The digits must be adjacent, and all X pairs (19, 28, 37, 46) are not, so it is a V. It can't be 14 (also not adjacent) so it must be a 23 pair.
2. Box 5 r3c5 - r6c5
X pairs split around 5 and can't contain a 5, so by themselves they can't made a consecutive sequence. V pairs also can't contain a 5, so any run that is purely pairs must have no Xs. Therefore this length 4 renban with 2 pairs on it must be two Vs, giving 1234 as the possible digits.
3. Boxes 1, 3 & 9 - length 5 renban lines with two pairs.
These can only be 14.23.5, 23.46.5, or 37.46.5 Either way, they contain a 5, and the 5 can't be on a pair, so there must be 5s in r1c5, r3c8 and r9c9
4. Boxes 6 & 7 - length 9 and length 7 renbans
Any consecutive sequence longer than 4 digits much include a 5, and there's only one non-pair space on both lines, so those must be 5s at r5c7 and r8c1
If the middle dot is an X, one of the neighbours must be higher than 5, which can't be part of a V pair, and if it is XX then that would bring you back to the same number (eg 2-X-8-X-2) which you can't have on a renban. So the central dot must be a V and similar reasoning prevents having VV so the dots on either side of the V must be Xs.
7. Boxes 7, 8 & 9 - bottom two renban lines
As seen in step 3, the highest digit the renban in box 9 can have is a 7. In step 6 we showed the renban in box 7 must have two X dots, implying 46.5.78.23 By elimination, the two spaces on the bottom row not in a line (r9c5, r9c6) must contain a 9, and either a 1 or a 6 depending on whether r9c7-r9c8 are 14 or 46
8. Boxes 9 & 6
The other dot in box 9 must be either 23 or 37, both of which contain a 3, which means r6c9-r5c9 can't be a 23 pair and so must be a 14 (putting a 96 in the column 8 spots on that line, and forcing the other end of the line to be a 23 surrounded by a 78).
9. Box 8 - The square of 4 cells r89,c56 containing 1469
We can't have a 1 or 4 in column 5, and we can't have a 4 in row 9, so r9c6 must be the 1, giving us a 9 at r9c5, a 6 at r8c5 and a 4 at r8c6
10. Box 5 - length 4 renban
Pencil marking in the information we have so far (r6c6 must be a 2 or a 3) lets us distinguish the 23 and the 14 (23 at the top) in column 5, which then combined with the 14 pair in column 9 let us resolve the 46 in column 1, which then bounces back to resolve the orientation of the 68 in column 8, followed by the orientations of the 14 in column 9 and the 14 in column 5.
11. Boxes 3, 6, 9 & 5
We can now follow through and fill in the remainder of the digits on the longer lines in those boxes, before we turn our attention to:
12. Boxes 1, 4 & 7 - length 9 renban
We can narrow it down 19.4.28.5.6.73 and resolve those pairs, which also gives us the rest of the length 7 line in box 7, the orientation of the 37 pair in r1c1-r1c2 and the orientation of the 23 in box 3.
13. Box 1, 2 & 3 - length 5 renban
We need a 6 because we have a 7, so the pair in r1c3-r1c4 must be a 64 not a 14, which then gives us the remaining digits in the grid.
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SudokuSolves/RomanDyslexia