IP (complexity)

In order to demonstrate that IP ⊆ PSPACE, we present a simulation of an interactive proof system by a polynomial space machine. Now, we can define:

${\displaystyle \Pr[V{\text{ accepts }}w{\text{ starting at }}M_{j}]=\max \nolimits _{P}\Pr \left[V\leftrightarrow P{\text{ accepts }}w{\text{ starting at }}M_{j}\right]}$

and for every 0 ≤ j ≤ p and every message history M_j, we inductively define the function N_Mj:

${\displaystyle N_{M_{j}}={\begin{cases}0&j=p{\text{ and }}m_{p}={\text{reject}}\\1&j=p{\text{ and }}m_{p}={\text{accept}}\\\max _{m_{j+1}}N_{M_{j+1}}&j<p{\text{ and }}j{\text{ is odd}}\\{\text{wt-avg}}_{m_{j+1}}N_{M_{j+1}}&j<p{\text{ and }}j{\text{ is even}}\\\end{cases}}}$

where:

${\displaystyle {\text{wt-avg}}_{m_{j+1}}N_{M_{j+1}}:=\sum \nolimits _{m_{j+1}}\Pr \nolimits _{r}[V(w,r,M_{j})=m_{j+1}]N_{M_{j+1}}}$

where Pr_r is the probability taken over the random string r of length p. This expression is the average of N_Mj+1, weighted by the probability that the verifier sent message m_j+1.

Take M₀ to be the empty message sequence, here we will show that N_M0 can be computed in polynomial space, and that N_M0 = Pr[V accepts w]. First, to compute N_M0, an algorithm can recursively calculate the values N_Mj for every j and M_j. Since the depth of the recursion is p, only polynomial space is necessary. The second requirement is that we need N_M0 = Pr[V accepts w], the value needed to determine whether w is in A. We use induction to prove this as follows.

We must show that for every 0 ≤ j ≤ p and every M_j, N_Mj = Pr[V accepts w starting at M_j], and we will do this using induction on j. The base case is to prove for j = p. Then we will use induction to go from p down to 0.

The base case of j = p is fairly simple. Since m_p is either accept or reject, if m_p is accept, N_Mp is defined to be 1 and Pr[V accepts w starting at M_j] = 1 since the message stream indicates acceptance, thus the claim is true. If m_p is reject, the argument is very similar.

For the inductive hypothesis, we assume that for some j+1 ≤ p and any message sequence M_j+1, N_Mj+1 = Pr[V accepts w starting at M_j+1] and then prove the hypothesis for j and any message sequence M_j.

If j is even, m_j+1 is a message from V to P. By the definition of N_Mj,

${\displaystyle N_{M_{j}}=\sum \nolimits _{m_{j+1}}\Pr \nolimits _{r}\left[V(w,r,M_{j})=m_{j+1}\right]N_{M_{j+1}}.}$

Then, by the inductive hypothesis, we can say this is equal to

${\displaystyle \sum \nolimits _{m_{j+1}}\Pr \nolimits _{r}\left[V(w,r,M_{j})=m_{j+1}\right]*\Pr \left[V{\text{ accepts }}w{\text{ starting at }}M_{j+1}\right].}$

Finally, by definition, we can see that this is equal to Pr[V accepts w starting at M_j].

If j is odd, m_j+1 is a message from P to V. By definition,

${\displaystyle N_{M_{j}}=\max \nolimits _{m_{j+1}}N_{M_{j+1}}.}$

Then, by the inductive hypothesis, this equals

${\displaystyle \max \nolimits _{m_{j+1}}*\Pr[V{\text{ accepts }}w{\text{ starting at }}M_{j+1}].}$

This is equal to Pr[V accepts w starting at M_j] since:

${\displaystyle \max \nolimits _{m_{j+1}}\Pr[V{\text{ accepts }}w{\text{ starting at }}M_{j+1}]\leq \Pr[V{\text{ accepts w starting at }}M_{j}]}$

because the prover on the right-hand side could send the message m_j+1 to maximize the expression on the left-hand side. And:

${\displaystyle \max \nolimits _{m_{j+1}}\Pr \left[V{\text{ accepts }}w{\text{ starting at }}M_{j+1}\right]\geq \Pr \left[V{\text{ accepts }}w{\text{ starting at }}M_{j}\right]}$

Since the same Prover cannot do any better than send that same message. Thus, this holds whether i is even or odd and the proof that IP ⊆ PSPACE is complete.

Here we have constructed a polynomial space machine that uses the best prover P for a particular string w in language A. We use this best prover in place of a prover with random input bits because we are able to try every set of random input bits in polynomial space. Since we have simulated an interactive proof system with a polynomial space machine, we have shown that IP ⊆ PSPACE, as desired.

In order to illustrate the technique that will be used to prove PSPACE ⊆ IP, we will first prove a weaker theorem, which was proven by Lund, et al.: #SAT ∈ IP. Then using the concepts from this proof we will extend it to show that TQBF ∈ IP. Since TQBF ∈ PSPACE-complete, and TQBF ∈ IP then PSPACE ⊆ IP.

We begin by showing that #SAT is in IP, where:

${\displaystyle \#{\text{SAT}}=\left\{\langle \varphi ,k\rangle \$ :\ \varphi {\text{ is a CNF-formula with exactly }}k{\text{ satisfying assignments}}\right\}.}

Note that this is different from the normal definition of #SAT, in that it is a decision problem, rather than a function.

First we use arithmetization to map the boolean formula with n variables, φ(b₁, ..., b_n) to a polynomial p_φ(x₁, ..., x_n), where p_φ mimics φ in that p_φ is 1 if φ is true and 0 otherwise provided that the variables of p_φ are assigned Boolean values. The Boolean operations ∨, ∧ and ¬ used in φ are simulated in p_φ by replacing the operators in φ as shown in the table below.

As an example, φ = a ∧ (b ∨ ¬c) would be converted into a polynomial as follows:

${\displaystyle {\begin{aligned}p_{\varphi }&=a\wedge (b\vee \neg c)\\&=a\wedge \left(b*(1-c)\right)\\&=a\wedge \left(1-(1-b)(1-(1-c))\right)\\&=a\left(1-(1-b)(1-(1-c))\right)\\&=a-(ac-abc)\end{aligned}}}$

The operations ab and a ∗ b each result in a polynomial with a degree bounded by the sum of the degrees of the polynomials for a and b and hence, the degree of any variable is at most the length of φ.

Now let F be a finite field with order q > 2ⁿ; also demand that q be at least 1000. For each 0 ≤ i ≤ n, define a function f_i on F, having parameters ${\displaystyle a_{1},\dots ,a_{i-1}\in F}$, and a single variable a_i in F: For 0 ≤ i ≤ n and for ${\displaystyle a_{1},\dots ,a_{i}\in F}$ let

${\displaystyle f_{i}(a_{1},\dots ,a_{i})=\sum \nolimits _{a_{i+1},\dots ,a_{n}\in \{0,1\}}p(a_{1},\dots ,a_{n}).}$

Note that the value of f₀ is the number of satisfying assignments of φ. f₀ is a void function, with no variables.

Now the protocol for #SAT works as follows:

• Phase 0: The prover P chooses a prime q > 2ⁿ and computes f₀, it then sends q and f₀ to the verifier V. V checks that q is a prime greater than max(1000, 2ⁿ) and that f₀() = k.
• Phase 1: P sends the coefficients of f₁(z) as a polynomial in z. V verifies that the degree of f₁ is less than n and that f₀ = f₁(0) + f₁(1). (If not V rejects). V now sends a random number r₁ from F to P.
• Phase i: P sends the coefficients of ${\displaystyle f_{i}(r_{1},\dots ,r_{i-1},z)}$ as a polynomial in z. V verifies that the degree of f_i is less than n and that ${\displaystyle f_{i-1}(r_{1},\dots ,r_{i-1})=f_{i}(r_{1},\dots ,r_{i-1},0)+f_{i}(r_{1},\dots ,r_{i-1},1)}$. (If not V rejects). V now sends a random number r_i from F to P.
• Phase n+1: V evaluates ${\displaystyle p(r_{1},\dots ,r_{n})}$ to compare to the value ${\displaystyle f_{n}(r_{1},\dots ,r_{n})}$. If they are equal V accepts, otherwise V rejects.

Note that this is a public-coin algorithm.

If φ has k satisfying assignments, clearly V will accept. If φ does not have k satisfying assignments we assume there is a prover ${\displaystyle {\tilde {P}}}$ that tries to convince V that φ does have k satisfying assignments. We show that this can only be done with low probability.

To prevent V from rejecting in phase 0, ${\displaystyle {\tilde {P}}}$ has to send an incorrect value ${\displaystyle {\tilde {f}}_{0}()}$ to P. Then, in phase 1, ${\displaystyle {\tilde {P}}}$ must send an incorrect polynomial ${\displaystyle {\tilde {f}}_{1}}$ with the property that ${\displaystyle {\tilde {f}}_{1}(0)+{\tilde {f}}_{1}(1)={\tilde {f}}_{0}()}$. When V chooses a random r₁ to send to P,

${\displaystyle \Pr \left[{\tilde {f}}_{1}(r_{1})=f_{1}(r_{1})\right]<{\tfrac {1}{n^{2}}}.}$

This is because a polynomial in a single variable of degree at most d can have no more than d roots (unless it always evaluates to 0). So, any two polynomials in a single variable of degree at most d can be equal only in d places. Since |F| > 2ⁿ the chances of r₁ being one of these values is at most ${\displaystyle n/2^{n}<n/n^{3}}$ if n > 10, or at most (n/1000) ≤ (n/n³) if n ≤ 10.

Generalizing this idea for the other phases we have for each 1 ≤ i ≤ n if

${\displaystyle {\tilde {f}}_{i-1}(r_{1},\dots ,r_{i-1})\neq f_{i-1}(r_{1},\dots ,r_{i-1}),}$

then for r_i chosen randomly from F,

${\displaystyle \Pr \left[{\tilde {f}}(r_{1},\dots ,r_{i})=f_{i}(r_{1},\dots ,r_{i})\right]\leq {\tfrac {1}{n^{2}}}.}$

There are n phases, so the probability that ${\displaystyle {\tilde {P}}}$ is lucky because V selects at some stage a convenient r_i is at most 1/n. So, no prover can make the verifier accept with probability greater than 1/n. We can also see from the definition that the verifier V operates in probabilistic polynomial time. Thus, #SAT ∈ IP.

In order to show that PSPACE is a subset of IP, we need to choose a PSPACE-complete problem and show that it is in IP. Once we show this, then it clear that PSPACE ⊆ IP. The proof technique demonstrated here is credited to Adi Shamir.

We know that TQBF is in PSPACE-Complete. So let ψ be a quantified boolean expression:

${\displaystyle \psi ={\mathsf {Q}}_{1}x_{1}\dots {\mathsf {Q}}_{m}x_{m}[\varphi ]}$

where φ is a CNF formula. Then Q_i is a quantifier, either ∃ or ∀. Now f_i is the same as in the previous proof, but now it also includes quantifiers.

${\displaystyle f_{i}(a_{1},\dots ,a_{i})={\begin{cases}f_{i}(a_{1},\dots ,a_{m})=1&{\mathsf {Q}}_{i+1}x_{i+1}\dots {\mathsf {Q}}_{m}x_{m}[\varphi (a_{1},\dots ,a_{i})]{\text{ is true}}\\0&{\text{otherwise}}\end{cases}}}$

Here, φ(a₁, ..., a_i) is φ with a₁ to a_i substituted for x₁ to x_i. Thus f₀ is the truth value of ψ. In order to arithmetize ψ we must use the following rules:

${\displaystyle f_{i}(a_{1},\dots ,a_{i})={\begin{cases}f_{i+1}(a_{1},\dots ,a_{i},0)\cdot f_{i+1}(a_{1},\dots ,a_{i},1)&{\mathsf {Q}}_{i+1}=\forall \\f_{i+1}(a_{1},\dots ,a_{i},0)*f_{i+1}(a_{1},\dots ,a_{i},1)&{\mathsf {Q}}_{i+1}=\exists \end{cases}}}$

where as before we define x ∗ y = 1 − (1 − x)(1 − y).

By using the method described in #SAT, we must face a problem that for any f_i the degree of the resulting polynomial may double with each quantifier. In order to prevent this, we must introduce a new reduction operator R which will reduce the degrees of the polynomial without changing their behavior on Boolean inputs.

So now before we arithmetize ${\displaystyle \psi ={\mathsf {Q}}_{1}x_{1}\dots {\mathsf {Q}}_{m}x_{m}[\varphi ]}$ we introduce a new expression:

${\displaystyle \psi '={\mathsf {Q}}_{1}\mathrm {R} x_{1}{\mathsf {Q}}_{2}\mathrm {R} x_{1}\mathrm {R} x_{2}\dots {\mathsf {Q}}_{m}\mathrm {R} x_{1}\dots \mathrm {R} x_{m}[\varphi ]}$

or put another way:

${\displaystyle \psi '={\mathsf {S}}_{1}y_{1}\dots {\mathsf {S}}_{k}y_{k}[\varphi ],\qquad {\text{ where }}{\mathsf {S}}_{i}\in \{\forall ,\exists ,\mathrm {R} \},\ y_{i}\in \{x_{1},\dots ,x_{m}\}}$

Now for every i ≤ k we define the function f_i. We also define ${\displaystyle f_{k}(x_{1},\dots ,x_{m})}$ to be the polynomial p(x₁, ..., x_m) which is obtained by arithmetizing φ. Now in order to keep the degree of the polynomial low, we define f_i in terms of f_i+1:

${\displaystyle {\text{If }}{\mathsf {S}}_{i+1}=\forall ,\quad f_{i}(a_{1},\dots ,a_{i})=f_{i+1}(a_{1},\dots ,a_{i},0)\cdot f_{i+1}(a_{1},\dots ,a_{i},1)}$

${\displaystyle {\text{If }}{\mathsf {S}}_{i+1}=\exists ,\quad f_{i}(a_{1},\dots ,a_{i})=f_{i+1}(a_{1},\dots ,a_{i},0)*f_{i+1}(a_{1},\dots ,a_{i},1)}$

${\displaystyle {\text{If }}{\mathsf {S}}_{i+1}=\mathrm {R} ,\quad f_{i}(a_{1},\dots ,a_{i},a)=(1-a)f_{i+1}(a_{1},\dots ,a_{i},0)+af_{i+1}(a_{1},\dots ,a_{i},1)}$

Now we can see that the reduction operation R, doesn't change the degree of the polynomial. Also it is important to see that the R_x operation doesn't change the value of the function on boolean inputs. So f₀ is still the truth value of ψ, but the R_x value produces a result that is linear in x. Also after any ${\displaystyle {\mathsf {Q}}_{i}x_{i}}$ we add ${\displaystyle \mathrm {R} _{x_{1}}\dots \mathrm {R} _{x_{i}}}$ in ψ′ in order to reduce the degree down to 1 after arithmetizing ${\displaystyle {\mathsf {Q}}_{i}}$.

Now let's describe the protocol. If n is the length of ψ, all arithmetic operations in the protocol are over a field of size at least n⁴ where n is the length of ψ.

• Phase 0: P → V: P sends f₀ to V. V checks that f₀= 1 and rejects if not.
• Phase 1: P → V: P sends f₁(z) to V. V uses coefficients to evaluate f₁(0) and f₁(1). Then it checks that the polynomial's degree is at most n and that the following identities are true:

${\displaystyle f_{0}(\varnothing )={\begin{cases}f_{1}(0)\cdot f_{1}(1)&{\text{ if }}{\mathsf {S}}=\forall \\f_{1}(0)*f_{1}(1)&{\text{ if }}{\mathsf {S}}=\exists .\\(1-r)f_{1}(0)+rf_{1}(1)&{\text{ if }}{\mathsf {S}}=\mathrm {R} .\end{cases}}}$

If either fails then reject.

• Phase i: P → V: P sends ${\displaystyle f_{i}(r_{1},\dots ,r_{i-1},z)}$ as a polynomial in z. r₁ denotes the previously set random values for ${\displaystyle r_{1},\dots ,r_{i-1}}$

V uses coefficients to evaluate ${\displaystyle f_{i}(r_{1},\dots ,r_{i-1},0)}$ and ${\displaystyle f_{i}(r_{1},\dots ,r_{i-1},1)}$. Then it checks that the polynomial degree is at most n and that the following identities are true:

${\displaystyle f_{i-1}(r_{1},\dots ,r_{i-1})={\begin{cases}f_{i}(r_{1},\dots ,r_{i-1},0)\cdot f_{i}(r_{1},\dots ,r_{i-1},1)&{\mathsf {S}}=\forall \\f_{i}(r_{1},\dots ,r_{i-1},0)*f_{i}(r_{1},\dots ,r_{i-1},1)&{\mathsf {S}}=\exists .\end{cases}}}$

${\displaystyle f_{i-1}(r_{1}\dots r)=(1-r)f_{i}(r_{1},\dots ,r_{i-1},0)+rf_{i}(r_{1},\dots ,r_{i-1},1){\text{ if }}{\mathsf {S}}=\mathrm {R} .}$

If either fails then reject.

V → P: V picks a random r in F and sends it to P. (If ${\displaystyle {\mathsf {S}}=\mathrm {R} }$ then this r replaces the previous r).

Goto phase i + 1 where P must persuade V that ${\displaystyle f_{i}(r_{1},\dots ,r)}$ is correct.

• Phase k + 1: V evaluates ${\displaystyle p(r_{1},\dots ,r_{m})}$. Then it checks if ${\displaystyle p(r_{1},\dots ,r_{m})=f_{k}(r_{1},\dots ,r_{m})}$ If they are equal then V accepts, otherwise V rejects.

This is the end of the protocol description.

If ψ is true then V will accept when P follows the protocol. Likewise if ${\displaystyle {\tilde {P}}}$ is a malicious prover which lies, and if ψ is false, then ${\displaystyle {\tilde {P}}}$ will need to lie at phase 0 and send some value for f₀. If at phase i, V has an incorrect value for ${\displaystyle f_{i-1}(r_{1},\dots )}$ then ${\displaystyle f_{i}(r_{1},\dots ,0)}$ and ${\displaystyle f_{i}(r_{1},\dots ,1)}$ will likely also be incorrect, and so forth. The probability for ${\displaystyle {\tilde {P}}}$ to get lucky on some random r is at most the degree of the polynomial divided by the field size: ${\displaystyle n/n^{4}}$. The protocol runs through O(n²) phases, so the probability that ${\displaystyle {\tilde {P}}}$ gets lucky at some phase is ≤ 1/n. If ${\displaystyle {\tilde {P}}}$ is never lucky, then V will reject at phase k+1.

Since we have now shown that both IP ⊆ PSPACE and PSPACE ⊆ IP, we can conclude that IP = PSPACE as desired. Moreover, we have shown that any IP algorithm may be taken to be public-coin, since the reduction from PSPACE to IP has this property.

A subset of IP is the deterministic Interactive Proof class, which is similar to IP but has a deterministic verifier (i.e. with no randomness). This class is equal to NP.

An equivalent definition of IP replaces the condition that the interaction succeeds with high probability on strings in the language with the requirement that it always succeeds:

${\displaystyle w\in L\Rightarrow \Pr[V\leftrightarrow P{\text{ accepts }}w]=1}$

This seemingly stronger criterion of "perfect completeness" does not change the complexity class IP, since any language with an interactive proof system may be given an interactive proof system with perfect completeness.[4]

In 1988, Goldwasser et al. created an even more powerful interactive proof system based on IP called MIP in which there are two independent provers. The two provers cannot communicate once the verifier has begun sending messages to them. Just as it's easier to tell if a criminal is lying if he and his partner are interrogated in separate rooms, it's considerably easier to detect a malicious prover trying to trick the verifier if there is another prover it can double-check with. In fact, this is so helpful that Babai, Fortnow, and Lund were able to show that MIP = NEXPTIME, the class of all problems solvable by a nondeterministic machine in exponential time, a very large class. Moreover, all languages in NP have zero-knowledge proofs in an MIP system, without any additional assumptions; this is only known for IP assuming the existence of one-way functions.

IPP (unbounded IP) is a variant of IP where we replace the BPP verifier by a PP verifier. More precisely, we modify the completeness and soundness conditions as follows:

• Completeness: if a string is in the language, the honest verifier will be convinced of this fact by an honest prover with probability at least 1/2.
• Soundness: if the string is not in the language, no prover can convince the honest verifier that it is in the language, except with probability less than 1/2.

Although IPP also equals PSPACE, IPP protocols behaves quite differently from IP with respect to oracles: IPP=PSPACE with respect to all oracles, while IP ≠ PSPACE with respect to almost all oracles.[5]

QIP is a version of IP replacing the BPP verifier by a BQP verifier, where BQP is the class of problems solvable by quantum computers in polynomial time. The messages are composed of qubits.[6] In 2009, Jain, Ji, Upadhyay, and Watrous proved that QIP also equals PSPACE,[7] implying that this change gives no additional power to the protocol. This subsumes a previous result of Kitaev and Watrous that QIP is contained in EXPTIME because QIP = QIP[3], so that more than three rounds are never necessary.[8]

Whereas IPP and QIP give more power to the verifier, a compIP system (competitive IP proof system) weakens the completeness condition in a way that weakens the prover:

• Completeness: if a string is in the language L, the honest verifier will be convinced of this fact by an honest prover with probability at least 2/3. Moreover, the prover will do so in probabilistic polynomial time given access to an oracle for the language L.

Essentially, this makes the prover a BPP machine with access to an oracle for the language, but only in the completeness case, not the soundness case. The concept is that if a language is in compIP, then interactively proving it is in some sense as easy as deciding it. With the oracle, the prover can easily solve the problem, but its limited power makes it much more difficult to convince the verifier of anything. In fact, compIP isn't even known or believed to contain NP.

On the other hand, such a system can solve some problems believed to be hard. Somewhat paradoxically, though such a system is not believed to be able to solve all of NP, it can easily solve all NP-complete problems due to self-reducibility. This stems from the fact that if the language L is not NP-hard, the prover is substantially limited in power (as it can no longer decide all NP problems with its oracle).

Additionally, the graph nonisomorphism problem (which is a classical problem in IP) is also in compIP, since the only hard operation the prover has to do is isomorphism testing, which it can use the oracle to solve. Quadratic non-residuosity and graph isomorphism are also in compIP.[9] Note, Quadratic non-residuosity (QNR) is likely an easier problem than graph isomorphism as QNR is in UP intersect co-UP.[10]