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No V2 this week : good thing, we have time for editing new walkthroughs for previous puzzles, but for people who search for new ones, here is JFFK3.
Not much difficult (the next Assassin might be by far more challenging ), I hope you will have fun with it.

JFFK3



PS code : 3x3::k:2304:2304:2304:6147:6147:6147:6147:7431:5128:3337:3337:4363:6147:5133:5133:5133:7431:5128:3602:2579:4363:4629:4629:4629:4629:7431:5128:3602:2579:4363:6430:6430:9504:9504:7431:5128:5412:5412:7718:6430:6430:9504:7431:7431:7468:5412:7718:7718:7718:6430:9504:2867:7468:7468:5412:4663:4663:2617:2617:9504:2867:7468:7468:5412:4663:4663:5186:9504:9504:2867:4678:4678:5186:5186:5186:5186:9504:3149:3149:3149:4678:

SSscore : 1.12

Solution :

Have fun !

Thanks Manu. Just like JFFK2 this one lived up to its name. I particularly enjoyed using the 37(8) cage.

I'll rate JFFK3 at Hard 1.0.

Here is my walkthrough.

Prelims

a) R2C12 = {49/58/67}, no 1,2,3
b) R34C1 = {59/68}
c) R34C2 = {19/28/37/46}, no 5
d) R7C45 = {19/28/37/46}, no 5
e) R1C123 = {126/135/234}, no 7,8,9
f) R2C567 = {389/479/569/578}, no 1,2
g) R678C7 = {128/137/146/236/245}, no 9
h) 30(4) cage at R5C3 = {6789}, CPE no 6,7,8,9 in R6C1
i) 37(8) cage at R4C6 = {12345679}, no 8

1. 45 rule on R1 2 innies R1C89 = 1 outie R2C4 + 12
1a. Max R1C89 = 17 -> max R2C4 = 5
1b. Min R1C89 = 13, no 1,2,3

2. 45 rule on R9 2 innies R9C59 = 1 outie R8C4 + 13
2a. Max R9C59 = 17 -> max R8C4 = 4
2b. Min R9C59 = 14, no 1,2,3,4

3. 45 rule on N14 1 outie R6C4 = 3 innies R5C12 + R6C1 + 3
3a. Min R5C12 + R6C1 = 6 -> R6C4 = 9, R5C12 + R6C1 = 6 = {123}, locked for N4 and 21(5) cage at R5C1, clean-up: no 7,8,9 in R3C2, no 1 in R7C5
3b. R5C3 + R6C23 = {678}, locked for N4, clean-up: no 6,8 in R3C1, no 2,3,4 in R3C2

4. Naked pair {59} in R34C1, locked for C1, clean-up: no 4,8 in R2C2
4a. Naked triple {459} in R4C123, locked for R4
4b. 9 in 37(8) cage at R4C6 locked in R7C6 + R8C56 + R9C5, locked for N8, clean-up: no 1 in R7C4
4c. 4 in 37(8) cage at R4C6 locked in R5678C6 + R8C5, CPE no 4 in R9C6

5. R5C12 + R6C1 = 6 -> R78C1 = 15 = {78}, locked for C1 and N7, clean-up: no 5,6 in R2C2

6. 45 rule on C123 2 remaining outies R89C4 = 8 = [17/26/35]

7. R1C123 = {135/234} (cannot be {126} which clashes with R3C2), no 6, 3 locked for R1 and N1

8. R2C567 = {389/578} (cannot be {479} which clashes with R2C2, cannot be {569} which clashes with R2C12), no 4,6, 8 locked for R2
8a. Killer pair 7,9 in R2C2 and R2C567, locked for R2

9. R234C3 = {179/269/458} (cannot be {278/467} which clash with R56C3)
9a. 9 of {179/269} must be in R4C3 -> no 9 in R3C3
9b. 7,8 of {179/458} must be in R3C3 -> no 1,4,5 in R3C3

10. 45 rule on C89 1 outie R5C7 = 1 innie R9C8 + 6, R5C7 = {789}, R9C8 = {123}

11. 45 rule on N58 1 remaining innie R9C6 = 1 outie R4C7 + 1, no 3 in R4C7, no 1,5,6 in R9C6
11a. 3 in 37(8) cage at R4C6 locked in R5678C6 + R8C5, CPE no 3 in R9C6, clean-up: no 2 in R4C7
11b. 2 in 37(8) cage at R4C6 locked in R45678C6 + R8C5, CPE no 2 in R9C6, clean-up: no 1 in R4C7

12. R9C678 = {138/147/237} (cannot be {129/156/246/345} because R9C6 only contains 7,8), no 5,6,9
12a. R9C6 = {78} -> no 7,8 in R9C7

13. Hidden killer pair 1,2 in R9C123 and R9C78 for R9 -> R9C123 must contain one of 1,2
13a. R89C4 = 8 (step 6) -> R9C123 = 12 = {156/246} (cannot be {129} which clashes with R9C678, cannot be {345} which doesn’t contain 1 or 2), no 3,9, 6 locked for R9 and N7, clean-up: no 2 in R8C4 (step 6)

14. 3 in R9 locked in R9C78, locked for N9
14a. R9C678 (step 12) = {138/237}, no 4
14b. 4 in R9 locked in R9C123 (step 13a) = {246} (only remaining combination), locked for R9 and N7, clean-up: no 8 in R5C7 (step 10)
14c. Naked pair {13} in R9C78, locked for N9, R9C6 = 8 (step 14a), R4C7 = 7 (step 11), clean-up: no 2 in R7C45
14d. R5C7 = 9, R9C8 = 3 (step 10), R9C7 = 1
14e. 2 in N8 locked in R7C6 + R8C56, locked for 37(8) cage at R4C6

15. R7C45 = {46} (cannot be {37} which clashes with R89C4), locked for R7 and N8
15a. 4,6 in 37(8) cage at R4C6 locked in R456C6, locked for C6 and N5

16. 45 rule on N5 3 remaining innies R456C6 = 11 = {146} (only remaining combination), locked for C6, N5 and 37(8) cage at R4C6

17. R8C4 = 1 (hidden single in N8), R9C4 = 7 (step 6)

18. R678C7 = {236/245}, no 8, 2 locked for C7
18a. 3 of {236} must be in R6C7 -> no 6 in R6C7
18b. 8 in C7 locked in R123C7, locked for N3

19. 18(3) cage in N9 = {279/459/567} (cannot be {468} because R9C9 only contains 5,9), no 8

20. 8 in N9 locked in R7C89, locked for R7 and 29(5) cage at R5C9 -> R78C1 = [78]

21. 9 in N3 locked in R13C9, locked for C9 -> R9C9 = 5, R7C7 = 2, R7C89 = [98], R9C5 = 9, clean-up: no 4 in R8C89 (step 19)
[9 in N3 locked in R13C9 has been there since step 14d but I’ve only just spotted it]
21a. R8C7 = 4 (hidden single in N9), R6C7 = 5 (step 18)
21b. Naked triple {368} in R123C7, locked for N3

22. R7C89 = [98] = 17 -> R5C9 + R6C89 = 12 = {246} (only remaining combination), locked for N6
22a. Naked pair {18} in R45C8, locked for C8 and N6 -> R4C9 = 3
22b. Naked pair {28} in R4C45, locked for R4 and N5 -> R45C8 = [18], R4C6 = 6

23. 7 in N5 locked in R56C5, locked for C5
23a. R2C26 = {79} (hidden pair in R2)

24. R4C9 = 3 -> R123C9 = 17 = {179} (only remaining combination) -> R2C9 = 1, R13C9 = {79}, locked for C9 and N3 -> R8C89 = [76]
24a. R6C8 = 6 (hidden single in N6)

25. Naked pair {78} in R6C23, locked for R6 and N4 -> R5C3 = 6, R6C5 = 3, R5C45 = [57]

26. Killer pair 4,5 in R1C123 and R1C8, locked for R1

27. R1C89 (step 1) = R2C4 + 12
27a. Max R1C89 = 14 -> R2C4 = 2, R1C89 = 14 = [59], R3C9 = 7, R12C6 = [79], R4C45 = [82], R1C4 = 6, R1C57 = [18], R23C7 = [36], R3C2 = 1, R4C2 = 9 , R34C1 = [95], R4C3 = 4, R2C2 = 7, R2C1 = 6

and the rest is naked singles

You make great puzzles manu! Here is a slightly different way to do this one without any killer pairs. Just to the cracked stage. I have NOT used a trick that could get a placement in step 1 since that would rate much higher than this puzzle needs Still, I'll give this WT way a 1.25 rating for step 9b and it takes 25 steps to crack it, even though the other 24 steps are all "easy" ones. Too much work for an optimized 1.0 WT!

Off Topic: I'll be quietening down on the forum for a little while with a busy patch coming up. Great that Tarek can take up some slack.

JFFK 3 WT (25 steps to crack)
[edit after Andrew's next two post: Thanks to Andrew for some typos and optimising ( ) step 22]
Prelims

i. 9(3)n1: no 7,8,9
ii. 13(2)n1: no 1,2,3
iii. 20(3)n2: no 1,2
iv. 14(2)n1 = {59/68}
v. 10(2)n1: no 5
vi. 37(8)n5: no 8, note: all other candidates are required.
vii. 30(4)n4 = {6789}
viii. 11(3)n6: no 9
ix. 10(2)n8: no 5

1. "45" n14: 1 outie r6c4 - 3 = 3 innies r5c12, r6c1
1a. min. 3 innies = {123} = 6 -> r6c4 = 9
1b. no 1 in r7c5
1c. 3 innies = {123}: all locked for n4 & 21(5)n4
1d. no 7,8,9 in r3c2

2. naked triple {678} in n4: all locked for n4
2a. r3c2 = (16)

3. 14(2)n1 = {59} (only valid combination): both locked for c1
3a. no 4,8 in r2c2

4. naked triple {459} in r4c123: all locked for r4

5. "45" n14: 2 remaining outies r78c1 = 15 = {78}: both locked for c1 & n7
5a. r2c2 = (79)

6. 4,5 & 9 which must be in the 37(8)n5 are all in c6 or n8 -> no 4,5,9 in r9c6 (Common Peer Elimination CPE)

7. "45" n2369: 1 innie r4c7 + 1 = r9c6
7a. no 3 in r4c7
7b. no 1,6 in r9c6

8. keep doing steps 6 & 7 recursively (thanks for that verb Andrew!): locked candidate 3 in 37(8)n5 -> no 3 in r9c6 -> no 2 in r4c7 -> no 2 in r9c6 -> no 1 in r4c7

9. So finally, we have r4c7 = (67) & r9c6 = (78)
9a. from step 7, the permutations are [67/78]
9b. -> 7 locked in the innie or outie -> no 7 in common peers r45678c6 & r89c5

10. 7 in 37(8)n5 now only in r4c7 = 7
10a. -> r9c6 = 8 (step 7)
10b. no 2 in 10(2)n8

11. r9c78 = 4 (cage sum) = {13}: both locked for r9 & n9

12. "45" c89: 1 outie r5c7 - 6 = r9c8
12a. -> r5c7 = 9
12b. r9c8 = 3, r9c7 = 1

13. "45" n147: 3 remaining innies r9c123 = 12 = {246}
13a. all locked for n7 & r9 & 20(5)n8

14. "45" c9: 3 outies r678c8 = 22 = {589/679}(no 1,2,4)
14a. 9 locked for c8 & n9

15. r9c5 = 9 (hidden single r9)
15a. no 1 in r7c4

16. split cage (from step 13) r89c4 = 8 = [17/35]
16a. = [3/7..]

17. 10(2)n8 = {46} ({37} blocked by r89c4 (step 16a)
17a. 4 & 6 locked for n8 & r7

18. 4 & 6 required in the 37(8)n5 now only in n5: both locked for n5 & c6

19. "45" n5: 3 remaining innies r456c6 = 11 and must have 4 & 6 -> 1 remaining innie = 1
19a. naked triple {146} in r456c6: 1 locked for c6, n5 & 37(8)n5

20. naked triple {235} in n8: locked for n8
20a. r89c4 = [17]
20b. r9c9 = 5

21. better get this hidden single: r6c2 = 8 (hidden for c2)
21a. r56c3 = {67}: both locked for c3

22. Naked triple {278} in r7c179: locked for r7 -> r7c8 = 9. [simpler than the naked quad I saw]

23. h22(3) r678c8 = {589/679}
23a. r6c8 = (56) -> no 6 in r8c8

24. r8c89 = 13 (cage sum)
24a. = [76] (last permutation)
24b. r6c8 = 6 (h22(3)r678c8)

25. 17(3)n1 must have 8 for c3 = {458}: all locked for c3

Now its smooth.
Cheers
Ed

Hi,

I would like to explain a particular technic that can be used fruitfully for this puzzleposted one week ago. It had be mentionned by Ed in his previous post. It consists in using both CPE and IO difference.
I think everyone who have tried this puzzle has noticed the role of cage 37(8) : the particular cell R9C6 sees almost all cells of that cage. If it had seen all cells of cage 37(8), one could have deduced R9C6=8 since 37(8)={12345679} (Isn't the technic called CPE ?). Here, I have made the cage pattern in order to enable the solver to deduce R9C6 <> R4C7 using IO for N2369 : R9C6 = R4C7+1 -> R9C6 <> R4C7 (Ed would say IOI : Innies Outies Inequality )

That enables to prove that R4C9 is different of all cells of 37(8) : R9C6=8, BTW R4C7=7.

One could use a similar technic for JFFK1 (I have posted sth about that in JFFK1 topic). I would be interested if someone could point out a previous Assassin for which that kind of technic can be used, because I have made the cage pattern of JFFK1/3 intentionally for using such a step.

Cheers,

manu

This is of course referring to the technique discussed in manu's message; I'll say a bit more about that in my next message.

Having gone through my WT again and then Ed's one, I think Ed's step 9b is harder than any of my moves. It's a neat move of a type that we don't often see. From memory the first time I saw such a move was in a WT by either Mike or Para on Ruud's site, before we moved here. Ed's step 9b is a very powerful one because it makes so many eliminations.

I've now done some minor editing to my WT and changed my rating from 1.0 to Hard 1.0. It might be argued that my step 9 should get a higher rating, because it uses a triple blocker, but it was such an obvious block that IMHO a Hard 1.0 rating is appropriate, just like I argue that rating is right for very simple (naked) killer triples; I always go to at least Easy 1.25 for hidden killer triples.

Ed has made an interesting point about giving a higher rating because of the number of steps. I give my ratings based on the types of steps and, where appropriate, increase the rating for difficulty of finding steps but I've never considered number of steps counting toward the rating; I think Ed and I will have to agree to disagree on that point.

I'm normally a fairly slow solver but I finished JFFK3 in one session; that's not something I would expect to do for a 1.25 rated puzzle so that's another reason why I didn't go for that rating.

Thanks manu for the discussion about that technique that you intentionally designed the cage patterns of JFFK1 and JFFK3 for.

The WTs posted by Ed and myself for JFFK3 have shown that it wasn't necessary to use that technique to solve JFFK3. I must admit I didn't spot it as described by manu although I did manage to get quite a lot out of that particular cage using recursion in my step 11. Ed took it further with his clever step 9b.

Manu told me about the technique after I'd posted my WT. I immediately realised that it was probably what I'd been missing when I was stuck on JFFK1. It was! Now I know about that key breakthrough I'll have another try at that puzzle.

The technique was clearly necessary to solve JFFK1.

There is one that's not quite the same but very similar. The original steps 1e and 1f of Afmob's WT for Ed's A138 are also based on a cell seeing all except one cell of an 8-cell cage.

After A138, JFFK1 and JFFK3 we'll know to look out for this type of move. At the moment it's probably at least a 1.5 rated move and almost certainly not programmed into the current version of SS. I've no idea whether this type of move is going to be sufficiently common to be programmed into software solvers.

One final thought prompted by this type of move. What is an optimised WT? Is it the most direct solving path, which may use technically harder and therefore higher rated steps, or is it a longer path that uses the lowest possible rated steps? IMHO it's probably a question that's answered according to personal preference rather than having one answer.

I don't usually make significant changes to my walkthroughs, except when necessary because of errors, but Ed's comment about the rating made me think about omitting step 9, my technically hardest step, which only gave a few candidate eliminations.

I'll stick with my rating of Hard 1.0.

Here is my improved walkthrough. I also found step 13b which significantly shortened step 14, have moved step 15a into step 16 and shortened step 27a.

Prelims

a) R2C12 = {49/58/67}, no 1,2,3
b) R34C1 = {59/68}
c) R34C2 = {19/28/37/46}, no 5
d) R7C45 = {19/28/37/46}, no 5
e) R1C123 = {126/135/234}, no 7,8,9
f) R2C567 = {389/479/569/578}, no 1,2
g) R678C7 = {128/137/146/236/245}, no 9
h) 30(4) cage at R5C3 = {6789}, CPE no 6,7,8,9 in R6C1
i) 37(8) cage at R4C6 = {12345679}, no 8

1. 45 rule on R1 2 innies R1C89 = 1 outie R2C4 + 12
1a. Max R1C89 = 17 -> max R2C4 = 5
1b. Min R1C89 = 13, no 1,2,3

2. 45 rule on R9 2 innies R9C59 = 1 outie R8C4 + 13
2a. Max R9C59 = 17 -> max R8C4 = 4
2b. Min R9C59 = 14, no 1,2,3,4

3. 45 rule on N14 1 outie R6C4 = 3 innies R5C12 + R6C1 + 3
3a. Min R5C12 + R6C1 = 6 -> R6C4 = 9, R5C12 + R6C1 = 6 = {123}, locked for N4 and 21(5) cage at R5C1, clean-up: no 7,8,9 in R3C2, no 1 in R7C5
3b. R5C3 + R6C23 = {678}, locked for N4, clean-up: no 6,8 in R3C1, no 2,3,4 in R3C2

4. Naked pair {59} in R34C1, locked for C1, clean-up: no 4,8 in R2C2
4a. Naked triple {459} in R4C123, locked for R4
4b. 9 in 37(8) cage at R4C6 locked in R7C6 + R8C56 + R9C5, locked for N8, clean-up: no 1 in R7C4
4c. 4 in 37(8) cage at R4C6 locked in R5678C6 + R8C5, CPE no 4 in R9C6

5. R5C12 + R6C1 = 6 -> R78C1 = 15 = {78}, locked for C1 and N7, clean-up: no 5,6 in R2C2

6. 45 rule on C123 2 remaining outies R89C4 = 8 = [17/26/35]

7. R1C123 = {135/234} (cannot be {126} which clashes with R3C2), no 6, 3 locked for R1 and N1

8. R2C567 = {389/578} (cannot be {479} which clashes with R2C2, cannot be {569} which clashes with R2C12), no 4,6, 8 locked for R2
8a. Killer pair 7,9 in R2C2 and R2C567, locked for R2

9. Omitted

10. 45 rule on C89 1 outie R5C7 = 1 innie R9C8 + 6, R5C7 = {789}, R9C8 = {123}

11. 45 rule on N58 1 remaining innie R9C6 = 1 outie R4C7 + 1, no 3 in R4C7, no 1,5,6 in R9C6
11a. 3 in 37(8) cage at R4C6 locked in R5678C6 + R8C5, CPE no 3 in R9C6, clean-up: no 2 in R4C7
11b. 2 in 37(8) cage at R4C6 locked in R45678C6 + R8C5, CPE no 2 in R9C6, clean-up: no 1 in R4C7

12. R9C678 = {138/147/237} (cannot be {129/156/246/345} because R9C6 only contains 7,8), no 5,6,9
12a. R9C6 = {78} -> no 7,8 in R9C7

13. Hidden killer pair 1,2 in R9C123 and R9C78 for R9 -> R9C123 must contain one of 1,2
13a. R89C4 = 8 (step 6) -> R9C123 = 12 = {156/246} (cannot be {129} which clashes with R9C678, cannot be {345} which doesn’t contain 1 or 2), no 3,9, 6 locked for R9 and N7, clean-up: no 2 in R8C4 (step 6)
13b. R9C123 = {246} (only remaining combination, cannot be {156} which clashes with R89C4), locked for R9 and N7, clean-up: no 8 in R5C7 (step 10)

14. Naked pair {13} in R9C78, locked for N9, R9C6 = 8 (step 12), R4C7 = 7 (step11), clean-up: no 2 in R7C45
14a. R5C7 = 9, R9C8 = 3 (step 10), R9C7 = 1
14b. 2 in N8 locked in R7C6 + R8C56, locked for 37(8) cage at R4C6

15. R7C45 = {46} (cannot be {37} which clashes with R89C4), locked for R7 and N8

16. 4,6 in 37(8) cage at R4C6 locked in R456C6, locked for C6 and N5
16a. 45 rule on N5 3 remaining innies R456C6 = 11 = {146} (only remaining combination), locked for C6, N5 and 37(8) cage at R4C6

[Alternatively steps 15 and 16 can be replaced by
15. 45 rule on N5 3 remaining innies R456C6 = 11 = {146} (only remaining combination), locked for C6, N5 and 37(8) cage at R4C6
16. R7C45 = {46} (hidden pair in N8), locked for R7]

17. R8C4 = 1 (hidden single in N8), R9C4 = 7 (step 6)
[Or hidden pair in N8, if preferred]

18. R678C7 = {236/245}, no 8, 2 locked for C7
18a. 3 of {236} must be in R6C7 -> no 6 in R6C7
18b. 8 in C7 locked in R123C7, locked for N3

19. 18(3) cage in N9 = {279/459/567} (cannot be {468} because R9C9 only contains 5,9), no 8

20. 8 in N9 locked in R7C89, locked for R7 and 29(5) cage at R5C9 -> R78C1 = [78]

21. 9 in N3 locked in R13C9, locked for C9 -> R9C9 = 5, R7C7 = 2, R7C89 = [98], R9C5 = 9, clean-up: no 4 in R8C89 (step 19)
[9 in N3 locked in R13C9 has been there since step 14 but I’ve only just spotted it]
21a. R8C7 = 4 (hidden single in N9), R6C7 = 5 (step 18)
21b. Naked triple {368} in R123C7, locked for N3

22. R7C89 = [98] = 17 -> R5C9 + R6C89 = 12 = {246} (only remaining combination), locked for N6
22a. Naked pair {18} in R45C8, locked for C8 and N6 -> R4C9 = 3
22b. Naked pair {28} in R4C45, locked for R4 and N5 -> R45C8 = [18], R4C6 = 6

23. 7 in N5 locked in R56C5, locked for C5
23a. R2C26 = {79} (hidden pair in R2)

24. R4C9 = 3 -> R123C9 = 17 = {179} (only remaining combination) -> R2C9 = 1, R13C9 = {79}, locked for C9 and N3 -> R8C89 = [76]
24a. R6C8 = 6 (hidden single in N6)

25. Naked pair {78} in R6C23, locked for R6 and N4 -> R5C3 = 6, R6C5 = 3, R5C45 = [57]

26. Killer pair 4,5 in R1C123 and R1C8, locked for R1

27. R1C89 (step 1) = R2C4 + 12
27a. Max R1C89 = 14 -> R2C4 = 2, R1C89 = 14 = [59], R2C8 = 4, R2C1 = 6, R3C2 = 1, R4C2 = 9

and the rest is naked singles

Probably more the second for me, though that may still use much harder steps than another walkthrough. In practice optimising my solution works like this.

1) Solve the puzzle once keeping rough notes

2) Think backwards from the first placement to guess which steps (in that area of the puzzle) were essential to achieve that first placement.

3) Solve it again, ticking those parts of the notes that help get to the first placement as directly as possible. Put brackets around the parts that don't appear to help.

4) While doing that, find simpler ways to do steps. Keep notes...possibly heading off into a completely new direction with something important missed the first time.

5) Work backwards from the second placement, then the 3rd etc.

6) Think about whether the 2nd placement could be more important than the 1st placement and can get to it straight away without the first placement. etc

7) Finally, get a clear picture in my mind of the whole solving path.

8) Type the WT. That can change the solution again if something new is found.

9) Check the WT

10) Triple check the WT and hope you don't find anything new!!

11) Decide how difficult that whole process was to keep "in the head". What rating does that feel like? 24 steps before I get to stop thinking feels like a 1.25 rating to me. Usually, 1.0 rated puzzles are around 5-15 with the remainder just getting it to singles.

Of course, deciding on the most difficult move also influences the rating. A very short WT with one difficult step still gets that steps rating. With JFFK3 I chose to go the longer route to get the lower rating.

12) Read other's WT to find what you missed.

Time consuming process but very satisfying when it's an interesting solution. But no wonder we don't get many new WT writers!

Cheers
Ed