SudokuSolver Forum

Assassin 123 "Roulette"





SSR: 2.51
"True" SSR: 1.26

Bear in mind all puzzles in this thread which I have posted or will post are valid ones. Use this information smartly this puzzle can be solved without any fish or chain (corresponding to the "true" SSR).

However, here is a v0.9 which can be solved without fishes/chains even if one doesn't use the info above.

(Warning: potentional spoiler for the default version)



Good luck! V2 will be posted when (if) this thread (ever) gets to page 2, or one week later.



(Added later: A hearty note of thanks to Ed, who helped immensely on the SSR department. Good on you mate!

Although personally I'm not into the "personal rating system", I don't oppose to anyone posting a (subjective) personal rating on this puzzle or his/her own walkthrough. )

Thanks for this week's Assassin!

It took me quite some time to find the first breakthrough move (step 2d) but after that I found the final cracking move (step 3d) in no time which is an interesting mix of IOU and combo analysis.

By the way, V2 shouldn't be posted later than this mid-week since in one week the next Assassin awaits.

A123 Walkthrough:

1. C456 !
a) 6(3) = {123} locked for R1+N2
b) 13(3) @ C5 must have one of (123) -> R4C5 = (123)
c) 24(3) = {789} locked for R9+N8
d) 17(3) @ C5 must have one of (789) -> R6C5 = (789)
e) 17(3) @ C5 <> 1 because R78C5 <> 7,9
f) 13(3) @ C5 <> 9 because R23C5 <> 1,3
g) Innies C5 = 15(3) must have one of (456) -> R5C5 = (456)
h) 13(3) @ C5 <> {238} since R23C5 <> 2,3
i) 17(3) @ C5 <> {278} since R78C5 <> 7,8
j) ! Hidden Killer triple (456) in 13(3) + R5C5 + 17(3) for C5
-> Each of them must have exactly one of (456)
-> 13(3) = {148/157/247} <> 3,6
-> 17(3) = {269/359/368} <> 4,7
k) Outies N2 = 7(2+1) <> 6,7,8,9

2. C123 !
a) 22(3) = 9{58/67} -> CPE: R5C2 <> 9
b) Innies+Outies C12: 7 = R1469C3 - R5C2; R5C2 = (45678)
- R5C2 <> 4 since R14C3 >= 9
- R1469C3 = 12/13/14/15(4) must have two of (123) -> R69C3 = (123)
- R1C3 <> 8,9 because R4C3 >= 5 and R4C3 <> 9 because R1C3 >= 4
c) 22(3) = 9{58/67} -> 9 locked for C2
d) ! Innies+Outies C12: 7 = R1469C3 - R5C2 -> R169C3 cannot be 7(3) = {124} (IOU @ N4)
e) Innies+Outies C12: 7 = R1469C3 - R5C2: R4C3 <> 8 since {1248} not placeable
f) 22(3): R34C2 <> 5 because R4C3 <> 8,9
g) Innies N7 = 19(3) <> 1; R78C3 <> 2,3 because R7C2 <= 6
h) Hidden Killer triple (123) in R23C3 for C3
i) Innies N1 = 17(3) = {179/269/278/359/368} <> 4
j) ! Innies+Outies C12: 7 = R1469C3 - R5C2; R5C2 = (5678)
- R1469C3 must have one of (35) since else smallest combo would be {1267} = 16(4) (using step 2d)
k) Innies N1 = 17(3) <> 5 because {359} blocked by Killer pair (35) of R1469C3

3. R123 !
a) Innies N1 = 17(3) can only have one of (123) -> R3C3 <> 1,2,3
b) Innies+Outies N1: R3C23 = R23C4 -> R3C23 and R23C4 cannot be {79} since R3C23 sees R3C4
-> 17(3) <> 1
c) 17(3) = {269/359/368} <> 4,7 because {278} blocked by Killer pair (78) of 13(3) and R2C3 = (23)
d) ! Innies+Outies N1: R3C23 = R23C4 (R3C23 sees R3C4)
- R3C23 cannot be {69} since it's only possible combo for R23C4 = 15(2)
- R23C4 cannot be {68} since it's only possible combo for R3C23 = 14(2)
-> R3C23 = 8{6/7} -> 8 locked for R3+N1
-> R23C4 = 9{5/6} -> 9 locked for C4+N2
e) Hidden Single: R4C2 = 9 @ C2, R5C9 = 9 @ R5
f) Innies N3 = 11(3) <> 9
g) Outies N2 = 7(2+1): R2C7 = (234) since R2C3 = (23) and R4C5 = (12)

4. N246
a) 21(3) @ N4 = {678} locked for R5
b) 21(3) @ N6 = {678} -> 8 locked for R4+N6
c) 16(3) can only have one of (234) -> R23C6 <> 4
d) 4 locked in R23C5 @ N2 for C5
e) R5C5 = 5
f) 9(3) @ R5 = {234} locked for R5
g) R5C1 = 1
h) 9(3) @ N4 = {234} -> 4 locked for C2 and 2 locked for R6+N4
i) 18(3) = 9{27/36/45} <> 1
j) Killer triple (234) locked in 18(3) + R5C78 for N6
k) 8(3) = {125} since R6C78 = (15) -> R7C8 = 2; {15} locked for R6+N6

5. R3
a) Naked triple (678) locked in R3C238 for R3
b) R3C6 = 5, R3C4 = 9, R2C4 = 6 -> R2C3 = 2

6. C789
a) Innies N9 = 11(2) <> 1,9
b) Hidden Single: R1C7 = 9 @ C7
c) Innies+Outies C89: 14 = R469C7 - R5C8; R5C8 = (34)
-> R469C7 = 17/18(3) = 5{48/67} -> R6C7 = 5
-> R49C7 = 12/13(2) = [76/84]
d) Hidden Single: R3C7 = 1 @ C7

7. N78
a) Hidden Single: R4C5 = 1 @ N5
b) R3C5 = 4 -> R2C5 = 8
c) R6C3 = 3, R7C2 = 4
d) 9 locked in R78C3 @ C3 for N7
e) Innies N7 = 19(3) = {469} -> 6,9 locked for C3+N7
f) 17(3) = {269} -> R6C5 = 9, R7C5 = 6, R8C5 = 2
g) 13(3) = 4{18/36} -> R8C7 = (68) and R8C6 = 4

8. Rest is singles.

Rating: (Hard) 1.25. I used Killer triples and some IOD Combo analysis.

Nice work Afmob.

I can understand every move you did in step 2, but I probably will never do it that way myself, just like you'll probably never do it my way. And that's perfectly fine because everyone has a unique style of his/her own, right?

Your step 3d is exactly how I picked as my "critical step". (Although I used a totally different way to get to that stage - you guys will see how later.) So congrats for spotting it!

I'll post v2 no earlier than Wednesday. Hopefully some others will have made more responses about v1 and v0.9 by then.

_________________
ADYFNC HJPLI BVSM GgK Oa m

This puzzle has a special property, which i can't really explain properly. Just something i noticed. Pretty sure udosuk had it in mind.



I hope someone understands this. Otherwise Udosuk will explain it.

Para

I got stuck on A123 so decided to have a go at the v0.9.

Even this wasn't an easy puzzle although breaking up the six-cell cages in the four corner nonets was a great help.

I'll rate A123 v0.9 at 1.25 because I used lots of triple - hidden triples, naked triples, killer triples and hidden killer triples - I don't think I've used all those types in a walkthrough before, probably not even as pairs let alone triples.

Here is my walkthrough for A123 v0.9. After comments in another thread, I've limited my use of strings of combinations/permutations as much as possible. However no promises that I won't revert to using them in later walkthroughs.

Prelims

a) 20(3) cage in N1 = {389/479/569/578}, no 1,2
b) 8(3) diagonal cage in N1 = {125/134}, 1 locked for N1
c) R1C456 = {123}, locked for R1 and N2
d) 20(3) cage in N3 = {389/479/569/578}, no 1,2
e) 22(3) cage at R3C2 = {589/679}, CPE no 9 in R56C2
f) 21(3) cage at R3C8 = {489/579/678}, no 1,2,3
g) R5C234 = {489/579/678}, no 1,2,3
h) R5C678 = {126/135/234}, no 7,8,9
i) 9(3) cage at R6C2 = {126/135/234}, no 7,8,9
j) 8(3) cage at R6C7 = {125/134}, CPE no 1 in R5C8
k) 10(3) cage in N7 = {127/136/145/235}, no 8,9
l) R9C456 = {789}, locked for R9 and N8
m) 22(3) cage in N9 = {589/679}, 9 locked for N9
n) 10(3) cage in N9 = {127/136/145/235}, no 8

1. 8(3) diagonal cage in N1 = {125/134}
1a. R1C3 = {45} -> no 4,5 in R2C2 + R3C1

2. 22(3) cage in N9 = {589/679}
2a. R9C7 = {56} -> no 5,6 in R8C8 + R7C9

3. Min R23C5 = {45} = 9 -> max R4C5 = 3 because R234 cannot be {45}4
3a. Max R23C5 = 12, no 9
3b. 45 rule on N2 3(2+1) outies R2C37 + R4C5 = 7, max R2C37 = 6 -> R2C3 = {2345}, R2C7 = {1234}

4. Max R78C5 = {56} = 11 -> min R6C5 = 7 because R678C5 cannot be 6{56}
4a. 45 rule on N8 3(2+1) outies R6C5 + R8C37 = 23 -> min R8C37 = 14, no 1,2,3,4, no 5 in R8C3
4b. Min R78C5 = 8, no 1

5. 45 rule on N1 3 innies R2C3 + R3C23 = 17 cannot contain 5 because {359/458} clash with 20(3) cage
5a. Max R2C3 + R3C2 = 13 -> min R3C3 = 6 because R2C3 + R3C23 cannot be [494]

6. Hidden triple {123} in R3C179 => R3C79 = {123}
6a. Max R3C9 + R1C7 = 12 -> no 1 in R2C8

7. 45 rule on N3 3 innies R2C7 + R3C78 = 11, no 9

8. 45 rule on N7 3 innies R7C23 + R8C3 = 19, no 1
8a. Max R7C2 + R8C3 = 15 -> min R7C3 = 4

9. Hidden triple {123} in R269C3 -> R2C3 = {23}, R69C3 = {123}
9a. Naked triple {123} in R2C23 + R3C1, locked for N1
9b. Max R9C3 = 3 -> min R7C1 + R8C2 = 13, no 1,2,3

10. 45 rule on C5 3 innies R159C5 = 15
10a. All combinations for 15(3) must contain at least one of 4,5,6 -> R5C5 = {456}
10b. Killer triple 4,5,6 in R5C234, R5C5 and R5C678, locked for R5

11. Hidden killer triple 7,8,9 in R23C5, R6C5 and R9C5 for C5 -> R23C5 must contain one of 7,8
11a. R234C5 cannot contain 6 and one of 7,8 -> no 6 in R23C5
11b. Min R23C5 containing one of 7,8 = 11 -> max R4C5 = 2

12. R2C37 + R4C5 = 7 (step 3b), max R2C3 + R4C5 = 5 -> min R2C7 = 2
12a. R2C37 = 5,6 = [23/24/32], 2 locked for R2
12b. R2C2 = 1 (hidden single in R2)

13. Hidden killer triple 1,2,3 in R1C5, R4C5 and R78C5 for C5 -> R78C5 must contain one of 2,3
13a. R678C5 cannot contain 4 and one of 2,3 -> no 4 in R78C5
13b. Max R78C5 containing one of 2,3 = 9 -> min R6C5 = 8

14. 45 rule on N9 3 innies R7C78 + R8C7 = 13

15. Hidden triple {789} in R1C7, R4C7 and R78C7 for C7, R78C7 can only contain one of 7,8 -> R14C7 = {789}
15a. Killer triple 7,8,9 in R78C7, R7C9 and R8C8, locked for N9
15b. 10(3) cage in N9 = {136/145/235}
15c. Killer pair 5,6 in 10(3) cage and R9C7, locked for N9
15d. Naked triple {789} in R148C7, locked for C7
15e. Min R8C7 = 7 -> max R78C6 = 6, no 6

16. R6C5 + R8C37 = 23 (step 4a)
16a. R6C5 = {89} -> R8C37 = 14,15 = [68/78/87], no 9 in R8C3, 8 locked for R8

17. R2C7 + R3C78 = 11 (step 7)
17a. Max R2C7 + R3C7 = 7 -> min R3C8 = 5 because R2C7 + R3C78 cannot be [434]
17b. 4 in R3 locked in R3C456, locked for N2, clean-up: no 8 in R3C5 (step 3a)

18. Hidden triple {789} in R7C139 -> R7C13 = {789}
18a. R7C23 + R8C3 = 19 (step 8)
18b. Min R78C3 = 13 -> max R7C2 = 5 because R7C23 + R8C3 cannot be [676]
18c. 6 in R7 locked in R7C45, locked for N8, clean-up: no 2 in R7C5 (step 4b)

19. Min R7C1 + R9C3 = 8 -> no 9 in R8C2
19a. R8C8 = 9 (hidden single in R8)

20. Min R1C7 + R3C9 = 8 -> max R2C8 = 6

21. Hidden killer pair 7,8 in R1C8 and R34C8 for C8, R34C8 can only contain one of 7,8 because it must contain one of 4,5,6 -> R1C8 = {78}

22. 20(3) cage in N3 = {389/479/578} (cannot be {569} because R1C8 only contains 7,8), no 6
22a. Killer triple 7,8,9 in R1C7 and 20(3) cage, locked for N3
22b. 6 in N3 locked in R23C8, locked for C8
22c. 21(3) cage at R3C8 = {579/678} (cannot be {489} because R3C8 only contains 5,6), no 4
22d. 5 of {578} must be in R3C8 -> no 5 in R4C8

23. 6 in R1 locked in R1C12, locked for N1
23a. 20(3) cage in N1 = {569}, locked for N1 -> R1C3 = 4, R3C1 = 3 (step 1), R2C3 = 2
23b. R2C3 = 2 -> R23C4 = 15 = {69} (cannot be {78} which clashes with R23C5), locked for C4 and N2
23c. Naked pair {78} in R3C23, locked for R3
23d. Naked pair {45} in R3C56, locked for R3 and N2 -> R3C8 = 6, R23C4 = [69]
23e. Naked pair {78} in R2C56, locked for R2

24. 16(3) cage at R2C7 = {358/457} -> R3C6 = 5, R3C5 = 4

25. R3C8 = 6 -> R4C78 = {78} (step 22c), locked for R4 and N6

26. R1C7 = 9 (hidden single in C7)
26a. 20(3) cage in N3 (step 22) = {578} (only remaining combination) -> R2C9 = 5, R2C1 = 9

27. R456C9 = {369} (only remaining combination), locked for C9 and N6

28. 4 in C9 locked in R89C9 -> 10(3) cage in N9 = {145} -> R9C8 = 5, R89C9 = {14}, locked for C9 and N9
28a. R3C9 = 2, R2C8 = 3 (cage sum), R23C7 = [41], R7C78 = [32], R56C8 = [41], R6C3 = 3, R9C3 = 1, R89C9 = [14]
28b. R9C7 = 6 (I could have made this a hidden single for N9 after step 27 but preferred to do the 10(3) cage first), R7C9 = 7 (step 2), R8C7 = 8, R1C89 = [78], R4C78 = [78], R7C13 = [89], R8C2 = 7 (cage sum), R8C3 = 6, R6C5 = 9 (step 4a)

29. R9C12 = [23], R8C1 = 5 (cage sum)

and the rest is naked singles

While working on this puzzle, I found a couple of interesting moves that I later found that I didn't need because easier moves were available. Maybe they will be useful when I try A123 again.

Thanks udosuk for a really interesting puzzle! Sounds like Para has worked out why the "true" SSR is 1.25! Don't really understand it yet!

Here's another way to solve Assassin 123. I really like this way . Even though it is basically creative combo work (no IOD though), it is much simpler than Afmob's really hard work in his step 2. I'll rate my way as Easy 1.50. I think Afmob's step 2 feels more like a 1.75 rating step [edit: changed my mind on that. See follow-up post] but some more subjective opinions on that would be useful feedback. Either way, we are way under the SSscore.

I really hope my way is valid! [edit: Afmob has verified it . He also found a really important omission. Thanks Afmob]. The puzzle is cracked after step 9. I've optimized it to make steps 5-9 as simple as possible. I originally saw a version of step 7 before step 5a.

Walkthrough for Assassin 123



Fingers crossed!
Ed

Guys, gone through all your nice walkthroughs. Excellent work all around!

Andrew, thanks for your considerate decision to limit your combo crunching. I mostly have problems with deep crunching of 4- or 5-cell groups so for this puzzle (mostly 3-cell groups) there shouldn't be any issues anyway. However you don't need to change your style for me, just go ahead with your most comfortable moves for all future puzzles! Note your walkthrough also demonstrates most the property as observed by Para, with most moves applied coming in "pairs".

Ed, really ingenious breakthrough move in your step 7!

This makes your walkthrough the most elegant one posted so far. The trick I used myself (as observed by Para), despite being simpler, has to make use of the uniqueness property so your approach should be the better one. Hopefully you can find a similar brilliant cracking move in v2 which I'll release some time on or after tomorrow.

Para, I'll do a pictorial explanation about that property later. Everyone should be able to see it clearly.

_________________
ADYFNC HJPLI BVSM GgK Oa m

Can't get to sleep so will post the v2 now (note it is a diagonal (X) puzzle):

Assassin 123 v2 "Eight T's"



Same solution as above.


SSR: 2.69
"True" SSR: 1.94

The property as observed by Para still stands in this one, but I think the big guns don't really need it. So fire away guys!

Wowwwww. I've been struggling with v0.9, and thought I'd drop by to see what the discussion was. I suspected exactly what you described, para (I even had also used the term "rotationally symmetric" yesterday); but, I hadn't been able to figure it out. I'm looking forward to your pictorial explanation, udo.

Thanks again, to all of you.

Since I'm stuck again on A123 I had a peek at Para's hidden message.

I can see what he's saying but I don't know that I'm convinced by the whole argument or that R5C5 is fixed immediately. It seems to me very much like UR, which some people use but by doing so avoid solving the whole puzzle.

I hope there are reasonable ways of solving A123 without needing to use this shortcut.

There are statements that the "true SS score" for this puzzle is about 1.25. Are they on the assumption that R5C5 is fixed immediately and working from there? I don't know what the rating of Para's suggestion is but I'd be surprised if it isn't considerably higher than 1.25. Of course if, in future, it becomes an established and well known technique then it might then get a lower rating but that wouldn't be appropriate at the moment with it not being widely known or understood.

Maybe udosuk will be able to show us how "rotationally symmetric", which azpaull uses in his message, leads logically to what Para said without relying in any way on uniqueness. That would certainly help it to become an established and well known technique.

BTW I did spot one interesting feature in the cage diagram. The outer cages total 180 making R19C19 = R28C28 but this doesn't seem to lead to anything useful.