SudokuSolves/SetToCatchAKiller

The puzzle = https://f-puzzles.com/?id=ykyl6vk7

Striped of lines and cages, leaving just the shaded grid, it looks like:
The grid = https://f-puzzles.com/?id=yzjwjvjg

We construct this by adding eight sets of 1-9:
r3, r4, r6, r7
c3, c4, c6, c7
then taking away a different eight sets of 1-9
r1, r2, r8, r9
c1, c2, c8, c9

This tells us that the digits that appear in the light blue area, plus two instances of each digits that appear in the dark blue area are the same digits making up the set formed by the digits in the light green area plus two instances of each digits that appear in the dark green area

Example = https://f-puzzles.com/?id=ygkfczer

Dark Blue = 2461689449821246
Light Blue = 18292387
Dark Green = 8492624828611429
Light Green = 16769443

Total Blue = Light Blue + Dark Blue + Dark Blue =
18292387 2461689449821246 2461689449821246

Total Green = Light Green + Dark Green + Dark Green =
16769443 8492624828611429 8492624828611429

5 "1"s, 8 "2"s, 1 "3", 8 "4"s, 0 "5"'s, 6 "6"s, 1 "7", 6 "8"s, 5 "9"s

The Solve

1. The light blues

These must be two sets of the digits 1-4

2. The light greens

These must be two sets of the digits 5-8

https://f-puzzles.com/?id=yfktgunl

3. SET (Set Equivalence Theory)

Even if the light greens contain no digits of 5-8, we know that the light greens contain some, and that the light blues don't THEREFORE the dark blues must contain at minimum a 5, a 6, a 7 and an 8

We've already been given the corner digits (1, 2, 3, 5)
So the question then becomes: Can we fit MORE than three digits larger than 4 into the remaining cages (8, 12, 14, and 11) ?

4. Cage sums

 8 in 3 (without a 5) = 134

11 in 3 (without a 1, and with at least one of 6, 7 or 8) = 236
12 in 3 (without a 2, and with at least one of 7 or 8) = 138 147 (so must contain a 1)
14 in 3 (without a 3, and with at least one of 7 or 8) = 158 167 248 257 (so must contain either 1 or 2)

5. Narrow down the possibilities

In box 1, the 3 in the 8 cage is forced into row 2, eliminating any possible 3 from the 12 cage in box 3
So that's now 147

In box 1, the 1 in the 8 cage and the 1 in the 3 cage in box 4, eliminates any possible 1 from the 14 cage in box 7
So that's now 248

https://f-puzzles.com/?id=yhnjvhhu

6. Return to SET

Now we know the numbers in the light blue, dark blue and light green (if not which one is where), we know exactly which digits must be in dark green, and which are not. All the 5678 required are accounted for, so the remainder (the dark greens) MUST all be 1-4

https://f-puzzles.com/?id=yegmbwp2

Furthermore, we can say how many there are of each: 1234 (from light blue, halved) + 134 + 124 + 234 + 123
means there are exactly four "1"s, four "2"s, four "3"S and four "4"s in the dark green areas.

All this, without using the clock lines at all, so far.

7. Fill in the remaining pencil marks

The white spaces are 56789, though we can eliminate 5 from r1 and c1

https://f-puzzles.com/?id=yhsw5fn2

8. Clock lines deductions

https://f-puzzles.com/?id=yejgudff

9. Continue from here?

NOTE:

The alternative approach which I tried this morning, using numeric totals, is:

Light blue = 2 x (1+2+3+4) = 2 x 10 = 20
Light green = 2 x (5+6+7+8) = 2 x 26 = 52
Dark green = ?
Dark blue = 5 + 8 + 2 + 12 + 3 + 14 + 1 + 11 = 13 + 14 + 17 + 12 = 56

LB + DB + DB = LG + DG + DG
DG = DB - ( ( LG - LB ) / 2 )

? = 56 - ( (52 - 20) / 2 ) = 40 = 4 x 10 = 4 sets of "1", "2", "3" & "4".

The solution = https://f-puzzles.com/?id=yhaoajpy