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Despite the hurry, I took my time to create this Killer because I wanted to make one like I did in the beginning: Generate a random sudoku and transform it into a Killer. The good thing about this approach is that the solution path is a bit broader since you don't focus on an unusual cage pattern but on getting the uniqueness right.

Assassin 148



3x3::k:5635:5635:4353:5128:1814:1814:1814:3352:3865:5635:4353:4353:4353:5128:4114:3865:3865:3352:5635:5893:5893:5128:2314:4114:4114:3093:3093:5893:3852:3852:5127:2314:2063:2063:4881:4881:5893:3600:3852:3852:5127:4877:4877:4881:3846:3600:3600:3342:3342:1803:5127:4877:4877:3846:788:788:4371:4371:1803:3593:3846:3846:3588:3354:3611:3611:4371:3593:6402:6402:6402:3588:3611:3354:5399:5399:5399:3593:6402:3588:3588:

Solution:

SS Score: 1.33
Estimated rating: Easy 1.25 - 1.25

Thanks Afmob for a new Assassin.
I had to work hard with combinations to solve this puzzle. I have seen an interesting particularity of the cage pattern (see PS remark), but I could not use it effectively...

Edit : Ed has pointed out some steps that should been clarified. Thanks Ed ! I have also added a final remark. Interesting puzzle that can be viewed using different ways


Walkthrough Assassin 148

1)cage 3(2) at r7 : {12} locked for r7 and n7

2)Innies for n78 : r7c5+r8c6=8 → r7c5=(356), r8c6=(532). Using cage 7(2), r6c5=(124)

3)Innies for n3 : r1c7+r3c7=5 → r13c7=[14/23/41]

4)a) Innies for r789 : r7c5+r7c7+r7c8=14. Combinations : r7c578={347/356}
b) Cells r7c78 contain at least one of {346} and at least one of {345} (because from step a) cells r7c578 contain two of {346} and two of {345}) : combinations
{1346/2345} for cage 14(4) are blocked → 14(4)=12{38/47/56}
c) cage 15(4) = {1347/1356/2346} from step a)

5)Cells r789c9 of cage 14(4) see all cells of cage 15(4)
a) 14(4) <>{1238} because all combinations of cage 15(4) contain two of {1238}
→ 14(4) =12{47/56}
b) Cage 14(4) contains two of {246} → cells r789c9 contain at least one of {246}
→ 15(4) <> {2346}, so 15(4) = 13{47/56}
c) Both cages 14(4) and 15(4) contain 1 → r9c8=1 and 1 is locked at r56c9 for c9 and n6.
d) 2 is locked for c9 and n9 at r789c9.

6)Outies for n6789 : we deduce a hcage 9(3) at r6c5+r5c6+r4c6.
a) Innies for n23 : r2c4+r3c5=7 → no 7,8,9 for r2c4 and r3c5, no 1,2 for r4c5.
b) Outies for n1 : r2c4+r4c1+r5c1=17. r2c4<=6 → r4c1+r5c1 >= 11 : no 1.
c) from steps a and b, digit 1 is locked for r4 at r4c23+r4c6 → r5c4<>1 (it sees all 1 of r4)
d) We deduce that 1 is locked for n5 at hcage 9(3) : {126/135}
→ r6c5<>4 (no combination with digit 4), r7c5=(56), r8c6=(32) (step 2)

7)a)Using steps 6)d) and 4)a), deduce that r7c578={356}, all locked for r7, and 3 is locked for
n9 at r7c78
b) cage 15(4) at n69 : {1356} (because it must contain two of {356}), with 3 locked at r7c78
→ cage 14(4) at n9 = {1247} (remind from step 5 that r789c9 see all cells of cage 15(4), so combinations of cage
14(4) cannot have two of {1356} : no {1256}) , with 2,4 an 7 locked for c9.

8)Innies for n7 : r7c3+r9c3=15 → r79c3=[78/87/96] (remind : 6 locked for r7)
a) 8 is locked for r7 at r7c34+r7c6. Using r7c3+r9c3=15 and cage 21(3), we deduce that
cells r9c45 cannot contain digit 8 unless r9c3 is 7 (↔ r7c3=8). We deduce that r9c3<>6 : no valid combination for cage 21(3).
b) r79c3={78} (locked for n7), cage 13(2) at n7 is {49}, cage 14(3) = {356}
c) 9 is locked for r7 and n8 at r7c46. We deduce there is no 9 in cage 21(3) → 21(3)={678} all locked for r9 with 6 locked at r9c45 for n8.
d) NS : r7c5=5 → r6c5=2, r8c6=3 (step 2)
e) Naked pair : r7c78={36} locked for r7,n9 and the rest of the cage 15(4)→ r56c9={15}
locked for c9 and n6.
f) HS for n7 : r9c1=3, naked pair {56} at r8c23 locked for r8 and n7.
g) HS for n9 : r9c7=5, r8c78={89} locked for r8 and n9
h) NS : r8c1=4, r9c2=9
i) Naked pair {24} at r9c69

9) a) hcage 9(3) at n5 : r45c6={16} (all locked for c6 and n5)
b) Naked pair {24} at r19c6 : 2,4 locked for c6 → r7c6=(789) (could been deduced using
killer pair {89} at r7c34+r7c6 and r7c23<>{89}, which proves r7c6=(89) )

10) No valid combination for cage 17(3) at n78 with digit 4 at r7c4 since r8c4<>5,6
→ r9c6=4 (HS for n8)

11) a) NS : r9c9=2. → r78c9=[47] → r8c45=[21] (all NS)
b) Cage combination for 17(3) : r7c34={78} locked for r7 → (NS) r7c6=9
c) NS : r1c6=2 → (NS) r1c57=[41]
d) r3c7=4 (step 3) and r23c6={57} (cage combination)

12) Cage 9(2) at c5 : [63] → (step 6)a) r2c4=1

13) Step 6)b → r4c1+r5c1=16 : r45c1={79} (locked for c1) → r3c23= {25} (locked for r3)
→ cage 12(2) at n3 : {39} locked for r3 and n3.

14) Singles and simple cage combinations





Remark 1) : cells r56789c9 and r7c78 see each over, so the total is at least 28. If we add these cells to r9c8, one must find 14+15=29, which proves (min-max argument) r8c1=1, and r56789c9 + r7c78 ={1234567}. But I did not find this interesting move really fruitful for the rest, so I did not use it.

Remark 2) : Step 8a) (and 8c) ) can also be deduced more clearly using IOE  : see steps 7 and 10 of Ed's WT

This solution is "a Para" - got so caught up in fun combo work (very different to manu's) that forgot to do many "45s"! Presumably, there is a simpler solution. Very enjoyable puzzle. Thanks Afmob!

This walkthrough has a number of crucial steps that, on their own, I would put in the 1.25 rating range. See steps 6, 7, 9 & 10. Perhaps step 10 should be in the 1.5 range. With this many crucial steps [edit: used to be 5 crucial steps], I'll make this solution an Easy 1.50 rating. [edit: Afmob found a flaw in the original step 10. The new step 10 is definitely a 1.50 rating]. I also like to rate "messy" cage patterns a bit higher. [edit: Afmob has pointed out, that it has rotational and diagonal symmetry. I was thinking of the crossing-over and diagonal cages. Personally, I find plain cage patterns a bit boring now - so "messy" is not a complaint from me!]. It takes 16 steps before it's cracked (new step 10 helped a lot!).


Assassin 148 Walkthrough
NOTE: this walkthrough is an optimised solution so some obvious eliminations are left out since they are not essential to the early solution. However, I still want to show most of the clean-up while going through. Please let me know of any corrections or clarifications. [Thanks Andrew for some typos and clarifications. Always appreciated.]

Prelims
i. 20(3)n2: no 1,2
ii. 7(3)n2 = {124}
iii. 13(2)n3: no 1,2,3
iv. 9(2)n2: no 9
v. 12(2)n3: no 1,2,6
vi. 20(3)n5: no 1,2
vii. 8(2)n5: no 4,8,9
viii. 19(3)n6: no 1
ix. 13(2)n4: no 1,2,3
x. 7(2)n5: no 7,8,9
xi. 3(2)n7 = {12}
xii. 14(4)n9: no 9
xiii. 13(2)n7: no 1,2,3
xiv. 21(3)n7: no 1,2,3

1. 3(2)n7 = {12}: both locked for r7 & n7
1a. no 5,6 in r6c5

2. "45" n7: 2 innies r79c3 = 15 = {69/78}

3. "45" n78: 2 innies r7c5 + r8c6 = 8 = [62]/{35}(no 4 in r7c5, no 789) = [2/3..]
3a. r8c6 = (235)
3b. no 3 in r6c5

4. "45" n9: 3 outies r56c9 + r8c6 = 9
4a. min. r8c6 = 2 -> max. r56c9 = 7 (no 7,8,9)
4b. max. r8c6 = 5 -> min. r56c9 = 4 (important for next step)

5. 15(4)n6 cannot have both {12} in r56c9 since must be min. 4 (step 4b) -> 12{39/48/57} all impossible
5a. = {1347/1356/2346}(no 8,9)
5b. = two of 1/2/3 (important for next step)

6. 14(4)n9, except for r9c8, sees all of the 15(4)n6-> the 14(4) can have at most two of 1/2/3 -> {1238} blocked. If that is not clear, I had to think of it like this. If every cell in the 14(4) "sees" the 15(4)n6 then it could only have one of 1/2/3 for the same house it shares since the 15(4) must have two of 1/2/3. This would make it a killer triple (123). However, in this puzzle, r9c8 hides from r56c9 -> a second 1/2/3 can hide here. But there is nowhere for a third 1/2/3 to hide.
6a. In summary, 14(4)n9 = {1247/1256/1346/2345}(no 8)
6b. = two of 1/2/3 -> r9c8 = (123) (NOTE: manu's PS comment at the end of his walkthrough is much better still!)

7. Hidden killer pair (89) in r7 -> r7c6 = (89). ie. r7c34 cannot be both 8 & 9 since it is in a 17(3) cage -> the only other place for (89) in r7 is r7c6
7a. r7c6 = (89) -> 14(3)n8 = {149/158/248}(no 3,6,7) ({239} blocked by h8(2)n8 step 3)
7b. no 8,9 in r8c5 nor r9c6 since the 14(3) can only have one of 8/9

8. "45" n789: 3 innies r7c578 = 14 = {347/356}
8a. 3 locked for r7

9. Hidden Killer triple (123) in n8 -> r8c4 = (123). ie. since 14(3)n8 = one of 1/2/3 (step 7a) & h8(2)n8 = one of 1/2/3 (step 3). The only other place for 1/2/3 in n8 is r8c4
9a. 17(3)n7 = {179/269/278/359/368}(no 4)

10. "45" n8: 1 outie r7c3 + 6 = 2 innies r9c45 (adding in the h8(2) in n8)
10a. r7c3 = (6789). For each of these, when the candidate is in r7c3, the only place for that digit in n8 is in the innies r9c45 -> the IOD (6) is locked in r9c45 (IOE)
10b. 6 locked in r9c45 for r9 & n8
10c. no 1 in r6c5

11. 21(3)n7 = {678}: 7 & 8 locked for r9
11a. no 5,6,7 in r8c1

12. h15(2)n7 = {78}: both locked for n7 & c3
12a. no 5,6 in r6c4

13. h8(2)n8 = {35}: both locked for n8

14. 17(3)n7 = {179/278}: 7 locked for r7

15. h14(3)r7c578 = {356}: 5 & 6 locked for r7 & 6 locked for n9 and 15(4)n6
15a. r7c9 = 4 (hsingle r7)

16. "45" n9: 3 outies = 9 -> r56c9 = 4/6 = {13/15}(no 2)
16a. 1 locked for c9 & n6
16b. 15(4) must have 1 = {1356}
16c. no 3 or 5 in r89c9: CPE on 3 & 5 in 15(4)

It's cracked now. Trying to get to singles ASAP from here. Excuse the lack of cleanup.

17. 14(4)n9 = [4]{127} -> r9c8 = 1, r89c9 = [72]

18. r9c6 = 4
18a. no 9 in r8c1
18b. r8c1 = 4, r9c2 = 9

19. naked triple {124} in r168c5: all locked for c5

20. 9(2)n2 = {36}: both locked for c5

21. r67c5 = [25]
21a. r8c56 = [13]
21b. r7c6 = 9 (cage sum)

22. r8c4 = 2

23. r9c4 = 6 (hsingle n8)

24. r9c7 = 5, r9c1 = 3

25. 2 remaining outies n9 r56c9 = 6 = {15}: 5 locked for c9 & n6

26. "45" n69: 2 remaining outies r45c6 = 7 = {16}: both locked for c6 & n5

27. r1c567 = [421]

28. 1 remaining innie n3 r3c7 = 4
28a. r23c6 = 12 = {57}: both locked for c6 & n2

29. r6c6 = 8

30. r2c4 = 1 (hsingle n2)
30a. r34c5 = [63]

31. 12(2)n3 = {39}: both locked for n3 & r3

32. naked pair {68} in r12c9: both locked for c9 & n3
32a. r4c9 = 9

33. "45" n1: 2 remaining outies r45c1 = 16 = [79]
33a. r3c23 = 7 (cage sum) = {25}: both locked for n1 & r3

34. naked pair {68} in r12c1: both locked for c1 & n1
34a. r3c1 = 1
34b. r1c2 = 7 (cage sum)

35. r7c12 = [21]
35a. r6c1 = 5 -> r56c2 = 9 = {36}: both locked for c2 & n4

36. 8(2)n5 = [62] (last permutation)
36a. r45c8 = 10 = [46]

37. r1c8 = 5 -> r2c9 = 8

Rest is singles.

Cheers
Ed

Great IOE find in your step 10, Ed! I used two Hidden Killer pairs (one was already mentioned by manu) to crack it.

A148 Walkthrough:

1. R789 !
a) 3(2) = {12} locked for R7+N7
b) Innies N7 = 15(2) = {69/78}
c) Innies N78 = 8(2) = [35/53/62]
d) Innies R789 = 14(3) = 3{47/56} -> 3 locked for R7
e) Outies N9 = 9(2+1): R56C9 <> 7,8,9 since R8C6 >= 2; R56C9 >= 4 since R8C6 <= 5
f) ! 15(4) = 3{147/156/246} because R56C9 >= 4, so it can't be {12} -> CPE: R789C9 <> 3
g) 25(4): R8C78+R9C7 <> 1,2 since R8C6 <= 5
h) 1,2 locked in 14(4) @ N9 = 12{38/47/56}

2. N89 !
a) 14(4): R9C8 = (12) because R789C9 = 12{...} blocked by Killer pair (12) of 15(4)
b) 14(4) = 12{47/56}
c) ! Hidden Killer pair (89) in R7C6 for R7 since 17(3) can only have one of (89) -> R7C6 = (89)
d) 14(3) = {149/158/248} <> 3,6,7 because R7C6 = (89) and {239} blocked by Killer pair (23) of Innies N78

3. C456 !
a) ! Hidden Killer pair (12) in Outies N14 = 14(3) + R8C4 for C4 since Outies N14 can only have one of (12)
-> R8C4 = (12)
b) 3 locked in Innies N78 @ N8 = 8(2) = {35} locked for N8
c) 14(3) = 4{19/28} -> 4 locked for N8
d) 21(3) = {678} locked for R9

4. R789
a) 8,9 locked in 25(4) @ N9 = {3589} -> 8 locked for R8
b) 13(2) = {49} locked for N7
c) Innies N7 = 15(2) = {78} locked for C7+N8
d) Hidden Single: R8C9 = 7 @ R8
e) 14(4) = {1247} -> R7C9 = 4; 1,2 locked for R9
f) 14(3) @ N8 = 4{19/28} -> R8C5 = (12), R9C6 = 4
g) 15(4) = {1356} -> 1 locked for C9+N6
h) R9C9 = 2, R9C8 = 1
i) 7(2) = [25/43]

5. C456
a) Naked triple (124) locked in R168C5 for C5
b) 9(2) = {36} locked for C5
c) R8C5 = 5 -> R7C5 = 2, R8C5 = 1 -> R7C6 = 9
d) Outies N69 = 10(3) = {136} -> R8C6 = 3; 1,6 locked for C6+N5
e) R1C5 = 4, R1C6 = 2, R1C7 = 1

6. R123
a) Innie N3 = R3C7 = 4
b) 16(3) = {457} -> 5,7 locked for C6+N2
c) 12(2) = {39} locked for R3+N3 since {57} blocked by R3C6 = (57)
d) Hidden Single: R2C4 = 1 @ N2
e) Outies N1 = 16(2) = {79} locked for C1+N4+23(4)
f) R3C5 = 6, R4C5 = 3, R6C6 = 8
g) 23(4) = {2579} because R45C1 = {79} -> 2,5 locked for R3+N1

7. R456
a) 20(3) = {578} -> R5C5 = 7, R4C4 = 5
b) R5C1 = 9, R4C1 = 7
c) 8(2) = {26} -> R4C6 = 6, R4C7 = 2
d) 19(3) = {469} because R4C89 <> 5,6 -> R4C9 = 9, R4C8 = 4, R5C8 = 6
e) 13(2) = {49} -> R6C3 = 4, R6C4 = 9

8. N13
a) R3C4 = 8, R3C1 = 1, R8C1 = 4
b) 22(4) = {1678} -> R1C2 = 7; 6,8 locked for C1+N1
c) 13(2) = {58} locked for N3

9. Rest is singles.

Another Assassin where I'd got stuck and came back to it later, then finding the breakthrough as I've commented under step 16.

The key to this puzzle is clearly the interaction between the 15(4) cage at R5C9 and the 14(4) cage in N9, which I took some time to find. The first remark after manu's walkthrough is a neat piece of logic, which would give the first placement immediately by using a technically harder step.

The IOE in Ed's step 10 is neat! I've suggest by PM that it ought to be included as another IOE example in the Killer Techniques topic because it's simpler than the two already posted by Ed and Afmob.

My solving path was much more like Afmob's one than the other two although my later stages could probably a lot shorter if I'd remembered to use step 11 again after I'd fixed R2C4.

I'll rate A148 at Easy 1.25 to 1.25 since, although longer, it's about the same difficulty level as Afmob's walkthrough; that is, of course, ignoring the fact that I got stuck.

Here is my walkthrough. Thanks Ed for your comments, particularly on step 18 where I've added a clarification and noted that I'd missed a combo elimination.

Prelims

a) 13(2) cage in N3 = {49/58/67}, no 1,2,3
b) R34C5 = {18/27/36/45}, no 9
c) R3C89 = {39/48/57}, no 1,2,6
d) R4C67 = {17/26/35}, no 4,8,9
e) R6C34 = {49/58/67}, no 1,2,3
f) R67C5 = {16/25/34}, no 7,8,9
g) R7C12 = {12}, locked for R7 and N7, clean-up: no 5,6 in R6C5
h) 13(2) cage in N7 = {49/58/67}, no 3
i) 20(3) cage in N2 = {389/479/569/578}, no 1,2
j) R1C567 = {124}, locked for R1, clean-up: no 9 in R2C9
k) 20(3) cage in N5 = {389/479/569/578}, no 1,2
l) 19(3) cage in N6 = {289/379/469/478/568}, no 1
m) R9C345 = {489/579/678}, no 1,2,3
n) 14(4) cage in N9 = {1238/1247/1256/1346/2345}, no 9

1. 45 rule on N3 2 innies R13C7 = 5 = [14/23/41]

2. 45 rule on N7 2 innies R79C3 = 15 = {69/78}, no 3,4,5
2a. 13(2) cage in N7 = {49/58} (cannot be {67} which clashes with R79C3), no 6,7
2b. Killer pair 8,9 in R79C3 and 13(2) cage, locked for N7

3. 45 rule on N23 2 innies R2C4 + R3C5 = 7 = {16/25/34}, no 7,8,9, clean-up: no 1,2 in R4C5
3a. Killer triple 1,2,4 in R1C56 and R2C4 + R3C5, locked for N2

4. 45 rule on N78 2 innies R7C5 + R8C6 = 8 = [35/53/62], clean-up: no 3 in R6C5

5. 45 rule on R789 3 innies R7C578 = 14 = {347/356}, no 8,9, 3 locked for R7

6. 45 rule on N9 2 innies R7C78 = 1 outie R8C6 + 6
6a. Min R8C6 = 2 -> min R7C78 = 8 -> max R56C9 = 7, no 7,8,9

7. 45 rule on R89 2 outies R7C69 = 1 innie R8C4 + 11
7a. Max R7C69 = 17 -> max R8C4 = 6

8. 45 rule on C6789 1 innie R6C6 = 2 outies R18C5 + 3
8a. Min R18C5 = 3 -> min R6C6 = 6
8b. Max R18C5 = 6, no 6,7,8,9 in R8C5

9. 45 rule on C1234 2 outies R29C5 = 1 innie R4C4 + 12
9a. Max R29C5 = 17 -> max R4C4 = 5
9b. Min R4C4 = 3 -> min R29C5 = 15, no 3,4,5

10. 20(3) cage in N5 = {389/479/569/578}
10a. R4C4 = {345} -> no 3,4,5 in R5C5

11. 45 rule on N1 2 innies R3C23 = 1 outie R2C4 + 6
11a. Max R2C4 = 6 -> max R3C23 = 12, min R45C1 = 11, no 1

12. 45 rule on N69 3 outies R458C6 = 10 = {127/136/145/235}, no 8,9

13. Hidden killer pair 1,2 in R13C7 and 15(3) cage for N3 -> 15(3) cage must contain one of 1,2
13a. 15(3) cage in N3 = {159/168/258/267} (cannot be {249/348} which clash with R13C7, cannot be {357/456} which don’t contain 1 or 2), no 3,4

14. 15(4) cage at R5C9 = {1347/1356/2346} (cannot be {1257} which clashes with combinations of R7C578), CPE no 3 in R89C9

15. 9 in N9 locked in 25(4) cage at R8C6 = {2689/3589/3679/4579} (cannot be {1789} because R8C6 only contains 2,3,5), no 1
15a. 2 of {2689} must be in R8C6 -> no 2 in R8C78 + R9C7

16. 1,2 in N9 locked in 14(4) cage
16a. Hidden killer pair 1,2 in R56C9 and R89C9 for C9 -> R89C9 cannot contain both of 1,2 -> R9C8 = {12}
16b. Killer pair 1,2 in R56C9 and R89C9, locked for C9
16c. 14(4) cage = {1247/1256}, no 8
[I got stuck a couple of steps later until I found step 16a so I’ve done a small rework. With hindsight, I could have done step 21 next.]

17. 8,9 in N9 locked in 25(4) cage at R8C6 (step 15) = {2689/3589}, no 4,7

18. 17(3) cage at R7C3 = {179/269/278/368/467} (cannot be {359} which clashes with R13C4 because 20(3) cage at R1C4 must contain one of 3,5 in R13C4, cannot be {458} which clashes with R13C4 + R4C4), no 5
18a. 4 of {467} must be in R7C4 (R7C34 cannot be {67} which clashes with R7C578), no 4 in R8C4
[Ed pointed that that 17(3) cage cannot be {368} which clashes with R7C5 + R8C6. This would eliminate 3 from R8C4.]

19. 45 rule on R89 4 outies R7C3469 = 28 = {4789/5689}
19a. Hidden killer pair 8,9 in R7C34 and R7C6 for R7, R7C34 cannot contain both of 8,9 -> R7C6 = {89}, R7C34 must contain one of 8,9
19b. 17(3) cage at R7C3 (step 18) = {179/269/278/368} (cannot be {467} which doesn’t contain 8 or 9), no 4
19c. 1,2,3 only in R8C4 -> R8C4 = {123}
19d. 4,5 of R7C3469 only in R7C9 -> R7C9 = {45}

20. 14(4) cage in N9 (step 16c) = {1247/1256}
20a. R7C9 = {45} -> no 4,5 in R89C9

21. 15(4) cage at R5C9 (step 14) = {1347/1356} (cannot be {2346} which clashes with 14(4) cage in N9 which must have 4 or 6 in C9), no 2, 1 locked in R56C9, locked for C9 and N6, clean-up: no 7 in R4C6
21a. R9C8 = 1 (hidden single in N9)

22. 14(3) cage in N8 = {149/158/248} (cannot be {167/257/347/356} because R7C6 only contains 8,9, cannot be {239} which clashes with R7C5 + R8C6), no 3,6,7
22a. R7C6 = {89} -> no 8,9 in R9C6
22b. 1 of {158} must be in R8C5 -> no 5 in R8C5

23. Naked triple {124} in R168C5, locked for C5, clean-up: no 5,7,8 in R34C5, no 2,3,5,6 in R2C4 (step 3)
23a. Naked pair {36} in R34C5, locked for C5, clean-up: no 1,4 in R6C5
23b. R67C5 = [25], R7C9 = 4, R8C6 = 3 (step 4), clean-up: no 9 in R1C8, no 8 in R3C8, no 5,6 in R4C7, no 7 in R7C78 (step 5)
23c. Naked pair {36} in R7C78, locked for R7, N9 and 15(4) cage at R5C9, clean-up: no 9 in R9C3 (step 2)
23d. Naked pair {15} in R56C9, locked for C9 and N6, clean-up: no 8 in R1C8, no 7 in R3C8
23e. Naked pair {27} in R89C9, locked for C9, clean-up: no 6 in R1C8, no 5 in R3C8

24. R1C6 = 2 (hidden single in N2), R9C6 = 4, R8C5 = 1, R7C6 = 9 (step 22), R1C5 = 4, R2C4 = 1, R1C7 = 1, R3C7 = 4 (step 1), R8C4 = 2, R89C9 = [72], clean-up: no 8 in R3C9, no 9 in R8C1, no 6 in R9C3 (step 2)

25. Naked pair {39} in R3C89, locked for R3 and N3 -> R34C5 = [63], clean-up: no 5 in R4C6
25a. Naked pair {68} in R12C9, locked for C9 and N3 -> R4C9 = 9, R3C89 = [93]
25b. 2 in N3 locked in R2C78, locked for R2

26. R9C1 = 3, R9C4 = 6 (hidden singles in R9), clean-up: no 7 in R6C3
26a. R9C2 = 9 (hidden single in N7), R8C1 = 4
26b. Naked pair {56} in R8C23, locked for R8 -> R8C78 = [98], R9C7 = 5
26c. Naked pair {78} in R79C3, locked for C3, clean-up: no 5 in R6C4

27. R1C4 = 3 (hidden single in C4), R2C5 + R3C4 = 17 = [98], R7C34 = [87], R9C35 = [78], R5C5 = 7, R4C4 + R6C6 = 13 = [58], clean-up: no 5,6 in R6C3
27a. Naked pair {49} in R6C34, locked for R6

28. Naked pair {27} in R24C7, locked for C7
28a. Naked pair {36} in R67C7, locked for C7 -> R5C7 = 8

29. 4 in C8 locked in R45C8
29a. R4C9 = 9 -> R45C8 = 10 = {46}, locked for C8 and N6 -> R6C78 = [37], R1C8 = 5, R2C9 = 8, R1C9 = 6, R1C3 = 9, R2C78 = [72], R23C6 = [57], R2C1 = 6, R6C34 = [49], R2C23 = [43], R5C4 = 4, R45C8 = [46], R5C6 = 1, R4C67 = [62], R4C3 = 1, R6C12 = [56], R5C123 = [932], R4C2 = 8 (cage sum), R3C1 = 1 (cage sum)

and the rest is naked singles