zoukankan      html  css  js  c++  java
  • Computability 7: Exercises

      Disjoint sets $A$, $B$ are said to be recursively inseparable if there is no recursive set $C$ such that $Asubseteq C$ and $Bsubseteq C$.

      Now we prove that $K_0={x ext{ | }phi_x(x)=0}$ and $K_1={x ext{ | }phi_x(x)=1}$ are recursively inseparable. Suppose there is a recursive set $C$ with characteristic function $phi_e$ such that $K_0subseteq C$ and $K_1subseteqoverline{C}$. Then both

        $ein CLeftrightarrowphi_e(e)=1Leftrightarrow ein K_1Rightarrow einoverline{C}$, and $einoverline{C}Leftrightarrowphi_e(e)=0Leftrightarrow ein K_0Rightarrow ein C$

    are contradictory. Therefore, there is no such recursive set $C$.

      Disjoint sets $A$, $B$ are said to be effectively recursively inseparable if there is a total computable function $f$ such that whenever $Asubseteq W_a$, $Bsubseteq W_b$ and $W_acap W_b=varnothing$, then $f(a,b) otin W_acup W_b$.

      For any $i eq j$, we prove $K_i={x ext{ | }phi_x(x)=i}$ and $K_j={x ext{ | }phi_x(x)=j}$ are effectively recursively inseparable as follow. Consider any $a$ and $b$ such that $K_isubseteq W_a$, $K_jsubseteq W_b$, $W_acap W_b=varnothing$. The following function $g$ should be computable. $$g(x,y,z)simeqegin{cases}j & zin W_a \ i & zin W_b \ uparrow & ext{otherwise } end{cases}$$ According to the s-m-n theorem, there exists a total computable function $f$ such that $phi_{f(x,y)}(z)simeq g(x,y,z)$, and hence

        $f(a,b)in W_aRightarrow phi_{f(a,b)}(f(a,b))=jLeftrightarrow f(a,b)in K_jsubseteq W_bsubseteqoverline{W_a}$,

        $f(a,b)in W_bRightarrow phi_{f(a,b)}(f(a,b))=iLeftrightarrow f(a,b)in K_isubseteq W_asubseteqoverline{W_b}$,

    both of which lead to a contradiction. Therefore, $f(a,b) otin W_acup W_b$, and $A$, $B$ are effectively recursively inseparable.

      If two sets are effectively recursively inseparable, then they must be recursively inseparable.

      If two r.e. sets are effectively recursively inseparable, then both of them are proved to be creative.

      Suppose $A$, $B$ are effectively recursively inseparable sets, and $A_1$, $B_1$ are disjoint sets. If there is a total computable function $f$ such that $xin ALeftrightarrow f(x)in A$, $xin BLeftrightarrow f(x)in B$, then $A_1$ and $B_1$ are effectively recursively inseparable.

      Given $A$, $B$ are effectively recursively inseparable, we denote total computable function $g'$ satisfying this restriction. Consider any $a$ and $b$ such that $A_1subseteq W_a$, $B_1subseteq W_b$, $W_acap W_b=varnothing$. According to the s-m-n theorem, there is a total computable function $varphi$ such that $phi_{varphi(e)}simeqpsi_U(e,f(x))$. We denote $a'=varphi(a)$, $b'=varphi(b)$, and hence $xin W_{a'}Leftrightarrow f(x)in W_a$, $xin W_{b'}Leftrightarrow f(x)in W_b$. It follows that $Asubseteq W_{a'}$, $Bsubseteq W_{b'}$, $W_{a'}cap W_{b'}=varnothing$, and hence $g'(a,b) otin W_{a'}cup W_{b'}$. Then there is a total computable function $g(x,y)simeq g'(varphi(x),varphi(y))$ such that $g(a,b) otin W_acup W_b$. Therefore, $A_1$, $B_1$ are effectively recursively inseparable.

      This is an extension of the reduction theorem regarding to productive sets.

    P.S.  补充两个构造 productive function 的例子。

      1. Given $B$ is recursive and $Aoplus B$ is creative, prove $A$ is creative.

      (1) $A$ is r.e. since $xin ALeftrightarrow 2xin Aoplus B$ is partially decidable.

      (2) There is a total computable function $S$ such that:

         $phi_{S(x)}(y)simeqegin{cases}1 & (y ext{ is even }wedge y/2in W_x)vee(y ext{ is odd }wedge (y-1)/2 otin B) \ uparrow & ext{ otherwise}end{cases}$

      and hence $W_{S(x)}=W_xoplusoverline{B}=overline{overline{W}_xoplus B}$.

       Suppose $g$ is the productive function of $overline{Aoplus B}$, and we have

        $W_xsubseteqoverline{A}Rightarrow W_{S(x)}subseteqoverline{Aoplus B}Rightarrow g(S(x))inoverline{Aoplus B}capoverline{W}_{S(x)}Rightarrow g(S(x))/2inoverline{A}capoverline{W}_x$

      to prove $overline{A}$ is productive.

      2. Given $B$ is recursive and $Aotimes B$ is creative, prove $A$ is creative.

       Just by the same token. The only difference is that

         $phi_{S(x)}(y)simeqegin{cases}1 &pi_1(y)in W_xveepi_2(y) otin B\ uparrow & ext{ otherwise}end{cases}$

      such that $W_{S(x)}=overline{overline{W}_xotimes B}$, and finally $g(S(x))inoverline{Aotimes B}capoverline{W}_{S(x)}Rightarrowpi_1(g(S(x)))inoverline{A}capoverline{W}_x$.

    References:

      1. Cutland, Nigel. Computability: an introduction to recursive function theory[M]. Cambridge: Cambridge University Press, 1980

  • 相关阅读:
    移动网页如何只调出数字键盘
    把HTML5网页封装成APP,APK的方法
    HTML5,微信开发原码社区
    display:table 水平居中
    input在苹果浏览器下变成圆角的解决方案
    四种方法解决DIV高度自适应问题
    jquery prop和attr的区别
    移动端网页JS框架-手机触摸事件框架,日历框架带滑动效果
    meta viewport标签的使用说明(手机浏览缩放控制)
    javascript json格式解析方法
  • 原文地址:https://www.cnblogs.com/DevinZ/p/4599758.html
Copyright © 2011-2022 走看看