# A redução de Karp é idêntica à redução de Levin

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### Definição: Redução de Karp

Uma linguagem A$A$ é Karp redutível a uma linguagem B$B$ se houver uma função computável em tempo polinomial f:{0,1}{0,1}$f:\{0,1\}^*\rightarrow\{0,1\}^*$ tal que para cada x$x$ , xA$x\in A$ se e somente se f(x)B$f(x)\in B$ .

### Definição: Redução de Levin

Um problema pesquisa VA$V_A$ é Levin redutível a uma pesquisa problema VB$V_B$ se houver função polinomial f$f$ que reduz Karp L(VA)$L(V_A)$ a L(VB)$L(V_B)$ e existem polinomial-tempo funções calculáveis g$g$ e h$h$ tal que

1. x,yVAf(x),g(x,y)VB$\langle x, y \rangle \in V_A \implies \langle f(x), g(x,y) \rangle \in V_B$ ,

2. f(x),zVBx,h(x,z)VA$\langle f(x), z \rangle \in V_B \implies \langle x, h(x,z) \rangle \in V_A$

Essas reduções são equivalentes?

Eu acho que as duas definições são equivalentes. Para quaisquer dois NP$\mathsf{NP}$ idiomas A$A$ e B$B$ , se A$A$ é Karp redutível para B$B$ , então A$A$ é Levin redutível a B$B$ .

Aqui está a minha prova:

Deixe- e haver casos arbitrárias de tempo ser que de . Suponha e são verificadores de e . Deixe- e ser certificados arbitrárias de e de acordo com . Deixe- ser que de de acordo com .x$x$x¯¯¯$\overline{x}$A$A$x$x'$B$B$VA$V_A$VB$V_B$A$A$B$B$y$y$y¯¯¯$\overline{y}$x$x$x¯¯¯$\overline{x}$VA$V_A$z$z$x$x'$VB$V_B$

Construa novos verificadores e com novos certificados e :VA$V'_A$VB$V'_B$y$y'$z$z'$

VA(x,y):$V'_A(x,y'):$

1. : Se , rejeitar. Caso contrário, imprima .y=0,x¯¯¯,y¯¯¯$y'=\langle 0,\overline{x},\overline{y}\rangle$f(x)f(x¯¯¯)$f(x)\ne f(\overline{x})$VA(x¯¯¯,y¯¯¯)$V_A(\overline{x},\overline{y})$
2. : Saída .y=1,z$y'=\langle 1,z\rangle$VB(f(x),z)$V_B(f(x),z)$

VB(x,z):$V'_B(x',z'):$

1. : Saída .z=0,z$z'=\langle 0,z\rangle$VB(x,z)$V_B(x',z)$

2. : Se , rejeitar. Caso contrário, imprima .z=1,x,y$z'=\langle 1,x,y\rangle$xf(x)$x'\ne f(x)$VA(x,y)$V_A(x,y)$

As funções computáveis ​​em tempo polinomial e são definidas como abaixo:g$g$h$h$

g(x,y)$g(x,y')$

1. y=0,x¯¯¯,y¯¯¯$y'=\langle 0,\overline{x},\overline{y}\rangle$: Output 1,x¯¯¯,y¯¯¯$\langle 1,\overline{x},\overline{y}\rangle$.

2. y=1,z$y'=\langle 1,z\rangle$: Output 0,z$\langle 0,z\rangle$.

h(x,z)$h(x',z')$

1. z=0,z$z'=\langle 0,z\rangle$: Output 1,z$\langle 1,z\rangle$.

2. z=1,x,y$z'=\langle 1,x,y\rangle$: Output 0,x,y$\langle 0,x,y\rangle$.

Let Yx$Y_x$ be the set of all certificates of x$x$ according to VA$V_A$ and Zx$Z_{x'}$ be the set of all certificates of x$x'$ according to VB$V_B$. Then the set of all certificates of x$x$ according to VA$V'_A$ is 0x¯¯¯Yx¯¯¯+1Zf(x)$0\overline{x}Y_\overline{x}+1Z_{f(x)}$ such that f(x)=f(x¯¯¯)$f(x)=f(\overline{x})$, and the set of all certificates of x$x'$ according to VB$V'_B$ is 0Zx+1x¯¯¯Yx¯¯¯$0Z_{x'}+1\overline{x}Y_\overline{x}$ such that x=f(x¯¯¯)$x'=f(\overline{x})$.

(This is derived from the accepting language of VA$V'_A$ and VB$V'_B$.)

Now let x=f(x)$x'=f(x)$, the rest part is easy to check.

Before proving your claim, you should define what it means by a language being Levin reducible to another.
Tsuyoshi Ito

Respostas:

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No. First note that Levin reduction only makes sense with respect to classes which certificate has a meaning, e.g. NP$\mathsf{NP}$ while Karp reduction is general. The word "certificate" for a problem makes sense only when a verifier is fixed. Levin's reduction assumes that the verifiers are fixed. You cannot change the verifiers. (In the following I assume that certificate verifiers are fixed as required by definition of Levin reduction.)

Second, Levin reduction means that one can find the certificate for one from the certificate for the other. It is true that the well-known Karp reductions between natural problems turn out to be Levin reduction but this does not need to be true in general.

If we can reduce instances of a problem A$A$ to problem B$B$ it doesn't mean we have a way of computing a certificate for one from a certificate for the other.

For this to be true we need the fact that a certificate search problem corresponding to the decision problem is polynomial time reducible to the decision problem. This is true for NP-complete$\mathsf{NP\text{-}complete}$ problems but not known to be true generally even for NP$\mathsf{NP}$ problems.

I agree on your first point that Karp reduction is more general than Karp reduction. According to it, I think my problem should be modified as "Let L1$L_1$ and L2$L_2$ be two languages in NP$NP$. If L1$L_1$ is Karp reducible to L2$L_2$, then L1$L_1$ is Levin reducible to L2$L_2$." However, I don't agree on your other comments.
c c

In my prove, first let L1$L_1$ and L2$L_2$ be arbitrary such two languages. Since they are in $NP$, there are P TM $M_1$ and $M_2$. For every instances $x\in L_1$, the set of all certificates are $Y_x$ for $TM_1$. Similarly, we define $Z_{x'}$ for $x'\in L_2$. Since $L_1$ is Karp reducible to $L_2$, there is such $f$ as defined.
c c

Now, we constructed new $M_1'$ and $M_2'$. For every instance $x\in L_1$, the new set of all certificates are $0Y_x+1Z_{f(x)}$, while for every instance $f(x)\in L_2$, the new set of all certificates are $0Z_{f(x)}+1xY_x$.(Here we use regular expressions) These are legal and all certificates for $M_1'$ and $M_2'$. By the way, there is P functions $g$ and $h$ as defined transform all possible certificate from one problem to the other.
c c

ps: We don't need to give a transformation from $x'\in L_2$ where there is no $x\in L_1$ such that $x'=f(x)$, since Karp and Levin reductions are both many to one mappings. I think this can answer the second last paragraph.
c c

@cc, it seems that you still think that you can change the verifiers, you cannot, the definition of Levin reduction is for search problems, i.e. the verifiers are fixed.
Kaveh

5

A quick counterexample for your proof: suppose that $x_1, x_2 \in L_1$, $f(x_1) = f(x_2) \in L_2$, and $w$ is a valid certificate for $x_1$ but not for $x_2$

$M_1'(x_1,\langle 0,w \rangle)= M_1(x_1,w)=1$

$M_1'(x_2,\langle 0,w \rangle)= M_1(x_2,w)=0$

By definition $g(x_1,\langle 0,w \rangle) = \langle 1,x_1,w \rangle$

$f(x_1)=f(x_2)$ so $M'_2(f(x_2),\langle 1,x_1, w \rangle)) = M_1(x_1,w) = 1$ so $\langle 1,x_1, w \rangle$ is a valid certificate for $f(x_2)$ but

$h(f(x_2), \langle 1,x_1, w \rangle) = \langle 0, w \rangle$ which is not a valid certificate for $x_2$

Thanks very much for pointing out the counterexample. I have modified the construction and I think it works now. Could you please have a look at it?
c c
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