3x3::k:4096:4096:5377:4098:4098:1795:1795:3588:3588:4096:5377:5377:4098:3845:5382:6151:6151:3588:5377:5377:3848:3848:3845:5382:6151:6151:2569:4106:4106:3848:9739:9739:5382:5382:5382:2569:4106:3340:3340:9739:9739:9739:1549:1549:3086:2831:5392:5392:5392:9739:9739:3601:3086:3086:2831:5650:5650:5392:2323:3601:3601:5908:5908:4373:5650:5650:5392:2323:4630:5908:5908:4887:4373:4373:1816:1816:4630:4630:5908:4887:4887:

Prelims

a) R1C67 = {16/25/34}
b) R23C5 = {69/78}
c) R34C9 = {19/28/37/46}, no 5
d) R5C23 = {49/58/67}, no 1,2,3
e) R5C78 = {15/24}
f) R67C1 = {29/38/47/56}, no 1
g) R78C5 = {18/27/36/45}, no 9
h) R9C34 = {16/25/34}
i) 19(3) cage at R8C9 = {289/379/469/478/568}, no 1

1a. 45 rule on N1 R3C3 = 8 -> R3C4 + R4C3 = 7 = {16/25/34}, no 7,9, clean-up: no 7 in R2C5, no 2 in R4C9, no 5 in R5C2
1b. 45 rule on C6789 4 innies R5689C6 = 30 = {6789}, locked for C6, clean-up: no 1 in R1C7
1c. Min R89C6 = 13 -> max R9C5 = 5
1d. 45 rule on N9 R7C7 = 3 -> R6C7 + R7C6 = 11 = [65/74/92], clean-up: no 4 in R1C6, no 6 in R8C5
1e. 45 rule on N3 2 innies R1C7 + R3C9 = 7 = [43/52/61], R1C6 = {123}, R4C9 = {789}
1f. 45 rule on N5 2 innies R4C6 + R6C4 = 7 = [16]/{25/34}
1g. 45 rule on N7 2 innies R7C1 + R9C3 = 6 = [24/42/51], R6C1 = {679}, R9C4 = {356}

2a. 12(3) cage at R5C9 = {138/156/237/246} (cannot be {129/147/345} which clash with R5C78), no 9
2b. Killer pair 1,2 in R5C78 and 12(3) cage, locked for N6
2c. Consider combinations for 12(3) cage
12(3) cage = {138} => R5C78 = {24}, 5 in N6 only in R4C78, locked for 21(4) cage at R2C6 => R7C6 = 5 (hidden single in C6) => R6C7 = 6 (cage sum)
or 12(3) cage = {156/246} => R4C8 = 3 (hidden single in N6) => R1C6 = 3 (hidden single in C6), R1C7 = 4, R3C9 = 3 (step 1e), R4C9 = 7
or 12(3) cage = {237}, 7 locked for N6
-> no 7 in R6C7, clean-up: no 4 in R7C6 (step 1d)
2d. 4 in C6 only in R234C6, locked for 21(4) cage at R2C6, no 4 in R4C78
2e. Consider permutations for R6C7 + R7C6 = [65/92]
R6C7 + R7C6 = [65]
or R6C7 + R7C6 = [92], no 9 in R4C9 => no 1 in R3C9, no 6 in R1C7, no 1 in R1C6 => R1C6 = 3, R234C6 = {145} = 10 => R4C78 = 11 = [83]
-> no 6 in R4C78
2f. 4 in N6 only in R5C78 = {24} or 12(3) cage = {246} -> 12(3) cage = {138/156/246} (cannot be {237}, locking-out cages), no 7
2g. 7 in N6 only in R4C789, locked for R4
2h. Consider combinations for 12(3) cage
12(3) cage = {138}, 8 locked for N6
or 12(3) cage = {156/246} => R4C8 = 3 (hidden single in N6) … => R4C9 = 7 (as in step 2c)
-> no 8 in R4C9, no 2 in R3C9, no 5 in R1C7 (step 1e), no 2 in R1C6
2i. 21(5) cage at R2C6 contains 4 for C6 but no 6 = {12459/13458/23457}

[With hindsight step 5 could have been found at this stage]

3a. 45 rule on N7 1 outie R9C4 = 1 innie R7C1 + 1
3b. R7C1 + R9C4 + R7C6 = [235/452/562], 2 locked for R7, clean-up: no 7 in R8C5
3c. R7C1 + R9C4 + R7C6 = [235/452/562] -> no 5 in R7C45, clean-up: no 4 in R8C5
3d. 2 in R7 only in R7C16, 2 in R7C1 => R6C1 = 9 or 2 in R7C6 => R6C7 = 9 (cage sum) -> 9 in R6C17 (locking cages), locked for R6
3e. 18(3) cage at R8C6 = {189/279/369/378/468} (cannot be {459} because 4,5 only in R9C5, cannot be {567} which clashes with R7C1 + R9C4 + R7C6), no 5

4a. 12(3) cage at R5C9 (step 2f) = {138/156/246}
4b. 45 rule on R6789 4 innies R6C5689 = 19 = {1378/1468/2368/2458} (cannot be {1567/2467} which clash with R6C16, ALS block, cannot be {3457} because 12(3) cage cannot contain two of 3,4,5), 8 locked for R6

[I was a bit slow to spot this, the first key step]
5a. R1C7 + R3C9 = 7 (step 1e)
5b. Consider placement for R6C7 = {69}
R6C7 = 6 => R1C7 = 4, R3C9 = 3 => R4C9 = 7
or R6C7 = 9 => R4C9 = 7
-> R4C9 = 7, R3C9 = 3, R1C7 = 4, R1C6 = 3, clean-up: no 4 in R4C3 (step 1a), no 2 in R5C8, no 4 in R6C4 (step 1f)
5c. 14(3) cage at R1C8 = {158/167/257}, no 9
5d. 9 in C9 only in R789C9, locked for N9
5e. 45 rule on N2 3 remaining innies R23C6 + R3C4 = 11 = {146/245}, 4 locked for N2
5f. 6 of {146} must be in R3C4 -> no 1 in R3C4, clean-up: no 6 in R4C3 (step 1a)

6a. 21(5) cage at R2C6 (step 2i) = {12459/13458} -> R4C78 = {59}/[83]
6b. 45 rule on R123 4 remaining outies R4C3678 = 17 = {12}{59}/{15}[83]/[24][83], no 3 in R4C3, clean-up: no 4 in R3C4 (step 1a)
6c. 4 in N2 only in R23C6, locked for C6, clean-up: no 3 in R6C4 (step 1f)
6d. R4C3678 = {12}{59}/{15}[83], 1,5 locked in R4
6e. 1 in R4 only in R4C36, R4C3 = 1 => R3C4 = 6 (cage sum) or R4C6 = 1 => R6C4 = 6 (step 1f) -> 6 in R36C4 (locking cages), locked for C4, clean-up: no 1 in R9C3, no 5 in R7C1 (step 1g), no 6 in R6C1
6f. Naked pair {24} in R7C1 + R9C3, locked for N7
6g. R7C1 + R9C4 + R7C6 (step 3c) = [235/452], 5 locked for N8, clean-up: no 4 in R7C5

7a. 18(3) cage at R8C6 (step 3e) = {189/279/369/468} (cannot be {378} which clashes with R78C5)
7b. Consider permutations for R67C1 = [74/92]
R67C1 = [74] => R6C6 = {68} => 18(3) cage = {189/279/369/378} (cannot be {468} which clashes with R6C6
or R67C1 = [92] => R9C3 = 4
-> 18(3) cage = {189/279/369}, no 4, 9 locked for C6 and N8
7c. 4 in N8 only in R78C4, locked for C4 and 21(5) cage at R6C2, no 4 in R6C23
7d. 45 rule on N8 4 innies R789C4 + R7C6 = 18 with 4 in R78C4 and 5 in R7C6 + R9C4 = {2457} (only possible combination) -> R7C6 = 2, R9C4 = 5, R7C1 = 4, R6C1 = 7, R9C3 = 2, R78C4 = [74], clean-up: no 6 in R5C23
[A lot easier from here]
7e. R78C4 = [74] = 11 -> R6C234 = 10 = {13}6/{35}2
7f. Naked triple {135} in R4C3 + R6C23, locked for N4, 3 locked for R6, clean-up: no 8 in R5C2
7g. Naked pair {49} in R5C23, locked for R5 and N4, clean-up: no 2 in R5C7
7h. Naked pair {15} in R5C78, locked for R5 and N6
7i. 12(3) cage at R5C9 = {246} (only remaining combination), 6 locked for N6 -> R6C9 = 9 (or cage total from step 7e), R4C78 = [83]

8a. Naked pair {26} in R4C12, locked for R4 and N4 -> R4C45 = [94], R5C1 = 8
8b. Naked pair {26} in R36C4, 2 locked for C4
8c. Naked pair {18} in R12C4, locked for N2, R1C5 = 7 (cage sum)
8d. Naked pair {69} in R23C5, 6 locked for C5 and N2, clean-up: no 3 in R8C5
8e. Naked pair {18} in R78C5, locked for C5 and N8, R9C5 = 3
8f. R3C4 = 2 -> R4C3 = 5 (cage sum)

[Now back to the corners]
9a. 14(3) cage at R1C8 = {158} (only remaining combination), locked for N3
9b. Naked triple {158} in R1C489, 1,5 locked for R1, 5 locked for N3
9c. 1 in R3 only in R3C12, locked for N1
9d. 2 in R1 only in 16(3) cage at R1C1 = {259} -> R1C12 = {29}, locked for N1, R2C1 = 5
9e. R1C3 = 6, R2C6 = 4, R2C23 = {37}, 7 locked for R2 and N1
9f. 17(3) cage at R8C1 = {368} (cannot be {179} which clashes with R7C3) -> R89C1 = [36], R9C2 = 8
9g. R9C6 = 9, R9C89 = [74] -> R8C9 = 8 (cage sum)

and the rest is naked singles.

Preliminaries by SudokuSolver
Cage 6(2) n6 - cells only uses 1245
Cage 15(2) n2 - cells only uses 6789
Cage 7(2) n78 - cells do not use 789
Cage 7(2) n23 - cells do not use 789
Cage 13(2) n4 - cells do not use 123
Cage 9(2) n8 - cells do not use 9
Cage 10(2) n36 - cells do not use 5
Cage 11(2) n47 - cells do not use 1
Cage 19(3) n9 - cells do not use 1

This is a highly optimised solution. Any clean-up needed will be stated.

1. "45" in n1: 1 innie r3c3 = 8
1a. r3c4 + r4c3 = 7 (no 7,9)

2. "45" on c6789: 4 innies r5689c6 = 30 = {6789}: all locked for c6
2a. min. r89c6 = 13 -> max. r9c5 = 5

3. "45" on n9: 1 innie r7c7 = 3
3a. r6c7 + r7c6 = 11 = [92/74/65]
3b. no 4 in r1c6

4. "45" on n3: 2 outies r1c6 + r4c9 = 10 = [19/28/37]

5. 4 in c6 in sp11(2)r6c7 + r7c6 or in 21(4)r2c6
5a. -> 4 in r4c78 must have 7 in r6c7 (Locking out cages)
5b. "45" on n6: 4 innies r4c789 + r6c7 = 27
5c. -> {4689} blocked
5d. = {3789/5679}(no 1,2,4) = 3 or 5 in r4c78
5e. 7 & 9 locked for n6

6. 21(5)r2c8....
6a. {123}{69/78} blocked by r1c6 = (123) (or doesn't have 3 or 5 in r4c78)
6b. {12468} blocked by no 3 or 5 for r4c78 (step 5d)
6c. {125}{67} blocked by outies n3 = 10 (or boring way, doesn't have 3 or 5 for r4c78)
6d. {23457} as {234}{57} only, blocked by [16]+[65] in r1c67 & r6c7+r7c6
6e. = {12459/13458/13467}
6f. 1 & 4 locked for c6
6g. -> no 9 in r4c9 (outiesn3=10)
6h. r6c7 = (69)

7. 5 in c6 in r7c6 -> 6 in r6c7; or 5 in c6 in 21(5)r2c6
7a. -> 6 in r4c78, 21(5) must also have 5 for c6 (locking out cages)
7b. -> 21(5)r2c6: {13467} blocked
7c. = {12459/13458}(no 6,7)

8. r4c9 = 7 (hsingle n6), r3c9 = 3
8a. -> r1c6 = 3 (outies n3=10), r1c7 = 4

more key steps
9. "45" on n7: 2 innies r7c1 + r9c4 = 6 = [24/42/51]
9a. r6c1 = (679)
9b. r9c4 = (356)
9c. note: two outies n7 = 12 = [66/75/93]
9d. note2 from r7c1 = (679)
i. {69} in r6c17 -> r6c6 = (78)
ii. or [75] in outies n7 -> 2 in r7c6

10. 18(3)n8: {78}[3] blocked since r6c6 = (69) -> [25] in r7c6 + r9c4 (step 9dii) leaves nothing for 9(2)n8
10a."45" on n7: 1 innie r7c1 + 1 = 1 outie r9c4
-> {567} blocked from 18(3) by [22] in r7c16
10b. {459} blocked by 4 & 5 only in r9c5
10c. = {189/279/369/468}(no 5) = 1 or 2 or 6
10d. -> [516] in r7c1 + r9c34 forces 2 into both r7c6 and 18(3)n8
10e. -> no 5 in r7c1, no 1 in r9c3, no 6 in r9c4, no 6 in r6c1

11. r6c17 from {679} -> must have 6 or 7 -> 5 in r7c6 (h11(2) or r9c4 (outiesn7=12)
11a. 5 locked for n8
11b. no 4 in 9(2)n8
11c. also 9 locked in r6c17 since can't have [55] in r7c6 + r9c4 -> 9 locked for r6
11d. -> 2 locked in r7c16 for r7 (locking cages)

12. "45" on n5: 2 innies r4c6 + r6c4 = 7 = [16]/{25}/[43] (r6c4 = (2356))

13. combined cage 24(4)r2378c5 = {69}{18/27}/{78}{36}
13a. 6 locked for c5
13b. note: has 6 in r23c5 or 3 in r78c5

Took a long time to find this "45". Final crack
14. "45" on n2369: 1 innie r3c4 + 1 = 2 outies r47c6
14a. but [515] blocked by 6 in both r6c4 (h7(2)n5) and r6c7(sp11(2))
14b. and [625] -> r39c4 = [63], blocked by combined 24(4)r2378c5 (step 13b)
14c. -> no 5 in r7c6
14d. -> r7c6 = 2, r6c7 = 9, r67c1 = [74], r9c34 = [25]
14e. no 7 in 9(2)n8
14f. 18(3)n8 = {189/369/468}(no 7)

15. "45" on n8: 2 remaining innies r78c4 = 11 and must have 7 for n8 = [74] only

16. killer pair 1,3 in 9(2)n8 and r9c5: both locked for c5

17. "45" on n1236: 2 remaining outies r4c36 = 6 = {15} only, both locked for r4
17a. no 3 in r6c4 (h7(2)n5)
17b. r4c78 = [83]
17c. no 1 in r3c4

18. naked pair {26} in r36c4: both locked for c4
18a. r4c4 = 9

19. naked pair {18} in r12c4: both locked for n2 & c4
19a. -> r1c5 = 7 (cage sum)

20. r4c6 = 1 (hsingle c6)
20a. r4c3 = 5, r3c4 = 2 (cage sum)
20b. r56c4 = [36]
20c. r56c6 = [78]

21. 13(2)n4 = {49}: both locked for n4, 4 for r5

22. r89c6 = {69} -> r9c5 = 3

23. no 3 in r2c1 since r1c12 cannot = {67/58/49} = 13

24. r8c1 = 3 (hsingle c1)
24a. -> r9c12 = 14 = {68}: both locked for n7 and r9
24b. -> r9c6789 = [9174]


much easier now

1. Innies c6789 = r5689c6 = +30(4) = {6789}
Innies n9 = r7c7 = 3
-> r7c6 from (245) and r6c7 from (976)
IOD n3 -> r1c6 = r3c9
Since r3c9 cannot be 5 -> r1c6 cannot be 5
Since 3 already placed in c7 -> r1c6 cannot be 4
-> r1c6 = r3c9 from (123) and r4c9 from (987)

2! Innies n6 = r4c789,r6c7 = +27(4)
(A) {3789}
(B) {4689}
(C) {5679}
Since r4c9 is Min 7 and r6c7 from (679) -> One of (345) in r4c78

(A) -> (HS 3 in c6) r1c6 = 3 -> 10(2)r3c9 = [37]
(B) Leaves no place for 4 in c6 (Since r1c6 is Max 3 and 14(3)r6c7 cannot be [743])
(C) -> (HS 5 in c6) 14(3)r6c7 = [653]
-> (Innies n3) [r1c7,r3c9] cannot be [61] -> 9 not in r4c9 -> r4c9 = 7

In all cases 10(2)r3c9 = [37] and 7(2)r1c6 = [34]

3! Consider Step 2 case (C). Innies n6 = [{59}76]
This puts (HS 3 in n6) r6c8 = 3 and r56c9 = {18}
This puts only solution for 14(3)n3 = [7{25}]
But this leaves remaining cells c9 = r789c9 = +19(3) contradicting 19(3) cage in n9.
-> Step 2 (A) must be correct.
-> (Since r6c7 cannot be 8) Innies n6 = [8379]
-> r7c6 = 2

4. Remaining cells c6 = r234c6 = {145}
Remaining Innies n2 = r2c6,r3c46 = +11(3). Cannot contain both (15)
-> Either r234c6 = [{14}5], r3c4 = 6, (Innies n5) r6c4 = 2
Or r234c6 = [{45}1], r3c4 = 2, r6c4 = 6.
Innies n1 -> r3c3 = 8
-> r4c36 = {15} and r3c6c4 = {26}

5. Innies n7 = r7c1,r9c3 = +6(2)
Given previous placements this can only be [r7c1,r9c3] = [42]
-> r6c1 = 7 and r9c4 = 5

6. Remaining Innies n8 = r78c4 = +11(2)
Since r89c6 is Min +13 -> Max r9c5 = 5
-> 9 in n8 in r89c6
-> Min r89c6 = +15 -> Max r9c5 = 3
-> (HS 4 in n8) r78c4 = [74]
-> (HS 7 in c6) r5c6 = 7

7. 3 in n8 only in c5
-> (HS 3 in n5) r5c4 = 3
-> 3 in n4 in r6c23
-> Remaining Innies n4 = r4c3,r6c23 = [5{13}] or [1{35}]
-> 13(2)n4 = {49}
-> 16(3)n4 = [{26}8]
-> r4c45 = [94]
Also 6(2)n6 = {15}
-> 12(3)n6 = {246}

8. 1 in n8 only in c5
-> (HS 1 in n5) r4c6 = 1
-> r23c6 = {45} and r3c4 = 2
-> r4c3 = 5 and r6c23 = {13}
-> r6c4 = 6
Also (HS 5 in n5) r6c5 = 5
-> r5c5 = 2
Also r6c6 = 8
-> r89c6 = {69} and r9c5 = 3
-> 9(2)n8 = {18}
Also 12(3)n6 = [6{24}]
Also 15(2)n2 = {69}
-> r1c5 = 7 and r12c4 = {18}

9. Given previous placements only solution for 14(3)n3 is {158} with 5 in r1.
-> 24(4)n3 = {2679} with 9 in c8
Also -> 2 in r1 in r1c12
-> 16(3)n1 = [{29}5]
-> r23c6 = [45]
-> 21(5)n1 = [6{37}14]
-> 13(2)n4 = [94]
-> r1c12 = [92]
-> r4c12 = [26]
-> 17(3)n7 = [368]
etc.