Overview: --------- - Get the DNA strand vibe - No Es on the DNA strands - Middle crossovers are are from [CG] - Rest of the crossovers - Interactions between strands ... DNA Strand Vibe --------------- Along the two DNA strands, I'll call the points where the two lines meet 'crossovers'. Vibe 1a: every third crossover is identical - it's 6 cells away on both lines 1b. Therefore both strands have a repeating sequence: let's call them PQRPQ (left) and UVWUV (right) (top to bottom) 1c. Note in passing that P & Q both see both of U and V. Vibe 2: Also note that the DNA strands - being only entropic, modular, and whisper - have no intrinsic polarity. I.e. you could replace x with 10-x on all cells on the strands and they're all just as satisfied. So this is screaming to be solved in A-I rather than 1-9. No Es on the DNA strands ------------------------ Obviously none on the GWs, but what about the crossovers? Chose one to put a 5 on. Move 3 steps along one way or the other along both lines. You're on a GW that connects a [28] (modular line) with a [46] (entropic line)! Good luck with that. The middle crossovers are from [CG] ----------------------------------- R and W are the middle twists. Stick an A on R or W, what's 3 steps away in both directions? On the entropic line it's [ABC], and on the modular line its' [ADG], but as they share a GW it can only be [AB]-G. Both of them need a G - good luck with that. Stick a B on R or W, what's 3 steps away? Two Hs! Stick a D in R or W - two As! So R and W are from [CG] or [37]. Rest of the crossovers ---------------------- PQR (and UVW) must each be different modularity, and different entropy. Therefore thet must be one of the diagonals in the number pad: \ AEI BFG CDH or / AFH BDI CEG. Nothing with an E is permitted, and there must be a C or F, so it's BFG or CDH (267 or 348). These are a polarity pair, so WLoG the left strand can be CDH By mutual visibility of PQ and UV, the right one must be BFG Interaction between strands --------------------------- You can pencil mark the whole strands, many down to 2 options, a fair few making pairs. The [DH] on r8c2 and r9c3 both see the [AD] on r8c5, making it A. Just from the on-line relations, you can resolve almost all of both strands, with just some ambiguity on the entropies. Simple sudoku lets you do a lot more, completing both strands(except r6c7) and box 8. Polarity disambiguation ----------------------- We've got a sum, that's clearly the best way of disambiguating high from low, right? F+H+[AE]+[EGH]+[CDE] minimise A-lo: 6+8+ 1 5 3 = 23 - only possibility maximise A-hi: 4+2+ 9 5 7 = 27 - need to drop 4 possible A-hi: 4+2+ 9 3 5 = 23 - proof follows: This fixes r6c4 from [AE] to A, filling a few cells, but more importantly removing H from [EGH]. So with A-hi, there's only one way to do make the sum - [EG]->G and [CDE]->E. With r7c3 now [CE], we can place the D in box 7, in r8c1 Because of the restrictions on the location of Gs and Es, r9c2 being G forces 2 Es into box 7. We now have a [BCE] triple in box 9, so r9c1 is G. The [CE] pair in row 7 helst tidy box 9 a bit too. Still no resolution to the genetic polarity question. Work out what X and Y do ------------------------ If we remove [BC] from X & Y and leave an unsolvable puzzle, then [BC] != [23], and A is 9 If we remove [GH] from X & Y and leave an unsolvable puzzle, then [GH] != [23], and A is 1 B and C are more represented on the X & Y, (11 instances, on cells with tighter constraints, if I've counted correctly), so that's the one that's more likely to break, so the one to tackle first. Time to colour - but don't put 'can be B' or 'can be C' colours on the X & Y, and also remember that r7c3 is forced to be C by the sum. Hypothesising they're not on the XY, B should be uniquely placeable, including r9c2. Hypothesising they're not on the XY, C isn't - but it's definitely on r7c3. So r8c2's [BC] can't be B or C. => [GH] is [23], and A is high! At this point, it's easier to just continue with A-I with that knowledge, rather than rewrite pencillings. => r6c3 is G, above E, resolving column 3. knowing r7c9 can't be H resolves the whole row knowing r8c8 can't be G puts G in r8c9 row 1's D is in r1c8 box 3's H is in r2c8, resolving column 1 box 3 now has an [AFI] triple, making r2c8 C, resolving row 5, and placing C in r3c2 It now also has a [BE] pair, placinf the G in r3c8 I now used what I call 'hot-cold' colouring, but there's probably a better name for the technique. r4c8 and r6c8 are both their higher option (hot) or their lower option (cold), and r8c8 and r9c8 are the opposite. Likewise r4c9 and r6c9 are the opposite But r8c7 is the same heat as r4c8. r3c7 and r3c9 are of opposite heat. if r4c8 is cold, r8c7 is cold, making r3c7 hot and r3c9 cold if r4c8 is hot, r4c9 is cold, making r3c9 cold and r3c7 hot => r3c7 is hot (E) and r3c9 is cold (B) This resolves all those cells, and lets us complete row 9 and column 9 And box 3, and everything. Now set A to 9... Proof of uniqueness: https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Whisper%7E5%7ER3C3%7ER4C3.Whisper%7E5%7ER5C1%7ER5C2.Whisper%7E5%7ER7C1%7ER7C2.Whisper%7E5%7ER8C3%7ER9C3.Whisper%7E5%7ER6C6%7ER6C7.Whisper%7E5%7ER4C7%7ER4C6.Whisper%7E5%7ER3C5%7ER2C5.Whisper%7E5%7ER7C5%7ER8C5.Modular%7E3%7ER3C4%7ER4C3%7ER4C2%7ER5C1%7ER6C1%7ER7C2%7ER8C2%7ER9C3%7ER9C4.Modular%7E3%7ER2C4%7ER3C5%7ER3C6%7ER4C7%7ER5C7%7ER6C6%7ER7C6%7ER8C5%7ER8C4.Entropic%7ER3C4%7ER3C3%7ER4C2%7ER5C2%7ER6C1%7ER7C1%7ER8C2%7ER8C3%7ER9C4.Entropic%7ER2C4%7ER2C5%7ER3C6%7ER4C6%7ER5C7%7ER6C7%7ER7C6%7ER7C5%7ER8C4.Sum%7E23%7ER7C3%7ER6C3%7ER6C4%7ER6C5%7ER5C5.%7ER1C7_1_4_5_6_7_8_9%7ER2C8_1_4_5_6_7_8_9%7ER3C9_1_4_5_6_7_8_9%7ER1C9_1_4_5_6_7_8_9%7ER3C7_1_4_5_6_7_8_9%7ER7C7_1_4_5_6_7_8_9%7ER8C8_1_4_5_6_7_8_9%7ER7C9_1_4_5_6_7_8_9%7ER9C7_1_4_5_6_7_8_9