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Author:

Andrew [ Mon May 06, 2019 7:54 pm ]

Post subject:

Re: Assassin 375

Thanks Ed for your milestone Assassin 375.

My most important steps were a batch of moderately heavy combination analysis, then a very short chain right at the end.

Thanks Ed for your comments. I've added some clarifications and corrected a few typos.

Here is my walkthrough for Assassin 375:

Rating Comment:

Author:	Ed [ Thu May 02, 2019 8:08 am ]
Post subject:	Assassin 375
Attachment: a375.JPG [ 59.78 KiB \| Viewed 6074 times ] Bump at the start and (right at the) end: but nothing too advanced. Kept looking for the normal stuff and eventually found the next elimination. Long, long WT though! But nice for a milestone number I think. SS gives it a 1.60. 11 years of Assassins here with Richard hosting us! 25 puzzles per year. code: 3x3::k:5120:5120:0000:0000:4354:4354:4354:2563:2563:3588:5120:0000:3333:2566:4354:2563:2563:3847:3588:3588:4360:3333:2566:0000:4362:4362:3847:4363:4360:4360:3333:0000:0000:4621:4621:3847:4363:4363:0000:0000:0000:5135:11024:4621:4621:2321:2321:2321:0000:0000:5135:11024:11024:11024:3346:3091:3091:3091:5135:5135:11024:4372:4372:3346:4373:4373:4105:4105:11024:11024:3084:3084:3346:3854:3854:3854:4105:11024:2305:2305:3084: solution: Code: +-------+-------+-------+ \| 7 9 6 \| 8 1 2 \| 5 3 4 \| \| 8 4 3 \| 5 6 9 \| 2 1 7 \| \| 1 5 2 \| 7 4 3 \| 9 8 6 \| +-------+-------+-------+ \| 9 7 8 \| 1 3 5 \| 4 6 2 \| \| 5 3 4 \| 2 9 6 \| 8 7 1 \| \| 2 6 1 \| 4 7 8 \| 3 5 9 \| +-------+-------+-------+ \| 4 2 7 \| 3 5 1 \| 6 9 8 \| \| 3 8 9 \| 6 2 7 \| 1 4 5 \| \| 6 1 5 \| 9 8 4 \| 7 2 3 \| +-------+-------+-------+ Cheers Ed

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Assassin 375 http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1491	Page 1 of 1

Author:	Ed [ Sat May 11, 2019 7:02 am ]
Post subject:	Re: Assassin 375
My start and Andrew's ending work best for me. Not sure what I missed - probably some pesky cage clean-up..... My key step had no eliminations! Anyway, overall another challenging puzzle! Many thanks to Andrew for some corrections and things I missed. Cheers! WT for a375: Preliminaries courtesy of SudokuSolver Cage 17(2) n3 - cells ={89} Cage 17(2) n7 - cells ={89} Cage 17(2) n9 - cells ={89} Cage 9(2) n9 - cells do not use 9 Cage 10(2) n2 - cells do not use 5 Cage 9(3) n4 - cells do not use 789 Cage 20(3) n1 - cells do not use 12 Cage 10(4) n3 - cells ={1234} Cage 43(8) n689 - cells ={13456789} 1. 10(4)n3 = {1234}: all locked for n3 2. 17(2)n3 = {89}: both locked for n3 and r3 2a. r23c9 = 11,12,13 -> r4c9 = 2,3,4 3. 17(2)n9 = {89}: both locked for r7 and n9 4. 17(2)n7 = {89}: both locked for r8 and n7 5. "45" on n9: 2 innies r78c7 = 7 = {16/34}(no 5,7) Crucial step for my solution 6. "45" on r7: 4 innies r7c1567 = 16 = {1267/1357/1456/2347/2356} 6a. note: {1357} can't have {13} in r7c56 since no 5,7 in r7c7 (no eliminations yet) 7. "45" on n6789: 3 innies r4c9 + r7c56 = 8. 7a. r7c56 <> 4 (step 6a) -> no 4 in r4c9 7b. r4c9 = 2,3 -> r7c56 = 5/6 = {14/23/15/24}(no 6,7) 7c. note: must have one of 3,4,5 8. "45" on n69: 1 innie r4c9 + 9 = 2 outies r89c6 8a. r4c9 = 2,3 -> r89c6 = 11/12 (no 1, no 3 in r9c6) 8b. note: must have one of 3,4,5 in r89c6 to get to 11 and not go over 12. 9. "45" on n7: 2 outies r79c4 = 12 = [39/48]{57}(no 1,2,6; no 3,4 in r9c4) 9a. note: must have one of 3,4,5 Loved finding this. 10. killer triple 3,4,5 in n8 in r79c4 + r7c56 + r89c6: all locked for n8 and no 3,4 in r7c7 since r7c7 sees all 3,4 in those cells in n8 (Common Peer Elimination CPE) 11. h7(2)r78c7 = {16} only: both locked for c7, n9 and for 43(8) cage 11a. no 3 in 9(2)n9 12. "45" on n3: 1 innie r1c7 - 3 = 1 outie r4c9 = [52] only 13. 16(3)n8: {16}[9] blocked by r8c7 = (16) 13a. = {178/268} Ooops, Andrew noticed that 6 in n8 is only in this cage so I could have been simpler 13b. must have 8 -> r9c5 = 8 14. "45" on n69: 2 remaining outies r89c6 = 11 = {47} only: both locked for n8 and c6, and for the rest of the 43(8)r5c7 14a. r5c7 + r6c789 = {3589}: all locked for n6; 5 locked for r6 15. r8c45 = 8 = {26} only: both locked for r8, 2 for n8 15a. r78c7 = [61] 16. r79c4 = h12(2) = [39] 17. naked pair {15} in r7c56: both locked for r7 and for 20(4)r5c6 17a. r7c56 = 6 -> r56c6 = 14 = {68} only: both locked for c6 and n5 18. r7c23 = 9 = {27} only: both locked for n7 18a. r7c1 = 4 18b. r89c1 = 9 = [36] only 18c. r9c23 = {15}: 5 locked for r9 19. 9(2)n9 = {27} only: 7 locked for r9 and n9 19a. r89c6 = [74], r9c9 = 3 20. 9(3)n4 = {126/234} 20a. must have 2: locked for n4 and r6 21. "45" on n2: 1 remaining outie r4c4 + 10 = 2 innies r1c4 + r3c6 21a. max. 2 innies = 13 -> r4c4 = 1 21b. -> two innies = 11 = [83/65] 22. 2 in c6 only in r12c6: 2 locked for n2 22a. 17(4)r1c7: {2456} blocked by 4,6 only in r1c5 22b. = [5]{129/237}(no 4,6) 23. "45" on r6789: 2 outies r5c67 - 3 = 2 innies r6c45 = [68]-{47} only (-> r5c67=[68]!) 23a. 4,7 locked for r6 and n5 23b. r3c78 = [98], r6c67 = [83], r7c89 = [98], r6c89 = [59], r8c89 = [45] 24. hidden single 6 in n6 -> r4c8 = 6 25. naked triple {126} in n4: 1 locked for n4 26. 17(3)n4 = {359/458}(no 7): must have 5: locked for n4 26a. {359} must have 3 in r5c2 -> no 9 in r5c2 I should have made r5c2 = (34) at this spot. As Andrew noticed, I could then have locked 5 for c1 26b. note: must have 9 in c1 or 4 in r5c2 (no more eliminations yet) 27. "45" on n14: 3 innies r125c3 = 13 27a. but {157} blocked by r9c3 27b. {247} blocked by r7c3 27c. {256} blocked by none of them in r5c3 27d. = {139/148/238/346}(no 5,7) 28. 7 in n4 only in 17(3)r3c3: 7 locked for r4 and no 7 in r3c3 28a. must have 7 = {179/278/467}(no 3,5) 28b. r4c7 = 4, r2c7 = 2, r9c78 = [72] 29. naked pair {13} in r12c8: 1 locked for c8 and n3 29a. r1c9 = 4, r5c89 = [71] 30. "45" on n4: 1 outie r3c3 + 2 = 1 innie r5c3 30a. = [13/24] 31. hidden pair 3,4 in n4 -> r5c23 = {34}: 3 locked for r5 32. 5 in n4 only in c1: 5 locked for c1 33. hidden single 2 in c6 -> r1c6 = 2 33a. r1c5 + r2c6 = 10 = {19}/[73] (no 3 in r1c5) A series of tricky steps now to finally crack this. 34. "45" on r1: 3 outies r2c268 = 2 innies r1c34 34a. r1c34 <> 13 -> no 3,9 in r2c2 34b. r1c34 <> 12 -> no 8 in r2c2 35. 20(3)n1: {389} blocked by none in r2c2 35a. = {479/569/578}(no 3) 35b. must have 4/5 which are only in r2c2 -> r2c2 = (45) 36. "45" on n1: 3 innies r123c3 = 11 36a. = {128/146/236}(no 9) 37. 14(3)n1: 37a. {149} as [914] only, blocked by 17(3)n4 = 9 in c1 or 4 in r5c2 (step 26a) 37b. {167} blocked by 1/6 needed for h11(3)n1 37c. {248} blocked by 2/4 needed for h11(3) 37d. {257} blocked by 5/7 needed for 20(3)n1 37e. {347} blocked by 3,4 only in r3c2 37f. {356} blocked by 3,6 only in r3c2 37g. = {158/239}(no 4,6,7) 38. hidden single 7 in c1 -> r1c1 = 7 38a. -> r12c2 = 14 = [85/94] 39. naked pair {12} in r3c13: both locked for n1 and 1 for r3 40. 14(3)n1: {239} blocked since [86] in r1c23 blocked by r1c4 40a. = [815] only cracked finally! Cheers Ed

Author:	wellbeback [ Thu May 16, 2019 4:43 pm ]
Post subject:	Re: Assassin 375
Back from my travels and caught up with work - so finally able to write a WT. Thanks for another fun puzzle Ed. Here's how I did it. Haven't yet looked at the other WTs. Assassin 375 WT: 1. 10(4)n3 = {1234} 17(2)n3 = {89} -> Remaining cells n3 = {567} -> r4c9 from (234) 2. 17(2)n7 = {89} 17(2)n9 = {89} -> (89) in r9 in n8 in r9c456 3. Innies n9 = r78c7 = +7(2) Since 43(8) has no 2 -> r78c7 = {16} or {34} 4. Innies n6789 = r4c9 + r7c56 = +8(3) -> r7c56 from +4, +5, +6 -> Remaining Innies r7 = r7c17 from +12, +11, +10 Since (89) already in r7 and since r7c7 from (1346) -> r7c17 cannot be +12 -> r7c56 cannot be +4 -> r4c9 cannot be 4 5. -> 13(3)r2c9 from [{76}2] or [{75}3] -> 12(3)n9 cannot be {237} -> at least one of (27) in 9(2)n9 -> 9(2)n9 = {27} 6. Outies n7 = r79c4 = +12(2) (No 1,2,6) r7c56 from +5 or +6 -> must include exactly one of (12) -> Innies n78 -> r89c6 from +12 or +11 -> No 1 Also 2 cannot go in 43(8) -> One of (12) in 16(3)n8 -> 16(3)n8 must contain an 8 or 9 (in r9c5) 7. Either r89c6 = +12 (No 6) or r89c6 = +11 -> r7c56 = +6 -> r56c6 = +14(2) -> r89c6 cannot be {56} -> 6 in n8 in 16(3) -> 16(3)n8 from [{16}9] or [{26}8] 8. r7c56 from +5 or +6 -> r7c17 either +11(2) or +10(2) -> Since Max r7c1 = 7 -> Min r7c7 = 3 -> r7c7 from (346) 9! Outies n7 = r79c4 = +12(2) = [75], [48], or [39] Whatever goes in r7c7 (346) must go in n8 in r89c45 Only possibility is r7c7 = 6 -> r8c7 = 1 -> 12(3)n9 = {345} 10. Also -> r1c7 = 5 -> 13(3)r2c9 = [{67}2] -> Remaining innies n6789 = r7c56 = +6 -> Remaining Innies r7 -> r7c1 = 4 -> r7c56 = {15} -> r56c6 = {68} -> r89c6 = [74] -> 16(3)n8 = [{26}8] -> r79c4 = [39] -> 12(3)r7c2 = [{27}3] -> 15(3)r9c2 = [{15}9] -> 13(3)n7 = [436] -> 12(3)n9 = [{45}3] 11. 18(4)n6 must be {1467} Remaining cells in n6 in 43(8) are {3589} Whichever of (89) is in r3c8 goes in n9 in r7c9 and in n6 in r56c7 -> Whichever of (89) is in r3c7 goes in n9 in r7c8 and in n6 in r6c9 -> HS r6c8 = 5 -> 3 in r56c7 Also 12(3)n9 = [453] 13. 8 in n2 in r123c4 -> Given previous placements only possibilities for 17(4)r1c5 are [{129}5] or [{237}5] -> 2 in n1/r3 in r3c123 Also 2 in n5 in r56c45 -> 2 in n2 in r12c6 14. Remaining IOD n2 -> r4c4 + 10 = r1c4 + r3c6 Max r4c1 = 8 and max r3c6 = 5 -> Max r4c4 = 3 Since 3 already in c4 and 2 already in r4 -> r4c4 = 1 -> [r1c4,r3c6] = [83] or [65] 15. Also -> 1 in n6 in r5c89 -> 1 in n4 in r6c123 => 9(3)n4 = {126} -> r56c6 = [68] -> 43(8) in n6 = [8359] 16! Remaining possibilities for 17(3)n4 are [{58}4] and [{59}3] Innies n1 = r123c3 = +11(3) Since 2 in r3c123 -> 5 not in r3c3 (would force r123c3 = [{24}5]) IOD n4 -> r5c3 = r3c3 + 2 -> Since 5 not in r3c3 -> 7 not in r5c3 -> 7 in n4 in r4c23 -> 18(4)n6 = [4671] -> r1c9 = 4, r12c8 = {13}, r2c7 = 2 -> r1c6 = 2 Also -> 9(2)n9 = [72] Also 17(2)n9 = [98] and 17(2)n3 = [98] 17! Since 7 in 17(3)r3c3 -> 3 in n4 in r5c23 -> r5c23 = {34} Whichever of (34) is in r5c2 goes in n1 in c3 -> r1235c3 = [{36}24] or [6413] 18. -> r6c2 = 6 -> r36c3 = {12} -> r7c23 = [27] and r9c23 = [15] -> r4c2 = 7 19! Either r12c3 = {36} -> 3 in n2 in r3c56 or r12c3 = [64] -> [r1c4,r3c6] = [83] Either way -> 3 in n2 in r3 -> 17(4)n2 = [{129}5] -> HS 7 in r1 -> r1c1 = 7 -> 3 in n1 only in c3 -> r123456c3 = [{36}2841] -> 17(3)n4 = [{59}3] Also 9(3)n4 = [261] 20. 3 cannot go in r3c5 since that puts 10(2)n2 = [73] and r23c9 = [67] ... ... which leaves no place for 6 in r3 -> r3c6 = 3 -> r1c4 = 8 -> r23c4 = {57} -> 10(2)n2 = {46} -> r12c3 = [63] -> r12c8 = [31] -> 17(4)n2 = [1259] -> 20(3)n1 = [794] etc.

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Author:	Andrew [ Mon May 06, 2019 7:54 pm ]
Post subject:	Re: Assassin 375
Thanks Ed for your milestone Assassin 375. My most important steps were a batch of moderately heavy combination analysis, then a very short chain right at the end. Thanks Ed for your comments. I've added some clarifications and corrected a few typos. Here is my walkthrough for Assassin 375: Prelims a) R23C5 = {19/28/37/46}, no 5 b) R3C78 = {89} c) R7C89 = {89} d) R8C23 = {89} e) R9C78 = {18/27/36/45}, no 9 f) 20(3) cage at R1C1 = {389/479/569/578}, no 1,2 g) 9(3) cage at R6C1 = {126/135/234}, no 7,8,9 h) 10(4) cage at R1C8 = {1234} i) 43(8) cage at R5C7 = {13456789}, no 2 Steps resulting from Prelims. 1a. Naked pair {89} in R3C78, locked for R3 and N3, clean-up: no 1,2 in R2C5 1b. Naked pair {89} in R7C89, locked for R7 and N9, clean-up: no 1 in R9C78 1c. Naked pair {89} in R8C23, locked for R8 and N7 1d. Naked quad {1234} in 10(4) cage at R1C8, locked for N3 2. 45 rule on N1 3 innies R123C3 = 11, no 9 in R12C3 3. 45 rule on N3 1 innie R1C7 = 1 outie R4C9 + 3, R1C7 = {567} -> R4C9 = {234} 4. 45 rule on N4 3 innies R4C23 + R5C3 = 19 = {289/379/469/478/568}, no 1 4a. 2 of {289} must be in R45C3 (R45C3 cannot be {89} which clashes with R8C3) -> no 2 in R4C2 5. 45 rule on N7 2 outies R79C4 = 12 = [39/48/57/75], no 1,2,6, no 3,4 in R9C4 6. 45 rule on N9 2 innies R78C7 = 7 = {16/34}, no 5,7 6a. R9C78 = {27/45} (cannot be {36} which clashes with R78C7) 7. 45 rule on C789 2 outies R89C6 = 1 innie R1C7 + 6 7a. Min R1C7 = 5 -> min R89C6 = 11, no 1 7b. Min R89C6 = 11, max R8C6 = 7 -> min R9C6 = 4 8. 45 rule on N36789 2 outies R56C6 = 1 innie R1C7 + 9 8a. Min R1C7 = 5 -> min R56C6 = 14, no 1,2,3,4 8b. Min R56C6 = 14 -> max R7C56 = 6, no 6,7 9. R1C7 = {567} -> R89C6 (step 7) = 11,12,13, R56C6 (step 8) = 14,15,16 -> R5689C6 = 25,27,29 = {3589/4678},{3789/4689/5679},{5789} (cannot be {3679} because cannot form 11 and 14 separately from this combination, cannot be {4579} = 14+11 = {59}{47} because 20(4) cage at R5C6 = {59}{24} clashes with R89C6 = {47}) 9a. {4678} = 14+11 must be {68}{47}, {4689} = 15+11 must be {69}[48], {5679} = 15+11 must be {69}{57} -> no 6 in R89C6 9b. R5689C6 = {3589/4678},{3789/4689/5679} (cannot be {5789} = 16+13 = {79}[58] because 20(4) cage at R5C6 = {79}{13} + R89C6 = [58] clash with R79C4) -> R56C6 = 14,15, R89C6 = 11,12 -> R1C7 = {56}, clean-up: no 4 in R4C9 (step 3) 9c. R78C7 (step 6) = {16/34} 9d. R1C7 + R89C6 = [5+11]/[6+12] = 5[38]/5{47}/6{57} (cannot be 6[39]/6[48] which clash with R78C7), no 9 in R9C6 9e. R178C7 = 5{16}/6{34} (cannot be 5{34} which clashes with R1C7 + R89C6), 6 locked for C7 9f. R56C6 = 14,15 = {59/68/69} (cannot be {78} which clashes with R89C6, while {79} has already been eliminated in step 9b), no 7 in R56C6 9g. R56C6 = 14,15 -> R7C56 = 5,6 = {14/23/15/24} 9h. Hidden killer pair 1,2 in R7C56 and 16(3) cage at R8C4 for N8, R7C56 contains one of 1,2 -> 16(3) cage must contain one of 1,2 9i. 6 in N8 only in 16(3) cage at R8C4 which must contain one of 1,2 = {169/268} -> R9C5 = {89}, R8C45 = {16/26}, 6 locked for R8, clean-up: no 1 in R7C7 (step 6) 9j. 43(8) cage at R5C7 = {13456789}, 9 locked for N6 9k. 43(8) cage at R5C7 = {13456789}, CPE no 1 in R4C7 10. 7 in N3 only in R23C9, locked for C9 10a. 12(3) cage at R8C8 = {156/345} (cannot be {147/246} which clash with R78C7, cannot be {237} which clashes with R4C9, no 2,7 10b. 6 of {156} must be in R9C9 -> no 1 in R9C9 [This elimination in R9C6 proved to be a very important one.] 10c. R9C78 = {27} (hidden pair in N9), locked for R9, clean-up: no 5 in R7C4 (step 5) 10d. 1 in N9 only in R8C789, locked for R8 10e. Naked pair {26} in R8C45, 2 locked for R8 and N8 10f. R8C45 = {26} -> R9C5 = 8 (cage sum), clean-up: no 2 in R3C5 10g. R9C4 = 9 (hidden single in N8) -> R7C4 = 3 (step 5) 10h. R8C6 = 7 (hidden single in N8) 10i. R9C4 = 9 -> R9C23 = 6 = {15}, locked for R9 and N7, R9C6 = 4, R7C7 = 6, R1C7 = 5 10j. R7C56 = {15} -> 20(4) cage at R5C6 = {1568}, 6,8 locked for C6 and N5 10k. Naked pair {67} in R23C9, 6 locked for C9, R9C19 = [63] -> R78C1 = 7 = [43], R4C9 = 2, R8C7 = 1 10l. Naked quad {3589} in R5C7 + R6C789, 3,5,8 locked for N6, 5 also locked for R6 10m. 9(3) cage at R6C1 = {126/234}, 2 locked for R6 and N4 10n. 8 in R4 only in R4C123, locked for N4 10o. 7 in R6 only in R6C45, locked for N5 10p. 2 in C6 only in R123C6, locked for N2 11. 17(4) cage at R1C5 contains 5 = {1259/2357} (cannot be {1457/2456} because 4,6,7 only in R1C5), no 4,6, 2 locked for N2 11a. 7 of {2357} must be in R1C5 -> no 3 in R1C5 12. 13(3) cage at R2C4 = {148/157}, no 6, 1 locked for C4 12a. 8 of {148} must be in R2C4 -> no 4 in R2C4 12b. Killer pair 4,7 in 13(3) cage and R6C4, locked for C4 13. 45 rule on N2 2 innies R1C4 + R3C6 = 1 remaining outie R4C4 + 10 13a. Max R1C4 + R3C6 = 13 -> R4C4 = 1, R1C4 + R3C6 = 11 = [65/83] 13b. 1 in R6 only in 9(3) cage at R6C1 (step 10m) = {126}, 6 locked for R6 and N4, 1 also locked for N4 -> R56C6 = [68] 13c. Naked triple {359} in R6C789, 3,9 locked for R6 and N6 -> R5C7 = 8 13d. Naked pair {47} in R6C45, 4 locked for N5 14. 17(3) cage at R4C1 = {359/458} -> R5C2 = {34}, R45C1 = {58/59}, 5 locked for C1 and N4 14a. 7 in C1 only in R123C1, locked for N1 15. Hidden killer pair 1,2 in R23C1 and R6C1 for C1, R6C1 = {12} -> R23C1 must contain one of 1,2, no 1,2 in R3C2 because 14(3) cage cannot contain both of 1,2 15a. R123C3 (step 2) = 11 = {146/236/245} (cannot be {128} which clashes with R23C1), no 8 16. 45 rule on N4 1 innie R5C3 = 1 outie R3C3 + 2 -> R3C3 = {125}, R5C3 = {347} 16a. 17(3) cage at R3C3 = {179/278/458} (cannot be {359} which clashes with R4C56, ALS block), no 3 16b. 3 in N4 only in R5C23, locked for R5 17. 14(3) cage at R2C1 must contain 1 or 2 (step 15) = {158/167/239/257} (cannot be {248} which clash with R123C3, cannot be {149} which clashes with 17(3) cage at R4C1), no 4 17a. 7,8 of {158/167} must be in R2C1 (14(3) cage cannot be [176] which clashes with R3C9) -> no 1 in R2C1 17b. R123C3 (step 15a) = {146/236/245} 17c. 1 of {146} must be in R3C3 -> no 1 in R12C3 17d. 1 in N1 only in R3C13, locked for R3, clean-up: no 9 in R2C5 17e. 9 in N2 only in 17(4) cage at R1C5 (step 11) = {1259}, no 3,7 18a. R3C13 = {12} (hidden pair in R3), 2 locked for N1, clean-up: no 7 in R5C3 (step 16) 18b. Naked pair {34} in R5C23, 4 locked for R5 and N4 18c. R15C9 = [41] 18d. R123C3 = {146/236} (cannot be {245} because 4,5 only in R2C3), no 5, 6 locked for C3 and N1 18e. Naked pair {12} in R36C3, locked for C3 -> R7C23 = [27] 18f. R4C2 = 7 (hidden single in N4) 18g. Naked pair {35} in R3C26, locked for R3, clean-up: no 7 in R2C5 18h. 13(3) cage at R2C4 (step 12) = {148/157} 18i. 5 of {157} must be in R2C4 -> no 7 in R2C4 18j. 7 in N2 only in R3C45, locked for R3 -> R23C9 = [76], clean-up: no 4 in R2C5 18k. R1C1 = 7 (hidden single in N1) -> R12C2 = 13 = [85/94] 19. R1C4 + R3C6 (step 13a) = [65/83] 19a. Consider placements for R3C2 = {35} R3C2 = 3 => R1C34 = [68] or R3C2 = 5 => R3C6 = 3 -> R1C4 + R3C6 = [83], R2C5 = 6 -> R3C5 = 4 19b. R13C2 = [95], R2C1 = 8 -> R3C1 = 1 (cage sum) and the rest is naked singles. Rating Comment: I'll rate my walkthrough for A375 at Hard 1.25. I would normally go for a slightly higher rating with a forcing chain, but the one in the final step was so short that this time I rated for my combination analysis.