Note $S_n$ the Sierpiński sieve graph of order $n$, which is obtained from the connectivity of the Sierpiński sieve.


For $n$ high enough, what is the treewidth of $S_n$?

I think that I can show that it is at least $4$ (by obtaining $K_5$ as a graph minor), and that it is at most $5$ (by obtained a tree-decomposition scheme involving bags of size $6$). My gut feeling is that it should be $5$, but I cannot find a proper database for forbidden minors at treewidth $4$ (and I can't access Sanders' thesis mentioned here).


1 Answer 1


You can recursively decompose each triangle into a smaller triangle and a trapezoid, and each trapezoid into two smaller triangles. The corresponding tree decomposition (whose bags contain the corners of a triangle at one level of the decomposition and a triangle or trapezoid at the next or previous level) has five vertices per bag and therefore has width four.

There is no $K_5$ minor, because the graph is planar. There is, however, a $K_{2,2,2}$ (octahedron) minor (actually, a subdivision), shown below. As this is one of the forbidden minors for treewidth three, the width is four.

enter image description here

  • $\begingroup$ Nice figure @David Eppstein. How did you draw it? $\endgroup$ Sep 6, 2016 at 22:33
  • $\begingroup$ Adobe Illustrator, mostly using their click, drag and snap copying operation to repeatedly triple a single equilateral triangle. $\endgroup$ Sep 6, 2016 at 22:47
  • $\begingroup$ Thanks for the clear explanation! I should have realized that there can't be a $K_5$ since it is planar... $\endgroup$ Sep 7, 2016 at 0:44

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