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Hi folks,

Wow, 7 weeks running without any Assassin being published! To partially make up for the loss, I'm posting no less than three versions here (all of which have been manually vetted of course ). Fortunately, detailed walkthroughs are only "required" for two of them (see below).


Assassin 188 (A188) (SudokuSolver v3.3.1 score: 1.45)

Like its offshoots, you'll get out of this Assassin if you can get into it. The proviso is not to be taken light-heartedly, especially for those who haven't been keeping up with all those great posts on Killer techniques...




Solution:

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



Assassin 188 Pencilmark Version (A188PM) (SudokuSolver v3.3.1 score: 1.40)

Despite its SS score, this version is intended to be solvable on paper without keeping track of every last candidate. In other words, it's for those people who prefer insertion solving to elimination solving. As such, a solving outline (rather than a detailed walkthrough) should be sufficient here. Relax and enjoy!




Solution:

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



Assassin 188 V2 (A188V2) (SudokuSolver v3.3.1 score: )

SudokuSolver insists that T&E is necessary for this one. Needless to say, I disagree...




Solution:

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



Good luck on finding that first placement!

Good to see you again, Mike!

Both A188 and A188 PM were fun to solve. I might post the crucial moves for A188 PM if no one else does.

A188 Walkthrough:

1. N3
a) Innies N3 = 15(3) can only have one of (123) -> R3C78 <> 1,2,3
b) Hidden Killer pair (12) in R1C7 for N3 since 5(2) can only have one of them
c) 6(3) = {123} -> 3 locked for N2+C6; CPE: R1C4 <> 2

2. R789
a) 17(2) = {89} locked for R9+N7
b) 13(2) = {67} locked for R8+N7
c) 7(2) <> 1
d) Innies R89 = 9(2) = {45} locked for R8
e) 11(2) <> 2,3
f) 19(5) = 123{49/58} since 145{27/36} blocked by R8C9 = (45) and {12367} blocked by
Killer pair (67) of 11(2); 3 locked for N9
g) Killer pair (45) locked in R8C9 + 19(5) for N9

3. R789
a) 7(2): R7C1 = (23)
b) 4,5 locked in 27(6) @ R6 for 27(6)
c) 11(2): R9C6 = (45)
d) Innies+Outies N78: -5 = R6C4 - R89C6: R6C4 <> 3,6 since R89C6 can't be 8 or 11

4. C456 !
a) 10(2) <> 4,6
b) ! Outies N6789 = 14(3) = {149/158/167/248/257}
c) Innies C5 = 5(2) = {14/23}; R7C5 <> 2
d) ! 16(3) <> 4 because {349} blocked by Killer pair (34) of Innies C5 and (457) is a Killer triple of Innies N6789
e) 4 locked in Innies C5 = 5(2) = {14} locked for C5
f) 10(2) = [37/82]

5. R789
a) 9 locked in 19(5) @ R8 = {12349} -> 4 locked for R9+N9
b) R9C6 = 5 -> R9C7 = 6, R8C9 = 5, R8C1 = 4 -> R7C1 = 3

6. R123
a) Innies R12 = 7(2) = [52/61]
b) 5(2): R3C9 = (34)
c) R2C6 <> 1,2 since it sees all 1,2 of N3
d) R2C6 = 3
e) Hidden Single: R3C9 = 3 @ N3 -> R2C9 = 2
f) R1C7 = 1, R1C6 = 2
g) Innie R12 = R2C1 = 5
h) 12(2) = {48} locked for R2+N3
i) 13(2) = {67} locked for R1+N3
j) 11(2) = {38} -> R1C4 = 8, R1C3 = 3
k) 14(2) = {59} -> R2C5 = 9, R1C5 = 5

7. R456+C9
a) 34(6) = {145789} -> R4C7 = 8; 9 locked for R3; R3C456 = {147} locked for R3+N2
b) 16(3) = {367} locked for C5+N5
c) Innes C9 = 20(3) = {479} -> R1C9 = 7, R9C9 = 4, R4C9 = 9
d) 13(2) = {49} -> R5C7 = 4, R5C6 = 9
e) 15(3) = {159} -> R4C7 = 5, R4C8 = 1
f) 5 locked in 21(3) @ R6 = {579} locked for R6+N4

8. Rest is singles.

Wonderful to have an Assassin from you Mike! My solution is quite different to Afmob's conventional way....but we end up in the same breakthrough area. At least my solution fits Mike's cryptic intro better . I've just done to the breakthrough.

Part walkthrough for A188
Cheers
Ed

Here is my partial walkthrough for A188.

Welcome back Mike! Many thanks for a challenging Assassin on your return.

As will be seen from my walkthrough, I didn't get the underlined part of the hint but only the part about killer techniques, which I consider refers to Afmob's step 2f and my step 11.

Afmob and Ed solved A188 very differently, with Afmob's steps 4b and 4d (I was impressed with step 4b since it doesn't give any eliminations but forms the basis for step 4d) and Ed's steps 4,5 and the more complicated 9,10. I struggled at one stage but eventually found another different breakthrough in my step 20.


Here is my walkthrough for A188.

Prelims

a) R1C34 = {29/38/47/56}, no 1
b) R12C5 = {59/68}
c) R1C89 = {49/58/67}, no 1,2,3
d) R2C78 = {39/48/57}, no 1,2,6
e) R23C9 = {14/23}
f) R5C34 = {16/25/34}, no 7,8,9
g) R5C67 = {49/58/67}, no 1,2,3
h) R78C1 = {16/25/34}, no 7,8,9
i) R8C23 = {49/58/67}, no 1,2,3
j) R89C5 = {19/28/37/46}, no 5
k) R9C12 = {89}
l) R9C67 = {29/38/47/56}, no 1
m) 6(3) cage at R1C6
n) 21(3) cage at R6C1 = {489/579/678}, no 1,2,3
o) 11(3) cage at R8C4 = {128/137/146/236/245}, no 9
p) 14(4) cage at R2C1 = {1238/1247/1256/1346/2345}, no 9
q) 19(5) cage at R8C6 = {12349/12358/12367/12457/13456}

Steps resulting from Prelims
1a. Naked pair {89} in R9C12, locked for R9 and N7, clean-up: no 4,5 in R8C23, no 1,2 in R8C5, no 2,3 in R9C67
1b. Naked pair {67} in R8C23, locked for R8 and N7, clean-up: no 1 in R78C1, no 3,4 in R9C5
1c. Naked triple 1,2,3 in 6(3) cage at R1C6, CPE no 2,3 in R1C4, clean-up: no 8,9 in R1C3
1d. 19(5) cage at R8C6 must contain 1, CPE no 1 in R8C9

2. 45 rule on C1 3 innies R169C1 = 24 = {789}, locked for C1

3. 45 rule on C9 3 innies R149C9 = 20 = {389/479/569/578}, no 1,2

4. 45 rule on C5 2 innies R37C5 = 5 = {14/23}

5. 45 rule on N3 3 innies R1C7 + R3C78 = 15 = {159/168/258/267/357} (cannot be {249/348} which clash with R23C9, cannot be {456} because R1C7 only contains 1,2,3), no 4
5a. R1C7 = {123} -> no 1,2,3 in R3C78
5b. Hidden killer pair 1,2 in R1C7 and R23C9, R23C9 contains one of 1,2 -> R1C7 = {12}

6. 3 in 6(3) cage at R1C6 only in R12C6, locked for C6 and N2, clean-up: no 2 in R7C5 (step 4)

7. 45 rule on C6789 1 innie R7C6 = 2 outies R3C34 + 1
7a. Min R3C34 = 5 (cannot be {12} which clashes with R12C6) -> min R7C6 = 6
7b. Max R3C34 = 8 -> no 8,9 in R3C3
7c. R3C34 must contain one of 1,2
7d. Killer pair 1,2 in R12C6 and R3C34, locked for N2

8. 34(6) cage at R3C4 can only contain one of 1,2 -> no 1,2 in R4C6

9. 45 rule on R12 1 innie R2C1 = 1 outie R3C9 + 2, no 1,2 in R2C1

10. 45 rule on R89 2 innies R8C19 = 9 = {45} (only remaining combination), locked for R8, clean-up: no 4,5 in R7C1, no 6 in R9C5

11. 19(5) cage at R8C6 = {12349/12358} (cannot be {12367} which clashes with R9C67, cannot be {12457/13456} which clash with R8C9 because all cells of the 19(5) cage “see” R8C9), no 6,7, 3 locked for N9
11a. Killer pair 4,5 in R8C9 and 19(5) cage (in R9C89), locked for N9, clean-up: no 6,7 in R9C6
11b. Killer pair 4,5 in R9C6 and 19(5) cage (in R9C89), locked for R9
11c. 4,5 in R7 only in R7C23456, locked for 27(6) cage at R6C4, no 4,5 in R6C4

12. Killer triple 3,4,5 in R23C9, R8C9 and R9C9, locked for C9, clean-up: no 8,9 in R1C8
12a. 5 in C9 only in R89C9, locked for N9

13. 45 rule on N78 2 innies R89C6 = 1 outie R6C4 + 5
13a. R8C6 = {1289}, R9C6 = {45} -> R89C6 cannot total 8,11 -> no 3,6 in R6C4

14. 16(3) cage at R4C5 = {259/268/367/457} (cannot be {169/358} which clash with R12C5, cannot be {178} which clashes with R89C5, cannot be {349} which clashes with R37C5), no 1

15. Hidden killer pair 6,7 in R9C45 and R9C7 for R9, R9C7 = {67} -> R9C45 must contain one of 6,7
15a. Hidden killer pair 6,7 in R7C46 and R9C45 for N8, R9C45 contains one of 6,7 -> R7C46 must contain one of 6,7
15b. 27(6) cage at R6C4 must contain both of 4,5 (step 11c) =
{124569/124578/134568} (cannot be {234567} because 2{34567} clashes with R9C45 and 7{23456} clashes with R7C1)
15c. 6,7 must be in R7C46 -> no 7 in R6C4

16. 34(6) cage at R3C4 = {145789/245689}
16a. R1C7 + R3C78 (step 5) = {159/168/258} (cannot be {267} because 34(6) cage at R3C4 only contains one of 6,7), no 7
16b. R1C89 = [49/67/76] (cannot be [58] which clashes with R1C7 + R3C78), no 5,8
16c. R1C34 = [29/38/56/65] (cannot be {47} which clashes with R1C89), no 4,7

17. 45 rule on N23 2 innies R12C4 = 1 outie R4C6 + 6
17a. R12C4 cannot total 11 (4,7 only in R2C4 and {56} clashes with R12C5) -> no 5 in R4C6
17b. 5 in 34(6) cage at R3C4 (step 16) only in R3C4678, locked for R3

18. Killer triple 7,8,9 in R1C1, R1C345 and R1C89 for R1 (R1C5 must be {89} if R1C34 = {56}), locked for R1

19. 8 in R1 only in R1C145, CPE no 8 in R2C4

[At this stage I was stuck for a while, then I looked at combinations for the 20(4) cage at R5C9 and interactions in C9, followed by gradually improving on that.
First I saw
45 rule on C9 2 innies R49C9 = 1 outie R1C8 + 7
R1C8 = {467} -> R49C9 = 11,13,14 = [65/74/85/94/95] (cannot be [83] which clashes with R23C9 = {23} when R1C8 = 4), no 3 in R9C9
and at that time I saw the alternative
R149C9 (step 3) cannot be {389} = [983] => R23C9 = {14} clashes with R1C89, CCC.
I continued using that result but hoped to avoid using a “chainy” step and later found the following]
20. R1C89 = [49]/{67}, R23C9 = {14/23} -> combined cage R1C89 + R23C9 = [49]{23}/{67}{14}/{67}{23}
20a. R149C9 {step 3) = {479/569/578} (cannot be {389} = [983] which clashes with R1C89 + R23C9 = [49]{23}, CCC), no 3

21. Naked pair {45} in R89C9, locked for C9 and N9, clean-up: no 1 in R23C9, no 3,6 in R2C1 (step 9)
21a. Naked pair {23} in R23C9, locked for C9 and N3 -> R1C7 = 1, clean-up: no 9 in R2C78
21b. Naked pair {23} in R12C6, locked for C6 and N2, clean-up: no 3 in R7C5 (step 4)
21c. Naked pair {14} in R37C5, locked for C5, clean-up: no 9 in R8C5
21d. Naked pair {45} in R28C1, locked for C1
21e. 1 in N2 only in R3C45, locked for R3
21f. 1 in C1 only in R45C1, locked for N4, clean-up: no 6 in R5C4

22. 9 in R8 only in R8C678 -> 19(5) cage at R8C6 (step 11) = {12349} (only remaining combination), no 5,8 -> R9C9 = 4, R8C9 = 5, R9C6 = 5, R9C7 = 6, R8C1 = 4, R7C1 = 3, R2C1 = 5, R3C9 = 3 (step 9), R2C9 = 2, R12C6 = [23], clean-up: no 6,9 in R1C4, no 9 in R1C5, no 7 in R2C78
22a. 19(5) cage at R8C6 = {12349}, 2 locked for N9
22b. 2 in R7 only in R7C234, locked for 27(6) cage at R6C4, no 2 in R6C4

23. Naked pair {48} in R2C78, locked for R2 and N3, clean-up: no 6 in R1C5, no 9 in R1C9
23a. Naked pair {67} in R1C89, locked for R1 and N3 -> R1C3 = 3, R1C4 = 8, R1C5 = 5, R2C5 = 9, R1C12 = [94], R9C12 = [89], R6C1 = 7, clean-up: no 4 in R5C4
23b. Naked pair {59} in R3C78, locked for 34(6) cage at R3C4, no 9 in R4C6

24. R6C1 = 7 -> R6C23 = 14 = [59/68/86], no 4,5 in R6C3

25. 11(3) cage at R8C4 = = {137} (only remaining combination) -> R9C4 = 7, R8C4 = 3, R9C3 = 1, R2C4 = 6, R2C23 = [17], R89C5 = [82], R9C8 = 3
25a. R7C6 = 6 (hidden single in N8)

26. Naked pair {14} in R3C45, locked for 34(6) cage at R3C4 -> R3C6 = 7, R4C6 = 8

27. R5C6 = {49} -> R5C67 = {49}, locked for R5

28. R45C4 = {25} (hidden pair in C4) -> R5C34 = {25}, locked for R5

29. R8C9 = 5 -> 20(4) cage at R6C9 = {1568} (only remaining combination), locked for C9 -> R1C89 = [67], R4C9 = 9, R5C67 = [94]
29a. R4C9 = 9 -> R4C78 = 6 = [51]

and the rest is naked singles.

Is Franks methodology OK? Or is picking a "weak spot" and showing there is only one way "trial and error" or something similar? I solved the insertion version in a similar way. The 9 cage in nonet 3 is "weak" and can only be 54 or 36.Actually only 36 works.That means the 13 cage in the same nonet can only be 58 or 49.Actually only 49 works after which the puzzle is straightforward.But is it "trial and error"?

That's a tough question but I try to answer it. It depends on how long the chain to reach a contradiction is, so if one can reach the contradiction by easy and short means it's alright. We (at least Andrew and me) often call such moves "combo analysis". But I try to avoid these types of moves, when other (easier) moves like in A188 and A188 PM are available.

But I can't tell from both of your messages whether they are T & E or not since they are too imprecise.

By the way, here is my way to crack A188 PM.

I won't disagree with that, since it should be possible to find Afmob's step 2c or my steps 10 and 10a that way. However I decided to use elimination solving, which I feel is the easier way.


Here is my walkthrough for A188PM; I decided to give the full walkthrough even though Mike indicated that an outline could be considered to be enough. Mike has pointed out that my original step 10 was incorrect. I think I had doubts about it at the time but had somehow (incorrectly) convinced myself about it. I've now re-worked step 10, using a move similar to Afmob's step 2c, then there are minor changes to steps 11 and 12 and to my final step. After re-working step 10 I can understand better why Mike called this puzzle a pencil mark version; this key step can be found that way.

Prelims

a) R1C34 = {19/28/37/46}, no 5
b) R12C5 = {39/48/57}, no 1,2,6
c) R1C89 = {49/58/67}, no 1,2,3
d) R2C78 = {18/27/36/45}, no 9
e) R23C9 = {19/28/37/46}, no 5
f) R5C34 = {49/58/67}, no 1,2,3
g) R5C67 = {39/48/57}, no 1,2,6
h) R78C1 = {69/78}
i) R8C23 = {12}
j) R89C5 = {17/26/35}, no 4,8,9
k) R9C12 = {49/58/67}, no 1,2,3
l) R9C67 = {49/58/67}, no 1,2,3
m) 8(3) cage at R1C6 = {125/134}
n) 10(3) cage at R4C5 = {127/136/145/235}, no 8,9
o) 11(3) cage at R6C1 = {128/137/146/236/245}, no 9
p) 11(3) cage at R8C4 = {128/137/146/236/245}, no 9
q) 12(4) cage at R2C1 = {1236/1245}, no 7,8,9
r) 35(5) cage at R1C1 = {56789}

Steps resulting from Prelims
1a. Naked pair {12} in R8C23, locked for R8 and N7, clean-up: no 6,7 in R9C5
1b. R9C12 = {49/58} (cannot be {67} which clashes with R78C1), no 6,7
1c. Killer pair 8,9 in R78C1 and R9C12, locked for N7
1d. 12(4) cage at R2C1 = {1236/1245}, 1,2 locked for C1
1e. 8(3) cage at R1C6 = {125/134}, CPE no 1 in R1C4, clean-up: no 9 in R1C3
1f. 35(5) cage at R1C1 = {56789}, CPE no 5,6 in R2C1

2. 11(3) cage at R8C4 = {137/146/236/245} (cannot be {128} because 1,2 only in R9C4), no 8
2a. 1,2 only in R9C4 -> R9C4 = {12}

3. 45 rule on R12 2 innies R2C19 = 3 = {12}, locked for R2, clean-up: no 7,8 in R2C78, R3C9 = {89}
3a. 8(3) cage at R1C6 = {125/134}
3b. 1 only in R1C67, locked for R1, clean-up: no 9 in R1C4
3c. 5 of {125} must be in R2C6 -> no 5 in R1C67

4. 45 rule on N3 3 innies R1C7 + R3C78 = 13 = {139/157/238/247} (cannot be {148} which clashes with R23C9, cannot be {256/346} which clash with R2C78), no 6

5. 45 rule on R89 2 innies R8C19 = 11 = [65/74/83], no 9, no 6,7,8 in R8C9, clean-up: no 6 in R7C1
5a. 9 in R8 only in R8C678, locked for 31(5) cage at R8C6, no 9 in R9C89

6. 45 rule on C5 2 innies R37C5 = 15 = {69/78}
6a. Hidden killer pair 8,9 in R12C5 and R37C5 for C5, R37C5 contains one of 8,9 -> R12C5 must contain one of 8,9 = {39/48} (cannot be {57} which doesn’t contain 8 or 9), no 5,7
6b. 7 in R2 only in R2C234, locked for 35(5) cage at R1C1, no 7 in R1C12

7. 45 rule on C1 3 innies R169C1 = 18 = {369/459} (cannot be {378} because 3,7 only in R6C1, cannot be{468/567} which clash with R78C1), no 7,8, 9 locked for C1, clean-up: no 6 in R8C1, no 5 in R8C9 (step 5), no 5 in R9C2
7a. 3 of {369} must be in R6C9 -> no 6 in R6C9

8. Naked pair {78} in R78C1, locked for N7, clean-up: no 5 in R9C1
8a. Naked pair {49} in R9C12, locked for R9 and N7

9. 11(3) cage at R6C1 = {137/146/236/245}, cannot be {128} because R6C1 only contains 3,4,5), no 8
9a. 3 of {137/236} must be in R6C1 -> no 3 in R6C23

[Mike pointed out that my original step 10 was flawed so I’ve re-worked it using a simpler, but not particularly obvious, step. After minor changes to the next two steps, I got back to my original steps again.]
10. 8(3) cage at R1C6 = {125} (cannot be [314/413] which clash with R12C5, cannot be [134/143] which clash with R2C78 which “see” both of R1C7 + R2C6) -> R2C6 = 5, R1C67 = {12}, locked for R1, clean-up: no 8 in R1C34, no 4 in R2C78, no 7 in R5C7, no 8 in R9C7

11. Naked pair {36} in R2C78, locked for R2 and N3, clean-up: no 9 in R1C5, no 7 in R1C89

12. Naked triple {789} in R2C234, locked for R2 and 35(5) cage at R1C1 -> R2C5 = 4, R1C5 = 8, clean-up: no 6 in R1C3, no 5 in R1C89, no 7 in R37C5 (step 6)

13. Naked pair {49} in R1C89, locked for R1 and N3 -> R3C9 = 8, R2C9 = 2, R1C67 = [21], R2C1 = 1, clean-up: no 6 in R1C4

14. Naked pair {56} in R1C12, locked for N1

15. Naked pair {57} in R3C78, locked for R3 and 30(6) cage at R3C4, no 7 in R4C6
15a. 30(6) cage at R3C4 = {135678} (only remaining combination) -> R3C5 = 6, R7C5 = 9, R4C6 = 8, R3C46 = {13}, locked for R3 and N2 -> R1C34 = [37], R2C4 = 9, clean-up: no 4,5,6 in R5C3, no 4 in R5C7, no 2 in R9C5, no 5 in R9C7

16. Naked pair {67} in R9C67, locked for R9 -> R9C3 = 5, R7C23 = [36], clean-up: no 3 in R8C5
16a. R9C3 = 5 -> R89C4 (step 2) = [42], R8C9 = 3, R8C1 = 8 (step 5), R7C1 = 7, R9C89 =[81], R9C5 = 3, R8C5 = 5, R7C6 = 1, R67C4 = [58], R5C4 = 6, R5C3 = 7, clean-up: no 5 in R5C7

17. Naked triple {127} in 10(3) cage at R4C5, locked for N5 -> R4C4 = 3

18. 11(3) cage at R6C1 (step 9) = {146/236} -> R6C2 = 6
18a. 1,2 only in R6C3 -> R6C3 = {12}
18b. Naked pair {12} in R68C3, locked for C3

19. 19(4) cage at R5C9 = {3457} (only remaining combination) -> R6C9 = 7, R57C9 = {45}, locked for C9 -> R1C89 = [49], R4C9 = 6

20. 12(4) cage at R2C1 = {1245} (only remaining combination), locked for C1 -> R6C1 = 3, R6C3 = 2 (step 18)

21. Naked pair {45} in R45C1, locked for C1 and N4 -> R3C1 = 2, R4C3 = 9

22. R4C2 = 1, clean-up: no 5 in R4C78
22a. R4C9 = 6 -> R4C78 = 6 = [42]

and the rest is naked singles.

As I think has been said before somewhere, it's T&E if you insert a number, chosen arbitrarily, to see what happens. But if there are few possibilities, e.g. 8/3 must be 134 or 125, then looking at the consequences of either choice is "enumerating the cases", which is a respectable technique in proving mathematical theorems (until someone comes up with a more elegant proof).

By the way, I only work by insertion, and avoid pencilling in more than three candidates for a square. I've never tried eliminating from nine candiates in each cell - I don't think I could read a display like that.

And thanks to MH for the puzzles - I found the two about equally hard, and similar in method.

cheers

_________________
Joe

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