23 hours and no takers? Here is my solution.
How to start:
- If A is green then B and C are gray because of 33 and D is gray because of 3. But then BCD are gray which is forbidden. Therefore A is gray and so is E.
- If F is gray, since E is gray, G and H must be green but that is not possible because of 24.
Therefore F must be green and 24 forces IJK to be gray.
- No 3 grays can joint at an intersection, so LMNO are green.
- L being green, with 33, forces P to be green and BCQQ to be gray.
- AB being gray, D must be green and because of 3, RS must be gray.
- Avoiding 3 grays forces TUV to be green.
- R and 2 force W to be green. N and 2 force X to be gray.
- If Y is gray, because of 2, Z must be gray also. But that is forbidden because of G. So Y is green.
And some more:
- The 4 forces A to be gray.
- The 5 forces the B's to be gray, which in turn force the C's to be green.
- If one of DEFG is green, the corresponding 23 couldn't be satisfied. They must be gray.
- If H is gray, I muat also be because of the 5. But that would be 3 grays. So H must be green. And J must be gray because of the 5 again.
- K is now forced to green.
- The remaining options for 23 force LM to be gray, which in turn forces N's to green.
- 1111 has 2 ways, Shade the O's or the P's. But if you shade the O's, the P's are green and one O is surrounde by green P's and N. So that cannot be. The P's are shaded.
- E and the P below along with the 3's at the right forbid to shade the Q's. If a Q is gray it must extend up or down resulting in a junction of 3 grays.
- This forces R to be green and S to be gray to satisfy the 3.
- T must be green to avoid a triple gray. This forces the grays and the greens around the 4.
- This forces U to be green, V to be gray to satisfy the 24 and the W's to be green.
- The X's are gray because of the 3.
- This forces Y's to be green.
- Z must be green, forcing the 2 grays and 6 green around the 2.
- 23 left forces A's to be gray and B to be green.
- 2 right forces C to be green.
- The placement of the 3 around 13 top left forces D to be gray and E to be green.
- F is now surrounded by green, it can only escape via G.
- Setting G forces the H's to be gray and the I's to be green to satisfy the 3's.
- Now J needs to escape via K. K is gray and the L's are green.
- If M is gray then it must extend to N, but 13 doesn't allow it. M is green and N is gray.
- O must be green to avoid a triple-gray.
- P is gray to satisfy the 2.
- Q can only connect via the R's. The R's are gray.
- This forces the gays and green around the 12 right of the R's.
- The remainng S's are forced green because of 2 grey neighbours.
- The T's are green because of a hint.
This completes the solution.