Assassin 51

Jean-Christophe · Post by **Jean-Christophe** » Fri May 18, 2007 9:26 am

It was enjoyable to solve. Just the right difficulty level for me. Definitly does not need T&E or big combination crushing like some others.

Here is my walkthrough & the solution:
1. Innies of R5 -> R5C5 = 6
2. Innies of R234 -> R4C5 = 9
3. Cage 20/3 in C5 -> R6C5 = 5

4. Innies of C5 -> R19C5 = 15 = {78}
5. Innies of C1234 -> R19C4 = 12
6. Innies of C6789 -> R19C6 = 6
7. Cage 18/3 in R9C456 = {9(18|27)} -> R9C4 = 9, R9C6 = {12}, R1C4 = 3
8. Cage 5/2 in R23C5 = {14} (NP @ N2, C5)
9. Cage 5/2 in R78C5 = {23} (NP @ N8)
10. R9C56 = [81], R1C56 = [75]

11. Cage 12/3 in R1C123 = {129|246} = {2..} -> no 2 elsewhere in N1, R1
12. 8 of R1 locked in Cage 18/3 = {8(19|46)} -> no 8 elsewhere in N3
13. Since R2C5 = {14} -> Cage 11/3 in R2C234 <> {146} = {128|236|245} = {2..} -> R2C4 = 2

14. Innies of C1 -> R159C1 = 6 = {123} (NT @ C1)
15. Cage 8/3 in R8C234 = {1(25|34)} -> no 1 elsewhere in N7, R8 -> R8C23 = {1(2|3)}, R8C4 = {45}
16. R8C23&R9C1 = NT {123} @ N7
17. R8C235 = NT {123} @ R8

18a. Cage 13/2 in R5C89 = {49|58}
18b. -> Cage 9/2 in R5C34 <> {45} = {18}|[27], Cage 12/2 in R5C67 <> {48} = [39|75]
19. 7 of R5 locked in R5C46 -> not elsewhere in N5

20a. Cage 16/3 in N125 = {169|178} = {1..}
20b. Since R3C4 <> {17} -> R3C3&R4C4 <> 8
20c. Since R4C4 <> {6789} -> R4C4 = 1, R3C34 = [78|96]

21. Cage 9/2 in R5C34 -> R5C3 = {12}
22. R5C3 & Cage 5/2 in R5C12 = complex naked pair on {12} -> not elsewhere in N4
23. R5C7 & Cage 13/2 in R5C89 = complex naked pair on {59} -> not elsewhere in N6

24a. Innies of R2 -> R2C159 = 14 = {149|158|167|347}
24b. Since each in R2C59 <= 6 -> R2C1 = {789}
25. Cage 18/3 in R234C1 -> R34C1 = {456}

26a. Outies of N1 -> R3C4+R4C123 = 25
26b. R3C4 = {68} -> R4C123 = 17|19 with 5 locked in R4C123 = {458|568} = {58(4|6)}
26c. Since R4C1 = {456} (step 25) -> 8 locked in R4C23 -> Cage 16/3 in N14 = {358}
26d. -> R4C23 = {58} (NP @ R4, N4), R3C2 = 3, R4C1 = {46}

27a. 7 of R4 locked in R4C78 -> Cage 14/3 in N36 = {167|257|347}
27b. Since R4C78 <> {57} -> R3C8 <> 2
28. Max R34C6=[94]=13 -> Min R3C7=17-13=4. Cannot repeat 4 in cage 17/3 -> Min R3C7=5
29. R3C9 = 2 (HS @ R3), R24C9 = [16]|{34}

30. 14/3 in R2C159 = {149|347} = {4(7|9)..} -> no 4 elsewhere in R2, R2C1 = {79}
31. R2C1&R3C4 = NP {79} @ N1
32. Cage 12/3 in R1 = {246} -> R159C1 = [213]
33. R5C234 = [427], R8C234 = [215], R78C5 = [23], R1C23 = [64]
...

Solution

Code: Select all

+-------+-------+-------+
| 2 6 4 | 3 7 5 | 1 9 8 |
| 7 1 8 | 2 4 9 | 6 5 3 |
| 5 3 9 | 6 1 8 | 7 4 2 |
+-------+-------+-------+
| 6 8 5 | 1 9 2 | 3 7 4 |
| 1 4 2 | 7 6 3 | 9 8 5 |
| 9 7 3 | 8 5 4 | 2 1 6 |
+-------+-------+-------+
| 8 9 7 | 4 2 6 | 5 3 1 |
| 4 2 1 | 5 3 7 | 8 6 9 |
| 3 5 6 | 9 8 1 | 4 2 7 |
+-------+-------+-------+

Edit revised from step 21 I don't remember how I eliminated the combination {167} for R2C159 = 14 which Para noticed.
However there was a nice X Wing on 5 in there

I'll list that alternating path with the X-Wing because I think it's interesting:
24a. Innies of R2 -> R2C159 = 14 = {149|158|167|347}
24b. Did not remember how I eliminated {167} here. Assume this is the case
24c. R2C159 = 14 = {149|158|347} = {(4|5)..}
24d. Since each in R2C59 <= 6 -> R2C1 = {789}
24e. Since R2C159 = {(4|5)..} -> Cage 11/3 in R2 <> {245} = {128|236}
25. Cage 18/3 in R234C1 -> R34C1 = {456}
26. 5 of R2 locked in R2C789 -> not elsewhere in N3

27a. 5 of C2 locked in R3479C2, 5 of C3 locked in R479C3
27b. This forms a grouped X Wing on 5 with Cage 16/3 & N7
27c. -> no 5 elsewhere in N7, Cage 16/3 must include a 5
27d. -> Cage 16/3 in N14 = {5(38|47)} (no {169})

27e. Alternatively, there is also a grouped X Wing on 5 with R34 * Cage 18/3 in C1 & Cage 16/3 in N14
27f. -> both cages must include a 5

28. 5 of C1 locked in R34C1 -> Cage 18/3 in C1 = {5(49|67)} (no 8)
29. Cage 21/3 in C1 = {8(49|67)}
30. R2C1&R3C4 = NP {79} @ N1
...

CathyW · Post by **CathyW** » Sat May 19, 2007 2:56 pm

Nothing elegant I'm afraid, using quite a lot of cage combination analysis and a couple of conflicting combinations, but this is how I managed A51:

1. Innies r5 -> r5c5 = 6 -> r46c5 = {59}, 59 not elsewhere in N5/c5
-> 9(2) r5c23 <>3, 13(2) r5c89 = {49/58}-> 12(2) r5c67 can’t be {48}, 9(2) can’t be {45}

2. r23c5, r78c5 = 5 = {14/23}, not elsewhere in c5 -> r19c5 = {78}

3. Innies c1: r159c1 = 6 = {123}, not elsewhere in c1 -> r5c2 <> 1.

4. Innies c9: r159c9 = 20 -> r159c9 <> 1,2

5. Innies c1234: r19c4 = 12 = {39/48/57}

6. Innies c6789: r19c6 = 6 = {15/24}

7. Innies r234: r4c5 = 9, -> r6c5 = 5 -> r5c6 = 3/7 -> r5c7 = 9/5 -> Killer pair {59} in r789c5, not elsewhere in N6.

8. 6(3) r6c78+r7c8 = {123}, r4c8 sees all cells <> 1,2,3

9. Innies r2: r2c159 = 14, r2c1 <> 4

10. Innies r8: r8c159 = 16 -> since r8c5 is max 4, r8c9 <> 1,2

11. Combinations for 18(3) r9c456 = 189/279 -> r9c4 = 789, r9c6 = 12, r1c4 = 345, r1c6 = 45
-> combinations for 15(3) in r1c456 = {357/348} -> r1c4 = 3, r9c4 = 9 -> 5(2) r23c5 = {14} not elsewhere in N2/c5, 5(2) r78c5 = {23} not elsewhere in N8 -> r1c6 = 5, r9c6 = 1, r9c5 = 8, r1c5 = 7
-> r8c4 = 4/5 -> r8c23 = {123} – forms naked triple within N7 and r8 -> r8c9 <> 3
-> 14(3) r9c123 = 257/347/356; 13(3) r9c789 = 247/256/346.

12. 12(3) r1c123 = 129/246, 2 not elsewhere in N1, r1
-> 18(3) r1c789 must have 8: {189/468} not elsewhere in N3
-> r2c4 <> 8

13. 11(3) r2c234 = {128/236/245}, not 146 since can’t have both 14 -> r2c23 <> 7 -> r2c4 = 2 -> r5c3 <> 7.

14. 16(3) r3c34+r4c4 = {169/178/367} -> r4c4 <> 4,8; r3c3 <> 4,5,6,8

15. 2 locked to r5c123, not elsewhere in N4.

16. 1 locked to r6c789, not elsewhere in N6.

17. 7 locked to r5c46, not elsewhere in N5 -> r4c4 = 1 -> r5c3 <> 8, r3c34 = 78/96.

18. 17(3) r3c67+r4c6: max from r34c6 = 14 -> r3c7 <> 1,2

19. Split 20(3) r159c9 = {389/479/569/578}. No 7 in r15c9 -> r9c9 <> 4

20. 21(3) r8c678 = {489/579/678} -> if {489}, r8c78 <> 4.

21. Split 16(3) r8c159 = {259/268/349/367}. {358} not possible since would leave no options for 8(3) r8c234.

22. 13(3) r9c789 = {247/256/346}. If {247}, r9c9 = 7 -> r9c78 <> 7.

23. Split 14(3) r2c159 = {149/158/167/347}. Combination analysis -> r2c1 <> 5,6
-> 18(3) r234c1 = {459/468/567} -> r34c1 = {456}

24. 14(3) r34c8+r4c7: since r4c78 = min 6 -> r3c8 <> 9.

25. 15(3) r6c6+r7c67 = {168/249/267/348/357/456}. Combination analysis -> r7c7 <> 2,4.

26. 14(3) r3c8+r4c78 = {167/248/257/347/356}. Combination analysis -> r3c8 <> 6.

27. 17(3) r3c67+r4c6 = {269/278/359/368/458/467}. Combination analysis -> r3c7 <> 4.

28. 19(3) r6c4+r7c34 = {469/478/568}
a) If {469}, r7c3 = 9, r6c4 = 4, r7c4 = 6
b) If {478}, r7c3 = 478, r6c4 = 48, r7c4 = 47
c) If {568}, r7c3 = 6, r6c4 = 8, r7c4 = 5
Conclusion: r7c3 <> 5, else would conflict with r3c4.

29. r3c34 = 78/96 -> r3c67 can’t be 67 -> 17(3) r3c67+r4c6 can’t be {467}
-> r3c67 can’t be 68 -> if 17(3) r3c67+r4c6 = {368}, r4c6 <> 3.

30. a) If r7c8 = 1: r6c67 = {23} -> r4c79, r6c9 <> 2
b) If r7c8 = 2: r6c67 = {13}, r3c8 <> 2, r3c9 = 2 -> r46c9 <> 2
c) If r7c8 = 3: r6c67 = {12} -> r4c79, r6c9 <> 2.
In each case r46c9 <> 2
-> combinations for 9(3) r234c9: r23c9 <> 6, r3c9 <> 3,4

31. Split 14(3) r2c159 = 743/815/914
11(3) r2c234 cannot be {45}2 else removes all options for split 14(3) -> r2c23 <> 4,5
-> 5 locked to r3c12 -> r3c789 <> 5
-> 17(3) r3c67+r4c6 now = {269/278/368} -> r4c6 <> 4
-> 4 locked to r6c46 -> r6c1239 <> 4
-> Killer pair (68) in r3c467: r3c4 = 6/8, r3c67 must have 6 or 8 -> r3c12 <> 6,8
-> 8 locked to r3c46 -> r2c6 <> 8
-> 20(3) r2c678 = 659/956/947 -> r2c78 <> 3.

32. 19(3) r6c23+r7c2 = {379/469/478/568}. Combination analysis -> r7c2 <> 6,8

33. In N1:
a) If r2c23 = {18} -> r1c2, r3c3 = {79} -> 12(3) = {246}, r3c1 = 5, r3c2 = 3 …
b) If r2c23 = {36} -> 12(3) = {129} -> r3c3 = 7, r2c1 = 8, r3c12 = {45} -> r3c5 = 1, r2c5 = 4 -> 20(3) r2c678 = {569} Conflict with r2c23
Conclusion: r2c23 = {18}

Straightforward from here on.

CathyW · Post by **CathyW** » Wed May 23, 2007 9:08 am

No V2 or 1.5 this week? Has anyone else solved the V1 and found an 'elegant' move? JC's walkthrough makes it seem easier!

Andrew · Post by **Andrew** » Thu May 24, 2007 5:21 am

Cathy

I've just managed to finish Assassin 51 but haven't yet had a chance to check my steps or to look at the posted walkthroughs.

A tough challenge. I eventually broke through by using multiple outies from the corner nonets.

Do you really want a V1.5 or V2 for this one? If J-C says that it's the right level for him, it's enough of a challenge for most of us.

Andrew

CathyW · Post by **CathyW** » Thu May 24, 2007 9:04 am

Not really - the V1 was definitely hard enough for me! It's just that Ruud normally posts a variation. No matter, looking forward to Assassin 52 tomorrow.

mhparker · Post by **mhparker** » Thu May 24, 2007 8:48 pm

Andrew wrote:Do you really want a V1.5 or V2 for this one? If J-C says that it's the right level for him, it's enough of a challenge for most of us.

Maybe JC just doesn't like torturing himself (yet!) as much as some of us on this forum do, although even we masochistic types drew the line when it came to the A50V2...

Just a reminder: if we get bored, we can always have fun doing the Texas Jig $\:D/$ (BTW, the latest one - number 30 - is more like a V2).

Andrew · Post by **Andrew** » Fri May 25, 2007 5:27 am

I've now looked at the walkthroughs posted by J-C and Cathy. We all used fairly similar steps, although not in the same order, but each had slightly different breakthrough moves.

I looked at more combinations and outies than the others. Some of them didn't contribute directly to the final solution but I've left them in because they were how I worked through the puzzle.

Here is my walkthrough

First the obvious steps

1. 45 rule on R5 1 innie R5C5 = 6
1a. 45 rule on R1234 1 innie R4C5 = 9 -> R6C5 = 5 (cage sum)

Now for the usual preliminary steps

2. R23C5 = {14/23}

3. R5C12 = {14/23}

4. R5C34 = {18/27}/[54], no 3,9, no 4 in R5C3

5. R5C67 = [39]/{48}/[75], no 1,2, no 3,7 in R5C7

6. R5C89 = {49/58}
6a. R5C67 (step 5) = [39/75] (cannot be {48} which clashes with R5C89), no 4,8

7. Killer pair 5/9 in R5C7 and R5C89 for R5 and N6, clean-up: no 4 in R5C4

8. R78C5 = {14/23}

9. Naked quad {1234} in R23C5 and R78C5, locked for C5

10. R2C234 = {128/137/146/236/245}, no 9

11. R2C678 = {389/479/569/578}, no 1,2

12. R234C9 = {126/135/234}, no 7,8,9

13. R678C1 = {489/579/678}, no 1,2,3

14. 19(3) cage in N47 = {289/379/469/478/568}, no 1

15. 19(3) cage in N587 = {289/379/469/478/568}, no 1

16. 6(3) cage in N69 = {123} -> no 1,2,3 in R4C8

17. R8C234 = 1{25/34}, 1 locked for R8, clean-up: no 4 in R7C5

18. R8C678 = {489/579/678}, no 2,3

19. R1C456 with R1C5 = {78} = {168/258/267/348/357}, no 7,8,9 in R1C46

20. 45 rule on C1 3 innies R159C1 = 6 = {123}, locked for C1, clean-up: no 1 in R5C2

21. 45 rule on C9 3 innies R159C9 = 20 = {389/479/569/578}, no 1,2

22. 45 rule on R2 3 innies R2C159 = 14 = {149/158/167/239/248/257/347/356}
22a. 7,8,9 only in R2C1 -> no 4 in R2C1

23. 45 rule on R8 3 innies R8C159 = 16 = {259/268/349/367} (cannot be {358/457} which clash with R8C234)
23a. No 5,6,8,9 in R8C5 -> no 2 in R8C9

24. 45 rule on C1234 2 innies R19C4 = 12 = [39/48/57]

25. 45 rule on C6789 2 innies R19C6 = 6 = {15/24}

26. R9C456 = [972/981] -> R9C4 = 9, R9C6 = {12}, R1C4 = 3 (step 24), clean-up: no 2 in R23C5, R1C6 = {45} (step 25)

27. Naked pair {14} in R23C5, locked for C5 and N2 -> R1C6 = 5, R1C5 = 7 (cage sum), R9C5 = 8, R9C6 = 1 (step 25 or cage sum)

28. Naked pair {23} in R78C5, locked for N8

29. 1 in R6 locked in R6C789, locked for N6

30. R8C234 = 1{25/34} (step 17), R8C4 = {45} -> no 4,5 in R8C23

31. Naked triple {123} in R8C23 and R8C5, locked for R8

32. Naked triple {123} in R8C23 and R9C1, locked for N7

33. R159C9 (step 21) = {389/479/569/578}
33a. 7 only in R9C9 -> no 4 in R9C9

34. R2C234 (step 10) = {128/146/236/245} (cannot be {137} because no 1,3,7 in R2C4), no 7

35. 16(3) cage in N125 = {169/178/268/367} (cannot be {259} because 5,9 only in R3C3, cannot be {349/457} because no 3,4,5,7,9 in R3C4, cannot be {358} because 3,5 only in R3C3), no 4,5

36. 15(3) cage in N689 = {168/249/267/348/357/456} (cannot be {159} because 1,5,9 only in R7C7, cannot be {258} because no 2,5,8 in R7C6)
36a. 5 only in R7C7 and 4 required in R7C6 for {249/348} -> no 4 in R7C7

37. 19(3) cage in N587 = {469/478/568} (cannot be {289} because no 2,8,9 in R7C4), no 2

38. R1C789 = {189/468} = 8{19/46}, no 2, 8 locked for R1 and N3
38a. R1C123 = 2{19/46}, 2 locked for N1

39. R2C234 (step 10) = {128/236/245} (cannot be {146} which clashes with R2C5) = 2{18/36/45}
39a. No 2 in R2C23 -> R2C4 = 2; clean-up: no 7 in R5C3
[At this stage I missed 7 locked in R5C46 for N5. This would have allowed R4C4 to be fixed in step 45a instead of step 51. A strange oversight considering my next step!]

40. 2 in R5 locked in R5C123, locked for N4

41. R2C159 (step 22) = {149/158/167/347}
41a. 7,8 only in R2C1 -> no 5,6 in R2C1

42. R234C1 = {459/468/567}
42a. R2C1 = {789} -> no 7,8,9 in R34C1

43. R8C678 (step 18) = {489/579/678}
43a. 8,9 only in R8C78 -> no 4 in R8C78

44. 14(3) cage in N36 = {158/167/248/257/347/356} (cannot be {149} because 1,9 only in R3C8, cannot be {239} because no 2,3,9 in R4C8), no 9

45. 16(3) cage in N125 (step 35) = {169/178/367}
45a. All combinations require {68} in R3C4 -> no 6,8 in R3C3 + R4C4

46. 45 rule on N7 2 innies R7C23 – 2 = 2 outies R6C1 + R8C4
46a. Max R6C1 + R8C4 = 14 -> max R7C23 = 16 -> R78C1 must contain 8 and/or 9

47. 45 rule on N1 2 innies R3C23 - 6 = 1 outie R4C1
47a. R3C3 cannot be the same as R4C1 -> no 6 in R3C2

48. 45 rule on N9 6 outies R6C6789 + R78C6 = 26, R78C6 = 10, 11 or 13 -> R6C6789 = 13, 15 or 16 and must contain 1,2 = 12{37/46/48/67}
48a. If R78C6 = {47}, R6C6789 = 8{12}4 -> no 8 in R6C9

49. 45 rule on N8 4 innies R78C4 + R78C6 = 22, R78C6 = 10, 11 or 13 -> R78C4 = 9, 11 or 12
49a. 45 rule on N7 6 outies R6C1234 + R78C4 = 36, R78C4 = 9, 11 or 12 -> R6C1234 = 24, 25, or 27 and must contain 9 in R6C123 = {3489/3679/3789/4689}

50. 14(3) cage in N36 = {167/248/257/347/356} (cannot be {158} because 1,5 only in R3C8)
50a. R4C78 cannot be {48} (clashes with R5C89) -> no 2 in R3C8
50b. 2 only in R4C7 -> no 8 in R4C7
50c. 1,5 only in R3C8 -> no 6 in R3C8

51. 45 rule on N3 6 outies R23C6 + R4C6789 = 33, R23C6 = 15 or 17 -> R4C6789 = 16 or 18 = {2347/2367}, no 8, 2,3,7 locked for R4 -> R4C4 = 1, clean-up: no 8 in R5C3
51a. R4C4 = 1 -> R3C34 = 15 = [78/96], no 3

52. 8 in R4 locked in R4C23, locked for N4
52a. 16(3) cage in N14 = {358} (only remaining combination) -> R3C2 = 3, R4C23 = {58}, clean-up: no 6 in R2C23, no 2 in R5C1
52b. Naked pair {58} in R4C23, locked for R4

53. 14(3) cage in N36 (step 50) = {167/257/347/356}
53a. 3 only in R4C7 -> no 4 in R4C7

54. 8 in N1 locked in R2C123, locked for R2

55. No 8 in R2C678 -> no 3 in R2C78 -> R2C9 = 3 (hidden single in R2)
55a. R2C9 = 3 -> R34C9 = 6 = {24} (cannot be {15} because 1,5 only in R3C9), locked for C9, clean-up: no 9 in R5C8

56. 1 in C9 locked in R67C9 -> R678C9 = 1{69/78}, no 5

57. R159C9 (step 33) = 5{69/78}
57a. 7 only in R9C9 and only other 5 in R5C9 -> no 8 in R5C9, clean-up: no 5 in R5C8

58. R5C8 = 8 (hidden single in N6), R5C9 = 5
58a. R159C9 (step 58) = 5{69/78}, R9C9 = {67} -> R1C9 = {89}

59. R5C4 = 7, R5C67 = [39], R5C1 = 1, R5C3 = 2, R5C2 = 4, R1C1 = 2, R9C1 = 3, R4C1 = 6, R8C3 = 1, R8C2 = 2, R78C5 = [23] (naked singles), clean-up: no 9 in R1C2, no 6 in R1C3 (both step 38a), no 8 in R2C2, no 5 in R2C3 (both step 39)

60. 4 in N9 locked in R9C78, locked for R9
60a. R9C789 = {247} (only remaining combination), locked for R9 and N9 -> R9C9 = 7, R1C9 = 8 (step 57)

61. Naked pair {56} in R9C23, locked for N7

62. 9 in C9 locked in R78C9, locked for N9

63. R6C3 = 3 (hidden single in N4) -> R67C2 = 16 = {79}, no 8

64. Naked pair {12} in R6C78, locked for R6, N6 and 6(3) cage -> R7C8 = 3, R678C9 = [619]

65. R4C2 = 8 (hidden single in C2)

and the rest is naked singles, naked pairs and cage sums