Overview of solve: ================== 0) Modular prep-work 1) Work out the possible numbers in the counting circles 2) Work out the shared number 3) Extend along the modular lines 4) Complete the circles 5) Leverage the cages and speedrun More detailed: ============== 0. Modular prep-work -------------------- The line including a fuse can be coloured - this one's useful from the start. The left curve may be coloured with other colours, as may the right, but there's not much point. 1. Possible numbers for counting circles ---------------------------------------- 9 = 1+2+6, 1+3+5, 2+3+4, 1+8, 2+7, 3+6, 4+5, 9 Trivial to remove are: 9 - needs to be in both sets 1+8 - ninth 8 is forced into ninth circle 3+6 - can't be placed on the modular lines two spaces apart 2+7 - neither value can be put in r5c5 due to circles two spaces apart 4+5 has several cases: - a shared 5 just won't fit at all, the circles cancel each others availability too quickly - blue being 4+5 forces 4 green 3s between the 4s and 5s along the modular line - the green 5 common 4 case forces 3 into blue r8c6, and 4s into blue r3c7 and r6c8, which fully defines the green set, and makes r2c4 unsatisfiable. So we're down to both blue and green being sets of 3 values. They can't be the same set as r2c4 and r8c4 must be the same modularity, but different value. So we're left with (26(1)35), (16(2)34), or (15(3)24). 2. Work out the shared value ---------------------------- Shared 1 can be dismissed as 6-?-2 on a modular line needs 1/4/7, not 3/5. Shared 2 with green 6s leaves nowhere for blue 3s. Shared 2 with blue 6s forces green 3s then green 4s, leaving nowhere for blue 1. So we're left with 1+3+5 and 2+3+4 3. Extend along the modular lines --------------------------------- The route-101 approach is just to pencil in [12345] in all the boxes, and then start removing the impossibles. Start with the trivial: r1c9 is on a kropke - not 5 r2c4 and r8c4 being the same residue from different sets means it can't be 3 This places two 3 one of two ways in columns 2 and 3. If blue has them, then 3 is excluded from all 3 cells in the line segment entering box 9 So r3c3 and r6c2 are 3 - remove 3 from pencil markings in all other greens and all neighbours This only leaves two more places for blue 3 - r8c6 and r9c9 (sing!), placing 3 in r7c1. We can now start painting [369] or subsets thereof all over teh modular line with the fuse. Now we have r6c2 and r6c3 the same modular residue, we can colour those two lines consistently. Sudoku lets us place 3 in r1c4, r4c7, and r2c8 4. Complete the circles ----------------------- Which of blue or green circles is 1+3+5, and which is 2+3+4? Could we fit five 5s on the remaining blue circles? Maybe, but r2c4 couldn't be one. So r8c4 isn't 2 either. Could we fit five 5s on the remaining green circles? No - the kropke not being 5 forces both r2c6 and r8c4 to be 5, r1c8 to be [14] by modularity, and r1c9 to be 2 by kropke but 1 by counting circles. So blue circles are fully enumberable as 1+3+5. After that, most green circles are enumberable as 2+3+4. The remaining 3 green circles can only have two 2s top right and one 4 bottom left by modularity This unifies the colouring across all three lines. Technically, you're already at the speedrun stage now. Pencilling in the [258] choices along the modular lines places 8, 5, and 2 in column 5. Pencilling in the [147] choices along the modular lines places 7 in r7c4. A bit of sudoku can be done, 3s, 4s, and 5s, even the [28] pair in row 5, and row 9's 1. 5. Leverage the cage and speedrun --------------------------------- The 15-cage must contain a 4, but not have a 3 or 5 be 2-[49]-[49], placing 2 in r4c1 and r3c2, 1s in r3c1, r4c3, and r5c6, and a 5 in r5c4, resolving the [58]-pair in row 9, and completing box 4. box 5 completes 6-7-8, which clears up a few other pencil markings. Most importantly, the naked 7 in r3c8 puts 4 in r3c6, which resolves the [49] pair in box 5, the [69] pair in box 8, and trivially completes boxes 2 and 3, then column 9, row 1, row 7, box 1 and box 7. Bosch! Proof of uniqueness: ==================== https://sigh.github.io/Interactive-Sudoku-Solver/index.html?q=.CountingCircles%7ER1C1%7ER2C4%7ER3C7%7ER4C2%7ER5C5%7ER6C8%7ER7C3%7ER8C6%7ER9C9.CountingCircles%7ER1C9%7ER2C6%7ER3C3%7ER4C8%7ER5C5%7ER6C2%7ER7C7%7ER8C4%7ER9C1.Modular%7E3%7ER2C5%7ER2C4%7ER3C3%7ER4C2%7ER5C2%7ER6C2%7ER7C3%7ER8C4%7ER8C5.Modular%7E3%7ER8C6%7ER7C7%7ER6C8%7ER5C8%7ER4C8%7ER3C7.Modular%7E3%7ER3C5%7ER3C4%7ER4C3%7ER5C3%7ER6C3%7ER7C4%7ER7C5%7ER7C6%7ER6C7%7ER5C7%7ER4C7%7ER3C6%7ER2C6%7ER1C7%7ER1C8.BlackDot%7ER1C8%7ER1C9.Cage%7E15%7ER6C4%7ER6C5%7ER6C6