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sudokuEd
Grandmaster

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

Posted: Tue Nov 28, 2006 12:20 pm    Post subject: Assassin 26

 Ruud wrote: it is one long stream of innie-outie differences.
More like an annoying drip - before the dam finally broke.

This solution is a joint effort between me and Peter. We had a terrible time with this puzzle - got tricked and totally stumped so many times. In the end, found a way through that avoided most of the innie-outie nightmares (step 20).

I'll be taking a break for a while. Going to NZ next Monday for a couple of weeks doing adventure stuff. Need a total break from anything to do with Ruud for a long while after the last month .

Please let me know if there is anything not valid, accurate or clear in this walk-through.

See you all anon. Ed

1. "45" n78 -> r6c1 - 8 = r9c6 -> r6c1 = 9, r9c6 = 1

2. n8: 17(2) = {89}, locked in N8 and r7

3. n8: 16(3) = {367/457} = 7{36/45} (no 2): 7 locked for n8,r8

4. r78c1 = 10 = [28/73/{46}] = [3/6/8] & (no 1,5)

5. c1:17(3) = {278/458/467} ({368} blocked by r78c1-step 4) (no 1,3)

6. 1 in c1 locked in 3 innies r129c1 = 9 = {126|135} (no 4,7,8)

7. 1 in c1 also locked in 20(4) n1 = {1289/1379/1469/1568} ({1478} blocked because no {478} in r12c1)
7a. ->1 locked for n1 in r12c1

8. c6:14(2) = [95/{68}] = [5/6,8/9..] -> 8 & 9 locked for c6 in r567c6

9. 45 on n5: r6c4 + r4c6 = 5 = [14/{23}]

10. n56: 18(3) must have {234}(r4c6) = {279|369|378|459|468} =[2/3/4]not two of -> r45c7 = {56789}

11.a. 8 in r7c5 -> no 8 in 14(3) in N5
b. 8 in r7c6 -> 14(2) in N5 = {59} -> 12(2) in N5 = {48}
-> no 8 in 14(3) in N5

12. n5:14(3) = {149/167/239/257} ({347} blocked by 12(2), {356} blocked by 14(2))

13. c7: 8(3) = {125|134} -> 1 locked in c7

14. r9c78 = 12 = {39/48/57} (no 2,6)

15. "45" n9 -> 3 innies: r78c7 + r7c9 = 8 = {125/134} = 1{25/34} no(67):1 locked for n9

16. n9:6 locked in 25(4) = 6{289/379/478} (no 5)

17. "45" n8 -> r9c3 - 3 = r7c4 -> r9c3 = {56789}
17a. when 6 in r9c3, 3 must be in r7c4 and rest of 14(3) in n8 must be {35}. Two 3's in n8 -> no 6 in r9c3, no 3 in r7c4

17b. In summary
[5] in r9c3 -> 2 in r7c4
-> rest of 14(3) = {36} ({27} blocked by 2 in r7c4)
[7] in r9c3 -> 4 in r7c4
-> rest of 14(3) = {25} ({34} blocked by 4 in r7c4)
[8] in r9c3 -> 5 in r7c4
-> rest of 14(3) = {24}
[9] in r9c3 -> 6 in r7c4
-> rest of 14(3) = {23}

18. "45" n14 -> r16c4 = 5 = [41/23/32]

19. "45" n147 -> 5 outies = 16

20. Combining steps 17b, 18 and 19 -> r1678c4 + r9c5 = 16 = [412{36}/41524/416{23}], ([23452/23542/32452/32542] all blocked by 12(2) c4)
-> r16c4 = [41], r7c4 = {256}, r9c3 = {589}, r9c4 = {236}, r9c5 = {2346}, 14(3) = {239/248/356} = [5{36}/824/9{23}]

21. n5: r4c6 = 4 (step 9), 12(2) = {39/57}, 14(3) must have 2 for n5 = 2{39/57}(no 6): 2 locked for c5 and {39/57} are locked in these 2 cages -> 14(2) = {68}: locked for c6 -> r7c56 = [89]

22. "45" n2 -> r1c7 - 2 = r3c6: r1c7 = {579}, r3c6 = {257}

23. r1:19(3). the only valid combination left is {379}: locked for r 1, and 3 also locked for n2. r3c6 = {57}(step 22)

24. "45" n2: 3 innies r1c56 + r3c6 = 17 and must have 3 and made up of candidates {3579} which sums to 24 -> 24 - 17 = 7.-> no 7 -> r3c6 = 5, r1c567 = [937], r8c6 = 7, r2c6 = 2, r2c45 = {67}:locked for n2, r2, r3c45 = {81}

25. n5:14(3) = {257}:locked for n5, c5 -> r2c45 = [76], r45c4 = {39}:locked for c4

25. n2:15(3) must have 2 or 7 (r2c6) = {267}

24. r1c23 = 11 = {56}:locked for r1,n1. 20(4) = 1{289/379} = 19{28/37}(no 4): 9 locked for n1, c2. r23c2 = {789},

25. "45" n1-> r4c3 + 2 = r3c1: r3c1 = {47}, r4c3 = {25}

26. c3:12(3) = {237/345} = 3{27/45} (no 8) with 3 locked in n1 -> r12c1 = [21], r23c2 = [89], r3c3 = 7 (single n1), r24c2 = [32], r3c1 = 4, r78c1 = [73], r1c89 = {18} -> r2c78 = {49}:locked for n3, r3c78 = {26}:locked for n3:....the rest is straight-forward
Andrew
Grandmaster

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

 Posted: Fri Jan 30, 2009 7:32 am    Post subject: As I commented in the A34 and A71 threads, I didn't manage to solve three of Ruud's Assassins when they first appeared. Now having caught up with my backlog of other walkthroughs I'm having another go at them. Having posted my walkthroughs for A31 and A34, here is the the last one. I thought the breakthrough in step 18 was difficult to spot but apart from that A26 wasn't particularly difficult although it took me quite a lot of moves after that to finish it. Here is my walkthrough. With hindsight the breakthrough step 18 could have been done after step 9. Prelims a) R3C45 = {18/27/36/45}, no 1 b) R45C4 = {39/48/57}, no 1,2,6 c) R56C6 = {59/68} d) R7C56 = {89}, locked for R7 and N8 e) R1C567 = {289/379/469/478/568}, no 1 f) R456C2 = {128/137/146/236/245}, no 9 g) R456C8 = {126/135/234}, no 7,8,9 h) 11(3) cage at R5C3 = {128/137/146/236/245}, no 9 i) R678C1 = {289/379/469/478/568}, no 1 j) R678C7 = {125/134}, 1 locked for C7 1. Killer pair 8,9 in R56C6 and R7C6, locked for C6 2. 45 rule on N78 1 outie R6C1 = 1 innie R9C6 + 8 -> R6C1 = 9, R9C6 = 1, clean-up: no 5 in R5C6 2a. R6C1 = 9 -> R78C1 = 10= [28]/{37/46}, no 5, no 2 in R8C1 2b. R9C6 = 1 -> R9C78 = 12 = {39/48/57}, no 2,6 3. 45 rule on N9 3 remaining innies R7C79 + R8C7 = 8 = {125/134}, 1 locked for N9 3a. 45 rule on N9 1 remaining outie R6C7 = 1 innie R7C9 3b. Max R67C9 = 13 -> min R5C9 = 2 4. R8C456 = {367/457}, no 2, 7 locked for R8 and N8, clean-up: no 3 in R7C1 (step 2a) 5. 45 rule on N23 1 outie R4C9 = 1 innie R1C4 + 3, no 7,8,9 in R1C4, no 1,2,3 in R4C9 6. 45 rule on N5 2 innies R4C6 + R6C4 = 5 = [23/32/41], R4C6 = {234}, R6C4 = {123} 6a. Max R4C6 = 4 -> min R45C7 = 14, no 2,3,4 in R45C7 7. 45 rule on N8 1 outie R9C3 = 1 remaining innie R7C4 + 3, no 2,3,4 in R9C3 8. 45 rule on C1 3 innies R129C1 = 9 = {126/135} (cannot be {234} which clashes with R78C1), no 4,7,8, 1 locked in R12C1 for C1 and N1 8a. Max R12C1 = 7 -> min R23C2 = 13, no 2,3 in R23C1 8b. Min R23C3 = 5 -> max R4C3 = 7 9. 45 rule on N14 2 remaining outies R16C4 = 5 = [23/32/41], R1C4 = {234}, clean-up: no 4,8,9 in R4C9 (step 5) 10. 1 in R1 locked in R1C189 10a. 45 rule on R1 3 innies R1C189 = 11 = {128/137/146}, no 5,9 11. R1C234 = {249/258/357/456} (cannot be {267/348} which clash with R1C189) 11a. 3 of {357} must be in R1C4 -> no 3 in R1C23 12. R1C567 = {289/379/469/478/568} 12a. 2 of {289} must be in R1C6 -> no 2 in R1C57 13. 45 rule on R9 3 innies R9C129 = 18 = {279/369/468/567} (cannot be {378/459} which clash with R9C78) 13a. 2 of {279} must be in R9C1 -> no 2 in R9C29 14. R345C1 = {278/458/467} (cannot be {368} which clash with R129C1), no 3 15. 45 rule on C9 2 outies R78C8 = 1 innie R1C9 + 10 15a. Max R78C8 = 16 -> max R1C9 = 6 16. 45 rule on N69 3 remaining innies R4C79 + R5C7 = 21 = {579/678}, 7 locked for N6 17. 45 rule on C9 3 innies R189C9 = 15 -> C9 has three 15(3) cages which must be {159/267/348} or {168/249/357} (cannot be {258/456} which clash with all other 15(3) combinations) 17a. R234C9 = {159/168/267/357} (cannot be {249/348} because R4C9 only contains 5,6,7), no 4 17b. R567C9 = {159/168/249/348} 17c. 9 of {159/249} must be in R5C9 -> no 2,5 in R5C9 18. 45 rule on C123 5(4+1) outies R1679C4 + R9C5 = 16 First I did 4+1 combination analysis, then I found 18a. Min R9C5 = 2 -> max R1679C4 = 14 which must contain 1 (cannot be {2345} which clashes with R45C4) -> R6C4 = 1, R1C4 = 4 (step 9), R4C6 = 4 (step 6), R4C9 = 7 (step 5), clean-up: no 6 in R1C189 (step 10a), no 5 in R3C4, no 5,8 in R3C5, no 8 in R4C4, no 5,8 in R5C4, no 7 in R9C3 (step 7) 18b. 8 in C4 locked in R23C4, locked for N2 18c. 2 in N5 locked in R456C5, locked for C5, clean-up: no 7 in R3C4 18d. 2 in N8 locked in R79C4, locked for C4, clean-up: no 7 in R3C5 18e. 1 in C7 locked in R78C7, locked for N9 19. 2 locked in R456C5 = {239/257}, no 6,8 19a. R56C6 = {68} (hidden pair for N5), locked for C6 -> R7C56 = [89] 20. R1C4 = 4 -> R1C23 (step 11) = {29/56}, no 7,8 21. R4C9 = 7 -> R23C9 (step 17a) = {26/35}, no 1,8,9 22. R1C9 = 1 (hidden single in C9) 22a. R1C18 (step 10a) = [28/37] 22b. R2C1 = 1 (hidden single in C1) 22c. R3C5 = 1 (hidden single in R3), R3C4 = 8 23. R12C1 = [21/31] = 3,4 -> R23C2 = 16,17 = {79/89}, 9 locked for C2 and N1, clean-up: no 2 in R1C23 (step 20) 23a. Naked pair {56} in R1C23, locked for R1 and N1 24. R6C4 = 1 -> R56C3 = 10 = {28/37/46}, no 5 25. R1C9 = 1 -> R89C9 (step 17) = {59/68}, no 2,3,4 25a. Killer pair 5,6 in R23C9 and R89C9, locked for C9, clean-up: no 5 in R6C7 (step 3a) 25b. R89C9 = 14 -> R78C8 = 11 = [29/38/74] (cannot be {56} which clashes with R89C9), R7C8 = {237}, R8C8 = {489} 26. 1 in N6 locked in R456C8 = {126/135}, no 4 27. R678C7 = {125/134} 27a. 2 of {125} must be in R6C7 -> no 2 in R78C7 27b. 2 in N9 locked in R7C89, locked for R7, clean-up: no 8 in R8C1 (step 2a), no 5 in R9C3 (step 7) 28. R9C4 = 2 (hidden single in C4) , R9C35 = [84/93], clean-up: no 3 in R7C4 (step 7) 29. 45 rule on N1 1 innie R3C1 = 1 remaining outie R4C3 + 2, no 2 in R3C1, no 1,3,6 in R4C3 29a. 1 in N4 locked in R45C2, locked for C2 29b. R456C2 = {128/137/146}, no 5 30. 8 in C1 locked in R45C1, locked for N4, clean-up: no 2 in R456C2 (step 29b), no 2 in R56C3 (step 24) 30a. Naked quint {13467} in R456C3 + R56C3, locked for N4 31. R7C234 = {156/345}, cannot be {147} because R7C4 only contains 5,6), no 7, 5 locked for R7 31a. 1 of {156} must be in R7C3, 5 of {345} must be in R7C4 -> no 5 in R7C3 32. 45 rule on N2 1 outie R1C7 = 1 remaining innie R3C6 + 2, no 3,8 in R1C7, no 2,3 in R3C6 33. R1C567 = {379} (only remaining combination), locked for R1, 3 locked in R1C56, locked for N2 -> R1C1 = 2, R1C8 = 8, R9C1 = 6 (step 8), clean-up: no 4 in R78C1 (step 2a), no 3 in R7C8 (step 25b), no 8 in R8C9 (step 25), no 4 in R9C7 (step 2b) 33a. R78C1 = [73], R3C1 = 4, R4C3 = 2 (step 29), R7C8 = 2, R8C8 = 9 (step 25b), clean-up: no 6 in R456C8 (step 26), no 2 in R6C7 (step 3a), no 5 in R89C9 (step 25), no 3 in R9C78 (step 2b) 35c. R89C9 = [68], R9C3 = 9, R9C5 = 3 (step 28), R7C4 = 6 (step 7), R7C23 = [51] (step 31), R1C23 = [65], R9C2 = 4, R8C23 = [28], clean-up: no 2 in R23C9 (step 21), no 9 in R45C5 (step 19) 36. R4C5 = 5, R45C1 = [85], R56C5 = {27}, locked for C5 and N5, R1C567 = [937], R2C5 = 6, R8C5 = 4, R8C7 = 1, R9C78 = [57] 37. R2C5 = 6 -> R2C46 = 9 = [72] and the rest is naked singles
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