There's so much symmetry here, most of the solve is working out what the jigsaw pieces are that are put around the ring of black kropkes. I'll refer to the possible pairs of boxes as either 'chair ' or 'table': Chair, e.g. boxes 1&2: | <- back back +-x-+- <- front leg -> | | <- front leg Table, e.g. boxes 2&3: left -> -+-x-+- <- right left leg -> | | <- right leg So what can go on the kropkes? It's a much simpler question than it first seems, if you ask the questions in the right order: Are there any 1x2 kropkes? -------------------------- table) No. You can't fill the 6 table-top cells with the only 5 available reachable digits. chair) No: the 3 and 4 on the back make the table 3x6 linking to impossible 4x8 or 8x4, or 4x8 linking to impossible 3x6 Therefore there are no 1x2 kropkes. Are 3x6 kropkes needed at all? ------------------------------ Obviously, as you can't put [24] and [48] in the same row Can 3x6 kropkes connect the not-obviously impossible (see above) way? --------------------------------------------------------------------- Edge chair-table being obviously impossible. Corner table\chair needs examining: Both 6x3 and 3x6 on the chair put [45] on the back With 6x3, the 6 forced off the leg can only connect to an impossible [48] on the next table kropke With 3x6 it has the ability to instead connect to a 2 on the next table kropke, but that can't pair with either 1 or 4. Does that mean it alternates [36]-[248]-[36]-[248] all the way round? --------------------------------------------------------------------- We didn't disprove [248]-[248] around a corner, but we don't need to. As soon as you have [248] on a chair, the table beside it must be [36]. And the table round the corner down its back leg must be [248]. Ad circulum. So, yes, [36] and [248] do alternate. Which way round? ---------------- a) a 3x6 on a chair puts a 4 on its back leg with an 8 on its table. As 8 can only renban to 6, that sequence must complete the full circle. All the table legs pointing into box 5 must be odd, 5,7,9 - there aren't 4 even numbers that can pair those on a kropke. Rejected, not even one may be 3x6. b) a 6x3 on a chair (and therefore all chairs) is only slightly trickier: the 3 does not reject the possibility of 2 on the adjoining table, pairing with a 4 - leaving [1245] on the shared leg, and [125] in the joint backwards deductions are no stronger than that, but that's enough to disprove the case. all the [12]-[1245]s line up, and the four white kropkes in box 5 can only be [36]. Again - rejected. Therefore it's [248] on the chairs and [36] on the tables. Can we do without any [48] kropkes? ----------------------------------- If there are only [24] and [36] kropkes, the 6s connect to at most 4 on the renbans, so the highest value on a renban is 7. There are 8 renban cells in a row. So clearly a value of at least 8 must be on all of the 4 lines of 8 (it may be 9, as the 8 may be off to the side). So, we definitely need some [48] kropkes. Can we do without any [24] kropes? ---------------------------------- Whichever way you put [48] on a chair, the 8 renbans to a 6x3, which renbans to a 4x8 and you get a cycle. But the white kropkes are skuppered either way. 4x8 leads to needing 4 partners for [579], and 8x4 leads to [125]-[1256]...[1256]-[125] quads similar to the above, you need 4 partners from [347]. So, definitely we need some [24] kropkes too. One golden rule to remember: a 2 always connects over a renban to a 3, and an 8 always connects to a 6. 4s permit order inversions from 3x6 to 6x3, but as there's always a 4 on the non-[36] kropke, that tells us nothing about how many 3x6 vs 6x3s there are. But we can do better - a chair can't be 2x4, as the 4 on the back of the chair would block both [24] and [48] on the next table down. Let's list the possible renbans, just to familiarise ourselves with all the playing pieces for this jigsaw: edge chair-table: 2x4 - 6x3 : impossible (3 can't be extended forward) 4x2 - 6x3 : impossible ( -"- ) 4x8 - 3x6 : impossible (8 and 3 can't join) 2x4 - 3x6 : impossible (2 can't be extended backwards) 4x2 - 3x6 : 2-1-3 2-5-3 4 4 8x4 - 3x6 : 4-1-3 4-2-3 4-5-3 (6) 2 5 2 (note: 1 is verboten because of white kropke) 8x4 - 6x3 : 4-5-6 4-5-6 4-7-6 3 7 5 4x8 - 6x3 : 8-5-6 8-7-6 8-7-6 8-9-6 7 5 9 7 corner table-chair: 3x6 / 2x4 : impossible (6 and 2 can't join) 6x3 / 8x4 : impossible (3 and 8 can't join) 3x6 / 4x2 : 6-5-7 6-7-5 4 4 However, because of what we know already, the rear 3 connects back to an 8x4, which connects back to a 3x6, whose 3 is blocked from connecting because of the 541 caused by the forward 2 connecting to a 3x6. So this is impossible. 3x6 / 4x8 : 6-5-7 6-7-5 4 4 (note: the 4 forces the 3 to connect to an 8x4, not a 4x2) 3x6 / 8x4 : 6-5-7 6-7-5 6-7-9 6-9-7 8 8 8 8 6x3 / 4x2 : 3-1-2 3-5-2 4 4 (note: must connect to a 3x6 via [15]/4, then an 8x4 via [79][79] - thus has an 8 on top) 6x3 / 4x8 : 3-1-2 3-2-1 3-2-5 3-5-2 (note: topped with [79] by forced connections) 4 4 4 4 Can all the [63] kropkes be the same orientation? ------------------------------------------------- They can't be all 6x3, as that's the no-2x4 situation. So there's at least one 3x6. 3x6's more fun, it needs to connect to something on the chair that an connect to another 3x6 on the next table. Namely one of these: 3x6 / 4x8 : 6-5-7 6-7-5 4 4 (note: the 4 forces the 3 to connect to an 8x4, not a 4x2) 3x6 / 8x4 : 6-5-7 6-7-5 6-7-9 6-9-7 8 8 8 8 Connecting to one of these: 4x2 - 3x6 : 2-1-3 2-5-3 4 4 8x4 - 3x6 : 4-1-3 4-2-3 4-5-3 (6) 2 5 2 So it must be an 8x4 in between every single one - a case already ruled out. So there must be at least one 6x3. But we can go further... Can there be three of one orientation of [63] kropke? ----------------------------------------------------- If we have three 3x6's, then we have a 6x3-?x?-3x6. So first, one of these: 6x3 / 4x2 : 3-1-2 3-5-2 4 4 (note: must connect to a 3x6 via [15]/4, then an 8x4 via [79][79] - thus has an 8 on top) 6x3 / 4x8 : 3-1-2 3-2-1 3-2-5 3-5-2 (note: topped with [79] by forced connections) 4 4 4 4 Connects to one of: 4x2 - 3x6 : 2-1-3 2-5-3 4 4 8x4 - 3x6 : 4-1-3 4-2-3 4-5-3 (6) 2 5 2 (note: 1 is verboten because of white kropke) So it's a 4x2 and the 2 connects to the 3 using a 2-[15]-3/4 renban However, it's oppoise a 8x4-3x6, whose renban is 4-[12]-3/[25]. You can't put [12]-[25]-[1346] [35]-4-[15] on the same row. So there must be at least *two* 6x3s. But can there be 3? Again, there must be a 6x3-?x?-3x6 So first, one of these: 6x3 / 4x2 : 3-1-2 3-5-2 4 4 Connects to one of: 4x2 - 3x6 : 2-1-3 2-5-3 4 4 And it's opposite 4x8-6x3, which must be 8-[579]-6/[579], which has [468] in its white kropke. However, the table before it has a 6x3, and the chair after it as a 4x8, which pushes 6 and 8 into the same column as the [468], and 4's also excluded. Therefore there are two 3x6s and two 6x3s. Can they alternate 6x3 3x6 6x3 3x6? ----------------------------------- No, as a 2-[15]-3/4 will be opposite a 2-[15]-3/4 Therefore there are two pairs. 6x3-4x2-3x6-8x4-3x6-[48]x[48]-6x3-4x8-6x3 From this we can learn some things about the cloned dominoes. What's on the cloned dominoes? ------------------------------ Just place the cycle of 8 kropke numbers in one hypothetical orientation. Then in turn pretend the cloned cells are in the four rotationally-symmetric positions. See what's excluded from the dominoes - usually [3468], but also what's forced into one of the dominoes (the edge ones are more informative) - usually [34]. You can't both have and not have the same numbers, so reject those orientations. That leaves 2 possible orientations, in one it's forced to be an 8 and a [79], in the other, it's forced to be a 1 and a [79]. However, the 1&[79] case requires an unsatisfiable 8x4-3x6 box because the 4-3 is a 4-1-3/2 whose 2 needs to pair with a 3 that's blocked by the 3x6 in the table back round the grid. Rotate your numbers if need be. So, can we start doing sudoku yet? ---------------------------------- Yes, go for it! Proof of uniqueness ------------------- https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Renban%7ER7C5%7ER8C5%7ER8C6%7ER8C4.Renban%7ER3C5%7ER2C5%7ER2C6%7ER2C4.Renban%7ER5C7%7ER5C8%7ER4C8%7ER6C8.Renban%7ER3C8%7ER2C8%7ER2C7%7ER2C9.Renban%7ER2C3%7ER2C2%7ER3C2%7ER1C2.Renban%7ER7C2%7ER8C2%7ER8C3%7ER8C1.Renban%7ER5C3%7ER5C2%7ER4C2%7ER6C2.Renban%7ER8C7%7ER8C8%7ER7C8%7ER9C8.BlackDot%7ER4C2%7ER3C2.BlackDot%7ER8C3%7ER8C4.BlackDot%7ER8C6%7ER8C7.BlackDot%7ER6C8%7ER7C8.BlackDot%7ER2C6%7ER2C7.BlackDot%7ER2C4%7ER2C3.BlackDot%7ER3C8%7ER4C8.BlackDot%7ER7C2%7ER6C2.WhiteDot%7ER3C5%7ER4C5.WhiteDot%7ER7C5%7ER6C5.WhiteDot%7ER5C3%7ER5C4.WhiteDot%7ER5C6%7ER5C7.SameValues%7E3%7ER3C3%7ER4C3%7ER1C8%7ER1C7%7ER8C9%7ER9C9