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3x3::k:3328:3585:8706:9219:8706:8706:8706:4612:2309:3328:3585:8706:9219:9219:9219:4612:4612:2309:3328:3585:3590:3590:9219:9219:4612:2311:2311:4104:4617:4617:3590:3590:9219:4362:4362:4362:4104:4617:2827:2828:2828:3597:4622:4622:5903:4104:2827:2827:2064:2064:3597:4622:5903:5903:4113:4113:4113:5394:3091:1044:4622:5903:5903:3093:3093:4113:5394:3091:1044:5142:5143:5143:3093:4376:4376:5394:5142:5142:5142:5142:5143:

1. 17(2)n7 = {89}
-> Innies c1 = r789c1 = +16(3) = {457} or {367}
-> Remaining cells in n7 = r78c23 = +12(4) = {1245} or {1236}

2. 4(2)n8 = {13}
-> Innies n8 = r9c56 = +8(2) = {26}
Also 9 in n8 in r78c4

3. 1 in n9/r9 must be in r9c78
2 in n9 not in 20(3)n9 so must be in r7c789
-> 2 in n7 in r8c23
Remaining innies n89 (two of r7c789) = +11(2) (No 9)
Remaining cells 20(5)n89 = +11(2) (No 9)
-> 9 in n9 in r8c89
-> r7c4 = 9

4. Innies n3 = r1c7 = 9
-> Remaining cells disjoint 34(5) = {4678}
Innies n1 = r123c3 = +18(3)
-> If any two of (468) in r12c3 -> r123c3 = {468}
Otherwise if 7 in r12c3 -> r123c3 = [{67}5] or {{78}3]
-> No (129) in r123c3
-> 9 in n1 in r23c12

5! For the numbers (89) in c1, one must be in 13(3)n1 and the other in 16(3)n4
Similarly for the numbers (12) in c1, one must be in 13(3)n1 and the other in 16(3)n4
-> for the numbers (12) in n1, one must be in 13(3)n1 and the other in 14(3)n1
IOD c1 -> r7c1 = r8c2 + 4
-> [r7c1,r8c2] from [51], [62], or [73]
Given all the above there is no solution for 12(3)n7 where [r7c1,r8c2] = [62]
(Since that would put 2 in 13(3)n1 = {238})
-> [r7c1,r8c2] from [51] or [73]
-> r8c3 = 2

6. -> r8c2 from (13)
Either [r7c1,r8c2] = [51] which puts r89c1 = {47} and 16(4)n7 = [5{36}2]
Or [r7c1,r8c2] = [73] which puts r89c1 = {45} and 16(4)n7 = [7{16}2]
-> 6 in r8 in r8c89
-> 20(3)n9 = [{69}5]
-> 3 in r9 in r9c78
-> 20(5)n89 = [8{26}{13}]
-> r7c789 = {247}
Also (HS 8 in r7) 12(2)n8 = [84]
-> 21(3)n8 = [957]
Also r789c1 = [574]
-> r8c2 = 1 and r7c23 = {36}
-> 4(2)n8 = [13]

7. Innies n5 = r4c456 = +12(3)
Whatever is in r4c6 cannot be in n2 in r3c4 since that would put r3c3 = 2.
-> r4c6 = r1c5
That value is also in 21(3)n8. I.e., be from (579)
Of these only 7 is possible since r1c5 cannot be 5 or 9.
-> r1c5 = r4c6 = 7
-> 1 in n5 in r4c45
-> r4c45 = [41]
Also r123c3 = {468}
-> 14(4)r3c3 = [6341]
-> r12c3 = {48}
-> r1c6 = 6
Also 17(2)n7 = [89]
-> 16(3)n4 = {268}
-> 13(3)n1 = {139}
-> 14(3)n1 = {257}
Also r7c23 = [63]

8. IOD r1234 -> r5c2 = r4c1 + 2
-> 16(3)n4 = [2{68}] and r5c2 = 4
-> r4c23 = [95]
-> r6c2 = 3 and r56c3 = {17}

9. Also (since r1c6 = 6) 14(2)n5 = {59}
Also r9c56 = [62]
-> 8(2)n5 = [62]
-> 11(2)n5 = [83]
Also r12c4 = {12}
Also r23c5 = {59}
Also r23c6 = {48}
Also 17(3)n6 = {368}

10. Since r7c789 = {247}
-> whichever two of those are in r7c89 are also in 18(4) in n6
-> 18(4}n69 = {2457} with 5 in n6 and 4 in r67c7
-> 23(5)n69 = {12479} with (19) in n6

11! r12c3 = {48} and r23c6 are also {48}
-> r2c36 = {48}
-> For (48) in n3 - one must be in r1 and one in r3
Since r3c7 cannot be either 4 or 8 -> 9(2)r3c8 = [54] or {18}
-> (HP in r3) r3c27 = {27}
Whichever of (27) is in r3c7 must also go in r5c8 and r7c9
-> r3567c7 = {2457} with r3c7 from (27), 5 in r56c7, and 4 in r67c7.
-> 5 in n3 in c8

12! Since r3c7 from (27) the other of (27) cannot be in either of the 9(2)s in n3, so must also be in 18(4)n3 in r12c8.
4 in n3 only in r1c8 or r3c9.
But 4 cannot be in r1c8 since that puts r2c8 from (27) leaving no place for 5 in n3.
-> 9(2)r3c8 = [54]

13. -> r23c5 = [59] and r23c6 = [48]
-> r12c3 = [48] and 13(3)n1 = [391]
-> Both of (36) in n3 in r2
-> 18(4)n3 can only be [2{36}7]
-> 9(2)r1c9 = [81]
etc.

Prelims

a) R12C9 = {18/27/36/45}, no 9
b) R3C89 = {18/27/36/45}, no 9
c) R5C45 = {29/38/47/56}, no 1
d) R56C6 = {59/68}
e) R6C45 = {17/26/35}, no 4,8,9
f) R78C5 = {39/48/57}, no 1,2,6
g) R78C6 = {13}
h) R9C23 = {89}
i) 11(3) cage at R5C3 = {128/137/146/236/245}, no 9
j) 21(3) cage at R7C4 = {489/578/678}, no 1,2,3
k) 20(3) cage at R8C8 = {389/479/569/578}, no 1,2
l) 14(4) cage at R3C3 = {1238/1247/1256/1346/2345}, no 9
m) Disjoint 34(5) cage at R1C3 = {46789}

1a. Naked pair {13} in R78C6, locked for C6 and N8, clean-up: no 9 in R78C5
1b. Naked pair {89} in R9C23, locked for R9 and N7
1c. R5C45 = {29/38/47} (cannot be {56} which clashes with R56C6), no 5,6
1d. 21(3) cage at R7C4 = {489/579} (cannot be {678} which clashes with R78C5), no 6, 9 locked for C4, clean-up: no 2 in R5C5
1e. R9C56 = {26} (hidden pair in N8), locked for R9 and 20(5) cage at R8C7
1f. 45 rule on N3 1 innie R1C7 = 9
1g. R1C356 + R2C3 = {4678}, CPE no 4,6,7,8 in R1C12
1h. 4,7 in C6 only in R1234C6, CPE no 4,7 in R1C4 + R2C45 + R3C5
1i. 9 in N2 only in R23C56, locked for 36(7) cage at R1C4, no 9 in R4C6
1j. R9C56 = {26} = 8 -> R8C7 + R9C78 = 12 = {138/147/345}
1k. 2 in N9 only in R7C789, locked for R7
1l. 45 rule on N36 3 outies R7C789 = 13 contains 2 = {238/247/256}, no 1,9
1m. R7C4 = 9 (hidden single in R7) -> R89C4 = 12 = {57}/[84], no 4 in R8C4
1n. 9 in N9 only in 20(3) cage at R8C8 = {389/479/569}
1o. 3 of {389} must be in R9C9 -> no 3 in R8C89
1p. 1 in N9 only in R8C7 + R9C78 = {138/147}, no 5
1q. 8 of {138} must be in R8C7 -> no 3 in R8C7
1r. 45 rule on N1 3 innies R123C3 = 18
1s. Max R12C3 = 15 -> min R3C3 = 3
1t. 45 rule on R123 3 outies R4C456 = 12 = {138/147/246/345} (cannot be {156} which clashes with R56C6, cannot be {237} which clashes with both R5C45 and R6C45)
1u. 45 rule on R1234 3 innies R4C123 = 16
[Note that this must have a common value with R4C1 but a different combination.]
1v. 45 rule on R1234 1 outie R5C2 = 1 innie R4C1 + 2, no 8,9 in R4C1, no 1,2 in R5C2
1w. 45 rule on C1 1 innie R7C1 = R8C2 + 4 -> R7C1 = {567}, R8C2 = {123}
1x. 12(3) cage at R8C1 = {147/237/345} (cannot be {156} = [615] which clashes with R7C1 + R8C2 = [51], cannot be {246} = [624] which clashes with R7C1 + R8C2 = [62]), no 6
[My original step 1ab was incorrect so I’ve reworked from here, the next step being simpler than my original steps 1y and 1z.]
1y. 45 rule on C1 3 innies R789C1 = 16 = {367/457}, no 1,2, 7 locked for C1 and N7, clean-up: no 9 in R5C2
1z. 16(4) cage at R7C1 contains 6 for N7 = {1267/1456/2356}
1aa. R7C789 = {238/247} (cannot be {256} which clashes with 16(4) cage, ALS block), no 5,6
1ab. 20(3) cage at R8C8 = {569} (hidden triple in N9) -> R9C9 = 5, R8C89 = {69}, 6 locked for R7, clean-up: no 4 in R12C9, no 4 in R3C8, no 7 in R8C4
1ac. Variable hidden killer triple 1,3,7 in R7C123, R7C6 and R7C789 for R7, R7C6 = {13}, R7C789 contains one of 3,7 -> R7C123 cannot contain more than one of 1,3,7
1ad. 16(4) cage = {1456/2356} (cannot be {1267} = 7{16}2 which contains both of 1,7 in R7), no 7, clean-up: no 3 in R8C2
1ae. Hidden killer pair 7,8 in R7C5 and R7C789, R7C789 contains one of 7,8 -> R7C5 = {78}, R8C5 = {45}
1af. 5 in N8 only in R8C45, locked for R8

2a. R7C1 + R8C2 (step 1w) = [51/62], 16(3) cage at R4C1 = {169/259/268/358} (cannot be {349} which clashes with R89C1)
2b. Consider placement for R7C1 = {56}
R7C1 = 5 => R8C2 = 1 => 1 in N1 only in 13(3) cage at R1C1, locked for C1
or R7C1 = 6 => 16(3) cage = {259/358}, no 1
-> 1 in C1 only in 13(3) cage = {139/148}, no 2,5,6, 1 locked for C1 and N1, clean-up: no 3 in R5C2 (step 1v)
2c. 2 in N1 only in 14(3) cage at R1C2 = {239/248/257}, no 6, 2 locked for C2
2d. R8C2 = 1 -> R7C1 = 5, R78C6 = [13], R8C3 = 2 (hidden single in N7)
2e. 2,6 in C1 only in 16(3) cage at R4C1 = {268}, 6,8 locked for N4, 8 locked for C1
2f. R4C1 + 2 = R5C2 (step 1v) -> R4C1 = 2, R5C2 = 4, clean-up: no 7 in R5C45
2g. 13(3) cage = {139}, 3,9 locked for N1, 3 locked for C1
2h. 14(3) cage = {257}, 5,7 locked for C2 and N1 -> R46C2 = [93], R4C3 = 5 (cage sum), R7C2 = 6, R7C3 = 3 (cage sum), clean-up: no 5 in R6C45
2i. R7C789 (step 1aa) = {247}, 4,7 locked for N9, 7 locked for R7 -> R7C5 = 8, R8C45 = [54], R9C4 = 7, R8C7 = 8, clean-up: no 3 in R5C4
2j. R6C45 = {26} (cannot be {17} which clashes with R6C3), locked for R6 and N5 -> R56C1 = [68], R5C4 = 8 -> R5C5 = 3
2k. Naked pair {59} in R56C6, locked for C6
2l. Naked triple {147} in R4C456, locked for R4, 1 locked for 14(4) cage at R3C3, no 1 in R3C4
2m. 45 rule on R123 2 innies R3C34 = 1 outie R4C6 + 2
2n. R4C6 = {47} -> R3C34 = 6,9 = [42/63]
2o. R4C4 = 4 (hidden single in C4) -> R3C34 = [63], R4C56 = [17]
2p. Naked pair {48} in R12C3, locked for disjoint 34(5) cage -> R1C56 = [76], R9C56 = [62], R6C45 = [62], clean-up: no 2,3 in R2C9
2q. Naked pair {48} in R2C36, locked for R2, clean-up: no 1 in R1C9
2r. 18(4) cage at R5C7 = {2457} (only remaining combination), 4 locked for C7, 5 locked for N6

3a. R1C7 = 9, R12C9 = [27/36/81], R3C89 = {18/27}/[54] -> 18(4) cage at R1C8 must contain two 9(2) pairs from {18/27/36/45}
3b. Hidden killer triple 1,3,6 in R23C7, R4C7 and R9C7 for C7, R4C7 = {36}, R9C7 = {13} -> R23C7 must contain one of 1,3,6 -> R12C8 must contain one of 3,6,8
3c. R23C7 must also contain one of 2,5,7 (other values in those cells or from hidden killer triple 2,5,7 in C7) -> R12C8 must contain one of 2,4,7 -> no 5 in R12C8
3d. R1C2 = 5 (hidden single in R1)
3e. R3C89 = {18/27}/[54] (cannot be {27} which clashes with R3C2), no 2,7
3f. R3C27 = {27} (hidden pair in R3)
3g. Naked quad {2457} in R3567C7, 2,5,7 locked for C7, clean-up: no 4 in R1C8
3h. R3C89 = [54] (hidden pair in N3) -> R23C5 = [59], R23C6 = [48], R123C1 = [391]
3i. R12C9 = [81] (cannot be [27] which clashes with R3C7)

and the rest is naked singles.

3. Hidden killer pair 4,8 in r3 -> 9(2)r3c8 must have one of 4,8 = {18}/[54]

4. hidden pair {27} in r3c27 -> both from (27)

5. naked {2457} in r3567c7: all locked for c7

6. 18(4)n3 must have one of (27) for r3c7 and one of (136) for r2c7 = {1278/1467/2358/2367}
6a. however, {1467} blocked by either 9(2) in n3
6b. = {1278/2358/2367}(no 4)
6c. must have 2: locked for n3

7. r3c9 = 4 (hsingle n3) -> r3c8 = 5

easy now