Can anyone help me with this?
I just wanted to clarify the thickness of $K_9$. The complete graph $K_9$ has 9 vertices and 36 edges. Then,$t(K_9)=\lceil{\frac{36}{3(9)-6}}\rceil=\lceil{\frac{36}{21}}\rceil=2$ and $t(K_9)=\lfloor{\frac{36+3(9)-7}{3(9)-6}}\rfloor=\lfloor{\frac{56}{21}}\rfloor=2$. But I found out that the thickness of $K_9$ is 3. This is easy to show if $t(K_9)\neq 2$.
How do I prove the thickness of $K_9$?
Any insight is appreciated. Thank you.
$\endgroup$ 11 Answer
$\begingroup$Your first equation should be $t(K_9) \geq \lceil(\frac{m}{3(n)-6})\rceil$, it only gives a lower bound.
I'm not sure where your equation which uses the floor function. The thickness of a complete graph can be given by $\lfloor \frac{n+7}{6} \rfloor$, except for when $n=9,10$.
From here, it can be seen that the minimum number of planar graphs which the edges of $K_9$ can be partitioned into is 3.
There is also a more in depth proof for why $K_9$ is non-biplanar and thus the thickness is not $=2$ here.
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