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Easter Eggs v2 Revisit http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1606 |
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Author: | Ed [ Tue Feb 16, 2021 6:25 am ] |
Post subject: | Easter Eggs v2 Revisit |
Attachment: eev2.JPG [ 63.51 KiB | Viewed 4496 times ] After a couple of easier Revisits (I could solve them!) back to a harder looking one. Up to page 7 of archive A! This gets a score of 1.90 JSudoku uses 5 'complex intersections' and some other techniques that it doesn't use for normal Assassins so this may be over my head. Anyway, worth a try. Code: Select, Copy & Paste into solver: 3x3::k:2304:4097:4354:4354:6916:4101:4101:2567:4360:2304:4097:4354:6916:6916:6916:4101:2567:4360:4882:4097:2068:2068:6916:3607:3607:2567:4122:4882:4882:2068:5918:2335:6176:3607:4122:4122:2852:2852:5918:5918:2335:6176:6176:2091:2091:2605:2605:3375:5918:2335:6176:2355:4404:4404:2605:5175:3375:3375:9018:2355:2355:3901:4404:1343:5175:3649:9018:9018:9018:4165:3901:1863:1343:5175:3649:3649:9018:4165:4165:3901:1863: Solution: +-------+-------+-------+ Ed |
Author: | Andrew [ Sat Feb 27, 2021 11:47 pm ] |
Post subject: | Re: Easter Eggs v2 Revisit |
Thanks Ed. That was a tough one, probably the hardest revisit which you've posted (so far); only SKX-2, which I posted as a revisit, may have been harder. Here's how I solved it this time. As usual I haven't looked at my earlier walkthrough or how others did it, including the original 'tag'-solution. My walkthrough for Easter Eggs v2 Revisit: Prelims a) R12C1 = {18/27/36/45}, no 1 b) R12C9 = {89} c) R5C12 = {29/38/47/56}, no 1 d) R5C89 = {17/26/35}, no 4,8,9 e) R89C1 = {14/23} f) R89C9 = {16/25/34}, no 7,8,9 g) 10(3) cage at R1C8 = {127/136/145/235}, no 8,9 h) 19(3) cage at R3C1 = {289/379/469/478/568}, no 1 i) 8(3) cage at R3C3 = {125/134} j) 9(3) cage at R4C5 = {126/135/234}, no 7,8,9 k) 10(3) cage at R6C1 = {127/136/145/235}, no 8,9 l) 9(3) cage at R6C7 = {126/135/234}, no 7,8,9 m) 20(3) cage at R7C2 = {389/479/569/578}, no 1,2 n) 35(5) cage at R7C5 = {56789} 1a. Naked pair {89} in R12C9, locked for C9 and N3 1b. Naked quint {56789} in 35(5) cage at R7C5, locked for N8 1c. Max R9C6 = 4 -> min R89C7 = 12, no 1,2 in R89C7 2a. 45 rule on R12 3 outies R3C258 = 22 = {589/679}, 9 locked for R3 2b. 5 of {589} must be in R3C8 -> no 5 in R3C25 2c. 10(3) cage at R1C8 = {127/136/145/235} 2d. R3C8 = {567} -> no 5,6,7 in R12C8 2e. 45 rule on R89 3 outies R7C258 = 22 = {589/679}, 9 locked for R7 2f. 9 in C1 only in R45C1, locked for N4, clean-up: no 2 in R5C1 3a. Hidden killer pair 8,9 in 15(3) cage at R7C8 and 16(3) cage at R8C7 for N9, neither can hold both of 8,9 -> they must each contain one of 8,9 3b. 15(3) cage = {168/249/258/348} (cannot be {159} which clashes with 10(3) cage at R1C8), no 7 3c. 10(3) cage at R1C8 = {127/136/235} (cannot be {145} which clashes with 15(3) cage), no 4 3d. 45 rule on C8 3 innies R456C8 = 20 = {479/569/578} (cannot be {389} which clashes with 15(3) cage), no 1,2,3, clean-up: no 5,6,7 in R5C9 3e. 45 rule on C2 3 innies R456C2 = 9 = {126/135/234}, no 7,8,9, clean-up: no 3,4 in R5C1 3f. 1 of {126/135} must be in R6C2 -> no 5,6 in R6C2 3g. 19(3) cage at R3C1 = {289/379/469/478/568} 3h. 2,3 of {289/379} must be in R4C2 -> no 2,3 in R34C1 3i. 10(3) cage at R6C1 = {127/136/145/235} 3j. R6C2 = {1234}, 10(3) cage contains two of 1,2,3,4 -> R67C1 must contain one of 1,2,3,4 3k. Killer quad 1,2,3,4 in R12C1, R67C1 and R89C1, locked for C1 4a. 45 rule on C34 2 innies R28C4 = 15 = {69/78} 4b. 45 rule on C67 2 innies R28C6 = 11 = [29/38/47/56/65] 5a. 45 rule on R1234 3 innies R4C456 = 11 = {128/137/146/236/245}, no 9 5b. 45 rule on R6789 3 innies R6C456 = 19 = {289/379/469/478/568}, no 1 5c. 2,3 of {289/379} must be in R6C5 -> no 2,3 in R6C46 5d. R4C456 = 11, R6C456 = 19 -> R5C456 = 15 5e. 45 rule on N5 2 outies R5C37 = 11 = [29]/{38/47/56}, no 1, no 2 in R5C7 5f. 9(3) cage at R4C5 = {126/135/234} 5g. 1 in R5 only in R5C456 = 15 = {159/168} or R5C89 = [71] -> R5C456 = 15 = {159/168/249/258/348/456} (cannot be {267/357}, blocking cages), no 7 5h. 1 in N5 only in R4C456 or in R5C456 = 15 = {159/168} 5i. R4C456 = 11 = {128/137/146/245} (cannot be {236} which clashes with 9(3) cage = {126/234} CCC (combination crossover clash) while after R4C456 = {236}, R5C456 = {159} clashes with 9(3) cage = {135}, CCC) 5j. R5C456 = {168/249/258/456} (cannot be {159/348} which clash with R4C456), no 3 5k. R6C456 = 19 = {289/379/469/568} (cannot be {478} which clashes with R4C456) 5l. 3 in R5 only in R5C12 = {38} or R5C37 = {38} or R5C89 = [53] -> R5C456 = {168/249/456} (cannot be {258} blocking cages) 5m. R4C456 = {128/137/245} (cannot be {146} which clashes with R5C456), no 6 5n. R6C456 = {289/379/568} (cannot be {469} which clashes with R5C456), no 4 5o. 1 in R5 only in R5C456 = 15 = {168} or R5C89 = [71] -> R5C89 = [53/71] (cannot be [62], blocking cages), no 2,6 5p. Consider combinations for 9(3) cage = {126/135/234} 9(3) cage = {126/135}, 1 locked for N5 or 9(3) cage = {234} => R5C456 = {456} (cannot be {249} which clashes with {234}, CCC), 5 locked for R5 => R5C89 = [71] -> no 1 in R5C46 5q. Consider combinations for R4C456 = {128/137/245} R4C456 = {128/137} or R4C456 = {245} => R5C456 = {168} => R5C5 = 1 => 9(3) cage = [513] (cannot be [216] which clashes with R5C456, CCC) -> no 5 in R4C46, no 4 in R4C5 [Taking this a bit further …] 5r. R4C456 = {128/137}, 1 locked for N5 => R5C456 = {249/456} or R4C456 = {245} => 9(3) cage = [513] => R5C46 = {68}, R6C46 = {79}, R5C89 = [53], R45C6 both even, R6C6 odd => R5C7 odd = {79}, R5C7 + R6C6 = {79} = 16 => R45C6 = 8 = [26] -> no 4 in R4C6, no 8 in R5C6 6a. 45 rule on N1 2 innies R3C13 = 1 outie R1C4 + 3 6b. Min R3C13 = 6 -> min R1C4 = 3 6c. 10(3) cage at R1C8 (step 3c) = {127/136/235} 6d. 45 rule on N3 2 innies R3C79 = 1 outie R1C6 + 2 6e. Min R3C79 = 6 = {24} (cannot be {12/13/15/23} which clash with 10(3) cage, cannot be {14} which combined with 10(3) cage = {23}5 clash with 8(3) cage at R3C3) -> min R1C6 = 4 6f. 45 rule on N9 2 innies R7C79 = 1 outie R9C6 + 7 6g. Min R7C79 = 8 -> no 1 in R7C9 7a. 15(3) cage at R7C8 (step 3b) = {168/249/258/348}, 10(3) cage at R1C8 (step 3c) = {127/136/235}, R5C37 = 11 (step 5e) 7b. Consider combinations for R456C8 (step 3d) = {479/569/578} R456C8 = {479/578} contains 7 or R456C8 = {569), locked for C8 => R5C89 = [53], no 8 in R5C7, 15(3) cage = {348}, locked for N9, 9 in N9 only in 16(3) cage at R8C7 = {169/259} (cannot be {349} because 3,4 only in R9C6) => R7C9 = 7 (hidden single in N9) => R45C7 = [87] (hidden pair in N6) -> 7 in N6 only in R456C8 + R5C7, locked for N6 [Just spotted.] 7c. 4 in N5 only in R4C4 + R5C456, CPE no 4 in R5C3, clean-up: no 7 in R5C7 7d. 7 in N6 only in R456C8 = {479/578}, no 6, 7 locked for C8 7e. 10(3) cage at R1C8 = {136/235}, 3 locked for C8 and N3 7f. R3C258 (step 2a) = {589/679} 7g. R3C8 = {56} -> no 6 in R3C25 [Now unexpected progress, thanks to the hard work in step 5.] 8a. R4C456 (step 5m) = {128/137/245}, R5C456 (step 5l) = {168/249/456}, R28C6 = 11 (step 4b) 8b. Consider placement for 5 in 35(5) cage at R7C5 5 in R789C5, locked for C5 => R4C456 = {128/137} or R8C6 = 5 => R2C6 = 6 => R5C456 = {249/456} (cannot be {168} = [816]), 4 locked for N5 -> R4C456 = {128/137}, no 4,5, 1 locked for R4 and N5 [Cracked. The rest is fairly straightforward.] 8c. R5C456 = {249/456}, no 8, 4 locked for R5 8d. R5C9 = 1 (hidden single in R5) -> R5C8 = 7, clean-up: no 6 in R89C9 8e. 8(3) cage at R3C3 = {125/134}, 1 locked for R3 8f. 15(3) cage at R7C8 (step 3b) = {168/258} (cannot be {249} which clashes with R89C9), no 4,9, 8 locked for C8 and N9 8g. R46C8 = {49} (hidden pair in C8), locked for N6, clean-up: no 2 in R5C3 (step 5e) 8h. 9 in N9 only in 16(3) cage at R8C7 = {169/259/349}, no 7 8i. R7C9 = 7 (hidden single in N9) -> R6C89 = 10 = [46], R4C8 = 9, clean-up: no 5 in R5C3 (step 5e) 8j. R7C258 (step 2e) = {589}, no 6, 5,8 locked for R7 8k. 45 rule on N9 1 remaining innie R7C7 = 1 outie R9C6 = {1234} 8l. 6 in R7 only in R7C13, locked for N7 8m. 20(3) cage at R7C2 = {389/479/578} 8n. Killer triple 7,8,9 in R3C2 and 20(3) cage, locked for C2 8o. 9 in R6 only in R6C46, locked for N5 8p. R5C456 = {456}, 5,6 locked for R5, 5 locked for N5 8q. R5C12 = [92] (hidden pair in R5) 8r. Max R7C34 = 10 -> min R6C3 = 3 8s. 1 in R6 only in R6C12, locked for 10(3) cage at R6C1 8t. 10(3) cage = {127/136/145} 8u. 6 of {136} only in R7C1 -> no 3 in R7C1 8v. R456C2 (step 3e) = 9, R5C2 = 2 -> R46C2 = 7 = [43/61] 8w. 19(3) cage at R3C1 = {478/568} 8x. 6 of {568} must be in R4C2 -> no 6 in R34C1 8y. R4C2 = 6 (hidden single in N4) -> R6C2 = 1, R34C1 = {58}, locked for C1, clean-up: no 1,4 in R12C1 8z. 1 in C1 only in R89C1 = {14}, locked for N7 9a. R4C3 = 4 (hidden single in N4) -> R3C34 = 4 = {13}, 3 locked for R3 9b. 20(3) cage at R7C2 (step 8m) = {389/578}, 8 locked for C2 and N7 9c. R3C258 (step 2a) = {679} (cannot be {589} which clashes with R3C1) -> R3C8 = 6, R3C25 = {79}, 7 locked for R3 9d. R3C8 = 6 -> R12C8 = 4 = {13}, 1 locked for C8 and N3 9e. Naked triple {258} in 15(3) cage at R7C8, 2,5 locked for N9 9f. Naked pair {34} in R89C9, locked for C9 and N9 9g. R89C7 = {69} -> R9C6 = 1 (cage sum), R89C1 = [14], R89C9 = [43] 9h. R9C4 = 2 -> R89C3 = 12 = [39/57/75], no 9 in R8C3 9i. 7 in R4 only in R4C456 (step 8b) = {137}, 3,7 locked for N5, 3 locked for R4 9j. R7C7 = 1 -> R6C7 + R7C6 = 8 = [53], R34C9 = [52], R34C1 = [85], R4C67 = [78], clean-up: no 4 in R2C6, no 8 in R8C6 (both step 4b) 9k. R6C5 = 2 -> R45C5 = 7 = [16/34] 9l. R4C6 = 7, R5C7 = 3 -> R56C6 = 14 = [59/68] 9m. R28C6 (step 4b) = [29] (cannot be {56} which clashes with R56C6) -> R56C6 = [68], R3C67 = [42], R1C6 = 5, R45C5 = [34] 9n. R6C4 = 9, clean-up: no 6 in R28C4 (step 4a) 9o. R1C4 = 6 (hidden single in C4) -> R12C3 = 11 = [29] and the rest is naked singles. |
Author: | wellbeback [ Sun Feb 28, 2021 6:19 pm ] |
Post subject: | Re: Easter Eggs v2 Revisit |
Somewhat similar conclusions in my steps as for Andrew's, but individual ways to get those conclusions quite different as usual. Here's how I did it. Easter Eggs v2 Revisit WT: 1. Innies/outies Outies r12 = r3c258 = =22(3) = {589} or {679} Innies r1234 = r4c456 = +11(3) Innies r6789 = r6c456 = +19(3) -> r5c456 = +15(3) Outies n5 = r5c37 = +11(2) Outies r89 = r7c258 = +22(3) = {589} or {679} 35(5)n8 = {56789} -> Innies n8 = r79c46 = +10(4) = {1234} Innies c1234 = r28c4 = +15(2) -> 5 not in r8c4 Innies c6789 = r28c6 = +11(2) Innies c2 = r456c2 = +9(3) Innies c8 = r456c8 = +20(3) 2. 17(2)n3 = {89} Of (89) in n9 - at most one in c7 and at most one in c8 -> One each of (89) in c7 and c8 in n9 -> One each of (89) in c7 and c8 in n6 3! Max r56c5 = +8(2) -> Min r56c46 = +26(4) -> 1 not in r56c46 -> 1 in r5 only in r5c5 or r5c9 Trying r5c5 = 1 (a) r5c456 = <519> puts r4c456 = {236} which leaves no solution for 9(3)n5 (b) r5c456 = <618> puts 8(2)n6 = {35} and r46c5 = {35} which puts r28c6 = [65] which leaves no solution for 24(4)r4c6 -> 1 in r5 only in r5c9 4. -> 8(2)n6 = [71] -> 7(2)n9 from {25} or {34} Also r46c8 from {49} or {58} But the latter case leaves no place for 9 in c8 -> r46c8 = {49} 5. -> 8 in n9 in 15(3)n9 and 9 in n9 in r89c7 -> HS 7 in n9 -> r7c9 = 7 -> r6c89 = [46] and r4c8 = 9 Also Outies r89 = r7c258 = {589} with 9 in r7c25 6. Since 7 already on r5 -> 4 not in 11(2)n4 or outies n5 = r5c37 = +11(2) -> 4 in r5 in r5c456 Also 1 in n5 in r4c456 -> Innies r1234 = r4c456 = +11(3) = {128} or {137} -> 6 in n5 in r5c456 -> r5c456 = {456} -> 6 in n4 in r4c12 and 4 in n4 in r4c123 7! Outies r12 = r3c258 = +22(3) contains a 9 -> 9 not in 19(3)r3c1 -> 19(3)r3c1 = {568} with r3c1 = 5 or 8 -> r4c3 = 4 Also -> Outies r12 = r3c258 = +22(3) = [{79}6] -> 10(3)n3 = [{13}6] -> 15(3)n9 = {258} -> (Since r9c6 is max 4) -> r89c7 = {69} and r7c7 = 1 -> r9c7 = 1 8. r456c2 = +9(3) (No 789) -> 8 not in 11(2)n4 -> 11(2)n4 = [92] -> r456c2 = [621] -> r34c1 = {58} Also 9(2)n1 = {27} or {36} -> 5(2)n7 = {14} 9. Since r7c6 is max 4 -> 9(3)r6c7 = [531] -> r34c9 = [52] -> r4c456 = +11(3) = {137} -> r4c7 = 8 -> r5c37 = [83] -> r6c456 = [928] -> r45c4 = [15], r45c5 = [34], r45c6 = [76] etc. |
Author: | Ed [ Wed Mar 03, 2021 5:40 am ] |
Post subject: | Re: Easter Eggs v2 Revisit |
Very tough. Had to use the n5 black-hole which I was desperately trying to avoid since I make mistakes very easily in that situation. And did. Hence, this WT took a long time. [edit: special thanks to Andrew for confirming my solution is correct and making some nice suggestions and finding some typos.] EEv2 start: Preliminaries Cage 17(2) n3 - cells ={89} Cage 5(2) n7 - cells only uses 1234 Cage 7(2) n9 - cells do not use 789 Cage 8(2) n6 - cells do not use 489 Cage 9(2) n1 - cells do not use 9 Cage 11(2) n4 - cells do not use 1 Cage 8(3) n124 - cells do not use 6789 Cage 9(3) n5 - cells do not use 789 Cage 9(3) n689 - cells do not use 789 Cage 10(3) n47 - cells do not use 89 Cage 10(3) n3 - cells do not use 89 Cage 20(3) n7 - cells do not use 12 Cage 19(3) n14 - cells do not use 1 Cage 35(5) n8 - cells ={56789} note: no clean-up done unless stated 1. 17(2)n3 = {89}: both locked for c9 and n3 2. "45" on r12: 3 outies r3c258 = 22 = {589/679}(no 1,2,3,4) 2a. must have 9: locked for r3 3. 9 in c1 only in r45c1: locked for n4 4. "45" on c34: 2 innies r28c4 = 15 = {69/78} 4a. -> r5c3 + r456c4 = [2]{489/579/678/} all blocked 4b. -> no 2 in r5c3 Andrew noticed another way to do this: "r28c4 = 15, max r24568c4 = 35 -> max r456c4 = 20 -> min r5c3 = 3" 5. "45" on n5: 2 outies r5c37 = 11 = {38/47/56}(no 1,2,9) 6. "45" on r6789: 3 innies r6c456 = 19 (no 1) 6a. "45" on r5: 3 remaining innies r5c456 = 15 6b. only combos with 1 are {159/168} (no eliminations yet) first key step. Very tough 7. 9(3)n5 = {126/135/234} 7a. and "45" on r1234: 3 innies r4c456 = 11 7b. but {236} can only have {135} in 9(3) since {126/234} have two common digits with {236} (Combo crossover clash, CCC) 7c. -> {236} + 9(3) = {26}[3]{26} + [315] 7d. but then h15(3)r5c456 (step 6b) can't make {159/168} 7e. -> {236} blocked 7f. -> r4c456 = {128/137/146/245} = 1 or 5, 7g. -> {159} blocked from h15(3)r5c456 8. 1 in r5 is only in h15(3)r5c456 = {168} or in 8(2)n6 8a. -> {26} blocked from 8(2)n6 (Locking-out cages) 8b. -> 8(2)n6 = {17/35} 9. if 1 in r5 in 8(2)n6 = {17} -> 4 blocked from the two 11(2) in r5 -> 4 in r5c456: locked for n5 9a. or 1 in r5 in h15(3) = {168} only -> h11(3)r4c456 = {245} only 9b. -> 4 locked for n5 in r45c456 9c. and {168} in h15(3) -> 9(3)r4c5 = {135} = [513] (can't be {126} because of CCC) 9d. -> no 4 in r4c5 9e. and no 1 in r5c46 10. h19(3)r6c456 = {289/379/568} 10a. 2 and 3 must be in r6c5 -> no 2,3 in r6c46 10b. h15(3)r5c456: {267/348} blocked by h19(3) 10c. {357} blocked by 8(2)n6 10d. {258} blocked by two 11(2) in r5 10e. = {168/249/456}(no 3,7) [oops. Andrew noticed a much easier way to do the following step since {146} is blocked by h15(3)r5c456 (step 10e)] 11. h11(3)r4c456 = {128/137/146/245} 11a. but {14}[6]{14} blocked by 9(3)r4c5 = [612] (CCC) 11b. -> no 6 in r4c5 11c. {46}[1]{46} blocked by 9(3) can only be [153] but 5 in r5c5 can only have {456} in r5c456 (step 10d) 11d. -> h11(3)r4c456 = {128/137/245} Another tough one to see 12. 4 in n5 only in r4c46 + r5c456 12a. -> 4 in r5c7 sees all 4 in n5 apart from r4c4 so must repeat there 12b. -> 23(4)r4c4 must have {47} = 11 12c. but then r56c4 = 12: blocked since no 3 in r56c4 12d. -> no 4 in r5c7 and the other side 12e. 4 in r5c3 sees all 4 in n5 apart from r4c6 so must repeat there 12f. -> 24(4)r4c6 must have {47} = 11 12g. -> r56c6 = 13 = {58} only 12h. but 4 in r4c6 has {25} in r4c45 (step 11d) 12i. -> no 4 in r5c3 13. h11(2)r5c37 = {38/56}(no 7) = 3 or 5 14. 8(2)n6: {35} blocked by r5c37 14a. = {17} only: both locked for r5 and n6 15. "45" on c9: 3 outies r456c8 = 20 (no 1,2) 15a. -> r5c89 = [71] 15b. -> r46c8 = 13 = {49/58}(no 3,6) 16. 7(2)n9 = {25/34}(no 6) 17. 17(3)r6c8: {359} as [9]{35} only, blocked by 7(2) 17a. {458} as [8]{45} only, blocked by 7(2) 17b. = {269/278/368/467}(no 5) 17c. no 8 in r4c8 18. 16(3)r3c9: must have 4,5,9 for r4c8 18a. = {259/349/457}(no 6) 19. h15(3)r5c456 = {249/456}(no 8) 19a. must have 4, locked for r5 and n5 20. 1 in n5 only in h11(3) = {128/137}(no 5,6,9) 20a. 1 locked for r4 21. 8(3)r3c3 must have 1 which is only in r3: locked for r3 22. "45" on c1, 3 outies r456c2 = 9 = {126/135/234}(no 7,8 22a. -> r6c2 = (1,2,3,4) 23. 19(3)r3c1 = {289/379/469/478/568} 23a. note: if it has 5, must also have 8 23b. -> h9(3)r456c2: {135} as [531] blocked since it will have two 8s in c1 23c. -> no 5 in r4c2 23d. -> no 6 in r4c1 in 19(3) (no valid permutations) 24. killer quad 1,2,3,4 in 9(2)n1, 10(3)r6c1 and 5(2)n7: 2,3,4 locked for c1 25. 6 in r4 only in r4c2 or r4c7 25a. if in r4c2 -> r34c1 = {58} only (step 23) 25b. if in r4c7 -> r3c67 = 8 = {35} 25c. -> r3c167 = 5 or 8 26. h22(3)r3c25: {589} blocked by r3c167 26a. -> = {679} only: all locked for r3 26a. -> r3c8 = 6 26b. -> r128 = 4 = {13}: both locked for n3 and c8 Much easier now Ed |
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