A visit to the polynomial-time hierarchy

Thm:

1. LexicoSAT is ${\Delta }_2^P$-complete
2. CircuitSAT is ${\Delta }_2^P$-complete

LexicoSAT:

the set of $\phi (x_1,\cdots ,x_n)$ s.t. $v(x_n)=\top$ in the first (lexicographically) valuation $v$ that satisfies $\phi$

CircuitSAT:

evaluate $x_n$ in the following program/circuit:

  x_1 = ∃ Y_1; F_1(Y_1)
  x_2 = ∃ Y_2; F_2(Y_2, x_1)
  ⋮
  x_n = ∃ Y_n; F_n(Y_n, x_1, ⋯, x_{n-1})

⟶ you call the SAT oracle $n$ times

BUT: you have to store the intermedite $x_i$ (not the the formulas, they’re part of the input), so it’s not trivially in $L^{NP}$

1. ask: is $\phi \in SAT$
2. then, set $x_1=\top$, and is $\phi (x_1\leftarrow \top )\in SAT$
3. ⋮
4. then, set $x_n=\top$, and is $\phi (x_1\leftarrow \top ,\cdots ,x_{n-1}\leftarrow \top )\in SAT$

But again, you have to recall the $x_i$ ⟶ not easily in $L^{NP}$

CircuitSAT is ${\Delta }_2$-hard

NB: Recall that $\exists \forall SAT$ is $\underset{=N{P}^{NP}}{\underbrace{{\Sigma }_2}}$-complete

Let $L\in {\Delta }_2$, $ℳ$ a PTIME det. TM calling SAT (or any other NP-complete problem), write its transition table.

Your $x_i$ are here of the form $x_{t,i}^a$, and

LexicoSAT is ${\Delta }_2$-hard

Let’s show that $CircuitSAT{\preccurlyeq }_{\mathrm{L}}LexicoSAT$

Given $S\stackrel{\scriptscriptstyle\mathrm{def}}{=}$

• $x_1\stackrel{\scriptscriptstyle\mathrm{def}}{=}\exists Y_1,F_1(Y_1)$
• ⋮
• $x_n\stackrel{\scriptscriptstyle\mathrm{def}}{=}\exists Y_n,F_n(Y_n,x_1,\cdots ,x_{n-1})$

where the $Y_i$ are all disjoint

$S^{\prime }$ is the conjunction of

• $\overline{x_1}\vee F_1(Y_1)$
• ⋮
• $\overline{x_n}\vee F_n(Y_n,x_1,\cdots ,x_{n-1})$
• $(\overline{x_n}\vee z)\wedge (x_n\wedge \bar{z})$

with the ordering

1. $S^{\prime }$ is satisifiable (set all the $x_i$ and $z$ to $\perp$ (and everything else as well)).
2. $v_S:x_1,\cdots ,x_n$ in $S$ (when you run the program) can be extended to a satisfying assignment (set $z$ to the same value as $x_n$)

3. any $v$ for $S^{\prime }$ is smaller or equal (lexicographically) than $v_S$

   By contradiction: assume that:

   • $v$ satisifies $S^{\prime }$
   • $v$ agrees with $v_S$ on $x_1,\cdots ,x_i$

   • $v>v_S$ on $x_{i+1}$ (the first position where they disagree):

     • $v(x_{i+1})=\top$
     • $v_S(x_{i+1})=\perp$

     then it means that $v(F_{i+1}(Y_{i+1},x_1,\cdots ,x_i))=\top$, so that there exists indeed a satisfying assignment for the formula $\exists Y_{i+1},F_{i+1}(Y_{i+1},x_1,\cdots ,x_i)$, and by definition of $v_S$, $x_{i+1}$ has to be set to $\top$: contradiction.

ParitySAT

ParitySAT

$$\{{\phi }_1,\cdots ,{\phi }_n\mid \text{ such that an even number of }{\phi }_i\text{ are satisifiable.}\}$$

Obviously:

• $ParitySAT\in {\Delta }_2$: you use the SAT oracle $n$ times and remember the result
• But you also have $ParitySAT\in {\theta }_2^P=L^{NP}$: you use the SAT oracle $n$ times and each time, you remember the last bit (parity).

NB: counting how many of these valuations are satisfying ones is also in ${\theta }_2^P=L^{NP}$, since counting can be done in logspace.

Thm:

1. ParitySAT is ${\Theta }_2^P$-complete
2. $${\Theta }_2^P=L^{NP}=P_{||}^{NP}=P^{NP[O(\log n)]}$$

$P_{||}^{NP}$:

there is just one call to the oracle, but we call all the questions (prepared in advance) in parallel (that is, we can’t use the answer of one question for another question)

$P^{NP[O(r(n))]}$:

you can only call the oracle $\le r(n)$ times

$ParitySAT\in P^{NP[O(\log (n))]}$: call the following oracle $\log n$ times (dichotomic search):

($L\in NP$: guess the $k$ formulas, call SAT)

Lemma: $P^{NP[O(\log (n))]}=P_{||}^{NP}$

$\subseteq$

Let $L\in P^{NP[O(\log (n))]}$, $ℳ$ does $p(n)$ queries at most $c\log n$ times.

${ℳ}^{\prime }$ can simulate the behavior of $ℳ$ by building a tree, branching each time $ℳ$ call the oracle instead of calling it as well ⟹ the overall size of the time is polynomial, since $ℳ$ calls the oracle a logarithmic number of times.

Then, ${ℳ}^{\prime }$ makes all the calls in parallel, and, as a result, knows which path to follow in the tree.

$\supseteq$

Let $L\in P^{NP}$, recognizes by $ℳ$.

With the same trick as the one used to show $ParitySAT\in P^{NP[O(\log (n))]}$, ${ℳ}^{\prime }$ computes the numbered of satisfied queries.

• guess which ones are part of the $k$ ones
• if the computation is accepting (in NP, à la Cook-Levin): check in polynomial time that the $k$ $w_i$ guessed are actually satisifiable.

NB: $N{P}^{NP[1]}=N{P}^{NP}$

indeed: as $N{P}^{NP}$ reduces to $\exists \forall SAT\in N{P}^{NP[1]}$ (guess the number of times you call the oracle)

Boolean Hierarchy (over $NP$) = Difference Hierarchy

• First stage: $NP$ and $coNP$
• Second one: $B_1$ intersection of languages in $NP$ with languages in $coNP$ (by abuse of notation, denoted by $NP\cap coNP$) and $co{B}_1$
• Third one: $B_2=B_1\cup \cdots \cup B_1$ and $co{B}_2$
• ⋮

Alternation of union and intersection (with complementary).