Cells adjacent to green marks must total less than 10, blue must total 10, red must total more than 10. Also NC so horizontally/vertically adjacent cells cannot be {12}, {23}, … {78}, {89}, so at least one of the cells adjacent to each green mark must contain one of 1,2,3 as they cannot be {45}; similarly at least one of the cells adjacent to each red mark must contain one of 7,8,9 as they cannot be {56}.
Prelims.
Delete 9 from cells either side of green marks.
Delete 5 from cells either side of blue marks.
Delete 1 from cells either side of red marks.
Clean-ups only as stated.
1. 1 in N9 only in R9C89, locked for R9
1a. 1 in R9C89 -> no 2 in R9C89 (NC)
1b. R7C5 = 1 (hidden single in N8), no 2 in R68C5 + R7C46 (NC)
1c. 1 in N7 only in R8C123 -> no 2 in R8C2 (NC)
2. 9 in N3 only in R3C789, locked for R3
2a. 9 in R3C789 -> no 8 in R3C8 (NC)
3. 1 in N6 only in R4C789, locked for R4
3a. 1 in R4C789 -> no 2 in R4C8 (NC)
4. 9 in N4 only in R4C13, locked for R4
4a. 9 in R4C13 -> no 8 in R4C2 (NC)
5. 9 in N7 only in R8C3 + R9C23 -> no 8 in R9C3 (NC)
6a. There are three groups of red marks in N1, 7,8,9 in one of R12C1 and in one of R12C3, then because there are two other red marks R3C2 = {78}, no other 7,8,9 in N1, no 7 in R4C2 (NC)
6b. There are five red marks in N6 which place 7,8,9 in three of R56C789, no 7,8 in R4C789
6c. There are three groups of red marks in N8, 7,8,9 in one of R78C4 and in one of R78C6, then because there are two other red marks R9C5 = {789}, no other 7,8,9 in N8
6d. There are five red marks in N9, 7,8,9 in one of R89C7, the three right-hand red marks require 7,8,9 in two of R78C89 but there’s another red mark between R7C78 -> 7,8,9 must be in R7C8 and R8C9 (they cannot be in both of R78C8 because of blue mark), no other 7,8,9 in N9, no 8 in R6C8 (NC)
6e. There are three groups of green marks in N3, 1,2,3 in one of R1C78, 1,2,3 in one of R12C9, 1,2,3 in one of R2C78 , no other 1,2,3 in N3
6f. There are three groups of green marks in N6, 1,2,3 in R4C78, 1,2,3 in R45C9 and 1,2,3 in R6C78, no other 1,2,3 in N6
[Even though there are six green marks in N4, I don’t think 1,2,3 can be placed at this stage. They may possibly be eliminated from R4C13 but I’ll leave that for now.]
[Time for some clean-ups.]
Starting with the blue marks
No 7,8 in R1C2 -> no 2,3 in R1C1
No 1 in R1C5 -> no 9 in R1C4
No 9 in R12C7 -> no 1 in R12C7
No 2,3 in R3C9 -> no 7,8 in R2C9
No 3 in R5C8 -> no 7 in R6C8
No 1 in R6C8 -> no 9 in R5C8
No 1,2,7,8 in R8C5 -> no 2,3,8,9 in R8C6
No 4,6 in R9C5 -> no 4,6 in R9C6
No 1 in R9C6 -> no 9 in R9C5
No 4,6 in R7C8 -> no 4,6 in R8C8
No 1 in R8C8 -> no 9 in R7C8
No 4,6 in R8C9 -> no 4,6 in R9C9
No 2 in R9C9 -> no 8 in R8C9
R8C8 = {23} -> no 2,3 in R8C7, no 3 in R9C8 (NC)
R9C6 = {23} -> no 2,3 in R9C7 (NC)
Next the green marks
No 1 in R1C7 -> no 8 in R1C8
No 1 in R2C7 -> no 8 in R2C8
No 1 in R4C2 -> no 8 in R5C2
No 1 in R6C8 -> no 8 in R6C7
No 1 in R7C12 -> no 8 in R78C12
And then the red marks
No 9 in R3C2 -> no 2 in R2C2 + R3C1
No 9 in R5C8 -> no 2 in R5C9
No 7,8,9 in R6C8 -> R6C9 = {789} -> no 8 in R5C9 (NC)
No 9 in R7C8 -> no 2 in R7C79
No 8,9 in R8C6 -> no 3 in R7C6
No 9 in R9C5 -> no 2 in R9C4
7a. R4C7 = 1 (hidden single in C7) -> no 2 in R4C6 (NC)
7b. R3C8 = 9 (hidden single in C8) -> no 8 in R3C79 (NC)
Clean-up: no 1,2 in R2C9 (blue)
7c. R8C8 = 2 (hidden single in N9) -> R7C8 = 8, (blue), no 1 in R9C8 (NC)
R9C8 = {456} -> no 5 in R9C7 (NC)
Clean-up: no 8 in R5C8 -> no 3 in R5C9 (red)
7d. R9C9 = 1 (hidden single in N9) -> R8C9 = 9 (blue)
7e. 3 in N9 only in R7C79, locked for R7
7f. 7 in N9 only in R89C7, locked for C7, no 6 in R89C7 (NC)
Clean-up: no 3 in R12C7 (blue)
7g. R5C7 = 9 (hidden single in N6) -> no 8 in R5C6 (NC)
7h. R6C9 = 8 (hidden single in N6) -> no 7 in R5C9 (NC)
7i. R5C8 = 7 (hidden single in N6) -> R6C8 = 3 (blue), no 6 in R4C8 + R5C9, no 2,4 in R6C7 (NC)
Clean-up: no 3 in R4C3 (blue)
7j. R4C9 = 2 (hidden single in N6)
Clean-up: no 8 in R5C3 (blue)
7k. R6C7 = 6 (hidden single in N6) -> no 5,7 in R6C6, no 5 in R7C7 (NC)
Clean-up: no 4 in R12C7 (blue)
7l. R7C7 = 3 (hidden single in C7) -> no 4 in R7C6 + R8C7 (NC)
7m. R8C7 = {57} -> no 6 in R8C6 (NC)
Clean-up: no 4 in R8C5 (blue)
8a. 7 in N3 only in R13C9 -> no 6 in R2C9 (NC)
Clean-up: no 4 in R3C9 (blue)
8b. R2C9 = {34} -> no 3,4 in R1C9, no 4 in R2C8 (NC)
8c. R2C9 = 3 (hidden single in N3) -> R3C9 = 7 (blue), R3C2 = 8
Clean-up: no 2 in R1C2 (blue)
8d. R1C9 = {56} -> no 5,6 in R1C8 (NC)
8e. R2C2 = {456} -> no 5 in R2C13 (NC)
8f. R3C7 = {45} -> no 4,5 in R3C6 (NC)
8g. R9C6 = 2 (hidden single in N8) -> R9C5 = 8 (blue)
Clean-up: no 2 in R1C4 (blue)
8h. 3 in N8 only in R8C45 + R9C4 -> no 4 in R8C4 (NC)
8i. R8C67 = [47/75] (cannot be [45] NC), 7 locked for R8
8j. R8C45 = [36/53/63] (cannot be [56] NC), 3 locked for R8 and N8
8k. R9C4 = {456} -> no 5 in R8C4 + R9C3 (NC)
8l. Naked pair {36} in R8C45, locked for R8 and N8
Clean-up: R8C4 = {36} -> R7C4 = {79} (red)
8m. R9C4 = {45} -> no 4 in R9C3 (NC)
8n. R8C3 = 8 (hidden single in R8) -> no 7 in R7C3, no 7,9 in R9C3 (NC)
Clean-up: no 2 in R5C3 (blue)
8o. R9C2 = 9 (hidden single in N7)
[I’d overlooked that it’s been hidden single in C8 for quite a while, since step 6a, but I don’t think it made much difference not spotting it earlier.]
8p. 7 in C2 only in R67C2 -> no 6 in R7C2 (NC)
8q. 2 in N5 only in R5C45 + R6C4 -> no 1,3 in R5C4 (NC)
8r. 2 in R3 only in R3C345 -> no 1,3 in R3C4 (NC)
9a. 8 in N4 only in R45C1 -> no 7,9 in R4C1 (NC)
9b. R4C3 = 9 (hidden single in N4) -> R5C1 = 1 (blue), no 8 in R4C4, no 2 in R5C24 + R6C3 (NC)
9c. R1C2 = 1 (hidden single in N1) -> R1C1 = 9 (blue), no 2 in R1C3 (NC)
Clean-up: no 8,9 in R1C3 -> no 2 in R2C3 (red)
No 8,9 in R2C3 -> no 3 in R1C3 (red)
[I forgot that R12C3 (red) must contain 7, with 8 and 9 already placed for N1, and therefore no 6 (NC), which would have simplified things slightly.]
9d. R8C1 = 1 (hidden single in N7), no 2 in R7C1 (NC)
9e. R8C2 = {45} -> no 4,5 in R7C2 (NC)
Clean-up: R8C2 = {45} -> no 7 in R7C2 (green)
9f. R7C2 = 2
9g. R6C2 = 7 (hidden single in C2) -> no 6 in R5C2 (NC)
Clean-up: R6C2 = 7 -> no 4,5 in R6C1 (green)
9h. R6C1 = 2 -> no 3 in R5C1 (NC)
Clean-up: R6C1 = 2 -> no 8 in R5C1 (green)
9i. R4C1 = 8 (hidden single in N4)
9j. R1C8 = 4, R3C7 = 5, R1C9 = 6, R2C8 = 1, R8C7 = 7, R8C6 = 4 -> R8C5 = 6 (blue)
9k. R6C3 = {45} -> no 4,5 in R6C4 + R7C3 (NC)
and the rest is naked singles, without using NC or coloured marks.