SudokuSolver Forum

Another milestone for us... the big 1-5-0

Thought this was a good time to return to our roots, with a simple(r) puzzle,
to entice the newer members and our guests!

I wanted to use a traditional cage pattern, no diagonals or remote cages,
and lots of I/O's that may or may not be useful, just like Ruud's early Assassins.
Many regulars will no doubt find this puzzle too easy, so I plan to post a V2 soon.

Enjoy!

Assassin 150




Code string:

3x3::k:1792:2561:2561:4099:4099:4099:3078:3078:2824:1792:3338:3851:3851:4099:1038:1038:4624:2824:4882:3338:3338:3861:4630:5399:4624:4624:2586:4882:3356:3861:3861:4630:5399:5399:2338:2586:4882:3356:2086:2086:4630:1577:1577:2338:2586:1069:4398:7215:7215:4145:7986:7986:4660:3381:1069:4398:7215:4145:4145:4145:7986:4660:3381:4398:4398:7215:7215:3395:7986:7986:4660:4660:5704:5704:5704:5704:3395:4685:4685:4685:4685:

Solution:

591327846
267841395
843569271
759138462
486295137
132476589
378612954
615984723
924753618

SS Score 1.03

Happy Easter everyone!

Ronnie

I found this one harder than SudokuSolver's rating suggested since it took me some time to find the breakthrough move. I hope someone finds an easier way to crack it.

A150 Walkthrough:
1. R789
a) Innie R9 = R9C5 = 5
b) Cage sum: R8C5 = 8
c) 4(2) = {13} locked for C1
d) Innies C1 = 15(2) = [69/78/96]

2. R123
a) 7(2) = {25} locked for C1+N1
b) 4(2) = {13} locked for R2
c) Innies N3 = 4(2) = {13} locked for N3
d) 12(2) <> 9
e) 11(2) = {29/56} since (47) is a Killer pair of 12(2)
f) Outies R1 = 11(3) = {245} -> R2C5 = 4; 2,5 locked for R2
g) 13(3) = 4{18/36} because {139} blocked by R3C9 = (13); R3C23 = 4{1/3} -> 4 locked for R3+N1

3. C123 !
a) 19(3) = 4{69/78} -> 4 locked for N4
b) 13(2) <> 9
c) Killer pair (68) locked in R2C2 + 13(2) for C2
d) 17(4): R678C2 <> 9 because R8C1 >= 6
e) ! Innies+Outies C12: -7 = R13C3 - R9C12: R9C2 <> 9 because R9C12 = [69] clashes with Innies C1 = 15(2)
and R9C12 = [89] implies R3C13 = 10 which clashes with 10(2) @ N1 (Double IOU?)
f) Hidden Single: R1C2 = 9 @ C2 -> R1C3 = 1
g) 13(3) = {346} -> R2C2 = 6; 3 locked for R3
h) 13(2) = {58} locked for C2+N4
i) 19(3) = {478} since R3C1 = (78) -> R3C1 = 8; 7 locked for C1+N4
j) 17(4) = {1367} because {1349} blocked by R3C2 = (34) -> R8C1 = 6; 1,3 locked for C2 and 7 locked for N7

4. C789
a) 11(2) = {56} -> R2C9 = 5, R1C9 = 6
b) 13(2) = {49} locked for C9
c) Innies C9 = 11(2) = {38} -> R8C9 = 3, R9C9 = 8
d) Innie R12 = R2C8 = 9
e) R3C9 = 1, R2C7 = 3
f) 3 locked in 9(2) @ N6 = {36} locked for C8+N6
g) 18(4) @ R6C8 = {2358} -> R6C8 = 8; 2,5 locked for C8+N9

5. C456
a) 16(4) @ N2 = {2347} -> 2,3,7 locked for R1+N2
b) Innies C1234 = Innies C6789 = 9(2): R7C46 = (267)
c) 22(4) = 49{27/36} -> R9C1 = 9; R9C4 = (67)
d) Hidden Single: R5C5 = 9 @ R5
e) R3C5 = 6 -> R4C5 = 3, R4C8 = 6, R5C8 = 3
f) 8(2) = {26} locked for R5
g) Hidden Single: R1C4 = 3 @ C4
h) Innie C1234 = R7C4 = 6

6. Rest is singles.

Rating: Easy 1.25. I used small IOD combo analysis.

Agreed - on the Casey scale this was "no pushover". Is your middle name Ruud? It looks as if it's going to be as easy as you said but then it jams up. Sorry I haven't kept track of how I got it as I was listening to Jimi Hendrix at the same time.

cheers

_________________
Joe

Well, first off let me apologize if I was misleading about the difficulty of V1. The last thing I wanted to do was scare away newbies or wannabes

When I solved it, I did use sort of a "trick" move, (actually more like a one-two punch), but the rest of the puzzle was straightforward, or at least seemed so to me. I assumed from the SS score that there was an easier way. I've never offered a personal "score" for any Assassin, and now I probably never will! Obviously I don't know how to properly rate a puzzle!

Moving on, here's the V2 I promised. I didn't really like this puzzle and searched for something better but I've run out of time. I had to resort to some fairly heavy combo-crunching to solve it, which I find tedious, but others might enjoy. It has a slightly altered cage pattern from V1 and also a different solution.

A150 V2



Code string: 3x3::k:1792:4097:4097:4099:4099:4099:2310:2310:3336:1792:2826:2827:2827:4099:2830:2830:2832:3336:5394:2826:2826:5653:4118:6167:2832:2832:4634:5394:3868:5653:5653:4118:6167:6167:4130:4634:5394:3868:3868:5653:4118:6167:4130:4130:4634:5394:3886:5935:5935:4145:6962:6962:4404:4634:3886:3886:5935:4145:4145:4145:6962:4404:4404:2623:2623:5935:5935:3395:6962:6962:1606:1606:4424:4424:4424:4424:3395:6221:6221:6221:6221:


Solution:
279581364
568324719
314976285
857269143
691438572
423157698
945713826
736892451
182645937


SS score: 1.67


P.S. While searching for a better V2, I did stumble across a fun puzzle, SS 1.31, which I will gladly send to anyone who asks!

Thanks for this assassin, Ronnie. It offers some interesting dependences between cells and cages, that are difficult to explain in a WT. I have tried to avoid hard combinations analysis and have prefered a short forcing chain that seems to me easier whereas it increases the puzzle rating. No way don't worry too much about the rating, Ronnie ; it is only an indication about the techniques needed, which is slightly different of the difficulty felt by a(n?) human solver. Combo analysis employed by SSolver are sometimes very hard to follow and could be viewed as a kind of contradiction step.

Edit : thanks to Ed for helpful comments
Walkthrough Assassin 150 V1

That was a tough Killer! Like V1 I hope that someone finds an easier way to solve it. Maybe by using a small forcing chain?

A150 V2 Walkthrough:

1. R6789
a) Innie R9 = R9C5 = 4
b) Cage sum: R8C5 = 9
c) Innies R6789 = 12(2) <> 1,2,6

2. R123 !
a) 16(2) = {79} locked for R1+N1
b) Innies N1 = 11(2) = {38/56}
c) 9(2) <> 2
d) Innies N3 = 12(2) <> 1,2,6
e) 11(2) @ N1 = {38/56}
f) 2 locked in 11(3) @ N3 = 2{18/36/45} <> 7
g) ! Hidden Killer pair (79) in Innies N3 + R2C9 since Innies N3 can only have one of them
-> R2C9 = (79) and Innies N3 = {39/57}
h) 13(2) = [49/67]
i) 11(2) @ N3: R2C6 = (2468)

3. R123
a) Outies R1 = 16(3): R2C5 <> 7 since R2C9 = (79)
b) 7,9 locked in R3C456 @ N2 for R3
c) Innies N3 = 12(2) = [79/35]
d) 11(2) @ N3 = [29/47]

4. C456
a) Innies C6789 = 4(2) = {13} locked for C6
b) Innies C1234 = 12(2) = [48/57]
c) 16(4) @ N8 = 1{258/267/357} because R7C4 = (78) -> R67C5 <> 7,8
d) 7 locked in 16(3) @ C5 = 7{18/36}

5. R123
a) 9(2) = {18/36} since (45) blocked by R1C4 = (45)
b) Killer pair (13) locked in R1C6 + 9(2) for R1
c) 7(2): R2C1 <> 4,6
d) Outies R1 = 16(3) = {169/178/259} <> 3 because R2C9 = (79) and {367} blocked by Killer pair (36) of 11(2) @ N1
e) 7(2) <> 4
f) 4 locked in 11(3) @ N1 = 4{16/25}
g) 3,8 locked in Innies N1 = 11(2) = {38}
h) 11(2) @ N1 = {38} locked for R2
i) Innies R12 = 7(2) <> 4

6. R123
a) Hidden Single: R2C6 = 4 @ R2 -> R2C7 = 7
b) R2C9 = 9 -> R1C9 = 4, R1C4 = 5
c) Innie N3 = R3C9 = 5
d) 18(4) = {2358} -> 2,3,8 locked for C9+N6
e) R8C9 = 1 -> R8C8 = 5
f) 16(4) = 25{18/36} -> 2 locked for C5+N2; R2C5 <> 1 because R1C6 = (13)
g) Killer pair (38) locked in 16(4) + R2C4 for N2

7. R6789
a) 16(4) = {1357} -> R7C4 = 7
b) R7C9 = 6, R9C9 = 7
c) 17(3) = [74/92]6
d) Innies R6789 = 12(2) = [48/93]
e) 24(4) = 7{269/359/368} -> R9C6 = (56)

8. R456
a) 16(3) @ N6 = {169/457}: R5C7 <> 4 since R45C8 <> 5
b) Killer pair (79) locked in 16(3) @ N6 + R6C8 for N6
c) 24(4) <> {4569} since it's blocked by R9C6 = (56)

9. C123
a) Innies C1 = 17(3) = {179/278/359/458/467} because 6{29/38} blocked by R1C1 = (26) and R1C3 = (38)
b) Innies C1 = 17(3): R79C1 <> 3 since R8C1 <> 5,9 and R8C1 <> 2,6 because 7 only possible there
c) 10(2): R8C2 <> 4,8

10. R6789 !
a) 27(5): R6C6 <> 2,6 because 7 only possible there and 39{168/258/456} with R78C7 = {39} blocked by Killer pair (39) of 24(4) @ N9
b) 27(5): R6C6 <> 8 since 7 only possible there, 389{16/25} with R78C7 = {39} blocked by Killer pair (39) of 24(4) @ R9 and R8C6 <> 1,4,5,9
c) ! Outies N9 = 29(3+2): R6C68 must have 9 because
- R6C678 <> [517] = 13 since R89C6 <= 14
- R6C678 <> [547] = 16 since it clashes with R89C6 = 13 = [85]
- R6C678 <> [567] = 18 since it clashes with R89C6 = 11 = [65]
-> 9 locked in R6C68 for R6
d) Innies R6789 = 12(2) = {48} -> R6C1 = 4, R6C9 = 8
e) Innies C1 = 17(3): R8C1 <> 8 since R79C1 <> 4,7
f) 10(2) = {37} locked for R8+N7
g) 15(3) <> 7 since R7C23 <> 3,6,7

11. C456+N6
a) 4,9 locked in 22(4) @ C4 for 22(4) -> 22(4) = 49{18/27/36} <> 5
b) Innies N6 = 16(3): R4C7 <> 5 since 4 only possible there
c) 24(4) @ N6 = {1689/2679/4578} because R4C7 = (146)
d) Hidden Killer pair (28) in R8C6 for C6 since 24(4) @ N6 can only have one of (28) -> R8C6 = (28)

12. C123+R8 !
a) 6 locked in 23(5) @ R8 for 23(5)
b) Innies N14 = 20(3+1) = 3+8{27/36} / 8+{138/156/237} -> 8 locked in R24C3 for C3
c) ! Innies N14 = 20(3+1) <> 3+{368} since R6C3 = 3 sees R2C3 = 3
d) ! Innies N14 = 20(3+1) <> 8+{138} since R4C3 = 8 sees R2C3 = 8
e) 15(3) @ R4C2 <> 3,7 because 7{26/35} are blocked by Killer pairs (57,67) of Innies N14
f) ! Killer pair (12) locked in 15(3) @ R4C2 + Innies N14 for N4
g) 21(4) = 34{59/68} -> 3 locked for C1
h) R8C1 = 7, R8C2 = 3
i) Hidden Single: R1C2 = 7 @ C2 -> R1C3 = 9
j) Hidden triple (378) locked in R246C3 @ C3 -> R46C3 = (378)
k) Innies N14 = 20(3+1) = {378}+2 -> R6C2 = 2

13. N69
a) Hidden Single: R6C7 = 6 @ R6
b) 16(3) = {457} -> R5C7 = 5; 4,7 locked for C8+N6
c) R4C7 = 1
d) 24(4) @ N6 = {1689} -> 6,8,9 locked for C6; 8 also locked for N5
e) 24(4) @ R9 = {3579} -> R9C6 = 5; 3,9 locked for R9+N9

14. Rest is singles.

Rating: (Easy?) 1.75. I used combo analysis of large Outies.

From these comments and my walkthrough I'll assume there probably isn't an easier way that we would consider to be a 1.0 solving path. When I've occasionally commented to Ed that I've looked for easier moves because of a SS score, he's replied that sometimes SS has used a harder step than one would expect from the score. This can happen if there's only one harder step and the solution is fairly short; that may have happened for A150. See also Ed's message later in this thread.

I took an overnight break after reaching innie-outie combo analysis. When I returned to the puzzle I found a better innie-outie but still needed combo analysis. Then after going through Afmob's walkthrough I found a simplified version of the key step which I've given after my walkthrough.

I'll rate A150 at Easy 1.25; that also applies to the simplified version because it still uses a combo overlap.

Here is my walkthrough.

Prelims

a) R12C1 = {16/25/34}, no 7,8,9
b) R1C23 = {19/28/37/46}, no 5
c) R1C78 = {39/48/57}, no 1,2,6
d) R12C9 = {29/38/47/56}, no 1
e) R2C34 = {69/78}
f) R2C67 = {13}, locked for R2, clean-up: no 4,6 in R1C1, no 8 in R1C9
g) R45C2 = {49/58/67}, no 1,2,3
h) R45C8 = {18/27/36/45}, no 9
i) R5C34 = {17/26/35}, no 4,8,9
j) R5C67 = {15/24}
k) R67C1 = {13}, locked for C1, clean-up: no 4,6 in R2C1
l) R67C9 = {49/58/67}, no 1,2,3
m) R89C5 = {49/58/67}, no 1,2,3
n) R345C1 = {289/379/469/478/568}, no 1
o) 21(3) cage at R3C6 = {489/579/678}, no 1,2,3
p) R345C9 = {127/136/145/235}, no 8,9

1. 45 rule on R9 1 innie R9C5 = 5, R8C5 = 8

2. Naked pair {25} in R12C1, locked for C1 and N1, clean-up: no 8 in R1C23

3. 45 rule on C1 2 innies R89C1 = 15 = [69/78/96]

4. 45 rule on C9 2 innies R89C9 = 11 = {29/47}/[38/56], no 1, no 6 in R8C9, no 3 in R9C9

5. 45 rule on N1 2 innies R2C3 + R3C1 = 15 = {69/78}, no 4
5a. 4 in C1 locked in R56C1, locked for N4, clean-up: no 9 in R45C2

6. 45 rule on N3 2 innies R2C7 + R3C9 = 4 = {13}, locked for N3, clean-up: no 9 in R1C78, no 8 in R2C9
6a. R12C9 = {29/56} (cannot be {47} which clashes with R1C78), no 4,7

7. 45 rule on R12 2 innies R2C28 = 15 = {69/78}
7a. Naked quad {6789} in R2C2348, locked for R2, clean-up: no 2,5 in R1C9
7b. R2C5 = 4 (hidden single in R2)

8. Min R2C2 = 6 -> max R3C23 = 7, no 6,7,8,9 (because cannot repeat 6 in cage)
8a. Naked triple {134} in R3C23 and R3C9, locked for R3
8b. 4 in R3 locked in R3C23, locked for N1, clean-up: no 6 in R1C23
8c. R3C23 = {14/34} = 5,7 -> R2C2 = {68}, clean-up: no 6,8 in R2C8 (step 7)
8d. Killer pair 6,8 in R2C2 and R45C2, locked for C2

9. 4 in R1 locked in R1C78 = {48}, locked for R1 and N3

10. 45 rule on C5 3 remaining innies R167C5 = 10 = {127/136}, no 9, 1 locked for C5

11. 45 rule on C1234 2 innies R17C4 = 9 = {27/36}/[54], no 1,9

12. 45 rule on C6789 2 innies R17C6 = 9 = {27/36}/[54], no 1,9

13. 17(4) cage at R6C2 = {1259/1367/1457/2357/2456} (cannot be {1349} which clashes with R3C2)
13a. 9 of {1259} must be in R8C1 -> no 9 in R678C2

[See at end for simplified version of step 14]
14. 45 rule on C12 3(1+2) innies R1C2 + R9C12 = 1 outie R3C3 + 17, IOU R9C12 cannot be [89] = 17
14a. Max R9C12 = 16 -> R1C2 must be greater than R3C3 -> no 1 in R1C2, clean-up: no 9 in R1C3
14b. R9C12 cannot be 15 (because of overlap clash with R89C1 = 15, step 3) -> R1C2 cannot be 2 greater than R3C3 -> no 3 in R1C2, clean-up: no 7 in R1C3

15. 45 rule on C12 2 innies R9C12 = 2 outies R13C3 + 7
15a. Min R13C3 = 4 -> min R9C12 = 11, no 1 in R9C2
15b. R13C3 = {13/14} (cannot be [34] because R9C12 cannot total 14), 1 locked for C3 and N1, clean-up: no 7 in R5C4
15c. R13C3 = 4,5 -> R9C12 = 11,12, no 6 in R9C1, no 7,9 in R9C2, clean-up: no 9 in R8C1

16. R1C2 = 9 (hidden single in C2), R1C3 = 1, R1C9 = 6, R2C9 = 5, R12C1 = [52], clean-up: no 6 in R2C3 + R3C1 (step 5), no 6,9 in R2C4, no 7,8 in R67C9

17. R345C9 = {127} (only remaining combination) -> R3C9 = 1, R45C9 = {27}, locked for C9 and N6, R2C67 = [13], clean-up: no 4 in R5C6, no 5 in R5C7

18. R89C9 = [38] (hidden pair in C9), R9C1 = 9, R8C1 = 6 (step 3)
18a. Naked pair {78} in R2C34, locked for R2 -> R2C2 = 6, R2C8 = 9, clean-up: no 7 in R45C2

19. Naked triple {237} in R1C456, locked for N2 -> R2C4 = 8, R2C3 = 7, R3C1 = 8, clean-up: no 1 in R5C4

20. Naked pair {58} in R45C2, locked for C2 and N4, clean-up: no 3 in R5C4

21. 7 in C2 locked in R678C2 -> 17(4) cage at R6C2 (step 13) = {1367} (only remaining combination), 3 locked for C2 -> R3C23 = [43], R9C2 = 2, R9C3 = 4, R9C4 = 7 (cage sum), R78C3 = [85], clean-up: no 5 in R5C4

22. Naked pair {26} in R5C34, locked for R5 -> R45C9 = [27], R45C1 = [74], R5C67 = [51], R45C2 = [58], R5C8 = 3, R4C8 = 6, R345C5 = [639], R9C78 = [61], R9C6 = 3, R4C3 = 9, R3C4 = 5, R4C4 = 1 (cage sum), R3C6 = 9

23. 16(4) cage at R6C5 = {1267} (only remaining combination) -> R6C5 = 7, R7C5 = 1, R7C46 = {26}, locked for R7 and N8 -> R8C6 = 4, R8C4 = 9

24. R7C3 + R8C34 = [859] = 22 -> R6C34 = 6 = [24]

25. Naked pair {27} in R8C78, locked for R8 and N9

and the rest is naked singles and a cage sum

Simplified version of step 14; in my original walkthrough I hadn’t spotted that steps 14 and 14b eliminate 9 from R9C2.

14. 45 rule on C12 3(1+2) innies R1C2 + R9C12 = 1 outie R3C3 + 17, IOU R9C12 cannot be 17 = [89]
14a. R9C12 cannot be [69] (because of overlap clash with R89C1 = 15, step 3)
14b. R9C12 not [69/89] -> no 9 in R9C2, R1C2 = 9 (hidden single in C2), R1C3 = 1, clean-up: no 7 in R5C4

Steps 15 and 16 are then simplified because of the placements in R1C23

I found this puzzle (A150) really tricky. I solved it using Andrew's alternate step 14 at the end of his WT. My first reaction on finding that way was, "Hmm - a composite move like that feels like at least a 1.50 rating" but then spend lots of frustration time trying to find an easier alternative. Wish I'd had manu's attitude with this puzzle and just been content with an interesting solution (and his has plenty of interest in it!!) and be "so-what" about the score and others' ratings.

I know SudokuSolver can't do that sort of composite move so thought I'd find out how it did it. I've explained everything out to make it so humans can follow how it works.

From here (at Afmob's step 3d or Andrew's step 13)

1. "45" on c12: innies r9c12 - 7 = 2 outies r13c3
1a. min. r13c3 = {13} = 4 -> min. r9c12 = 11 (no 1)

2. "45" on c2: 2 outies r3c3 + r8c1 + 2 = 2 innies r19c2
2a. 2 outies = [16] = 7 -> 2 innies = 9 = [72]
2b. 2 outies = [17] = 8 -> 2 innies = 10 = [73] ([19] blocked by 1 in r3c3; [37] blocked by 7 in r8c1)
2c. 2 outies = [19] = 10 -> 2 innies = 12 = [93] ([39] blocked by 9 in r8c1)
2d. 2 outies = [36] = 9 -> 2 innies = 11 = [74/92]
2e. 2 outies = [37] = 10 -> 2 innies = 12 = [93] ([39] blocked by 3 in r3c3)
2f. 2 outies = [39] = 12 -> 2 innies = 14 blocked
2g. 2 outies = [46] = 10 -> 2 innies = 12 = {39}
2h. 2 outies = [47] = 11 -> 2 innies = 13 = [94]
2i. 2 outies = [49] = 13 -> 2 innies = 15 blocked
2j. In summary, no 1 in r1c2, no 7 in r9c2 (this is the crucial one)
2k. -> No 9 in r1c3

3. "45" on c2: 3 outies r13c3 + r8c1 - 8 = 1 innie r9c2
3a. Looking at the 3 outies: with 9 in r8c1 and min. r13c3 = 4 which means a min. of 13 in the outies -> min. in r9c2 = 5. However, 9 is not available in r9c2 because it's in r8c1. -> no 9 in r8c1
3b. no 6 in r9c1 (h15(2))

4. "45" on c12: 3 innies r1c2 + r9c12 - 17 = 1 outie r3c3
4a. Now Andrew's nice 4-cell IOU move works in a simple way. r9c12 cannot equal the IOD of 17 since that would force r1c2 & r3c3 to be equal -> no 9 in r9c2 since r9c12 could only be [89] = 17.

Step 2 looks very much like hypotheticals so we'd probably give it a rating of 2.00. Can anyone see a pattern method to get rid of the 7 from r9c2?

How did SS get such a low score?
1) Unlike humans, it's "45" routines don't just work on IOU and max./min. but all permutations. Unfortunate in this case that it found the key one (no 7 in r9c2) so early in it's solution.
2) The SSscore is just the sum of the step count with weightings given to each step type. It found step 4 above by step 60 which is unusually early for most Assassins.
3) The first key placement at r1c2 has an unusual property. It basically reduces the whole puzzle to singles in one stroke. This reduces the step count even further and hence, gives a lower score. This type of placement is apparently called a "backdoor single". udosuk mentions this here and was well known in vanilla sudoku circles at that time (Sept '06).

It would be interesting to see what would happen to the overall correlation for all Assassins of the SSscore if the main "45" routines just did IOU and max/min. eliminations. Worth a try Richard?

So the moral is, the SSscore and human ratings can be quite different!

Another interesting week. Thanks Ronnie!

Cheers
Ed

Just tidying up my files for A150 and I realised that I hadn't commented on Ed's interesting post about how SS solved A150.

As I said to Ed at the time, I didn't see my alternate step 14 as a composite move; to me there are two separate and independent parts in this step, each of which eliminates one permutation. I don't see that as being any different to gradually eliminating combinations/permutations from a fixed cage. I'm therefore surprised that SS can't do what I did as two steps.

I'll agree that SS's step 2 looks very much like hypotheticals, which I'd rate at least 1.75 and possibly 2.0; in this case more likely 2.0 because some of the I-O permutation analysis involved tricky blocking.

It's interesting that the elimination of 7 from r9c2 was the key in SS's solving path. I felt that eliminating 9 from r9c2, which made r1c2 = 9 a hidden single in c2, was the key elimination.

I know it's way out of date but been meaning to have a look at how JSudoku does A-1-5-0. For another reason (to be revealed in about a week ), I've finally taken an interest in how JSudoku solves killers.



From marks above (copy and "Paste Into" A150 in SudokuSolver)
1. 3 of n1 locked in cage 10(2) in r1c23 or h16(4) in r3c2378 -> 7 also locked in Cage 10(2) or h16(4) -> no 7 in r3c1 (Locked cages)
1a. no 8 in r2c3 (h15(2)n1)
1b. no 7 in r2c4 (cage sum)

2. 19(3)n1 = {469/478}
2a. 8 of {478} must be in r3c1 -> no 8 in r45c1

3. Grouped XY-Chain -> no 6 in r2c3. Like this
3a. if r2c2 is not 6 it is 8 -> 8 in c1 is in r9c1 in h15(2)r89c1 (no 6) -> 6 in n7 in c789c3 -> no 6 in r2c3
3b. if r2c2 = 6 -> no 6 in r2c3
3c. no 9 in r2c4
3d. no 9 in r3c1 (h15(2)n1)

4. XY-X Chain -> no 6 in r3c78. Like this.
4a. If r1c9 is not 6 it is 9 -> r2c8 = 7 -> r1c3 = 9 -> r3c1 = 6 (h15(2)n1) -> no 6 in r3c78
4b. if r1c9 = 6 -> no 6 in r3c78

5. r1c9 = 6 (hsingle n3)

Now it's cracked.

Some nice moves in there.

Cheers
Ed