Hints: Graph Cutting (Three Vertices, Fixed Size)

Answer to Hint 1: Zero — you cannot have fewer than $L$ vertices while staying connected and containing $a,b,c$.

For a side branch at attachment $x$, define $F_x(z)=\sum_{s\ge 1}\mathrm{ways}_s(x)\,z^s$ as in E1. How does a fork center $r\in T$ with three side subtrees differ from a path vertex with one side leaf?

Answer to Hint 3: The formula is identical: at each $v\in T$, multiply factors $(1+F_x(z))$ over side neighbors $x$. A fork center may have zero side neighbors (all three branches are part of $T$), so its factor is just $1$.

Combine side contributions at all vertices of $T$ into one polynomial $G(z)$. What coefficient gives the answer for size $d$?

Answer to Hint 5: The answer is $[z^{d-L}]\,G(z)$ — the same as E1 with $L=|T(a,b,c)|$ instead of $|P(a,b)|$.

On the star, triple $(2,3,4)$: what is $L$? For $d=4$, how many supersets of size $4$ contain all three leaves?

Answer to Hint 7: $L=4$ (center plus three leaves). Only $S=V$ works; answer $1$.

Brute force all triples. You could run this E4 DP for every triple $(a,b,c)$ and add $1$ when the count is positive. What complexity does that give, and why is E smarter?

Answer to Hint 9: About $O(n^3 d^2)$ — too slow for $n=2000$. E reuses one root DP per vertex to count all valid triples at once, splitting into path triples and fork triples (three vertices in three different branches).

Preview of E. Fix root $r$. A fork triple picks one vertex from each of three different child subtrees at depths $d_1,d_2,d_3$ with $1+d_1+d_2+d_3=d$. How does that resemble convolving three E1-style branch polynomials — but indexed by depth instead of size?

Answer to Hint 11: Each child subtree contributes a histogram $\mathrm{cnt}[\mathrm{depth}]$ = number of vertices at that depth. Merge two branches into dp[1] (pairs of depths), then multiply by a third branch with need = d - j. Path triples on a long branch are counted separately (the dub/2 term in the model solution).