dep k with m nodes on cycle:

Rule: Number of leaves is equal to sum a_i - 2.
Rule: Must have more than one cycle vertex.
All cycle vertices must have deg at least 3.

m nodes on cycle, k depth: need at least 3 vertices in cycle.
Need at least m(k-1) extenders.

Seems a little easier to just depthmax?
Two vertices on cycle.  Even # of extenders: Okay.
Odd # of extenders: Maybe invalid.

We can def handle:
All 2s
2 or more 3+'s, even number of extenders (assuming balance constraint).
No 2's, balance constraint respected (build cycle)

Wait balance constraint is just sum of degrees is 2n lol.
We def can't handle:
- Sum of degs != 2n
- 3322211, 33211, etc

Iterate over # of vertices in cycle?

33: Even # of 2's ok
333: one 2 okay, two 2's bad, three 2's okay, four 2's ok.  Generally anything except two 2's is ok.
43: One 2 bad, two 2's ok, three 2's ok, four 2's ok, five 2's ok, generally anything we want is okay.
44: One 2 bad, 2+ is always fine.

two cycle vertices, one 2 is always bad.

Can we do a greedy assignment or something?

Generally: with two cycle vertices, 2+ 2's is always fine UNLESS 33 and odd number of 2's.


No 2's: Okay
2+ cycle vertices:
	Ok unless (a) only one extra and it's a 2 or (b) 33 and odd # of 2's.
		Aside from these cases, there must be at least two extras.

		Case 1.  Even # of extras.

If sum is not 2n: Bad
If all 2's: Ok
Otherwise, there is at least one 1.  If only one cycle vertex, bad.
If no 2's, okay (build cycle out of them, connect others).
If there's an even # of 2+ vertices, then we can split them.
If there are just three vertices with deg >= 1 and one is a 2 (note that we must have two cycle vertices), then bad.
If there's a 4 or above, we can add one extra connector to the 4 and then split the rest.
If there are 33 and then odd # of 2's, OR 33322111, then bad.
Last case: Odd # of connectors with at least three 3's.  Build 33 cycle, connect a 3 to the first one and an extra to that 3.  This is definitely possible (must have a third 3 bc otherwise it would be odd # of 2's, must have at least one 2 giving four total vertices, must have odd number of connectors so at least one more

What if 33322111?

At least one 1, at least 2 >2
at least one 2
odd # of 2+s
At least five 2+s
No 4+s
At least three 3's