Overview ======== - permissible triplets - which is which - polarity relations - resolve parity - sudoku - thermo to finish Nomenclature ------------ The three chevrons are "LEFT", "DOWN", and "UP". There are 6 possible triplets of Between lines "B", German Whispers "G", and Lockout lines "L": B>G>L, B>L>G, G>B>L, B>L>G, L>B>G, and L>G>B, or BGL, BLG, GBL, BLG, LBG, and LGB for short. "forward" and "next", and "connecting to" mean looking at the next line in that direction. Handy hints: as colours are all over the place on teh lines, use corner-marking 'B', 'G', and 'L' on teh midpoints of the line as a mnemonic. All of B, G, and L are polarity-reversible, so you can't tell N from 10-N. The box-aligned chevrons are symmetrically reversible too Because of this, solving for A-I, and mappeing A-I onto 9-1 or 1-9 can be useful. Despite the presence of a lot of [ABC]s ([123]/[789]s), I think polarity is a better colouring than entropy. Permissible Triplets -------------------- As the connection points on a G forward must be equal polarity, they are a maximum of 3 apart. => G cannot connect to L, ever => B>G>L and G>L>B are excluded If the chevron is box aligned, and the either it, or the line that follws is wholly within a box, the line that follows cannot be a B => Box-alinged, only B>L>G and L>B>G are permitted => Therefore the non-box-aligned chevron is maybe G>B>L or L>G>B For both cases, there must be a [AD] or [FI] on the G>B connection sandwiching an I or A G>B>L fails immediately, as the B>L connection a [BC] or [GH], so not a valid lockout. => The only three possible chevrons are B>L>G, L>B>G, and non-box-aligned L>G>B. Which is Which -------------- There's only one non-box-aligned chevron. => UP is L>G>B How to differentiate between B>L>G and L>B>G? The only functional difference is that RIGHT spans 3 boxes, and DOWN spans 2, so the latter might seem harder to fit all the numbers into, but that's not a strong bias. B>G cannot have [AI] on the connection points, which might add more pressure to the G. L>G requires the connection points to have teh same polarity, which might add pressure to the L The most constrained line is the G, so let's try and add more pressure to that until it breaks. Hypothesising L>B>G in both possible chevrons The L line cannot have either of [AI] on the endpoints, as the B must have a wide range. => L endpoints are [BG]/[CH], [BF]/[DH], or [CG], and in DOWN only [BH]. [BH] in DOWN can easily be satisfied, so easy disproof doesn't lie there [BG]/[CH] in RIGHT runs out of [AHI]/[ABI] in box 2, putting 6 digits in 7 cells. [BF]/[DH] in RIGHT runs out of [GH]/[BC] in box 2 just as quickly [CG] in RIGHT runs out of [CG]s in box 2 just as quickly => You can't L>B>G in RIGHT => RIGHT is B>L>G and DOWN is L>B>G Polarity relations ------------------ Arbitrarily selecting UP's GW to have its [ABCD]s in row 7. Coordinating UP and DOWN: If the polarities of the German whispers lined up a foursome, then: - the G line being [GHI] makes the B line impossible, and it's [GHI] because UP's G line has the D, and steals all the Is from DOWN, obliging it to have a F on its own GW. => You can colour the polarity of the GWs in UP and DOWN. => You can colour the polarity of the B line in UP the same colour as its endpoints Coordinating UP and RIGHT: If the polarities of the GWs in column 5 lined up as a foursome, then: - the UP L has at most a G in r9c5, so at most C in r9c3, making an [ABC] triple in column 3. => RIGHT's B line can have no low letter on its endpoints => You can colour RIGHT's G line to not make that polarity foursome You can also remove B from the endpoints of the L line in column 4, as the A it pushes onto the B line and the C it forces into UP's B line together force three cells in column 5 to be B, A, and [AB]. => the other endpoint from C on an L line can't be F, so it's a [CG] pair. => You can remove G from RIGHT's L innards, making r2c6 [AB]. => there's a D on RIGHT's G line, so r2c7 is I => there's a [CG] pair in column 4 and r5c4 is B and r7c4 is H => RIGHT's [HI] pair on its L line forces G and I onto UP's G/L connection => resolving the [AD] on the G line and forcing a F onto the L. => Up to polarity, UP is now solved, columns 3, 4, and 5 being FIDB|HIB|BGAC upwards. => RIGHT's B endpoints are an [AH] pair. => r2c4 is A, pushing A into r4c4 => DOWN's G line has [FG] endpoints connecting to [AB]s => r7c8 is I and r9c8 is H => DOWN's B has [AH] endpoints => DOWN's L cannot have F on its endpoints, so is [CG] Box 3 is missing an H - and it can not go in rows 1 or 3 because of RIGHT, and may not go in column 8 because of DOWN. => r2c8 is H => DOWN almost fully resolves up to polarity, columns GHFA|I?H|CAGB In reality, r2c8's H is a 2 or an 8, and it's on a kropke dot it can't be a 2 paired with a 1 as we already have an I in the box it can be a 2 paired with a 4 (represented as 'F'), it can be an 8 paired with a 4 (represented as a 'D') likewise => r4c9 is an I, leaving a slightly restricted [DEF] in column 9 If it's a 2 paired with a 4, the other kropke has to be a [36], represented as '[DG]' And if it's an 8 paired with a 4, that 4's represented by a 'D'. => Either way the 'D' in row 1 is in box 3 => RIGHT cleans up to rows A[CG]HB|[EF]AI|H[CG]ID => r3c8 becomes A => r5c[12] magically becomes a [AI] pair, but we can't resolve polarity as we don't know if A It's a 2 paired with a 4 and A is high 9 and I is low 1 Replacing letters with numbers is a drag, so take care... Sudoku ------ There's some immediate cleanup, in fact this is pretty much a sudoku speedrun apart from the fact that there is some variant furniture that's unresolved. r1c[78] become 6-3, r1c4 becomes 7, r3c9 becomes 5, r9c9 and r8c4 become 6, r5c6 becomes 2 box 8 reveals its 1, 4, 5, then 7 box 9 resovles its [57] box 1 resolves its [78] box 6 resolves its [58], but leaves a [46] unresolved column 6's resolves its [35] box 2 reolves its [45] column 3 resolves its [37] row 9 completes with a 3 in r9c2 Almost everything apart fom columns 1 & 2 is completed. Thermo to finish ---------------- row 8 completes 7-2, row 7 completes 8-5 row 6 is restricted to [34]-6, resolving box 6 and 5's [46]s, and leaving 3 in r6c1 row 4 completes 2-4 row 3 completes 4-7 row 2 completes 6-8 row 1 completes 5-1 Proof of uniqueness given knowledge about the triplets: https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Between%7ER1C3%7ER1C4%7ER2C5%7ER3C4%7ER3C3.Between%7ER6C7%7ER7C7%7ER8C8%7ER7C9%7ER6C9.Between%7ER7C3%7ER6C3%7ER5C4%7ER6C5%7ER7C5.Lockout%7E4%7ER1C4%7ER1C5%7ER2C6%7ER3C5%7ER3C4.Lockout%7E4%7ER5C7%7ER6C7%7ER7C8%7ER6C9%7ER5C9.Lockout%7E4%7ER9C3%7ER8C3%7ER7C4%7ER8C5%7ER9C5.Whisper%7E5%7ER1C5%7ER1C6%7ER2C7%7ER3C6%7ER3C5.Whisper%7E5%7ER7C7%7ER8C7%7ER9C8%7ER8C9%7ER7C9.Whisper%7E5%7ER8C3%7ER7C3%7ER6C4%7ER7C5%7ER8C5.Thermo%7ER5C1%7ER5C2.Thermo%7ER5C3%7ER5C2.Thermo%7ER8C2%7ER7C2%7ER6C2%7ER5C2.BlackDot%7ER1C9%7ER2C9.BlackDot%7ER1C7%7ER1C8