0
$\begingroup$

I have proven that when $\alpha > n^2$ every equilibrium must be a tree and that the lower bound for the social cost is $\Omega(\alpha n + n^2)$. I know that the PoA = $O(1)$, so the upper bound must be $O(\alpha n + n^2)$ but I can't prove it. Any kind of help is appreciated.

$\endgroup$
0
$\begingroup$

Since $\alpha > n^2$, the lower bound actually is $\Omega(\alpha n)$. As for the upper bound, overestimating the distance for every node we get also $O(\alpha n)$. So, PoA=O(1).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.