Proof overview: =============== - Ascertain pairing of line types, thence each one's possible contents - Solve, bit by bit Ascertain line type pairing: ---------------------------- ent mod gw arrow n Y (y) darrow Y n y lockout n y Y Immediately, entropy is double-arrow. Arrow can't be GW, because box 1 would need two 1s, so is mod. Therefore lockout is modular arrow + modular 6-[12]-[12]-3 9-[1245]-[1245]-[36] double arrow + entropic [45]-[12]-[87]-[45] [46]-[789]-[123]-[46] [56]-[23]-[89]-[56] lockout + GW 2916 2917 2918 3917 3918 4918 3927 3928 4928 2817 3817 2816 3827 4938 2716 Solve ----- Colouring for entropy works better than the alternatives, I found. Pencil/colour in the ends of the modularrows (M/As) and entropic double-arrows (E/DAs) Colour ambiguously the middles of the E/DAs and the column 2 GW/L as low/high Column 1's arrow can't be 6--3, or 9--6, because that puts 4 lows in the column. Column 9's and row 1's M/As are similar, as the E/DAs can't be -36-. Row 2's 3 is on the E/DA, as it can't be on the M/A. Column 2's 3 is also on the E/DA, because putting it on the GW/L makes it 4938 and the E/DA sum would also need a 9 in the middle. => r7c2 is 3 => column 1's and row 1's E/DAs sum to 3+[78] and have a 6 in their circles Row 3's 6 is excluded from all furniture, and box 2, so r3c7 is 6 => box 1's 6 is in r2c2 => box 2's 7 and 8 are both in columns 5 and 6, so r3c4 is low and r3c3 is high => the GW/L there can't have an 8 at the ends, so can't have a 4 at the ends either => r3c4 is 1 and r2c4 and r4c3 are both 2 => r2c4 is low and r2c3 is middle => neither M/A can have a 2 on row 4, and therefore can't have a 4 on row 3 => Row 3's 4 must be in box 2 => row 1's E/DA is a [56] pair summing to 3+8 => box 3 row 3 has a [47] pair, and r3c3 is 8 Column 2 has a virtual [12] pair in the GW/L and M/A, so r1c2 is 7 and r6c2 is 8 => column 1's E/DA is a [56] pair => r2c1 is 1, r1c1 is 2, and r3c2 is 5 => r5c2 is 4, and r4c2 is 1, leaving a [29] pair in box 7. => r5c1 is 7, leaving a [48] pair in box 7 => column 9's M/A is 9-4-2-3 Box 8's E/DA pencils to [456]-[78]-[12]-[456], but summing to <=10 must have a 4 => r9c1 is 8 and r9c1 is 4 => r9c2 is 9 and r8c2 is 2 row 9's GW/L can't have a 4 at the end, so can't have a 3 in middle, or a 2, so has a 1 => seeing 1248 left and 39 above, r8c9 is [567], creating a [567] triple with r2c9 and r6c9 => r7c9 is 4 => r1c7 is 4, leaving [18] in r1c[89] The inner sum of column 8's E/DA can't be 1+8 as that breaks r1c8, so the sum is 10 and r6c9 is 6 r8c[3459] form a [4567] quad, so r8c[67] is a [19] pair and r8c8 is 3 => column 9's E/DA is 6-1-9-4 => r8c7 is 1 and r8c6 is 9, leaving the GW [23]-9-1-[78] => r1c8 is 8, r1c9 is 1, and r2c7 is 9 => r6c1 is 5 and r7c1 is 6 => r6c3 is 9 and r5c3 is 6 => r5c5 is 9 and r5c8 is 5 => r2c8 is 7, r2c9 is 5, r8c9 is 7, and r8c2 is 5 => r9c7 is 8 and r9c9 is 2, making r5c9 8 and r4c7 7, then r4c4 becomes 6 => row 8's E/DA has 4 and 6 ends and therefore an -8-2- middle => r9c6 and r5c4 are 3 => box 2's E/DA resolves its [56], resolving row 4's [58], resolving row 2's [38] At this point, you're just speedrunning Proof of uniquess (once line pairing is concluded): https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Arrow%7ER1C4%7ER2C4%7ER2C3%7ER1C3.Arrow%7ER3C1%7ER3C2%7ER4C2%7ER4C1.Arrow%7ER4C9%7ER4C8%7ER3C8%7ER3C9.Modular%7E3%7ER1C4%7ER2C4%7ER2C3%7ER1C3.Modular%7E3%7ER3C1%7ER3C2%7ER4C2%7ER4C1.Modular%7E3%7ER4C9%7ER4C8%7ER3C8%7ER3C9.DoubleArrow%7ER7C9%7ER7C8%7ER6C8%7ER6C9.DoubleArrow%7ER7C1%7ER7C2%7ER6C2%7ER6C1.DoubleArrow%7ER1C5%7ER2C5%7ER2C6%7ER1C6.DoubleArrow%7ER8C4%7ER7C4%7ER7C5%7ER8C5.Entropic%7ER7C9%7ER7C8%7ER6C8%7ER6C9.Entropic%7ER6C1%7ER6C2%7ER7C2%7ER7C1.Entropic%7ER1C5%7ER2C5%7ER2C6%7ER1C6.Entropic%7ER8C4%7ER7C4%7ER7C5%7ER8C5.Lockout%7E4%7ER4C3%7ER3C3%7ER3C4%7ER4C4.Lockout%7E4%7ER8C1%7ER8C2%7ER9C2%7ER9C1.Lockout%7E4%7ER9C6%7ER8C6%7ER8C7%7ER9C7.Whisper%7E5%7ER4C3%7ER3C3%7ER3C4%7ER4C4.Whisper%7E5%7ER8C1%7ER8C2%7ER9C2%7ER9C1.Whisper%7E5%7ER9C6%7ER8C6%7ER8C7%7ER9C7