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# 数学代写|计算复杂度理论代写Computational Complexity Theory代考|ENCM517 Time versus alternations: time-space tradeoffs for SAT

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## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|Time versus alternations: time-space tradeoffs for SAT

Despite the fact that SAT is widely believed to require exponential (or at least super-polynomial) time to solve, and to require linear (or at least super-logarithmic) space, we currently have no way to prove these conjectures. In fact, as far as we know, SAT may have both a linear time algorithm and a logarithmic space one. Nevertheless, we can prove that SAT does not have an algorithm that runs simultaneously in linear time and logarithmic space. In fact, we can prove the following stronger theorem:
THEOREM $5.13$ (??)
For every two functions $S, T: \mathbb{N} \rightarrow \mathbb{N}$, define $\operatorname{TISP}(T(n), S(n))$ to be the set of languages decided by a $T M M$ that on every input $x$ takes at most $O(T(|x|))$ steps and uses at most $O(S(|x|))$ cells of its read/write tapes. Then, SAT $\notin \mathbf{T I S P}\left(n^{1.1}, n^{0.1}\right)$.

The class $\operatorname{TISP}(T(n), S(n))$ is typically defined with respect to TM’s with RAM memory (i.e., TM’s that have random access to their tapes; such machines can be defined in a similar way to the definition of oracle TM’s in Section 3.5). Theorem $5.13$ and its proof carries over for that model as well. We also note that a stronger result is known for both models: for every $c<(\sqrt{5}+1) / 2$, there exists $d>0$ such that SAT $\notin \mathbf{T I S P}\left(n^c, n^d\right)$ and furthermore, $d$ approaches 1 from below as $c$ approaches 1 from above.
ProOF: We will show that
$$\boldsymbol{N T I M E}(n) \nsubseteq \mathbb{\operatorname { T I S }}\left(n^{1.2}, n^{0.2}\right)$$

## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|Defining the hierarchy via oracle machines

Recall the definition of oracle machines from Section 3.5. These are machines that are executed with access to a special tape they can use to make queries of the form “is $q \in O$ ” for some language $O$. For every $O \subseteq{0,1}^*$, oracle TM $M$ and input $x$, we denote by $M^O(x)$ the output of $M$ on $x$ with access to $O$ as an oracle. We have the following characterization of the polynomial hierarchy:
THEOREM $5.15$
For every $i \geq 2, \boldsymbol{\Sigma}i^p=\mathbf{N P}^{\boldsymbol{\Sigma}{i-1} \mathrm{SAT}}$, where the latter class denotes the set of languages decided by polynomial-time NDTMs with access to the oracle $\boldsymbol{\Sigma}_{i-1}$ SAT.

Proof: We showcase the idea by proving that $\boldsymbol{\Sigma}_2^p=\mathbf{N P}^{\mathrm{SAT}}$. Suppose that $L \in \boldsymbol{\Sigma}_2^p$. Then, there is a polynomial-time TM $M$ and a polynomial $q$ such that
$$x \in L \Leftrightarrow \exists u_1 \in{0,1}^{q(|x|)} \forall u_2 \in{0,1}^{q(|x|)} M\left(x, u_1, u_2\right)=1$$
yet for every fixed $u_1$ and $x$, the statement “for every $u_2, M\left(x, u_1, u_2\right)=1$ ” is the negation of an NP-statement and hence its truth can be determined using an oracle for SAT. We get that there is a simple NDTM $N$ that given oracle access for SAT can decide $L$ : on input $x$, non-deterministically guess $u_1$ and use the oracle to decide if $\forall u_2 M\left(x, u_1, u_2\right)=1$. We see that $x \in L$ iff there exists a choice $u_1$ that makes $N$ accept.

## 数学代写|计算复杂度理论代写Computational Complexity Theory代 考|Time versus alternations: time-space tradeoffs for SAT

ProOF: 我们将证明
$$\boldsymbol{N T I M E}(n) \nsubseteq \operatorname{TIS}\left(n^{1.2}, n^{0.2}\right)$$

## 数学代写|计算复杂度理论代写Computational Complexity Theory代 考|Defining the hierarchy via oracle machines

: $5.15$

$$x \in L \Leftrightarrow \exists u_1 \in 0,1^{q(|x|)} \forall u_2 \in 0,1^{q(|x|)} M\left(x, u_1, u_2\right)=1$$

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