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Hi,

there is no S.S score mentionned here for this one, and it's not a forgetting, because I
am tired with questions of rating . I'd like you feel free about all kind of moves you
could use with this puzzle. The considerations of rating deprive me of pleasure with solving killers,
since I generally search for the most direct and neat road solving killer, without other goal. But no solving
path or approach is better than another one, so please, no competition for a "best" WT.

The most important : have pleasure with this puzzle.








3x3::k:4096:2049:3074:3074:2052:1285:1285:3335:3848:2313:4096:2049:2060:2052:3598:3335:3848:3089:3858:2313:4096:2060:7702:3598:3848:3089:3866:3858:3858:3869:7702:7702:7702:4641:3866:3866:3108:3108:3869:7702:7720:7702:4641:2347:2347:2093:3108:3869:7720:7720:7720:4641:2347:3381:2093:2871:3384:9017:7720:9017:4412:1853:3381:2871:3384:3393:9017:9017:9017:2885:4412:1853:3384:3393:2122:2122:9017:1357:1357:2885:4412:


Solution :

Hi manu

I've noted your comments about rating. For some time now I've been posting my rating comments in hidden text and leaving them like that when I converted my walkthroughs from TT to normal text. I've been doing that after Ed asked us to post rating comments in hidden text; I hope all others will do the same. Then Ed, manu or anyone else who doesn't want to see rating comments can avoid them. It also means that hints about the types of moved used for a walkthrough will be hidden.

I hope this is acceptable.

Let's forget about ratings and forget about elegancy.

Bottom line: it's a very enjoyable puzzle. Thanks!




Edited: revised and shortened step 3

_________________
ADYFNC HJPLI BVSM GgK Oa m

Thanks for this WT.



Why about elegancy ? Being "so what" about the ratings does not mean forgetting about elegancy .

Your WT is really different to mine, and there is an interesting move you did not seem to have seen, which could avoid you some combo analysis

Happy that you like this assassin, and that several solving paths are possible here : this is a pleasure for a puzzle maker who does not like competition.

I take it from what you said above that you wanted to stop talks about elegancy, as in no WT is better/more elegant than others.


First I can't see I've used any combo analysis in my walkin. (For me "combo analysis" means analysing three or more combinations of 4-cell or 5-cell given/hidden cages, and in my walkin I only analysed 2-cell or 3-cell given/hidden cages, and never had more than two possible combinations in all analyses.)

Also, if "no solving path or approach is better than another one", then all my moves should be of the same interesting level as all yours, and you also haven't seen the interesting moves I found before reading my walkin, which could avoid your whatever analyses.

_________________
ADYFNC HJPLI BVSM GgK Oa m

You did not understand. Nobody can say that a walkthrough is better than another one. I did not like the way you have criticized Andrew's WT. It is not a question of elegancy : elegancy is only a personal point of view.End.




In this forum, you must accept ideas different of yours. Yes, I think you have done combo analysis, because the way you explain your steps is sometimes hazy, and I consider that what you have done is a kind of combo analysis. But it's not a criticism. I have no problem with combo analysis. To be honest, there are very nice moves in your WT that I was not able to find. I have only said my approach was different to yours : not better. Why don't you accept that ??



I do agree ! I have never said my moves were more interesting than yours !





Why are you so aggressive ??? Once again, I did not say my WT was better to yours. It's a different one. Yes, there is sthg interesting with this puzzle you have not used. Yes, I had not found some of the nice moves you have written in your WT. No WT is better to another one.

I cannot accept the word "whatever analyses" you have used, unless you are a kind of "show off" : in this case, your place should not be in this forum.

END OF DISCUSSION

Thanks for this Assassin, manu!

I used the same cracking moves as udosuk though with a different reasoning.

Edit: Fixed a mistake in former step 4a. Thanks for finding this, Andrew!

A177 Walkthrough:

1. R789+N5 !
a) Outies N5 = 15(2) = {69/78}
b) Innies N8 = 10(3) = 1{27/36} since R7C5 >= 6 -> R9C46 = 1{2/3} -> 1 locked for R9+N8
c) ! Innies N9 = 10(2) = [64/82] because {37} blocked by Killer pair (37) of Innies N8
d) Innies R789 = 15(3) = [168/276] since R7C59 = (678) -> R7C1 = (12); 6 locked for R7
e) ! 7(2) <> {16} since it's a Killer pair of Innies R789 (R7C1 = 1 xor R7C9 = 6)
f) 1 locked in 17(3) @ N9 = {179} locked for N9
g) 13(2) @ N9 = [58/76]
h) Innies+Outies N89: 6 = R6C9+R9C3 - R7C5: R7C5 = (67) -> R6C9+R9C3 = 12/13 = [57/75/76] <> 2,3 -> CPE: R6C3+R9C9 <> 7

2. R789+C789 !
a) Outies N6 = 10(2) = [28/46]
b) Innies N3 = 5(2) = [14/32]
c) ! Outies N1245 = 10(2+1) must have one of 1 since R7C5 >= 6 and R1C7 <> 2 -> CPE: R1C1+R7C7 <> 1
d) R9C9 = 9, R7C7 = 7, R7C5 = 6, R7C9 = 8 -> R6C9 = 5
e) Innie R789 = R7C1 = 1 -> R6C9 = 7
f) Innie N7 = R9C3 = 7 -> R9C4 = 1
g) 5(2) = {23} -> R9C7 = 2, R9C6 = 3

3. R123
a) Outie N4 = R3C1 = 4
b) Innie N1 = R1C3 = 8 -> R1C4 = 4
c) 5(2) = {23} -> R1C6 = 2, R1C7 = 3
d) Outie N5 = R3C5 = 9
e) 8(2) @ R2C4 = {35} locked for C4+N2
f) 8(2) @ R1C5 = {17} locked for C5

4. C789
a) R3C9 = 2 -> 15(3) = 2[67/76/94]
b) Killer pair (46) locked in 15(3) + 9(3) for N6
c) 18(3) = {189} locked for C7+N6
d) Hidden Single: R1C9 = 1 @ C9 -> 15(3) = 1[86/95]
e) 15(3) @ R3C9 = {267} -> 6,7 locked for R4+N6
f) Hidden Single: R4C9 = 6 @ C9, R2C9 = 7 @ C9 -> R3C8 = 5
g) 11(2) = {56} -> R9C8 = 6, R8C7 = 5

5. C123
a) 13(2) = {49} -> R9C2 = 4, R8C3 = 9
b) 15(3) = {456} locked for C3+N4
c) 16(3) = {169} since R3C3 = (13) -> R3C3 = 1; 6 locked for N1
d) 8(3) @ N1 = {35} -> R1C2 = 5, R2C3 = 3
e) 11(2) = {38} -> R7C2 = 3, R8C1 = 8

6. N58
a) 35(6) = {245789} -> R9C5 = 8, R7C6 = 5, R7C4 = 9
b) 30(5) = 689{25/34} -> R6C6 = 9, R6C4 = 8

7. Rest is singles.

Rating: Easy 1.25. I used two unusual Killer pairs.

Thanks manu for a really challenging Assassin. I found it much harder than udosuk and Afmob and, having now read their posts, I see that I missed their nice interactions between Innies for R789 and the 7(2) cage in N9 and I also missed Outies for N1245. Without those steps I think it's a much harder puzzle. However it did allow me to make some interesting "clone" steps.


Here is my walkthrough.

Prelims

a) 8(2) cage in N1 = {17/26/35}, no 4,8,9
b) R1C34 = {39/48/57}, no 1,2,6
c) R12C5 = {17/26/35}, no 4,8,9
d) R1C67 = {14/23}
e) 13(2) cage in N3 = {49/58/67}, no 1,2,3
f) 9(2) cage in N1 = {18/27/36/45}, no 9
g) R23C4 = {17/26/35}, no 4,8,9
h) R23C6 = {59/68}
i) 12(2) cage in N3 = {39/48/57}, no 1,2,6
j) R67C1 = {17/26/35}, no 4,8,9
k) R67C9 = {49/58/67}, no 1,2,3
l) 11(2) cage in N7 = {29/38/47/56}, no 1
m) 7(2) cage in N9 = {16/25/34}, no 7,8,9
n) 13(2) cage in N7 = {49/58/67}, no 1,2,3
o) 11(2) cage in N9 = {29/38/47/56}, no 1
p) R9C34 = {17/26/35}, no 4,8,9
q) R9C67 = {14/23}
r) 9(3) cage in N6 = {126/135/234}, no 7,8,9
s) 35(6) cage in N8 = {146789/236789/245789/345689}, 8,9 locked for N8

1. 45 rule on C123 2 outies R19C4 = 5 = [32/41], clean-up: R1C3 = {89}, R9C3 = {67}
1a. Killer pair 3,4 in R1C4 and R1C67, locked for R1, clean-up: no 5 in R2C3, no 5 in R2C5, no 9 in R2C7
1b. Killer pair 1,2 in R9C4 and R9C67, locked for R9, clean-up: no 9 in R8C7
1c. 13(2) cage in N7 = {49/58} (cannot be {67} which clashes with R9C3), no 6,7

2. 45 rule on N1 2 innies R1C3 + R3C1 = 12 = [84/93]
2a. Max R3C1 = 4 -> min R4C12 = 11, no 1

3. 45 rule on N7 2 innies R7C1 + R9C3 = 8 = [17/26], clean-up: R6C1 = {67}

4. 45 rule on N3 2 innies R1C7 + R3C9 = 5 = {14/23}
4a. Max R3C9 = 4 -> min R4C89 = 11, no 1

5. 45 rule on N9 2 innies R7C9 + R9C7 = 10 = [64/73/82/91], clean-up: no 8,9 in R6C9

[And now the 45 I spotted when I was setting up and colouring the diagram, although I only saw the CPE when I reached this step.]
6. 45 rule on N5 2 outies R37C5 = 15 = [87/96]
6a. 8,9 in N2 locked in R3C5 + R23C6, CPE no 8,9 in R45C6

7. 45 rule on N8 3 innies R7C5 + R9C46 = 10 = {127/136}, no 4, 1 locked for R9 and N8, clean-up: no 9 in R7C9 (step 5), no 4 in R6C9

8. 45 rule on C789 2 outies R19C6 = 5 = [23/32/41], no 1 in R1C6, clean-up: no 4 in R1C7, no 1 in R3C9 (step 4)

9. 45 rule on R123 3 innies R3C159 = 15 = {249/348}, 4 locked for R3, clean-up: no 5 in R2C1, no 8 in R2C9

10. 45 rule on R789 3 innies R7C159 = 15 = {168/267}, 6 locked for R7, clean-up: no 5 in R8C1, no 1 in R8C9
10a. 7 of {267} must be in R7C5 (R7C159 cannot be [267] because R6C19 cannot be [66]), no 7 in R7C9, clean-up: no 6 in R6C9, no 3 in R9C7 (step 5), no 2 in R9C6, no 3 in R1C6 (step 8), no 2 in R1C7, no 3 in R3C9 (step 4)

11. 45 rule on R789 2 outies R6C19 = 1 innie R7C5 + 6
11a. R7C6 = {67} -> R6C19 = 12,13 = {57/67}, 7 locked for R6

12. Hidden pair 1,2 in R1C7 + R3C9 and 15(3) cage for N3, R1C7 + R3C9 contains one of 1,2 -> 15(3) cage must contain one of 1,2 = {159/168/258/267} (cannot be {249} which clashes with R3C9), no 3,4

13. Combined cage R1239C4 = 13 = {1237/1345} (cannot be {1246} which clashes with R1C6 because {1246} must be 4{26}1), no 6, 1,3 locked for C4, 3 also locked for N2, clean-up: no 5 in R1C5, no 2 in R23C4
13a. Killer pair 6,7 in R12C5 and R7C5, locked for C5

14. 45 rule on C12 4 outies R2378C3 = 15 = {1239/1248/1257/1347/1356/2346}
14a. 9 of {1239} must be in R8C3 -> no 9 in R37C3
14b. 5 of {1257/1356} must be in R8C3 -> no 5 in R37C3

15. 45 rule on N6 3 innies R4C89 + R6C9 = 18 = {279/378/459/567} (cannot be {369/468} because R6C9 only contains 5,7)
15a. 18(3) cage at R4C7 = {189/378/459/468} (cannot be {279/369/567} which clash with R4C89 + R6C9), no 2

16. R1C3 + R3C1 = [84/93]
16a. 16(3) cage in N1 = {169/178/259/268/457} (cannot be {349} which clashes with R3C1, cannot be {358} which clashes with R1C3 + R3C1, cannot be {367} which clashes with 8(2) cage), no 3

17. 45 rule on C89 4 outies R2378C7 = 22 = {1579/1678/2569/4567} (other combinations clash with 18(3) cage at R4C7), no 3, clean-up: no 8 in R9C8

18. 7(2) cage in N9 = {16/25/34}, R7C9 + R9C7 (step 5) = [64/82] -> combined cage R7C89 + R8C9 + R9C7 = {1628/2546/3428}, 2 locked for N9, clean-up: no 9 in R9C8
18a. 9 in N9 locked in 17(3) cage = {179/359}, no 4,6,8

[At this stage I remembered that this Assassin had been created by manu. First I spotted “clones” connected with the 30(5) and 30(6) cages in N5, R3C5 must equal one of R6C46 and R7C5 must equal one of R45C46, but I was never able to use these.
Then I spotted that the “clones” around the edges were more helpful at this stage.]

19. 45 rule on N69 1 outie R3C9 = 1 innie R9C7
19a. 2 in C7 locked in R39C7, R3C9 = R9C7 -> 2 locked in R3C79, locked for R3 and N3, clean-up: no 7 in R2C1
19b. 15(3) cage in N3 (step 12) = {159/168/258/267}
19c. 2 of {267} must be in R3C7 -> no 7 in R3C7

20. 45 rule on N7 1 outie R6C1 = 1 innie R9C3
20a. 7 in R6 locked in R6C19 (step 11a), R6C1 = R9C3 -> 7 locked in R6C9 + R9C3, CPE no 7 in R9C9

21. 7 in N5 locked in 30(6) cage at R3C5
21a. 30(5) cage at R5C5 = {15789/24789/25689/34689/45678} (cannot be {35679} because not possible to make 30(6) cage at R3C5 with 1,2,4,7), 8 locked for N5

22. R7C5 = {67} -> R6C19 = 12,13 = {57/67} (step 11a)
22a. R7C5 = 6 => no 6 in R6C46
22b. R7C5 = 7 => R6C19 = 13 = {67}, locked for R6
22c. -> no 6 in R6C46

23. 30(5) cage at R5C5 = {15789/24789/25689/34689} (cannot be {45678} because 6,7 only in R7C5), 9 locked for N5

24. R3C9 = R9C7 (step 19) = {24}
24a. All cells in N6 except for R56C8 “see” at least one of R3C9 or R9C7 -> R56C8 must contain at least one of 2,4 -> 9(3) cage in N6 = {126/234} (cannot be {135} which doesn’t contain 2 or 4), no 5, 2 locked for N6

25. R4C89 + R6C9 (step 15) = {378/459/567}
25a. 18(3) cage at R4C7 (step 15a) = {189/378/459} (cannot be {468} which clashes with R4C89 + R6C9), no 6

26. 6 in C7 locked in R2378C7 (step 17) = {1678/2569/4567}
26a. 2 of {2569} must be in R3C7 -> no 9 in R3C7

27. 45 rule on N1 1 outie R1C4 = 1 innie R3C1 = {34}
27a. All cells in N3 except for R2C789 “see” either R1C4 or R3C1 -> R2C789 must contain one of 3,4 (R2C789 cannot contain both because R1C7 + R3C9 contains one of 3,4)
27b. Killer pair 3,4 in R1C7 + R3C9 and R2C789, locked for N3, clean-up: no 9 in R2C9
[Note that 3,4 have already been eliminated from R2C8 but the logic above reads better when saying R2C789.]

28. 16(3) cage in N1 (step 16a) = {169/178/268/457} (cannot be {259} because R3C3 only contains 1,6,7,8)
28a. 4 of {457} must be in R2C2 -> no 5 in R2C2

29. 45 rule on N7 1 outie R9C4 = 1 innie R7C1 = {12}
29a. All cells in N9 except for R8C789 “see” either R7C1 or R9C4 -> R8C789 must contain at least one of 1,2
[Similar comment to note after step 27, no 1,2 in R8C7.]

30. R8C89 contains at least one of 1,2
30a. R8C8 = 1 => 17(3) cage in N9 = {179}
30b. R8C9 = 2 => R7C8 = 5 => 17(3) cage in N9 (step 18a) = {179} (cannot be {359} when R7C8 = 5)
30c. -> 17(3) cage in N9 = {179} -> R9C9 = 9, R7C7 + R8C8 = {17}, locked for N9, clean-up: no 4 in R8C3, no 6 in R8C9, no 4 in 11(2) cage in N9

31. R7C159 (step 10) = {168/267}
31a. Killer pair 1,7 in R7C159 and R7C7, locked for R7, clean-up: no 4 in R8C1

32. 9 in C7 locked in 18(3) cage at R4C7, locked for N6
32a. 18(3) cage at R4C7 (step 25a) = {189/459}, no 3,7

33. R1C7 = 3 (hidden single in C7), R1C4 = 4, R1C3 = 8, R1C6 = 2, R9C6 = 3 (step 8), R9C7 = 2, R9C4 = 1, R9C3 = 7, R7C1 = 1 (step 29), R6C1 = 7, R6C9 = 5, R7C9 = 8, R7C7 = 7, R8C8 = 1, R7C5 = 6, R3C5 = 9 (step 6), R3C9 = 2 (step 4), R3C1 = 4 (step 2), clean-up (working across in nonets, rather than rows): no 1 in R1C2, no 6 in R2C3, no 1,5 in R3C2, no 7 in R23C4, no 5 in R23C6, no 6 in R1C8, no 5 in R2C7, no 7 in R3C8, no 4 in 18(3) cage at R4C7 (step 32a), no 5 in 7(2) cage in N9

34. Naked triple {189} in 18(3) cage at R4C7, locked for C7 and N6, clean-up: no 6 in 9(3) cage in N6 (step 24a)

35. R4C89 = {67} (hidden pair in N6), locked for R4

36. R2C7 = 4 (hidden single in C7), R1C8 = 9, R2C9 = 7, R3C8 = 5, R12C5 = [71], R3C7 = 6, R1C9 = 1, R2C8 = 8, R23C4 = [53], R23C6 = [68], R3C2 = 7, R2C1 = 2, R2C3 = 3, R1C2 = 5, R1C1 = 6, R2C2 = 9, R3C3 = 1, R4C89 = [76], R8C7 = 5, R9C8 = 6, R8C3 = 9, R9C2 = 4

and the rest is naked singles.

Hi,

thanks for yours WT's that show several interesting solving roads were possible here. I am happy when I make a puzzle for which its possible
Here is another way

The main step is 4)e). The general idea is the following : proving that all digits of a cage X(n) belong to a cage Y(n+1) . The last digit of Y(n+1) must be Y-X, and Y-X does not belong to the cage X(n). Here, X=5, Y=9, n=2, 5(2)=r19c7, 9(3) at n6.

(prelims from Sudocue)

1)Outies for n5 : r3c5+r7c5= 16 : {6789}

2)Innies for n8 : r7c5+r9c4+r9c6=10.
=> r7c5=(67), r9c46={12/13}
=> 1 locked at r9c46 for n8 and r9

3)Innies for n9 : r7c9+r9c7= h10(2) = [82/73/64]
=> (cage sum for 13(2)) :r6c9=(567)

4)a) Innies-outies for n6+n9 : r3c9=r9c7= (234) (cloned cells)
=> digit at r3c9=r9c7 locked for c8 and n6 at r56c8.
b) Innies for n3 : r1c7+r3c9=h5(2)=[14/23/32]
c) Innies-outies for n3 : r4c8+r4c9=10+r1c7, so :
=> r4c8 – r1c7 = 10 – r4c9 > 0 : r4c8 <> r1c7
=> r4c9 – r1c7 = 10 – r4c9 > 0 : r4c8 <> r1c7
d) Since r6c9 >=5 > r1c7, we deduce from step c) that digit at r1c7 is locked for n6 at cage 9(3)
e) We deduce from steps a)+d) that both digits of h-cage h5(2)n3 are locked at cage 9(3)n6: the third cell of 9(3)n6 is 9 – 5 = 4.
=> h5(2)n3 <> 4 : h5(2)n3={23} locked for n3
=> 9(3)n6 = {234} locked for n6

The puzzle is cracked now.


5)a) Cloned cells : r9c7=r3c9=(23)
=> naked pair {23} locked for c7 at r19c7
=> r7c9<>6, r6c9<>7
b) 5(2)n8={23} locked for r9
c) 8(2)n7=[71]
d) Innie for n7 : r7c1=1 => r6c1=7
e) Outie for n4 : r3c1=4.
f) Innie for n1 : r1c3= 8 => r1c4=4

6)a) 5(2)n2={23} locked for r1
b) 1 locked for n2 at r12c5 : r12c5=8(2)={17} locked for c5 and n2.
c) r7c5=6 => r9c6=3 (step 2)), r9c7=2=r3c9 (cloned), r1c67=[23]
d) Step 3) => r7c9=8. R6c9=5.
e) Last combo : 18(3)n6={189} locked for n6 and c7
f) 9(2)n1<>{18} : r3c2<>1
g) Hidden single for r3 : r3c3=1
=> 16(3)n1={169} locked for n1.
h) Last combos : 8(2)n1=[53], 9(2)n1=[27]

7) a) Outie for n5 : r3c5=9
=> 14(2)n2={68} locked for 2 and c6
b) Last combo : 15(3)n3={267}, {67} locked for r4+n6
c) 7(2)n9={34} locked for n9
=> 11(2)n9= {56} locked for n9
=> 17(3)n9=[719]
d) 30(5)n5 <> 1 since {15789} is impossible since r7c5=6
e) 4 locked for n4 and c3 at r456c3

Singles for the rest