27
$\begingroup$

Usually one constructs a graph and then asks questions about the adjacency matrix's (or some close relative like the Laplacian) eigenvalue decomposition (also called the spectra of a graph).

But what about the reverse problem? Given $n$ eigenvalues, can one (efficiently) find a graph that has this spectra?

I suspect that in general this is hard to do (and might be equivalent to GI) but what if you relax some of the conditions a bit? What if you make conditions that there is no multiplicity of eigenvalues? What about allowing graphs that have "close" spectra by some distance metric?

Any references or ideas would be welcome.

EDIT:

As Suresh points out, if you allow undirected weighted graphs with self loops, this problem becomes pretty trivial. I was hoping to get answers on the set of undirected, unweighted simple graphs but I would be happy with simple unweighted directed graphs as well.

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ I think you might need to modify the question to 'unweighted undirected graphs with no self loops' or something like that ? I can imagine constructing a diagonal matrix with the required eigenvalues and declaring it to be a disconnected graph with weighted selfloops ? $\endgroup$
    – Suresh Venkat
    Commented Dec 11, 2010 at 19:47
  • 6
    $\begingroup$ Even simpler question (I don't know the answer) is how to construct simple connected graphs whose top few eigenvalues are given $\endgroup$
    – Yaroslav Bulatov
    Commented Dec 12, 2010 at 1:07
  • 6
    $\begingroup$ An alternative way of stating the question (the version with simple undirected graphs) is: given n real numbers (in some format), decide whether there exists an n×n symmetric 0/1 matrix with zero diagonals such that its n eigenvalues are the given numbers. $\endgroup$
    – Tsuyoshi Ito
    Commented Dec 12, 2010 at 1:12
  • 1
    $\begingroup$ @Yaroslav: I am not sure, but that problem sounds harder to me than the case where all n eigenvalues are given. $\endgroup$
    – Tsuyoshi Ito
    Commented Dec 12, 2010 at 12:57
  • 8
    $\begingroup$ Tiny observation: If we have no restrictions on the eigenvalues, the problem is really hard (even not include the algorithmic part) since this will implies the (non-)existence to the 57-regular Moore graph, which the eigenvalues are all known. $\endgroup$
    – Hsien-Chih Chang 張顯之
    Commented Dec 12, 2010 at 17:41

3 Answers 3

Reset to default
11
$\begingroup$

Cvetcovic et all in Section 3.3 of "Recent results in the theory of graph spectra" goes over algorithms for constructing graphs given spectrum in some special cases

Improve this answer
$\endgroup$
11
$\begingroup$

Even asking whether a graph with a given spectrum exists is a tough question. This is witnessed by the open problem of determining whether there exists a graph of girth 5 diameter 2 and order 3250 whose spectrum (if it exists) is known.

Improve this answer
$\endgroup$
5
$\begingroup$

One other obstacle in defining your question is that the are isospectral (same eigenvalues) but non-isomorphic graphs. So given a list of eigenvalues in such a case, which graph do you want? Maybe you just want an algorithm to return one random element of the set of such non-isomorphic graphs?

Improve this answer
$\endgroup$
6
  • $\begingroup$ I was thinking something along the lines of sampling from the space of graphs that are isospectral but this seems like we're quickly descending into a GI equivalent problem (thus my comment above). For simplicity, we could restrict to all distinct eigenvalues (which, if IIC ensures a unique graph) but I'm really just trying to see what's known or out there. $\endgroup$
    – user834
    Commented Dec 12, 2010 at 21:17
  • 6
    $\begingroup$ I don't think distinct eigenvalues ensures reconstructibility, here are some spectra of isospectral graphs over 7 nodes yaroslavvb.com/upload/save/cstheory-isospectral.png $\endgroup$
    – Yaroslav Bulatov
    Commented Dec 12, 2010 at 22:57
  • 3
    $\begingroup$ I like the random element formulation. I would be interested to know if it's equivalent to GI. One reason I'm interested in the random element formulation, is the question, raised in my paper with Arora and Steurer on unique games, of whether graphs with certain spectrum can be expanders for small sets. Intuitively, one may hope that a random graph with this spectrum will be the best expander possible for all set sizes, and so insight on reverse spectra may be useful there. $\endgroup$
    – Boaz Barak
    Commented Dec 13, 2010 at 2:37
  • $\begingroup$ @Yaroslav: Thank you for that link and thanks for correcting me! $\endgroup$
    – user834
    Commented Dec 13, 2010 at 20:09
  • 1
    $\begingroup$ @user834: Re: your comment about a GI-equivalent problem. Note that determining isomorphism of graphs with bounded eigenvalue multiplicity (in particular, graphs with no multiple eigenvalues) can be done in polynomial time. $\endgroup$
    – Joshua Grochow
    Commented Oct 30, 2013 at 17:58

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