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The $k$-bounded spanning tree problem is where you have an undirected graph $G(V,E)$ and you have to decide whether or not it has a spanning tree such that each vertex has a degree of at most $k$.

I realize that for the case $k=2$, this is the Hamiltonian path problem. However I'm having trouble with cases where $k>2$. I tried thinking about it in the sense that you can add more nodes onto an existing spanning tree where $k=2$ and maybe since the base is NP complete, adding things on will make it NP-complete as well, but that doesn't seem right. I'm self-studying CS and am having trouble with theory, so any help will be appreciated!

FrankW
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user17199
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2 Answers2

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The question has been asked before on stackoverflow, where it has also been answered. The idea is to connect each vertex to $k-2$ new vertices. The new graph has a $k$-bounded spanning tree iff the original graph has a hamiltonian path.

Mohit Singh and Lap Chi Lau gave a polytime algorithm which find a $(k+1)$-bounded spanning tree if a $k$-bounded spanning tree exists. So we can determine the minimum degree of a spanning tree up to an uncertainty of $1$.

Community
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Yuval Filmus
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My understanding is if you have an algorithm that can solve k-bounded spanning tree problem with any k, you can use that algorithm to solve a special case with k=2, which is essentially a Hamiltonian path. So if your algorithm can achieve polynomial time, then it can be used to solved Hamilton path in polynomial time, which is equivalent to solving any np-complete problems in polynomial time. So the k-bounded spanning tree problem must be np-complete. Note that this is a general idea, not a complete proof.

Also note that being np-complete doesn't mean that there's no polynomial time algorithms that can solve the problem. No one has proven this yet. It only means that all problems that are np-complete are equally hard and if one can be solved in polynomial time then all can be solved in polynomial time.

Sam G
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