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To say that graph isomorphism is downward self reducible means the following: There is an algorithm which decided graph isomorhpism for two given graphs of n vertices in polynomial time by accessing an oracle that solves graph isomorphism for (n-1) vertices.

To be clear, the oracle only tells 'Yes' or 'No'. It does not give us the isomorphism.

This problem appears as an exercise in Arora-Barak's chapter on Interactive Proofs.

How do I show that Graph Isomorphism is downward self-reducible?

My attempt is to start with an arbitrary vertex in $G_1$ and guess its image in $G_2$, and remove them and make an oracle call on the remaining graph. But now, I do not know how to proceed. These graphs being isomorphic may not necessarily mean that the original graphs are isomorphic.

Agnishom Chattopadhyay
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  • I removed your "NP-complete" tag, since graph isomorphism isn't known to be NP-complete (and seems not to be -- if it were, the polynomial hierarchy would collapse). – David Richerby Apr 24 '19 at 13:30
  • @Apass.Jack clarfied – Agnishom Chattopadhyay Apr 25 '19 at 19:18
  • Does the "Chapter on Interactive Proofs" imply probabilistic methods? Although I believe graph isomorphism is deterministic downward self polynomial-time reducible, can you clarify that the definition in the book is indeed having nothing to do randomization or probability? I do not have the book. – John L. May 07 '19 at 04:52

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This answer shows how to prove an easier variation, the case of (undirected loopless) multigraph isomorphism.

The problem of multigraph isomorphism is downward self-reducible

The simple idea is to merge two vertices in a way that distinguish those two vertices and edges involving them.

Let $G$ be a multigraph. If the sum of degrees of two vertices reaches the maximum, we will call an ordered pair of those two vertices a degree-maximum pair. For a degree-maximum pair $(u,v)$, construct $G_{u,v}$ as follows.

  • Its vertices are all vertices of $G$ except $u$ and $v$, together with a new vertex $u_v$ which is understood as $u$ merged by $v$. Note that $G_{u,v}$ has one less vertice than $G$.

  • Its edges consists of two parts.

    • the edges of $G$ whose endpoints are neither $u$ nor $v$ with the same multiplicity. Let $m$ be the maximum multiplicity of these edges.
    • for all vertex $w$ of $G$ that is neither $u$ nor $v$, edge $\{w, u_v\}$ with multiplicity that is the sum of the multiplicity of $\{w,u\}$ in $G$ and the product of $m+1$ and the multiplicity of $\{w,v\}$ in $G$.

It is easy to see that if we are given a multigraph $G'$ that is isomorphic to $G_{u,v}$ for some multigraph $G$, we can reconstruct $G$ from $G'$ up to isomorphism.

Suppose we are given two graphs of $n$ vertices, $G$ and $H$. Use the oracle to compare $G_{u,v}$ with $H_{x,y}$ for all tuples $(u,v,x,y)$, where $(u,v)$ is a degree-maximum pair in $G$ and $(x,y)$ is a degree-maximum pair in $H$. Once the oracle return yes, we conclude $G$ and $H$ are isomorphic; otherwise the oracle always return no, in which case we conclude $G$ and $H$ are not isomorphic. It take polynomial time to construct $G_{u,v}$ from $G$ or $H_{x,y}$ from $H$. There are less than $n^2$ tuples. So the whole algorithm takes polynomial time. QED.

I am yet to find a proof that shows the problem of (non-multi)graph isomorphism is downward self-reducible.

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John L.
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  • As my effort to simplify is too aggressive, there is a minor/critical flaw in my algorithm above. I should have selected a degree-minimum connected pair $(u,v)$. – John L. Apr 26 '19 at 05:00