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Hi,

I had first wanted to post another puzzle (ssscore : 1.94), but did not manage to find a satisfying solving path : it will certainly be posted the next weeks because I think it offers some possibilties I did not manage to use, and hope you will find something interesting with it.
Anyway, here is an "easier" puzzle for A174 ; I know that (at least !) one of you guys will love the kind of technical steps it involves. I wish hat I did not say too much about this puzzle, .Would l be happy if there are several ways to tackle this puzzle, I guess.

Assassin 174



3x3::k:4352:4352:4352:2307:3332:3332:2566:2566:4616:3593:1290:2315:2307:3341:782:782:4616:4113:3593:1290:2315:3341:7958:3351:4616:3097:4113:2843:3356:3356:3356:7958:3351:3351:3097:4113:2843:3365:3365:7958:7958:7958:2346:2346:1836:3885:2606:4399:4399:7958:5426:5426:5426:1836:3885:2606:4152:4399:7958:1339:3644:1085:2878:3885:4152:2625:2625:1339:2884:3644:1085:2878:4152:2377:2377:2635:2635:2884:4174:4174:4174:


Solution :




SSscore : 1.48

Thanks for the new Assassin, manu!

I've enjoyed solving this Assassin since it didn't involve any combo analysis but just some Killer subsets. And one certain Killer quad (there are several ones that can be used!) cracks this Killer. By the way, I hope that my path differs from the one you were referring to.

A174 Walkthrough:

1. R1+C9
a) Innies R1 = 5(2) = {14/23}
b) Innies C9 = 11(2) = [29/38/47]

2. C789 !
a) 4(2) = {13} locked for C8+N9
b) 16(3) @ N9: R9C78 <> 7,8,9 since R9C9 = (789)
c) 7 locked in R789C9 @ N9 for C9
d) 11(2) = {29/47} since (56) is a Killer pair of 14(2)
e) Hidden Killer pair (15) in 16(3) @ N3 + 7(2) for C9 since none of them can have both
-> 7(2) <> {34} and 16(3) = {169/259/358} <> 4
f) 12(2) <> 9
g) Outies N3 = 14(2+1): R4C9 <> 1,2,3 since R2C6+R4C8 <= 10
h) ! Killer quad (1234) locked in 10(2) + 16(3) + R1C9 + R2C7 for N3
i) 4 locked in R1C789 @ N3 for R1

3. R123
a) 13(2) @ R1 <> 9
b) 17(3) = 9{17/35} since other combos blocked by Killer pairs (69) of 14(2) and (68,78) of 13(2) @ R1
-> 9 locked for R1+N1
c) 14(2) = {68} locked for C1+N1
d) Innies N2 = 10(3) <> 8,9
e) 9 locked in 13(2) @ R2C5 @ N2 = {49} locked for N2

4. C1+R789
a) Innies C1 = 5(2) = [14/32]
b) Killer pair (24) in R9C1 + 16(3) @ N9 for R9
c) 9(2) <> 5,7
d) 16(3) @ R9 = 5{29/47} since {268} blocked by Killer pair (68) of 9(2) -> 5 locked for R9+N9
e) 14(2) @ N9 = {68} locked for C7
f) 9 locked in R789C9 @ N9 for C9

5. C789
a) 16(3) @ N3 = {358} locked for C9; 3 also locked for N3
b) 7(2) = {16} locked for N6
c) 9(2) <> 3,8
d) Hidden Single: R4C7 = 3 @ N6
e) 2 locked in 9(2) @ N6 = {27} locked for R5+N6

6. R456
a) 11(2) = [29/74]
b) 13(2) = {58} locked for R5+N4 since {49} blocked by R5C1 = (49)
c) 21(3) = 9{48/57} since R6C78 <> 6,7 -> 9 locked for R6
d) 13(3) @ N4 <> 9
e) 9 locked in 11(2) @ N4 = {29} -> R5C1 = 9, R4C1 = 2
f) 15(3) = {357} locked for C1; 5 also locked for N7

7. R789
a) R9C1 = 4
b) 16(3) @ N7 = {349} locked for N7
c) 9(2) = {18} locked for R9+N7
d) 10(2) @ R6C2 = [37/46]
e) 10(2) @ R9 = {37} locked for R9+N8

8. N1
a) R1C1 = 1
b) 17(3) = {179} -> 7 locked for R1+N1

9. Rest is singles.

Rating: 1.25. I used a Killer quad.

That's not true. Your step 3b is a pretty significant step of combo analysis:


Meanwhile, I use a different path which involves a bigger cage, but the complexity of combo analysis required is considerably less than yours (2 possible combos instead of 5 in yours):

_________________
ADYFNC HJPLI BVSM GgK Oa m

I think the difference is one of terminology.

In addition to using "Killer" in the conventional sense, Afmob also uses "Killer subsets" where others would say "blocks", "blocked by" or "clashes with"; we all have our own walkthrough styles.

I'm inclined to feel that combo analysis is the stage where combos are eliminated either by the above or because a candidate value has been completely eliminated from the cage (real, hidden or split); it also includes elimination of combos because one or more cells don't contain any of the required candidates.

After that there can be a further stage due to the positioning of candidates within a cage. That is really permutation analysis although it's often referred to as combo analysis and included in that term.

It may well be that the term combo analysis, when writing walkthroughs, now includes permutation analysis but it does also include the earlier stage where complete combos are eliminated.

Thanks Andrew for your (elaborated) explanation on the terminology of "combo analysis".

To me, what I regard "combo analysis" is when you somehow need to list out all the combinations of a certain cage (given or implied). Usually this is quite simple for 2-cell cages, and often not too hard for 3-cell ones (like the step 3b used by Afmob here), but for 4-cell and 5-cell ones it could get very tedious, and it's likely to take a long time for players to perform the analysis because (1) they need to generate all possible combinations for the cage, and (2) there are usually a large number of combinations (more than 5) to be explored and studied.

I also have a theory that all killer pairs (might be even triples/quads) can be expressed as combo analyses on the innies of a given row/column/nonet, so usually when one rates a walkthrough based on the usage of killer subsets but overlooks on the complexity of combo analyses used, I have a problem with that. Pity I really don't have time to post a good explanation of this manoeuvre in the killer technique section. But I'm sure you guys have seen some of these in my past walkthroughs.

Perhaps you guys can give me some insights on what you consider good practice of combo analysis when you try my A175 this weekend. (Hint hint .)

_________________
ADYFNC HJPLI BVSM GgK Oa m

Maybe the one other thing I could say about Afmob's step 3b is that it's routine combo analysis; it started from a fairly long string of combos but all the clashes were with two-cell cages, each of which only had two remaining combos, so each combo elimination from the 3-cell cage was straightforward. I did something similar with my step 12, except that I hadn't eliminated 4 from the cage.

I've now gone through Afmob's walkthrough and udosuk's walk-in. Afmob's killer quad was a more powerful one than the one I used in R1. I think if udosuk goes through my walkthrough there will be the feeling after my early steps "umm, could have gone a bit deeper in C9" . Udosuk's hidden killer pair for 5,6 in C5 was another nice way to crack this Assassin.

Here is my walkthrough. Thanks udosuk for pointing out my typos and Ed for commenting that there are 23 Prelims; I've now added the one I originally missed.

Prelims

a) R12C4 = {18/27/36/45}, no 9
b) R1C56 = {49/58/67}, no 1,2,3
c) R1C78 = {19/28/37/46}, no 5
d) R23C1 = {59/68}
e) R23C2 = {14/23}
f) R23C3 = {18/27/36/45}, no 9
g) 13(2) cage at R2C5 = {49/58/67}, no 1,2,3
h) R2C67 = {12}
i) R34C8 = {39/48/57}, no 1,2,6
j) R45C1 = {29/38/47/56}, no 1
k) R5C23 = {49/58/67}, no 1,2,3
l) R5C78 = {18/27/36/45}, no 9
m) R56C9 = {16/25/34}, no 7,8,9
n) R67C2 = {19/28/37/46}, no 5
o) 5(2) cage at R7C5 = {14/23}
p) R78C7 = {59/68}
q) R78C8 = {13}
r) R78C9 = {29/38/47/56}, no 1
s) R89C6 = {29/38/47/56}, no 1
t) R8C34 = {19/28/37/46}, no 5
u) R9C23 = {18/27/36/45}, no 9
v) R9C45 = {19/28/37/46}, no 5
w) R6C678 = {489/579/678}, no 1,2,3

Steps resulting from Prelims
1a. Naked pair {12} in R2C67, locked for R2, clean-up: no 7,8 in R1C4, no 3,4 in R3C2, no 7,8 in R3C3
1b. Naked pair {13} in R78C8, locked for C8 and N9, clean-up: no 7,9 in R1C7, no 9 in R34C8, no 6,8 in R5C7, no 8 in R78C9
1c. R45C1 = {29/38/47} (cannot be {56} which clashes with R23C1), no 5,6
1d. R78C9 = {29/47} (cannot be {56} which clashes with R78C7), no 5,6

2. 45 rule on R1 2 innies R1C49 = 5 = {14/23}, clean-up: no 3,4 in R2C4

3. 45 rule on R9 2 innies R9C16 = 10 = [19]/{28/37/46}, no 5, no 9 in R9C1, clean-up: no 6 in R8C6

4. 45 rule on C1 2 innies R19C1 = 5 = {14/23}, clean-up: no 2,3,4 in R9C6 (step 3), no 7,8,9 in R8C6

5. 45 rule on C9 2 innies R19C9 = 11 = [29/38/47], clean-up: no 4 in R1C4 (step 2), no 5 in R2C4
5a. Killer triple 7,8,9 in R78C7, R78C9 and R9C9, locked for N9

6. 45 rule on N2 3 innies R2C6 + R3C56 = 10 = {127/136/145/235}, no 8,9

7. Killer quad 1,2,3,4 in R1C1, R1C49 and R1C78, locked for R1, clean-up: no 9 in R1C56
7a. 9 in N2 locked in 13(2) cage at R2C5 = {49}, 4 locked for N2

8. 45 rule on N8 3 innies R7C45 + R8C4 = 19, no 1, clean-up: no 9 in R8C3

9. R56C9 = {16/25} (cannot be {34} which clashes with R1C9 and R78C9, ALS block), no 3,4
[Ed has suggested that this might be better described as Killer ALS block.]

10. Max R1C9 + R2C8 = 13 -> min R3C7 = 5
10a. Max R1C9 + R3C7 = 13 -> min R2C8 = 5

11. 45 rule on N3 2 outies R4C89 = 1 innie R2C7 + 11
11a. R2C7 = {12} -> R4C89 = 12,13, no 1,2,3 in R4C9

12. R1C123 = {179/359/467} (cannot be {269/458} which clash with R23C1, cannot be {278/368} which clash with R1C56), no 2,8, clean-up: no 3 in R9C1 (step 4), no 7 in R9C6 (step 3), no 4 in R8C6
12a. Killer pair 5,7 in R1C123 and R1C56, locked for R1, clean-up: no 3 in R1C7
12b. 2 in N1 locked in R3C23, locked for R3

13. 16(3) cage in N7 can only contain one of 1,2,4
13a. R9C1 = {124} -> no 1,2,4 in R7C3 + R8C2

14. 45 rule on C789 1 outie R6C6 = 2 innies R24C7 + 3
14a. Min R24C7 = 3 -> min R6C6 = 6
14b. Max R6C6 = 9 -> max R24C7 = 6, no 6,7,8,9 in R4C7

15. 45 rule on C1234 3 innies R359C4 = 12 = {129/147/246/345} (cannot be {138/156/237} because R3C4 only contains 4,9), no 8, clean-up: no 2 in R9C5
15a. R3C4 = {49} -> no 4,9 in R59C4, clean-up: no 1,6 in R9C5
15b. 5 of {345} must be in R5C4 -> no 3 in R5C4

16. 45 rule on C6789 3 innies R157C6 = 15
16a. Max R17C6 = 12 -> min R5C6 = 3

17. R9C789 = {259/268/457}
17a. R9C23 = {18/36/45} (cannot be {27} which clashes with R9C789), no 2,7

18. R7C45 + R8C4 = 19 (step 8) = {289/379/478/568} (cannot be {469} because cannot then place 5 for N8)
18a. R89C6 = [29/56] (cannot be [38] which clashes with R7C45 + R8C4), clean-up: no 2 in R9C1 (step 3), no 3 in R1C1 (step 4)
[I was trying to find a better way to write step 18. I suppose it’s a sort of variable hidden killer pair for 5,6 in N8. Because 5 is either in R7C45 + R8C4 or in R8C6 then if R7C45 + R8C4 contains 6 it must also contain 5.
With hindsight this step has been available from the position after the Prelims.]

19. Naked pair {14} in R19C1, locked for C1, clean-up: no 7 in R45C1
19a. Killer pair 8,9 in R23C1 and R45C1, locked for C1

20. R23C2 = [32] (cannot be [41] which clashes with R1C1), clean-up: no 6,7 in R2C3, no 6 in R3C3, no 7,8 in R67C2, no 6 in R9C3

21. 7 in N1 locked in R1C23, locked for R1, clean-up: no 6 in R1C56
21a. Naked pair {58} in R1C56, locked for R1 and N2, clean-up: no 1 in R1C4, no 2 in R1C78, no 4 in R1C9 (step 2), no 7 in R9C9 (step 5), no 4 in R9C78 (step 17)

22. R9C789 (step 17) = {259/268}, 2 locked for R9 and N9, clean-up: no 9 in R78C9, no 8 in R9C5
22a. Killer pair 6,9 in R9C6 and R9C789, locked for R9, clean-up: no 3 in R9C3, no 1 in R9C4, no 4 in R9C5
22b. Killer pair 3,7 in R12C4 and R9C4, locked for C4, clean-up: no 3,7 in R8C3
22c. Killer pair 1,4 in R9C1 and R9C23, locked for N7, clean-up: no 6,9 in R6C2, no 6,9 in R8C4

23. Naked pair {37} in R9C45, locked for N8, clean-up: no 2 in 5(2) cage at R7C5
23a. Naked pair {14} in 5(2) cage at R7C5, locked for N8, clean-up: no 6 in R8C3

24. Naked pair {28} in R8C34, locked for R8 -> R8C6 = 5, R9C6 = 6, R1C56 = [58], R9C1 = 4 (step 3), R1C1 = 1, clean-up: no 6 in R1C23 (step 12), no 9 in R1C8, no 8 in R2C3, no 6,9 in R7C7, no 5 in R9C23

25. R157C6 = 15 (step 16), R1C6 = 8 -> R57C6 = 7 = [34], R8C5 = 1, clean-up: no 8 in R4C1, no 6 in R5C8

26. 31(7) cage at R3C5 must contain 1,2,3,4,7 -> R5C4 = 1, 4,7 locked in R3456C5 for C5, 4 also locked for N5 -> 13(2) cage at R2C5 = [94], R9C45 = [73], R2C4 = 6, R1C4 = 3, R1C9 = 2, R2C67 = [21], R3C56 = [71], R23C3 = [45], R2C1 = 8, R3C1 = 6, R3C8 = 8, R4C8 = 4, R3C7 = 9, R2C8 = 7 (cage sum), R1C78 = [46], R23C9 = [53], R4C9 = 8 (cage sum), R56C9 = [61], R6C2 = 4, R7C2 = 6, R8C7 = 6, R7C7 = 8, R7C5 = 2, R4C5 = 6, R6C5 = 8, R5C5 = 4, R7C4 = 9, R8C34 = [28], R7C9 = 7, R7C13 = [53], R8C1 = 7, R6C1 = 3 (cage sum), clean-up: no 7,8,9 in R5C2, no 7,9 in R5C3, no 2,5 in R5C7, no 5 in R5C8

and the rest is naked singles.

Nah, you go at your own pace. Because you don't optimise your WT it is pointless to compare your approach with mine and others.

_________________
ADYFNC HJPLI BVSM GgK Oa m

Me . Loved Afmob's killer quad since I found it (but then missed his step 3b ). Loved Andrew's step 18 since it's technical and it gives me more grist for a follow-up techniques post I've been planning. Loved udosuk's end of para. 3 because it's the one I was least likely to find. Well done. But instead, I managed to crack this puzzle the hard way with a move that's Way Over The Top (WOTT!) so think it wins manu's technical challenge. Found a grouped killer single using Locking Cages....

Partial Walkthrough for A17414 steps
This is an optimised WT with just the essential steps included to get to my breakthrough. However, I still try and include clean-up as I go. Please let me know if anything is wrong or needs to be clearer. Many thanks to Andrew for an important clarification.

Prelims - 23!
i. 9(2)n2: no 9
ii. 13(2)n2: no 1,2,3
iii. 10(2)n3: no 5
iv. 14(2)n1 = {59/68}
v. 5(2)n1 = {14/23}
vi. 9(2)n1: no 9
vii. 13(2)n2: no 1,2,3
viii. 3(2)n2 = {13}
ix. 12(2)n3: no 1,2,6
x. 11(2)n4: no 1
xi. 13(2)n4: no 1,2,3
xii. 9(2)n6: no 9
xiii. 7(2)n6: no 7,8,9
xiv. 10(2)n4: no 5
xv. 21(3)n5: no 1,2,3
xvi. 5(2)n8 = {14/23}
xvii. 14(2)n9 = {59/68}
xviii. 4(2)n9 = {13}
xix. 11(2)n9: no 1
xx. 10(2)n7: no 5
xxi. 11(2)n8: no 1
xxii. 9(2)n7: no 9
xxiii. 10(2)n8: no 5

1. 4(3)n9 = {13}: both locked for c8 & n9
1a. no 8 in 11(2)n9
1b. no 9 in 12(2)n3
1c. no 6,8 in r5c7
1d. no 7,9 in r1c7

2. "45" r1: 2 innies r1c49 = 5 = {14/23}
2a. r2c4 = (5..8)

3. "45" c9: 2 innies r19c9 = 11 (no 1)
3a. r9c9 = (789)
3b. no 4 in r1c4 (h5(2))
3c. no 5 in r2c4
3d. min. r9c9 = 7 -> max. r9c78 = 9 (no 7,8,9 since can't have {27}[7])

4. 7 in n9 only in c9: 7 locked for c9

5. "45" n3: 3 outies r2c6 + r4c89 = 14
5a. max. r2c6 = 2 -> min. r4c89 = 12
5b. max. r4c8 = 8 -> min. r4c9 = 4

6. 16(3)n3 = {169/259/268/349/358} which must have one of 1,2,3 in r23c9
6a. Killer quad 1,2,3,4 in n3 in 10(2), r1c9, r2c7 and r23c9: all locked for n3
6b. The 10(2), r1c9 and r2c7 take up three of the four spots for 1..4 -> only one spot left in r23c4 which must be from 1,2,3 (step 6) -> no 4 in r23c9
6c. no 8 in r4c8

7. 4 in n3 only in r1: 4 locked for r1
7a. no 9 in 13(2)n2

NOTE that Afmob's breakthrough move is now available! Another 6 steps to get to mine.
8. "45" on c1: 2 innies r19c1 = 5 = [14]/{23}

9. "45" on r9: 2 innies r9c16 = 10 = [28/37/46] ie r9c6 = (678)
9a. r8c6 = (345)

10. Hidden Killer Triple 1,2,3 in n6 in 7(2), 9(2) and r4c7
10a. 9(2) must have one of 1,2,3 for n6 -> no {45}: no 4 or 5 in 9(2)
10b. r4c7 = (123)

11. "45" on n36: 1 outie r6c6 - 3 = 2 innies r24c7
11a. max. r24c7 = 5 -> max. r6c6 = 8
11b. min. r24c7 = 3 -> min. r6c6 = 6

12. "45" c6789: 3 innies r157c6 = 15
12a. = {159/168/258/267/348/357/456} ({249} invalid by r1c6)
12b. NOTE: only one combo has 9 so 9 forces a direct link with 1 in this cage. Important!

Note that udosuk's way shows that most of the combos in 12a are redundant with his hidden killer pair move at end of paragraph 3.

Now the WOTT! move.
13. Grouped Killer Single 1 in c67 in 3(2)n2, 13(3)n2 & r57c6. Like this.
13a. 9 in c6 is in h15(3) and forces a direct link to 1 (step 12b) or 9 is in 13(3)n2 and forces a direct link to 1 (can only be {139}) -> between them, a 1 is taken for c67 (Locking Cages - follow-up post on this).
13b. 3(2)n2 must have a 1 for c67
13c. -> 1 locked for c67!
13d. no 9 in r1c8
13e. no 8 in r5c8

14. 9 in r1 only in n1: 9 locked for n1

Much easier now. Note: Andrew's neat step 18, can also be seen as a direct-links move (like my WOTT! move). Will do a follow-up post on this.

Cheers
Ed

Thanks Ed for your interesting WOTT move.

Thanks also udosuk for your alternative way to get the same result as my step 18. I use combined cages at times. They can be a useful way to take account of interactions between cages in a row/column/nonet.


I guess it was a bit cheeky to use udosuk's phrase but I think the self-criticism was valid; I ought to have got more from C9, and maybe then N3, at that stage. However if I'd done that I might never have found my step 18.

This post is about Locking Cages and it's inverse, Blocking Cages. They both use cage combinations to either lock candidates or block permutations/combinations respectively. This technique is called Locked Cages in sudopedia but I feel that Locking Cages is a better name since the inverse Blocking Cages makes more sense than "Blocked Cages". But, take your pick.

Locking Cages is when two cages contain the only candidate for a house which then "locks" a second candidate if that second candidate has a direct link to the first. It is easiest to see when two cages in the same house have the same cage total.



In this example above from A167, 1 in c1 must be in one of the 10(2) cages -> one of those cages must also have 9 -> 9 locked for c1.

It is much more common that one of the cages is hidden.



In MessyOne#1 above, r56c3 (in yellow) is a h9(2) cage. Candidate 8 in this nonet must be in this hidden 9(2) or in the 9(2) cage -> 1 locked for the nonet. These become easy to find since the cage totals are the same.

Much more difficult is when the cage totals are different, ie, when at least one of the cages is more than 2 cells.

.

This pic above is an alternative way to do Andrew's step 18 in A174 (a previous post in this thread), using Locking Cages. 5 in n8 must be in the 11(2) = {56} or the hidden 19(3) at r7c45+r8c4 = {568}. In both cages, 5 forces candidate 6 -> 6 locked for n8. Also, other combinations in the h19(3) with the second candidate (6) can be eliminated since they don't also have the forcing candidate (5). So, the 19(3) is now = {289/379/478/568}. This second part is essential to Andrew's step 18. Andrew notes that his move was available at the beginning of the puzzle. A table may make it easier to find Locking Cages when the two cages do NOT have the same total. See below.

My WOTT! step 14 in A174 (a previous post in this thread) using Locking Cages is much more complex since it involves two 3-cell cages and locks one candidate for 2 columns!

Now, a look at the inverse, Blocking Cages.



This is from A171 ALT and is an alternative way to do manu's step 5b here. There is a hidden 12(2) at r23c3 which overlaps a 13(2) & a 10(3) -> [84] permutation is blocked from the h12(2) since 5 is also required in both cages ie, 5 is forced into both the 13(2)=[58] and 10(3)={15}[4]. So, the 13(2) and 10(3) work as Blocking Cages to h12(2) permutations. This leaves the h12(2) = {57} and cracks the puzzle.

Further applications of these techniques are available but those can wait for a separate post sometime. For example, almost Locking Cages can form killer subsets with other cages or almost Blocking Cages will eliminate combinations in other cages.

Direct-link candidate's table
This table is of combinations in cages (above 2 cells and no single combination cages) where one candidate forces a second candidate in the same cage. For example, an 11(3) cage = {128/137/146/236/245}. 8, 7 and (surprisingly) 5 only occur in one combination -> each one has a direct-link with the other candidates in that one combination. So if 5 is in an 11(3) it must also have 4 and 2.

This table has very limited use since it gets longer once combinations from any cage are eliminated during solving.

3-cell combinations with direct-link candidates: note, a direct-link only works one way! The first couple have the links listed for illustration.
.8(3) = {125/134}: 2,3,4,5 ie (2-5),(3-4),(4-3),(5-2)
.9(3) = {126/135/234}: 4,5,6 ie (4-2)(4-3),(5-1)(5-3),(6-1)(6-2)
10(3) = {127/136/145/..}: 4,6,7
11(3) = {128/137/245/...}: 5,7,8
12(3) = {129/138/...}: 8,9
13(3) = {139/...}: 9
.
.
17(3) = {179/...}: 1
18(3) = {189/279/...}: 1,2
19(3) = {289/379/568/...}: 2,3,5
20(3) = {389/479/569/...}: 3,4,6
21(3) = {489/579/678}: 4,5,6
22(3) = {589/679}: 5,6,7,8

4-cell combinations with direct-link candidates (! = 2 combinations still force a direct-link with one candidate)
12(4) = {1236/1245}: 3,4,5,6
13(4) = {1237/1246/1345}: 5,6,7
14(4) = {1238/1247/1256/1346/2345}: 5(5-2)!,6(6-1)!,7,8
15(4) = {1239/1248/1257/1347/1356/2346}: 6(6-3)!,7(7-1)!,8,9,
16(4) = {1249/1258/1348/...}: 8(8-1)!,9
17(4) = {1259/1349/...}: 9(9-1)!
.
.
23(4) = {1589/1679/...}: 1(1-9)!
24(4) = {1689/2589/2679/...}: 1,2(2-9)!
25(4) = {1789/2689/3589/3679/4579/4678}: 1,2,3(3-9)!,4(4-7)!
26(4) = {2789/3689/4589/4679/5678}: 2,3,4(4-9)!,5(5-8)!
27(4) = {3789/4689/5679}: 3,4,5
28(4) = {4789/5689}: 4,5,6,7

In digit form summary
1:
17(3) = {179}
18(3) = {189}
23(4) = {1589/1679}: (1-9)!
24(4) = {1689}
25(4) = {1789}

2:
.8(3) = {125}
18(3) = {279}
19(3) = {289}
24(4) = {2589/2679}: (2-9)!
25(4) = {2689}
26(4) = {2789}

3:
.8(3) = {134}
19(3) = {379}
20(3) = {389}
12(4) = {1236}
25(4) = {3589/3679}: (3-9)
26(4) = {3689}
27(4) = {3789}

4:
.8(3) = {134}
.9(3) = {234}
10(3) = {145}
20(3) = {479}
21(3) = {489}
12(4) = {1245}
25(4) = {4579/4678}: (4-7)!
26(4) = {4589/4679}: (4-9)!
27(4) = {4689}
28(4) = {4789}

5:
.8(3) = {125}
.9(3) = {135}
11(3) = {245}
19(3) = {568}
21(3) = {579}
22(3) = {589}
12(4) = {1245}
13(4) = {1345}
14(4) = {1256/2345}: (5-2)!
26(4) = {4589/5678}: (5-8)!
27(4) = {5679}
28(4) = {5689}

6:
.9(3) = {126}
10(3) = {136}
20(3) = {569}
21(3) = {678}
22(3) = {679}
12(4) = {1236}
13(4) = {1246}
14(4) = {1256/1346}: (6-1)!
15(4) = {1356/2346}: (6-3)!
28(4) = {5689}

7:
10(3) = {127}
11(3) = {137}
22(3) = {679}
13(4) = {1237}
14(4) = {1247}
15(4) = {1257/1347}: (7-1)!
28(4) = {4789}

8:
11(3) = {128}
12(3) = {138}
22(3) = {589}
14(4) = {1238}
15(4) = {1248}
16(4) = {1258/1348}: (8-1)!

9:
12(3) = {129}
13(3) = {139}
15(4) = {1239}
16(4) = {1249}
17(4) = {1259/1349}: (9-1)!

Many thanks to Andrew for supplying these combinations in text format. There are sure to be some mistakes in this so please check it out and let me know of any corrections or clarifications. [edit: Huge thanks to Afmob for checking all this and verifying it. I'll put it in the Killer Technique's forum.]

Cheers
Ed