Automaton recognizing ambiguously accepted words of another automatonSynchronizing sequence and...

Was it really inappropriate to write a pull request for the company I interviewed with?

Did Amazon pay $0 in taxes last year?

What was so special about The Piano that Ada was willing to do anything to have it?

What is this tube in a jet engine's air intake?

When an outsider describes family relationships, which point of view are they using?

-1 to the power of a irrational number

What does *dead* mean in *What do you mean, dead?*?

Is there a way to make cleveref distinguish two environments with the same counter?

PTIJ: Sport in the Torah

If nine coins are tossed, what is the probability that the number of heads is even?

Cycles on the torus

What is Tony Stark injecting into himself in Iron Man 3?

How to copy the rest of lines of a file to another file

Why do we say 'Pairwise Disjoint', rather than 'Disjoint'?

Is it appropriate to ask a former professor to order a book for me through an inter-library loan?

Converting from "matrix" data into "coordinate" data

I am the person who abides by rules, but breaks the rules. Who am I?

ESPP--any reason not to go all in?

School performs periodic password audits. Is my password compromised?

Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?

Too soon for a plot twist?

Why aren't there more Gauls like Obelix?

Would those living in a "perfect society" not understand satire

How to install round brake pads



Automaton recognizing ambiguously accepted words of another automaton


Synchronizing sequence and Synchronizable DFAShow that the language of words with even sum of positions of a letter is regularConverting generalized NFAs to NFAsDeleting useless (dead) states from a finite automatonfinite automata that accepts integers divided by 3?The upper bound on a Nondeterministic Finite Automata's configurations numberFiguring out the language accepted by a DFA?How to transform Nondeterministic finite automaton (NFA) to regular expression equivalentMyhill-Nerode Theorem DFA MinimizationUnderstanding application of Arden's theorem to find regular expression













2












$begingroup$


Let $A$ be a nondeterministic automaton. Let $X(A)$ the set of words for which there at least two accepting paths.



In one of the previous exam, for which no answers are available, it is required to prove that there exist a deterministic automaton whose language is $X(A)$. Furthermore, if the original automaton has $k$ states, then the new automaton should have $3^k$ states.



I have tried multiples approaches to no avail:




  • Make an automata that keeps track of which path was taken, but the size of the automata grew infinitely because of cycles...

  • An automaton for each state I also kept track of which state it came from, but it was just a scaled down version of previous attempt. Which didn't work if the path started to converge early.

  • Tried an automaton where I encoded each state with the set of all the states from automaton $A$. And for each state I included whether I went through that state 0, 1 or more than 2 times.










share|cite|improve this question









New contributor




truvaking is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    2












    $begingroup$


    Let $A$ be a nondeterministic automaton. Let $X(A)$ the set of words for which there at least two accepting paths.



    In one of the previous exam, for which no answers are available, it is required to prove that there exist a deterministic automaton whose language is $X(A)$. Furthermore, if the original automaton has $k$ states, then the new automaton should have $3^k$ states.



    I have tried multiples approaches to no avail:




    • Make an automata that keeps track of which path was taken, but the size of the automata grew infinitely because of cycles...

    • An automaton for each state I also kept track of which state it came from, but it was just a scaled down version of previous attempt. Which didn't work if the path started to converge early.

    • Tried an automaton where I encoded each state with the set of all the states from automaton $A$. And for each state I included whether I went through that state 0, 1 or more than 2 times.










    share|cite|improve this question









    New contributor




    truvaking is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2





      $begingroup$


      Let $A$ be a nondeterministic automaton. Let $X(A)$ the set of words for which there at least two accepting paths.



      In one of the previous exam, for which no answers are available, it is required to prove that there exist a deterministic automaton whose language is $X(A)$. Furthermore, if the original automaton has $k$ states, then the new automaton should have $3^k$ states.



      I have tried multiples approaches to no avail:




      • Make an automata that keeps track of which path was taken, but the size of the automata grew infinitely because of cycles...

      • An automaton for each state I also kept track of which state it came from, but it was just a scaled down version of previous attempt. Which didn't work if the path started to converge early.

      • Tried an automaton where I encoded each state with the set of all the states from automaton $A$. And for each state I included whether I went through that state 0, 1 or more than 2 times.










      share|cite|improve this question









      New contributor




      truvaking is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Let $A$ be a nondeterministic automaton. Let $X(A)$ the set of words for which there at least two accepting paths.



      In one of the previous exam, for which no answers are available, it is required to prove that there exist a deterministic automaton whose language is $X(A)$. Furthermore, if the original automaton has $k$ states, then the new automaton should have $3^k$ states.



      I have tried multiples approaches to no avail:




      • Make an automata that keeps track of which path was taken, but the size of the automata grew infinitely because of cycles...

      • An automaton for each state I also kept track of which state it came from, but it was just a scaled down version of previous attempt. Which didn't work if the path started to converge early.

      • Tried an automaton where I encoded each state with the set of all the states from automaton $A$. And for each state I included whether I went through that state 0, 1 or more than 2 times.







      automata regular-languages finite-automata






      share|cite|improve this question









      New contributor




      truvaking is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      truvaking is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      Yuval Filmus

      194k14183347




      194k14183347






      New contributor




      truvaking is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 2 hours ago









      truvakingtruvaking

      284




      284




      New contributor




      truvaking is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      truvaking is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      truvaking is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Construct an NFA which will simulate (nodeterministically) two runs of the original NFA on the input. The NFA also remembers one bit – whether the two runs have diverged so far or not. If the bit is off and the NFA chooses two different transitions, then you turn it on. The NFA accepts if both runs accept and the bit is on.



          If the original NFA has $k$ states, then the new NFA will have $2k^2$ states. You can convert it to a DFA with $4^{k^2}$ states.



          We can reduce the number of states in the corresponding DFA by performing the powerset construction directly. At each step, we will remember, for each step, whether (i) it is impossible to be at that state, (ii) it is possible to be at that state, but there is a unique such run, (iii) it is possible to be at the state, and there are at least two such runs. A state is accepting if there is an accepting states of type (iii), or at least two accepting states of type (ii). This construction uses only $3^k$ states, and so matches your hint.






          share|cite|improve this answer









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "419"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });






            truvaking is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f105377%2fautomaton-recognizing-ambiguously-accepted-words-of-another-automaton%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Construct an NFA which will simulate (nodeterministically) two runs of the original NFA on the input. The NFA also remembers one bit – whether the two runs have diverged so far or not. If the bit is off and the NFA chooses two different transitions, then you turn it on. The NFA accepts if both runs accept and the bit is on.



            If the original NFA has $k$ states, then the new NFA will have $2k^2$ states. You can convert it to a DFA with $4^{k^2}$ states.



            We can reduce the number of states in the corresponding DFA by performing the powerset construction directly. At each step, we will remember, for each step, whether (i) it is impossible to be at that state, (ii) it is possible to be at that state, but there is a unique such run, (iii) it is possible to be at the state, and there are at least two such runs. A state is accepting if there is an accepting states of type (iii), or at least two accepting states of type (ii). This construction uses only $3^k$ states, and so matches your hint.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Construct an NFA which will simulate (nodeterministically) two runs of the original NFA on the input. The NFA also remembers one bit – whether the two runs have diverged so far or not. If the bit is off and the NFA chooses two different transitions, then you turn it on. The NFA accepts if both runs accept and the bit is on.



              If the original NFA has $k$ states, then the new NFA will have $2k^2$ states. You can convert it to a DFA with $4^{k^2}$ states.



              We can reduce the number of states in the corresponding DFA by performing the powerset construction directly. At each step, we will remember, for each step, whether (i) it is impossible to be at that state, (ii) it is possible to be at that state, but there is a unique such run, (iii) it is possible to be at the state, and there are at least two such runs. A state is accepting if there is an accepting states of type (iii), or at least two accepting states of type (ii). This construction uses only $3^k$ states, and so matches your hint.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Construct an NFA which will simulate (nodeterministically) two runs of the original NFA on the input. The NFA also remembers one bit – whether the two runs have diverged so far or not. If the bit is off and the NFA chooses two different transitions, then you turn it on. The NFA accepts if both runs accept and the bit is on.



                If the original NFA has $k$ states, then the new NFA will have $2k^2$ states. You can convert it to a DFA with $4^{k^2}$ states.



                We can reduce the number of states in the corresponding DFA by performing the powerset construction directly. At each step, we will remember, for each step, whether (i) it is impossible to be at that state, (ii) it is possible to be at that state, but there is a unique such run, (iii) it is possible to be at the state, and there are at least two such runs. A state is accepting if there is an accepting states of type (iii), or at least two accepting states of type (ii). This construction uses only $3^k$ states, and so matches your hint.






                share|cite|improve this answer









                $endgroup$



                Construct an NFA which will simulate (nodeterministically) two runs of the original NFA on the input. The NFA also remembers one bit – whether the two runs have diverged so far or not. If the bit is off and the NFA chooses two different transitions, then you turn it on. The NFA accepts if both runs accept and the bit is on.



                If the original NFA has $k$ states, then the new NFA will have $2k^2$ states. You can convert it to a DFA with $4^{k^2}$ states.



                We can reduce the number of states in the corresponding DFA by performing the powerset construction directly. At each step, we will remember, for each step, whether (i) it is impossible to be at that state, (ii) it is possible to be at that state, but there is a unique such run, (iii) it is possible to be at the state, and there are at least two such runs. A state is accepting if there is an accepting states of type (iii), or at least two accepting states of type (ii). This construction uses only $3^k$ states, and so matches your hint.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Yuval FilmusYuval Filmus

                194k14183347




                194k14183347






















                    truvaking is a new contributor. Be nice, and check out our Code of Conduct.










                    draft saved

                    draft discarded


















                    truvaking is a new contributor. Be nice, and check out our Code of Conduct.













                    truvaking is a new contributor. Be nice, and check out our Code of Conduct.












                    truvaking is a new contributor. Be nice, and check out our Code of Conduct.
















                    Thanks for contributing an answer to Computer Science Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f105377%2fautomaton-recognizing-ambiguously-accepted-words-of-another-automaton%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    VNC viewer RFB protocol error: bad desktop size 0x0I Cannot Type the Key 'd' (lowercase) in VNC Viewer...

                    Couldn't open a raw socket. Error: Permission denied (13) (nmap)Is it possible to run networking commands...

                    Why not use the yoke to control yaw, as well as pitch and roll? Announcing the arrival of...