For solving https://sudokupad.app/zfluc388zd

1. The given 3 in R9C5 and the given 3 in R3C9 leave only three places for a 3 in box 9, but two of those are eliminated by R3C9 and the large clone area, so 3 goes into R9C7.

2. The given 6s in columns 8 and 9 leave only two places for a 6 in box 6, one on the large clone area, and the other on the medium clone area. Consider the large clone area first: if R6C7 is 6, then so is R2C3 by clone, and then 6 in row 3 must be in column 5, and 6 in row 7 must also be in column 5, which won't work. So 6 goes into R4C7, and also R7C4 by clone, and those two digits combine with the given 6 in R9C2 to eliminate 6 from the large clone area entirely. That gives us 6 in R2C1 and then in R6C3, and also R3C6 by clone, forcing a 6 into the center cell of the puzzle, R5C5. We now have all of the 6s.

3. Where can 4 go in box 4? It cannot be in column 2 because of the given 4 in R2C1, and it cannot be in column 3 because R4C3 is a clone of R8C7, which shares a column with the given 4 in R5C7. So it appears in either R4C1 or R6C1. With that in mind, what is the value of R9C1? It cannot be 1, 2, 3, 4, 5, 6, 7, or 8, so it is 9.

4. Other cells also have very few possibilities: R3C3 and R7C7 are clones and share a row, column, or box with 1, 2, 3, 4, 5, 6, and 8, so they can be only 7 or 9. The small single-cell clone are with R2C8 and R8C2 can contain only 1 or 7. But consider where that clone cell goes in box 6: The digit in R2C8 cannot repeat in column 8. It also cannot appear in column 7 of box 6, because R4C7 and R5C7 are the wrong digits and R6C7 is a clone from row 2. So the digit in R2C8 appears in box 6 in column 9, and therefore in column 7 in box 9. It can't be a 3, and it can't be on row 8 because of the clone in R8C2. So R2C8, R3C3, R7C7, and R8C2 are the same digit, and the digit they have in common is 7.

5. We have given 2s in R5C3 and R8C1, so 2 going into either R2C2 or R3C2 in box 1. Both are clone cells, so 2 also goes into either R6C6 or R7C6, and in no other cells in column 6. Now consider R3C7. It shares a row or column with 1, 3, 4, 6, and 7, so it can be only 2, 5, 8, or 9. But if it were 2, that would push the 2 in box 1 up into R2C2 as well as its clone in R6C6, which shares a row and box with R6C4, a clone of R3C7, so that cell and its clone can be only 5, 8, or 9. There is now a hidden pair in column 7. From the bottom up: R8C7 shares space with 2, 3, 4, 5, 6, and 7, but its clone at R4C3 also cannot be 1 because that would leave no place for 1 to go in box 7, so both cells are 8 or 9. The cells at R6C7 and R2C3 together share space with every digit except 5, 8, and 9, just like R3C7. That reveals a 5-8-9 triple, but also leaves only two cells in column into which 1 and 2 can go, so R2C7 and R1C7 are a 1-2 pair.

6. Consider R2C2 and its R6C6 clone. It can only be 2, 3, 5, or 9 based on shared spaces, but if R6C6 is 5, that forces the 5 in column 7 into R3C7 where its clone in R6C4 makes a second 5 on that row. Similarly, if R3C2 is 5, that forces the 5 in column 7 into R6C7 where its clone in R2C3 makes second 5 in that box. So neither R2C2 nor R3C2 nor their clones can be 5. Consider R2C2 and its clone again: if R2C2 is 3, that forces 3 in box 4 into column 1, but R6C6 eliminates 3 from the R4C6/R7C3 clones, leaving no place for a 3 in column 3. With both 3 and 5 eliminated, R2C2 and R3C2 make a 29 pair, as do their clones. This forces the 9 in column 3 into R4C3 and into its clone R8C7.

7. We can fill in some smaller candidate sets. We know in box 1 where 1, 2, 4, 6, 7, and 9 go, even if we don't know which of the 2-9 pair goes above the other, so the remaining three cells must be a 3-5-8 triple. However, there's no place for an 8 in column 3 box 4 or 7, so it must be in box 1, leaving R1C1 to be only 3 or 5. That pairs it up with R7C1, which can also be only 3 or 5 (remembering that 4 is in either R4C1 or R6C1), so the other three cells in column 1 are 4, 7, and 8, while the three cells in column 2 in box 4 are 1, 3, and 5. R6C4 and R6C7 together are a 5-8 pair, so there are no other 5s or 8s on that row, but both are also clones, so there are also no 5s or 8s in R2C9 (which shares a row with R2C3 and a box with R3C7), and also no 5 or 8 in the R4C6/R7C3 clones because R7C3 shares a column with a clone of R6C7 while R4C6 shares a box with R6C4. Coloring clone pairs may help to make these clear.

8. The 8 on row 8 is either in R8C4 or in R8C8, eliminated from both R8C3 and R8C6 (by clone) by the given 8 in R7C2. If R6C4 were 8, it would eliminate 8 from both R8C4 and R4C4, but that is a clone of R8C8, so there would be no place for 8 in row 8 at all. Therefore R6C4 and its clone are 5, while R6C7 and its clone are 8.

9. More reductions: Where can 9 be in row 5? Only R5C4, R5C8, or R5C9. Can the R2C4/R6C8 clone be a 9? Between them, they would leave no place for 9 in row 5, so those clone cells can be only 1, 2, or 3. Now there are only two places for a 9 in column 8. This is also true of columns 2 and 6, where the 9 must be in a clone cell. Colors will help with this: if 9 is in the lower of the two cells in column 2, it must also also be in the lower of the two cells in column 8, R5C8. If 9 is instead in the upper of the two cells in column 2, it is also in the upper of the two cells in column 6, because they are cloned. In either case, there cannot be a 9 in R5C4, nor in R6C9. Column 4 is the important one, because in both cases, R1C4 is 9, which means R1C9 can only be 8. That forces the 8 in box 9 into R8C8 and into its clone in R4C4. In fact, there are enough 8s in the grid now to complete them, filling in R3C5, R5C1, and R9C6.

10. Column 4 contains a 2, and it can't be anywhere other than R3C4 or R4C4. That means there is no 2 in row 1 of box 2, forcing a 2 into R1C7 and a 1 into R2C7.

11. R6C9 can be either 2 or 7, except row 2 tells us that the large clone area contains the two for that row, so it also contains the two for row 6, forcing R6C9 to be a 7. This forces F6C1 to be a 4, and R4C1 to be a 7. Now the only cell in box 5 that can contain a 7 is R5C6, which makes R1C6 a 1 and R1C5 a 7, but also makes R8C6 a 5, and its clone also. This kicks an unwinding into high gear: R4C9 is a 2, so R6C8 is a 3, as is its clone, which means R2C6 is a 4, so R4C6 is a 3, along with its clone, and so on.