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# 数学代写|计算复杂度理论代写Computational Complexity Theory代考|CS58400 PSPACE completeness

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## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|PSPACE completeness

As already indicated, we do not know if $\mathbf{P} \neq$ PSPACE, though we strongly believe that the answer is YES. Notice, $\mathbf{P}=$ PSPACE implies $\mathbf{P}=\mathbf{N P}$. Since complete problems can help capture the essence of a complexity class, we now present some complete problems for PSPACE.
DEFINITION $4.8$
A language $A$ is PSPACE-hard if for every $L \in$ PSPACE, $L \leq_p A$. If in addition $A \in$ PSPACE then $A$ is PSPACE-complete.

Using our observations about polynomial-time reductions from Chapter ?? we see that if any PSPACE-complete language is in $\mathbf{P}$ then so is every other language in PSPACE. Viewed contrapostively, if PSPACE $\neq \mathbf{P}$ then a PSPACE-complete language is not in $\mathbf{P}$. Intuitively, a PSPACE-complete language is the “most difficult” problem of PSPACE. Just as NP trivially contains NP-complete problems, so does PSPACE. The following is one (Exercise 3):
SPACETM $=\left{\left\langle M, w, 1^n\right\rangle:\right.$ DTM $M$ accepts $w$ in space $\left.n\right}$.

## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|Savitch’s theore

The astute reader may notice that because the above proof uses the notion of a configuration graph and does not require this graph to have out-degree one, it actually yields a stronger statement: that TQBF is not just hard for PSPACE but in fact even for NPSPACE!. Since TQBF $\in$ PSPACE this implies that PSPACE = NSPACE, which is quite surprising since our intuition is that the corresponding classes for time ( $\mathbf{P}$ and $\mathbf{N P})$ are different. In fact, using the ideas of the above proof, one can obtain the following theorem:
THEOREM $4.12$ (SAVITCH [SAV70])
For any space-constructible $S: \mathbb{N} \rightarrow \mathbb{N}$ with $S(n) \geq \log n, \operatorname{NSPACE}(S(n)) \subseteq \mathbf{S P A C E}\left(S(n)^2\right)$
We remark that the running time of the algorithm obtained from this theorem can be as high as $2^{\Omega\left(s(n)^2\right)}$.
Proof: The proof closely follows the proof that TQBF is PSPACE-complete. Let $L \in \operatorname{NSPACE}(S(n))$ be a language decided by a TM $M$ such that for every $x \in{0,1}^n$, the configuration graph $G=G_{M, x}$ has at most $M=2^{O(S(n))}$ vertices. We describe a recursive procedure REACH? $(u, v, i)$ that returns “YES” if there is a path from $u$ to $v$ of length at most $2^i$ and “NO” otherwise. Note that REACH? $(s, t,\lceil\log M\rceil)$ is “YES” iff $t$ is reachable from $s$. Again, the main observation is that there is a path from $u$ to $v$ of length at most $2^i$ iff there’s a vertex $z$ with paths from $u$ to $z$ and from $z$ to $v$ of lengths at most $2^{i-1}$. Thus, on input $u, v, i$, REACH? will enumerate over all vertices $z$ (at a cost of $O(\log M)$ space) and output “YES” if it finds one $z$ such that REACH? $(u, z, i-1)=$ “YES” and REACH? $(z, v, i-1)=$ “YES”. Once again, the crucial observation is that although the algorithm makes $n$ recursive invocations, it can reuse the space in each of these invocations. Thus, if we let $s_{M, i}$ be the space complexity of REACH? $(u, v, i)$ on an $M$-vertex graph we get that $s_{M, i}=s_{M, i-1}+O(\log M)$ and thus $s_{M, \log M}=O\left(\log ^2 M\right)=O\left(S(n)^2\right)$.

## MATLAB代写

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