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Caida
Hooked

Joined: 03 Nov 2007
Posts: 38

 Posted: Fri Jan 04, 2008 1:33 am    Post subject: Assassin 84 With the time zone advantage I get to be first this time. Edited to add: I would rate this a 1.25 - I don't think it was a particularly narrow solving path nor did I use any advanced techniques. Here is my walkthrough for A84 - it is how I solved it so a bit messy at the end. Hope it makes sense! I have added some notes for clarity - thanks Andrew and Ed! Assassin 84 Walkthrough Preliminaries: a. 11(2)n1 = {29/38/47/56} (no 1) b. 11(3)n23 = {128/137/146/236/245} (no 9) c. 19(3)n3 and n45 and n56 and n89 = {289/379/469/478/568} (no 1) d. 7(2)n3 and n4 = {16/25/34} (no 7..9) e. 15(5)n5 = {12345} (no 6..9) -> 1,2,3,4,5 locked for n5 f. 12(2)n6 and n7 = {39/48/57} (no 1,2,6) g. 3(2)n9 = {12} (no 3..9) -> 1,2 locked for n9 and r9 1. Innies c1234: r258c4 = 7(3) = {124} -> locked for c4 2. Outies n4: r46c4 = 17(3) = {89} -> locked for c4 and n5 2a. pair {67} locked for n5 and c6 in r46c6 -> no 6,7 elsewhere in c6 3. Outies n1: r13c4 = 12(2) = {57} -> locked for c4 and n2 3a. pair {36} locked for n8 and c4 in r79c4 -> no 3,6 elsewhere in n8 3b. 12(3)n12 = {16/34}[5]/{14/23}[7] 3c ->r1c23 no 5,7,8,9 4. Outies n3: r13c6 = 10(2) = [28/82/91] 4a. -> r13c6 no 3,4 4b. -> r1c6 no 1 5. Outies n9: r79c6 = 6(2) = [15/24] 5a. -> r7c6 no 4,5,8,9 5b. -> r9c6 no 8,9 5c. killer pair {12} in h10(2)n2 and h6(2)n8(r7c6) -> no 1,2 elsewhere in c6 5d. 1 locked in n8 in h6(2)n8(r7c6) and r8c4 -> no 1 elsewhere in n8 Note another way to say this is if r7c6 <>1 then h6(2)n8(r79c6) = {24} -> r8c4 = 1 6. 12(3)n56: r4c78 no 6,7,8,9 (min r4c6 = 6) 6a. 19(3)n56 = {379/469/478/568}: r56c7 no 2,6,7 6b. -> 6 in n6 locked in 15(3) 6c. 15(3)n6 = {168/267/456} (no 3,9) 7. 12(3)n89: r7c78 no 9 (needs both 1,2) 7a. 19(3)n89 = {469/478/568}: r89c7 no 3,4,5 7b. killer pair {89} in c7 in 19(3)n56 and 19(3)n89 -> no 8,9 elsewhere in c7 7c. 17(3)n9 = {359/368/458} (no 7) Note: combo {467} blocked by 19(3)n89 7d. killer pair {89} in n9 in 19(3)n89 and 17(3)n9 -> no 8,9 elsewhere in n9 7e. 12(3)n89 = [1]{47}/[2]{37/46} (no 5) Note: combo [1]{56} blocked by 17(3)n9 7f. -> 5 locked in n9 in 17(3)n9 -> 17(3)n9 = {359/458} (no 6) 7g. 18(3)n23: r1c8 no 1 (need {89}) 8. Innies c12: r147c2 = 9(3) = {126/135/234} (no 7,8,9) 8a. 18(3)n45: r4c3 no 1,2 (needs {89} or {79}) 9. Innies c89: r147c8 = 17(3) = [917/926/827/953/836/854/647/746] 9a. -> r1c8 no 2,3,4,5 9b. 18(3)n23: r1c7 no 5,6 10. 13(3)n78@r7c2 = [193/283/256/526/346/436/463/643] 10a. -> r7c3 no 1,7 10b. 13(3)n78@r8c3 = [193/283/256/346/436/463/643] 10c. -> r8c3 no 5,7,8,9 10d. -> r9c3 no 7 11. Innies r7: r7c159 = 20(3) = {389/569/578} (no 1,2,4) Note: combo {479} blocked by 12(3)n89 11a. 12(3)n89 = [1]{47}/[2]{46} -> all other combinations blocked by h20(3) and r7c4 11b. 4 locked in r7c78 in r7 and n9 -> no 4 elsewhere in r7 and n9 11c. r7c78 no 3 11d. 17(3)n9 = {359} -> locked for n9 11e. -> 8 locked in n9 in c7 -> no 8 elsewhere in c7 11f. -> 9 locked in c7 in n6 -> no 9 elsewhere in n6 11g. -> r6c89 no 3 11h. 19(3)n56 = {379/469}: r56c7 no 5 11i. 13(3)n78@r7c2 = [193/283/256/526]: r7c23 no 3,6 11j. 15(3)n6 = {168/267} (no 4,5) Note: combo {456} blocked by 12(2)n6 12. Innies r4: r4c159 = 15(3) = {159/168/249/258/348/357/456} Note: combo {267} blocked by r4c6 12a. -> r4c1 no 1,7 13. Innie and Outties r9: r8c37 – r9c5 = 2 13a. -> min r8c37 = 7 -> min r9c5 = 5 (r9c5 no 4) 13b. -> max r9c5 = 9 -> max r8c37 = 11 (r8c3 no 6) 14. 18(3)n23 = [279/918/819] Note: combo [936] blocked by r23c7 = {37/46} combo [837] blocked by r23c7 = {36} OR r23c7 = {45} AND r56c7 = {39} combo [846] blocked by r23c7 = {36/45} combo [927] blocked by 7(2)n3 (r1c78 = 27, r23c7 = 46 -> no possible combo for 7(2)) 14a. -> r1c7 no 2,3,4 14b. -> r1c8 no 6,7 14c. 19(3)n3 no 2 (combo {289} blocked by r1c8) 14d. combo {16} blocked from 7(2)n3 by 18(3)n23 and 11(3)n23 -> if r1c7 = 1 -> combo blocked; if r1c7 = 7 -> r23c7 = {12} because r789c7 would be 4{68} -> combo blocked 14e. -> r3c89 no 1,6 14f. 5 locked in n3 in 19(3) and 7(2) -> no 5 elsewhere in n3 14g. hidden single c7: r4c7 = 5 14h. 12(3)n56 = [651] (only possible combination) 14i. 19(3)n56 = 7{39} -> {39} locked for n6 and c7 14j. single: r9c8 = 2; r9c9 = 1 14k. pair: 12(2)n6 = {48} -> locked for n6 and r6 14l. single: r6c4 = 9; r4c4 = 8 14m. r56c7 = [93] 15. 18(3)n45 = [378] -> only possible combination 15a. single r4c9 = 2 15b. pair {67} locked in n6 in r5 -> no 6,7 elsewhere in r5 15c. single: r4c5 = 4; r4c1 = 9 15d. r5c12 no 2 15e. r56c3 no 5; r5c3 no 2 16. 4 locked in n2 in r2 -> no 4 elsewhere in r2 16a. 7(2)n3 = {34} -> locked for n3 and r3 16b. r3c12 no 7,8; r3c2 no 2 16c. 2 locked in r5 in n5 -> no 2 elsewhere in n5 16d. hidden single c7: r7c7 = 4 17. 11(3)n23 = {128} -> all other combinations blocked 17a. r3c6 = 8; r23c7 = {12} -> locked for n3 and c7 17b. r1c6 = 2 (step 4) 17c. 18(3)n23 = [279] 17d. single: r1c4 = 5; r3c4 = 7; r7c6 = 1 17e. r9c6 = 5 (step 5) 17f. r5c6 = 3 17g. r1c3 no 4 17h. r2c3 no 2,3,5,6; r3c3 no 5,6 17i. hidden single c3: r7c3 = 5 Now it is singles and cage sums to the end Cheers, CaidaLast edited by Caida on Fri Jan 11, 2008 6:06 pm; edited 3 times in total
Afmob
Expert

Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

 Posted: Fri Jan 04, 2008 10:07 pm    Post subject: Wow, this assassin had a lot of Killer pairs you could use and it was pretty stubborn before only singles were needed to solve it. A84 Walkthrough: 1. C1234 a) Outies = 7(3) = {124} locked for C4 b) Outies N4 = 17(2) = {89} locked for C4+N5 c) Outies N7 = 9(2) = {36} locked for C4+N8 d) Outies N1 = 12(2) = {57} locked for N2 e) 12(3) must have 5 xor 7 because R1C4 = (57) -> R1C23 <> 5,7,8,9 f) 17(3) @ R2C3 must have 5 xor 7 because R3C4 = (57) -> R23C3 <> 5,7 g) Both 13(3) <> 7 because R9C46 = (36) h) 7 locked in R456C3 for N4 i) 18(3) @ R4C1 <> 2 j) Innies C12 = 9(3) <> 8,9 2. C6789 a) 3(2) = {12} locked for R9+N9 b) Outies N6 = 13(2) = {67} locked for C6 c) Outies N9 = 6(2) = [15/24] d) Outies N3 = 10(2) = [28/82/91] e) Killer pair (12) locked in Outies N3 + R7C6 for C6 f) 12(3) @ N6: R4C78 <> 6,7,8,9 because R4C6 >= 6 g) 19(3) @ N6 must have 6 xor 7 because R6C6 = (67) -> R56C7 <> 2,6,7 h) 15(3) must have 1 xor 2 because 12(3) @ N6 can't have both -> 15(3) <> 3 i) 6 locked in 15(3) for N6 -> 15(3) = 6{18/27} j) 19(3) @ N9 must have 2 of (6789) and it's only possible @ R89C7 -> R89C7 <> 3,4,5 k) Killer pair (89) locked in both 19(3) for C7 3. N9 a) 17(3) <> 6,7 because 6{47/38} blocked by Killer pairs (67,68) of 19(3) @ N9 b) 5 locked in 17(3) for N9 c) Killer pair (89) locked in 19(3) + 17(3) 4. R789 a) Killer pair (34) of 12(3) blocks {346} of 13(3) @ R7 -> R7C23 <> 3,4,6 b) Killer pair (12) locked in 12(3) + R7C6 for R7 c) Innies+Outies R9: 2 = R8C37 - R9C5 -> R9C5 <> 4 because R8C37 >= 7 d) Killer pair (23) of 13(3) @ R7 blocks {237} of 12(3) e) 12(3) = 4{17/26} -> 4 locked for R7+N9 f) 17(3) = {359} locked for N9 g) 19(3) = 8{47/56} -> 8 locked for C7 5. R456 a) 9 locked in 19(3) @ C7 for N6 -> 19(3) = 9{37/46} b) 12(2) <> 3 c) Innies+Outies R6: 12 = R5C37 - R6C5 -> R5C37 <> 2,3 d) Innies+Outies R6: 12 = R5C37 - R6C5 -> 9 locked in R5C37 = 9{4/5/6/7/8} because R5C7 = (49) e) 18(3) @ R4C1: R4C1 <> 1,3,5 because R5C12 <> 9 6. R123 ! a) Innies N1 = 17(4) = {1268/1349/2348} -> R23C3 <> 6 because it forces 4 in 17(3) @ R23C3 b) ! 17(3) @ R1C1 <> 4 because 4{58/67} blocked by Killer pairs (46,48) of Innies N1 c) Killer pair (89) locked in both 17(3) for N1 d) 11(2) <> 2,3 e) Killer pair (57) locked in 11(2) + R3C4 for R3 f) 7(2) <> 2 g) Killer pair (46) locked in 11(2) + 7(2) for R3 h) 11(3) <> 4,5 because R3C67 <> 4,5,6 i) 11(3) = {128} because other combos blocked by Killer pairs (13,36) of 7(2) -> R3C6 = 8, {12} locked for C7+N3 j) 7(2) = {34} locked for R3+N3 k) 19(3) = {568} locked for N3 l) 18(3) = {279} -> R1C6 = 2, R1C7 = 7, R1C8 = 9 7. N1 a) 11(2) = {56} locked for R3+N1 b) 12(3) = {345} -> R1C4 = 5, {34} locked for R1+N1 c) 17(3) @ R2C3 = 7{19/28} -> R3C4 = 7, R2C3 <> 2 8. R789 a) Outies N9 = 6(2) = [15] -> R7C6 = 1, R9C6 = 5 b) 12(3) = {147} -> R7C8 = 7, R7C7 = 4 c) 12(2) <> 7 d) 4 locked in R8C456 for R8 9. R456 a) 19(3) @ N6 = {379} -> R5C7 = 9, R6C7 = 3, R6C6 = 7 b) 12(2) = {48} locked for R6+N6 c) 15(3) = {267} locked for N6 d) 12(3) = {156} -> R4C6 = 6, R4C7 = 5, R4C8 = 1 e) 18(3) @ R4C2 = 7{29/38} -> R4C3 = 7 f) Outies N4 = 17(2) = {89} -> R4C4 = 8, R6C4 = 9 g) R4C2 = 3, R1C2 = 4, R1C3 = 3 10. N47 a) Killer pair (34) of 12(2) blocks {346} of 13(3) @ R8C3 -> 13(3) <> 4 b) Hidden Single: R9C1 = 4 @ R9 -> R9C2 = 8 c) 18(3) @ R4C1 = {189} -> R4C1 = 9, R5C1 = 8, R5C2 = 1 11. Rest is singles. Rating: 1.25Last edited by Afmob on Thu Jan 10, 2008 8:52 pm; edited 2 times in total
gary w

Joined: 07 Sep 2007
Posts: 84
Location: south wales

 Posted: Fri Jan 04, 2008 11:08 pm    Post subject: Found this to be one of the easier assassins..took about 35-40 mins only. r46c4=[89] so r13c4={57}.But then the combos of the two 17(3) cages in N1 means that r1c4=7,r3c4=5 is not possible so r1c4=5,r3c4=7. r79c4 and r79c6 fall almost as easily and it was really just straightforward combo work to the end. Regards Gary
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

 Posted: Sat Jan 05, 2008 10:47 am    Post subject: I seem to have found this Assassin a bit easier than Caida and Afmob - [EDIT2: step 9 clarified. Thanks Andrew] though haven't looked at Afmob's WT yet. I haven't had much practice at gut-feel ratings for a while so will be very interested in what the SSscore is for this one. I'll give it a high 1.0. I never seemed to get stuck anywhere which I often do for 1.25 puzzles. Still haven't finished A78. Gary - could you explain in more detail how you made those first two placements? Thanks. Cheers Ed A84 WT Any comments or corrections very welcome. Prelims i. 19(3)n3: no 1 ii. 11(3)n3: no 9 iii. 11(2)n1: no 1 iv. 15(4)n5: no 6789 v. 7(2)n3 & n4: no 789 vi. 19(3)n4 & n6 & n9: no 1 vii. 12(2)n6 & n7: no 126 viii. 3(2)n9 = {12} 1. 3(2)n9 = {12}: both locked for r9 & n9 1a. min r9c34 = {34} = 7 -> max r7c3 = 6 1b. min r7c78 = {34} = 7 -> max r7c6 = 5 2. 15(4)n5 = {12345}: all locked for n5 2a. min r4c6 = 6 -> max r4c78 = 6 (no 6..9) 3. "45" c12: 3 innies r147c2 = h9(3) = {126/135/234}(no 789) 3a. max r4c2 = 6 -> min r4c34 = 12 (no 1,2) 4. "45" c1234: 3 innies r258c4 = h7(3) = {124}:all locked for c4 4a. min r1c4 = 3 -> max r1c23 = 9 (no 9) 5. "45" n4: 2 outies r46c4 = h17(2) = {89}: both locked for c4 and n5 6. Naked pair (NP) {67} in r46c6: both locked for c6 6a. 12(3)n5 must have 6/7 = [6]{15/24}/[7]{14/23} = [1/2..but not both] 6b. 19(3)n6 must have 6/7 = [6]{49/58}/[7]{39/48}(no 2, no 67 in r56c6) 7. 6 in n6 only in 15(3) and must have 1/2 for n6 = 6{18/27}(no 3459) 8. "45" n9: 2 outies r79c7 = h6(2) = [15/24](no 36789) 8a. r7c6 = {12}, r9c6 = {45} 9. 17(3)n9 = {359/368/458/467} = [5/6..] 9a. 19(3)n9 must have 4/5 = [4]{69/78}/[5]{68}(no 4/5 in r89c7) 9b. EDIT: r89c7 = {69/78/68} = [6/7,8/9...] -> {467} blocked from 17(3)n9 9c. 17(3) = {359/368/458}(no 7) = [8/9..]: 9d. Killer pair {89} between 17(3) & r89c7: both locked for n9 10.12(3)n8 must have 1/2 = [1]{47}/[2]{37/46}(no 5) ([1]{56} blocked by 17(3)n9 step 9) = [3/4..] 11. 5 in n9 only in 17(3) = 5{39/48}(no 67) 12. "45" n7: 2 outies r79c4 = h9(2) = {36}: both locked for n8 & c4 13. 13(3)r7c2 must have 3/6 = [3]{19/28}/[6]{25}(no 47) = [1/2..] ({346} clashes with 12(3)n8 step 10) 13a. r7c23 = {19/28/25}(no 3467) 14. Killer pair {12} in 13(3)r7c2 (step 13) & r7c6: both locked for r7 15. NP {57} in r13c4: both locked for n2 15a. 12(3)n1 must have 5/7 = [5]{16/34}/[7]{14/23}(no 8; no 57 in r1c23) 15b. 17(3)r2c3 must have 5/7 = [5]{39/48}/[7]{19/28/46}(no 57 in r23c3) 16. In n1, 5 & 7 have to be shared between 17(3)r1c1 & 11(2). Since neither can have both they must each have 1 16a. 11(2) = {47/56}(no 2389) = [4/6,5/7..] 16b. 17(3)r1c1 = {179/278/359/458/467} 17. Killer pair {57} between 11(2)n1 & r3c4: both locked for r3 17a. no 2 in r3c89 18. from step 15b. r23c3 = {39/48/19/28}(no 6) ({46} clashes with 11(2)n1) 19. "45" n3: r13c6 = h10(2) = [91]/{28} 19a. r1c6 = {289}, r3c6 = {128} 20. 18(3)n2 = {189/279/378/459}(no 6) ([9]{36}/[8]{46} clash with 7(2)n3;{567} clashes with r1c4) 21. Killer pair {46} in 11(2)n1 & 7(2)n3: both locked for r3 22. 11(3)n3 = {128} only ({137}/[2]{36} clashes with 7(2)n3;other combo's have no valid permutation) 22a. ->no 1 in r3c89 22b. no 6 r3c89 23. 7(2)n3 = {34}: both locked for r3 & n3 23a. no 7 in r3c12 24. 11(2)n1 = {56}: both locked for r3 & n1 25. r13c4 = [57] 25a. r23c3 = {19/28}(no 34) 25b. r1c23 = {34}: both locked for n1 & r1 26. 5 in r2 only in c89 in 19(3)n3 = {568}: all locked for n3 26a. generalized X-wing on 5 in 19(3)n3 & 17(3)n9: 5 locked for c89 [ALT: 5 in c7 only in n6: 5 locked for n6. Thanks Andrew] 26b. no 7 in 12(2)n6 27. r23c7 = {12}: both locked for n3 & c7 & not in r3c6 28. r13c6 = [28] (h10(2)r13c6) 29. r79c6 = [15] (h6(2)r79c6) 29a. no 7 in r9c12 29b. r89c7 = {68}: both locked for n9 & c7 30. r7c78 = {47}: both locked for n9 & r7 31. deleted 32. "45" c89: r147c8 = h17(3) = [917] (only valid permutation) rest is naked/hidden/cage sumsLast edited by sudokuEd on Fri Jan 11, 2008 6:43 am; edited 5 times in total
gary w

Joined: 07 Sep 2007
Posts: 84
Location: south wales

 Posted: Sun Jan 06, 2008 12:20 am    Post subject: Hi sudokuEd, Yes... you know r13c4=12 (outies of N1) and that therefore they must be {57} because outies of N4-> r46c4={89}.If r3c4=5 and r1c3=7 in N1 the two 17(3) cages combined with the 7 and 5 placements are very restrictive and mean that the 11(2) cage must contain the 7 and thus is {47}.You cannot now place the 1 in N1.Therefore r3c4=7 and r1c3=5.After that it's really an extended mop-up. Regards Gary
Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

Posted: Sun Jan 06, 2008 1:22 am    Post subject:

 Ed wrote: I seem to have found this Assassin a bit easier than Caida and Afmob ... I never seemed to get stuck anywhere which I often do for 1.25 puzzles.

I must admit that I did get stuck. After making good progress but no placements on Thursday evening I took a break at about step 30 or 31 because I couldn't see what to do next. Step 32, which I did the next day, was initially just looking to see what I hadn't yet tried but turned out to be the key breakthrough for this puzzle. I hadn't looked at combinations for the 17(3) cage in N1 earlier because that cage with all candidates still present didn't look very promising. In fact step 32 could be done much earlier, possibly straight after step 14.

For the reason given above, I rate A84 at 1.25.

Here is my walkthrough

Prelims

a) R3C12 = {29/38/47/56}, no 1
b) R3C89 = {16/25/34}, no 7,8,9
c) R6C12 = {16/25/34}, no 7,8,9
d) R6C89 = {39/48/57}, no 1,2,6
e) R9C12 = {39/48/57}, no 1,2,6
f) R9C89 = {12}, locked for R9 and N9
g) 11(3) cage at R2C7 = {128/137/146/236/245}, no 9
h) 19(3) cage in N3 = {289/379/469/478/568}, no 1
i) 19(3) cage at R5C3 = {289/379/469/478/568}, no 1
j) 19(3) cage at R5C7 = {289/379/469/478/568}, no 1
k) 19(3) cage at R8C7 = {289/379/469/478/568}, no 1
l) 15(5) cage in N5 = {12345}, locked for N5

1. 45 rule on R7 3 innies R7C159 = 20 = {389/479/569/578}, no 1,2

2. 45 rule on N1 2 outies R13C4 = 12 = {39/48/57}, no 1,2,6

3. 45 rule on N3 2 outies R13C6 = 10 = {19/28/37/46}, no 5, no 1 in R1C6

4. 45 rule on N4 2 outies R46C4 = 17 = {89}, locked for C4 and N5, clean-up: no 3,4 in R13C4 (step 2)
4a. Naked pair {67} in R46C6, locked for C6, clean-up: no 3,4 in R13C6 (step 3)
4b. Naked pair {57} in R13C4, locked for C4 and N2

5. 45 rule on N9 2 outies R79C6 = 6 = [15/24]

6. Killer pair 1,2 in R13C6 and R7C6, locked for C6

7. 45 rule on N7 2 outies R79C4 = 9 = {36} (only remaining combination)
7a. Naked pair {36} in R79C4, locked for C4 and N8

8. 45 rule on C12 3 innies R147C2 = 9 = {126/135/234}, no 7,8,9

9. 19(3) cage at R8C7 = {469/478/568} (cannot be {379} because R9C6 only contains 4,5), no 3
9a. R9C6 = {45} -> no 4,5 in R89C7

10. 19(3) cage at R5C7 = {379/469/478/568} (cannot be {289} because R6C6 only contains 6,7), no 2
10a. R6C6 = {67} -> no 6,7 in R56C7

11. Killer pair 8,9 in R56C7 and R89C7, locked for C7

12. R1C234 = {147/156/237/345} (cannot be {129/138/246} because R1C4 only contains 5,7), no 8,9
12a. R1C4 = {57} -> no 5,7 in R1C23

13. 17(3) at R2C3 = {179/278/359/458/467} (cannot be {269/368} because R3C4 only contains 5,7)
13a. R3C4 = {57} -> no 5,7 in R23C3

14. 45 rule on N1 4 innies R1C23 + R23C3 = 17 = {1268/1349/2348}
14a. 6 of {1268} must be in R1C23 (only way to make 12 in R1C234) -> no 6 in R23C3

15. 13(3) cage at R8C3 = {139/238/256/346} (cannot be {148/157/247} because R9C4 only contains 3,6), no 7
15a. 5,8,9 of {139/238/256} must be in R9C3 -> no 5,8,9 in R8C3

16. 17(3) cage in N9 = {359/368/458} (cannot be {467} which clashes with R89C7), no 7
16a. Hidden killer pair 3,4 in R7C78 and 17(3) cage for N9 -> R7C78 must contain 3/4
16b. R7C678 = {138/147/237/246}, cannot be {129/156} because R7C78 must contain 3 or 4, cannot be {345} because R7C6 only contains 1,2), no 5,9
16c. 5 in N9 locked in 17(3) cage = {359/458}, no 6
16d. Killer pair 8,9 in 17(3) cage and R89C7, locked for N9

17. R4C678 = {147/156/237/246} (cannot be {129/138/345} because R4C6 only contains 6,7), no 8,9
17a. R4C6 = {67} -> no 6,7 in R4C78

18. Hidden killer pair 1,2 in R4C78 and 15(3) cage for N6 -> 15(3) cage must contain 1/2
18a. 6 in N6 locked in 15(3) cage = {168/267}, no 3,4,5,9

19. R1C678 = {189/279/369/378/459/468} (cannot be {567} because R1C6 only contains 2,8,9)
19a. 1 of {189} must be in R1C7 -> no 1 in R1C8

20. Hidden killer pair 1,2 in R7C23 and R7C6 for R7 -> R7C23 must contain 1/2
20a. R7C234 = {139/238/256} (cannot be {148/157/247} because R7C4 only contains 3,6), no 4,7
20b. R7C4 = {36} -> no 3,6 in R7C23
20c. 1 of {139} must be in R7C2 -> no 1 in R7C3

21. 45 rule on R3 2 outies R2C37 = 1 innie R3C5 + 1, min R2C37 = 3 -> min R3C5 = 2

22. 45 rule on R6 2 outies R5C37 = 1 innie R6C5 + 12, min R5C37 = 13, no 2,3

23. 45 rule on R9 2 outies R8C37 = 1 innie R9C5 + 2, min R8C37 = 7 -> min R9C5 = 5
23a. Max R8C37 = 11 -> no 6 in R8C3

24. 45 rule on N7 4 innies R7C23 + R89C3 = 17 = {1259/1349/2348/2456} (cannot be {1268/1358} because cannot make the two 13(3) cages)
24a. {2348} must have {28} in R7C23 because no 3,4 in R7C23 -> no 8 in R9C3

25. 13(3) cage at R8C3 (step 15) = {139/256/346} must contain at least one of 3,4,5 in R9C34
25a. Killer triple 3,4,5 in R9C12, R9C34 and R9C6, locked for R9

26. 45 rule on N6 4 innies R4C78 + R56C7 = 18 = {1359/1458/2349} (cannot be {2358} which clashes with 15(3) cage)

27. 45 rule on C89 3 innies R147C8 = 17 = {179/269/278/359/368/458/467}
27a. 2 of {269/278} must be in R4C8 -> no 2 in R1C8
27b. 8,9 of {359/368/458} must be in R1C8 -> no 3,5 in R1C8
27c. 8 of {458} must be in R1C8, 4 of {467} must be in R4C8 -> no 4 in R1C8

28. R1C678 (step 19) = {189/279/369/378/468} (cannot be {459} because 4,5 only in R1C7), no 5
28a. 3,4 of {369/468} must be in R1C7 -> no 6 in R1C7

29. 45 rule on R4 3 innies R4C159 = 15 = {159/168/249/258/267/348/357/456}
29a. 9 of {159} must be in R4C1, 1 of {168} must be in R4C5 -> no 1 in R4C1

30. R4C234 = {189/279/369/378/459/468} (cannot be {567} which clashes with R4C6)
30a. 1,2 of {189/279} must be in R4C2 -> no 1,2 in R4C3

31. 16(3) in N7 = {169/178/268/367} (cannot be {259/349/457} which clash with R7C23 + R89C3 (step 24), cannot be {358} which clashes with R9C12), no 4,5

32. 17(3) cage in N1 = {179/278/359} (cannot be {269/368/458/467} which clash with R1C23 + R23C3), no 4,6
32a. R1C23 + R23C3 (step 14) = {1268/1349/2348}
32b. Killer pair 8,9 in R1C23 + R23C3 and 17(3) cage, locked for N1, clean-up: no 2,3 in R3C12

33. Killer pair 5,7 in R3C12 and R3C4, locked for R3, clean-up: no 2 in R3C89
33a. Killer pair 4,6 in R3C12 and R3C89, locked for R3

34. 11(3) cage at R2C7 = {128} (only remaining combination, cannot be {137} because R3C67 = [13] clashes with R3C89, cannot be {146/245} because no 4,5,6 in R3C67, cannot be {236} because R23C7 = [63] clashes with R3C89) -> R3C6 = 8, R1C6 = 2 (step 3), R7C6 = 1, R9C6 = 5 (step 5), clean-up: no 7 in R9C12
34a. Naked pair {12} in R23C7, locked for C7 and N3, clean-up: no 6 in R3C89
34b. Naked pair {34} in R3C89, locked for R3 and N3 -> R1C7 = 7, R1C8 = 9, R1C4 = 5, R3C4 = 7, R3C5 = 9, clean-up: no 3 in R6C9
34c. Naked pair {56} in R3C12, locked for N1

35. R1C4 = 5 -> R1C23 = 7 = {34}, locked for R1 and N1
35a. R9C6 = 5 -> R89C7 = 14 = {68}, locked for C7 and N9
35b. R7C6 = 1 -> R7C78 = 11 = [47], R7C5 = 8, R9C5 = 7

36. R8C6 = 9 (hidden single in C6)
36a. Naked pair {24} in R8C45, locked for R8

37. Naked pair {35} in R8C89, locked for R8 and N9 -> R7C9 = 9, R8C3 = 1, R9C34 = [93] (step 25), R3C3 = 2, R23C7 = [21], R7C234 = [256], R7C1 = 3, R2C3 = 8, R1C1 = 1, R1C59 = [68], clean-up: no 5 in R6C1, no 4,6 in R6C2, no 3,4 in R6C8
37a. Naked pair {48} in R9C12, locked for R7 and N7 -> R89C7 = [86]

38. 19(3) cage at R5C7 (step 10) = {379} (only remaining combination) -> R5C7 = 9, R6C67 = [73], R4C678 = [651], R6C89 = [84], R3C89 = [43], R8C89 = [35], R9C89 = [21], R2C89 = [56], R5C8 = 6, R6C4 = 9, R4C4 = 8

39. R4C4 = 8 -> R4C23 = 10 = [37]

and the rest is naked singles and a cage sum

Last edited by Andrew on Fri Jan 11, 2008 5:39 am; edited 2 times in total
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Mon Jan 07, 2008 9:02 am    Post subject:

 Gary wrote: Found this to be one of the easier assassins..took about 35-40 mins only.

Gary is correct. If you find the trick, you can start placing digits for this Assassin very quickly. And the interesting thing is that it can all be done without any detailed combination work whatsoever. Here's how:

 Quote: After the trivial steps (R46C4 (outies N4) = {89} -> R13C4 (outies N1) = {57})... 1. None of the 4 cages in N1 can have both of {57} 1a. -> no 5,7 in R1C23+R23C3 1b. -> 17(3)r1c1 and 11(2) must have 1 each of {57} (hidden killer pair) 1c. -> 11(2) = {47/56} (no 2,3,8,9) 2. Only places for {89} in N1 now are the 2 17(3) cages 2a. Neither can have both of {89}, so each must have 1 of {89} (hidden killer pair again) 2b. So each of 17(3) must have 1 of {89} and 1 of {57} (step 1) 2c. -> neither of the two 17(3) cages can also contain a 6 3. Only places for 6 in N1 are thus the 12(3) and 11(2) cages 3a. -> Either 12(3) contains a 6 -> R1C234 = {16}[5] 3b. Or 11(2) contains a 6 -> R3C12 = {56} 3c. -> 5 locked in R1C4+R3C12 3d. -> no 5 in R3C4 (CPE) 3e. -> R13C4 = [57] 4. The rest is fairly trivial now. R3C12 (step 1c) must be {56}, R1C23 and R3C89 must both be {34}, 11(3)n23 must be {128} (all other combos blocked by 7(2)n3) -> 5 in N3 is locked in 19(3) = {568} -> R3C6 = 8, R23C7 = {12} -> R1C678 = [2]{79} -> R79C6 (outies N9) = [15], and so on... Furthermore, the whole puzzle can easily be done without using any 45 tests other than the obvious 6 outie cages for N1..6.

Nice to have a puzzle with some strong interactions waiting to be discovered. Thanks, Ruud.
_________________
Cheers,
Mike

Last edited by mhparker on Sun Jan 13, 2008 7:19 pm; edited 1 time in total
goooders
Regular

Joined: 18 Feb 2007
Posts: 16
Location: london

 Posted: Mon Jan 07, 2008 9:30 am    Post subject: ok so how do you rate this? unlike gary i stared at this puzzle for an hour then realised the "trick" and finished it very quickly so overall if it took an hour and a half i would say 1.25 on the "back of an envelope" however doesnt gary rate this as 0.75?
Afmob
Expert

Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

 Posted: Mon Jan 07, 2008 12:19 pm    Post subject: The rating is not necessarily determined by the amount of time it took you to solve a Killer but the most important factor is the difficulty of your moves though the amount of time might hint how difficult it is to spot those moves which might the change the rating by about 0.25. For example, you can solve an easy Killer which only needs Innies/Outies of small size, locked candidates, pairs and singles by using forcing or contradiction chains though they aren't necessary. So you can use technique you might need to solve Killers of rating 1.5 but if a Killer only requires techniques of rating 0.75/1.0 then the rating of the Killer should be 0.75/1.0. So the rating is mainly defined by the most difficult techniques you had to use to solve it and the official rating should be the walkthrough with moves of lowest difficulty. E.g. for A74 Brickwall I had to use a big contradiction chain to solve it so my rating would be 2.5 but Para managed to solve it without such moves so he rated it 2.0 and that's the official rating of this assassin which is fine with me. But like I said there are other factors which change the rating (difficulty of seeing a move, how narrow is the solving path, is it a "one-trick-pony" or does it need lots of difficult moves) but in my opinion they aren't as important as the difficulty of the moves. Gary used an impressive move which quite shortens the walkthrough but Killers of rating 0.75 don't need those moves. But that's just my opinion .
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Mon Jan 07, 2008 1:45 pm    Post subject:

 goooders wrote: ok so how do you rate this? unlike gary i stared at this puzzle for an hour then realised the "trick" and finished it very quickly so overall if it took an hour and a half i would say 1.25 on the "back of an envelope" however doesnt gary rate this as 0.75?

 Afmob wrote: The rating is not necessarily determined by the amount of time it took you to solve a Killer but the most important factor is the difficulty of your moves

Agreed. My personal gut feeling is that it's probably an easy 1.25, although I guess that Gary will rate it as 1.0 max. Maybe similar in technical difficulty to the A82?

IMO, the thing that makes it an "easy" 1.25, and which would also explain why Ed found it easier than most 1.25-rated Assassins, is that the solving path is quite wide.

P.S. Where's our resident Yokozuna? I'll soon be catching him up at this rate!
_________________
Cheers,
Mike
Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

 Posted: Mon Jan 07, 2008 9:00 pm    Post subject: I found it interesting that Gary's trick, which originally appeared to contain some T&E (maybe that's just because of Gary's concise style), could be presented logically. Thanks Mike! I must admit that I still feel it's a bit too early to have jumped in with such complex logic when there were plenty of easier normal killer steps that could be used first. Maybe that's just my more methodical approach to solving. I prefer to use easier steps first. As for rating A84 using Mike's explanation of Gary's trick, it's probably still an easy 1.25. Looking at Mike's 3 steps I don't see how it can be rated any lower. I think in all the solutions this was the key area so there was probably one fairly narrow part of the solution path even though it was broken open in different ways.
gary w

Joined: 07 Sep 2007
Posts: 84
Location: south wales

 Posted: Thu Jan 10, 2008 1:09 am    Post subject: As so often I have to thank Mike for putting my very brief observations into a formal solution.And although it didn't take me long to do Afmob is also quite right in saying that the difficulty doesn't necessarily correlate with the time taken.So I think a 1.25 rating is reasonable..I think I was just lucky to spot the short cut. Regards Gary
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