Suggest edit — Non-deterministic Turing Machine

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title: "Non-deterministic Turing Machine"
source: https://www.jemoka.com/posts/kbhnon_deterministic_turing_machines/
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We have multiple transitions for a state, symbol pair in non-deterministic TMs.
the machine may proceed according to several possible transitions the machine accepts an input if there exists an accepting computation history for the machine on the string Here&rsquo;s the
basically a turing machine except the transition is a subset instead of a single transition.
$L$ is in $NP$ IFF there are polynomial-length proofs (&ldquo;witnesses&rdquo;) that can be decided efficiently for membership in $L$ There exists some constant $k$ and a polynomial-time Turing-machine $V$,
\begin{equation} L = \qty {x \mid \exists y \in \Sigma^{*}, |y| \leq |x|^{k}, V(x,y) \text{ accepts}} \end{equation}
IFF $L \in NP$
We can call the $V$ the polynomial length proof; we can call $y$ the &ldquo;witness&rdquo;
Proof
if $L$ exists, we create an $N$ to guess $y$ of length at most $|x|^{k}$ nondeterministically, run $V(x,y)$ on each (which is polynomial time), and output; this means that $L(N)$ is $L$ if there is an NTM $N$ that decides $L$ (so $L \in NP$), we define $V(x,y)$ to accept IFF $y$ encodes an accepting computation history of $N$ on $x$ (now, $|y| \leq |x|^{k}$ because $N$ is an NP NTM, and $V(x,y)$ is polynomial because we are just iterating over the computation history) Boolean Formula Satisfiability Boolean formulas are logical expressions, like:
\begin{equation} \phi = \qty(\neg x \wedge { y}) \vee z \end{equation}
A satisfying assignment is a setting of variables which makes the formula above true.
A Boolean formula is satisfiable if there is a true/false setting to the variables that makes the formula true.
\begin{equation} SAT = \qty {\phi \mid \phi\text{ is satisfiable}} \end{equation}
3cnf-formula This is a particular type of satisfiable formula
\begin{equation} \qty(x_1 \vee \neg x_2 \vee x_3) \wedge\qty(x_4 \vee x_2 \vee x_5) \wedge\qty(x_3 \vee x_2 \vee \neg x_1) \end{equation}
\begin{equation} 3SAT = \qty {\phi \mid \phi\text{ is satisfiable and 3cnf}} \end{equation}
We claim that $3SAT \in NP$; we can express $3SAT$ as:
\begin{equation} 3SAT = \qty {\phi \mid \phi\text{ is in 3cnf and $\exists$ a string $y$ \hat{t} encodes an satisfyingly assignment}} \end{equation}
in particular,
\begin{equation} 3SAT-CHECK = \qty {(\phi, y) \mid \phi\text{ is in 3cnf and $y$ is a satisfying assignment}} \end{equation}
is in $P$.
(actually all of SAT is in $NP$)
Time Complexity of Non-deterministic Turing Machines First,
\begin{equation} \text{TIME}(t(n)) \subseteq \text{NTIME}(t(n)) \end{equation}
where $\text{NTIME}$ is the set of languages such that there exists a Non-deterministic Turing Machine which decides the language within $t(n)$ time.
accepting accepting on Non-deterministic Turing Machines is a history of $N$ on $w$ such that we have a sequence $c_0, &hellip;, c_{t}$ where
$C_{0}$ is the start configuration of $q_0 w$ $C_{t}$ is an accepting configuration (we are in an accepting state) each configuration $C_{i}$ yields $C_{i+1}$ accepting in time $t$ means that such a history exists.
$N$ has a time complexity of $T(n)$ if for all $n$, for all inputs of length $n$ and for all histories, $N$ halts in $T(n)$ time. (we can get it to halt for all strings by having a clock, and computing $n$ and stopping $T$ as needed; if $n$ is correct, then we don&rsquo;t actually change the language that $T$ recognized)
non deterministic turing machines are not more powerful than turing machines Theorem: every non-deterministic Turing machine $N$ can be transformed into a Turing machine $M$ that accepts only strings $N$ accepts.
(We don&rsquo;t care about rejection in this case, because &ldquo;some rejection state&rdquo; seems odd if there are some that&rsquo;s not a rejection state)
Proof:
We pick an ordering of strings over $\qty {Q \cup \Gamma \cup \#}^{*}$.
To check if we accept $w$ with $M$: for all strings in $\qty {Q \cup \Gamma \cup \#}^{*}$, check if for $D \in \qty {Q \cup \Gamma \cup \#}$, and $D = C_0 \# &hellip; \# C_{k}$, if the sequence $C_0 &hellip; C_{k}$ is an accepting computation history in $N$ (that is, $C_0$ is a valid start configuration, and $C_{k}$ is (1) a valid accepting one) that (2) $C_{0}$ corresponds to the string $w$. If so, accept.
(i.e. we precompute paths)
clique problem a $k$ clique is a subgraph which is &ldquo;complete&rdquo; (all possible edges are connected) with $k$ nodes
assume some encoding of graphs (adjacency matricies):
the language
\begin{equation} \text{CLIQUE} = \qty {(G, k) \mid G\text{ is undirected with $k$ clique}} \end{equation}
In particular, we claim that $\text{CLIQUE} \in \text{NTIME}\qty (n^{c})$ for some $c&gt;1$. We do this by&hellip;.
nondeterministically guess a subset $S \subseteq V$ with $|S| = k$ for all pairs of $u,v \in S$, if $(u,v)$ is not in $E$, then reject accept this is NP-Complete, see also many, many NP-Complete thins
Hamiltonian path problem A Hamiltonian path is a path which goes through an node exactly once
\begin{equation} \text{HAMPATH} = \qty {(G, s,t) \mid G\text{ is directed with path $s,k$}} \end{equation}
this is also $\text{HAMPATH} \in \text{NTIME}\qty (n^{c})$ for some $c&gt;1$. We do this by&hellip;.
Again find all paths and guess and check: guess a sequence of paths $v_1, &hellip;, v_{|v|}$, etc.