amount of the puzzle solved with just that:
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:
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.
But there's a more interesting deduction involving parity.
We have to visit 55 interior vertices total. That's an odd number.
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.)
...and this is where I'm stuck.
## clue on top tells us that R2C2 must be shaded. It needs to escape downwards in order to not break the
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
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.
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.
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):
To make this unique, we need to set F and G to 2.