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As well as the cages there are pale green, green, pink and grey groups

Prelims

a) R1C34 = {49/58/67}, no 1,2,3
b) R23C8 = {13}
c) R34C3 = {15/24}
d) R45C4 = {29/38/47/56}, no 1
e) R56C2 = {69/78}
f) R67C1 = {15/24}
g) R67C9 = {69/78}
h) R9C12 = {39/48/57}, no 1,2,6
i) R9C89 = {15/24}
j) 19(3) cage at R4C1 = {289/379/469/478/568}, no1
k) 10(3) cage at R4C8 = {127/136/145/235}, no 8,9

1. Naked pair {13} in R23C8, locked for C8 and N3, clean-up: no 5 in R9C9

2. 45 rule on N4 4 innies R456C3 + R6C1 = 11 = {1235}, locked for N4, clean-up: no 2 in R3C3, no 2 in R7C1
2a. 3 in N4 only in R56C3, locked for C3 and green group, clean-up: no 8 in R4C4
2b. R6C3 = 3 (hidden single in R6)
2c. 3 in N5 only in R4C456 + R5C5, locked for pale green group, no 3 in R12C5 + R3C456

3. 45 rule on R9 2 innies R9C46 = 12 = {39/48/57}, no 1,2,6

4. 6 in R9 only in 15(3) disjoint cage at R9C3 = {168/267} (cannot be {456} which clashes with R9C89), no 3,4,5,9

5. 3 in R7 only in R7C24567, locked for pink group, no 3 in R8C7
5a. 3 in R8 only in R8C124569, locked for grey group, no 3 in R9C46
5b. 3 in R9 only in R9C12 = {39}, locked for R9 and N7
5c. 9 in grey group only in R8C45689, locked for R8
5d. 9 in pink group only in R7C45678, locked for R7, clean-up: no 6 in R6C9
5e. 9 in C3 only in R12C3, locked for N1

6. R34C3 = {15}/[42], R5C3 = {125} -> combined half cage R345C3 = {15}2/[42]1}/[42]5, 2 locked for C3 and N4, clean-up: no 4 in R7C1
6a. Naked pair {15} in R67C1, locked for C1

7. R7C1 “sees” all cells of the pink group except for R8C7 -> R7C1 = R8C7 = {15}
7a. R9C89 = {24} (cannot be [51] which clashes with R8C7), locked for R9 and N9, clean-up: no 8 in R9C46 (step 3)
7b. Naked pair {57} in R9C46, locked for R9, N8 and grey group
7c. R8C37 = [75] (hidden pair in R8), 7 placed for pink group -> R7C1 = 5, R6C1 = 1
7d. Naked pair {25} in R45C3, locked for C3
7e. 1 in green group only in R5C67, locked for R5
7f. Clean-up: no 6,8 in R1C4

8. R7C9 = 7 (hidden single in R7) -> R6C9 = 8, clean-up: no 7 in R5C2
8a. 8 in green group only in R5C46, locked for R5 and N5, clean-up: no 7 in R6C2
8b. 8 in pale green group only in R12C5 + R3C456, locked for N2
8c. Naked pair {69} in R56C2, locked for C2 and N4 -> R9C12 = [93]

9. 45 rule on N6 2 remaining innies R56C7 = 1 outie R3C7 + 8
9a. Max R56C7 = {79} = 16 -> max R3C7 = 8
9a. R56C7 cannot total 12,14 -> no 4,6 in R3C7
9b. R3C7 = {278} -> R56C7 = 10,15,16 = [19]/{46/69/79}, no 2 in R56C7

10. 15(3) cage at R7C4 = {249/348} (cannot be {168} which clashes with R9C5), no 1,6, 4 locked for R7 and N8
10a. 4 in C3 only in R123C3, locked for N1

11. Caged X-Wing for 6,9 in R56C8 and green group, no other 6,9 in R56

12. 10(3) cage at R4C8 = {127/136/145/235}
12a. 1 of {136} must be in R4C9 -> no 6 in R4C9

13. 19(4) cage at R3C7 = {2458/2467/3457} (cannot be {1279/1369/1378/1459/1468/1567/2359/2368} because 1,3,6,9 only in R4C7), no 1,9, 4 locked for N6
13a. 3,6 of {2467/3457} must be in R4C7 -> no 7 in R4C7
13b. 9 in N6 only in R56C7, locked for C7 and green group, clean-up: no 2 in R4C4
13c. 9 in N5 only in R4C456, locked for pale green group, no 9 in R12C5 + R3C456
13d. 4 in green group only in R5C46 + R6C456, locked for N5, clean-up: no 7 in R5C4
13e. 4 in pale green group only in R12C5 + R3C456, locked for N2, clean-up: no 9 in R1C3
13f. R2C3 = 9 (hidden single in C3)
13g. R3C9 = 9 (hidden single in R3)
[It looks like R56C7 = {69} can be eliminated because R56C27 would be {69}{69}, which would eliminate 19(4) cage = {3457}, but I refuse to use Unique Rectangle because it doesn’t solve the whole puzzle; I prefer to show uniqueness by eliminating all other possibilities.]

14. 25(4) cage at R1C6 = {1789/2689/3589/3679} (cannot be {4579} which clashes with R9C6, cannot be {4678} because 4,8 only in R3C6), no 4, 9 locked for C6
14a. 8 of {1789/2689/3589} must be in R3C6, 6,7 of {3679} must be in R3C6 -> R3C6 = {678}

[It took me some time to see a way forward, then I found …]
15. 10(3) cage at R4C8 = {127/136/235}
15a. Consider combinations for R45C4 = [38/56/65/74/92]
R45C4 = [38] => R5C9 = 3 (hidden single in N6) => R4C89 = 7 = [61] (cannot be {25} which clashes with R4C3)
or R45C4 = [56/65], grouped X-wing for 5 in R45C3 and R45C4, no other 5 in R45 => 10(3) cage at R4C8 = {127/136}
or R45C4 = [74] => R5C1 = 7 => R5C8 = {25} => 10(3) cage = {127/136} (cannot be {235} which clashes with R5C8)
or R45C4 = [92] => R5C3 = 5 => R5C9 = 3 => R4C89 = 7 = [61] (cannot be {25} which clashes with R4C3)
-> 10(3) cage = {127/136} -> R4C9 = 1
15b. 6,7 of {127/136} only in R4C8 -> R4C8 = {67}
15c. Naked triple {679} in R4C8 and R56C7, locked for N6
15d. Naked triple {245} in R56C8 and R9C8, locked for C8
[Cracked. The rest is fairly straightforward.]

16. R5C6 = 1 (hidden single in R5)
16a. 1 in pale green group only in R12C5 + R3C45, locked for N2
16b. R5C4 = 8 (hidden single in R5) -> R4C4 = 3
16c. R5C9 = 3 (hidden single in N6), R4C9 = 1 -> R4C8 = 6 (cage sum)
16d. R4C7 + R5C56 = {245} = 11 -> R3C7 = 8 (cage sum) -> R1C8 = 7, clean-up: no 6 in R1C3

17. R79C7 = [31] (hidden pair in N9)
17a. 15(3) cage at R7C4 (step 10) = {249} (only remaining combination), locked for R7 and N8 -> R7C8 = 8, R7C23 = [16], R9C35 = [86], R1C3 = 4 -> R1C4 = 9, R3C3 = 1 -> R4C3 = 5

18. R5C3 = 2, placed for green group, R56C7 = {79}, locked for green group
18a. Naked triple {456} in R6C456, locked for R6 and N5 -> R5C5 = 7, placed for pale green group

19. R3C6 = 6, 3 in N2 only in 25(4) cage at R1C6 (step 14) = {3679} (only remaining combination) -> R4C6 = 9, R12C6 = [37], R4C5 = 2, placed for pale green group

and the rest is naked singles, without using the coloured groups.

I'll rate my walkthrough at 1.5. My key steps, in the order I used them, were a combined half-cage, a simple "clone" and a moderate forcing chain for the final breakthrough.

Colours I use I abbreviate as: LG (Lighter Green) and DG (Darker Green).

1. Innies r9 -> r9c46 = +12
Whatever those values are must go in r8 in r8c37 ...
... and in r7 in r7c19.
-> 6(2)c1 from [15] or [24]

2. Innies n4 -> r456c3 + r6c1 = +11 = {1235}
-> 3 in r56c3 and 5 in r456c3
-> 6(2)c3 from [42] or [15].
-> Both of those -> r6c1 = 1
-> r7c1 = 5
-> r67c9 = [87]
-> r8c37 = [75]
-> r9c46 = {57}
-> 6(2)r9 = {24}
-> 12(2)r9 = {39}
-> 15(3)r9 = {168}

3. 8 in r6c9 -> 8 in DG in r5c46 -> 8 in n4 in r4c12
-> 15(2)n4 = {69}
-> 19(3)n4 = {478}
Also 12(2)r9 = [93]

4. 4(3)n3 = {13}
-> HS 3 in r6 -> r6c3 = 3
-> r5c3 from (25)
-> whichever r5c3 is (2 or 5), it can only go in r6 in r6c8
-> 10(3)n6 cannot be {235} -> must contain a 1
-> HS 1 in DG -> r5c6 = 1
-> HS 8 in DG -> r5c4 = 8
-> r4c4 = 3
-> HS 3 in r5 -> r5c9 = 3
-> 10(3)n6 = [613]
Also HS 3 in c7 -> r7c7 = 3
Also HS 1 in c7 -> r9c7 = 1
Also HS 6 in n9 -> r8c9 = 6
-> r78c8 = {89}

5. 6 in DG only in r6c456
-> 15(2)n4 = [69]
-> HS 9 in DG -> r5c7 = 9
Note that 19(4)r3c7 cannot contain any of (139) and must contain 5 in r56c8
-> 19(4) = {2458} with 8 in r3c7 and 5 in r56c8.
-> NS r6c7 = 7

6. 15(3)r7 cannot be same as 15(3)r9 (i.e., not {168})
Only possibility left is 15(3)r7 = {249}
-> r78c8 = [89]
-> r7c23 = [16]
-> r8c12 = {24}
-> r8c456 = [1{38}]
-> 15(3)r9 = [861]

7. 3 in n2 only in r12c6.
-> r8c56 = [38]
-> 25(4)c6 = {3679}
-> r9c46 = [75]
Since 7 and 3 in n5 are in LG -> 25(4) = [{37}69]

8. 13(2)r1 can only be {49}
Whichever of (25) goes in r5c3 can only go in n5 in r4c5 and in n2 in r2c4
-> HS 9 in n2 -> r1c4 = 9
-> r1c3 = 4
-> 6(2)c3 = [15]
-> r5c3 = r4c5 = r2c4 = 2

etc.