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eliminations around the outer circuit, a design feature of this puzzle.

I'll rate my walkthrough for the Minimum Cages: 14 Cages Killer at Easy 1.5 because I used a short forcing chain. I'm not sure how I'd rate steps 6 and 7 but not more than in the 1.25 range, possibly in the 1.0 range, so my rating is based on step 10.

Prelims

a) 28(7) cage at R1C1 = {1234567}
b) 39(6) cage at R1C5 = {456789}
c) 42(7) cage at R1C6 = {3456789}
d) 22(6) cage at R2C6 = {123457}
e) 29(7) cage at R4C4 = {1234568}
f) 21(6) cage at R5C1 = {123456}
g) 41(7) cage at R6C1 = {2456789}
h) 38(6) cage at R6C8 = {356789}
i) 29(7) cage at R6C9 = {1234568}

1. R6C46 = {79} (hidden killer in N5), locked for R6 and 21(4) cage at R5C7
1a. R6C46 = 16 -> R5C7 + R7C5 = 5 = {14/23}

2. 1,2 in R1 only in R1C1234, locked for 28(7) cage at R1C1, no 1,2 in R234C1

3. 8,9 in C1 only in R789C1, locked for 41(7) cage at R6C1, no 8,9 in R9C234, 9 also locked for N7

4. 9 in C9 only in R1234C9, locked for 42(7) cage at R1C6, no 9 in R1C678

5. 1 in R9 only in R9C6789, locked for 29(7) cage at R6C9, no 1 in R678C9

6. 9 in R1234 only in 39(6) cage at R1C5, 42(7) cage at R1C6, 35(6) cage at R2C5 and 18(3) cage at R3C6 -> all four of these cages must contain 9 in R1234, no 9 in R5C3
6a. 18(3) cage at R3C6 = {189/279/369/459}

7. Using the same logic, 23(5) cage at R5C2, 19(4) cage at R5C8, 41(7) cage at R6C1 and 38(6) cage at R6C8 must each contain 9 for R5789
7a. 19(4) cage cannot contain both of 8,9 -> no 8 in 19(4) cage at R5C8
7b. 9 in 19(4) cage at R5C8 only in R5C8 + R7C67, CPE no 9 in R7C8
7c. 9 in 38(6) cage at R6C8 only in R8C678 + R9C5, CPE no 9 in R8C5

8. 9 in R89 only in 41(7) cage at R6C1 and 38(6) cage at R6C8 -> both of these cages must contain 9 in R89, no 9 in R7C1

9. 45 rule on R1234 2 outies R5C39 = 3 innies R4C456 + 4
9a. Min R4C456 = 6 -> min R5C39 = 10, no 1,2 in R5C3, no 1 in R5C9
9b. R9C9 = 1 (hidden single in C9)

10. 18(3) cage at R3C6 and 19(4) cage both contain 9 in C678 -> either 21(4) cage at R5C7 or 38(6) cage at R6C8 must contain the third 9 in C678
10a. Consider the placements for 9 in 21(4) cage at R5C7 and 38(6) cage at R6C8
R6C6 = 9 => no 9 in R8C678 => R9C5 = 9
or R6C4 = 9
-> 9 must be in R6C4 or R9C5, CPE no 9 in R7C4

11. 23(5) cage at R5C2 must contain 9 (step 7) -> R5C2 = 9
11a. 8 in C2 only in R234C2, locked for 39(6) cage at R1C5, no 8 in R1C5 + R2C34
11b. 9 in 39(6) cage at R1C5 only in R1C5 + R2C34, CPE no 9 in R2C5
11c. 8 in R1 only in R1C6789, locked for 42(7) cage at R1C6, no 8 in R234C9

12. 9 in 35(6) cage at R2C5 only in R3C345, locked for R3

13. 18(3) cage at R3C6 (step 6a) = {189/279/369/459} -> R4C7 = 9
13a. R3C67 = {18/27/36/45}

14. R7C6 = 9 (hidden single in R7), R6C6 = 7, R6C4 = 9, clean-up: no 2 in R3C7 (step 13a)
14a. R8C8 = 9 (hidden single in C8)
14b. R9C1 = 9 (hidden single in R9)

15. 8 in N6 only in R6C89, locked for R6
15a. 8 in N69 only in 38(6) cage at R6C8 and 29(7) cage at R6C9, locked for those two cages, no 8 in R8C6 + R9C56

16. 8 in 38(6) cage at R6C8 only in R67C8 + R8C7, CPE no 8 in R9C8
16a. R9C7 = 8 (hidden single in R9), clean-up: no 1 in R3C6 (step 13a)
16b. R6C8 = 8 (hidden single in R6)
16c. R1C9 = 8 (hidden single in C9)

17. R1C5 = 9 (hidden single in R1)
17a. R2C9 = 9 (hidden single in R2)
17c. R3C3 = 9 (hidden single in R3)

18. 8 in N8 only in R7C4 + R8C5, locked for 23(5) cage at R5C2, no 8 in R7C3
18a. 23(5) cage at R5C2 contains both of 8,9 = {12389} (only remaining combination), no 4,5,6,7

19. 7 in N8 only in R9C45, locked for R9
19a. 7 in N7 only in R78C1, locked for C1 and 41(7) cage at R6C1, no 7 in R9C4
19b. 7 in 28(7) cage at R1C1 only in R1C2345, locked for R1
19c. 7 in 42(7) cage at R1C6 only in R34C9, locked for C9, clean-up: no 3,4,5 in R5C3 (min R5C39 = 10, step 9a)
19d. 7 in C2 only in R1234C2, CPE no 7 in R2C3

20. R9C5 = 7 (hidden single in R9)
20a. Naked triple {356} in 38(6) cage at R6C8, CPE no 3,5,6 in R8C9
20b. R7C7 = 7 (hidden single in N9), clean-up: no 2 in R3C6 (step 13a)
20c. R5C3 = 7 (hidden single in R5)
20d. R8C1 = 7 (hidden single in R8)
20e. R7C1 = 8 (hidden single in C1)
20f. R8C5 = 8 (hidden single in R8)

21. R7C67 = [97] = 16 -> R5C8 + R6C7 = 3 = {12}, locked for N6, clean-up: no 3,4 in R7C5 (step 1a)

22. 2 in C9 only in R78C9, locked for 29(7) cage at R6C9, no 2 in R9C68
22a. 3 in R9 only in R9C68, locked for 29(7) cage at R6C9, no 3 in R67C9
22b. 2 in R9 only in R9C234, locked for 41(7) cage at R6C1, no 2 in R6C1

23. R15C1 = {12} (hidden pair in C1)
23a. Naked pair {12} in R5C18, locked for R5

24. 3 in C1 only in R234C1, locked for 28(7) cage at R1C1, no 3 in R1C234
24a. 3 in N1 only in R23C1, locked for C1
24b. 3 in R1 only in R1C678, locked for 42(7) cage at R1C6, no 3 in R34C9

25. R5C9 = 3 (hidden single in C9), R5C7 = 4, R7C5 = 1 (step 1a), clean-up: no 5 in R3C6 (step 13a)
25a. Naked pair {23} in R7C67, locked for R7 and 23(5) cage at R5C2 -> R6C3 = 1, R5C1 = 2, R1C1 = 1, R5C8 = 1, R6C7 = 2

26. R5C39 = R4C456 + 4 (step 9)
26a. R5C39 = [73] = 10 -> R4C456 = 6 = {123}, locked for R4 and N5
26b. R6C5 = 4 (hidden single in N5)

27. R6C2 = 3 (hidden single in R6)
27a. R8C2 = 1 (hidden single in C2)
27b. R7C3 = 3 (hidden single in C3), R7C4 = 2

28. R2C2 = 8 (hidden single in R2)
28a. R8C3 = 8 (hidden single in C3)
28b. R5C4 = 8 (hidden single in C4)
28c. R3C6 = 8 (hidden single in C6), R3C7 = 1 (step 13a)

29. R2C7 = 5, R4C8 = 7
29a. Naked pair {24} in R23C8, locked for N3 and 22(6) cage at R2C6) -> R2C6 = 1
29b. Naked pair {36} in R1C78, locked for R1 and 42(7) cage at R1C6 -> R34C9 = [75], R6C9 = 6, R6C1 = 5

30. R4C6 = 2 (hidden single in C6), R4C45 = [13]

31. R8C3 = 5 (hidden single in C3)

and the rest is naked singles.

1 7 2 5 9 4 3 6 8
3 8 6 7 2 1 5 4 9
4 5 9 3 6 8 1 2 7
6 4 8 1 3 2 9 7 5
2 9 7 8 5 6 4 1 3
5 3 1 9 4 7 2 8 6
8 6 3 2 1 9 7 5 4
7 1 5 4 8 3 6 9 2
9 2 4 6 7 5 8 3 1

Treat it as a Jigsaw puzzle! And use Jigsaw techniques. Especially the Law of Leftovers.

Firstly - a couple of abbreviations:
Outer ring cage on each corner nonet - called OC<Nonet> - i.e., OC1, OC3, OC7, OC9
Inner ring cage on each corner nonet - called IC<nonet> - i.e., IC1, IC3, IC7, IC9


1. Initial Observations

a) Every OC contains (456)
-> No corner cell can contain any of (456)
Also -> no cell at the middle of each edge can contain any of (456).

I.e., None of r19c159 and r5c19 can contain 456.

b) Any values common to both IC and OC in a corner nonet must be in the outies for the IC or OC from that nonet.

c) Innies n5 -> r6c46 = {79}


2. 456

OC1, IC1, OC3, IC3 all contain (45)
-> (45) in r4c1289

OC3, IC3, OC9, IC9 all contain 5
-> 5 in r1289c6

OC9, IC9, OC7, IC7 all contain (56)
-> (56) in r6c1289

OC7, IC7, OC1, IC1 all contain (456)
-> (456) in r1289c4

Given the above:
-> HS 5 in n5 -> r5c5 = 5

-> OC1, IC1, OC7, OC7 between them contain (456) in c1234
-> (Since r5c5 = 5) No 5 in 35/6@r2c5
-> 35/6@r2c5 must contain a 6 in r23c5
-> 6 in r89c4 and also 6 in r4c12
-> HS 6 in n5 -> r5c6 = 6
-> HS 4 in n5 -> r6c5 = 4

-> No 4 in 35/6@r2c5
-> 35/6@r2c5 = {236789}

Also -> No (456) in 23/5@r5c2
-> 23/5@r5c2 = {12389}


3. Remaining cages

All 3's accounted for (in OC1, OC3, 35/6@r2c5, IC3, 29/7@r4c4, IC7, 23/5@r5c2, IC9, OC9)
-> 21/4@r5c7 cannot contain a 3
-> r5c7 = 4, r7c5 = 1

1 in seven cages OC1, IC3, 29/7@r4c4, IC7, 23/5@r5c2, 21/4@r5c7, OC7
and NOT in five cages IC1, OC3, 35/6@r2c5, OC7, IC9

-> 1 in only other cages which are 18/3@r3c6 and 19/4@r5c8
-> 18/3@r3c6 = {189}

-> All cages contents known except 19/4@r5c8. Easy to figure out it must be {1279}


Now all cage contents are known it becomes much easier to solve.