Overview: ========= Understand the power-parity difference between {1,4,6,9} and {2,3,8} Dabble with 5s and 7s to find the first in Polarity of row 9 Power parity in column 2 Identify next highest pressure area - apply pressure there Bring it home Understand power parity ----------------------- Draw a graph of possible adjacent values along the tendrils. Each step either adds or removes a power of 2 or 3, so their parity alternates. 1 has zero, 2 and 3 have one, 4, 6, and 9 have two, 8 has three. Also note that 9 and 8 only have one way out (OWO), and 4 only one next place to go. Dabble with 5s and 7s --------------------- corner-mark them in - notice how box 6 row 6 constrains box 4 => 7 in r4c1 and 5 in r5c3 => 5 in r9c2 pencil the GW now you know its polarity It looks like boxes 4 and 7 are the "in" Power parity in column 2 ------------------------ If r7c1 has odd power parity (2 or 3), so must r[68]c2. One of r[45]c2 must be the third. By OWO, the 8 is one of the forks, needing to connect to a 4, making r7c1 and r6c2 a 2. The tendril on box 4 must start 1-3-, but no longer can be completed => r7c1 is even power parity, therefore is 1 => r5c1's tendril is 2-[16]-3-9 => r7c1's forked tendril is 1-2-[146]/[46] Polarity of rows 1 and 9 ------------------------ column 2 leaves a [789] triple in box 1, resolving the GW's polarity => there's a [34] pain in column 1, and thus a [56] pair too. row 9's tendrils can't begin with a high polarity, as they'd need a second 2 in row 7. Column 9 -------- Box 4 puts a lot of restrictions on box 6 In fact, a tendril of length three that has no 2 on it must have 3 in the middle => r5c8 is 3, and r5c9 is [19], and one of r5c[79] is 9 to not break r5c2 The tendril above can only be 2-4-8, or 8-4-2, because it can't contain 1 and 2, as that leaves r6c9 unfillable => r4c8 is 4 , flanked by a [28] pair. The pressure on the GW is now immense, with 2, 3, and 4 all in box 6: If r6c9's 1, that GW's unsatisfiable, beginning 1-9-2. => r5c9's the 1 And that's huge! r4c9's tendril resolves 2-4-8 r5c2 being 6 makes r5c7 9, and r8c8 4 => that's 3 in the corner => box 4 is resolved => Row 1 begins 4-9-, and row 9 ends -1-9-4 => the tendril from r9c7 is 1-3-6 => the tendril from r9c5 is 2-[16]-3 => 9 goes in r6c5 and 1 in r1c5, then r8c5 and r9c3 are 6, and r4c5 is 5 => column 5's 7 is in r5c5, as the tendril can only have [48] on it. => r1c4 is 7, needing a 2 to its left on the GW, and putting 6 in r1c6 => that's another 3 in the corner, need a 9 below it, and 2 below that. => the [16] is row 4 resolves, ditto the [78] in row 9 which resolves the [48] in row 5 => OWO makes box 2's 4 above 8, which resolves teh [78] in box 1 => column 9's 6 is in r6c9, leavinga [57] => box 9's [28] are 8 above 2, resolving box 7's [78], box 97' [57], and box 3's [58], and box 6's [57] => box 8 fully resolves The top boxes also all fully resolve as teh tendril can only be 2-1-2-4-8 Bosh! Proof of uniqueness: https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Whisper%7E5%7ER1C1%7ER1C2%7ER1C3%7ER1C4%7ER1C5%7ER1C6.Whisper%7E5%7ER1C9%7ER2C9%7ER3C9%7ER4C9%7ER5C9%7ER6C9.Whisper%7E5%7ER9C9%7ER9C8%7ER9C7%7ER9C6%7ER9C5%7ER9C4.Whisper%7E5%7ER9C1%7ER8C1%7ER7C1%7ER6C1%7ER5C1%7ER4C1.Binary%7EGIFhUQAwAAABE%7E_ambiguous%2520kropke%7ER7C1%7ER7C2%7E%7ER7C2%7ER8C2%7E%7ER5C1%7ER5C2%7E%7ER5C2%7ER4C2%7E%7ER9C5%7ER8C5%7E%7ER8C5%7ER7C5%7E%7ER9C7%7ER8C7%7E%7ER8C7%7ER7C7%7E%7ER5C9%7ER5C8%7E%7ER5C8%7ER5C7%7E%7ER4C9%7ER4C8%7E%7ER4C8%7ER4C7%7E%7ER1C3%7ER2C3%7E%7ER2C3%7ER2C4%7E%7ER2C4%7ER2C5%7E%7ER2C5%7ER3C5%7E%7ER7C2%7ER6C2%7E%7ER4C2%7ER4C3