I have read the definition of treewidth/tree-decomposition both in Wikipedia and in here: https://medium.com/@karlrombauts/treewidth-how-all-graphs-are-trees-in-disguise-ec699b69e2fb

I'm finding something hard to understand why the treewidth of a tree is $1$. By the definition of tree-decomposition, I may select to have a single $X_1$ to contain all vertices. All requirements hold.

If we are referring to the $|X_1|$ it will be $n$ and then the tree width is $n-1$. If we are referring to the quantity of such sets, so we have only $X_1$, therefore the treewidth is $1-1=0$.

Likewise, this can be for any graph, not only trees. We can always decide $X_1$ will contain all vertices...

So I am not sure why it is claimed that the treewidth of a tree is exactly $1$? I think I'm missing out something in the understanding, and I'm looking for an explanation to this, and maybe one more example for a non-tree graph when the treewidth isn't trivially $1$.

Also, I read that $k$-treewidth do not contain $K_{k+2}$ as a minor. I'm not looking for a proof,but merely an intuition.

Edit: I understand now the width is the minimal maximal bag size. However, I still fail to see why any tree has width $1$. If the idea is $1$ bag for $1$ edge, wouldn't this be the case for any graph too? Why will it only work for a tree?

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    $\begingroup$ The width of a decomposition is the size of the largest bag. The width of a graph is the smallest possible width over all tree decompositions of the graph. Just because you choose a bad decomposition, doesn't mean that the width is large. It is always possible to make a tree decomposition of width 1 of any forest. $\endgroup$
    – Pål GD
    Jun 9, 2023 at 18:17
  • $\begingroup$ Thanks for the clarification. I still don't see what is the tree-decomposition of a tree/forest. Is it simply a bag per edge? If so, why can't we always construct a bag for an edge in a generic graph,and get width $1$? $\endgroup$ Jun 13, 2023 at 18:34
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    $\begingroup$ If you do that then the decomposition won't form a tree. There will be cycles in the decomposition. $\endgroup$ Jun 13, 2023 at 19:21

1 Answer 1


Yes, for a tree, the decomposition corresponds to one bag per edge. However, if you consider the graph $C_6$, the cycle on 6 vertices, you don't get a tree decomposition if you follow the same scheme.

What you would do for a cycle is essentially to pick an arbitrary vertex, $v_1$, remove it from the graph. Perform your trick with one bag per edge for the remaining graph (a $P_5$), and then put $v_1$ into every bag.

This naive trick works for every graph, i.e., find a smallest set $X$ such that $G-X$ is a forest, and then add $X$ to every bag. Warning: does not guarantee that the width is optimum, but does actually give you a valid tree decomposition, and perhaps some intuition. That also explains how adding one vertex can only increase the width of a graph with at most one.

(Ps, this set $X$ is called a feedback vertex set).

  • $\begingroup$ Thank you! I still don't see how having $K_{k+2}$ as a minor prevents the treewidth from being $k$. Is it perhaps a non-trivial result? Because your trick, as you said, doesn't guarantee we have no better way to decompose. $\endgroup$ Jun 19, 2023 at 15:47
  • $\begingroup$ There is a theorem starting that every treewidth-k graph has a k+1 sized separator. $\endgroup$
    – Pål GD
    Jun 19, 2023 at 17:38

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