7
$\begingroup$

Circle packing theorem is a famous result stating that for every connected simple planar graph $G$ there is a circle packing in the plane whose intersection graph is $G$ https://en.wikipedia.org/wiki/Circle_packing_theorem.

I know that this result has many proofs and I want to read one of them, but don't understand how to start (for quite a while). The article in wiki gives a reference to Thurston notes, but the proof comes only in the last section and I am not sure if this is the simplest approach. I like these notes very much, but was never able to read them till the end. So I wonder if there are some simple proofs of this result nowadays. Can you advise something?

$\endgroup$
4
  • $\begingroup$ may be the fact that there exists a triangulation of every maximal planar graph may come handy in the proof $\endgroup$ – vidyarthi Aug 25 '20 at 10:28
  • 2
    $\begingroup$ Does this answer to your question?mathoverflow.net/q/187845/90655 $\endgroup$ – C.F.G Sep 3 '20 at 8:23
  • $\begingroup$ Thanks a lot C.F.G! I have not spotted this question. It looks like mine is a duplicate. I'll study the answers $\endgroup$ – aglearner Sep 3 '20 at 13:39
  • 1
    $\begingroup$ Although your question is close to a duplicate to "Koebe–Andreev–Thurston theorem - where can I find a proof?," additional expositions have appeared in the ~6 yrs since that post. $\endgroup$ – Joseph O'Rourke Sep 3 '20 at 17:26
7
$\begingroup$

I can recommend Sariel Har-Peled's exposition in supplemental Chapter 15 of his book Geometric Approximation Algorithms. Ch15 PDF download. He emphasizes angles via a "whac-an-angle" game. He acknowledges that

Our presentation follows Pach and Agarwal [pa-cg-95].




$\endgroup$
3
+100
$\begingroup$

Books are written on the subject, so, finding a proof (which are many by now) shouldn't be a problem. I also enjoyed greatly Rohde's tribute to Schramm that explains in very nice way some ideas that Schramm introduced into the area; following references from there one should be able to find more detailed accounts.

$\endgroup$

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.