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  • 计算理论导引答案

    2014-10-31 22:25:49
    计算理论导引答案Michael Sipser(2-9章)分章节的
  • 计算理论导引第二版 作者: Michael Sipser 课后答案
  • 计算理论导引 答案

    热门讨论 2008-01-02 23:43:08
    本书由计算理论领域的知名权威Michael Sipser所撰写。他以独特的视角,系统地介绍了计算理论的三个主要内容:自动机与语言、可计算性理论和计算复杂性理论。绝大部分内容是基本的,同时对可计算性和计算复杂性理论中...
  • 详细答案,计算机理论导引第三版答案,Introduction-to-the-Theory-of-Computation-Solutions
  • 计算理论导引第三版课后习题答案Introduction-to-the-Theory-of-Computation-Solutions
  • 最新最全答案计算理论导引第三版课后习题答案Introduction-to-the-Theory-of-Computation-Solutions
  • 计算理论导引》唐常杰译第二版课后1-9章题答案
  • 计算理论导论的答案~~~大家可以参考看看
  • 英文版 计算理论导引 课后习题答案
  • 计算理论导引 第二版课后答案
  • 计算理论导引课后答案(中文版)

    热门讨论 2011-05-01 09:15:01
    中文版的计算理论导引的课后答案,有些题目不是很全,可以去找英文的看看
  • 计算理论导引 答案 计算引论 计算理论导引答案
  • 计算理论导引2017年期末考试题,基本上都是课后题里的内容
  • 计算理论导引答案第二版

    热门讨论 2009-09-14 13:15:15
    本书由计算理论领域的知名权威Michael Sipser所撰写。他以独特的视角,系统地介绍了计算理论的三个主要内容:自动机与语言、可计算性理论和计算复杂性理论。绝大部分内容是基本的,同时对可计算性和计算复杂性理论中...
  • 最近在学这本书,无奈课后习题很多,答案又找不到,没想到在github上找到了 直接上链接 https://github.com/gaurangsaini/sipser-computation-3rd-solutions

    最近在学这本书,无奈课后习题很多,答案又找不到,没想到在github上找到了

    直接上链接

    https://github.com/gaurangsaini/sipser-computation-3rd-solutions

    展开全文
  • 计算理论导引答案中文(第二版)

    热门讨论 2011-12-28 11:26:01
    整理的资源,基本上全了。最新版的答案,你懂的。
  • 计算理论导引第二版的课后习题答案,英文版的,但是很全
  • 第四章:可判定性4 Decidability关于其他章节的内容,请点这:《计算理论导引》学习笔记4.1 Decidable Languages几个可判定的语言acceptance problem for DFAthe acceptane problem for DFAs of testing whether a ...

    第四章:可判定性

    4 Decidability

    关于其他章节的内容,请点这:《计算理论导引》学习笔记

    4.1 Decidable Languages

    几个可判定的语言

    acceptance problem for DFA

    the acceptane problem for DFAs of testing whether a particular deterministic finite automaton accepts a given string can be expressed as a language\( A_{DFA}\).

    \( A_{DFA} = \{\langle B,w \rangle | \text{ B is a DFA that accepts input string w}\}\).

    The problem of testing whether a DFA B accepts an input w is the same as the problem of testing whether is a member of the language \( A_{DFA}\).

    We simply need to present a TM M that decides \( A_{DFA}\).

    M = “on input , where B is a DFA and w is string:

    1. Simulate B on input w.

    2. If the simulation ends in an accept state, accept. If it ends in a nonaccepting state, reject.”

    acceptance problem for NFA

    \( A_{NFA} = \{\langle B,w \rangle | \text{ B is a NFA that accepts input string w}\}\).

    N = “on input , where B is a NFA and w is string:

    1. Convert NFA B to an equivalent DFA C.

    2. Run TM M on input .

    2. If M accepts, accept; otherwise, reject.”

    acceptance problem for REX

    \( A_{REX} = \{\langle B,w \rangle | \text{ B is a regular expression that generates string w}\}\).

    P = “On input , where R is a regular expression and w is a string:

    1. Convert regular expression R to an equivalent NFA A.

    2. Run TM N on input .

    3. If N accepts, accept; if N rejects, reject.”

    emptiness testing for DFA

    \( E_{DFA} = \{\langle A \rangle | \text{ A is a DFA and } L(A) = \emptyset\}\).

    T = “On input , where A is a DFA:

    1. Mark the start state of A.

    2. Repeat until no new states get marked:

    3. Mark any state that has a transition coming ino it from any state that is already marked.

    4. If no accept state is marked, accept; otherwise, reject.”

    whether two DFAs recognize the same language

    \( EQ_{DFA} = \{\langle A,B \rangle | \text{ A and B are DFA and } L(A) = L(B)\}\).

    \( L(C) = (L(A) \cap \overline{L(B)}) \cup (\overline{L(A)} \cap L(B))\).

    We can construct C form A and B with the sonstructions for proving the class of regular languages closed under complementation, union, and intersection.

    \( L(C) = \emptyset \iff L(A) = L(B)\).

    F = “On input , where A and B are DFAs:

    1. Condtruct DFA C as described.

    2. Run TM T on input .

    3. If T accepts, accept. If T rejects, reject.”

    acceptance problem for CFG

    \( A_{CFG} = \{\langle G,w \rangle\} | \text{ G is a CFG that generates string w }\}\).

    直接枚举会得到recognizer而不是decider

    If G were in Chomsky normal form, any derivation of w has 2n-1 steps, where n is the length of w.

    In that case, checking only derivations with 2n-1 steps to determine whether G generates w would be sufficient.

    S = “on input , where G is a CFG and w is a string:

    1. Convert G to an equivalent grammar in Chomsky normal form.

    2. List all derivations with 2n-1 steps, where n is the length of w; except if n = 0, then instead list all derivations with one step.

    3. If any of these deravation generate w, accept; if not, reject.”

    emptiness testing for CFG

    \( E_{CFG} = \{\langle G \rangle | \text{ G is a CFG and } L(G) = \emptyset\}\).

    转化为可达性问题,目的地是terminal symbols,起点是start variable

    R = “On input , where G is a CFG:

    1. Mark all terminal symbols in G.

    2. Repeat until no new variables get marked:

    3. Mark any variable A where G has a rule \( A \longrightarrow U_1U_2\dots U_k\) and each symbol \( U_1,\dots,U_k\) has already been marked.

    4. If the start variable is not marked, accept; otherwise, reject.

    whether two CFGs generate the same language

    \( EQ_{CFG} = \{\langle G,H \rangle | \text{ G and H are CFGs and } L(G) = L(H)\}\).

    在这里,\( EQ_{DFA}\)的套路用不了了

    因为,The class of context-free languages is not closed under complementation or intersection.(Exercise 2.2)

    事实上,这是不可判定的🙂

    every context-free language is decidable

    \( M_G\) = “On input w:

    1. Run TM S on input .

    2. If this machine accepts, accepts, if it rejects, reject.”

    13f3b6d54e435b67d00bec2384d12b5d.pngrelationship-among-classes-of-languages

    4.2 Undecidability

    对角线法

    接下来讲了一波无穷大集合之间怎么比较大小,就是为了引入对角线法?

    简单的说,如果两个集合之间存在一个双射函数使其一一对应,那么两个集合大小相等

    比如偶数集与自然数集、直线与平面🙂,局部与整体等大,是不是很神奇,这就是无穷大的魅力

    在这基础上定义了可数(countable)这个概念:

    A set A is countable if either it is finite or it has the same size as N.

    我们日常在数数(count)的时候“1,2,3,。。。”其实就是在被数的对象与自然数集N之间建立一一对应关系

    于是按一个套路把有理数排列起来后,发现可以一个不漏地数下去,就证明有理数可数了

    很自然地,我们会想,无理数呢?实数呢?复数呢?

    很遗憾,这些都是不可数的,这时就要用对角线法证明了

    反证法:

    假设R与N之间存在双射函数f(n)

    那么我们可以构造出一个不在f(n)值域的实数出来,这就破坏假设了,证毕。

    这个数是这样的:

    小数点后第i位与f(i)的的小数点后第i位不同,如图

    44d9014728094f0b25203221758f0179.pngdiagonalization-R

    这个数的每一位都和对角线上对应位不同,所以叫对角线法。

    很容易证明,代数数是可数的,而复数是不可数的,复数又由代数数与超越数组成

    于是就证明了超越数的存在性,却没证明一个数是超越数,康托尔就是迪奥

    类似的思路,图灵机是可数的,语言是不可数的

    所以必定存在some languages are not Turing-recognizalbe

    一个不可判定的语言

    \( A_{TM} = \{\langle M,w \rangle | \text{ M is a TM and M accepts w }\}\) is undecidable.

    不妨先证一下\( A_{TM}\)是可被图灵识别的:

    只需构造一个图灵机出来就可以了:

    U = “On input , where M is a TM and w is a string:

    1. Simulate M on input w.

    2. If M ever enters its accept state, accept; if M ever enters its reject state, reject.”

    不可判定的证明:

    (还有个更简洁的证明在第六章)

    假设\( A_{TM}\)是可判定的 \( \longrightarrow\) 存在decider H \( \longrightarrow\) 存在decider D \( \longrightarrow\) 存在悖论

    所以假设不成立,证毕。

    其中,

    H就是\( A_{TM}\)的decider

    \( H(\langle M,w \rangle) =

    \begin{cases}

    accept & \text{if M accepts w} \\

    reject & \text{if M does not accept w}

    \end{cases}\)

    D = “On input , where M is a TM:

    1. Run H on input >.

    2. Output the opposite of what H outputs. That is, if H accepts, reject; and if H rejects, accept.”

    \( D(\langle M \rangle) =

    \begin{cases}

    accept & \text{if M does not accepts }\langle M \rangle \\

    reject & \text{if M accepts }\langle M \rangle

    \end{cases}\)

    是不是很像“理发师悖论”?所以接下来:

    \( D(\langle D \rangle) =

    \begin{cases}

    accept & \text{if D does not accepts }\langle D \rangle \\

    reject & \text{if D accepts }\langle D \rangle

    \end{cases}\)

    当对D输入的时候,D既不可以accept,又不可以reject,也不可以其他,所以矛盾出现。

    是不是感觉没用对角线法?其实人家用了,你看不出来而已,有图有真相:

    977a0d7f00aac87e074bc6d2eb17fa1f.pngdiagonalization-TM

    另外,有个问题:

    如果D里面不运行H,而是运行U,好像也有矛盾啊,那不就证明了U不存在?

    这个问题我纠结了几个钟ORZ。。。最后我的答案是这样的:(欢迎到评论区和我讨论)

    首先明确D修改后是这样的:

    E = “On input , where M is a TM:

    1. Run U on input >.

    2. If U ever enters its accept state, reject; if U ever enters its reject state, accept.”

    也就是说,与D不同的是,E有可能会死循环,这就是关键

    现在再对E输入看看:

    同样,E既不可以accept,又不可以reject

    但是,现在E可以死循环,所以我的结论是:E()会死循环

    所以E并不像D那样是一个矛盾的不可能存在的图灵机,答毕。

    定理4.0

    A language is decidable iff it is Turing-recognizable and co-Turing-recognizable.

    (We say that a language is co-Turing-recognizable if it is the complement of a Turing-recognizable language.)

    证明很简单,略去

    推论:

    \( \overline{A_{TM}}\) is not Turing-recognizable.

    证明很简单,略去

    但是,不妨设法构造一个识别\( \overline{A_{TM}}\)的图灵机出来,看看问题在哪

    \( \overline{A_{TM}} = \{\langle M,w \rangle | \text{ M is a TM and M does not accept w }\}\)

    Z = “On input , where M is a TM and w is a string:

    1. Simulate M on input w.

    2. if M ever enters its accept state, reject; if M ever enters its reject state, accept.

    3. if M loops, accept.”

    这个并不能作为证明,但是,从第三点可以看出,因为停机问题不可判定,所以\( \overline{A_{TM}}\)不可被图灵识别

    展开全文
  • 计算机理论导引课后答案,word,pdf
  • 本书系统地介绍了计算理论的三个主要内容:自动机与语言、可计算性和计算复杂性。绝大部分内容是基本的,同时对可计算性和计算复杂性理论中的某些高级内容作了重点介绍。作者以清闲的笔触、生动的语言给出了宽泛的...
  • 计算理论导引答案 计算理论这门课忒难了,所以这个很有帮助。
  • 计算理论导引电子版教材与答案

    热门讨论 2010-05-17 10:44:05
    计算理论导引中文电子版 英文版原版答案 第二版整理中文答案
  • 计算理论导引第二版答案

    热门讨论 2009-11-08 14:19:11
    作者将复杂的概念讲解的非常清晰透彻。而且在对定理进行相应的证明时,总是能让读者感觉到一点点的激动与兴奋。这里给大家提供的是它的源代码,希望大家可以提高自己的学习~~~~~~~~~~~
  • 本书是计算理论领域的经典著作,被国外多所大学选用用为教材。本书以注重思路、深入引导为特色,系统地介绍计算机理论的三大主要内容:自动机与语言、可计算性理论和计算复杂性理论。同时,对可计算性和计算复杂理论...
  • sipser的计算理论导引这本书很经典,课后题也不简单,这是第五章的课后答案,很全面。
  • 计算理论导引 第一版 课后题答案

    热门讨论 2008-11-10 22:36:51
    计算理论导引 第一版 一书的课后习题答案,是英文版的,欢迎下载!!
  • 计算理论导引 英文版答案 Introduction to the Theory of Computation Michael Sipser
  • 麻省理工CS教材,讲述计算理论的经典教材,英文通俗易懂,由麻省理工教授MICHAEL SIPSER所著。该电子版为高清版,非扫描,带有分章节的目录模块,方便查询。

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