Define $K\_SCC = \{ \langle G, k \rangle \,:\, G \text{ has at least $k$ strongly connected components} \}$

I want to show that $K\_SCC \in NSPACE(\log n)$, using that $st-CONN$ and $\overline{st-CONN}$ are both in NL, where $st-CONN = \{\langle G,s,t \rangle \,:\, \text{there is a path from $s$ to $t$ in $G$} \}$.

Would appreciate any help

  • $\begingroup$ Any strongly connected component $C$ is uniquely represented by the vertex $x\in C$ that has the smallest numerical label. Show that you can recognize such vertices in NL, and then you can just count them in increasing order. $\endgroup$ Jun 17 at 15:22

Ask the prover to give you any node from $k$ distinct connected components.

You have only to verify that the nodes are not in the same connected components (hence, they are in $k$ different components, meaning that $\langle G, k\rangle \in K_{SCC}$)

Also, ask for the proof of $st-CON$ between any two of them.

Notice that even though the proof is gigantic, at every point in time the verifier will need to only verify a small portion of the proof: only one $st-CON$ is being processed at a time, hence the verifier can be constructed in such a way that will require only $O(\log(n))$ space.

The pseudocode for the verifier should look similar to this:

  • For every $i\neq j$ with $1\le i,j\le k$, do:
    • Take a look at the next $O(\log(n))$ bits of proof, and verify $st-CON$ for the $i$'th and the $j$'th nodes
  • $\begingroup$ As the question is stated, $k$ is not constant. It is part of the input. $\endgroup$ Jun 17 at 18:42
  • $\begingroup$ Oops :p I totally missed this important thing... $\endgroup$
    – nir shahar
    Jun 17 at 19:01
  • $\begingroup$ When I think about it, since at every point in time the verifier only checks one $ST-con$ proof of size $O(\log(n))$, then the verifier can just check the proofs one by one in some predefined order, and the verifier wont use more than $O(\log(n))$ space. Yes, the proof is larger, but if I remember correctly thats allowed for as long as the verifier requires small space. $\endgroup$
    – nir shahar
    Jun 17 at 19:15
  • $\begingroup$ This is true except for how you specify which $k$ nodes you want to show are in different components. This will take $O(k\log(n))$ memory, which is not logarithmic in the size of $k$... $\endgroup$
    – nir shahar
    Jun 17 at 19:23
  • $\begingroup$ The definition of NL does not allow you to generate a polynomial-size “proof” and subsequently verify it in logarithmic space. This would in fact give you all of NP. See my comment below the question how to do this correctly. $\endgroup$ Jun 17 at 19:27

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