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Assassin 74
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sublue
Regular
Regular


Joined: 12 Oct 2007
Posts: 21
Location: Champaign, Illinois, USA

PostPosted: Fri Oct 26, 2007 1:52 am    Post subject: Assassin 74 Reply with quote

Hi guys! I'm a newbie, so please be merciful. Please forgive any notation errors (turns out it's harder to explain one's thinking than to just solve these).

Editted thrice for corrections and clarifications (thanks to Mike and Andrew).

Thank you for your patience with the "New Kid." My apologies to everyone for using the wrong brackets throughout. I was a math major in a former life, and my brain has yet to adjust. I also apologize for mixed case in labeling nonets, rows and columns. I typed everything lower case, but my computer insists on capitalizing the first letter of every sentence.


Prelims:
1) 6(3) n1 = [1,2,3] locked for n1.
2) 20(3) n12: r1c4 => no 1,2
3) 24(3) n7 = [7,8,9] locked for n7
4) 7(3) n7 = [1,2,4] locked for n7
a) R7c123 = [3,5,6] locked for n7, r7.
5) Both 11(3) n2 no 9
6) 19(3) n23 no 1
7) 20(3) n8 no 1,2
8) 21(3) n3 no 1,2,3
9) 22(3) n6 no 1,2,3,4. 9 is locked into 22(3) cage for n6.
10) 7(3) n6 = [1,2,4] locked for n6
11) 20(3) n69 no 1,2

Solving:
Draw in Hidden cages with 45 and Outies
12) R123c3 is 21(3)

13) R123c4 is 12(3)

14) R123c7 is 8(3)
a) R123c7 no 6,7,8,9 and 1 is locked to 8(3) cage and r12c7 for n3, c7.

15) R123c6 is 22(3)
a) R1c6 no 1
b) R123c6 no 2,3,4
c) 9 is locked r23c6 for n2, 19(3) and c6.
i) 19(3) n23 no 5.

16) R6c123 is 12(3)

17) R7c123 is 14(3)

18) R6c456 is 17(3)

19) R7c456 is 13(3)

20) R6c789 is 16(3)

21) R7c789 is 18(3)

Using Venn Diagram logic and algebra, eliminate some choices with “paired” up hidden cages:

(Note: Venn Diagrams are the math major in me leaking out again. Think of the MasterCard symbol. One circle contains all left-handed persons. The other circle contains all blue-eyed people. The overlap area is left-handed blue-eyed people. The area on the card outside the circles is everybody else. I'm told that normal people call this a variant of innies and outies. Para can attest that my brain is having a hard time with that concept.)
22) R7c3 – 1 = r6c1 = (2,4,5)
a) If r6c1 is 5, conflict with n47 13(3), so r6c1 <> 5 and r7c3 <> 6.
b) R6c23 = (1,2,4,6,7,8,9)

23) R7c6 +2 = r6c4 = (3,4,6,9); r7c6 has no 8.

24) R6c7 + 4 = r7c9 = (7,9); r 6c7 is (3,5)

25) R3c7 + 3 = r1c6 = (5,6,7)

Innies & Outies:
26) Innies c89: r579c8 hidden 11(3)
a) No 9 in r579c8
b) See step 9), 9 is now locked to r45c7 for 22(3), n6 and c7.

27) Outies c6789: r468c5 hidden 20(3)
a) R468c5 no 1,2

28) Outies c1234: r579c5 hidden 14(3)

29) Innies c12: r468c2 hidden 13(3)

Picking off choices:
30) N6 hidden 16(3) r6c789 has locked 3, since n/e in r6.
a) No 3 in r6c456 => r7c6 no 1

31) N9 12(3) r89c7 min 5 => r9c8 max 7, so no 8.

32) Step 14a) locks 1 to n23 11(3) => no 2,5 in 11(3)

33) Hidden 8(3) from step 14a) => can’t have 2; r123c7 = (1,3,4) => Naked Triple (1,3,4) for hidden 8(3), n3 and c7.
a) Naked Single 5 at r6c7
i) Cleanup 5s from r7, c7 and n6.
ii) N69 14(3) r7c8 no 4,8.

34) N6 hidden 16(3): since r6c7 = 5, r6c89 = {3,8} locked to N6 hidden 16(3), n69 20(3) and r6.
a) Cleanup: n69 20(3): r7c9 = 9.
i) Cleanup 9s from r7, n9,c9.

35) Step 15c) has 9 locked in n2 19(3). Since r3c7 = (3,4), r23c6 <> 8.
a) N2 hidden 22(3) r123c6 is now a locked triple (6,7,9).
i) Cleanup 6s and 7s at n2, c6:
ii) R6c4 <> 9.

36) N58 15(3) at r6c56 + r7c6 must be r6c5 = 9 and r67c6 = {2,4} NP locked for N4.
a) NP {2,4} at r6c1 and r6c6 locked for R6.

b) Cleanup 9s r6, c5, n5:
c) Cleanup 2s and 4s r6, c6: NS r6c4 = 6.
d) Cleanup 6s r6, c4, n5.
e) N47 13(3) has r6c23 = {1,7} NP locked for N4, thus r7c3 = 5.
f) Cleanup 1s,7s n4, 5s r7, c3: r6c1 = 4 => r6c6 =2 => r7c6 = 4.

37) N58 15(3) at r67c4 + r7c5 has locked 2s in r7c45 => r7c45 = (2,7).
a) Cleanup of 2s and 7s => r7c7 = 8 and r7c8 = 1.

38) N9 12(3) r89c7 = (2,6,7) => r9c8 = (3,4)

39) N58 split hidden 20(3) r6c5 = 9 => r4c5 = (3,5,8) and r8c5 = (3,6,8)

40) N8 12(3) has locked 1 in r89c6 since n/e.
a) Cleanup c5: no 1 in r45c6 => NT at n5 16(3) = (3,5,8)
b) Cleanup 3s,5s,8s from N5: NT at n5 12(3) = (1,4,7)

41) N2 r3c45 locked 1 since n/e in r3.

42) N3 16(3) has locked 2 => no 7 in 16(3)

43) N1 hidden 21(3) must have an 8, so 8 locked in n1 r123c3 and c3.

44) Cannot be a 3 in r4c2.

45) R1c4 + 1 = r3c3 = (4,6,9) and r1c4 = (3,5,8)

46) N1 hidden 21(3) now has 8 locked in r12c3 => must be an 8 in n12 20(3)=> no 8 in r1c4 and no 4 or 6 in r12c3.

47) Repeat of step 45 => now no 9 in r3c3.

48) Check combinations in n12 13(3) => no 8 in r3c4.

49) N2 11(3) r12c5 min 5 => max 6 for r3c5 => no 8.

50) 8 now locked in n3 16(3) in r3c89 since n/e in row.

51) N3 check combos in 21(3) => no 6.=> r1c8 = 9 and r12c9 = (5,7) locked.

52) N1 6(3) has 1 locked in r12c1 since n/e in col.

53) N9 HS r8c8 = 5.

54) N2 HS r1c6 = 6.
a) Thus n23 11(3) r12c7 = (1,4) and r3c7 = 3.

55) N4 17(3) has 3 locked in r45c3 since n/e in col. => no 2 in 17(3).

56) N7 7(3) has 2 locked in r89c3 since n/e in col.

57) Check combos on n47 split hidden 13(3) r468c2 => no 6 or 9 in r4c2.

58) Naked Triple r4c256 = (3,5,8).

59) Cleanup results in HS at r5c3 = 3.

60) N9 12(3) has 7 locked in r89c7 since n/e in n9.
a) R89c7 = (2,7) and r9c8 = 3
b) Cleanup n6 22(3) r45c6 = (6,9) and r5c8 = 7.

61) Cleaning up now getting wild:

62) R4c4 = 7 Hidden Single.

63) NP (6,9) at r4c37 => r4c1 = 2

The rest just seems to fall out as hidden singles and naked singles with iterative cleanups.
Susan
_________________
Susan


Last edited by sublue on Sat Nov 03, 2007 6:33 am; edited 3 times in total
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Andrew
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Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

PostPosted: Fri Oct 26, 2007 7:07 am    Post subject: Reply with quote

Hi Susan

Welcome to the forum! As a retired engineer it's good to welcome another engineer. It's even better that a newbie posts the first walkthrough for a new Assassin. Very Happy Keep up the good work and we look forward to more of your walkthroughs.

This one flowed easily for quite a time, then got a bit stubborn so I'll rate it at 1.0.

Ruud wrote:
Computer programs would tell you there's no symmetry in this Assassin. We humans know better.

There was certainly plenty of pattern repetition giving repeating themes for the solving path. That made it fairly easy with the given cage totals but I suspect it could have been extremely difficult with different totals. Oops, have I invited a variant for the weekend? Wink I suspect that there was going to be either a variant or a new puzzle this weekend anyway.

Here is my walkthrough. I'll have another look at it tomorrow (well later today since it's already after midnight) to see if I can move step 31 earlier and if that simplifies the later steps.

OK, done that. It didn't make much difference but I said I'd do it.
Previous steps 31 and 32 are now 25 and 26 with resulting renumbering and a few minor corrections.


Prelims

a) 6(3) cage in N1 = {123}, locked for N1
b) 20(3) cage at R1C3 = {389/479/569/578}, no 1,2
c) R123C5 = {128/137/146/236/245}, no 9
d) 11(3) cage at R1C6 = {128/137/146/236/245}, no 9
e) 19(3) cage at R2C6 = {289/379/469/478/568}, no 1
f) 21(3) cage in N3 = {489/579/678}, no 1,2,3
g) 22(3) cage in N6 = 9{58/67}, 9 locked for N6
h) 7(3) cage in N6 = {124}, locked for N6
i) 20(3) cage at R6C8 = {389/479/569/578}, no 1,2
j) 24(3) cage in N7 = {789}, locked for N7
k) 7(3) cage in N7 = {124}, locked for N7
l) 20(3) cage in N8 = {389/479/569/578}, no 1,2

1. Naked triple {356} in R7C123, locked for R7

2. 3 in N6 locked in R6C789, locked for R6
2a. R6C789 = 3{58/67}

3. 45 rule on N9 3 innies R7C789 = 18 = {189/279} = 9{18/27}, no 4, 9 locked for R7 and N9
3a. 45 rule on N8 3 innies R7C456 = 13 = {148/247} = 4{18/27}, 4 locked for N8

4. 20(3) cage in N8 = {389/569/578} -> 12(3) cage in N8 must have 3/5 = {138/156/237}, no 9
[Alternatively 12(3) cage must have 1/2 from step 3a.]
4a. 9 in N8 locked in 20(3) cage = {389/569}, no 7

5. 45 rule on N7 1 innie R7C3 – 1 = 1 outie R6C1 -> R6C1 = {245}
5a. 13(3) cage at R6C1 = {256/346} (only remaining combinations) = 6{25/34}, 6 locked in R7C12, locked for R7, clean-up: no 5 in R6C1 (step 5)
[Alternatively 2,4 only in R6C1 -> R6C1 = {24}, R7C3 = {35} (step 5).]

6. 13(3) cage at R6C2 = {139/157/256/346} (cannot be {148/247} because R7C3 only contains 3,5, cannot be {238} which clashes with R6C1 = 2 because 3 of {238} is in R7C3), no 8
6a. 5 of {157/256} must be in R7C3 -> no 5 in R6C23

7. 45 rule on N8 1 outie R6C4 – 2 = 1 innie R7C6 -> R6C4 = {469}, R7C6 = {247}

8. 15(3) cage at R6C4 = {168/249/267}
8a. 9 of {249} must be in R6C4 -> no 4 in R6C4, clean-up: no 2 in R7C6 (step 7)

9. 15(3) cage at R6C5 = {249/267} (cannot be {159/168/258} because R7C6 only contains 4,7, cannot be {456} which clashes with R6C4 = 6 because 4 of {456} is in R7C6) = 2{49/67}, no 1,5,8, 2 locked for R6 and N5 -> R6C1 = 4, R7C3 = 5 (step 5)
9a. 7 of {267} must be in R7C6 -> no 7 in R6C56

10. Grouped X-wing 3 in 6(3) cage at R1C1 and split-cage R7C12, locked for C12
10a. 3 in C3 locked in 17(3) cage at R4C2 = 3{59/68}, no 1,2,7
10b. 5 of {359} must be in R4C2 -> no 9 in R4C2

11. Naked triple {269} in R6C456, locked for R6 and N4, clean-up: no 7 in R6C789 (step 2a)

12. Naked pair R6C23 = {17}, locked for N4

13. Naked triple {358} in R6C789, locked for N6

14. 45 rule on N3 3 innies R123C7 = 8 = 1{25/34}, 1 locked for C7 and N3
14a. 1 locked in R12C7 for 11(3) cage -> no 1 in R1C6
14b. 11(3) cage = 1{28/37/46}, no 5
14c. 6,7,8 only in R1C6 -> R1C6 = {678}

15. 45 rule on N3 1 outie R1C6 – 3 = 1 innie R3C7 -> no 2 in R3C7

16. 45 rule on N9 1 innie R7C9 – 4 = 1 outie R6C7 -> R6C7 = {35}, R7C9 = {79}
16a. R7C789 (step 3) = {189/279}
16b. 1 of {189} must be in R7C8 -> no 8 in R7C8

17. 19(3) cage at R2C6 = {379/469/478/568} (cannot be {289} because R3C7 only contains 3,4,5), no 2
17a. R3C7 = {345} -> no 3,4,5 in R23C6

18. 45 rule on N3 3 outies R123C6 = 22 = {679} (only remaining combination), no 8, clean-up: no 2 in R12C7 (step 14b), no 5 in R3C7 (step 15)
18a. Naked triple {679} in R123C6, locked for C6 and N2 -> R67C6 = [24], R6C5 = 9, R6C4 = 6

19. Naked triple {134} in R123C7, locked for C7 and N3 -> R6C7 = 5, R7C9 = 9 (step 16)

20. R89C7 must contain 2 (cannot make 12(3) with other combinations of 6,7,8 in R89C7) -> 2 locked in R89C7 for N9, clean-up: no 7 in R7C78 (step 3) -> R7C78 = [81]
20a. 12(3) cage in N9 = {237/246}, no 5
20b. 3,4 only in R9C8 -> R9C8 = {34}

21. 45 rule on N1 3 innies R123C3 = 21 = {489/678} = 8{49/67}, 8 locked for C3 and N1, clean-up: no 6 in R4C2 (step 10a)

22. 45 rule on N1 3 outies R123C4 = 12 = {138/345} = 3{18/45}, no 2, 3 locked for C4 and N2

23. R7C4 = 2 (hidden single in C4), R7C5 = 7

24. 7 in C4 locked in 12(3) cage at R4C4 = {147} (only remaining combination), locked for N5

25. Naked triple {358} in R4C256, locked for R4

26. R5C3 = 3 (hidden single in C3)

27. 20(3) cage in N8 (step 4a) = {389/569}
27a. 3,6 only in R9C5 -> R9C5 = {36}

28. 21(3) cage in N3 = {579/678}, 7 locked for N3
28a. 9 of {579} locked in R1C8 -> no 5 in R1C8

29. 45 rule on N1 1 innie R3C3 – 1 = 1 outie R1C4 -> R1C4 = {358}, R3C3 = {469}

30. 8 in C3 locked in R12C3 for 20(3) cage at R1C3 -> no 8 in R1C4, clean-up: no 9 in R3C3 (step 29)
30a. 20(3) cage = 8{39/57}, no 4,6

31. 13(3) cage at R2C4 = {148/346} = 4{18/36}, no 5
31a. CPE no 4 in R3C5

32. 1 in C6 locked in R89C6 -> no 1 in R8C5
32a. 12(3) cage in N8 = 1{38/56}
32b. 6 of {156} must be in R8C5 -> no 5 in R8C5

33. 2 in C3 locked in R89C3, locked for N7
[Alternatively Grouped X-wing 2 in 6(3) cage at R1C1 and 16(3) cage at R4C1, locked for C12.]

34. 9 in C7 locked in R45C7, locked for N6

35. 1 in R3 locked in R3C45, locked for N2

36. R123C5 = {128/245}
36a. 1 of {128} must be in R3C5 -> no 8 in R3C5

37. 13(3) cage at R2C4 (step 31) = {148/346}
37a. 1 of {148} must be in R3C4 -> no 8 in R3C4

38. 8 in R3 locked in R3C89, locked for N3, clean-up: no 6 in 21(3) cage (step 28)
38a. 21(3) cage in N3 = {579} (only remaining combination) -> R1C8 = 9
38b. Naked pair {57} in R12C9, locked for C9 and N3

39. R1C6 = 6 (hidden single in R1), clean-up: no 3 in R12C7 (step 14b)

40. R3C7 = 3 (hidden single in C7)

41. 13(3) cage at R2C4 (step 31) = {148/346}
41a. 3 of {346} must be in R2C4 -> no 4 in R2C4

42. R8C8 = 5 (hidden single in C8)
42a. R89C9 = 10 = {46} (only remaining combination)
42b. Naked pair {46} in R89C9, locked for C9 and N9 -> R9C8 = 3, R6C89 = [83], R9C5 = 6, R89C9 = [64], clean-up: no 8 in R89C4 (step 4a)
42c. R89C4 = [95]

43. R1C4 = 3, R2C4 = 8, clean-up: no 7 in R12C3 (step 30a), no 6 in R3C3 (step 29), no 4 in R3C4 (step 31)
43a. R12C3 = [89]
43b. R3C34 = [41]

44. R4C3 = 6 (naked single), R4C2 = 8

45. R8C2 = 4 (hidden single in C2)

46. Naked pair {26} in R23C8, locked for C8 and N3

and the rest is naked singles


Last edited by Andrew on Thu Nov 01, 2007 10:33 pm; edited 2 times in total
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CathyW
Master
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Joined: 31 Jan 2007
Posts: 161
Location: Hertfordshire, UK

PostPosted: Fri Oct 26, 2007 9:49 am    Post subject: Reply with quote

Hi Susan and welcome from me too. It's really good to have another female here!!

I've only just printed out this week's puzzle to look at in lunch break later. Also got some catching up to do after being away for a few days.

ttfn

Cathy x

Edit: 13 placements during lunch break. Hopefully can finish off this evening.
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Afmob
Expert
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Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

PostPosted: Fri Oct 26, 2007 5:55 pm    Post subject: Reply with quote

This assassin was pretty straightforward but it took me quite a while to finish it since I found no obvious shortcuts. So to my surprise the walkthrough for A74 is much longer than the one for M1 although M1 was more difficult.

Walkthrough for A74:

1. N7
a) 7(3) = {124} locked
b) 24(3) = {789} locked
c) Naked triple {356} locked in R7C123 for R7
d) 13(3) @ R6C1: R7C12 = {35/36/56} -> R6C1 = (245)
-> R6C1 <> 5 since R7C12 would be {35}
-> 13(3) = 6{25/34} -> 6 locked for N7

2. N89
a) Innies N8 = 13(3) = 4{18/27} because of step 1c -> 4 locked for R7+N8
b) Killer pair (78) of Innies of N8 blocks {578} of 20(3)
c) 20(3) = 9{38/56} -> 9 locked for N8
d) Innies N9 = 18(3) = 9{18/27} -> 9 locked for N9

3. N6
a) 22(3) = 9{58/67} -> 9 locked
b) 7(3) = {124} locked
c) Innies = 16(3) = 3{58/67} -> 3 locked for R6
d) 14(3): R7C78 must be at least 8 (1+7) because 3 (1+2) is too small -> R6C7 <> 7,8

4. N3
a) Innies = 8(3) = 1{25/34} -> 1 locked for C7+N3
b) Outies = 22(3) = 9{58/67} -> 9 locked for C6+N2
c) 19(3) must have 2,3,4 xor 5 and R3C7 = (2345) -> R23C6 <> 5
d) 19(3) <> 5 since R23C6 would be {68} -> no combo for Outies
e) 11(3) must have 5,6,7 xor 8 and R1C6 = (5678) -> R12C7 <> 5
f) Innies = 8(3) = {134} locked for C7+N3
g) 16(3) = 2{59/68}
h) 11(3) = 1{37/46}
i) Outies = 22(3) = {679} because R1C7 = (67) -> locked for C6+N2

5. N2
a) 11(2) = 2{18/45} -> 2 locked for C5+N2
b) 3 locked in R123C4 for C4

6. N1
a) 6(3) = {123} locked for N1

7. N69
a) 3 locked in R6C89 @ 20(3) -> 20(3) = {389} -> R7C9 = 9, R6C89 = {38} locked for R6+N6
b) 22(3) = {679} locked
c) R6C7 = 5
d) 2,8 locked in R789C7 for N9
e) 14(3) = 5[27/81]
f) 15(3) = 5{37/46} -> 5 locked for N9
g) 1 locked in R79C8 for C8
h) 9 locked in R45C7 for N6

8. N2
a) 1 locked in R3C45
b) 20(3): R12C3 <> 4 since R1C4 <> 7,9

9. C89
a) Innies = 11(3): R5C8 = (67) -> R79C8 <> 6,7
b) R7C8 = 1 -> R7C7 = 8

10. R67
a) Both 15(3) = 2{49/67} -> no 1
b) Naked pair (24) locked in R67C6 for C6
c) 15(3) @ R6C5 = {249} -> R6C5 = 9
d) 15(3) @ R6C4 = {267} -> R6C4 = 6, R7C4 = 2, R7C5 = 7
e) R7C6 = 4 -> R6C6 = 2, R6C1 = 4
f) 13(3) @ R6C2 = {157} -> R7C3 = 5
g) Naked pair (17) locked in R6C23 for N4

11. C45
a) 16(3) = {358} locked for N5
b) 20(3) @ N2 must have 3,4 xor 5 -> only possible @ R1C4 -> R1C4 = (345)
c) 20(3) @ N2: Killer pair (69) of 18(3) @ N1 blocks {569} -> 20(3) <> 6

12. N1
a) 7,8,9 locked in R12C3 + 18(3)
b) 18(3) <> 8 because {468} is blocked by R3C3 = (46)
c) 13(3) = 4{18/36}
d) 8 locked in 20(3) = 8{39/57} -> 8 locked for C3

13. C123
a) 3 locked in R34C3 for N4
b) 16(3) = 2{59/68} -> 2 locked for N4
c) 1 locked in R12C1 for N1
d) 2 locked in R89C3 for N7
e) 17(3) must have 5 xor 8 -> only possible @ R4C2 -> R4C2 = (58)

14. R4
a) Naked triple (358) locked in R4C256
b) Hidden Single: R5C3 = 3 @ N4

15. N8
a) 1 locked in R89C6
b) 20(3) must have 3 xor 6 -> only possible @ R9C5 -> R9C5 = (36)
c) 12(3): R8C5 <> 5 since R89C6 has no 6

16. N2
a) 11(3): R3C5 <> 8 since R12C5 has no 1
b) 11(3): R3C5 <> 4 because 4 locked in 13(3)
c) 13(3): R3C4 <> 8 because R2C4 and R3C3 <> 1

17. N3
a) 8 locked in R3C89
b) 21(3) = {579} -> R1C8 = 9, {57} locked for C9+N3
c) Hidden Single: R1C6 = 6 @ R1, R8C8 = 5 @ N9, R5C8 = 7 @ C8, R4C4 = 7 @ R4
d) 11(3) = {146} -> 1,4 locked for C7
e) R3C7 = 3

18. N69
a) 6 locked in R45C7 for C7
b) 12(3) = {237} -> R9C8 = 3
c) R6C8 = 8, R6C9 = 3
d) R9C5 = 6, R9C9 = 4, R8C9 = 6

19. C4
a) 20(3) @ N8 = {569} -> R8C4 = 9, R9C4 = 5
b) 20(3) @ N2 = {389} -> R1C4 = 3, R1C3 = 8, R2C3 = 9

20. Rest is clean-up and singles.

Rating: 1.0, I might try to tackle this one again to shorten up my WT but I can't promise it Confused.


Last edited by Afmob on Fri Nov 02, 2007 5:43 am; edited 3 times in total
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PostPosted: Sat Oct 27, 2007 12:25 am    Post subject: assassin 74 Reply with quote

Hi all,

Haven't been through the posted wts in detail but doubt if I've done anything very different.I'm not sure I can rate very accurately but perhaps this was about a 1.0? Took me about 2.5 hours.

A brief outline of my solving path..



Various prelims include

1.r468c5=20
2.r6c7=r7c9-4

from which r6c7=3/5 r7c9=7/9

3. x-wing work on c89 placed r6c7=5 r7c9=9

This leads to a number of placements in r67 and a large number of either/ors in many other cages.Combination work -> r3c3=4/6 ( thus r1c4=3/5) at which point I got stuck for a while.Then noticed that the 6/5 pairing also forced a 5 into r4c2 and hence could not place a 5 in c5.

4.Thus r3c3=4 r1c4=3 etc and then it was straightforward.

Because there was nothing other than x-wings and combination work involved in solving this killer I'd rate it about 1.0.



I notice Ruud likes to use 3 cages but he excelled himself here..every cage was a 3 cage and all but one was L shaped.Interesting.The killersudokuonline "mindbending" series typically employ much larger cages and are certainly easier than the assassin series.Any reason ,Ruud,why 3 cages figure so predominantly in your excellent puzzles??

Like Andrew I suspect that this cage pattern may accommodate some very difficult variants?? Gulp!!

Regards

Gary
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PostPosted: Sat Oct 27, 2007 11:19 am    Post subject: Reply with quote

Nice observation. I do like the size 3 cages because they can be used to make interesting patterns, create above-average puzzles which are also fast to check with my software.

A few variants:

A74 V2

Harder or not? This variant requires a few steps which are more advanced than those required for the original, but in total, the solving path is shorter.



3x3::k:3584:3584:5378:5378:2308:5893:5893:2567:2567:3584:2058:5378:4620:2308:4622:5893:3600:2567:2058:2058:4620:4620:2308:4622:4622:3600:3600:4379:3356:3356:3358:4639:4639:3873:3618:3618:4379:4379:3356:3358:3358:4639:3873:3873:3618:3117:5422:5422:3376:3633:3633:3379:4404:4404:3117:3117:5422:3376:3376:3633:3379:3379:4404:5183:1856:1856:3906:4419:4419:3397:4678:4678:5183:5183:1856:3906:3906:4419:3397:3397:4678:


A74 Brick Wall

I tried this cage pattern first, but the only valid puzzle that came out of it was this one, a little off the Assassing scale:



3x3::k:3840:3840:3840:3843:3843:3843:3846:3846:3846:4873:2826:2826:3852:3853:3853:3343:4368:4368:4873:4873:2826:3852:3852:3853:3343:3343:4368:4123:4636:4636:4382:2591:2591:4897:2594:2594:4123:4123:4636:4382:4382:2591:4897:4897:2594:4909:2862:2862:1584:5681:5681:3635:4660:4660:4909:4909:2862:1584:1584:5681:3635:3635:4660:1599:5184:5184:3906:5187:5187:2373:5190:5190:1599:1599:5184:3906:3906:5187:2373:2373:5190:

A solving tip: Brick wall

Ruud
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PostPosted: Sat Oct 27, 2007 6:42 pm    Post subject: Reply with quote

This one required some more complicated combo analysis. I can't say that the walkthrough was shorter than my WT for V1.
Let's see how devilish the Brick Wall is. Twisted Evil

Walkthrough for A74 V2:

1. N1
a) Innies = 23(3) = {689} locked for C3+N1
b) 14(3) = 7{25/34} -> no 1
c) 21(3) <> 5 since R12C3 <> 5,7
d) 21(3) = 8{49/67} -> R1C4 = (47), 8 locked for N1

2. C12
a) Innies = 19(3) -> no 1
b) Innies = 19(3) = {289/469/478} because R8C2 = (24)
c) 7(3) = {124} locked for N7 and 1 locked for C3
d) 21(3) <> 4 because R67C3 <> 8,9
e) 21(3) = {579} because R67C3 = {57} -> R6C2 = 9, {57} locked for C3
f) 13(3): R4C2 = (678) because R45C3 = {23/24/34}
g) Innies = 19(3) = 9[64/82]
h) 13(3) = 3{28/46} -> 3 locked for N4
i) 17(3) <> 1 since it has no 9
j) Hidden Single: R6C1 = 1 @ N4

3. N7
a) 12(3) = 1{38/56}
b) 20(3) <> 7 since {578} blocked by R7C3 = (57)
c) Hidden Single: R7C3 = 7 -> R6C3 = 5

4. N3
a) Innies = 21(3) -> no 1,2,3
b) Innies = 21(3) <> 5 since R12C7 <> 5,7
c) Innies = 21(3) = 8{49/67} -> 8 locked for C7+N3 and R3C7 = (47), R1C6 <> 8
d) Outies = 20(3) -> no 1,2
e) Outies = 20(3) = 9{38/47/56} since {578} impossible because of R1C6 = (69)
-> 9 locked for C6+N2

5. R123
a) Naked triple (689) locked in R1C367 for R1
b) 8 locked in R3C46 for N2
c) 18(3) @ R2C4 <> 2 because of Innies+Outies N1:
If 18(3) = {279} -> R3C3 = 9 -> R1C4 = 7 so 18(3) can't be {279}
d) 9(3) = 2{16/34} -> 2 locked for C5

6. N56
a) Innies N6 = 16(3) = 6{28/37} -> 6 locked for R6+N6
b) Innies N5 = 14(3) = 4{28/37} -> 4 locked for N5
c) 13(3) @ R4C4 <> 8 since {238} blocked by Killer pair (23) of Innies of N5
d) 18(3) <> 2 because {279} blocked by Killer pair (27) of Innies of N5
e) Outies N6 = 14(3): has no 6 since {356} blocked by Killer pair (35) of 12(3) @ N7
f) Outies N6 = 14(3): has no 5 since {158} blocked by Killer pair (58) of 12(3) @ N7
g) 13(3) @ R6C7 <> 6 since R7C78 would be {34} -> no combo for Outies of N6
h) 6 locked in 17(3) = 6{29/38/47}

7. N2
a) 18(3) @ R2C6 must have 4 xor 7 since R3C7 = (47) -> R23C6 <> 4,7
b) 7 locked in R123C4 for C4

8. C6789
a) Outies C6789 = R468C5 = 24(3) = {789} locked for C5

9. N58
a) Innies N5 = 14(3) must have 7 xor 8 and R6C5 = (78) -> R6C46 <> 7,8
b) 14(3) must have 7 xor 8 since R6C5 = (78) -> R7C6 <> 8
c) 18(3) <> 3 since {378} blocked by R6C5 = (78) and {369} blocked by Killer pair (36) of 13(3) @ R4C4
d) 14(3) <> 1 because R6C6 <> 1,5,6,7,8
e) 14(3) <> 6 because R6C5 = (78) -> 14(3) = {248/257/347}
f) 7 locked in 17(3) @ N8 -> 17(3) = 7{19/28/46}
g) 18(3) = {189/567} -> R45C6 <> 7 since R4C5 <> 5,6

10. N58
a) 7 locked in R46C5 for C5
b) 7 locked in 17(3) = 7{19/28} because R8C5 = (89)
c) 4 locked in 14(3) = 4{28/37}

11. N6 !
a) ! 17(3) <> 9 since R6C89 would be {26} -> no combo for Outies because R6C7 <> 8
b) 17(3) = 6{38/47}
c) ! Outies N6 = 14(3) = {149/239/248} -> {239} impossible since R7C9 would be 3
and R7C78 would be {29} -> no combo for 13(3)
-> R7C789 <> 3 and 4 locked in R7C789 = 4{19/28} for R7+N9
d) Hidden Single: R6C6 = 4 @ C6
e) 17(3) must have 4 xor 8 and R7C9 = (48) -> R6C89 <> 8
f) Innies = {367} locked for R6+N6
g) R6C4 = 2, R6C5 = 8 -> R7C6 = 2, R8C5 = 9, R4C5 = 7
h) 13(3) = {139} since R6C7 = (37) -> R6C7 = 3 and {19} locked for R7+N9

12. N9
a) Hidden Single: R7C9 = 4
b) 13(3) @ R8C7 = 2{38/56} -> 2 locked
c) Hidden Single: R9C1 = 9 @ R9, R3C7 = 7 @ C7

13. N2
a) 18(3) @ R2C6 = 7[38/56/65]
b) Hidden Single: R1C6 = 9, R2C6 = 3 @ C6 -> R3C6 = 8
c) 9(3) = {126} locked for C5+N2
d) 23(3) = {689} -> 6,8 locked for C7+N3

14. C789
a) 13(3) @ R8C7 = {256} -> R9C8 = 6 and {25} locked for C7+N9
b) R6C8 = 7, R6C9 = 6
c) 15(3) = 9{15/24} since R45C7 have no 2,5,8 -> 9 locked for N6
d) Hidden Single: R3C9 = 9 @ C9
e) R3C3 = 6, R1C3 = 8, R2C3 = 9, R1C7 = 6, R2C7 = 8

15. C456
a) 21(3) = {489} -> R1C4 = 4
b) R3C4 = 5, R2C4 = 7
c) 13(3) @ R4C4 = {139} -> R5C5 = 3 and {19} locked for C4+N5
d) 13(3) @ R6C4 = {256} -> R7C4 = 6, R7C5 = 5
e) R9C5 = 4
f) Hidden Single: R8C1 = 6 @ N7 -> R9C2 = 5
g) R9C7 = 2, R8C7 = 5, R9C3 = 1, R9C6 = 7, R8C6 = 1

16. N13
a) 8(3) = {134} locked, 3 locked for R3
b) 14(3) @ R3C8 = {149} -> 1,4 locked for C8+N3
c) R7C8 = 9, R7C7 = 1

17. N6
a) 15(3) = {249} -> R5C8 = 2

18. Rest is singles.

Rating: 1.25-1.5


Last edited by Afmob on Fri Nov 02, 2007 5:42 am; edited 2 times in total
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PostPosted: Sat Oct 27, 2007 8:58 pm    Post subject: Reply with quote

Ruud wrote:

A solving tip: Brick wall


Interesting new technique, but i can't really get the hang of it. I have been practising it for hours now but nothing productive came from it. I think i might take a break for a while because i am getting a headache. What am i doing wrong?

Para
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PostPosted: Sun Oct 28, 2007 4:29 pm    Post subject: assassin 74 v2 Reply with quote

Another excellent puzzle.

I solved this one via



Some early combo work enabled r6c123 to be filled in (=195) and -> r7c3=7
This placed a 7 in the 3 cage r12c1r2c1 N1.
In N2 the 9(3) cage must contain a 2 as the combos with the 21(3) cage N12 won't permit a 2 in r23c4 so the 9(3) cage is {126} or {234}
1-O on N3 -> r1c6=6/9 (can't be 8 as this puts an impermissible 6 into r3c7.
The key move for me... if r1c6=6 -> r123c5={234} -> r23c4=1. -> r1c4=7 and then x-wings on the 1,6 and 7 in N1/2 means the 14(3) cage in N3 cannot be filled.
Therefore r1c6=9 r3c7=7





This together with outies r6-9 -> r468c5=24={789} -> r6c5=7/8 lead (eventually) to the solution.



Now I'm boldly off to the brick wall too. Rolling Eyes

Regards

Gary

P.S How do you access Ruud's "solving tip"....just in case I should need it!!!
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PostPosted: Sun Oct 28, 2007 10:48 pm    Post subject: Reply with quote

I didn't find A74V2 much harder than A74 so I can't rate it any higher than 1.25, having rated A74 as 1.0. However, as can be seen from my comment after the Prelims, I missed something while solving A74 which might have made that one easier. Therefore I'll stick at 1.25 for A74V2.

Ruud was right that the solving path is shorter than for the original, but not by much.

Here is my walkthrough for A74V2.

Prelims

a) 8(3) in N1 = 1{25/34}, 1 locked for N1
b) 21(3) cage at R1C3 = {489/579/678}, no 1,2,3
c) R123C5 = {126/135/234}, no 7,8,9
d) 23(3) cage R1C6 = {689}, CPE no 6,8,9 in R1C89
e) 10(3) cage in N3 = {127/136/145/235}, no 8,9
f) 21(3) cage at R6C2 = {489/579/678}, no 1,2,3
g) 20(3) cage in N7 = {389/479/569/578}, no 1,2
h) 7(3) cage in N7 = {124}, locked for N7

There were so many innies and innies/outies on nonets in A74 that I completely forgot to look for innies in columns, so let’s put that right and start with one.

1. 45 rule on C12 3 innies R468C2 = 19 = {289/469/478} (cannot be {379/568} because R8C2 only contains 1,2,4), no 1,3,5
1a. 2,4 must be in R8C2 -> no 2,4, in R46C2

2. 1 in N7 locked in R89C3, locked for C3

3. 45 rule on C6789 3 outies R468C5 = 24 = {789}
3a. Naked triple {789} in R468C5, locked for C5

4. 14(3) cage at R6C5 = {149/158/167/239/248/257/347} (cannot be {356} because R6C5 only contains 7,8,9)
4a. 7,8,9 must be in R6C5 -> no 7,8,9 in R67C6

5. 12(3) cage at R6C1 = {138/156/237/345} (cannot be {129/147/246} because 1,2,4 only in R6C1), no 9
5a. 1,2,4 only in R6C1 -> R6C1 = {124}

6. 45 rule on N7 1 innie R7C3 – 6 = 1 outie R6C1 -> R6C1 = {12}, R7C3 = {78}

7. 21(3) cage at R6C2 = {489/579/678}
7a. 4,5 of {489/579} must be in R6C3 -> no 9 in R6C3

8. 9 in N7 locked in 20(3) cage = {389/569}, no 7

9. 7 in N7 locked in R7C123, locked for R7

10. 45 rule on N1 3 innies R123C3 = 23 = {689}, locked for C3 and N1 -> R7C3 = 7, R6C1 = 1 (step 6), R6C23 = [95] (only remaining permutation)
10a. R468C2 (step 1) = 9{28/46}, no 7
10b. R7C12 = {38/56} (step 5)

11. 3 in C3 locked in R45C3, locked for N4
11a. 13(3) cage at R4C2 = 3{28/46}
11b. 17(3) cage in N3 = 7{28/46}

12. 45 rule on N1 1 innie R3C3 – 2 = 1 outie R1C4 -> R1C4 = {467}

13. 21(3) cage at R1C3 = {489/678}, 8 locked in R12C3, locked for C3, clean-up: no 6 in R1C4 (step 12)

14. 45 rule on N3 3 innies R123C7 = 21 = {489/678} (cannot be {579} because 5,7 only in R3C7) = 8{49/67}, no 1,2,3,5, 8 locked for C7 and N3
14a. 4 of {489} must be in R3C7 -> no 9 in R3C7

15. 45 rule on N3 1 outie R1C6 – 2 = 1 innie R3C7 -> R3C7 = {467}
15a. 8 locked in R12C7, locked for 23(3) cage -> no 8 in R1C6, clean-up: no 6 in R3C7 (step 15)

16. 18(3) cage at R2C6 = {279/378/459/468/567} (cannot be {189/369} because R3C7 only contains 4,7), no 1
16a. 4,7 must be in R3C7 -> no 4,7 in R23C6

17. 45 rule on N5 3 innies R6C456 = 14 = {248/347} = 4{28/37}, no 6, 4 locked for R6 and N5
17a. 7,8 must be in R6C5 -> no 7,8 in R6C4
17b. 6 in R6 locked in R6C789, locked for N6
17c. R6C789 = 6{28/37}

18. 14(3) cage at R6C5 (step 4) = {248/257/347} (cannot be {158/167} because 1,5,6 only in R7C6), no 1,6

19. 13(3) cage at R6C4 = {139/148/238/256/346}
19a. 8,9 of {139/148} must be in R7C4 -> no 1 in R7C4
19b. 8 of {238} must be in R7C4, 2 of {256} must be in R6C4 -> no 2 in R7C4

20. 45 rule on N8 1 outie R6C4 = 1 innie R7C6 -> no 5 in R7C6

21. 14(3) cage at R6C5 (step 18) = {248/347} = 4{28/37}, 4 locked in R67C6, locked for C6

22. 45 rule on N9 1 innie R7C9 – 1 = 1 outie R6C7, R6C7 = {237}, R7C9 = {348}

23. 6 locked in R6C89 -> 17(3) cage at R6C8 = 6{38/47}, no 2

24. 45 rule on N9 3 innies R7C789 = 14 = {149/248} (cannot be {158/356} which clash with R7C12, cannot be {239} because R7C78 = {29} would give 2{29} in 13(3) cage at R6C7) = 4{19/28}, no 3,5,6, clean-up: no 2 in R6C7 (step 22)
24a. 4 locked in R7C789, locked for R7 and N9, clean-up: no 4 in R6C4 (step 20)

25. R6C789 (step 17c) = {367} (only remaining combination)
25a. Naked triple {367} in R6C789, locked for R6 and N6 -> R6C456 = [284], R7C6 = 2 (step 20)
[Forgot to do clean-up for R7C789 using step 24 but that didn’t matter; R7C9 was fixed in step 27.]

26. 13(3) cage at R6C4 (step 19) = 2{38/56}, no 1,9
26a. 8 of {238} must be in R7C4 -> no 3 in R7C4

27. Naked quad {3568} in R7C1245, locked for R7 -> R7C9 = 4, R6C7 = 3 (step 22)
27a. Naked pair {19} in R7C78, locked for N9

28. 13(3) cage in N9 = {256} (only remaining combination, cannot be {238} because 3,8 only in R9C8), locked for N9

29. 45 rule on C1234 3 outies R579C5 = 12 = {156/246/345}
29a. 4 of {345} must be in R9C5 -> no 3 in R9C5

30. 45 rule on C89 3 innies R579C8 = 17 = {269} (only remaining combination, cannot be {458} because R7C8 only contains 1,9) -> R7C78 = [19], R59C8 = [26], R6C89 = [76]

31. Naked pair {25} in R89C7, locked for C7

32. Naked pair {49} in R45C7, locked for C7 and N6 -> R3C7 = 7

33. R3C9 = 9 (hidden single in C9), R3C3 = 6, R1C4 = 4 (step 12)
33a. R3C9 = 9 -> R23C8 = 5 = {14}
33b. Naked pair {14} in R23C8, locked for C8 and N3

34. R12C7 = {68} -> R1C6 = 9, R12C3 = [89], R12C7 = [68]

35. R2C4 = 7 (hidden single in N2), R3C4 = 5, clean-up: no 6 in R7C5 (step 26)

36. R3C7 = 7 -> R23C6 = 11 = [38]

37. R1C5 = 1 (hidden single in R1), R23C5 = [62]

38. R9C5 = 4 (hidden single in C5)

39. 3 in R3 locked in R3C12, locked for N1
39a. 8(3) cage in N1 = {134}, locked for N1

40. 7 in N8 locked in 17(3) cage = {179} (only remaining combination), locked for N8 -> R8C5 = 9, R4C5 = 7
40a. Naked pair {17} in R89C6, locked for C6

41. R7C5 = 5 (hidden single in N8), R7C4 = 6 (cage sum), R5C5 = 3, R5C3 = 4, R45C7 = [49]

42. Naked pair {56} in R45C6, locked for N5 -> R45C4 = [91]

43. R4C3 = 3 (hidden single in C3), R4C2 = 6 (cage sum)

44. R4C1 = 2 (hidden single in R4)

and the rest is naked singles


7 2 8 4 1 9 6 5 3
5 1 9 7 6 3 8 4 2
4 3 6 5 2 8 7 1 9
2 6 3 9 7 5 4 8 1
8 7 4 1 3 6 9 2 5
1 9 5 2 8 4 3 7 6
3 8 7 6 5 2 1 9 4
6 4 2 8 9 1 5 3 7
9 5 1 3 4 7 2 6 8


Last edited by Andrew on Thu Nov 01, 2007 10:39 pm; edited 2 times in total
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PostPosted: Mon Oct 29, 2007 3:47 pm    Post subject: the brick wall Reply with quote

Well,how are you doing guys???
I've made a little bit of progress but am now stuck in a morass of either/ors which just don't want to seem to yield.
This is what I've got so far:



1. r1c4=7/8/9 if it's 7/8 then r1c123={159} In N1 the 19(3) cage does not contain a 5.
2. r3c5=1/3 if it's 3 it puts the 9 into r1c4 if it's 1 then r1c4=7/8
3.In N 2 2 is in c6 and in N3 r1c789<>9
4.In N 4 r6c1=4/5/8 and 2 is in r6c2/3
5. In N5 r5c5=7/8 r45c4=27 or 28 the 10(3) cage ={145} and r6=3{69}
6.In N6 the 19(3) cage contains a 9 and the 10(3) cage a 3 and 7 is in r6 and r6c7=1/7
7. In N7 6 is in r7c1/2 r7c3=4/5/8 r8c1=1/2 r9c3=7/8/9 3 at r9c1/2
8. In N8 r7=127 r8c4=6 3 is in r8c5/6 r9c45={45} r9c6=8/9
9. In N9 3 is in r7 and r7c9=3/9 r8c7=1/2 r9c78=61 or 62 r9c9=7/8/9



Perhaps someone has got further now.Look forward to seeing a wt.

Regards and good luck!!!

Gary
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PostPosted: Tue Oct 30, 2007 3:14 pm    Post subject: Re: the brick wall Reply with quote

gary w wrote:
Well,how are you doing guys???

To: Gary W,
Assassin forum,
Sudocue.net

October 30, 2007


Dear Gary,

Nice to hear from you!

Well, I've managed to get to the Brick Wall (BTW, writing it in capital letters makes it sound like the name of a british pub! Smile ). However, there's no sight of Afmob. Probably he'd already been and gone by the time I got here. Or, who knows, maybe he's already managed to climb over it?

So far, I haven't managed to get very far. The ladder I brought with me is too short, and I unfortunately don't possess a longer one! Still, by using it, I can get high enough up the wall to begin to "appreciate" it, even if I can't scale it right now.

BTW, I'm wondering whether you and I are looking at the same wall, because I can't seem to correlate or make sense of the notes (graffiti??) you've written on it. For example, how did you manage to get down to just {789} in r1c4 (if I understand you correctly)? Maybe you can explain?

Hopefully, by combining our (metaphorical) ladders, we'll be able to clamber over this formidable obstacle. I hope so.

Until then, all the best.
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Cheers,
Mike
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PostPosted: Tue Oct 30, 2007 3:47 pm    Post subject: Reply with quote

What a monster! That was the most difficult Killer Sudoku I've solved so far. But I got no idea about the rating since the hardest I've solved was about 1.75.

Edit: Step 11a is wrong/can not be applied at that stage since I forgot the combo [5214]. Thanks Andrew for finding this error. I won't change this wt since I replaced it with an improved version (in a new post).

Walkthrough for A74 Brick Wall:

1. N5
a) Outies = 10(3) -> no 8,9
b) 22(3) = 9{58/67} -> 9 locked for R6+N5
c) 22(3): R6C56 <> 5 since R7C6 <> 8,9

2. N7
a) 6(3) = {123} locked
b) 11(3): R6C23 <> 7,8 since R7C3 >= 4
c) 19(3): R6C1 <> 2,3 because R7C12 would be {79/89} -> Both are Killer pairs of 20(3)
d) Innies = 19(3) = {469/478/568}

3. N89
a) Innies N9 = 16(3) <> 2 since Killer pair (59) of 20(3) @ N9 blocks {259}
and Killer pair (68) of Innies of N7 blocks {268}
b) 6(3) = {123} -> 2 locked for N8, R6C4 <> 2
c) Killer pair (58) of 15(3) blocks {578} of 20(3) @ N8
d) 20(3) @ N8 = 9{38/47/56} -> 9 locked for N8
e) Because of step 3b Innies of N8 = 10(3) = {127/235}
f) 22(3) must have 5 xor 7 and R7C6 = (57) -> R6C56 <> 7
g) 6(3) = {123} -> R89C4 <> 1,3
h) 15(3): R9C5 <> 7,8 since R89C4 is at least 9
i) 15(3) <> 7 because R7C6 = (57) blocks {357}

4. N25
a) 9 locked in R123C4 for N2
b) 17(3) <> 1,3 since Killer pair (68) of 22(3) blocks {368}
c) Killer pair (68) locked in 17(3) + 22(3) for N5

5. R89
a) Hidden pair (12) in R8C17 -> no other possibilities
b) 3 locked in R9C12 for R9
c) 9(3) = {126} locked for N9, 6 locked for R9
d) 15(3) = 6{18/45} -> R8C4 = 6
e) 20(3) @ N8 = 9{38/47}
f) 20(3) @ N7 = 7{49/58} -> 7 locked for N7

6. N9
a) Innies = 16(3) = {349/358/457}: R7C6 = (57) blocks {457} -> no 7
b) Innies = 16(3) = 3{49/58} -> 3 locked
c) Hidden Single: R7C6 = 7 @ R7

7. N8
a) 20(3) = {389} locked
b) 6(3) = {123} -> R6C4 = 3
c) 15(3) = {456} -> 4,5 locked for R9
d) 22(3) = {679} -> 6 locked for R6+N5

8. N6
a) 9 locked in 19(3) = 9{28/37/46}
b) Innies = 16(3) = 7{18/45} -> 7 locked for R6+N6
c) 19(3) = 9{28/46}
d) 3 locked in 10(3) = 3{16/25}
e) 18(3) = {189/378/459} -> must have 3 xor 9 and is only possible @ R7C9 -> R7C9 = (39)
f) Killer pair (48) locked in 18(3) + 19(3)
g) 14(3): R6C7 <> 5 since R7C78 has no 1

9. N4
a) 2 locked R6C23
b) 11(3) = 2{18/45} = Innies
c) 19(3): R7C12 <> 4 because R6C1 <> 6,9

10. C456
a) Innies+Outies C56: -8 = R1C4 - R3579C5 -> R1C4 <> 1 since R3579 is at least 10
b) 10(3): R4C5 <> 2 since R45C6 has no 7
c) 15(3) @ R2C4: R3C5 <> 7 since R23C4 <> 3,6
d) Innies+Outies C56: -8 = R1C4 - R3579C5 -> R1C4 <> 2:
If R1C4 = 2 then R3579C5 = 10(4) = {1234}
-> R9C5 = 4 -> R7C5 = 2 (because R1C4 = 2 -> R7C4 = 1) but R5C5 <> 1,3 -> {1234} impossible

e) Consider positions of 1 in C4: Either 15(3) @ R2C4 = 1{59/68} -> R3C5 <> 2
or R7C4 = 1 -> R7C5 = 2 -> R3C5 <> 2
-> R3C5 <> 2
f) All 15(3) of N2 can't be {456} since it's a Killer triple of 15(3)
g) Innies+Outies C45: -22 = R1C6 - R2468C5 -> R1C6 <> 8 because
-> R2468C5 would be 30(4) = [6798] -> 10(3) = {127} with R45C6 = {12}
and 15(3) @ R2C5 = 6{18/27} -> impossible because of R45C6 = {12}

11. C456
a) Innies+Outies C56: -8 = R1C4 - R3579C5 -> R1C4 <> 4:
If R1C4 = 4 -> 15(3) @ R2C4 = 9{15} -> R3579C5 = 12(4) = [2514] (because R9C5 = (45))
but this means R5C5 = 5 so that R45C4 @ 17(3) = {48} -> conflict with R1C4 = 4

b) Innies+Outies C56: -8 = R1C4 - R3579C5 -> R1C4 <> 5:
If R1C4 = 4 -> 15(3) @ R2C4 = 9{24} -> 15(3) @ R1C4 = [573] -> R8C5 = 3 (HS @ N8) therefore
R3579C5 = 13(4) = {1247} but that's impossible because R1C5 = 7

c) 15(3) @ R1C4 must have 7,8 xor 9 and R1C4 = (789) -> R1C5 <> 7,8
d) Innies+Outies C56: -8 = R1C4 - R3579C5 and R1C4 = (789)
-> R3579C5 = 15/16/17 and R579C5 >= 10 -> R3C5 <> 8
e) ! 15(3) @ R2C4: R23C4 <> 1:
If R23C4 has 1 -> 15(3) = 1{59/68} -> R7C4 = 2 -> 17(3) = {458} but 8 can't be @ R5C5
because R9C4 = (45) -> R123C45 <> 4,5 because R9C45 builds Naked pair (45) with 17(3)
-> 15(3) @ R1 = {168} with 8 @ C4 -> Conflict with 17(3) because 8 of 17(3) must be @ C4

12. C456
a) Hidden Single: R7C4 = 1 @ C4 -> R7C5 = 2
b) Innies+Outies C56: -8 = R1C4 - R3579C5 and R1C4 = (789)
-> R3579C5 = 15/16/17 -> R3C5 <> 4,5 because R379C5 would 11 and R5C5 >= 7
c) 15(3) @ R2C4 = {159/267/348/357} because R3C5 = (136) and R23C4 <> 6
d) Killer triples (258), (357) of 15(3) @ R2C4 blocks {258},{357} of other 15(3)s @ N2
e) 15(3) @ R2C5 = {168/267/348} -> no 5
-> Killer pair (68) blocks {168} of 15(3) @ R1C4
f) 15(3) @ R1C4 = {159/249/267/348} -> R1C6 <> 6 since R1C45 <> 2

13. N12+R1 !
a) All of 15(3) @ R1 can't be {456} since it's a Killer triple of 15(3)
b) Killer triple (267) of 11(3) blocks {267} of 15(3) @ R1C1
c) Killer triple (348) of 19(3) blocks {348} of 15(3) @ R1C1
d) ! Killer triple (159) of 15(3) @ R1C1 blocks {159} of both 15(3) @ R1
e) Clean up: 15(3) @ R1C4 <> 1,5
f) 5 locked in R23C4 for C4 -> 15(3) @ R2C4 = 5{19/37}
g) R9C4 = 4, R9C5 = 5

14. N5
a) 17(3) = {278} locked

15. R1+N1
a) Killer pair (24) of 15(3) @ R1C4 blocks {249} of both 15(3) @ R1
b) Clean up: 15(3) @ R1C1 <> 4 and 15(3) @ R1C7 <> 9
c) Killer pair (28) of 15(3) @ R1C4 blocks {258} of both 15(3) @ R1
d) Clean up: 15(3) @ R1C1 <> 2
e) 19(3) <> 5 since {568} blocked by Killer pair (56) of 15(3) @ R1C1
f) 11(3) <> 7 since {137} blocked by Killer pair (13) of 15(3) @ R1C1

16. N3
a) Killer pair (78) of 15(3) blocks {278} of 17(3)
b) Killer pair (36) of 15(3) blocks {368} of 17(3) and {346} of 13(3)
c) Killer triple (238) of 15(3) blocks {238} of 13(3)
d) Killer triple (467) of 15(3) blocks {467} of 17(3)
e) 13(3): R3C8 <> 5 since R23C7 would be {17/26} -> {17} blocked by R6C7 = (17)
and {26} blocked by 9(3) = {126}

17. C456
a) Innies+Outies C45: -22 = R1C6 - R2468C5 and R1C6 = (234)
-> R2468C5 = 24/25/26 -> R2C5 <> 1 since R468C5 <= 21
-> R2C5 <> 4 since R468C5 would be <= 18
b) 15(3): R1C6 <> 3:
If R1C6 = 3 -> 15(3) = [843] -> R3C5 = 1 -> no candidate for R4C5

18. N3 !
a) ! 13(3) <> 6 because it would be {256} -> R23C7 = 5{2/6} and R3C8 = (26)
together with 9(3) = {126} builds a Naked pair (26) in C78 -> no combo for 19(3)
-> 13(3) = {139/148/157/247}
b) 17(3) <> 1,7 since {179} blocked by Killer pair (17) of 13(3)
c) ! 17(3) <> 3 because 10(3) @ N6 must have 3 therefore:
If 17(3) has 3 -> 3 locked in 17(3) + 10(3) for C89 -> R7C7 = 3 (Hidden Single @ N9)
-> 5 locked in R123C7 for N6 -> no 5 in 17(3) -> {359} impossible
-> 17(3) = {269/458}
d) 13(3) <> 2 since Killer pair (24) of 17(3) blocks {247}
e) 1 locked in 13(3) for N3

19. N1
a) 1 locked in 15(3) = 1{59/68}
b) 2 locked in 11(3) = 2{36/45}
c) 19(3) <> 6 because Killer pair (46) of 11(3) blocks {469}

20. N6
a) 1 locked in R456C9
b) 14(3) = 7{34} -> R6C7=7 and 3,4 locked for R7+N9
c) R7C9 = 9

21. C123
a) Naked triple (568) locked in R7C123 for N7
b) 11(3) @ N7 must have 5 xor 8 and R7C3 = (58) -> R6C23 <> 5
c) 19(3) @ N7 = {568}

22. N3
a) 5 locked in R123C7
b) 17(3) = {269} -> R2C8 = 9 and {26} locked for C9+N3
c) 13(3) = 1{48/57}
d) 15(3) = 3{48/57} -> 3 locked for R1

23. N2
a) Hidden Single: R1C6 = 2 @ R1
b) 15(3) @ R1C4 = 2[94/76]
c) 8 locked in 15(3) @ R2C5 = 8{16/34}
d) Hidden Single: R5C5 = 7 @ C5

24. N6
a) 18(3) = 9{18/45} -> R6C9 <> 8
b) 10(3): R45C9 = {13/15/35} -> R4C8 = (246) but R4C8 <> 4 -> R4C8 = (26)
c) 3 locked in 10(3) for C9

25. R12
a) Coloring 1 @ R2 -> R4C1 <> 1:
- i) R2C6 = 1 -> R4C5 = 1 @ N5 -> R4C1 <> 1
- ii) R2C7 = 1 -> R9C8 = 1 @ N9 -> R8C1 = 1 @ N7 -> R4C1 <> 1
b) 15(3) @ R1C7: R1C8 <> 7 because R1C9 <> 3,5

26. N47
a) 16(3) <> {358} because R6C1 = (58)
b) Killer triple (468) of 16(3) blocks {468} of 18(3)
c) 16(3): R4C1 <> 5,8 because R5C12 <> 7
d) 2 locked in R89C1 for N7

27. N1
a) 11(3): R3C3 <> 3 because R2C23 would be {26} blocked by R2C9 = (26)

28. N46
a) 19(3): R5C7 <> 2,8 because:
If R5C7 = (28) then R4C7 must be 9 and R5C8 = (28) but that's impossible because
of R5C4 = (28)
b) 18(3) @ R4C2: R4C23 <> 8 because:
If R4C23 has 8 -> R4C4 = 2, R5C4 = 8 -> 19(3) @ R4C7 = {469} -> no candidates for R4C8
c) 18(3) @ R4C2: R5C3 <> 1 because only R5C3 can be 8

29. N12 !
a) 19(3): R3C12 <> 3: If R3C12 has 3 than 19(3) = 7{39} with R3C12 = {39}
-> blocked by Killer pair (39) in R3 of 15(3) @ R2C4
b) 3 locked in R2C123 for R2
c) 15(3) @ R2C5: R3C6 <> 4 because R2C56 <> 3
d) ! Consider both candidates of R1C5 = (46) -> R6C5 <> 6:
- i) R1C5 = 4 -> 15(3) @ R1C7 = [537] -> R9C9 = 8 -> R9C6 = 9 -> R6C6 = 6 -> R6C5 <> 6
- ii) R1C5 = 6 -> R6C5 <> 6
e) R6C5 = 9, R6C6 = 6

30. R23
a) Hidden pair (26) @ R3 in R3C39 -> no other possibilities
b) 15(3) @ R2C4: R3C4 <> 7 because if R3C4 = 7 -> 13(3) = {148} -> no 5 in R3
c) Consider both placements for R123C4 = [759/975] -> R2C1 = (37):
- i) [759] -> 15(3) @ R1C4 = [762] -> 15(3) @ R1C1 = {159} -> 19(3) = {478}
-> 15(3) @ R1C7 = {348} -> 13(3) = {157} -> R3C8 = 7 -> R2C1 = 7
- ii) [975] -> 15(3) @ R1C1 = {168} -> 19(3) = {379} -> R2C1 = 3
-> R2C1 = (37)
d) 13(3): R2C567 is only place where 1,8 @ R2 is possible and R2C56 can't have both
because R3C6 <> 6 -> R2C7 = (18)
e) Consider both placements for R2C1 = (37) -> 18(3) @ N4 <> 1,4:
- i) If R2C1 = 3 -> R9C2 @ 6(3) = 3 -> 3 locked for 18(3) @ N4 = 3{69/78}
- ii) If R2C1 = 7 -> 9 locked in 16(3) for N4 -> 18(3) @ N4 = 7{38/56}
-> 18(3) = {369/378/567}

31. N1 !
a) ! 19(3) <> 3 because of following conflict, if R2C1 = 3:
- i) R9C2 = 3 @ 6(3), 4 locked in R45C1 for N4 -> Innies N4 = 11(3) = {128} -> 18(3) @ N4 = {369}
- ii) 15(3) @ R2C4 = [753] -> R4C5 = 1; 13(3) @ N3 = {148} -> R8C8 = 7
- iii) 5 locked in R89C9 @ 20(3) for C9 -> 10(3) @ N6 = {136} -> {36} locked for R4
-> no combo for 18(3) @ N4

b) 19(3) = {478}, R2C1 = 7, {48} locked for R3+N1
c) 11(3) = {236} locked for N1
d) Hidden Single: R1C4 = 7 @ N2, R3C8 = 7 @ N3
e) R2C4 = 5, R3C4 = 9 -> R3C5 = 1, R3C6 = 3, R3C7 = 5 -> R2C7 = 1

32. N2
a) 15(3) @ R1C4 = [762] -> R1C5 = 6
b) 15(3) @ R2C5 = [843] -> R2C5 = 8, R2C6 = 4
c) R4C5 = 4, R8C5 = 3, R8C7 = 2, R9C7 = 6, R9C8 = 1, R8C1 = 1, R9C2 = 3, R9C1 = 2
d) Hidden Single: R2C3 = 3 @ N1

33. N4
a) 3 locked in 16(3) = {349}, 4 locked for R5
b) 18(3) = {567} locked

34. Rest is singles.

Finally finished but I'm not too happy about my last big contradiction move on R2C1, but I haven't find a way around it.


Last edited by Afmob on Thu Mar 20, 2008 5:28 pm; edited 8 times in total
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CathyW
Master
Master


Joined: 31 Jan 2007
Posts: 161
Location: Hertfordshire, UK

PostPosted: Tue Oct 30, 2007 5:28 pm    Post subject: Reply with quote

Back-tracking to the original A74 - I finally finished it on Saturday. Haven't had time to do a WT so far (still catching up on other puzzles Laughing ) but I did make use of a grouped turbot fish (4) to place 3 in r3c7.
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Para
Yokozuna
Yokozuna


Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

PostPosted: Tue Oct 30, 2007 8:45 pm    Post subject: Reply with quote

Hi

This is what i have done so far on the brick wall, some cracks and a few holes but it is still not falling down. About the same as Afmob's progress i think by quickly checking his walkthrough.

[Edit] Some moves added. Now got 12 digits placed.

[Edit] Another few moves added. Now 13 digits placed.

[Edit] Another set of moves added. Now 81 digits placed Wink

Walk-through Assassin 74 V3 "Brick Wall"

1. 19(3) at R2C1, R4C7 and R6C1 = {289/379/469/478/568}: no 1

2. 11(3) at R2C2 and R6C2 = {128/137/136/236/245}: no 9

3. 10(3) at R4C5 and R4C8 = {127/136/145/235}: no 8,9

4. 22(3) at R6C5 = {589/679}: no 1,2,3,4

5. 20(3) at R8C2, R8C5 and R8C8 = {389/479/569/578}: no 1,2

6. 9(3) at R8C7 = {126/135/234}: no 7,8,9

7. 6(3) at R8C1 = {123} -> locked for N7

8. 6(3) at R6C4 = {123} -> pointing eliminates {123} from R89C4

9. 45 on N7: 1 innie and 1 outie: R6C1 = R7C3: R6C1: no 2,3,9
9a. 3 outies: R6C123 = 11 = {128/137/146/236/245}
9b. 3 innies: R7C123 = 19 = {469/478/568}

10. 45 on N8: 1 innie and 1 outie: R6C4 + 4 = R7C6: R7C6: no 8,9
10a. 3 outies: R6C456 = 18 = {189/279/369/378}
10b. 3 innies: R7C456 = 10 = {127/136/145}

11. 9 in 22(3) at R6C5 locked within R6C56 -> locked for R6 and N5
11a. 22(3) = {89}[5]/{679}: R6C56: no 5
11b. R6c456: {378} blocked

12. 9 in N6 locked within 19(3) cage at R4C7: 19(3) = {289/379/469}: no 5

13. 45 on N9: 1 innie and 1 outie: R6C7 + 2 = R7C9: R6C7: no 8; R7C9: no 1,2
13a. 3 outies: R6C789 = 16 = {178/268/349/358/367/457} = {1/2/3/7..}
13b. 3 innies: R7C789 = 16 = {169/178/259/268/349/358/367/457}

14. R6C4789 needs at least 2 of {1237} -> R6C123: {137} blocked: no 7
14a. Clean up: R7C3: no 7

15. 45 on C6: 1 innie + 4 outies: R1C6 + 22 = R2468C5: R1C6: no 9

16. 45 on C4: 1 innie + 4 outies: R1C4 + 8 = R3579C5: R1C4: no 1

17. 17(3) at R4C4 = {278/368/458/467}: no 1

18. Hidden Pair {12} in R8C17: R8C17 = {12}
18a. 3 in N7 locked for R9

19. 9(3) at R8C7 = {126}(last combo) -> locked for N9
19a. 6 in N9 locked for R9
19b. Clean up: R6C7: no 4

20. 3 in R8 locked within 20(3) cage at R9C5 and R8C8: one of these cage is {389}
20a. 3 20(3) cages in R89: One is {389} and both others need one of {89}; the 3 20(3) cages in R89 need 4 of {89}: no {89} anywhere else in R89
This would probably have been easier if i has spotted the hidden pair in R7C45 instead, but it looks nicer.

21. Both {89} in N8 locked within 20(3) cage at R8C5: 20(3) = {389} -> locked for N8
21a. R6C4 = 3(hidden in 6(3) at R6C4)
21b. R7C6 = 7(step 10); R8C4 = 6(hidden)
21c. 3 in N8 locked for R8
21d. Naked Pair {12} in R7C45 -> locked for N8
21e. Naked Pair {45} in R9C45 -> locked for R9
21f. Clean up: R6C56 = {69} -> locked for R6 and N5; R6C7: no 5; R7C9: no 5,8; R7C3: no 6

22. 11(3) at R6C2 = {12}[8]/245}: R6C23: no 8; 2 locked within R6C23 -> locked for R6 and N4
22a. Clean up: R7C9: R7C9: no 4

23. 7 in R6 locked in N6
23a. Clean up: 19(3) in R4C7 = {289/469}: no 3

24. 3 in N6 locked in 10(3) at R4C8: 10(3) = {136/235}: no 4

25. 9 in C4 locked for N2

26. 19(3) at R6C1 = [4]{69}/{568}: R7C12: no 4

27. 10(3) = [7]{12}/{145}: R4C5: no 2

28. 45 on C4: 2 innies + 3 outies: R17C4 + 5 = R359C5: Min R17C4 = 3 -> Min R359C5 = 8: R359C5 = 8 = [125] blocked by R7C5
28a. Min R359C5 = 9 -> Min R17C4 = 4: R1C4: no 2

Ok realised this a bit late.
29. 15(3) cages can have max 1 of {123}.
29a. 3 15(3) cages in R1 need one of {123}: {456} blocked in all 3
29b. 3 15(3) cages in N2 {456} blocked through same reasoning

Some more reasoning between 3 15(3) cages in one Nonet.
30. 1 in a 15(3) cage = {159/168} = {5/8..}
30a. 3 15(3) cages in R1 or N2 can't be {258}

31. 45 on C4: 3 innies and 3 outies: R123C4 = R579C5 + 7
31a. Analysis
Valid combos: R579C5/R123C4
214=7 - 14 = {149}=[4]19
425=11 - 18 = {279}=[9]27
524=11 - 18 = {279}=[9]27
725=14 - 21 = {579}=[9]57/[7]59
824=14 - 21 = {489}=[9]48
825=15 - 22 = {589}=[8]59
Blocked by C4 + R3C5
215=8 - 15 = {159} ---
415=10 - 17 = {179}=[7]19 --
514=10 - 17 = {179}=[7]19 --
724=13 - 20 = {479}=[7]49 --
Blocked by C4
--714=12 - 19 = {478}=[7]48
--715=13 - 20 = {578}=[8]57
--814=13 - 20 = {479}=[7]49
--815=14 - 21 = {579}=[9]57
31b. Conclusions: R1C4: no 5; R23C4 = {19/27/49/57/59} -> R3C5 = {1356}
31c. 15(3) at R2C4 = {159/267/348/357}
31d. 15(3) at R1C4 = [4]{38}/[7]{26}/[8]{34}/[9]{24}/[9]{15}: R1C5: no 7

32. 15(3) at R2C5: [4]{38} blocked by R89C6: R2C5: no 4

33. R123C4 = [4]{19} -> R2C5 = 7(hidden): R3579 = [5214] blocked by R4C5
33a. R1C4: no 4; R3C5: no 5
33b. R123C4 = [9]{27}/[9]{57}/[9]{48}/[7]{59}/[8]{59}: no 1
33c. R7C4 = 1(hidden); R7C5 = 2
33d. Clean up: R1C56: no 8(step 31d)

34. 11(3) at R2C2 = {128/137/146/236/245} = {2/6/7..};{3/4/8..}
34a. 15(3) at R1C1 = {159/168/249/357} = {1/5/9..}: {267/348} blocked by 11(3) at R2C2
34b. 15(3) at R1C4 = [9][42]/[8]{34}/[762]: [9]{15} blocked by 15(3) at R1C1: no 1,5; R1C6: no 6
34b. 15(3) at R1C7: {159} blocked

35. 1's in N2: R3C5 = 1 or 15(3) at R2C5 = {168}: R3C5: no 6

36. 15(3) at R2C4 = {59}[1]/{48}[3]/{57}[3]= {3/5..}: R23C4: no 2
36a. 2 in N2 locked for C6

37. 10(3) at R4C5 = {145} -> locked for N5

38. 15(3) at R2C5 = {168/267/348}: no 5
38a. 15(3): [1]{68} blocked by R689C6: R2C5: no 1

39. 5 in N2 locked for C4
39a. R9C45 = [45]
39b. 15(3) at R1C4: [843] blocked by R34C5: R1C6: no 3

40. 15(3) at R2C4 = {59}[1]/{57}[3]: no 8

41. 15(3) at R1C7 = {168/267/348/357} = {2/3/8..},{2/7/8..},{3/4/6..},{4/6/7..}: {249} blocked by R1C6: no 9
41a. 13(3) at R2C7 = {139/148/157/247/256}: {238/346} blocked
41b. 17(3) at R2C8 = {179/269/359/368/458}: {278/467} blocked

42. 15(3) at R1C1 = {159/168/357} = {5/6..},{1/7..}: {249} blocked by R1C6: no 2,4
42a. 19(3) at R2C1 = {289/379/469/478}: {568} blocked: no 5
42b. 11(3) at R2C2 = {128/146/236/245}: {137} blocked: no 7

43. 13(3) at R2C7:[9]{13} blocked by R3C5: R2C7: no 9
43a. 13(3): {17}[5] blocked by R6C7 and {26}[5] blocked by R89C7: R3C8: no 5

44. 2 and 6 in C9 either in 10(3) at R4C8 or R123C9. 10(3) can't have both {26} so R123C9 needs one of {26}
44a. 13(3) at R2C7 = {139/148/157/247} = {1/7..}: {256} blocked by R123C9: no 6

45. 17(3) at R2C8 = {269/359/368/458}: {179} = {3/6/8..} blocked by 13(3) at R2C7: no 1

46. 17(3) at R2C8 = {3/6/8..}
46a. 17(3) has 3 -> R7C7 = 3(hidden) -> R6C7 <> 1
46b. 17(3) has 6/8 -> R1C789 <> {168}: no 1 -> R456C9 = 1 -> R6C7<>1
46c. Conclusion: R6C7: no 1
46d. R6C7 = 7; R7C9 = 9

47. R7C78 = {34}(last combo) -> locked for R7 and N9
47a. Clean up: R6C1: no 4; R6C23: no 5

48. Naked Triple {568} in R7C123 -> locked for N7

49. 5 in C7 locked for N3

50. 17(3) at R2C8 = {269/368}: no 4,7; 6 locked for N3

51. 15(3) at R1C7 = {348/357}: no 1,2: 3 locked for R1 and N3
51a. R1C6 = 2(hidden)
51b. R1C45 = [76/94]

52. 15(3) at R2C5 = {168/348}: no 7
52a. R5C5 = 7(hidden)

53. 17(3) at R2C8 = {269}: R2C8 = 9; R23C9 = {26} -> locked for N3 and C9

54. 15(3) at R1C1 = {159/168} = {5/6..}: no 7; 1 locked for N1

55. 11(3) at R2C2 = {236/245} = {5/6..}: no 8; 2 locked for N1
55a. Killer Pair {56} in 15(3) at R1C1 and 11(3) at R2C2 -> locked for N1

56. 1 in C9 locked for N6

57. 10(3) at R4C8 = [2]{35}/[6]{13}: R4C8: no 3,5; 3 locked in R45C9 for C9

58. R6C89 = {45}/[81]: R6C9: no 8

59. 2 in C1 locked for N7

60. Killer xy-wing: Pivot cell(s): R1C45 = {47}; Wings: R1C789 = {45}, R2C5 = {57} -> Common Peers can't have 5: R2C7: no 5

61. 15(3) at R1C7 = [537]/[3]{48}/[438/843]: R1C8: no 7

62. 7's in N3: 15(3) at R1C7 = [537] or R3C8 = 7
62a. 45 on C7: 1 innie and 4 outies: R1C7 + 10 = R3579C8
62b. Valid combos: [3]-[7231]/[4]-[7241]/[5]-[8241]/[5]-[1842]/[8]-[7641](other combos blocked by step 61 and 62)
62c. R3C8 = {178}; R5C8 = {268}; R9C8 = {12}
62d. R9C7 = 6(hidden)

63. 19(3) at R4C7 = {49}[6]/{2[9]8}: [9]{28} blocked by R5C4: R5C7: no {28}

64. ER on 4: R4C5 = 4 or R1C5 = 4 -> R23C7 = 4: R4C7: no 4

65. Combination analysis 19(3) at R4C7 + 10(3) at R4C8: R4C78 = [26/86/92] = {2/8..},{6/9..}
65a. Killer Pair {28} in R4C4 + R4C78 -> locked for R4

66. R6C5: no 6, the following chain explains how.
66a. R1C5 = 6 -> R6C5 <> 6
66b. R1C5 = 4 -> 15(3) at R1C7 = [537] -> R9C9 = 8 -> R9C6 = 9 -> R6C6 = 6 -> R6C5<>6
66c. R6C56 = [96]

67. Hidden Pair {26} in R3C39: R3C3 = {26}

68. 18(3) at R4C2 = {19}[8]/{37}[8]/[9]{45}/{49}[5]/[954]/{36}[9]/{39}[6]/{57}[6]/{67}[5]: {45}[9] blocked by R4C56, [5]{49} blocked by R89C3, {69}[3] blocked by R4C78: R5C3: no 1,3
68a. {19}[8] blocked by following chain: R5C3 = 8 -> R6C8 = 8 -> R4C7 = 9 -> R4C23 <> 9: 18(3): {19}[8] blocked: no 1

69. Hidden Killer Pair {69} in R4C123 + R4C78: R4C123 needs one of {69}
69a. 16(3) at R4C1: [1]{69} blocked by 68; [1]{78} not possible: R4C1: no 1

70. 8 in R23 locked within cages 15(3) at R2C5 and 19(3) at R2C1 or 13(3) at R2C7
70a. When 19(3) = {478}, 13(3) has no 8, so 13(3) = [157] -> 19(3) = [7]{48}: R2C1: no {48}

71. 11(3) at R2C2 = {23}[6]/{36}[2]/{45}[2] = {3/5..}
71a. Killer Triple {357} in R2C1 + R2C23 + R2C4 -> locked for R2
71b. 3 in R2 locked for N1

72. 15(3) at R2C5 = [843]/[6]{18}: R3C6: no 4

73. 13(3) = {1[4]8}/[157]: [4]{18} blocked by R3C56: R2C7: no 4; R3C7: no 1,8

74. 16(3) at R4C1 = [6]{19}/[9]{16}/[781]/{349}/[763]/[7]{45}: [718] blocked by R289C1, [736] blocked by R2C1: R4C1: no 5; R5C2: no 8

75. 5's in R3: R3C4: no 7
75a. R3C4 = 5 : R3C4 <> 7
75b. R3C7 = 5 -> R3C8 = 7: R3C4 <> 7

76. R9C1: no 3
76a. R9C2 = 3: R9C1 <> 3
76b. R9C2 = 1 -> R3C8 = 1 -> R3C5 = 3 -> R2C4 = 7 -> R2C1 = 3: R9C1 <> 3
76c. R9C2 = 3(hidden)
76d. Naked Pair {12} in R89C1 -> locked for C1

77. 11(3) at R2C2 = {2[3]6}/{45}[2]: R2C3: no 2,6

78. 16(3) at R4C1 = {69}[1]/[781]/[439]/[934]/[7]{45}: [739] blocked by R2C1; [3]{49} blocked by R5C7: R4C1: no 3; R5C2: no 6

79. 15(3) at R1C1 = [5]{19}/[6]{18}/[8]{16}/[9]{15}
79a. 45 on C1: 1 innie and 3 outies: R1C1 + 12 = R357C2
79b. [5]-[458/746/845]
79c. [6]-[495/756/846/945]: [918] blocked by 15(3) at R1C1
79d. [8]-[758/956]
79e. [9]-[498/795]
79f. R5C2: no 1

80. Hidden Pair {12} in N4 in R6C23 = {12} -> locked for R6
80a. Clean up: R6C8: no 8

81. R6C1 = 8(hidden); R7C3 = 8(hidden)

82. R6C89 = {45} -> locked for N6
82a. R5C7 = 9
82b. Clean up: R5C8: no 6

83. R4C8 = 6(hidden)

84. 16(3) at R4C1 = [934]/[7]{45}: no 6; R4C1: no 4; 4 locked for N4 and R5
84a. R5C3 = 6(hidden); R3C3 = 2; R23C9 = [26]; R6C23 = [21]
84b. R1C2 = 1(hidden); R3C2 = 8(hidden)

85. R1C13 = {59}(last combo in 15(3)) -> locked for R1 and N1

And the rest is naked singles.

greetings

Para


Last edited by Para on Wed Nov 07, 2007 10:50 pm; edited 6 times in total
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