Question

Let $\mathit{U}\mathbf{=}\mathbf{\backslash }\mathit{l}\mathit{e}\mathit{f}\mathit{t}\mathbf{\backslash }\mathbf{\{}\mathit{l}\mathit{e}\mathit{f}\mathit{t}\mathbf{\backslash }\mathit{l}\mathit{a}\mathit{n}\mathit{g}\mathit{l}\mathit{e}\mathbf{}\mathit{M}\mathbf{,}\mathit{x}\mathbf{,}\mathbf{\backslash }\mathbf{\#}\mathbf{^}\mathbf{\{}\mathit{t}\mathbf{\}}\mathbf{\backslash }\mathit{r}\mathit{i}\mathit{g}\mathit{h}\mathit{t}\mathbf{\backslash }\mathit{r}\mathit{a}\mathit{n}\mathit{g}\mathit{l}\mathit{e}\mathbf{\backslash }\mathit{m}\mathit{i}\mathit{d}\mathbf{\backslash }\mathit{r}\mathit{i}\mathit{g}\mathit{h}\mathit{t}$$\mathit{N}\mathit{T}\mathit{M}\mathbf{}\mathit{M}$accepts $\mathit{x}$within steps on at least one branch}. Note thatlocalid="1663241578125" $\mathit{M}$isn’t required to halt on all branches. Show that$\mathit{U}$is$\mathit{N}\mathit{P}$-complete.

Short answer

It can be easily seen that the language$U$ is exactly accepted by $M_{U}$, therefore, $U \in N P$ Hence,$U$ is NP-complete.

Step-by-step solution

Introduction

User contains an$NDTM\$M\_\left\{L\right\}\$$ , for any given language, in such a way that $\$\backslash forallx\backslash inL$, $M\_\left\{L\right\}\$$acceptson minimum single branch in maximum $p_{L}(|x|)$ steps. Here, $p_{L}() {\text { denotes a fixed polynomial depending }} on the machine.

It is also known that any $\$x\backslash notinl\$$will not be accepted by$\$M\_\left\{L\right\}\$$. Then, user can use a polynomial time to create$\$y=\backslash left\backslash langleM\_\left\{L\right\},x,\backslash \#^\{p\_\{L\left\}\right(\left(x\right)\left)\right\}\backslash right\backslash rangle\$$ for the given x.

Consider the previous argument, according to this argument,$\$x\backslash inL\$$ if and only if $\$y\backslash inU\$$. Therefore,$U$ is $NP$-hard.

Explanation

To show that is also in , users have to create an $NDTM\$M\_\left\{U\right\}\$$, which given an input $\$\{y=\backslash left\backslash langleM\_\{L\},x,\backslash \#^\{t\}\backslash right\backslash rangle\}\$$, simulates$M$ on$x$ for$t$ steps.

Every branches of are guessed by $M_{U}$ non-deterministically and accepts if and only if is accepted by .

Since the minimum length of input is $t$and user simulate $M$for maximum$t$ steps, the running time is polynomial in the length of the input.

Now it can be easily seen that the language is exactly accepted by $M_{U}$, therefore, $U \in N P$ Hence, is $NP$-complete.