SudokuSolver Forum

Don't be afraid of these hexagonal cages because they could be very useful !
I have made different versions with the same cage pattern and hope that this one
has the right difficulty. It is the kind of killer I like : no early cage combination fixed, and the whole cage pattern must be used. I have an other version that could offer a nice V2 : it will be sent later
according to the activity on this V1 and JFFK6 as soon as A154 is claimed.

*****************


ASSASSIN 153





PS Code :3x3::k:2560:6913:2306:2306:3332:4101:4101:7431:2560:6913:3850:6913:2306:3332:4101:7431:2320:7431:6913:3850:6913:3349:2070:1303:7431:2320:7431:3611:6913:3357:3349:2070:1303:4129:7431:3875:3611:3357:3357:6951:6951:6951:4129:4129:3875:2349:2862:2607:6951:7729:6951:2611:2612:3125:2349:2862:2607:7729:2106:7729:2611:2612:3125:3135:3135:5697:7729:2106:7729:4421:3910:3910:3135:5697:5697:5697:7729:4421:4421:4421:3910:


Solution :

SSscore : 1.27


PS : r1c19 is a 10(2) cage

I really liked the cage pattern of this Killer, thanks manu!

There were lots of moves I omitted to shorten my wt which would have probably led to a different solution path, so I guess there should be different ways to tackle it.

A153 Walkthrough:

1. N689
a) 8(2) <> {17} since it's a Killer pair of 30(6)
b) 30(6) = 1489{26/35} since other combos blocked by Killer pairs (23,25,36) of 8(2)
c) 7 locked in R9C46 @ N8 for R9
d) Innies+Outies N69: 5 = R9C6 - R4C8 -> R9C6 = (6789), R4C8 = (1234)

2. N1
a) 27(6) = 1245{69/78} since other combos blocked by Killer pairs (67,68,79,89) of 15(2)
b) 3 locked in R1C13 @ N1 for R1

3. R789+N46
a) Outies R89 = 14(4) <> 9
b) 30(6) = 1489{26/35} -> 9 locked for N8
c) Innies+Outies N47: 1 = R9C4 - R4C2 -> R9C4 <> 1,4; R4C2 <> 8,9
d) 1,4 in N8 locked in 30(6) for 30(6)
e) Innies+Outies N69: 5 = R9C6 - R4C8: R4C8 <> 4

4. R123
a) Killer pair (89) locked in 27(6) + 15(2) for N1
b) 10(2): R1C1 <> 7; R1C9 <> 1,2
c) R4C8 <> 3 since it sees all 3 of N3
d) Innies+Outies N3: -7 = R4C8 - R1C79: R1C79 <> 9 since R4C8 <= 2
e) Innies+Outies N3: -7 = R4C8 - R1C79: R1C7 <> 4,6,7,8 since R1C9 >= 4, R1C9 <> 5 and R4C8 <= 2

5. C789 !
a) ! Hidden Killer pair (12) in R23C9 + R89C9 for C9 since 15(3) can only have one of (12) and R23C9 = (12) blocked by R4C8 = (12)
-> 15(3) = {159/168/249/258/267} <> 3 and Killer pair (12) locked in R23C9 + R4C8 for 29(6); CPE: R23C8 <> 1,2
b) 9(2) <> 7,8
c) 9 locked in 29(6) @ N3 = 1289{36/45} because {124679} blocked by Killer pair (46) of 9(2) -> 8 locked for N3
d) Hidden Single: R1C9 = 7 @ N3 -> R1C1 = 3
e) 15(2) = {69} locked for C9+N6
f) 12(2) = {48} locked for C9
g) 3 locked in 29(6) @ C9 = {123689} for N3
h) 9(2) = {45} locked for C8+N3
i) 15(3) = 5{19/28} because R89C9 = (125) -> 5 locked for N9; R8C8 = (89)
j) Innies+Outies N69: 5 = R9C6 - R4C8: R9C6 <> 8

6. C789
a) 17(4) = 16{28/37} because R9C6 = (67) -> 1 locked for N9
b) 15(3) = {258} -> R8C8 = 8; 2 locked for C9+N9
c) R7C9 = 4, R6C9 = 8
d) 16(3) @ N6 = {457} -> R5C8 = 7; 4 locked for C7
e) 10(2) @ C8 = {19} -> R6C8 = 1, R7C8 = 9
f) 10(2) @ C7 = {37} -> R6C7 = 3, R7C7 = 7
g) Innies+Outies N69: 5 = R9C6 - R4C8 -> R4C8 = 2, R9C6 = 7
h) 17(4) = {1367} -> R9C8 = 3; 1,6 locked for C7
i) R1C7 = 2, R1C8 = 6

7. C456
a) 13(2) @ C5 <> {67}
b) 16(3) @ N2 = {268} because {259} blocked by Killer pair (59) of 13(2) @ N2 -> R1C6 = 8, R2C6 = 6
c) 13(2) @ C5 = {49} locked for C5+N2
d) 9(3) = {135} because R1C34 <> 2,3 -> R2C4 = 3; 1,5 locked for R1
e) Innies R6789 = 14(2) = {59} locked for R6+N5
f) Both 8(2) <> {26} because it's blocked by R6C5 = (26)
g) Hidden Single: R3C6 = 2 @ N2 -> R4C6 = 3
h) 8(2) @ N2 = {17} locked for C5
i) 27(5) = {14589} -> R5C5 = 8; 1,4 locked for R5+N5
j) R5C7 = 5, R4C7 = 4, R4C5 = 7, R4C4 = 6, R4C9 = 9, R5C9 = 6, R5C1 = 9 -> R4C1 = 5

8. N14
a) 9(2) = {27} -> R7C1 = 2, R6C1 = 7
b) R4C2 = 1, R4C3 = 8
c) 10(2) = {46} -> R7C3 = 6, R6C3 = 4
d) 11(2) = {56} -> R6C2 = 6, R7C2 = 5
e) 15(2) = {78} locked for C2+N1

9. Rest is singles.

Rating: Hard 1.0. I used a Hidden Killer pair.

Thanks manu for a great puzzle with an interesting cage pattern!

As will be seen from my comment before steps 11, 12 and 13, I took some time to get going on this Assassin. I also ought to have spotted step 19 earlier. I had seen my other key move, step 24, much earlier but it wasn't useful until there was a cage sum limit on R3C456 and therefore on R1C37.

I'll rate my walkthrough for A153 at least 1.25 because I used several combined cages and the hidden killer pair in step 24 also feels like that rating.

Here is my walkthrough. Thanks Afmob for your comments about step 26; I hope it's clearer now. I've also added a note at the end of step 19.

Prelims

a) R1C19 = {19/28/37/46}, no 5
b) R12C5 = {49/58/67}, no 1,2,3
c) R23C2 = {69/78}
d) R23C8 = {18/27/36/45}, no 9
e) R34C4 = {49/58/67}, no 1,2,3
f) R34C5 = {17/26/35}, no 4,8,9
g) R34C6 = {14/23}
h) R45C1 = {59/68}
i) R45C9 = {69/78}
j) R67C1 = {18/27/36/45}, no 9
k) R67C2 = {29/38/47/56}, no 1
l) R67C3 = {19/28/37/46}, no 5
m) R67C7 = {19/28/37/46}, no 5
n) R67C8 = {19/28/37/46}, no 5
o) R67C9 = {39/48/57}, no 1,2,6
p) R78C5 = {17/26/35}, no 4,8,9
q) 9(3) cage at R1C3 = {126/135/234}, no 7,8,9

1. 45 rule on R6789 2 innies R6C46 = 14 = {59/68}

2. 45 rule on R1234 4 innies R4C1379 = 26 = {2789/3689/4589/4679/5678}, no 1

3. 45 rule on N6 4 innies R4C8 + R6C789 = 14 = {1238/1247/1256/1346/2345}, no 9, clean-up: no 1 in R7C78, no 3 in R7C9

4. 45 rule on N47 1 outie R9C4 = 1 innie R4C2 + 1, no 9 in R4C2, no 1 in R9C4

5. 45 rule on N69 1 outie R9C6 = 1 innie R4C8 + 5, R4C8 = {1234}, R9C6 = {6789}
5a. Min R9C6 = 6 -> max R8C7 + R9C78 = 11, no 9

6. 45 rule on N1 2 outies R1C9 + R4C2 = 1 innie R1C3 + 7, IOU no 7 in R4C2, clean-up: no 8 in R9C4 (step 4)

7. 45 rule on N3 2 outies R1C1 + R4C8 = 1 innie R1C7 + 3, IOU no 3 in R4C8, clean-up: no 8 in R9C6 (step 5)

8. Combined cage R4567C9 = 27 = {3789/4679/5679}, 9 locked for C9, clean-up: no 1 in R1C1

9. 45 rule on C9 3 innies R123C9 = 1 outie R8C8 + 3
9a. Min R123C9 = 6 -> min R8C8 = 3

10. 45 rule on N1235 2 outies R4C28 = 1 innie R6C5 + 1
10a. Min R4C28 = 3 -> min R6C5 = 2

I ought to have spotted the next three steps earlier; manu gave a clue about the hexagons. The central 2-cell cages “see” all the cells of the surrounding hexagons so each hexagon plus central cage forms a 8-cell cage.

11. 27(6) cage at R1C2 + R23C2 = 42 = {12456789}, no 3, clean-up: no 4 in R9C4 (step 4)
11a. 3 in N1 locked in R1C13, locked for R1, clean-up: no 7 in R1C1
11b. 45 rule on N1 2 innies R1C13 = 1 outie R4C2 + 3
11c. R4C2 = {124568} -> R1C13 {13/23/34/35/36/38} (because R1C13 must contain 3), no 9 in R1C1, clean-up: no 1 in R1C9

12. 29(6) cage at R1C8 + R23C8 = 38 = {12345689}, no 7, clean-up: no 2 in R23C8
12a. 7 in N3 locked in R1C79, locked for R1, clean-up: no 6 in R2C5
12b. 45 rule on N3 2 innies R1C79 = 1 outie R4C8 + 7
12c. R4C8 = {124} -> R1C79 = {17/27/47} (because R1C79 must contain 7), no 5,6,8,9, clean-up: no 2,4 in R1C1

13. 30(6) cage at R6C5 + R78C5 = 38 = {12345689}, no 7, clean-up: no 1 in R78C5
13a. 7 in N8 locked in R9C46, locked for R9
13b. 45 rule on N8 2 innies R9C46 = 1 outie R6C5 + 7
13c. R9C46 = {27/37/57/67/79} (because R9C46 must contain 7) -> no 4,8 in R6C5

14. Combined cage R3478C5 = 16 = {1267/1357/2356}
14a. 45 rule on C5 3 innies R569C5 = 16 = {268/349/358} (cannot be {169/259/457} which clash with R3478C5, cannot be {178} because no 1,7,8 in R6C5, cannot be {367} which clashes with R78C5), no 1,7

15. 1 in C5 locked in R34C5 = {17}, locked for C5, clean-up: no 6 in R1C5

16. R4C28 = R6C5 + 1 (step 10)
16a. R4C28 cannot total 4 -> no 3 in R6C5 -> no 3 in R9C4 (step 13c), clean-up: no 2 in R4C2 (step 4)
16b. R4C2 = {14568} -> R1C13 (step 11b) = {13/34/35/36/38}, no 2

17. R569C5 (step 14) = {268/349/358}
17a. 5,9 of {349/358} must be in R6C5 -> no 5,9 in R59C5

18. R6C46 = 14 (step 1) -> R5C456 = 13 = {139/148/238/247/346} (cannot be {157} which clashes with R4C5, cannot be {256} which clashes with R6C46), no 5

19. 45 rule on R89 4 outies R6C5 + R7C456 = 14 = {1238/1256/1346/2345}, no 9
[These are 4 outies, rather than 4(3+1) outies, because they can all “see” each other.]
19a. R6C5 = {256} -> R9C46 (step 13c) = {27/57/67}, no 9, clean-up: no 8 in R4C2 (step 4), no 4 in R4C8 (step 5)
[In hindsight, after seeing Ed’s partial walkthrough, I ought now to have continued with
19b. R4C2 = {1456} -> R1C13 (step 11b) = {13/34/35/36}, no 8, clean-up: no 2 in R1C9
19c. R4C8 = {12} -> R1C79 (step 12b) = {17/27}, no 4, clean-up: no 6 in R1C1
19d.R1C19 = [37]
This would have avoided the need for the hidden killer pair in step 24.
I can only think that I must have been distracted by spotting 9 locked in C5.]

20. 9 in C5 locked in R12C5 = {49}, locked for C5 and N2, clean-up: no 4,9 in R4C4, no 1 in R4C6

21. 45 rule on N5 4 innies R4C456 + R6C5 = 18 = {1368/1467/2367/2457} (cannot be {1278} because 1,7,8 only in R4C45, cannot be {1458/2358/3456} which clash with R6C46)
21a. 3,4 of {2367/2457} must be in R4C6 -> no 2 in R4C6, clean-up: no 3 in R3C6
21b. 7 of {2457} must be in R4C5 with 5 in R4C4 -> no 5 in R6C5
21c. R6C5 = {26} -> R9C46 (step 13c) = {27/67}, no 5, clean-up: no 4 in R4C2 (step 4)

22. R78C5 = {35} (cannot be {26} which clashes with R6C5), locked for C5 and N8

23. R5C456 (step 18) = {148/238/247} (cannot be {139} because R5C5 only contains 2,6,8, cannot be {346} which clashes with R4C6), no 6,9
23a. 9 in N5 locked in R6C46 = {59}, locked for R6 and N5, clean-up: no 8 in R3C4, no 4 in R7C1, no 2,6 in R7C2, no 1 in R7C3, no 7 in R7C9

24. Hidden killer pair 3,7 in R1C19 and R1C37 for R1 -> either R1C19 = [37] or R1C37 = [37] (R1C19 is a 10(2) cage so must either contain both of 3,7 or neither of them)
24a. 45 rule on N2 3 innies R3C456 = 2 outies R1C37 + 7
24b. Max R3C456 = 15 -> max R1C37 = 8 so cannot be [37]
24c. R1C19 = [37], clean-up: no 8 in R45C9, no 6 in R67C1, no 5 in R7C9
24d. R1C13 = R4C2 + 3 (step 11b), R1C1 = 3 -> R1C3 = R4C2, no 4 in R1C3
24e. R1C79 = R4C8 + 7 (step 12b), R1C9 = 7 -> R1C7 = R4C8, no 4 in R1C7

25. Naked pair {69} in R45C9, locked for C9 and N6, clean-up: no 3 in R6C9, no 4 in R7C78
25a. Naked pair {48} in R67C9, locked for C9

26. Hidden killer pair 1,2 in R23C9 and R89C9 for C9, R23C9 cannot contain both of 1,2 because they would clash with R1C7 (or R4C8), R89C9 cannot contain both of 1,2 -> R23C9 and R89C9 must each contain one of 1,2
26a. Killer pair 1,2 in R1C7 and R23C9, locked for N3, clean-up: no 8 in R23C8

27. 8 in N2 locked in R12C6, locked for C6
27a. 16(3) cage at R1C6 = {178/268}, no 3,5
27b. R1C7 = 1,2 -> no 1,2 in R12C6
27c. Naked triple {678} in R129C6, locked for C6

28. R6C6 = 5 (hidden single in C6), R6C4 = 9
28a. R8C6 = 9 (hidden single in C6)

29. R2C4 = 3 (hidden single in N2), R1C34 = 6 = {15}, locked for R1 -> R1C7 = 2, R4C8 = 2 (step 24e), R9C6 = 7 (step 5), clean-up: no 6 in R3C8, no 6 in R4C2 (step 24d), no 8 in R67C7, no 8 in R67C8
29a. Naked pair {68} in R12C6, locked for N2, clean-up: no 7 in R4C4

30. R3C6 = 2 (hidden single in N2), R4C6 = 3
30a. 4 in N5 locked in R5C46, locked for R5

31. 45 rule on N6 3 remaining innies R6C789 = 12 = {138/147}, 1 locked for R6 and N6, clean-up: no 8 in R7C1, no 9 in R7C3
31a. 4 of {147} must be in R6C9, no 4 in R6C78, clean-up: no 6 in R7C78

32. 1 in N3 locked in R23C9, locked for C9
32a. 2 in C9 locked in R89C9 -> 15(3) cage at R8C8 = {258} (only remaining combination, cannot be {267} because 6,7 only in R8C8) -> R8C8 = 8, R89C9 = {25}, locked for C9 and N9, R23C9 = [13], R67C9 = [84], R7C6 = 1, R5C6 = 4, clean-up: no 6 in R2C8, no 7 in R6C2, no 6 in R6C3, no 7 in R6C78 (step 31), no 3 in R7C2, no 2 in R7C3, no 3 in R7C78

33. Naked pair {45} in R23C8, locked for C8 and N3
33a. Naked pair {13} in R6C78, locked for R6 and N6 -> R5C78 = [57], R4C7 = 4, R7C8 = 9, R6C8 = 1, R67C7 = [37], R1C8 = 6, R12C6 = [86], R9C8 = 3, clean-up: no 9 in R3C2, no 9 in R4C1, no 2 in R6C1, no 2,4 in R6C2, no 8 in R7C2
33b. R67C2 = [65], R6C5 = 2, R5C45 = [18], R1C34 = [15], R34C4 = [76], R34C5 = [17], R45C9 = [96], R9C45 = [26], R78C4 = [84], R78C5 = [35], R89C9 = [25], R5C1 = 9, R4C1 = 5, R4C23 = [18], R89C7 = [61]

35. R7C1 = 2, R6C1 = 7

and the rest is naked singles

manu,

I think this is your best puzzle yet! I loved all the interesting interactions you created: the hexagonal cages combined with the X(2) 's made so many CPE's.
It was a really fun puzzle. Thanks

Ronnie

Thanks for all these kind words about this puzzle. I recall I have a V2 ready (with a similar cage pattern) for you but I wait for A154 to be claimed, so ...

Here's a completely different way to get into this puzzle. No combo work just "45"s! More suitable for manu's V2 really, if he ever gets to post it.

Just to the first couple of placements. [edit: a couple of sub-steps shifted around for clarity; edit2: some important clarifications - thanks Andrew!]

Walk-partway for Assassin 153 [7 steps]
Prelims
i. 10(2)r1c19: no 5
ii. 9(3)n1: no 7,8,9
iii. 13(2)n2: no 1,2,3
iv. 15(2)n1 = {69/78}
v. 9(2)n3: no 9
vi. 13(2)n2: no 1,2,3
vii. 8(2)n2: no 4,8,9
viii. 5(2)n2 = {14/23}
ix. 14(2)n4 = {59/68}
x. 15(2)n6 = {69/78}
xi. 9(2)n4: no 9
xii. 11(2)n4: no 1
xiii. 10(2): no 5
xiv. 10(2) at r6c7 & r6c8: no 5
xv. 12(2)n6: no 1,2,6
xvi. 8(2)n8: no 4,8,9

1. "45" r89: 4 outies r6c5 + r7c456 = 14
1a. min. r7c456 = 6 -> max. r6c5 = 8

2. "45" n8: r6c5 + 7 = r9c46
2a. since the outie r6c5 sees all cells in n8 except the two innies -> the IOD (7) is locked in r9c46 (IOE: see here for a full explanation of this technique)
2b. 7 locked in r9c46 for r9 & n8
2c. no 7 in r6c5 (since r9c46 cannot be [77]
2d. no 9 in r9c46 since the other one of r9c46 = r6c5 (IOE)
2e. no 1 in 8(2)n8

3. "45" n3: 2 outies r1c1 + r4c8 - 3 = r1c7
3a. -> no 3 in r4c8 (IOU)

4. "45" n69: 1 outie r9c6 - 5 = 1 innie r4c8
4a. r9c6 = (67)
4b. r4c8 = (12)

5. "45" n3: 1 outie r4c8 + 7 = 2 innies r1c79
5a. since the one outie r4c8 see's all cells in n3 except the 2 innies -> IOD (7) locked in r1c79 (IOE)
5b. 7 locked for r1 & n3
5c. and r1c79 = 7{1/2} (since the other cell of r1c79 = r4c8 IOE)
5d. r1c1 = (389)
5e. no 2 in 9(2)n3

6. "45" n47: 1 outie r9c4 - 1 = 1 innie r4c2
6a. -> max r4c2 = 7, min. r9c4 = 2

7. "45" n1: 1 outie r4c2 + 3 = 2 innies r1c13
7a. and again! -> IOD (3) locked in r1c13 (IOE)
7b. 3 locked for r1 & n1
7c. no 8 or 9 in r1c1 (since the other cell of r1c13 = r4c2 IOE)
7d. r1c1 = 3
7e. r1c9 = 7 (cage sum)

On you go from there.

Cheers
Ed

Of course, the ideas you have used for the previous V1 will be useful, but you will certainly need other ones for cracking this one...

Assassin 153 V2




SSscore : 2.37

PS code : 3x3::k:2048:7169:2818:2818:3076:4357:4357:7431:2048:7169:3594:7169:2818:3076:4357:7431:2832:7431:7169:3594:7169:2837:1558:1303:7431:2832:7431:3867:7169:5149:2837:1558:1303:4385:7431:3619:3867:5149:5149:7463:7463:7463:4385:4385:3619:3629:5149:2607:7463:7729:7463:2099:4385:5429:3629:3629:2607:7729:2618:7729:2099:5429:5429:3391:3391:4929:7729:2618:7729:3909:4678:4678:3391:4929:4929:4929:7729:3909:3909:3909:4678:

Solution :

Got this V2 out way quicker than Ronnie's A154! So thank you manu! It just takes a single composite move to do crack it (step 7) and no, it's nothing to do with IO anything . I felt reluctant to use it because of how many elements are in it but couldn't find a simpler way.

Please let me know of any corrections or clarifications. Thanks.

Assassin 153 V2

Prelims
i. 8(2)r1c19: no 4,8,9
ii. 11(3)n1: no 9
iii. 12(2)n2: no 1,2,6
iv. 14(2)n1 = {59/68}
v. 11(2)n3: no 1
vi. 11(2)n2: no 1
vii. 6(2)n2 = {15/24}
viii. 5(2)n2 = {14/23}
ix. 15(2)n4 = {69/78}
x. 14(2)n6 = {59/68}
xi. 10(2)n4: no 5
xii. 8(2)n6: no 4,8,9
xiii. 21(3)n6: no 1,2,3
xiv. 10(2)n8: no 5

1. "45" n3: 2 outies r1c1 + r4c8 - 3 = 1 innie r1c7
1a. -> no 3 in r4c8 (IOU)

2. the 30(6)n5 and 10(2)n8 both "see" all of each other -> as a combined cage they make a 40(8) cage -> no 5

3. 5 in n8 only in r9c46: locked for r9.

4. "45" n69: 1 outie r9c6 - 3 = r4c8
4a. r9c6 no 1,2,3,6
4b. r4c8 no 7,8,9

5. "45" r123: 5 outies r4c24568 = 17 = 123{47/56}(no 8,9)
5a. 1,2 & 3 all locked for r4

6. "45" n47: 1 outie r9c4 - 1 = r4c2
6a. no 1,9 in r9c4

7. No 4 in r4c2. Like this.
7a. since 5 is locked at r9c46 and each of these two cells is linked to r4c28 -> r4c2 = 4 (step 6) OR r4c8 = 2 (step 4) and cannot be both
7b. for r4c2 to equal 4, the h17(5) in r5 can only be {12347} (step 5)
7c. and with 1 only in r4c8 (the only other available candidate is 2 but this is blocked see 7a.)
7c. -> the only permutation is [47231]
7d. but r4c45 as [72] is blocked by r3c45 = [44] (cage sums)
7e. -> no 4 in r4c2
7f. no 5 in r9c4 (i/o n47)

8. r9c6 = 5 (hsingle r9)
8a. other 3 cells in 15(4)n8 = 10 = 1{27/36}(no 4,8,9)
8b. 1 locked for n9
8c. no 7 in r6c7

The puzzle is cracked. I'll miss lots of clean-up since just getting it to singles ASAP.
9. "45" n69: 1 innie r4c8 = 2

10. "45" n6: 2 remaining innies r6c79 = 12 = [39/57] = [5/9...]

11. 14(2)n6 = {68} ({59} blocked by h12(2)n6 step 10)
11a. both locked for c9 and n6

12. 8(2)n6 = {35}: both locked for c7

13. "45" n3: 2 innnies r1c79 = 7 and must have 2 for n3 = [25]
13a. r1c1 = 3 (cage sum)

14. 15(4)n9 = {1356} only combination
14a. -> r9c8 = 3
14b. r89c7 = {16}: both locked for c7 & n9

15. r67c7 = [35]
15a. -> r6c9 = 9 (h12(2)n6)

16. h17(5)r4 = {1356}[2] ({1347}[2] blocked by r4c7)
16a. all locked for r4

17. r45c9 = [86]

18. r45c1 = [78] (last permutation)

19. "45" n1: 1 innie r1c3 = 1 outie r4c2 = (16)

20. "45" n4: 3 innies r4c2 + r6c13 = 10 = [1][54] (only permutation)
20a. -> r1c3 = 1 (step 19)

21. r7c89 = 12 = [84] (only permutation)

22. "45" n47: 1 outie r9c4 = 2

23. r7c3 = 6 (cage sum)

24. r7c12 = 9 = [27] (only permutation)

25. r4c34567 = [96534]
25a. r3c456 = [512] (cage sums)

26. r12c4 = 10 = [73] (last permutation)

27. r6c2 = 6 (hsingle n4)

28. r23c2 = [59] (last permutation)

29. r23c1 = {46}: both locked for n1 & c1
29a. r1c2 = 8

30. r12c5 = [48] (last permutation)

rest is singles.

Cheers
Ed

In my case I was stuck on Afmob's A155, a fairly low rated puzzle that is clearly a lot harder than it rating, so I had a go at A153 V2.

Thanks manu for another excellent puzzle!

As you said the ideas used for the V1 were useful, this time I started with the hexagons, but it needed more. My key breakthrough was step 15a, which is fairly similar to Ed's step 7a; I think spotting that step is almost certainly essential for a reasonable length solution. The SSscore is much higher than my rating so it's possible that it didn't find this step and used something more difficult.

Until I found that breakthrough move I looked several times at the hidden killer pair 3,5 in R1C19 and R1C37 for R1, either R1C19 = [35] or R1C37 = [35] but I was never able to find anything that stopped R1C37 being [35].

I'll rank A153 V2 at 1.5 because of step 15a.

Here is my walkthrough. I've deleted the original step 12 and rewritten some of the other steps for clarity.

Prelims

a) R1C19 = {17/26/35}, no 4,8,9
b) R12C5 = {39/48/57}, no 1,2,6
c) R23C2 = {59/68}
d) R23C8 = {29/38/47/56}, no 1
e) R34C4 = {29/38/47/56}, no 1
f) R34C5 = {15/24}
g) R34C6 = {14/23}
h) R45C1 = {69/78}
i) R45C9 = {59/68}
j) R67C3 = {19/28/37/46}, no 1
k) R67C7 = {17/26/35}, no 4,8,9
l) R78C5 = {19/28/37/46}, no 5
m) 11(3) cage at R1C3 = {128/137/146/236/245}, no 9
n) 21(3) cage at R6C9 = {489/579/678}, no 1,2,3

In the first 3 steps, the central 2-cell cages “see” all the cells of the surrounding hexagons so each hexagon plus central cage forms a 8-cell cage.

1. 28(6) cage at R1C2 + R23C2 = 42(8) = {12456789}, no 3
1a. 3 in N1 locked in R1C13, locked for R1, clean-up: no 5 in R1C1, no 9 in R2C5
1b. 45 rule on N1 2 innies R1C13 = 1 outie R4C2 + 3
1c. R1C13 contains 3 and one of {1245678} -> R4C2 = {1245678}, no 9
[Steps 1b and 1c, and the corresponding parts of steps 2 and 3, can also been seen as IOEs.]

2. 29(6) cage at R1C8 + R23C8 = 40(8) = {12346789}, no 5, clean-up: no 6 in R23C8
2a. 5 in N3 locked in R1C79, locked for R1, clean-up: no 7 in R2C5, no 5 in R4C2 (step 1c)
2b. 45 rule on N3 2 innies R1C79 = 1 outie R4C8 + 5
2c. R1C79 contains 5 and one of {1246789} -> R4C8 = {1246789}, no 3

3. 30(6) cage at R6C5 + R78C5 = 40(8) = {12346789}, no 5
3a. 5 in N8 locked in R9C46, locked for R9
3b. 45 rule on N8 2 innies R9C46 = 1 outie R6C5 + 5
3c. R9C46 = contains 5 and one of {12346789} -> R6C5 = {12346789}
[No eliminations at this stage but step 3c will be used for clean-ups later.]

4. 45 rule on N4 3 innies R4C2 + R6C13 = 10 = {127/136/145/235}, no 8,9, clean-up: no 8 in R1C3 (step 1c), no 1,2 in R7C3
4a. 5 of {145} must be in R6C1 -> no 4 in R6C1

5. 45 rule on N47 1 outie R9C4 = 1 innie R4C2 + 1, no 1,4,6,9 in R9C4

6. 45 rule on N69 1 outie R9C6 = 1 innie R4C8 + 3, no 7,8,9 in R4C8, no 1,2,3,6,8 in R9C6, clean-up: no 7,8,9 in R1C79 (step 2c), no 1 in R1C1, no 1,6 in R6C5 (step 3c)

7. 45 rule on R123 5 outies R4C24568 = 17 = {12347/12356}, no 8,9, 1,2,3 locked for R4, 3 locked for N5, clean-up: no 2,3 in R3C4
7a. R6C5 = {24789} -> R9C46 (step 3c) must contain 5 and one of {24789}, no 3, clean-up: no 2 in R4C2 (step 5)
7b. R4C2 = {1467} -> R1C13 (step 1c) must contain 3 and one of {1467}, no 2, clean-up: no 6 in R1C9

8. 3 in N5 locked in R4C46
8a. 45 rule on N5 4 innies R4C456 + R6C5 = 16 = {1348/1357/2347/2356}, no 9
8b. R6C5 = {2478} -> R9C46 (step 3c) must contain 5 and one of {2478}, no 9, clean-up: no 6 in R4C8 (step 6)
8c. R4C8 = {124} -> R1C79 (step 2c) = must contain 5 and one of {124}, no 6

9. R4C2 + R6C13 (step 4) = {127/136/145} (cannot be {235} because R4C2 only contains 1,4,6,7), 1 locked for N4

10. 45 rule on N6 3 innies R4C8 + R6C79 = 14 = {149/167/239/248/257/347} (cannot be {158} which clashes with R45C9, cannot be {356} because R4C8 only contains 1,2,4)
10a. 8,9 of {149/248} must be in R6C9, 4 of {347} must be in R4C8 -> no 4 in R6C9

11. 21(3) cage at R6C9 = {489/579/678}
11a. 4 of {489} must be in R7C9 (R67C9 cannot be {89} which clashes with R45C9), no 4 in R7C8

12. Deleted. This step, using 45 rule on C5 and then combined cage R34569C5, was unnecessarily complicated. Step 23b has been added to make the same eliminations.

13. Max R1C7 = 5 -> min R12C6 = 12, no 1,2
13a. Max R9C6 = 4 -> min R8C7 + R9C78 = 11, no 9

14. 45 rule on C9 2 outies R78C8 = 3 innies R123C9 + 8
14a. Min R123C9 = 6 -> min R78C8 = 14, no 1,2,3,4
14b. Max R78C8 = 17 -> max R123C9 = 9, no 7,8,9
14c. 7 in C9 locked in R6789C9, CPE no 7 in R7C8

15. 5 in N8 locked in R9C46
15a. Either R9C4 = 5 or R9C6 = 5 -> either R4C2 = 4 (step 5) or R4C8 = 2 (step 6) -> no 4 in R4C8, clean-up: no 7 in R9C6 (step 6)
15b. R4C8 = {12} -> R1C79 (step 2c) must contain 5 and one of {12}, no 4

16. R4C8 + R6C79 (step 10) = {167/239/257}, no 8
16a. R4C8 = {12} -> no 1,2 in R6C7, clean-up: no 6,7 in R7C7

17. 4 in N6 locked in 17(4) cage = {1349/1457/2348} (cannot be {2456} which clashes with R45C9), no 6

18. 15(4) cage at R8C7 = {1248/1257/1347/1356/2346}
18a. R9C6 = {45} -> no 4,5 in R8C7 + R9C78
18b. 4 in N9 locked in R789C9, locked for C9

19. Max R123C9 = 9 (step 14b) -> R123C9 = {123/125/126/135}, 1 locked for C9 and N3

20. R4C24568 (step 7) = {12347/12356}
20a. Either R4C2 = 4 or R4C8 = 2 (step 15a)
20aa. {12347} can only be [47231]
20ab. 2 of {12356} must be in R4C8
20b. -> no 7 in R4C2, no 2,4 in R4C46, no 4 in R4C5, clean-up: no 7,9 in R3C4, no 2 in R3C5, no 1,3 in R3C6, no 8 in R9C4 (step 5)
[I ought to have spotted this step when I did step 15a.
After step 20aa can eliminate [47231] because that would make R3C45 = [44], which is a multidirectional clash. However this elimination comes quickly by more normal steps so I haven’t used that clash.]

21. R9C46 contains 5 and one of {247} -> R6C5 = {247}, no 8 (step 3c)

22. R4C2 = {146} -> R1C13 (step 1c) must contain 3 and one of {146}, no 7, clean-up: no 1 in R1C9
22a. Naked pair {25} in R1C79, locked for R1 and N3, clean-up: no 9 in R23C8

23. 1 in C9 locked in R23C9, locked for 29(6) cage at R1C8 -> R4C8 = 2, R9C6 = 5 (step 6), clean-up: no 4 in R4C2 (step 15a), no 7 in R4C4 (step 20aa), no 4 in R3C4
23a. Naked quad {1356} locked in R4C2456, locked for R4, 5 locked for N5, clean-up: no 9 in R5C1, no 8,9 in R5C9
23b. Naked pair {15} in R34C5, locked for C5, clean-up: no 9 in R78C5

24. R4C2 = {16} -> R1C13 (step 1c) must contain 3 and one of {16}, no 4

25. R9C46 = [25/75] -> R6C5 (step 3c) = {27}, no 4

26. R9C6 = 5 -> 15(4) cage at R8C7 (step 18) = {1257/1356}, no 8, 1 locked for N9, clean-up: no 7 in R6C7

27. R4C8 = 2 -> R4C8 + R6C79 (step 16) = {239/257}, no 6, clean-up: no 2 in R7C7
27a. 5 of {257} must be in R6C7 -> no 5 in R6C9

28. Naked pair {35} in R67C7, locked for C7 -> R1C79 = [25], R1C1 = 3, R5C9 = 6, R4C9 = 8, clean-up: no 9 in R4C1, no 7 in R5C1
28a. R45C1 = [78], clean-up: no 3 in R7C3

29. 15(4) cage at R8C7 (step 26) = {1356} (only remaining combination) -> R9C8 = 3, R89C7 = {16}, locked for C7 and N9, R67C7 = [35], clean-up: no 8 in R23C8, no 7 in R7C3

30. 21(3) cage at R6C9 = {489} (only remaining combination) -> R6C9 = 9, R7C89 = [84], R45C7 = [47], R4C3 = 9, R7C3 = 6, R6C3 = 4, R1C3 = 1, clean-up: no 2 in R3C5, no 2,4,6 in R8C5

31. Naked pair {47} in R23C8, locked for C8 -> R8C8 = 9, R1C8 = 6

32. R1C13 = [31] -> R4C2 = 1 (step 1c), R34C5 = [15], R4C6 = 3, R3C6 = 2, R4C4 = 6, R3C4 = 5, R6C1 = 5 (step 4), R56C8 = [51], R6C5 = 2 (step 8a), R9C4 = 2 (step 5), R89C9 = [27], R9C3 = 8, R3C3 = 7, R23C8 = [74], R23C9 = [13], clean-up: no 7 in R1C5, no 9 in R2C2, no 8 in R8C5

33. R9C34 = [82] = 10 -> R8C3 + R9C2 = 9 = [54], R2C3 = 2, R5C23 = [23], R6C2 = 6, clean-up: no 8 in R23C2
33a. R23C2 = [59]

34. R6C1 = 5 -> R7C12 = 9 = [27]

and the rest is naked singles and a cage sum

I post the way I have solved that V2, especially for step 2 that offers another way for cracking this puzzle.




1)a) Cells of 28(6) and 14(2) at n1 see each other : 28(6)+14(2) = 42(7) = {12456789} : no 3
→ 3 locked for n1/r1 at r1c13
b) Cells of 29(6) and 11(2) at n3 see each other : 29(6)+11(2) = 40(7) = {12346789} : no 5
→ 5 locked for n3/r1 at r1c79
c) Cells of 30(6) and 10(2) at n8 see each other : 30(6)+10(2) = 40(7) = {12346789} : no 5
→ 5 locked for n8/r9 at r9c46
d) Digit at r4c8 is locked for n3 at r1c79 → r4c8<>3
e) IO for n69 : r9c6=3+r4c8 → r9c6<>1,2,3 and r4c8<>7,8,9 → r4c8=(1246), r9c6=(4579)

2)a) IO for c9 : r7c8+r8c8 = 8 + r1c9+r2c9+r3c9
b) r7c8+r8c8 >= 14 : no 1,2,3,4
c) digits at r78c8 can't be at r123c9. Like this : if one of r78c8 was locked at r123c9, the other one would be equal to (step a) 8 + two of r123c9 >= 11.
d) We deduce from c) that r7c8 is locked at 14(2) at c9 since r7c8 sees r6789c9: r7c8=(5689)
e) From step a), r1c9+r2c9+r3c9=r7c8+r8c8 - 8 <= 17-8=9 : no 7,8,9 in r123c9
f) r67c9 <> 7. Like this : if one of r67c9 was 7, the other one and r7c8 would total 14 with the same combination as 14(2) at n9 since r7c8 is locked at 14(2), and there would be a clash between 14(2) and 21(3)
→ 7 locked for n9 at r89c9, clean up : r6c7<>1
g) At n6 : 21(3) <> 7 : {489}, 4 locked at r67c9 for c9
h) Innies for n6 : r4c8+r6c7+r6c9=14=h14(3) : permutations [239/428] ([158] is not possible since {58} blocks 14(2) at n6)
→ r4c8=(24), r6c7=(23) and r6c9=(89)
i) Last cell for 4 at 21(3) : r7c9=4

3)a) Outies for r89 : r6c5+r7c4+r7c5+r7c6 = h15(4)
→ r6c5 <> 7 since all combinations of h15(4) would be blocked by r7c9=4 and digit 5 locked at r9c46
b) Digit 7 of 40(7) (step 1)b)) is locked at n8 → r9c6<>7 → (step 1)e)) r4c8 <> 4
(that was proven in a different way in Andrew'WT)
c) From step 2)h), h14(3) at n6 is [239] → r7c7=5, r7c8=8
d) 18(3) at n9 (remind 7 is locked in) : last combination {279} → r8c8=9, r89c9= {27} locked

4)a) At n6 : 14(2)= {68} locked . At n3 : r1c9=5, r1c1=3, r23c9={13} locked and 11(2)={47} locked
b) Hidden pair : r45c7= {47} locked, r56c8={15} locked → r1c8=6 and r9c8=3. Hidden single : r1c7=2
c) r4c2 is locked for n1 at r1c3 : → r4c2 = (1478)
d) Outies for r123 : r4c24568 total 17 : {12356} since {12347} is blocked by r4c7=(47)
→ only valid candidate : r4c2 = 1
→ only valid candidate : r4c5=5
→ For the same reason : r4c6=3, and r4c4=6

Everything easy from here (singles, last cage combination)