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This one is very late. Not worried about an occasional blank but couldn't bear a milestone number being missed.

I found this puzzle perfect as a milestone Assassin with a couple of mind-twisters available but hoping there are other ways. In grey since still mourning the loss of the Ashes .

Assassin 170


SSscore 1.30

Cheers
Ed

Thanks Ed for another great puzzle : it is really delightful to solve with many little steps.
A missing Assassin would be a pity !

Assassin 170 WT

Thanks Ed for another challenging Assassin. I found it hard going at one stage but that was mainly because of a silly oversight by me; once I reworked that area it got a lot easier.


Here is my walkthrough; simplified a bit more than usual but I won't claim that it's optimised.

Prelims

a) R23C1 = {39/48/57}, no 1,2,6
b) R23C9 = {29/38/47/56}, no 1
c) 17(2) cage at R3C3 = {89}, CPE no 8,9 inR3C4 + R4C3
d) 10(2) cage at R3C7 = {19/28/37/46}, no 5
e) R56C5 = {59/68}
f) 10(2) cage at R6C4 = {19/28/37/46}, no 5
g) 12(2) cage at R6C6 = {39/48/57}, no 1,2,6
h) R9C23 = {29/38/47/56}, no 1
i) R9C78 = {19/28/37/46}, no 5
j) R1C123 = {125/134}, 1 locked for R1 and N1
k) 26(4) cage in N2 = {2789/3689/4589/4679/5678}, no 1
l) 13(4) cage at R3C4 = {1237/1246/1345}, no 8,9
m) 27(4) cage in N4 = {3789/4689/5679}, no 1,2

1. 45 rule on R1 1 outie R2C5 = 7, clean-up: no 5 in R3C1, no 4 in R3C9
1a. R1C456 = 19 = {289/469/568}, no 3
1b. 13(4) cage at R3C4 = {1246/1345}, CPE no 4 in R1C5
1c. 7 in R1 locked in R1C789, locked for N3, clean-up: no 4 in R2C9, no 3 in R4C6
1d. R1C789 = {279/378/567}, no 4

2. 45 rule on C1 1 outie R5C2 = 1 innie R1C1 + 8 -> R1C1 = 1, R5C2 = 9, clean-up: no 5 in R6C5, no 2 in R9C3
2a. R456C1 = 18 = {468/567} (cannot be {378} which clashes with R23C1), no 3, 6 locked for C1 and N4
2b. 2 in C1 locked in R789C1, locked for N7, clean-up: no 8 in R6C4, no 9 in R9C3

3. Killer pair 8,9 in R4C4 and R56C5, locked for N5, clean-up: no 1,2 in R3C7, no 1 in R7C3, no 3,4 in R7C7

4. 45 rule on C9 1 outie R5C8 = 1 innie R1C9 + 1, no 1,2,5 in R5C8, no 8,9 in R1C9

5. 45 rule on N1 1 innie R3C3 = 2 outies R2C4 + R4C2 + 5
5a. R3C3 = {89} -> R2C4 + R4C2 = 3,4 = {12/13} (cannot be [22] because R2C4 + R4C2 both in 20(5) cage at R2C2)

6. 45 rule on N8 2 innies R8C46 = 8 = {17/26/35}, no 4,8,9

7. 45 rule on R1234 4 innies R4C1379 = 26 = {4679/5678} (cannot be {2789/3689/4589} which clash with R4C4), no 1,2,3, 6,7 locked for R4, clean-up: no 3,4 in R3C7
7a. Killer pair 8,9 in R4C1379 and R4C4, locked for R4

8. 6 in N1 locked in R2C23 + R3C2
8a. 45 rule on N1 4 innies R23C23 = 25 = {2689/3679/4678}, no 5

9. 20(5) cage at R2C2 must contain both of 1,6 = {12368/12467}, no 9
9a. 7 of {12467} must be in R3C2 -> no 4 in R3C2
9b. R23C23 (step 8a) = {2689/4678} (cannot be {3679} which clashes with combinations for 20(5) cage at R2C2), no 3, 8 locked for N1, clean-up: no 4 in R23C1
[When considering the rating for this walkthrough, step 9b can be ignored since step 13 gives the same eliminations using simpler steps.]

10. 1,4 in N3 locked in R2C78 + R3C8, locked for 19(5) cage at R2C6, no 1,4 in R2C6 + R4C8
10a. 45 rule on N3 4 innies R23C78 = 16 = {1249/1348/1456}
10b. R3C7 = {689} -> no 6,8,9 in R2C78 + R3C8

11. 1,4 locked in 19(5) cage at R2C6 = {12349/13456}, no 8
11a. 6,9 only in R2C6 -> R2C6 = {69}
11b. CPE no 3 in R1C8

12. 8 in N2 locked in R1C456, locked for R1, clean-up: no 3 in R1C79 (step 1d), no 4 in R5C8 (step 4)
12a. 8 locked in R1C456 (step 1a) = {289/568}, no 4

13. R1C23 = {34} (hidden pair in R1), locked for N1, clean-up: no 9 in R23C1
13a. R23C1 = [57], clean-up: no 6 in R3C9
13b. Naked triple {468} in R456C1, locked for C1 and N4
13c. Naked triple {239} in R789C1, locked for N7, clean-up: no 1,7 in R6C4, no 8 in R9C23

14. Killer pair 6,9 in R1C456 and R2C6, locked for N2
14a. 13(4) cage at R3C4 (step 1b) = {1345} (only remaining combination), no 2, CPE no 5 in R1C5
14b. 4 in N2 locked in R3C456, locked for R3 and 13(4) cage at R3C4, no 4 in R4C5

15. 1 in N7 locked in R7C2 + R8C23, locked for 23(5) cage at R6C2, no 1 in R6C2+ R8C4, clean-up: no 7 in R8C6 (step 6)

16. 45 rule on N9 2 outies R6C8 + R8C6 = 1 innie R7C7 + 6
16a. Min R7C7 = 5 -> min R6C8 + R8C6 = 11, no 1,2,3,4 in R6C8, no 1 in R8C6, clean-up: no 7 in R8C4 (step 6)

17. 45 rule on N6 2 innies R46C8 = 1 outie R5C6 + 9
17a. Max R46C8 = 14 -> max R5C6 = 5

[At this stage I spotted a nice “clone and non-clone” move
Hidden killer pair 6,9 in R1C456 and R2C6 for N2, hidden killer pair 6,9 in R1C456 and R1C789 for R1 -> the value of R2C6 must be in R1C789 -> R3C7 not equal to R2C6
45 rule on N3 2 outies R2C6 + R4C8 = 1 innie R3C7 + 3
R2C6 not equal to R3C7 -> no 3 in R4C8
However I’ve decided to see if I could find simpler steps to make this elimination.]

18. 45 rule on N3 3(2+1) outies R2C6 + R4C68 = 13 = 6[25/43]//9[13], no 2 in R4C8

[Due to an oversight I had to re-work step 11 and add steps 12 and 13 but I’ve left the next step until now so that I could include the comment about the “clone and non-clone” move.]

19. R3C3 = 9 (hidden single in N1), R4C4 = 8, clean-up: no 2 in R2C9, no 1 in R4C6, no 6 in R56C5
19a. R56C5 = [59], clean-up: no 7 in R7C7

20. R2C6 + R4C68 = 13 (step 18)
20a. R4C6 is even, R4C8 is odd -> R2C6 must be even -> R2C6 = 6 (or from permutations in step 18), clean-up: no 5 in R1C46 (step 12a), no 5 in R3C9
20b. Naked pair {28} in R2C23, locked for R2 and 20(5) cage at R2C2 -> R3C2 = 6, R3C7 = 8, R4C6 = 2, R4C8 = 5 (step 18), R4C3 = 7, clean-up: no 3 in R23C9, no 3 in R6C4, no 4 in R6C6, no 8 in R7C3, no 2,6 in R8C4 (step 6), no 4 in R9C2, no 5 in R9C3, no 2 in R9C8
20c. R23C9 = [92]

21. Naked pair {46} in R4C19, locked for R4 -> R4C7 = 9, R7C7 = 5, R6C6 = 7, clean-up: no 1 in R9C8
21a. Naked pair {46} in R6C4 + R7C3, CPE no 4,6 in R7C4
21b. Naked pair {35} in R8C46, locked for R8 and N8

22. R1C9 = 5 (hidden single in R1), R5C8 = 6 (step 4), R1C78 = [67], R4C9 = 4, R4C1= 6, R6C8 = 8, R56C1 = [84], R6C4 = 6, R7C3 = 4, R1C23 = [43], R9C3 = 6, R9C2 = 5, clean-up: no 2,3,4 in R9C7, no 4 in R9C8

23. R6C8 + R8C6 = R7C7 + 6 (step 16)
23a. R7C7 = 5, R6C8 = 8 -> R8C6 = 3, R8C4 = 5

24. R4C9 + R5C8 = [46] = 10 -> R56C9 = 10 = [73]

25. Naked triple {168} in R789C9, locked for N9 -> R9C7 = 7, R9C8 = 3

26. R6C2 = 2 -> R56C3 = [15], R8C3 = 8, R56C7 = [21], R8C7 = 4, R5C46 = [34], R4C5 = 1, R23C4 = [14]

27. Naked pair {29} in R8C18, locked for R8 -> R8C5 = 6, R89C9 = [18]
27a. Naked pair {29} in R9C14, locked for R9 -> R9C56 = [41], R9C4 = 9 (cage sum)

and the rest is naked singles.

Thanks manu for getting A171 ready for us. In the mean-time, here is (nearly) an old-style V2. When one of Ruud's Assassins felt a bit easy I used to like to find a way to toughen it up a bit by changing the cage pattern slightly. The following puzzle is one of these.

manu's (step 7) and Andrew's (step 11) shows the V1 is pretty straightforward for a milestone puzzle and could do with toughening up. This V2 is actually the original puzzle I was hoping to be the Assassin. However, I couldn't solve it after multiple attempts. But had one last look tonight and finally saw the key, which is not too difficult [edit: invalid key, so it is difficult]. So, turns out, it's really a V1.5.

I found a really nice move in the V1 which I'll post about another time. unfortunately, it doesn't work for this new one. A bit of pink returning to the cheeks in the puzzle pic.

Assassin 170 v1.5

Solution: same as V1
SSscore 1.58

Cheers
Ed

Thanks for giving us two Killers this week, Ed!

I didn't optimize this walkthrough since the breakthrough move was the same as the one from SudokuSolver. So I hope someone comes up with a better path.

A170 V1.5 Walkthrough:

1. C123
a) Innies+Outies C1: 8 = R5C2 - R1C1 -> R5C2 = 9, R1C1 = 1
b) 2 locked in 14(3) @ C1 = 2{39/48/57} <> 6 for N7
c) 6 locked in 27(4) @ C1 = 69{48/57} for N4
d) 11(2) <> 9
e) Outies N1 = 12(2+1): R2C4+R4C2 <= 4 = 1{2/3} since R4C4 >= 8
f) Hidden Single: R2C6 = 6 @ N2, R3C2 = 6 @ N1
g) 20(5) = 126{38/47} (from step 1e)
h) 17(2) = {89} -> CPE: R3C4+R4C3 <> 8,9
i) 12(2) <> {48} since it's a Killer pair of 20(5)
j) Killer pair (35) locked in 8(3) + 12(2) for N1

2. R1234
a) 6 locked in 18(3) @ R1 = 6{39/48/57} <> 2
b) Innies R1 = 19(3) <> 5
c) 10(2) <> 4
d) Outies N3 = 7(2) = [25/34]
e) 10(2) = [73/82]
f) 19(5) = {13456} since R4C8 = (45) -> 3 locked for N3; CPE: R1C8 <> 4,5
g) 2 locked in 11(2) @ N3 = {29} locked for C3+N9
h) 9 locked in Innies R1 @ R1 = 19(3) = 9{28/37} for 39(8)
i) Innies+Outies N2: R4C5 = R2C4 = (123)
j) Naked triple (123) locked in R4C256 for R4

3. C789
a) Innies+Outies C9: 1 = R5C8 - R1C9 -> R1C9 <> 8; R5C8 = (5678)
b) 20(4) = {1478/1568/3467} since {3458} blocked by R4C8 = (45)
c) Killer pair (45) locked in R4C8 + 20(4) for N6
d) Hidden Killer pair (13) in 15(3) for C9 since 20(4) can only have one of (13) -> 15(3) = {168/348/357}
e) Innies+Outies N36: -2 = R45C6 - R6C8 -> R6C8 <> 1,2,3 and R5C6 <> 7,8
f) Hidden Killer triple (123) in R56C7 for N6 since 20(4) must have exactly one of (13)
-> R56C7 = 2{1/3} -> 2 locked for C7+16(4)
g) 16(4) can only have two of (123) -> R5C6 = (45)
h) 16(4) = 2{149/158/347/356} since R5C6 = (45) and because of step 3f
i) Innies+Outies N36: -2 = R45C6 - R6C8: R6C8 = (89) since R45C6 >= 6
j) Outies N9 = 18(2+1): R8C6 <> 8,9 since R6C8 = (89) and R6C6 >= 3
k) 10(2) @ N9: R9C8 <> 8

4. R789
a) 31(5) must have 9 -> CPE: R89C5 <> 9
b) 11(2): R9C3 <> 5
c) 10(2) @ N7: R6C4 <> 8

5. C6789 !
a) Innies C6789 = 23(4) = {1589/2489/2579/3479/3578}
b) Killer pair (45) locked in Innies C6789 + R5C6 for C6
c) 12(2): R7C7 <> 7,8
d) 10(2) @ N9 <> {46} since it's blocked by Killer triple (456) of 15(3) + 26(5)
e) R8C7 <> 9 since it sees all 9 of N6
f) ! Consider placement of 10(2) + 15(3) @ N9 -> 3 locked in 10(2) + 15(3) for N9:
- i) 15(3) = 3{48/57} -> 3 locked for N9
- ii) 15(3) = {168} locked for N9 -> 10(2) = {37} locked for N9
g) 12(2): R6C6 <> 9
h) Outies C9 = 19(2+1): R1C7 <> 7 since R5C8 <> 4

6. N458
a) R7C4 <> 6,9 since it sees all 6,9 of N5
b) 16(4): R5C4 <> 2 because 6 only possible there and 2{158/347} blocked by Killer pairs (47,58) of 27(4)
c) 16(4) <> {1456} since it's blocked by Killer pair (45) of 27(4)
d) R56C3+R6C2 <> 4 since R4C3 = (457) and (45,47) are Killer pairs of 27(4)

7. N36 !
a) Innies N36 = 28(4+1): R4C7 <> 8 because 78{13}+9 unplaceable since 2 locked in R56C7 @ N6
b) ! Innies N36 = 28(4+1): R3C7 <> 7 since
- i) 6 in R4C7 is too small to reach cage size
- ii) [79138] unplaceable because 2 locked in R56C7 @ N6
c) R3C7 = 8 -> R4C6 = 2, R3C3 = 9, R4C4 = 8
d) 18(3) = {567} locked for R1+N3
e) 19(5) = {13456} -> R4C8 = 5
f) Hidden Single: R4C7 = 9 @ R4
g) 16(4) = {1249} -> R5C6 = 4; 1 locked for C7+N6

8. R789
a) 12(2) = {57} -> R6C6 = 7, R7C7 = 5
b) 10(2) @ N9 = {37} locked for R9+N9
c) 11(2) = {56} -> R9C2 = 5, R9C3 = 6
d) 15(3) = {168} locked for C9+N9
e) 26(5) = {23489} -> R6C8 = 8, R8C7 = 4, R8C6 = 3

9. C123
a) 12(2) = {57} locked for C1+N1
b) 27(4) = {4689} -> R5C1 = 8; 4 locked for C1+N4
c) R4C3 = 7
d) 5 locked in 16(4) @ N4 = {1357} for 16(4)

10. N58
a) 5 locked in 31(5) @ N5 = 579{28/46} for C5 since R7C4 = (1247); 7 locked for R7+N8
b) 4 locked in 20(4) @ R9 for N8
c) 31(5) = {25789} -> R5C5 = 5, R6C5 = 9, R7C6 = 8; 2 locked for R7+N8

11. Rest is singles.

Rating: Easy 1.5 - 1.5. I used a small forcing chain (combined cages) and small combo analysis of large Innies though the chain was probably unnecessary.

Edit: Ed improved my walkthrough and made it quite quick and still easy to understand. Thanks Ed!

Ooops! Sorry guys, especially Afmob. I realize my way of solving A170 v1.5 isn't valid and can't find a way around it. As punishment, I'll optimize Afmob's WT if he'd like to include it in his post, perhaps continuing a new tradition Afmob started with A169 by posting both types of solutions.

While I'm here, a nice move I found in the A170 original. The first step both manu and Andrew used.

1. "45" n1: 3 outies r24c4 + r4c2 = 12
1a. r4c4 = (89) -> r2c4 + r4c2 = 3/4 = {12/13} ([22] blocked by the two outies being in the same cage)
1b. 1 locked for 20(5)n1

2. Generalized Killer single 1 in r234. Like this.
2a. 19(5)n2 must have 1
2b. 13(4)n2 must have 1
2c. r2c4 + r4c2 must have 1 (step 1 above)
2d. -> three spots for 1 in r234 taken: no 1 elsewhere in r234
2e. -> no 9 in 10(2)n3

Of course, this move is completely over-the-top but I enjoyed seeing it.

Cheers
Ed

You can feel free to optimize my walkthrough if you want to. I only started this tradition because of my current lack of time, but I'll be back soon.

Here I've written a very optimised complete walkthrough for the v1.5, hope you guys would have a look and tell me what you think about it:



Minor typo fixed, thanks reminder from Andrew.

_________________
ADYFNC HJPLI BVSM GgK Oa m

Time to try and bring some closure to this V1.5 so I can move on.

First, congrats to Afmob and udosuk for finding a short way to crack this one. Basically, one tough bit of combo analysis did it (Afmob and my optimized Walkthrough (WT) step 4a & b; udosuk's WT paragraph 2, last 3 lines). I never thought to look at that spot so once again, really well done Afmob and udosuk!

Very glad I didn't end up making it the V1 (by mistake). I guess the reason why they both picked on that spot was because of the locked candidates in that area. udosuk's locking two candidates is really neat. I don't know how many times I looked at that those individually but never saw how to combine them. Afmob's way shows that only the one locked candidate was necessary with careful analysis.
Seems you'd like more than just some cursory comments about your breakthrough move. As you'll see, in many ways I'm the wrong person to be doing this but, here goes.

First, your walkthrough appears to be incredibly optimised with only the essential eliminations to get to the breakthrough and what looks like a very quick finish! I didn't need to optimise Afmob's WT after all! It takes a lot of work to get to that sort of solution so really appreciate it when a WT poster makes the effort. Thanks.

Second, I couldn't read it. Not a word by word reading anyway. Started reading but then got frustrated so just skimmed. This is the usual way I "read" Afmob's WTs too. You guys pack it in too much for me. I'm more of a words person and can't cope with too many symbols. I like to have all the "trivial" stuff in so I can stay focused on the logic. I also like a spread-out look to the WT. I prefer things like; step numbers, blank line between steps, locked candidates and locked naked subsets spelled out, missing candidates from cages or hidden cages named, cage locations, r and c locations for innies and outies. I can't seem to handle too many symbols. I still have to slow down every single time I come to an "@" to work out if it means "at" or "for" or "in". I keep flowing along if the word is used rather than the symbol.

Finally, I need the harder steps explained fully. For example, I couldn't see how udosuk did his last 3 lines of Para. 2 and had to break it down more:

One breakdown of udosuk's Para. 2, last 3 lines:
5 remaining Innies for n36: r3456c7+r6c8=90-17=28 with 2 and 9 for n6 locked there
-> 3 remaining innies equal 28-9-2=17
->h17(2+1) must come from {13678} with 8 the only possible repeat in the two nonets and only one of 1 or 3 possible in r56c7 (must have a 2 there)
->h17(2+1) = {188/368}(no 7 in r34c7)

If I have to think too hard about steps with missing information then I give up.

Unfortunately, skimming means I can never be sure if the WT is valid but I know you guys very rarely make a fatal mistake!!!!!! Also, we have a couple of really hard working forum members who habitually read and validate nearly every walk-through posted here (thanks Afmob and Andrew!). The rule-of-thumb for WT posters is that no news is good news. No feedback? It's valid. If it's not valid, you'll hear about it (eventually)!!

It would be really great to have some new members who are willing to read WTs. I for one, love getting any sort of quick feedback confirming there are no flaws. Also, it's nice to know for sure that at least one person has read it.

Alternate way to crack the V1.5
Enough about WT styles. Now, a completely different topic for this puzzle. I played around with the settings on SudokuSolver to see if it could find a different (and hopefully easier) way to crack it (Afmob pointed out in his intro that SS, with the "Scoring" settings, cracks it in the same area as his WT). Found one. Certainly different and I like to think I had a chance of finding it but can't really say it's easier. You be the judge.

See the attachment for the SudokuSolver settings. (Import them from your computer into SS through Solver-Options-Import)

From A170 v1.5, Afmob's optimized WT step 3h with all the 9's highlighted:



4. XY-Chain -> no 7 in r4c7. Like this.
4a. If r3c7 is not 7 it's 8. When it's 8 -> r3c3 = 9 -> r4c4 = 8 -> 9 in r4 must be in r4c7 -> no 7 in r4c7
4b. If r3c7 = 7 -> no 7 in r4c7

5. 7 in n6 must be in 20(4)n6
5a. 20(4)n6 = 7{148/346}(no 5)

6. Hidden single 5 found at r4c8 in n6

Cracked.

Cheers
Ed

Thanks for you thorough post, Ed!

I guess from a technical point of view the forcing chain is as difficult as our combo analysis though from my gut feeling I think the chain is easier (to find and understand).