High and Tight
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For solving: https://sudokupad.app/nddxlely2q

1. Where to begin? Let's focus on the top grid to start, since we know that a five-cell or longer renban line must have a 5 on it. Boxes 1, 2, and 4 therefore each have 5s on lines, and 2 and 4 are entirely self-contained, so R3C3 is 5.
1b. Boxes 6 and 8 also have long lines, and the just-placed 5 in box 1 ensures the lines don't overflow into boxes 3 or 7. If either line tried to have a 5 in box 9, it would eliminate any chance for 5 on the other, so the 5s are on the center three cells of both lines.
1c. That means the only digit possibly contiguous with a 23-cage (which must be 689) is 7, so R8C4 is 7.

2. Speaking of the 23-cage, it exists entirely with box 8 of grid 1, but also entirely within box 2 of grid 2, where it is accompanied by a 16-cage. Although there are eight ways to make 16 in three cells, only one does not involve 6, 8, or 9: the 16-cage is made of 457.
2b. The remaining three cells in grid 2's box 2 are from 123.
2c. Up in grid 1, the 5 in box 6 is in column 7, so R9C7 cannot be 5, and is from 47. Note that that elimination stops at the thick black border, so 5 could still be in grid 2 in either R3C5 or R3C6.
2d. Also in grid 1, we have 6789 placed in box 8, and still need 12345, with 5 in row 7.

3. R9C4 is part of an 8-cage that descends into grid 2. The two options for a three-cell 8-cage are 125 and 134, but there is no place for a 5 within the cage. In grid 1, the 5 is in row 7, not row 9. In grid 2, the 5 is in column 5 or 6, not column 3 or 4. So the 8-cage is made of 134, and the 4 can't be in grid 2's R3C4, because grid 2's box 2 already has a 4. Also, grid 1's R9C5 can no longer be 1. It might be really tempting to say R9C5 can't be 3 or 4 either, BUT it doesn't share a row, column or box in either grid with the "elbow" of the 8-cage, so it could repeat whatever is there. The reason is can't be a 1 is that it would need to be on the renban line with a 2.
3b. Another cage that crosses the border is the 22-cage in grid 2's box 1. As part of grid 2, it shares a box with 7, so it made from 589. As part of grid 2, R9C3 cannot be 5 because of the 5 in R3C3.
3c. The rest of grid 2's box 1, including the tip of the long renban line, are from 12346. But the 4 is locked in either R2C3 or R3C3. Further, grid 2's R3C2 shares either row or box with all three cells of the 8-cage, and so cannot be 1, 3, or 4, and is from 26.
3d. In the same way, R8C3 is on a renban line with the top row of box 8, and via grid 2, shares a box with R9C4. Since that's all the possible options for 1 in box 8, R8C3 cannot be 1, lest there be no option for 1 in box 8. So it's from 236.
3e. More odd geometry: in grid 2's box 2, there are two options for 2. In grid 1, either of those cells would mean there is no 2 in row 7 on the renban line, so those cells are now just from 1345. That means R9C5 is 2.
3f. That 2 means R9C4 can no longer be 4, so the 4 in the 8-cage has to be in grid 2's R3C3.
3g. That 4 pushes the 4 in the 16-cage up into R2C6, aka grid 1's R9C7.
3h. We've placed 4 in grid 2's rows 2 and 3, so where does it go in row 1? To answer this, consider that in grid 1, we've placed 4 in row 9, so where does it go in row 8? It's the same question, but crossing the grid edges as it does, the answer is that 4 is already in grid 1's box 9, as well as grid 2's box 1, and so 4 must go outside of both, in grid 1's R8C1, and also in grid 2's R1C9.

4. Some easier-to-see steps: grid 2's 22-cage cannot have 5 in row 3 because of the 57 pointing pair in the 16-cage, so R2C1 is 5.
4b. This combines with the 5 in grid 1's box 1 to leave only one option for 5 on box 4's long renban line: R6C1.
4c. Where does 5 go in grid 2's box 3? Not in row 3, because of the same 57 pointing pair. Not in row 2, because of the 5 in box 1. So it goes into the 18-cage in row 1, and to avoid conflict on grid 1's negative diagonal, it goes into R1C8.
4d. Since there's already a 4 in box 3, the 18-cage must be from 567. Since there's already a 7 in row 1, R1C7 is 6, and R2C7 is 7.
4e. The forces the 6 in the 23-cage down into R2C5, and the 6 in box 1 into R3C2.
4f. The three-cell renban line in grid 1's box 8 now includes 6 and 7, so R8C5 must be 8, and R8C6 is 9.

5. The other 18-cage in box 3 now cannot have 5, 6, or 7 in it, so it is from 189. That means R2C8 is 3 and R3C7 is 2, which in turns means R2C3 is 1, R3C4 is 3, and R1C6 is 1, while R2C9 is not 1.
5b. The renban line primarily in grid 1's box 8 now has 1 placed, and 345 as candidates on row 7, so R8C3 must be 2 to connect them, and R8C2 must be 3. This completes grid 1's row 8.
5c. Without 2 available, the two-cell renban line on row 7 cannot contain 1, so 1 in column 1 is in R7C1, since there's already a 1 on row 9.
5d. That two-cell renban line also cannot contain 9, since that would force 8 into R9C3, making the renban line broken. In fact, row 9 already contains every digit but 8 and 9, so R9C1 and R9C3 are an 89 pair.
5e. That means the two-cell renban line is a 67 pair.
5f. Taking stock: row 7 is now 1, a 67 pair, a 345 triple, and a 289 triple. For now, we've gotten most of what we can from the boxes where the grids overlap, so we will need to go either up or down from here.

6. Let's go up. Box 6 could be interesting, since we have a tiny bit of information about it, in R7C8. That cell is either 2 (making the renban line 12345 or 23456) or from 89 (making the renban line 45678 or 56789). If it were 2, the line would need 4, which can't go in column 7, so 4 would have to go in R3C8. It would need 3 and 5 and either 1 or 6, but 1 also is already in column 7, so it would be 356 on the line. In that case, the question would be: where does 4 go in box 6? Not in column 7, and not in column 8, and putting it in column 9 would require it to be on a renban line which cannot contain 3 nor 5. All of which to say: R7C8 is not 2.
6b. So R7C8 is 8 or 9, and the line is either 45678 or 56789. If the former, then again the 4 would need to go into R3C8, with column 7 holding 567, while if the latter, we again would need to ask: where does the digit in R7C8 go in box 6? It cannot repeat on the line, nor in the column, so it must in column 9, where it must be on a short renban line next to either 8 (if 9) or 7 or 9 (if 8). In either case, R3C8 shares a row with 5 and a column with 6 and 7, so it's the other of 89 (if it's not 4), and the three cell on the long line in box 6 are always a triple: 567, with no 5 in R6C7.
6c. We do need an 8 in box 6, and it can't go in columns 7 nor 8, and it can't go on the three-cell renban line because 7 is not available. So the two-cell renban line in box 6 is an 89 pair. That makes R7C9 a 2.
6d. We still need 1234 in box 6, with the renban line either 123 or 234, so R6C8 is from 14, and since there's already both 2 and 3 in column 9, R6C9 is also from 14, making a 14 pair.
6e. That makes R4C8 and R5C8 a 23 pair.

7. There's a two-cell renban line in box 3, and five known digits in the column, so the line can be 23, sneaking between the placed 1 and 4, or it can be above the 567 triple and be 89--except it can't, because that would leave no option for R7C7. So R1C7 and R2C7 are a 23 pair, and R3C7 is from 89, paired with R7C7 in the column and R9C1 on the positive diagonal.
7b. Speaking of the diagonal, what else do we have on it? 1 can be in four places, but there are now 2s in boxes 3 and 7, so the 2 on the diagonal must be in box 5, where R9C5 rules it out of satisfying both diagonals at once, and limiting 2 to R4C6 or R6C4. That also means the 2 on the negative diagonal must be in box 1, in either R1C1 or R2C2. That also pushes 2 in box 2 onto the long renban line, since it can't go in R3C5.
7c. We have 3 on the positive diagonal already, and 4 can be in four different places. 5 can only be in R2C8 or R4C6. 6 has three options, 7 has four options, and we started with 89 as a pair.
7d. On the negative diagonal, we've had only two options for 1, but requiring 2 to be on the diagonal in box 1 means that R2C2 can't actually be 1, since it's only sequential with 2. That means there's only one place 1 can be: R5C5, satisfying the 1 for both diagonals.
7e. To make sure our candidates are up to date: The positive diagonal, from R9C1 to R1C9, is: 89, 3, 67, 26, 1, 2457, 89, 45, 467. The negative diagonal, from R1C1 to R9C9, is: 2789, 24789, 5, 489, 1, 78, 89, 6, 3.

8. Where does 3 go on row 6? Not in box 6, and we just determined that R6C6 is from 78, which makes that renban line pretty high, so not in box 5 either. In box 4, there's already a 5 in column 1, and it can't be column 2 because there's a 3 in that column in box 7. So that's a hidden single: R6C3 is 3.
8b. There will need to be a 4 on that renban line to connect the 3 and 5, but it could be in either R4C3 or R5C2.
8c. The rest of row 6 is R6C2, which can be from 267, and R6C5, which can be from 679. As soon as we filled R6C2, we should have seen a 267 triple with C2, C5, and C7. That means R6C6 cannot be 7, and so is 8. It also means R6C5 cannot be 6 or 7, and so is 9. On a renban line, 89 must be joined by 7 in R5C6.
8d. Placing the 8 made R7C7 9, and placing that 9 or the 9 in box 5 made R4C4 4.
8e. R1C1 and R2C2 are now a 27 pair.
8f. We just said moments ago that the 5-cell renban line in box 4 needs a 4, and now we know that R5C2 is 4.
8g. The 9 we placed in R7C7 means that R7C8 is 8, as is R3C7.
8h. R9C1 is now 9 via grid 1's positive diagonal, so R9C3 is 8, which leaks down into grid 2, where R3C1 is 9. Also in grid 2, R2C9 is 9.

9. After cleaning up pencil marks for placed digits and focusing back on grid 1, we might ask what is remaining in box 5. We have 25 and 26 in the corners, but need 356 in R4C5 and R5C4.
9b. The 1s in R9C4 and R5C5 push the 1 in box 2 onto the 5-cell renban line, making it 12345.
9c. Since the renban line is low, now column 4 has a 2356 quadruple on rows 3, 5, 6, and 7, or we can ask where 8 and 9 go in that column, the answer to which is R4C1 and R4C2.
9d. In box 2, 6 and 7 go into R1C5 and R3C5 as a 67 pair.

10. Where does 1 go in box 4? The 1s in box 7 and 5, plus the pointing pair in box 6, give us two options: R4C2 and R4C3. But the 5-cell line in boxes 4, 1, and 2 now includes a digit from 89 and a digit from 67, so it cannot also include a 1, so R4C3 is 1, and that is a 12345 line, so R6C2 is 2.
10b. With 2 placed, R6C4 is 6, so R7C3 and R6C7 are both 7, and R7C2 is 6.
10c. That also resolves a pair in box 1: R2C2 is 7, so R1C1 is 2.
10d. Since we've now placed both 6 and 7 on the positive diagonal, R1C9 is 4. That makes R2C8 5, which makes R4C6 2.
10e. Placing that 4 also unwinds the long box 6 renban, which is now a 56789, with 9 in R3C8.
10f. It also makes R6C9 1, so R6C8 is 4. The 2 means R4C8 is 3, and R5C8 is 2.
10g. In box 5, R4C5 is 5 and R5C4 is 3. In box 6, R4C7 is 6 and R5C7 is 5.

11. The 2 we placed in R1C1 resolved the two-cell renban line in box 3, where R1C7 is 3 and R2C7 is 2.
11b. The effect of all this activity on box 2 is dramatic. We know that R3C4 is 2 and R1C6 is 5.
11c. Down on row 7, R7C4 is 5.
11d. In column 8, we have placed eight digits, so R1C8 is 1.
11e. In box 1, to answer whether 7 is paired with 6 or 8, we should first look to box 3, which is missing 6 and 7, since the 7 in row 2 tells us that R2C9 is 6 and R3C9 is 7, and that in turn tells us that R2C1 is 8.
11d. These resolve some pairs in box 2: R2C4 is 9, so R1C4 is 8. R3C5 is 6, so R1C5 is 7.

12. To get from 5 to 9 sequentially, we need more than just 7, so R5C1 is 6 and R4C2 is 8.
12b. This resolves R4C9 to 9 and R5C9 to 8.
12c. Our missing digits in box 4 are 7 in R4C1 and 9 in R5C3.
12d. In row 1, R1C2 is 9 and R1C3 is 6.
12e. In column 3, R2C3 is 4, which gives us R2C5 as 3, R2C6 as 1, and R3C6 as 4.
12f. That resolves R7C5 to 4 and R7C6 to 3.
12g. Speaking of 3, R3C1 is 3, and R3C2 is 1. Grid 1 is now done! We made use of cages from grid 2, and even some funky box geometry to get started, but grid 1 is now done, and six more boxes plus two unresolved pairs remain. And not a ton of cages, but I see some known values.

13. We started with a 23 cage back in step 1, and box 9 has a 24 cage, which must be 789. Except R8C9 can't be 9 because of R2C9, and R8C8 can't be 8 because of the negative diagonal.

14. The 6-cage in box 8 is from 123, and 2 and 3 are already in column 4, so R8C4 is 1.
14b. The 8-cage in box 4 is limited. It must contain a 1, which must be in column 2. It then also contains either 2 (in R4C3) and 5, or 3 (in R4C3) and 4. After 24, 6, and 8, the other cages are less immediately helpful.
14c. Where does 3 go in box 5? The diagonals push it into the four edge cells, three of which are in columns which already contains 3, so R5C6 is 3.
14d. What about 2? As discussed, 2 and 3 are in column 4 already, and there's a 23 pointing pair in column 5, so it's just above or just below the 3 we just placed. The 2 in R3C7 is on the positive diagonal, so 2 goes into R6C6.
14e. That takes care of the 2 on the negative diagonal, so where does 2 go on row 9? Not in C1 because of the diagonal, and there's already a 2 in C2. The pointing pair is enough to eliminate C4, C5, and C6, even if there weren't also 2s in C4 and C6. C7 has the same 2 that eliminated the positive diagonal, and C8 is a high digit. We *just* placed 2 on the negative diagonal, so C9 is out, leaving only R9C3 to be 2.
14f. That brings us back to the 8-cage, which can no longer be 125, and so must be 134, with 3 in R4C3.

15. Remember the trouble we had placing 2 in row 9? Let's try it in row 8. We can't place a second 2 in box 7, and box 9 is high digits except for column 7 with a 2 in it. So it must be in box 8, where we've already noted it's part of a pair, so we can resolve it: R8C5 is 2 and R7C5 is 3.
15b. On row 7, 2 is now over in box 9, but it has two options there, so let's think instead about 3. We've placed it on row 7, where does it go in row 8? Not in box 7, where two placed 3s eliminate columns 1 and 3, while the positive diagonal eliminates column 2. Not in box 8, where we just placed a 3. Since 3 isn't high enough to be part of the 24-cage, R8C7 is 3.
15c. Back over in box 1, on row 9 the diagonal doesn't affect column 2, so R9C2 is 3.
15d. We've got so many 3s placed already, we need only one more: R6C9 is 3, and that's all the 3s.

16. It's tempting to set up the 14-cage with its eight options, point out that the 3 in the cage and the 2 in the box reduce the options to two, and move on, but instead, consider: where does 4 go in row 8? We've filled in a number of cells, and the remaining cells are either too high, have placed 4s in the column, or have a pointing pair including a 4 in the column, so R8C1 is 4. Which resolves the 14-cage much more directly, so R8C2 is 7.
16b. Now that 7 does some work. It means the 24-cage is completely resolved: R9C8 is 7, R8C8 is 9, and R8C9 is 8.
16c. That bounces up to box 3 and puts 8 in R3C8 and 1 in R3C9.

17. Let's look at the diagonals for a moment. We just placed 7 on the positive diagonal, so where does it go on the negative diagonal? It can only go in R4C4.
17b. Although there are eight ways to make 16 in three cells, a fact that has not changed since step 2, only one does not involve 2, 3, or 7: the cage is made up of 169, with the 1 in R6C5, and a 69 pair filling out the cage.
17c. The other cells in box 5 are from 458, but the center cell is on a diagonal with 4 and 8 already on it, so R5C5 is 5, which resolves R3C5 to be 7 and R3C6 to be 5.
17d. Now the remaining cells in box 5 are from 48, and the *positive* diagonal comes in again, so R4C6 is 8 and R4C5 is 4.
17e. That makes R4C2 to be 1, and R5C2 is 4.
17f. It also means R9C5 is 8, the only digit missing from the column.
17g. We need only two more digits on the negative diagonal, 1 and 6. The 6 in row 1 means R7C7 is 1 and R9C9 is 6.

18. Moving quickly now. Row 8 needs only 5 and 6, and there's a 5 already in column 6, so R8C3 is 5 and R8C6 is 6.
18b. Column 6 still needs 7 and 9, and you can either notice that 7 is in row 9 already, or you can observe that placing 9 in the 12-cage with a 1 would require a 2 adjacent to a 2 in row 6. Either way, R9C6 is 9, R7C6 is 7, and to complete the cage, R6C7 is 4.
18c. Row 9 looks almost full. It needs only 4 and 5, and we have a 4 in column 7, so R9C4 is 4 and R9C7 and R7C4 are both 5.
18d. To finish off box 9 we need 2 and 4, and we have a 4 in column 9 already, so R7C8 is 4 and R7C9 is 2. And hey, we've placed all the 4s!

19. Box 7 needs 689, with 68 in column 2, and 8 on the positive diagonal, so R7C2 is 9, R7C3 is 6, and R7C1 is 8.
19b. That resolves the 16-cage, so R6C4 is 9 and R5C4 is 6.
19c. Here's a fun cell: R6C8 is immediately surrounded by 1234 on the row, and more distantly in line with 5789 in the column, so R6C8 is 6.
19d. The rest of row 6 has so many placed digits affecting it, it's settled. Left to right, 7, 5, 8. Column 3 needs 9 in R5C3, and that's all the 9s.
19e. The box needs 2 and 6, so R5C1 is 2, and R4C1 is 6.

20. The remaining six digits are all resolved and need only to be typed, so I will leave that for you.

Trivia: The tool I used to create this puzzle could not solve it logically, even with all options turned on. It could handle the bottom grid well enough, but made almost no progress at all on the top grid. Step 6, eliminating the 2 from R7C8, is all the help the solver needs to finish the puzzle. So that's officially the hardest step in the puzzle.