Solution path for Horse Windoku by jneen

Puzzle at https://f-puzzles.com/?id=y5csr6m4

Note: We will refer to the panes in lexical order 1-4.
We will refer to the four "shadow" regions surrounding pairs of panes by S12, S13, S24, and S34 - indicating which two panes imply the shadow region.

=== Phase 1: The 4 mirror ===
* Notice how powerful the given 4 is with respect to the shadow regions S12 and S13. In fact, it locks a 4 into one of r24c9 and another into r9c24.
* This has an effect on pane 4: the 4 in box 9 must appear within the pane, which eliminates 4 from the rest of the pane.
* In particular, where can 4 go in row 8 or col 8? r8c24 would eliminate both possibilities in the shadow region. r8c15 would eliminate all possibilities in c2 and c4 respectively. So 4 in r8 must be in box 9. But the same exact logic holds for c8 as well. So 4 must be in r8c8.

=== Phase 2: Horse window mirrors ===
* Notice that all the logic from phase 1 was done without relying on any givens other than the given 4. That means the 4 placement is *so powerful* it creates a full mirror across the grid. This mirror must hold for the other 4 panes as well!
* For example, whatever we were to put in r7c3 would have to be mirrored in exactly the same way in r2c8.
* But r2c8 is a 6, so r7c3 cannot be anything but a 6.

=== Phase 3: Edge placements ===
* The next-most-powerful placement of givens are the 7 and the 8. Where can 8 appear in r8? r8c19 are eliminated by the virtual region S34 (and a given 9). r8c37 are eliminated by knight's move. The only place left for an 8 is r8c2.
* Copy this 8 across the mirror into r3c7.
* Apply exactly the same logic to the 7, asking where it can appear in c2. Place it in r2c2 and copy it across the mirror to r7c7. All the mirrors are now complete.
* 8 is now a hidden single in two shadow regions S12 and S24, allowing us to place 8s in r4c1 and r9c8. Similarly, place 6 in r4c9.
* 7 in box 6 are confied to r4c89 by pane 4, and by knight's move and the given 7. This fixes a 7 in r6c4.
* This in turn fixes the 7 in boxes 2 and 8, then box 7. The remaining 7s are somewhere in r34c89.
* Recalling the logic from phase 2, whatever goes in r3c5 must e mirrored in r2c28. But it cannot be a 7, so it must be a 6.
* Similarly, r5c7 must be 4 or 6 to mirror one of r28c8, but it cannot be a 6, so it is a 4.

=== Phase 4: Four corners ===
* 6 in box 5 can only go in r4c4. This fixes a hidden single 6 in box 4, then in box 1. The remaining 6s are in r89c67 in some order.
* Hidden single 4 in box 5 (using knight's move twice as well as pane 4), then in boxes 2 and 3.
* Hidden single 8 in box 5 (or pane 4), in the final remaining corner, then box 6.

=== Phase 5 ===
* 5 in box 3 must be in r1c89, which fixes a hidden single 5 in box 1.
* 8 in box 1 is restricted to r1c13, fixing a hidden single 8 in box 2. Using pane 1, this fixes the rest of the 8s.
* Complete row 2 with 3 and 9.
* Hidden single 9 in box 6.
* (Bonus:) Naked single in the central cell r5c5. All the even numbers as well as 7 are in the column, 5 is eliminated by knight's move, and 1 and 9 would eliminate all possibilities for 1/9 in panes 1 and 4, respectively. It must be 3.

=== Phase 6: pane-pointing 9s ===
* Hidden single 9 in pane 4, must go in r7c6.
* 9 in pane 3 is confined to r6c23, which fixes a hidden single 9 in box 5.
* 2, 3, and 9 left to go in box 1. But r1c3 can't be a 2 or a 9 due to the shadow region S12. So it is a 3.

=== Phase 7: The last stretch! ===
* 5 in r4 (or S12) can now only be in r4c9. This fixes the remaining 7s, as well as a 5 in box 3.
* 3 in r3 can only go in r4c2.
* Hidden single 3 in box 6, then box 5. Then 3 in r7 must be in r7c9.
* 12 pair in c9 implies a 12 pair in the corner shadow region. So r1c1 cannot be a 2. This fixes the 29 pair in box 1 as well as the 19 pair in box 2 and a number of 12 pairs around the grid.

=== Phase 8: Cleanup ===
* Finish box 5 with 1, 2, and 5, then row 5 with 5 and col 5 with a 1.
* Finish pane 4 with 5 and 6, then columns 6 and 7
* Finish row 8 with 2 and 3.
* Finish row 6 with 2, 4, and 9.
* Finish pane 3 with 1 and 5, then row 7 with 4
* Finish box 7 with 5, 9, and 3.

SOLUTION

9 6 8  1 4 7  3 5 2
1 7 5  8 2 3  9 6 4
3 2 4  9 6 5  8 1 7

8 3 1  6 9 4  2 7 5
6 5 7  2 3 1  4 9 8
2 4 9  7 5 8  1 3 6

4 1 6  5 8 9  7 2 3
7 8 2  3 1 6  5 4 9
5 9 3  4 7 2  6 8 1