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SumoCueV1=9J0=14J0+1J0=15J0+3J1+3J2=14J2+6J2=14J2+0J0+1J0=20J0+11J1+3J1=19J1+14J2+6J2+8J2=14J0+18J3+11J1+11J1=30J1+14J1+14J1=20J4+25J2+18J0=14J3=15J3+29J5+22J5=6J5+32J4=15J4+25J2+28J3+28J3=11J3+38J5+22J5=14J5+41J4+34J4+34J4=17J6+28J3=15J3+47J5+22J5=7J5+50J4+34J4=14J7+45J6+45J3=17J8+56J8+22J8=25J8+59J8+53J4+53J7=15J6=14J6+56J6+56J8=15J8+59J8+59J7=15J7=7J7+63J6+64J6+64J6+67J6+67J8+67J7+70J7+70J7+71J7

265473198
739518246
154296387
816932475
473865921
328741659
591684732
687329514
942157863

I have used some useful LOL's and clones cells (again !) , but the decisive step seems to be the hidden killer quad {6789} at n6 (step 4)d)) which leads to r6c7=6

1)a) Outies for c12 : r19c3=h7(2) at c3 : no 7,8,9
b) Innies for c1234 : r19c4=h5(2) at c4 : no 5,6,7,8,9.
c) L.O.L for n147 : {r1c4, r9c4} = {r3c3, r7c3} : r37c3=h5(2)at c3 ={14/23}
d) h7(2)c3<>{34} since this combination would block combination of h5(2)c3.
e) Killer pair {12} locked for c3 at h5(2)+h7(2)

2)a) 11(2)r5 <> {56} since {56} block combinations of 14(2)r5
b) Innies-outies for c1 : r3c2+r7c2=10+r5c1 >= 11 : r37c2<>1.
c) « 45 »-rule for n4 : r4c3+r5c3+r6c3=31 – (r3c2+r7c2) <= 20 :since the sum of 3 of {6789} is more than 20, r456c3 must contain at
least one of (34), only possible at r5c3 since r46c3=(6789)
=> r5c34=[38/47]
d) Killer pair {34} locked for c3 at r5c3+ h5(2)

3)a) Innies-outies for n7 : r7c2=1+r8c3+r9c4 >= 1+5+1=7
=> r7c2=(789)
=> r8c3<>8,9
=> r9c4<>4, r1c4<>1
b) L.O.L for n123 : {r4c1, r4c9}={r3c2, r3c8}. r4c1 cannot equal r3c2 since they share a cage, idem for r3c8 and r4c9, so
=> r4c1=r3c8 : no 1,2
=> r3c2=r4c9 : no 2
c) 14(4)n4<>{2345} blocked by r5c3=(34)
=> 14(4) contains one of {6789}
d) Killer quad {6789} locked for n4 at 14(4)+r46c3+r7c2
=> r3c2<>6,7,8,9 : r3c2=(345)=r4c9 (step 3)b) )
e) Max r4c9=5 => Min (r3c8+r3c9)=15 : r3c89=(6789)
=> r4c1=(6789)

4)a) L.O.L for n789 : {r6c1, r6c9}={r7c2, r7c8}
=> r6c1=r7c8, and r6c9=r7c2=(789)
b) Innies-outies for r789 : r6c1+r6c9=4+r7c5 <= 13
c) r6c19<>{67} which would block combinations of 15(2)r6
Since r6c1+r6c9 <=13 and r6c19<>{67}, r6c19 contain one of {12345}
=> r6c1=(12345), so r7c8=(12345)
d) 15(4)n6 contain exactly one of {6789}
=> Hidden killer quad {6789} locked for n6 at 15(4)+r3c8+r5c7+r6c7
=> r6c7=6
The puzzle is now cracked

5)a) 7(2)r6=[16], 15(2)r6={78} locked for r6
b) r6c9=9 => r7c2=9 (step 4)a))
c) Naked pair {78} locked for c4 and n5 at r56c4
=> 15(2)r4=[69]
d) 14(2)n3={68} locked for n3 and c9
=> r3c9=7
e) h7(2)c3={25}=r19c3 locked for c3
f) h5(2)c3={14}=r37c3 locked for c3
g) 11(2)n4=[38], 15(2)r6=[87], 14(2)r5=[59]
i) Innie for n4 : r3c2=5=r4c9
j) r3c8=8=r4c1
=> cage sum r3c1=1
k) Hidden single for c3 : r2c3=9 => r8c3=7
l) Innie for n1 : r1c4=4 => r9c4=1
m) 15(2)n7={69} locked for n7 and c1
n) 9(2)n1={27} locked for n1 and c1
o) r7c8+r7c9=5 : r7c8<>5 => r6c1<>5
p) Last combo : r67c1=[35] => r7c8=3

The rest is straightforward

Prelims
i. 9(2)r1c1: no 9
ii. 14(2)r1c9 & r5c6 = {59/68}
iii. 20(3)r3c8: no 1,2
iv. 14(4)r4c2: no 9
v. 15(2)r4c3 & r6c3 & r8c1 = {69/78}
vi. 6(2)r4c6 = {15/24}
vii. 11(2)r5c3: no 1
viii. 7(2)r6c6 & r8c9: no 7,8,9

1. "45" on c123: 2 innies r19c4 = 5 = {14/23}

2. "45" on Nr1c1: 1 outie r3c2 + 8 = 2 innies r1c4 + r2c3
2a. max. 2 innies = 13 -> max r3c2 = 5

3. LOL on nonets lined up with r123: 2 outies r4c19 = 2 innies r3c28
3a. but r3c2 cannot equal r4c1 since they have a common cage -> r3c2 = r4c9 = (345)
3b. same deal with r3c8 = r4c1 (no 1,2)

4. 20(3)r3c8: min. r3c89 = 15 (no 3,4,5)
4a. no 3,4,5 in r4c1 (step 3b)

5. We know that r3c2=r4c9 and r4c1=r3c8 (step 3a.b) -> the only cells left in the 14(3)r3c1 and 20(3)r3c8 to make up the difference in the cage totals (6) are r3c1 and r3c9 -> r3c1 + 6 = r3c9
5a. ->r3c1 = (123), r3c9 = (789)

6. "45" on r123: 2 outies r4c19 - 4 = 1 innie r3c5
6a. min. 2 outies = 9 -> min. r3c5 = 5

7. Combining steps 3b and 6.: -> r3c8 + r4c9 - 4 = r3c5
7a. -> no 4 in r4c9 (Clone IOU)!
7b. -> no 4 in r3c2 (step 3a)

8. "45" on r12
8a. four outies r3c3467 = 15 and must have 4 for r3 = {1248/1347/2346}(no 5,9)
8b. note: has two of 1,2,3

9. Killer triple 1,2,3 in r3c1 & h15(4)r3 (step 8b): locked for r3

The puzzle is cracked.
10. r3c2 = 5, r4c9 = 5 (step 3a)

11. 14(2)r1c9 = {68}: both locked for c9 & Nr1c6
11a. r3c1, no 2 (step 5)

12. r3c89 = 15 = [69/87]: r3c8 = (68)
12a. ->r4c1 = (68) (step 3b)

13. "45" on Nr1c1: 2 remaining innies r1c4 + r2c3 = 13 = [49]
13a. no 5 in 9(2)r1c1

14. Killer pair 6,8 in r4c1 & 15(2)r8c1: both locked for c1

15. 9(2)r1c1 = {27}: both locked for Nr1c1 & c1

16. 15(2)r8c1 = {69}: both locked for c1 & Nr6c1

17. r4c1 = 8, r3c1 = 1 (cage sum)

18. r3c8 = 8 (step 3b), r3c9 = 7 (cage sum)

19. r3c5 = 9 (hidden single r3)

Enjoy the rest!

Thanks Ed for your comment and corrections; I've added some more I found while going through my walkthrough again.

Prelims

a) R12C1 = {18/27/36/45}, no 9
b) R12C9 = {59/68}
c) R4C34 = {69/78}
d) R4C67 = {15/24}
e) R5C34 = {29/38/47/56}, no 1
f) R5C67 = {59/68}
g) R6C34 = {69/78}
h) R6C67 = {16/25/34}, no 7,8,9
i) R89C1 = {69/78}
j) R89C9 = {16/25/34}, no 7,8,9
k) 20(3) cage at R3C8 = {389/479/569/578}, no 1,2
l) 14(4) cage at R4C2 = {1238/1247/1256/1346/2345}, no 9

1. 45 rule on C12 2 outies R19C3 = 7 = {16/25/34}, no 7,8,9

2. 45 rule on C1234 2 innies R19C4 = 5 = {14/23}

3. 45 rule on C6789 2 innies R19C6 = 10 = {19/28/37/46}, no 5

4. 45 rule on C89 2 outies R19C7 = 9 = {18/27/36/45}, no 9

5. 45 rule on C1 2 outies R37C2 = 1 innie R5C1 + 10
5a. Max R37C2 = 17 -> max R5C1 = 7
5b. Min R37C2 = 11, no 1 in R37C2

6. 45 rule on C9 2 outies R37C8 = 1 innie R5C9 + 10
6a. Max R37C8 = 17 -> max R5C9 = 7
6b. Min R37C8 = 11, no 1 in R7C8

7. Law of Leftovers (LoL) for C123 two outies R19C4 must exactly equal two innies R37C3, R19C4 = {14/23} -> R37C3 = {14/23}
7a. R19C3 = 7 = {16/25} (cannot be {34} which clashes with R37C3), no 3,4
7b. Killer pair 1,2 in R19C3 and R37C3, locked for C3, clean-up: no 9 in R5C4
7c. R5C34 = {38/47}/[92] (cannot be {56} which clashes with R5C67), no 5,6 in R5C34

8. LoL for C789 two outies R19C6 must exactly equal two innies R37C7, no 5 in R19C6 -> no 5 in R37C7

9. LoL for R123 two outies R4C19 must exactly equal two innies R3C28 but R3C2 + R4C1 are in the same cage and R3C8 + R4C9 are in the same cage -> R3C2 = R4C9 and R3C8 = R4C1, no 1,2 in R3C2 + R4C1
[There’s the obvious Min R3C2 + R4C1 = 7 -> max R3C1 = 7, however one can get more out of this LoL …]
9a. R3C2 = R4C9 and R3C8 = R4C1 -> R3C9 = R3C1 + 6 (difference between 14(3) and 20(3) cages) -> R3C1 = {123}, R3C9 = {789}
9b. 20(3) cage at R3C8 = {389/479/569/578}
9c. 3 of {389} must be in R4C9 (R34C9 cannot be {89} which clashes with R12C9) -> no 3 in R3C8, clean-up: no 3 in R4C1 (LoL)
9d. 6 of {569} must in R3C8 (R34C9 cannot be [96] which clashes with R12C9) -> no 6 in R4C9, clean-up: no 6 in R3C2 (LoL)

10. 45 rule on NR1C1 2 innies R1C4 + R2C3 = 1 outie R3C2 + 8
10a. Max R1C4 + R2C3 = 13 -> max R3C2 = 5, clean-up: max R4C9 = 5 (LoL, step 9)
10b. Min R3C2 = 3 -> min R1C4 + R2C3 = 11 -> min R1C4 = 2, min R2C3 = 7, clean-up: no 4 in R9C4 (step 2)
10c. Max R3C12 = 8 -> min R4C1 = 6, clean-up: min R3C8 = 6 (LoL, step 9)

11. 45 rule on NR1C6 1 outie R3C8 = 2 innies R1C6 + R2C7 + 3
11a. Max R1C6 + R2C7 = 6, no 6,7,8,9 in R1C6 + R2C7, clean-up: no 1,2,3,4 in R9C6 (step 3)

12. LoL for R789 two outies R6C19 must exactly equal two innies R7C28 but R6C1 + R7C2 are in the same cage and R6C9 + R7C8 are in the same cage -> R6C1 = R7C2 and R6C9 = R7C8, no 1 in R6C19
12a. R6C1 = R7C8 and R6C9 = R7C2 -> R7C1 = R7C9 + 3 (difference between 17(3) and 14(3) cages), no 1,2,3 in R7C1, no 7,8,9 in R7C9

13. R37C2 = 1 innie R5C1 + 10 (step 5)
13a. Max R37C2 = 14 -> max R5C1 = 4
13b. Min R37C2 = 11 -> min R7C2 = 6, clean-up: min R6C9 = 6 (LoL, step 12)
13c. Min R7C12 = 10 -> max R6C1 = 7, clean-up: max R7C8 = 7 (LoL, step 12)
13d. 5 in NR3C2 only in R3456C2, locked for C2

14. 45 rule on NR6C1 1 outie R7C2 = 2 innies R8C3 + R9C4 + 1
14a. Max R8C3 + R9C4 = 8, no 8,9 in R8C3

15. 45 rule on NR6C9 2 innies R8C7 + R9C6 = 1 outie R7C8 + 9
15a. Min R7C8 = 2 -> min R8C7 + R9C6 = 11, no 1 in R8C7

16. 1,2 in NR3C2 only in 14(4) cage at R4C2 = {1238/1247/1256}
16a. Killer quad 6,7,8,9 in 14(4) cage, R46C3 and R7C2, locked for NR3C2, clean-up: no 2,3,4 in R5C4
16b. Killer pair {34} in R37C3 and R5C3, locked for C3

17. R7C2 = R8C3 + R9C4 + 1 (step 14)
17a. Min R8C3 + R9C4 = 6 -> min R7C2 = 7, clean-up: min R6C9 = 7 (LoL, step 12)
17b. Min R6C1 + R7C2 = 9 -> max R7C1 = 8, clean-up: max R7C9 = 5 (step 12a)
17c. Min R7C12 = 11 -> max R6C1 = 6, clean-up: max R7C8 = 6 (LoL, step 12)
17d. R37C8 = R5C9 + 10 (step 6)
17e. Max R37C8 = 15 -> max R5C9 = 5

18. 45 rule on R12 4 outies R3C3467 = 15 = {1248/1257/1347/1356/2346} (cannot be {1239} which clashes with R3C1), no 9
18a. Killer quint 1,2,3,4,5 in R3C12 and R3C3467, locked for R3

19. Hidden killer quad 1,2,3,4 in R12C1, 14(3) cage at R1C2, R1C4 and R3C1 for NR1C1, R12C1 contains one of 1,2,3,4, R1C4 = {234}, R3C1 = {123} -> 14(3) cage must contain one of 1,2,3,4 = {158/167/257/356} (other combinations contain two of 1,2,3,4), no 4,9
19a. 5 of {257} must be in R1C3 -> no 2 in R1C3, clean-up: no 5 in R9C3 (step 1)
19b. 9 in NR1 only in R2C3 + R4C1, CPE no 9 in R4C3, clean-up: no 6 in R4C4

20. 14(3) cage at R8C2 = {149/239/248} (cannot be {167} which clashes with R89C1, cannot be {347} because R9C3 only contains 1,2,6), no 6,7, clean-up: no 1 in R1C3 (step 1)
20a. R9C3 = {12} -> no 1,2 in R89C2
20b. Killer pair 8,9 in R89C1 and 14(3) cage, locked for NR6C1, clean-up: no 5 in R7C9 (step 12a)
20c. 1 in NR6C1 only in R9C34, locked for R9, clean-up: no 8 in R1C7 (step 4), no 6 in R8C9

21. 9 in C2 only in R789C2 -> 17(3) cage at R6C1 and 14(3) cage at R8C2 form a combined cage without any repeated candidates (but note that R7C2 doesn’t “see” R89C1)
21a. 14(3) cage at R8C2 (step 20) = {149/239/248}
21b. 17(3) cage at R6C1 = {359/368/458/467} (cannot be {269} which clashes with 14(3) cage, cannot be {278} because {278} + 14(3) cage = {149} clashes with R89C1), no 2, clean-up: no 2 in R7C8 (LoL, step 12)
[Ed commented that an alternative way to eliminate {278} is 7 in R7C1 clashes with R89C1 = {78} since R7C2 repeats in R89C1.]
21c. 7 of {467} must be in R7C2 -> no 7 in R7C1, clean-up: no 4 in R7C9 (step 12a)
21d. R9C34 = {12} (hidden pair in NR6C1), locked for R9, clean-up: no 2 in R1C4 (step 2), no 7 in R1C7 (step 4), no 5 in R8C9
21e. 3 in NR6C1 only in R6C1 + R89C2, CPE no 3 in R6C2
[Can also eliminate {368}from the 17(3) cage, using LoL interaction with 14(3) cage at R6C9, but I’ll leave that for now in the hope that there’s a simpler way and because it doesn’t eliminate any candidates.]

22. 17(3) cage at R6C1 (step 21b) = {359/368/458/467}
22a. 5 in C1 only in R12C1 = {45} or in 17(3) cage = {359/458} -> 17(3) cage = {359/368/458} (cannot be {467}, locking-out cages), no 7, clean-up: no 7 in R6C9 (LoL, step 12)
22b. Killer pair 8,9 in R6C34 and R6C9, locked for R6
22c. Killer pair 8,9 in R7C2 and 14(3) cage at R8C2, locked for C2

23. R3C9 = 7 (hidden single in C9), placed for NR1C6, R3C1 = 1 (step 9a), placed for NR1C1, clean-up: no 8 in R12C1, no 7 in R4C1 (LoL, step 9)
23a. 20(3) cage at R3C8 = {479/578}, no 3,6
23b. 14(3) cage at R3C1 = {149/158}, no 3,6
23c. R2C3 + R4C1 = {89} (hidden pair in NR1C1)
23d. Killer pair 4,5 in R4C67 and R4C9, locked for R4

24. 14(4) cage at R4C2 (step 16) = {1247/1256}, no 3
24a. R5C3 = 3 (hidden single in NR3C2), R5C4 = 8, placed for NR4C4, clean-up: no 7 in R4C3, no 6 in R5C67, no 7 in R6C3
24b. Naked pair {59} in R5C67, locked for R5
[Cracked. The rest is fairly straightforward.]

25. LoL (step 7), no 3 in R37C3 -> no 3 in R1C4
25a. R1C4 = 4, placed for NR1C1, R9C4 = 1 (step 2), clean-up: no 5 in R12C1, no 6 in R9C6 (step 3), no 5 in R9C7 (step 4)
25b. R9C3 = 2, R3C3 = 4, placed for NR1C5, R7C3 = 1, placed for NR7C3, R3C2 = 5, R4C1 = 8 (cage sum), placed for NR1C1, R2C3 = 9, R4C3 = 6, R6C3 = 8, both placed for NR3C2, R4C4 = 9, R6C4 = 7, both placed for NR4C4, R5C6 = 5, placed for NR4C4, R5C7 = 9, placed for NR3C8, R3C8 = 8, R4C9 = 5 (cage sum), placed for NR1C6, R6C9 = 9, placed for NR6C9, R7C2 = 9, R1C3 = 5, R8C3 = 7, placed for NR6C1, clean-up: no 1 in R1C6 (step 3), no 1 in R4C67, no 2 in R6C7, no 2 in R8C9, no 4 in R9C7 (step 4)
25c. LoL (step 8), R9C6 = {78} -> R7C7 = {78} and R1C6 = {23} -> R3C7 = {23}

26. Naked pair {69} in R89C1, locked for C1, clean-up: no 3 in R12C1
26a. Naked pair {27} in R12C1, locked for C1 and NR1C1 -> R5C1 = 4, R7C1 = 5 R6C1 = 3, placed for NR6C1, clean-up: no 4 in R6C67

27. R6C9 = 9 -> R7C89 = 5 = {23} -> R7C8 = 3, R7C9 = 2, placed for NR6C9, R7C4 = 6, R8C4 = 3 (cage sum), both placed for NR7C3, R3C4 = 2, placed for NR1C5, R2C4 = 5, R5C9 = 1, placed for NR3C8, R8C9 = 4, placed for NR6C9, R9C9 = 3, clean-up: no 6 in R6C6

28. R3C7 = 3, placed for NR1C5
28a. R1C6 = 3 (hidden single in NR1C6), R9C6 = 7 (step 3), R1C2 = 6, R12C9 = [86]

29. 14(3) cage at R1C7 = {149} (only remaining combination) -> R2C8 = 4, R1C8 = 9, R1C7 = 1, R9C7 = 8 (step 4)

30. R23C7 = [23] = 5 -> R23C6 = 14 = [86]

and the rest is naked singles, without using the nonets.