Assassin 48

[Andrew]

Congratulations to the tag team on the Hevvie!

I must have a look at your moves when I've got time.

What's the situation with the Lite version? Is anybody planning on writing a walkthrough for it?

I've been doing it while the Hevvie was in progress. I felt that it wasn't much lighter than A48 with the exception that it finished smoothly without needing a breakthrough.

Here is my walkthrough

Para. Thanks for your comments and corrections. I must remember not to refer to naked pairs as killer pairs.

1. R1C34 = {79}, locked for R1

2. R12C5 = {12}, locked for C1 and N2

3. R1C67 = [34/43/52/61], no 8, no 5,6 in R1C7

4. R23C4 = {36/45}, no 7,8,9

5. R23C6 = {69/78}, no 3,4,5

6. R5C12 = {89}, locked for R5 and N4

7. R5C89 = {17/26/35}, no 4

8. R78C4 = {16/25/34}, no 7,8,9

9. R78C6 = {16/25/34}, no 7,8,9

10. R89C5 = {49/58/67}, no 3

11. R9C34 = {69/78}

12. R9C67 = {16/25/34}, no 7,8,9

13. 19(3) cage in N36 = {289/378/469/478/568}, no 1

14. 9(3) cage in N47 = {126/135/234}, no 7,8,9

15. R678C9 = {289/378/469/478/568}, no 1

16. 17(5) cage in N9 = 123{47/56}, no 8,9, 1,2,3 locked for N9, clean-up: no 4,5,6 in R9C6

17. 45 rule on C9 3 innies R159C9 = 10 = {127/136/145/235}, no 8

18. 45 rule on C123 3 innies R159C3 = 19 = {289/378/469/478} (cannot be {568} because no 5,6,8 in R1C3), no 1,5,6,7 in R5C3, no 7,9 in R9C3, clean-up: no 6,8 in R9C4

19. Naked pair {79} in R19C4, locked for C4

20. 8 in C4 locked in R46C4, locked for N5

21. 45 rule on C123 2 outies R19C4 – 12 = 1 innie R5C3, R19C4 = 16 -> R5C3 = 4

22. 4,8 locked in 17(4) cage in N45 (steps 21 and 20) = {2348}, no 1,5,6, 2,3 locked for C4 and N5, clean-up: no 6 in R23C4, no 4,5 in R78C4

23. 1 in N5 locked in R456C6, locked for C6 and 21(4) cage -> no 1 in R5C7, clean-up: no 6 in R78C6, no 6 in R9C7

24. Naked pair {45} in R23C4, locked for N2, clean-up: no 2,3 in R1C7

25. Naked pair {16} in R78C4, locked for N8, clean-up: no 7 in R89C5

26. Killer pair 2/3 in R78C6 and R9C6 for C6 and N8

27. R1C67 = [61] (naked singles) -> R12C5 = [21], clean-up: no 9 in R23C6

28. Naked pair {78} in R23C6, locked for C6 and N2

29. R1C34 = [79], R3C5 = 3, R9C34 = [87] (naked singles), clean-up: no 5 in R8C5

30. 9 in C6 locked in R46C6, locked for N5

31. 45 rule on C789 2 outies R19C6 – 2 = 1 innie R5C7, min R19C6 = 8 -> min R5C7 = 6
31a. 21(4) cage in N56 = 19{47/56}

32. Killer triple 5/6/7 in R5C5, R5C7 and R5C89 for R5 -> R5C6 = 1, clean-up: no 7 in R5C89

33. R34567C5 = 367{49/58}
33a. 8,9 only in R7C5 -> R7C5 = {89}

34. R159C9 (step 17) = {136/145/235}
34a. 1 only in R9C9 -> no 4,6 in R9C9

35. 45 rule on C1 3 innies R159C1 = 16 = [385/394/493/583/592], no 8 in R1C1, no 1,6,9 in R9C9

36. 45 rule on N7 3 outies R6C123 = 9 = {126/135} = 1{26/35}, no 7, 1 locked for R6 and N4

37. 7 in N4 locked in R4C12, locked for R4
37a. 45 rule on N1 3 outies R4C123 = 15 = 7{26/35}

38. R4C9 = 1 (hidden single in R4)
38a. R23C9 = 15 = {69/78}, no 2,3,4,5

39. 45 rule on N3 1 remaining innie R3C8 = 2, clean-up: no 6 in R5C9
39a. R4C78 = 17 = {89}, locked for R4 and N6

40. R6C4 = 8, R6C6 = 9 (hidden singles in R6)

41. R159C9 (step 34) = {235}, locked for C9

42. 4 in C9 locked in R678C9 = 4{69/78}

43. 45 rule in N1 1 outie R4C1 – 1 = 1 innie R3C2, no 8,9 in R3C2, no 3 in R4C1

44. 45 rule on N7 1 outie R6C1 = 1 innie R7C2, no 4 in R7C2
44a. R6C1 and R7C2 cannot contain any candidates that aren’t in R123C3; in this case in R23C3 since R1C3 = 7 which isn’t in R6C1/R7C2

45. 4 in R4 locked in R4C56, locked for N5

46. 45 rule on N9 1 innie R7C8 – 2 = 2 outies R6C9 + R9C6, min R6C9 + R9C6 = 6 -> min R7C8 = 8, max R6C9 + R9C6 = 7 -> R6C9 = 4

47. 17(3) cage in N69 = [269/278/359/368/539/638], no 7 in R6C7

48. Naked pair {89} in R47C8, locked for C8

49. Naked pair {89} in R7C58, locked for R7, clean-up: no 6,7 in R8C9

50. R1C2 = 8 (hidden single in R1) -> R5C12 = [89]

51. R9C5 = 9 (hidden single in R9) -> R7C58 = [89], R4C78 = [98], R8C5 = 4, R8C9 = 8, R7C9 = 7, clean-up: no 3 in R78C6

52. Naked pair {25} in R78C6 -> R4C6 = 4, R9C6 = 3, R9C7 = 4

53. Naked pair {69} in R23C9, locked for N3

54. 17(3) cage in N69 (step 47) = [26/35/53]9 -> no 6 in R6C7, no 7 in R6C8

55. R5C7 = 7 (hidden single in N6) -> R2C8 = 7 (hidden single in N3) -> R23C6 = [87] -> R1C8 = 4, R3C7 = 8 (hidden singles in N3), R6C5 = 7 (hidden single in N5)
[Para pointed out that I had missed R7C1 = 4 (hidden single in N7). This would have simplified step 59.]

56. 6 in C7 locked in R78C7, locked for N9

57. R9C2 = 6, R9C8 = 1 (hidden singles in R9), clean-up: no 6 in R6C1 (step 44)

58. 6 in C1 locked in R234C1 = 6{29/47}, no 1,3,5

59. 1 in C1 locked in R678C1 = 1{29/47}, no 3,5, clean-up: no 3,5 in R7C2 (step 44)
59a. 9 only in R8C1 -> no 2 in R8C1

60. R7C1 = 4 (hidden single in R7) -> R68C1 = [17] (step 59), R7C2 = 1 (step 44)
60a. R6C23 = [26]/{35}, no 2 in R6C3

61. R4C2 = 7 (hidden single in R4)
61a. R3C2 + R4C3 = 7 = [43/52], no 5,6 in R4C3

62. R234C1 (step 58) = {269}, locked for C1

and the rest is naked singles and cage sums

[Glyn]

I wondered whether anything could be done to crack the problem of the Hevvie using colouring. Starting from Mikes Grid I found the following.


1(a)If R1C7=5 requires R1C6=6 to satisfy 11(2) cage. R1C6 forces R9C5=6.
1(b)If R1C7=5 forces R1C3=4 forces R5C3=5. The 14(2) cage in N5 must be {68} forces R6C7=6.
1(a) and 1(b) together force R8C8=6. (Nonet 9 as blocked for 6 in R9 and C7).

2) R1C7=5 forces R9C7=3 forces R8C8=5.

3) R8C8 cannot equal both 5 and 6 therefore R1C7<>5.

Unfortunately it doesn't entirely avoid killer info, but the puzzle is cracked at this point, wonder how early it might have worked.

All the best

Glyn

[mhparker]

I wondered whether anything could be done to crack the problem of the Hevvie using colouring.

It's not coloring, of course. More like a forcing chain, but - yes - it works all the same, so congratulations on finding it!

Unfortunately it doesn't entirely avoid killer info

Actually, I don't see this as a disadvantage at all. Quite the opposite, in fact.

wonder how early it might have worked.

After step 92 (of 108).

Thanks for sharing this information, Glyn. See you on the Assassin 49 thread!

[Glyn]

Hi everyone

I found the chain as a result of some coloring in Sudocue. I'll try it again if I get stuck.

Shame it is trickier to do coloring on the Jigsaws.

By the way Pete Ty has posted a nasty Jigsaw Killer, but it is hidden away in General Website Comments at the end of the Tiny Font post. I'm sure it would be of interest to readers of this forum. A sort of unofficial Texas Jigsaw to keep us busy until Ruud has another one.

All the best

Glyn

[sudokuEd]

..(a move) to crack the problem of the Hevvie ...wonder how early it might have worked.

Step 57. Kind of.

Here is a condensed and simplified, though STALE (Still-Takes-A-Lotof-Effort) walk-through for Assassin 48 Hevvie.

Managed to get it down to using "just" 57 of the original 110 steps before Glyn's fine forcing chain could be used to crack it. Have included a bunch of new clean-up-at-the-end steps to muck it.

Many of the original steps have been changed/re-ordered to make them simpler and to take account of short-cuts found later. Have left the step numbers the same in case you want to check something. Let me know if its too confusing doing it this way.

Am still in shock from Richard's 8 outtie's move to get a placement (step 102). Another world record for Richard. Too r-rated for a simplified walk-through though.

Any puzzle raters out there who can give this puzzle a rating? Now that its easier to digest, can any one find a shortcut?


Assassin 48 Hevvie: Condensed/simplified walk-through.

1. 3(2)n8 = {12}:locked for c4, n8

2. 9(2)n8 = {36/45}(no 789) = [3/5..]

3. 8(2)n2 = {17/26}(no 3589) ({35} blocked by 9(2)n8 step 2)
3a. 8(2)n2 = [1/6..]

4. 7(2)n2 = {25/34}(no 16789} ({16} blocked by 8(2)n2 step 3a)
4a. 7(2)n2 = [2/3,4/5..]

5. 1 in c6 in 12(4)n5 = 12{36/45}(no 789)
5a. 1 locked for n5 and no 1 r5c7
5b. 12(4) must have 2 -> no 2 r5c5

6. complex hidden Killer pair 2/3 r23456c6. Here's how.
6a. 12(4)n5 = {1236} ->2/3 must be in r5c7 so r456c6 won't clash with 7(2)n2 (step 4b.) -> [2/3] both locked in 7(2) and r456c6 for c6
6b. 12(4)n5 = {1245} -> 4 must be in r5c7 so r456c6 won't clash with 7(2) (step 4b) -> [2/3] both locked in 7(2) and r456c6
6c. r5c7 = {234}
6d. no 4 r456c6
6d. 2 and 3 locked for c6 in r23456

7. 11(2)n2 = [47/56/65/74/83/92]
7a. r1c7 = 2..7

8. 12(2)n8 = [48/57/75/84/73](no 1,2,6)
8a. r9c7 no 9

9. "45"c789 -> 3 innies r159c7 = 12 = h12(3)c7 = {237/[624]/345}(no 8)
9a. no 4 r9c6

10. A neat little contradiction chain eliminates [624] from h12(3)c7
10a. from step 6a, 2 in r5c7 -> r456c6 = {136} -> 7(2)n2 = {25}
10b. 6 in r1c7 -> 5 in r1c6: but 2 5's c6
10c. h12(3) = {237/345} = 3{27/45}(no 6)
10d. no 5 r1c6
10e. 3 locked c7

11. 8 and 9 in c5 only in 28(5)
11a. 28(5) {14689} blocked by 8(2)n2
11b. 28(5) = {13789/23689/24589}
11c. 1 only in r3c5 -> no 7 r3c5

12. 4 in c6 only in n2: 4 locked for n2

13. 14(2)n2 = {59/68}

14. 12(2)n1 = {39/[48]/57}(no 1,2,6, no 8 r1c3)

15. "45"n2: 3 innies = 16 = h16(3)n2 = {178/259/268/349}
15a. {169/367} blocked by 8(2)
15b. {358/457} blocked by 7(2)
15c. 1 and 2 are only available in r3c5 for 3 innies -> no 5,6,8 r3c5
15d. r1c46 + r3c5 = {[78]1/[59]2/[86]2/[34]9/[94]3}. Note: blocked combo's
i.[87]1 :blocked 2 4's r1c37
ii.[39]4 :blocked 2 9's r1c37
15e. no 7 r1c6 -> no 4 r1c7

16. 7 in c6 only in n8: 7 locked for n8
16a. r9c3, no 1,4,9

17. "45"n8 -> 3 innies = 18 = h18(3)n8
17a. h18(3) = {369/378/459} ({468/567} blocked by 15(2))
17b. {369} combo: if r9c46 = [39] means r9c467 = [393](2 3's r9) -> {369} combo must have 3 in r7c5 -> no 6 r7c5 (since no other combo with 6)
17c. {378} combo: 7 only in r9c6 -> no 8 r9c6 (since only combo with 8)
17d. no 4 r9c7
17e. {459} combo:
i.if r9c46 = [45] -> r9c3467 = [7457] clash -> 9 must be in r9c46 -> no 9 r7c5
ii.if r7c5 + r9c46 = 4[59] -> 9(2)n8 = {36} and r9c37 = [63]: clash in r9!
iii. -> can only have r7c5 + r9c46 = 4[95]/5[49]
iv. no 5 r9c4

17f. In summary: h18(3)n8 = r5c5 + r9c46 = [3[69]/3[87]/8[37]/4[95]/5[49]
17g. no 6 r7c5,
17h. no 8 r9c6 -> no 4 r9c7
17i. no 9 r7c5
17j. no 5 r9c4 -> no 6 r9c3

20. 19(3)n3: no 1
20a.14(2)n4 = {59/68}
20b. 8(2)n6: no 4,8,9

The first of the clever "45"s with another chain move.
23. 45 on n147 & n5 (!!) - outies: r1c4 r3c5 r7c5 r5c7 r9c4 = 23
i. -> no 3 at r3c5. Here's how
23a. from step 15d, 3 in r3c5 -> 9 in r1c6 -> r3c5 + r1c4 = [39] = 12
23b. 3 in r3c5 -> 28(5) = {13789/13689}(no 4,5) -> r7c5 can only be 8
23c. from step 17f. the only combo in h18(3)n8 with 8 in r7c5 is
23d. 8[37]: -> r7c5 + r9c4 = [83] = 11
23e. These 4 outies of n147 & 5 with r3c5 = 3 sums to 12 + 11 = 23. But the 5 outies must = 23, not just the 4.
23g. -> no 3 at r3c5
23h. ->h16(3)n2 = r1c46 + r3c5 = {[78]1/[59]2/[86]2/[34]9}
23i. no 9 r1c4
23j. no 3 r1c3

A big contradiction chain - but essential to solve this puzzle.
26. no 9 r1c3. Here's how.
26a. r1c3 = 9 -> r1c46 = [34] (step 23h.)-> r1c7 = 7
26b. -> h12(3)c7: r159c7 = [723]
26c. -> r9c46 = [69/49](step 17f.) -> r9c3 = 5/7
26d. "45"c123: r159c3 = 16 = h16(3)c3
26e. but 7 is blocked from r9c3 because cannot have r19c3 = [97] = 16 in the h16(3)c3
26f. -> h16(3)c3 = [925]
26g. however,this last combo is also blocked since the 2 in r5c3 clashes with the 2 already in c7 (step 26c). Dizzy yet?
26h. conclusion: no 9 in r1c3 -> no 3 r1c4

27. 3 now locked in r23c6 for n2 - locked for c6
27a. 7(2)n2={34} - locked for n2 and c6
27b. 2 in c6 only in n5: 2 locked for n5 and no 2 r5c7
27d. cleanup 11(2)n23 - no 7 at r1c7

28. 5 now locked in r123c4 for n2 - 5 locked for c4
28a. cleanup: 21(4)n45 - no 7 possible at r5c3: since no 1,2,5 in r456c4

29. cleanup 28(5)c5=1{3789}/2{3689}/2{4589} = [1/2]
29a. 1 and 2 only available in r3c5 -> no 9 at r3c5

25. no 9 r5c3. Here's how - with a short chain.
25a. "45"c123: r19c4 - 7 = r5c3
25b. r5c3 = 9 -> r19c4 = 16 = [79]
25c. r19c4 = [79] -> 14(2)n2 = {68}
25d. but this leaves no 5 for c4
25e. -> no 9 r5c3

30. no 9 at r9c4. Here's how.
30a. "45"c123: r19c4 - 7 = r5c3
30b. max. r5c3 = 8 -> max. r19c4 = 15
30c. ->if r9c4 = 9 can only have r19c4 = [59] = 14:
30d. -> r5c3 must = 7: but no 7 in r5c3
30b. -> no 9 at r9c4
30c. cleanup 11(2)n78 - no 2 at r9c3

31. 11(2) & 12(2) r9 must use 3 (i.e. {38}{75}/[56][93]/[74][93]) -
31a. 3 locked for r9
31b. cleanup 9(2)n8 - no 6 at r8c5

32. Cleanup from step 30b.
32a. from step 17f. h18(3)n8 now = [3[69]/3[87]/8[37]/5[49]
32b. no 4 r7c5
32c. no 5 r9c6
32d. no 7 r9c7

35. 9 in n8 now only in c6: 9 locked for c6
35a. no 2 r1c7

41. naked pair {35} at r19c7: 3 and 5 locked for c7
41a. r5c7=4
41b. 12(4)n56 = {125}4 - no 6

38. 9 now locked in 14(2) for n2 = {59}: locked for c4
37a. 12(2)n12 no 7 at r1c3
38a. 21(4)n45 - no 5 or 9 r456c4 so cannot have 1 at r5c3

39. 8 locked in r1c46 for n2: 8 locked for r1

40. 5 locked in r789c6 for c6 - 5 locked for n5

45. "45" on c123: r159c3 = 16 = h16(3) = [457]/5{38} = 5{47/38}
45a. r5c3 = {358}
45b. r9c3 = {378}
45c. no 6 r9c4
45d. 5 locked for c3

42. 5 & 8 locked in 12(2)/11(2) for r1 - no 5 in rest of r1

43. 21(4) n45 = 3{468/567}
43a. must use 3 - eliminates 3 at r5c5

46. 6 locked in n5 for c4 - 6 nowhere else in n5

Now: a very clever "45"
34. 45 on r5: innies = h19(4) = {1279}/{1378}/{2359}/{2368}
i.combos with 89, 56 or 69 blocked by 14(2)n3
47a. {1279} blocked by 1,2 only in r5c6
47b. -> h19(4)r5 = 3{178/259/268} = [1/2..]
47c. must use 3 - locked for r5
47d. cleanup - no 5 in 8(2)n6
47e. 1 or 2 required in h19(4) only in r5c6 -> no 5 r5c6
47f. h19(4) = {378}[1]/[5392]/[3682] ([3691]blocked by 14(2)n4)

49. from 47c. 3 locked in r5 is actually locked in c34 - eliminate 3 from r46c4 for 21(4) cage
49a. 21(4) now = {3468}/53{67}(no other combo with 5 at r5c3 step 47f.)
49b. no 7 at r5c4
49c. h19(4)r5 now = [{38}71/5392/3682]

48. 5 only in n4 for r5 - 5 nowhere else in n4

The next steps require careful unpicking of combinations.
57. No 6 in r5c4 because of a hard to spot contradiction.
57a. from step 49c, 6 in r5c4 -> h19(4)r5 must have 8 in r5c5
57b. the only combo for the 21(4)n45 with r5c4 = 6 is {3468} with 3 in r5c3 -> 8 in r46c4
57c. but this means 2 8's n5
57d. -> no 6 r5c4
57e. h19(4) now = {38}[71]/[5392]
57f. no 8 r5c5


58. 21(4)n45 - {3468} combo is only combo with 8 with {38} only in r5c34
58a. no 8 at r46c4

Another very clever "45" which only works because of the combo work above.
63. 4 Innies r1234: r1c5+r2c456 = 21. r3c5 = 1/2 -> r4c456 = 19 or 20
63a. At least one of r4c456 must be >= 8 (else max. = {567} = 18) -> r4c5 = {89} only
63b. {89} in r4c5 -> no {12} possible in r4c456
63c. -> r4c6 = 5
63d. r4c5 = {89}
63e. 3 remaining innies r4c4 + r34c5 = 16 = 6[19]/7[18]/6[28] (no 4 r4c4)
63f. -> r34c5 =[19/18/28]
63g. -> 28(5) = [197{38}/18973/28945]

65a. combining parts of steps 63e and g: r4c4 = 7 -> r3456c5 = [1897]
65b. but this means 2 7's n5
65c. -> r4c4 != 7
65d. r4c4 = 6
65e. from 63g: r34567c5 = [197]{38} or [28945] -> no 7,9 in r6c5
65f. Cleanup: no 4 in r3c8 (due to {456} unavailable in r4c78}


Now: it's finally time to move the focus: to n469.
64. 4 in r4 now locked in n4 -> not elsewhere in n4
64a. Cleanup: no 8 in r7c2 (due to {45} unavailable in r6c23)

54. 45 on n3 - outies total 20(4). r1c6 = {68} -> r4c789 = 12 or 14
54a. if r4c789 = 12(3) = {138} ({129/237} blocked by 8(2)n6
54b. if r4c789 = 14(3) = [239]
54c. In summary, r4c789 = {138}/[239] = [8/9] not both
54d. no 7 r4c789
54e. no 2 r4c89

55. cleanup 11(3) n36 = {128/137/236}
55a. no 1,3,5,8 at r3c8

66. 7 in r4 now locked in n4 -> not elsewhere in n4
66a. Cleanup: no 1,6 in r7c2 (due to {457} unavailable in r6c23)

68. Outies n1: r1c4+r4c123 = 20
68a. r1c4 = {78} -> r4c123 = 12 or 13
68b. 7 already locked in r4c123 (step 66) -> no 8,9 in r4c123
68c. max. of r4c23 = 4 + 7 = 11 -> no 1, 2 in r3c2

69. "45"n9: r6c789 - 16 = r9c7
69a. r9c7 = {35} -> r6c789 = 19/21
69a. r6c789 must contain 1 of {89} for n6 (only place in n6 outside of r4c789 step 54c),
69b. r6c789 must contain exactly one of {67} - only place in n6 outside of 8/2 cage
69c. r6c789 must have 5 for n6
69d. if r9c7 = 3 -> r6c789 = 19 = {568} only
i. -> r6c78 + r7c8 = 16(3) = [{68}2] ([{58}3] blocked by 3 in r9c7 step 69a:[655] not valid for 16(3) cage)
ii. -> r6c789 = {68}[5]
69e. if r9c7 = 5 -> r6c789 = 21 = {579}
i. -> r6c78 + r7c8 = 16(3) = [754/952]
ii. -> r6c789 = [759/957]
69f. r7c8 = 2,4
69g. r6c9 = 5,7,9
69h. no 123 r7c78


70. 3 in n6 now locked in r4 -> not elsewhere in r4

53. 4 innies on n9 = 14 = h14(4).
i. from step 69d and e, r9c7 + r7c8 = [32/52/54]
53a. if r9c7 + r7c8 = [32] = 5 -> r6c9 = 5 (step 69dii) and r78c9 = 9 = {18} only ({27/36} blocked by [23])
53b. if r9c7 + r7c8 = [54] = 9 and r6c9 = 9 (step 69e) -> r78c9 = 5 = {23} ({14} blocked by r7c8)
53c. if r9c7 + r7c8 = [52] = 7 and r6c9 = 7 (step 69e) ->r78c9 = 7 = {34} ({16} means 14(3) = {167}-> 19(3)n3 = {289} which clashes with r5c9)

53d. In summary: r78c9 = {18/23/34}(no 5,6,7,9)
53f. h14(4)n9 = {1238/2345} = 23{18/45}
53g. 2 and 3 locked for n9
53h. 14(3)n6 = {158/239/347}

73. 2 in r9 locked in n7 -> not elsewhere in n7

75. Common Peer Elimination (CPE): r3c8 can see all candidate positions with digit 2 in c7
-> no 2 in r3c8
75a. 11/3 at r3c8 = {137/236} = 3{17/26} ({128} blocked by r3c8)
75b. r4c8 = 3
75c. r4c7 = {12}

This step uses a clever overlap of innies and combinations.
87. no 5 r9c89. Here's how.
87a. "45" on r9: 5 innies = 22 = h22(5) and must have 1 and 2 for r9 (and no 3) = {12469}/{12568} ({12478} blocked by 11(2)r9)
87b. 31(5)n9 = {16789}/{45679}
87c. Only combo with 5 in h22(5)r9 is {12568} -
i. 2 locked at c12
ii. c89={16}/{18}/{68}(from combos in 31(5)n9) (note:{56} blocked by r9c5 which must have 5 or 6)
87d. no 5 in c89

105a. i/o difference(n6): r4c9 + r6c789 = r3c8 + 22
-> no 6 in r6c8
(reason: would force r3c8 to 7 -> r4c9 + r6c79 = 23 = {689} -> 2 6's in n6)
105b. Only combo with 6 in 16(3) = {268} -> No 8 in r6c7

A final forcing chain to crack open this puzzle then some new mucking steps
1(a)If R1C7=5 requires R1C6=6 to satisfy 11(2) cage. R1C6 forces R9C5=6.
1(b)If R1C7=5 forces R1C3=4 forces R5C3=5 (since must have h16(3)c3 = [457])
1(c) ->The 14(2) cage in N5 must be {68}
1(d) forces R6C7=6 (single n6).
1(a) and 1(d) together force R8C8=6. (Nonet 9 is blocked for 6 in R9 and C7).

2) R1C7=5 forces R9C7=3 forces R8C8=5.

3) R8C8 cannot equal both 5 and 6 therefore R1C7<>5.

4)r1c67 = [83], r1c34 = [57], r9c67 = [75], r6c4 = 4
4a. r5c34 = {38}: locked for r5

5) 14(2)n4 = {59}:locked for n4, r5

6) r5c5 = 7 -> rest of 28(5) = {1389}
6a.34c5 = [19]

7) 15(2) n8 = {69}: locked for n8, c6

8) r89c5 = [54]

9) r67c5 = {38}

10. 8(2)n6 = {26}: locked for n6

11. r4c79 = [18], r3c8 = 7

12. r56c6 = [12]

13. 11(2) n7 = {38}:locked for r9

14. "45" n9: r6c9 - 5 = r7c8
14a. no 5 r6c9
14b. r6c8 = 5

15. 24(6)n3 = {12489}(no 6):locked for n3
15a. r23c9 = {56}:locked for c9

16. r5c89 = [62]

17. 14(3)n6 = [7]{34} ({149} blocked by r1c9)
17a. {34} locked for c9, n9

18. r6c7 = 9, r7c8 = 2, r78c4 = [12]

19. r23c7 = {28}:locked for c7
19a. r8c8 = 8 (hsingle n9)

20. r9c89 = {19}:locked for r9

21. r9c12 = {26}:locked for n7
21a. 22(5)n7 = 246{19/37}
21b. 4 also locked for n7

22. naked pair {38} r59c6:locked for c3

23. 14(3)n4: = {158/167/356} = [5/7..]
23a. 5 and 7 only in r7c2 = {57}

24. deleted

25. 14(3)n1 r4c23 = {247} -> 14(3) = {248/347} = [3/8..]
25a. 3 or 8 only in r3c2 = {38}

26. "45" n1: 3 innies = 14
26a. r3c2 + r23c1 = 14 = 3[74]/8{24}
26b. r23c1 = {247}
26b. 4 must be in r23c1: locked for c1, n1
26c. 13(3)n1 = [742]/{24}7 = {247}:locked for c1
26d. r4c1 = {27}

27. r9c12 = [62]

28. 16(3)n4 = {358} = [853]

29. r5c12 = [95], r5c34 = [38], r67c5 = [38], r9c34 = [83]

30. r78c9 = [34], r7c2 = 7, r4c2 = 4, r78c7 = [67], r78c6 = [96]

31. r7c3 = 4, r1c19 = [19], r12c8 = [41], r1c25 = [62], r2c5 = 6

32. r23c9 = [56], r23c4 = [95]

33. split 26(5)n1: {289} blocked by r2c7
33. = {379} only

Time for a beer.