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\lecture{12}{Wednesday, March 19, 2003}{Madhu Sudan}{Nishanth Sastry}
Today:
\begin{itemize}
\item
IP and AM proofs (Graph Non-isomorphism)
\item
complexity classes IP, AM, MA etc.
\item
Some properties:
\begin{itemize}
\item Power of the prover ($IP \subseteq PSPACE$)
\item Goldwasser-Sipser Protocol to prove ``largeness''
\item Private coins = Public coins
\end{itemize}
\end{itemize}
\section{Example Interactive Proof: Graph Non-Isomorphism}
Recall from last class that interactive proofs involve a
computationally unbounded prover $P$ trying to prove (language
membership) to a polytime verifier $V$. What do interactive proofs
look like?
We say two graphs $G, H$ are isomorphic to each other if a relabeling
of the vertices of $G$ (i.e., a permutatation of the labels) yields
$H$ and vice-versa. Determining whether two graphs are isomorphic is in
NP (The certificate is the permutation of the labels). But how do we
prove that two graphs are \emph{not} isomorphic? %% Define the language:
%% \begin{displaymath}
%% GNI = \{ \langle G, H \rangle | \forall \pi \in S_{|G|}, the set of
%% all permutatations of the labels of G, G \ne \pi(H)\}
%% \end{displaymath}
We can list all permutations of $G$ and prove that none of them will
make it equal to $H$, but this is computationally infeasible. The
following shows that graph non-isomorphism admits an interactive proof:
Verifier $V$ picks a random graph $F \in \{G, H\}$ and a random
permutation of its labels $F' = \pi(F)$ which it sends to the prover
$P$. $V$ accepts if $P$ can guess whether $F'$ came from $G$ or $H$. If $G, H$ are
indeed non-isomorphic, $P$ can correctly guess the original graph of $F$
since it is assumed to be computationally unbounded. Otherwise, $F, G$
and $H$ are isomorphic to each other and even $P$ cannot distinguish
them apart (At best it can guess correctly with a probability of
$1/2$.).
%% \section{Resources for Interactive Proofs}
%% Later, we will show that secrecy
\section{Complexity classes: Interactive Proofs and Arthur-Merlin Games}
The idea of proofs by interaction was independently developed by
Goldwasser, Micali and Rackoff who called it ``Interactive Proofs''
and by Babai and Moran who called it ``Arthur-Merlin games''. In
current notation, we use IP to denote all languages with poly-round
interactive proofs. In Arthur-Merlin games, Arthur is a
polynomial-time bounded entity analogous to the Verifier in IP and
Merlin is capable of unbounded computational power, just like the
Prover in IP. Arthur is allowed to toss coins and challenge Merlin but
the key difference from IP is that the coins tossed are public. We
denote 1-round games where Arthur goes first and Merlin responds by
AM. Similarly MA denotes 1-round games where Merlin goes first and
Arthur responds. We can extend this notation to AMAMA and so on. For
convenience, we will also use the notation IP[$k$] to denote an
interactive proof (private coins) with $k$ rounds of
interaction. Similarly define AM[$k$] and MA[$k$].
Some of the factors that could affect the computational power of
interactive proofs are:
\begin{itemize}
\item Interaction: Number of rounds (constant/polynomial)
\item Verifier's Randomness: whether the coin tosses are made public
kept private.
\item Error: One-sided error vs. two sided error
\end{itemize}
In the rest of the lecture, we will focus on some of these aspects.
\section{Power of the prover}
For any fixed verifier, what should the prover do to maximize the
probability of acceptance? Consider the quantity $p_{q_1a_1\ldots
q_i}$, which denotes the max probability at round $i$ over all
future answers that verifier accepts. This is simply the probability
that the verifier accepts if the prover gives the best possible
answer:
$$
p_{q_1a_1\ldots q_i} = \mathop{max}_{a_i} \{p_{q_1a_1\ldots q_i a_i}\}
$$
We can compute this recursively in terms of all possible questions
$q_{i+1}$ that can be asked by the verifier in round $i+1$.
$$
p_{q_1a_1\ldots q_ia_i} = \sum_{q_{i+1}} Pr[q_{i+1}|\mbox{past}]p_{q_1a_1\ldots q_ia_iq_{i+1}}
$$
Since the prover is unbounded computationally, it can compute this by
induction over the number of remaining rounds. Since this computation
can be done in PSPACE, we have the following theorem:
\begin{theorem}
$IP \subseteq PSPACE$
\end{theorem}
In a later lecture, it will be shown that IP = PSPACE.
\section{Amplification: sequential and parallel (round-preserving)}
The IP verifier was defined to accept good strings with a probability
of at least 2/3 and bad strings could be accepted with a probability
of no more than 1/3. Analogous to the amplification lemma for the
randomized complexity classes, we can reduce the error to any
polynomially small value by repeating the interactive proof over many
trials and accepting only if a majority of the trials accept. However,
notice that the questions asked in each round in a given trial depend
only on the previous round(s) of that trial. Thus, the questions asked
in round $i$ of each trial are independent of each other and we might
as well run the trials in parallel. Thus, we can amplify the error
while preserving the number of trials.
\section{Private coins are equivalent to public coins}
It seems that IP should be more powerful than the arthur-merlin model
because the coin-flips made by the verifier are secret whereas Arthur
has no secrets from Merlin. However, we
can convert one to the other: Consider an IP proof. Suppose Arthur
asks Merlin to prove that there are a large number of random strings
on which to accept. If Merlin can successfully show this, then IP
reduces to Arthur-Merlin games, because regardless of the random coins
flipped, Merlin can make Arthur accept with high probability.
\subsection{Goldwasser-Sipser Protocol}
This is a protocol that demonstrates the largeness of sets by
approximate counting. We will use this as a ``subroutine'' in proving
the equivalence of IP[$k$] and AM[$k$]. Consider a
problem defined by the following
inputs: $S \subseteq \{0,1\}^{n}$, and $m = O(\sqrt n)$. We would
like the verifier to accept with high probability if S is BIG, defined
as $|S| > 2^m$ and reject (accept with probability less than a third)
if S is SMALL, defined as $|S| < \frac{2^m}{100}$. Furthermore, it is
guaranteed that S is either BIG or SMALL.
The solution is similar to the Valiant-Vazirani reduction from a
previous class. The idea is to use a hash function to map $\{0,1\}^m$
to a sufficiently small space, say $\mathcal(H) = \{0,1\}^{m-4}$.If
$S$ is large enough, then no matter which $y \in \mathcal(H)$ we pick;
we can find an element in $S$ that maps to it.
\begin{claim} If $|S| > 2^m$, and $h$ is chosen from a family of pairwise
independent hash functions, then for
at least $2/3$ of the $y$'s, $\exists x \in S$ such that $h(x)=y$ .
\end{claim}
\begin{claim} If $|S| < \frac{2^m}{100}$, for any hash function
$h$ at most $16/100 < 1/6$ of the y's have $x \in S$ such that $h(x)
= y$.
\end{claim}
Thus the following protocol verifies that $S$ is large: Merlin sends a
random hash function $h:\{0,1\}^m \rightarrow \{0,1\}^{m-4}$ to the
Arthur. Arthur sends a random $y \in \{0,1\}^{m-4}$ back to Merlin. If
$S$ is large enough, then Merlin is able to find $x \in S$ such that
$h(x) = y$ to send to Arthur. Otherwise, Merlin will succeed in
finding a suitable $x$ with a probability less than a third (actually,
less than $1/6$) as required.
Notice that Arthur does not even need to be able to verify that $x \in
S$ efficiently (i.e., we do not need $S \in BPP$). As long as $S \in
IP$, we can verify the largeness of S by continuing the above protocol
by executing the protocol that verifies membership in $S$!
\subsection{1-round protocols: equivalence of public and private coins}
We will show that a 1-round IP protocol can be simulated by an
Arthur-Merlin game with a constant number of rounds. This can be
extended to any number of rounds of IP.
The goal is to provide an AM[$O(1)$] protocol to decide an arbitrary IP[1]
protocol. We consider what an ``optimal'' prover would be able to
do in a given round. The prover should be able to commit to an answer from
the question asked such that for a large number of (private) random
coin tosses in the future, this answer will lead the verifier to accept.
We first fix a verifier $V$ with 2/3 completeness and a soundness
1/poly. Let $Q$ be the set of questions $V$ can ask. For each $q \in
Q$, let $S_q$ be the set of (private) random strings that lead to $q$
being asked. Let $r$ be the length of the random strings. If the IP[1]
protocol accepts an input, we want our AM protocol to
accept on at least $(2/3) 2^r$ random strings. Otherwise, we want it to
accept over a very small number of strings, say
1/poly($r$). In other words,
$$
\sum_{q \in Q} |S_q|
\left \{
\begin{array}{ll}
\geq (2/3)2^r & \mbox {if verifier accepts}\\
\leq 1/\textrm{poly}(r) & \mbox {if verifier rejects}\\
\end{array}
\right.
$$
For simplicity assume that some of the questions are
never asked, and for all other questions, exactly $2^l$ random strings
lead to that question: $|S_q| = 0$ or $|S_q| = 2^l$. For this
we need at least $(2/3) 2^{r-l}$ questions for which prover has an
answer that the verifier will accept. To show this, we use the
Goldwasser-Sipser protocol: Merlin sends a
hash funchtion h; Arthur responds with a $y$; Merlin replies with $q \in Q$
such that $h(q)=y$. To complete the proof, Arthur needs to verify that
$|S_q| \geq 2^l$. This is accomplished by another Goldwasser-Sipser
protocol (with $q$, now known to both sides, as input).
\section{$AM[k]=AM$}
The idea is that Arthur corresponds to the BP operator; Merlin
corresponds to the $\exists$ operator. Thus AMAM$(x)$ =
BP$\exists$BP$\exists(x)$. We can exchange the BP and $\exists$
operators as in the proof of Toda's theorem.
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