SudokuSolver Forum :: View topic - Assassin 2X Revisit

Author:

Andrew [ Sat Sep 19, 2020 4:22 am ]

Post subject:

Re: Assassin 2X Revisit

Thanks Ed for reposting this excellent variant by Mike Parker from over a decade ago. If it was posted now it would certainly still qualify as an Assassin.

I haven't looked at how I solved it back then but I think I've almost certainly found a more interesting way this time. My key breakthrough move was a bit like a wellbeback type of step.

Here is my walkthrough for Assassin 2X Revisited:

Same rating as first time, which Ed has already posted.

Author:	Ed [ Mon Sep 14, 2020 8:12 pm ]
Post subject:	Assassin 2X Revisit
Thought it might be worthwhile to have a go at an old puzzle to see if the new techniques we've been finding give an interesting alternative solution. Just picked the first puzzle that gets a 1.50 rating in the archives. However, it has so many small cages it may not yield anything but could be fun trying. Assassin 2X Attachment: a2x.JPG [ 106.09 KiB \| Viewed 6119 times ] Archive post here SSscore = 1.60, JSudoku (not default routines) uses 3 "complex intersections" triple click code: 3x3:d:k:4352:4352:3841:3841:4610:3587:3587:3332:3332:4352:4613:4613:3841:4610:3587:4358:4358:3332:4615:4613:3848:3848:4610:3081:3081:4358:3338:4615:4615:3848:3595:3595:2572:3081:3338:3338:2573:2573:2573:3595:2572:5390:3599:3599:3599:3344:3344:4113:2572:5390:5390:4370:3347:3347:3344:4116:4113:4113:3093:4370:4370:6166:3347:4375:4116:4116:3864:3093:2585:6166:6166:3354:4375:4375:3864:3864:3093:2585:2585:3354:3354: solution: Code: +-------+-------+-------+ \| 8 6 7 \| 3 9 2 \| 5 1 4 \| \| 3 4 9 \| 5 1 7 \| 6 2 8 \| \| 2 5 1 \| 6 8 4 \| 7 9 3 \| +-------+-------+-------+ \| 7 9 8 \| 2 5 3 \| 1 4 6 \| \| 5 1 4 \| 7 6 8 \| 2 3 9 \| \| 6 3 2 \| 1 4 9 \| 8 5 7 \| +-------+-------+-------+ \| 4 2 5 \| 9 7 6 \| 3 8 1 \| \| 1 8 6 \| 4 3 5 \| 9 7 2 \| \| 9 7 3 \| 8 2 1 \| 4 6 5 \| +-------+-------+-------+ Cheers Ed

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Assassin 2X Revisit http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=3&t=1581	Page 1 of 1

Author:	Ed [ Fri Sep 25, 2020 7:02 pm ]
Post subject:	Re: Assassin 2X Revisit
Wonderful Andrew! Really like 4e and 5g. Both are definitely interesting. Mike would have been really pleased I'm sure. I couldn't solve it and never noticed that "45" behind 4e. However, I did enjoy trying. Loved the 4 13(3) cages in c9. They must all have different combinations since they see 2 cells of each of the others. The top three 13(3) have 1 for both c8 & c9. Those 9 cells sum to 39 so the repeat 1 could be replacing a 7 to get the 9 cells down to 39 (ie 45 - 39 = 6). However, those 3 cages must have 7 for c9 so must have at least two repeats. But couldn't take it any further. I haven't looked at how it was done previously but plan to. Cheers Ed

Author:	wellbeback [ Sun Oct 04, 2020 8:03 pm ]
Post subject:	Re: Assassin 2X Revisit
I have only the vaguest of memories that I might have looked at this before. I presume I must have done but, if so, no memory of how I did it back then and doubt I solved it like I did now. Nicely spotted about all those 13(3)s in c89. I didn't use that at all. I did find some cool moves though! (Similar in feel to some of the early HSs!). Here is how I started this time around... Assassin 2X WT: 1. Innies r1234 = r4c456 = +10(3) (No 89) Innies c1234 = r456c4 = +10(3) (No 89) 10(3) on D/ in n5 (No 89) -> 21(3)n5 = {489} -> The three 10(3)s in n5 = {127}, {136}, {235} Since the three corners except for r6c6 in n5 are each in two of the 10(3)s - they must contain the numbers which appear twice in the 10(3)s I.e., they must be (123) 2. Innies c6789 = r456c6 = +20(3) -> r456c6 = [3{89}} and r6c5 = 4 Also r456c4 = [172] or [271] and r45c5 = {56} 3. Outies n9 = +32(6) Max r789c6 = +16 (7 + 9(2)) -> Min r6c789 = +16 4. 10(3)n4 cannot be the same as H+10(3)r4c456 since that would put r6c789 = the same +10(3) contradicting Min r6789 = +16 -> 10(3)n4 cannot be the same as any of the three +10(3)s in n5. -> 10(3)n4 = {145} -> (Since Innies r5 = r5c456 = +21(3)) n5 = [253][768][149] -> 14(3)n6 = {239} -> r4c123 = 9 + two of (678) and r4c789 = (14) and one of (678) -> r6c123 = (23) + one of (678) and r6c789 = 5 and two of (678) 5. Outies n3 = +24(6) Since r4c789 from +11, +12, or +13 -> r123c6 from +13, +12, or +11 -> (Since (89) already in c6) r123c6 cannot contain both (12) Also 18(3)n2 cannot contain both (12) -> one each of (12) in c5 and c6 in n2 -> 18(3)n2 from {189} or {279) -> Must contain a 9 -> 9 in c4 in n8 Also 3 in c5 in n8 and 3 in c4 in n2 6. Innies n9 = +8(3) = {125} or {134} 9 in r6c6 prevents 1 in r7c7 -> 1 in r7c9 or 79c7 D\ already contains (26) -> Either 5 in Innies n9 puts 5 in r7c7 or 5 in 13(3)n9 = {256} puts 5 in r9c9 -> 5 locked in D\ in n9 7. 1 in D\ in n1 Innies n1 = +10(3) -> Either r3c3 = 1 or Innies n1 = {235} with 3 in r3c3. -> r3c4 cannot be 3 -> 3 in n2 in r12c4 -> 15(3)r1c3 = [7{35}] or [4{38}] -> Innies n1 from [721] or [451] But the latter puts r4c123 = [{67}9] which puts another 5 in r3 in r3c4 -> Innies n1 = [721] and 15(3)r1c3 = [7{35}] 8. 15(3)r3c3 only [168] or [186] -> (Since 7 in r1c3 and r5c3 from (45)) Outies c12 = r258c3 = +19(3) can only be [946] or [649] -> 15(3)r3c3 = [168] -> r789c4 = {489} -> r123c5 = {189} -> r789c5 = {237} -> r789c6 = {156} -> r123c6 = {247} Also r4c123 = [{79}8] -> r4c789 = {146} -> r6c789 = {578} -> r6c123 = {236} Straightforward from here.

Author:	Andrew [ Mon Oct 05, 2020 8:28 pm ]
Post subject:	Re: Assassin 2X Revisit
Loved wellbeback's solving path particularly the way it got straight to the heart of the puzzle, then how his step 4 makes use of that start, using a similar method to my steps 4d and 4e but on different cages!

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Author:	Andrew [ Sat Sep 19, 2020 4:22 am ]
Post subject:	Re: Assassin 2X Revisit
Thanks Ed for reposting this excellent variant by Mike Parker from over a decade ago. If it was posted now it would certainly still qualify as an Assassin. I haven't looked at how I solved it back then but I think I've almost certainly found a more interesting way this time. My key breakthrough move was a bit like a wellbeback type of step. Here is my walkthrough for Assassin 2X Revisited: Prelims a) 10(3) cage at R4C6 = {127/136/145/235}, no 8,9 b) 10(3) cage at R5C1 = {127/136/145/235}, no 8,9 c) 21(3) cage at R5C6 = {489/579/678}, no 1,2,3 d) 24(3) cage at R7C8 = {789} e) 10(3) cage at R8C6 = {127/136/145/235}, no 8,9 1a. 45 rule on N1 3 innies R1C3 + R3C13 = 10 = {127/136/145/235}, no 8,9 1b. Naked triple {789} in 24(3) cage at R7C8, locked for N9 1c. 45 rule on N9 3 innies R7C79 + R9C7 = 8 = {125/134}, no 6, 1 locked for N9 1d. 45 rule on R5 3 innies R5C456 = 21 = {489/579/678}, no 1,2,3 1e. 4 of {489} must be in R5C5 -> no 4 in R5C46 1f. 45 rule on R12 3 outies R3C258 = 22 = {589/679}, 9 locked for R3 1g. 45 rule on C12 3 outies R258C3 = 19 = {289/379/469/478/568}, no 1 1h. 2 of {289} must be in R5C3 -> no 2 in R28C3 1i. Max R3C34 = 13 (cannot be {68} which clashes with R3C258) -> min R4C4 = 2 2a. 45 rule on C6789 3 outies R5C5 + R6C45 = 11 = {146/245} (cannot be {128/137/236} because 1,2,3 only in R6C4), no 3,7,8,9 2b. 1,2 only in R6C4 -> R6C4 = {12} 2c. R5C5 + R6C45 = 11 = {146/245}, 4 locked for C5 and N5 2d. 10(3) cage at R4C6 = {136/145/235} (cannot be {127} because no 1,2,7 in R5C5), no 7 2e. 45 rule on C5 3 innies R456C5 = 15 = {456} (only remaining combination), 5,6 locked for C5 and N5 2f. 10(3) cage at R4C6 = {136/235} (cannot be {145} because 4,5 only in R5C5) -> R4C6 = 3, placed for D/ 2g. R6C5 = 4 (hidden single in C5) -> R56C6 = 17 = {89}, locked for C6 and N5 2h. R5C4 = 7 2i. Naked pair {12} in R46C4, locked for C4 2j. 10(3) cage at R5C1 = {136/145/235} 2k. Killer pair 5,6 in 10(3) cage and R5C5, locked for R5 2l. R258C3 (step 1g) = {289/379/469/478/568} 2m. 3 of {379} must be in R5C3 -> no 3 in R28C3 2n. Max R7C67 = [75] -> min R6C7 = 6 2o. Min R89C4 = 7 -> max R9C3 = 8 3a. 45 rule on N8 2 innies R7C46 = 2 outies R9C37 + 8 3b. Min R9C37 = 3 -> min R7C46 = 11, no 3 in R7C4, no 1 in R7C6 3c. Max R7C46 = 16 -> max R9C37 = 8, no 8 in R9C3 4a. R5C4 = 7 -> R4C45 = 7 = [16/25] 4b. 45 rule on R5 2 remaining innies R5C56 = 14 = [59/68] 4c. 45 rule on R6789 2 remaining innies R6C46 = 10 = [19/28] 4d. 45 rule on N1 6(3+3) outies R123C4 + R4C123 = 38, max R123C4 = {689} = 23 -> min R4C123 = 15 4e. 14(3) cage at R5C7 = {149/239} (cannot be {248} which forces R5C56 = [59], R6C456 = [248] and R4C123 = {248} = 14 but R4C123 must total at least 15), 9 locked for R5 and N6 4f. R56C6 = [89] -> R5C5 = 6, placed for both diagonals, 9 placed for D\, R4C5 = 5, R4C4 = 2, placed for D\, R6C4 = 1, placed for D/ 4g. 10(3) cage at R5C1 = {145/235}, 5 locked for N4 4h. 5 in R6 only in R6C89, locked for 13(3) cage at R6C8 4i. 13(3) cage at R6C8 = {157/256}, no 3,4,8 4j. 3 in N6 only in 14(3) cage = {239}, 2,3 locked for R5, 2 locked for N6 4k. Naked triple {145} in 10(3) cage at R5C1, 1,4 locked for N4 4l. R258C3 (step 1g) = {469/478/568} 4m. R5C3 = {45} -> no 4,5 in R28C3 4n. Max R6C7 + R7C6 = 15 -> no 1 in R7C7 4o. 1 on D\ only in R1C1 + R2C2 + R3C3, locked for N1 4p. Min R4C12 = 13 -> max R3C1 = 5 4q. Consider placement for 9 in R4 9 in R4C12 => 18(3) cage at R3C1 = 2{79}/3{69} or 9 in R4C3 => R3C34 = 6 = [15] -> no 5 in R3C1 4r. R1C3 + R3C13 = 10 (step 1a), min R1C3 + R3C1 = 5 -> max R3C3 = 5 4s. Min R2C3 + R3C2 = 11 -> max R2C2 = 7 4t. 8 on D\ only in R1C1 + R8C8, CPE no 8 in R1C8 + R8C1 4u. R7C79 + R9C7 (step 1c) = {125/134} 4v. 5 of {125} must be in R7C7 -> no 5 in R9C7 4w. 12(3) cage at R3C6 = {147/156/246}, no 8 5a. 12(3) cage at R7C5 = {129/138/237} 5b. 45 rule on N47 3 innies R4C123 = 3 outies R789C4 + 3 5c. 9 in R4 only in R4C123, R4C123 = 22, 23, 24 -> R789C4 = 19, 20, 21 5d. R789C4 = 19, 20, 21 = {469/568/569/489} (cannot be {389} = 20 which clashes with 12(3) cage), no 3 5e. 3 in N8 only in 12(3) cage = {138/237}, no 9, 3 locked for C5 5f. 9 in N8 only in R789C4 = {469/569/489}, 9 locked for C4 5g. 15(3) cage at R8C4, min R789C4 = 19 -> R7C4 at least 4 more than R9C3, no 4 in R7C4, no 6,7 in R9C3 5h. R789C4 = 19, 20, 21, 12(3) cage at R7C5 -> R789C6 = 12, 13, 14 5i. 10(3) cage at R8C6, min R789C6 = 12 -> R7C6 at least 2 more than R9C7, no 2 in R7C6 5j. R1C3 + R3C13 (step 1a) = {127/136/145/235} 5k. 4 of {145} must be in R3C1 -> no 4 in R13C3 6a. 6 in N9 only in 13(3) cage at R8C9 = {256/346} 6b. 5 of {256} must be in R9C9 -> no 5 in R8C9 + R9C8 6c. 5 in N9 only in R7C7 + R9C9, locked for D\ 6d. 15(3) cage at R3C3 = {159/168/348/357} = [159/168/186/348/357], no 3 in R3C4 6e. 3 in C4 only in R12C4, locked for 15(3) cage at R1C3 6f. 15(3) cage at R1C3 = {357} (only possible combination, cannot be {348} because no 3,4,8 in R1C3) -> R12C4 = {35}, 5 locked for C4 and N2, R1C3 = 7 6g. R789C4 (step 5f) = {469/489}, 4 locked for C4, N8 and 15(3) cage at R8C4 6h. 15(3) cage at R3C3 = [168/186] -> R3C3 = 1, R4C3 = {68} 6i. R258C3 (step 4l) = {469} (only remaining combination, cannot be {568} which clashes with R4C3) -> R5C3 = 4, R28C3 = {69}, locked for C3 -> R4C3 = 8, R3C4 = 6 7a. R1C3 + R3C13 = 10 (step 1a), R1C3 = 7, R3C3 = 1 -> R3C1 = 2 (cage sum) 7b. R3C1 = 2 -> R4C12 = 16 = {79}, 7 locked for R4 and N4 7c. R6C7 = 8 (hidden single on R6) -> R7C67 = 9 = [54/63] 7d. R8C8 = 7 (hidden single on D\) -> R8C7 = 9, R7C8 = 8, R7C4 = 9, R28C3 = [96] 7e. Naked pair {48} in R89C4, 8 locked for N8 -> R9C3 = 3 (cage sum), R67C3 = [25], 5 placed for D/ 7f. Naked pair {36} in R6C12, 6 locked for R6 -> R7C1 = 4 (cage sum) 7g. R6C89 = [57] -> R7C9 = 1 (cage sum) 7h. R8C3 = 6 -> R78C2 = 10 = [28], 8 placed for D/, R8C1 = 1 7i. 12(3) cage at R7C5 = {237} (only remaining combination), 2,7 locked for C5 and N8 7j. R789C6 = [651] -> R9C7 = 4 (cage sum), R7C7 = 3, placed for D\ 7k. R3C67 = [47], 7 placed for D/, R4C7 = 1 (cage sum) 7l. R3C8 = 9 -> R2C78 = 8 = [62] and the rest is naked singles, without using the diagonals. Same rating as first time, which Ed has already posted.