• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>

            woaidongmao

            文章均收錄自他人博客,但不喜標(biāo)題前加-[轉(zhuǎn)貼],因其丑陋,見諒!~
            隨筆 - 1469, 文章 - 0, 評(píng)論 - 661, 引用 - 0
            數(shù)據(jù)加載中……

            DFA求補(bǔ)集

            Comlement and Intersection of Regular Language

             

            Subjects to be Learned

            • Complement of Regular Language
            • Complement of DFA
            • Intersection of Regular Languages

            Contents


            Complement

            Let M = < Q , clip_image001, q0 , clip_image002, A > be a DFA that accepts a language L. Then a DFA that accepts the complement of L, i.e. clip_image001* - L, can be obtained by swapping its accepting states with its non-accepting states, that is Mc = < Q , clip_image001, q0 , clip_image002, Q - A > is a DFA that accepts clip_image001* - L .

            For example the following DFA accepts the language a+ over clip_image001= { a , b }.



                        clip_image003



            A DFA that accepts its complement is obtained from the above DFA by changing all single circles to double circles and vice versa as shown below.



                        clip_image004



            Remark 1: If we have NFA rather than DFA, we must first convert it to DFA before swapping states to get its complement.

            Remark 2: Since a language is regular if and only if it is accepted by some NFA, the complement of a regular language is also regular.


            Intersection of Regular Languages

            Langauges are sets. Therefore all the properties of sets are inherited by languages. In particular De Morgan's law also applies to languages. By Remark 2 above, if L1 and L2 are regular languages, then their complements are regular languages. Since L1 clip_image005L2 = clip_image006by De Morgan's law, L1 clip_image005L2 is regular.

            Thus summing all this up we can say that the set of regular languages over an alphabet is closed with respect to union, intersection, difference, concatenation and Kleene star operations.

            Test Your Understanding of Complemnent and Intersection of FAs

            Indicate which of the following statements are correct and which are not.
            Click True or Fals , then Submit.

            posted on 2009-11-05 12:14 肥仔 閱讀(1872) 評(píng)論(0)  編輯 收藏 引用 所屬分類: 狀態(tài)機(jī) & 自動(dòng)機(jī) & 形式語(yǔ)言

            久久综合伊人77777| 国产精品一久久香蕉国产线看观看| 欧美激情精品久久久久| 久久久久久综合一区中文字幕| 九九热久久免费视频| 久久亚洲欧美国产精品| 国产99久久久久久免费看| 久久AV无码精品人妻糸列| 伊人丁香狠狠色综合久久| 久久无码专区国产精品发布| 精品久久一区二区三区| 久久久久久久97| 久久久久婷婷| 99久久免费国产精品热| 久久综合亚洲鲁鲁五月天| 国产免费福利体检区久久| 97久久香蕉国产线看观看| 久久99热这里只有精品国产| 精品乱码久久久久久夜夜嗨 | 成人精品一区二区久久| 精品国产日韩久久亚洲| 久久久久久久亚洲精品| 久久国产精品成人免费| 久久午夜无码鲁丝片| 精品久久久无码人妻中文字幕| 久久久WWW成人| 久久精品国产亚洲5555| 中文字幕亚洲综合久久| 久久精品国产99久久久| 韩国免费A级毛片久久| 色妞色综合久久夜夜| 久久夜色精品国产欧美乱| 亚洲AV日韩AV天堂久久| 亚洲精品午夜国产VA久久成人| 青青草原综合久久大伊人| 欧美精品福利视频一区二区三区久久久精品 | 久久夜色精品国产噜噜噜亚洲AV| 久久人人爽人人爽人人片AV麻烦| 麻豆国内精品久久久久久| 久久亚洲中文字幕精品一区四 | 浪潮AV色综合久久天堂|