# Evidence that Graph Isomorphism problem is not $NP$-complete

Graph isomorphism problem is one of the longest standing problems that resisted classification into $P$ or $NP$-complete problems. We have evidences that it can not be $NP$-complete. Firstly, Graph Isomorphism can not be $NP$-complete unless the polynomial hierarchy [1] collapses to the second level. Also, the counting[2] version of GI is polynomial-time Turing equivalent to its decision version which does not hold for any known $NP$-complete problem. The counting version of natural $NP$-complete problems seems to have much higher complexity since they are $\#P$-complete. Finally, the lowness result [3] of GI with respect to $PP$ ($PP^{GI}=PP$) is not known to hold for any $NP$-complete problem. The lowness result of GI has been improved to $SPP^{GI}=SPP$ after Arvind and Kurur proved that GI is in $SPP$ [4].

Edit: It is known that SAT is truth-table equivalent to USAT (set of Boolean formulas with exactly one satisfying assignment) while it is not known whether GI and UGI are equivalent under polynomial time truth-table reductions.

What other (recent) results can provide further evidence that GI can not be $NP$-complete?

[1]: Uwe Schöning, "Graph isomorphism is in the low hierarchy", Proceedings of the 4th Annual Symposium on Theoretical Aspects of Computer Science, 1987, 114–124

[2]: R. Mathon, "A note on the graph isomorphism counting problem", Information Processing Letters, 8 (1979) pp. 131–132

[3]: Köbler, Johannes; Schöning, Uwe; Torán, Jacobo (1992), "Graph isomorphism is low for PP", Computational Complexity 2 (4): 301–330

[4]: V. Arvind and P. Kurur. Graph isomorphism is in SPP, ECCC TR02-037, 2002.

• The Hidden Subgroup Problem contains as a special case GI, but I don't believe it contains any problems known to be NP-complete: see, en.wikipedia.org/wiki/Hidden_subgroup_problem. Put another way, it is plausible that a quantum computer could efficiently solve GI, but few people think a quantum computer can efficiently solve an NP-complete problem. – Sam Hopkins Jan 3 '15 at 1:51
• Integer factorization has also resisted classification. Is graph isomorphism a "longer standing" problem than factorization? – Gerry Myerson Jan 3 '15 at 3:28
• @GerryMyerson I think stating that GI is "longest standing" is not accurate. I have edited the post accordingly. – Mohammad Al-Turkistany Jan 3 '15 at 3:58
• reposted (with proper reference to this question) at cstheory.stackexchange.com/questions/29266/… – Thomas Klimpel Aug 13 '15 at 8:13

Theorem 2 Graph Isomorphism is in ${\mathsf{RP}^{\mathrm{MCSP}}}$. Moreover, every promise problem in ${\mathsf{SZK}}$ belongs to ${\mathsf{BPP}^{\mathrm{MCSP}}}$ as defined for promise problems.
Here $\mathsf{RP}$ stands for randomized polynomial time, MCSP is the Minimum Circuit Size Problem, and $\mathsf{SZK}$ stands for Statistical Zero Knowledge. So we learn that if MCSP is not NP-complete and RP!=NP, then graph isomorphism is not NP-complete either.