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Hi folks,

The deadline's getting ever-nearer, and no-one's volunteered for A190 . So let's turn the clock back to a time over three years ago to an Assassin that didn't exactly get a huge forum response. Indeed, it may well be the shortest Assassin thread ever, with just one short post, and nd declaring it as "pretty straightforward after the past couple of weeks' puzzles". Therefore, I have always felt that it would be nice to have a variant that did some justice to Ruud's cage pattern. And so, here it is, a variant I wanted to post in 2008, but somehow never got around to:


Assassin 35 V1.5 (SudokuSolver v3.3.1 score: 1.18)




Solution:

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


Enjoy!

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Thanks Mike for a challenging variant using Ruud's original cage pattern. Maybe the A35 thread on Ruud's site was so short because nd posted a short walkthrough for it within hours of the puzzle appearing. That's probably why I never posted my walkthrough which had 48 steps although the first 18 of those were Prelims for 2-cell cages.

Maybe I was too much influenced by the SS score but for some time I resisted using steps 17 and 21 until I found that the only alternatives I could see were harder techniques.


Here is my walkthrough for A35 V1.5.

Prelims

a) R1C12 = {29/38/47/56}, no 1
b) R1C56 = {13}
c) R2C23 = {29/38/47/56}, no 1
d) R3C34 = {59/68}
e) R4C45 = {16/25/34}, no 7,8,9
f) R45C9 = {17/26/35}, no 4,8,9
g) R56C1 = {59/68}
h) R5C56 = {89}
i) R56C8 = {49/58/67}, no 1,2,3
j) R6C67 = {17/26/35}, no 4,8,9
k) R67C9 = {29/38/47/56}, no 1
l) R7C56 = {59/68}
m) R7C78 = {19/28/37/46}, no 5
n) R8C34 = {29/38/47/56}, no 1
o) R8C56 = {15/24}
p) R8C89 = {18/27/36/45}, no 9
q) R9C45 = {19/28/37/46}, no 5
r) R9C89 = {18/27/36/45}, no 9
s) 19(3) cage at R1C3 = {289/379/469/478/568}, no 1
t) 10(3) cage at R2C1 = {127/136/145/235}, no 8,9
u) 20(3) cage at R2C5 = {389/479/569/578}, no 1,2
v) 11(3) cage at R3C5 = {128/137/146/236/245}, no 9
w) 10(3) cage at R6C2 = {127/136/145/235}, no 8,9

Steps resulting from Prelims
1a. Naked pair {13} in R1C56, locked for R1 and N2, clean-up: no 8 in R1C12
1b. Naked pair {89} in R5C56, locked for R5 and N5, clean-up: no 5,6 in R6C1, no 4,5 in R6C8
1c. R8C89 = {18/27/36} (cannot be {45} which clashes with R8C56), no 4,5

2. 45 rule on C12 2 outies R29C3 = 7 = {25/34}/[61], no 7,8,9, no 6 in R9C3, clean-up: no 2,3,4 in R2C2

3. 45 rule on C89 2 outies R17C7 = 12 = [48/57/84/93], no 1,2,6, no 7 in R1C7, no 9 in R7C7, clean-up: no 1,4,8,9 in R7C8

4. 9 in N9 only in R7C9 + R89C7
4a. 45 rule on N9 3 innies R7C9 + R89C7 = 17 = {179/269/359}, no 4,8, clean-up: no 3,7 in R6C9

5. 45 rule on N9 1 innie R7C9 = 1 outie R9C6 + 1, no 3,7,9 in R9C6

6. 45 rule on C2 3 innies R129C2 = 21 = {489/579/678}, no 1,2,3, clean-up: no 9 in R1C1

7. 45 rule on R89 1 innie R8C2 = 1 outie R7C1 + 2, no 6,7,8,9 in R7C1, no 1,2 in R8C2
7a. 10(3) cage at R6C2 = {127/136/145/235}
7b. 7 of {127} must be in R8C2 -> no 7 in R67C2

8. 45 rule on R12 1 outie R3C9 = 2 innies R2C18 + 2
8a. Min R2C18 = 3 -> min R3C9 = 5
8b. Max R2C18 = 7, no 7,8,9 in R2C18

9. 1 in R2 only in R2C189
9a. 45 rule on R2 4 innies R2C1489 = 14 = {1238/1247/1256/1346}, no 9

10. 45 rule on R8 3 innies R8C127 = 19 = {379/469/478/568} (cannot be {289} because R8C2 only contains 3,4,5,6,7), no 1,2

11. 45 rule on C1234 2 innies R49C4 = 1 outie R6C5 + 3, IOU no 3 in R9C4, clean-up: no 7 in R9C5
11a. 7 in N8 only in R789C4, locked for C4

12. 45 rule on N5 2 innies R46C6 = 9 = {27/36}/[45], no 1, no 5 in R4C6, clean-up: no 7 in R6C7

13. 11(3) cage at R3C5 = {236/245}, no 7,8, clean-up: no 2 in R6C6 (step 12), no 6 in R6C7
13a. 3 of {236} must be in R4C6 -> no 6 in R4C6, clean-up: no 3 in R6C6 (step 12), no 5 in R6C7
13b. 7 in N5 only in R6C56, locked for R6, clean-up: no 6 in R5C8

14. 45 rule on N5 2 outies R3C56 = 1 innie R6C6 +2, IOU no 2 in R3C5
14a. 11(3) cage at R3C5 (step 13) = {236/245}
14b. 2 in 11(3) cage only in R34C6, locked for C6, clean-up: no 4 in R8C5, no 3 in R7C9 (step 5), clean-up: no 8 in R6C9
14c. 6 of {236} must be in R3C5 -> no 6 in R3C6

[At this stage I spotted the interesting
Hidden grouped killer pair 8,9 in R2C56, R5C56 and R7C56 (plus others in C56) for C56, R5C56 contains both of 8,9, R7C56 contains one of 8,9 -> R2C56 cannot contain both of 8,9 -> no 3 in R2C7.
I didn’t use it because the SS score suggests there must be an easier way. Also eliminating 3 from R2C7 doesn’t achieve much at this stage.]

15. Killer pair 5,6 in R3C34 and R3C56, locked for R3

16. 45 rule on R1 3(2+1) outies R2C4 + R23C9 = 16, min R2C4 = 2 -> max R23C9 = 14, no 7,8 in R2C9
[Alternatively Min R1C789 = 13 (cannot be {245/246} which clash with R1C12) -> max R23C9 = 14, no 7,8 in R2C9.]

17. R7C56 = {59/68}, R8C56 = {15}/[24] -> combined cage R78C56 = {59}[24]/{68/15}/{68}[24]
17a. R9C45 = [19/73/91] (cannot be {28/46} which clash with combined cage R78C56), no 2,4,6,8

18. 45 rule on N8 3 innies R78C4 + R9C6 = 15 = {159/168/249/357} (cannot be {258/456} which clash with R8C56, cannot be {267/348} which clash with R78C56)
18a. 4 of {249} must be in R9C6 -> no 4 in R78C4, clean-up: no 7 in R8C3

19. 4 in N8 only in R89C6, locked for C6, clean-up: no 5 in R6C6 (step 12), no 3 in R6C7
19a. 11(3) cage at R3C5 (step 13) = {236/245}
19b. 4 of {245} must be in R3C5 -> no 5 in R3C5

20. 12(3) cage in N5 = {147/156/345} (cannot be {237} which clashes with R4C6, cannot be {246} which clashes with R46C6), no 2
20a. 2 in N5 only in R4C456, locked for R4, clean-up: no 6 in R5C9

21. R8C56 = {15}[24], R8C89 = {18/27/36} -> combined cage R8C5689 = {15}{27}/{15}{36}/[24]{18}/[24]{36}
21a. R8C127 (step 10) = {379/469/478} (cannot be {568} which clashes with R8C5689), no 5, clean-up: no 3 in R7C1 (step 7)
21b. R8C34 = {29/38/56} (cannot be [47] which clashes with R8C127), no 4,7

22. 45 rule on C1234 3 outies R469C5 = 14 = {149/239/356} (cannot be {167} which clashes with R6C6, cannot be {257} because R9C5 only contains 1,3,9, cannot be {347} = [473] which clashes with 12(3) cage = {14}7), no 7

23. R6C6 = 7 (hidden single in N6), R6C7 = 1, clean-up: no 7 in R45C9

24. R2C5 = 7 (hidden single in C5), clean-up: no 4 in R2C3
24a. 20(3) cage at R2C5 = {479/578}, no 3,6
24b. 4 of {479} must be in R2C7 -> no 9 in R2C7

25. R7C9 + R89C7 (step 4a) = {269/359}, no 7, clean-up: no 4 in R6C9, no 6 in R9C6 (step 5)
25a. R67C9 = {29} (cannot be {56} which clashes with R45C9), locked for C9, clean-up: no 6 in R4C9, no 7 in R8C8, no 4,5 in R9C6 (step 5), no 7 in R9C8
25b. Naked pair {35} in R45C9, locked for C9 and N6, clean-up: no 8 in R6C8, no 6 in R8C8, no 4,6 in R9C8
25c. 8 in N6 only in R4C78, locked for R4

26. 12(3) cage in N5 (step 20) = {156/345}, 5 locked for N5, clean-up: no 2 in R4C45

27. R4C6 = 2 (hidden single in N4), R3C6 = 5, R3C5 = 4 (step 13), clean-up: no 9 in R3C34, no 3 in R4C4, no 9 in R7C5, no 1 in R8C5
[I could have got R4C6 after step 23 using R46C6 = 9.]

28. Naked pair {68} in R3C34, locked for R3 -> R3C9 = 7, clean-up: no 2 in R8C8, no 2 in R9C8

29. Naked pair {89} in R25C6, locked for C6 -> R7C6 = 6, R7C5 = 8, R5C56 = [98], R2C6 = 9, R2C7 = 4 (step 24a), R9C6 = 1, R7C9 = 2 (step 5), R6C9 = 9, R6C1 = 8, R5C1 = 6, R6C8 = 6, R5C8 = 7, R45C7 = [82], R3C7 = 3 (cage sum), R7C8 = 3, R7C7 = 7, R1C7 = 5 (step 3), R1C56 = [13], R9C5 = 3, R9C4 = 7, R4C5 = 6, R4C4 = 1, R6C5 = 5, R8C56 = [24], R4C8 = 4, clean-up: no 6 in R1C2, no 2 in R2C3, no 3,5,8,9 in R8C3, no 9 in R8C4, no 6 in R8C9, no 8 in R9C8, no 6,8 in R9C9

30. R8C34 = [65], R3C34 = [86], R7C4 = 9, R89C7 = [96], R9C89 = [54], R9C3 = 2, R9C12 = [98], clean-up: no 5 in R2C2, no 3 in R2C3, no 4 in R7C1 (step 7)

and the rest is naked singles.

You were still a newbie then (that week was your first post in the Assassin forum, about Hidden killer pairs) so perhaps felt a bit shy. Thanks for sharing it now!

The disadvantage of solving this puzzle (and HATMAN'S Paper Killer) is that it means I haven't been able to finish my next Assassin. If I put off doing this A35V1.5 it would have gone stale. I still believe that our tradition of "no new killers until the next Assassin has been claimed" is the simplest way to keep our bread-and-butter Assassin series going. Even if it means we only have an Assassin every two weeks (with so few willing to make them) at least we could have regularity.

Now to this variant. I found some Ridiculous killer subsets [edit: Killer ALS XZ no less! See from here] in this variant (steps 12,14). Afmob will be proud! I hope they are valid (are! but wrong names). Got so involved in just one area of the puzzle that I forgot to look anywhere else. So, I went a different way to Andrew but we shared one key area (Andrew's step 17).

If you haven't tried this puzzle yet then do so. It's a ripper.

Start to A35V1.5
A very big thankyou to Afmob, Andrew and Mike for verifying steps 12 & 14 plus many typos.

Prelims
i. 11(2)cages n1: no 1
ii. 19(3)n1: no 1
iii. 4(2)n2 = {13}
iv. 10(3)n1: no 8,9
v. 20(3)n2: no 1,2
vi. 14(2)n1 = {59/68}
vii. 11(3)n2: no 9
viii. 7(2)n5: no 7,8,9
ix. 8(2)n6: no 4,8,9
x. 14(2)n4 = {59/68}
xi. 17(2)n5 = {89}
xii. 13(2)n6: no 1,2,3
xiii. 10(3)n4: no 8,9
xiv. 8(2)n5: no 4,8,9
xv. 11(2)n6: no 1
xvi. 14(2)n8 = {59/68}
xvii. 10(2)n9: no 5
xviii. 11(2)n7: no 1
xix. 6(2)n8 = {15/24}
xx. 9(2) cages n9: no 9
xxi. 10(2)n8: no 5

1. naked pair {13} in r1c56: both locked for r1 and n2
1a. no 8 in 11(2)r1c1

2. naked pair {89} in r5c56: both locked for r5 and n5
2a. no 4,5 in r6c8
2b. no 5,6 in r6c1

3. "45" on n5: 2 innies r46c6 = 9: no 1, no 5 in r4c6
3a. no 7 in r6c7

4. "45" on n5: 2 outies r3c56 - 2 = 1 innie r6c6
4a. -> no 2 in r3c5 (IOU)

5. 11(3)n2 = {236/245}(no 7,8)
5a. 2 locked for c6
5b. {236} must have 6 in r3c5 -> no 6 in r34c6
5c. no 3 in r6c6 (h9(2)r46c6)
5d. no 5,6 in r6c7,
5e. no 4 in r8c5

6. "45" on c89: 2 outies r17c7 = 12; no 1,2,6; no 9 in r7c7, no 7 in r1c7
6a. r7c8: no 1,4,8,9

7. "45" on n9: 1 outie r9c6 + 1 = 1 innie r7c9
7a. -> no 9 in r9c6, no 3 in r7c9
7b. but r9c6 + r7c9 [56/89] blocked by 14(2)n8 = [5/6,8/9]
7c. no 5,8 in r9c6
7d. no 6 or 9 in r7c9
7e. no 2,5,8 in r6c9

double step 8 deleted

8. 9 in n9 in r89c7: 9 locked for c7
8a. 16(3)n9 must have 9 = 9{16/34}(no 2,5,7,8)({259} blocked by no 2,5 in r9c6)
8b. no 3 in r7c7 (h12(2)r17c7) -> no 7 in r7c8
8c. no 3 in r6c9 (outies n9 = 10) -> no 8 in r7c9
8d. deleted

9. {45} blocked from 9(2)r8c8 by 6(2)n8 = [4/5..]
9a. no 4,5

10. no 4 in r7c9 since 5 in n9 is only in 9(2)r9c8 or r7c9 (Locking-out Cages)
10a. no 7 in r6c9
10b. no 3 in r9c6 (outies n9 = 10)
10c. {349} in 16(3)n9 (step 8a) must have 4 in r9c6 -> no 4 in r89c7

11. combined cage 20(4) at r78c56 = {1568/2459/2468}: ie, must have both 1,6 or have 4 -> 4 locked for n8 and c6 in r89c6
11a. no 6 in 10(2)n8
11b. no 7 in r8c3
11c. no 5 in r6c6(h9(2)r46c6)
11d. no 3 in r6c7

12. Outies n9 r9c6 + r6c9 = {46}/[19] = [6/[1]..]; 8(2) at r6c6 = [71/62] = [6]/[1]: -> Almost(?) Killer pair (discussion on this to follow here)1,6 -> no 1 in common peers in r89c7!
12a. {169} in 16(3)n9 (step 8a) must have 1 in r9c6 -> no 6 in r9c6
12b. no 4 in r6c9 (outies n9 = 10)
12c. no 7 in r7c9

13. 1 in n9 in one of the two 9(2) cages -> 8 locked for n9 (Locking Cages)
13a. no 2 in r7c8
13b. r17c7 = [84/57] (h12(2))

14. "45" on n9: 3 innies r7c9 + r89c7 = 17 and must have 9 for n9 = [2]{69}/[5]{39} = [5/6..]
14a. 14(2)n8 = [5/6..] -> Almost(?) Killer pair 5,6 with h17(3) -> no 6 in common peers in r7c8!
14b. r7c8 = 3


On from there.
Ed

Thanks, Andrew and Ed, for providing walkthroughs for this puzzle, which were both a pleasure to read through. I was amazed how much progress Ed managed to make just using N9 and a handful of other cells!

Meanwhile, Andrew has asked me to share a shortcut of mine based on his WT, since this is not fully covered in either of his or Ed's solving paths.

From Andrew's step 5:


5a. R7C9+R9C6 cannot be [65/98] (blocked by 14(2) at R7C5); no 6,9 in R7C9, no 5,8 in R9C6, clean-up: no 2,5 in R6C9

6. Innies N89: R78C4+R7C9 = 16(3)
6a. -> R78C4 cannot sum to 10, because no 6 in R7C9
6b. -> cannot have both of {37}
6c. only other place for {37} in N8 is 10(2) at R9C4
6d. 10(2) at R9C4 cannot have just one of {37}, so must have both
6e. -> 10(2) at R9C4 = {37}, locked for R9 and N8
6f. cleanup: no 4,8 in R8C3; no 2,6 in R9C89

7. Now continue with Andrew's step 11 (IOU, C1234), which directly fixes R9C45

_________________
Cheers,
Mike