Odds and Evens walkthrough

For solving https://sudokupad.app/fypwpu4e55

1. Column 5 has all odd digits as givens, so the remaining open cells are even digits. In box 5, both cells are within a Knight's move of R5C3 and R5C7, so R4C5 and R6C5 are a 2-8 pair. That makes R2C5 and R8C5 a 4-6 pair. Row 5 is all even digits plus a 5, so the remaining open cells are odd digits: 1, 3, 7, and 9. In box 5, both cells are within a knight's move of R3C5 and R7C5, so R5C4 and R5C6 are a 1-9 pair, and R5C2 and R5C8 are a 3-7 pair. The rest of box 5 can be filled in: R4C4 and R6C6 are a 4-6 pair, and R4C6 and R6C4 are a 3-7 pair.

2. Speaking of the anti-knight constraint, the given 1 and 3 (and the two even squares) mean those digits can only go in two places in row 2, so R2C1 and R2C9 are a 1-3 pair. Similarly, the given 7 and 9 mean those digits can only go in two places in row 8, so R8C1 and R8C9 are a 7-9 pair.

3. Time to start coloring!
3a. If R2C5 is blue, then so is R6C6, while R4C4 and R8C5 are orange. The puts a blue in either R8C4 or R9C4, and an orange in either R1C6 or R2C6. Make R4C5 green and R6C5 purple.
3b. Now R2C8 "sees" both blue and orange (including by Knight's move), so it can be neither 4 nor 6, and is either 2 or 8. Similarly, R8C2 "sees" both blue and orange (again including by Knight's move), so it is also either 2 or 8.
3c. Now the question is: are they the same, or different? Let's check 2 first. Because of the Knight's move constraint and the given digits, the 2 in box 2 could only be in R1C4 or R3C4, while the 2 in box 1 could only be in R1C3 or R3C3. Either choice in either box 1 or box 2 eliminates both possibilities in the other box, so our newly-identified 2-8 cells are not both 2. For thoroughness, let's check 8 as well. The 8 in box 8 could only be in R7C6 or R9C6, while the 8 in box 9 could only be in R7C7 or R9C7, and either choice in either box eliminates both options in the other box. So R2C8 and R8C2 are different! We'll resolve which is which in a few more steps.
3d. That means R2C2 and R8C8 each see both 2 and 8, regardless of which is which, they are both 4-6 pairs. R2C2 is orange since there's already a blue on row 2, and R8C8 is blue since there's already an orange on row 8.
3e. That resolves a few things: R2C2 and R2C5 push the orange 4-6 pair into R1C6, while R8C5 and R8C8 push the blue 4-6 pair into R9C4.
3f. R3C7 "sees" both blue and orange by Knight's move, so it is also either 2 or 8, and it is also a Knight's move away from R4C5, so it is purple and R2C8 is green. Unsurprisingly, I hope, R7C3 also can't be either blue or orange, nor can it be purple, so it is a green 2 or 8, and R8C2 is purple.

4. More coloring: We have four blue 46 cells. Consider where blue goes in box 3: R2C5 and R8C8 force blue into one of the corners, but the two corners in column 7 are a knight's move from R2C5, and R1C9 is odd, while blue is even. That leaves only R3C9, which is blue. Now consider box 6: blue cannot be in columns 8 or 9, and in column 7, it can only be R5C7, so blue is 6.

5. We have four orange cells, which are 4s. We can turn the given 4 in R5C3 orange. In box 3, R3C8 is an orange 4. In box 9, the anti-knight rule eliminates column 7, while the 4 we just placed eliminates column 8, so our orange 4 is either R7C9 or R9C9. Now in box 6, the 4 can't go in column 8 or 9, nor in row 4, so R6C7 is an orange 4. The anti-knight constraint again helps us, pushing the orange 4 down into R9C9. In row 7, R7C1 is 4, and that's all nine 4s placed.

6. Back to the blue 6s. In row 7, R7C2 is 6. In box 1, R1C1 is 6. That leaves R4C3 to be the ninth 6.

7. We have three green 28 cells, and it's still true that R2C8's value in box 2 can only be in R1C4 or R3C4. That means green in column 6 is in either R8C6 or R9C6. In box 6, green can be either R6C9 or (if green is 8) R5C9. That means in box 9, green is either R8C7 or R9C7. In box 4, green can be either R6C2 or (if green is 2) R5C1. Both of those cells "see" up into box 1, so green in that box can only be R1C2 or R3C1.

8. We have three purple 28 cells, and it's still true that R8C2's value in box 8 can only be in R7C6 or R9C6. That means purple in column 4 is in either R1C4 or R2C4. In box 4, purple can be either R4C1 or (if purple is 2) R5C1. That means in box 1, purple is either R1C3 or R2C3. In box 6, purple can be either R4C8 or (if purple is 8) R5C9. Both of those cells "see" down into box 9, so purple in that box can be only R7C9 or R9C8.

9. Boxes 3 and 7 now have all even digits, even if the 28 pairs are unresolved. The remaining cells are all odd, but consider the anti-knight constraint and it's effect on 1 in box 3 and 9 in box 7. That gives us 3 in R2C1 and 1 in R2C9, as well as 9 in R8C1 and 7 in R8C9.

10. Let's color the 37 pairs. Make R4C6 yellow and R6C4 red. Because of the anti-knight rule, R5C2 is yellow and R5C8 is red. Yellow in box 6 must be in either R6C8 or R6C9, both of which "see" the 7 in R8C9, so yellow is 3 and red is 7.

11. Let's focus on the top of the puzzle, although the same logic should work the same way along the bottom.
11a. Think about row 2. Our four even digits are spread across five possible cells, with purple in either R2C3 or R2C4. We've filled in 1 and 3, but that means R2C6 and R2C7 are from 5, 7, and 9, with the remaining digit of that trio being whichever of R2C3 or R2C4 is not purple. In fact, because we have four 7s in columns 4, 5, 8, and 9, we know that 7 appears twice in boxes 2 and 3. If 7 is in R1C7, the it must also be in R2C6. Otherwise it's in R2C7 and R3C6. Either way, the 7 in row 2 is column 6 or 7. In fact, R2C6 and R2C7 are both either 5, 7, or 9.
11b. R3C6 might also be 5, 7, or 9, since it see all even digits as well and 1 and 3, but it's a knight's move away from the given 5, so it's only 7 or 9. In fact, the given 5 pushes 5 in row 3 over into box 1, and the given 1 and R2C9 push the 1 in row over into box 1 as well.
11c. So what is in R2C3? It's not 4 or 6, and now we know it's not 1 or 3 or 5 or 7. That leaves only 2, 8, or 9. Let's come back to this cell in a bit.
11d. R3C4 could be 2, 8, or 9, but we've marked it in green but not purple for a reason: It shares a row with purple. In fact, if it's 8, then green is 8, which would mean the green in box 4 is in R6C2, forcing the green in box 1 into R3C1, which would conflict with R3C4 being green at all. So R3C4 is either 2 or 9.
11d. Looking at R2C3 again, can it be 2? If it is 2, then R2C8 (and therefore green) would be 8 and purple would be 2. But with purple in row 3 going to R3C7, and in row 2 going to R2C3, it would in the row 1 go to R1C4. That would leave R3C4 to be green and 8, but we just worked out that cannot be. So R2C3 is either 8 (and purple) or 9.
11e. Where is 9 in box 4? It is in either R4C2 or R6C3. Both of those see R2C3, so we finally reduce R2C3 to a single digit: 8.

12. Purple is 8, and green is 2, which should resolve all nine 8s, while leaving unresolved a quartet of cells in boxes 8 and 9, two of which are green 2s, but we don't know which two.

13. Now there's only one place remaining for 7 in box 4: R4C2. That in turn leave only place for 9 in box 4: R6C3. With a 9 in column 1 and the 9 we just placed in column 3, the 9 in column 2 must be in R3C2. That makes R3C6 a 7, and with the anti-knight constraint, R2C7 is also 7, pushing 7 in row 1 over into R1C3.

14. In row 2, we have two empty cells for 5 and 9, but R2C4 sees the just-placed 9 in R3C2, so R2C4 is 5 and R2C6 is 9.

15. That resolves the 1 and 9 in row 5. Now the 1 in R5C6 and the 1 in R2C9 leave only one spot in box 6 where 1 may go, so R4C7 is 1.

16. We could have picked this up sooner, but R6C8 is 3. With that placed, R1C8 can no longer be 3, and it already can't be even, or 1, or 7, or 9, so it's 5. That means R4C8 is 9, and R4C9 is 5.

17. There's only digit missing from column 8, so R7C8 is 1, which means R7C4 is 3, R8C4 is 1, and R8C3 is 3. Now the just-placed 1 is a knight's move away from R9C2, so R9C2 is 5 and R9C3 is 1.

18. It's column 3's turn it be missing a single a digit, so R3C3 is 5 and R3C1 is 1. That makes R6C1 a 5 and R6C2 a 1.

19. Over in column 9, we're missing two digits, but there's already a 3 in row 7, so R7C9 is 9 and R1C9 is 3. We can finish out box 3 by putting 9 in R1C7.

20. We come at last to our unresolved quartet of cells. Row 7 still needs a 5, so it goes in R7C7. That forces a 5 into R8C6. We also need a 3 in box 9, and the only place it can go is R9C7. That leaves two last cells, and we can put 2 into both R8C7 and R9C6.