How to compute the thickness of $K_{5,5}$?
By Euler formula it is greater than 1, and 3 is constructible($K_{2,5}*3$), but how to prove it is not 2?
2026-05-02 05:09:13.1777698553
Thickness of $K_{5,5}$
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Actually, the thickness is $2$. Note that we can easily see it's not $1$ because $K_{5,5}$ contains $K_{3,3}$.
According to wolfram math world, the general formula for the thickness of $K_{n,n}$ is $\lfloor\frac{n+5}{4}\rfloor$, giving a reference to Harary 1994, p.121.
However, it's easy enough to give an explicit (if messy) construction to show that the thickness is two. For an example, see the following (I'm afraid I don't have good graph building software, so I just made it work with Geogebra):