Ttyjfyk

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

Are all players supposed to be able to see each others' character sheets?

Are these two graphs isomorphic? Why/Why not?

Are E natural minor and B harmonic minor related?

Logistic regression BIC: what's the right N?

Is it possible to clone a polymorphic object without manually adding overridden clone method into each derived class in C++?

How can I portion out frozen cookie dough?

How should I solve this integral with changing parameters?

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

Is this Paypal Github SDK reference really a dangerous site?

Is divide-by-zero a security vulnerability?

Movie: boy escapes the real world and goes to a fantasy world with big furry trolls

Automaton recognizing ambiguously accepted words of another automaton

The preposition for the verb (avenge) - avenge sb/sth (on OR from) sb

How do I increase the number of TTY consoles?

Strange opamp's output impedance in spice

How do I raise a figure (placed with wrapfig) to be flush with the top of a paragraph?

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

"If + would" conditional in present perfect tense

Why does Central Limit Theorem break down in my simulation?

Why restrict private health insurance?

Smooth vector fields on a surface modulo diffeomorphisms

Create chunks from an array

Do black holes violate the conservation of mass?

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.

automata regular-languages finite-automata

share|cite|improve this question

edited 2 hours ago

Yuval Filmus

194k14183347

asked 2 hours ago

truvakingtruvaking

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.

$endgroup$

add a comment | 

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.

automata regular-languages finite-automata

share|cite|improve this question

edited 2 hours ago

Yuval Filmus

194k14183347

asked 2 hours ago

truvakingtruvaking

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.

$endgroup$

add a comment | 

$begingroup$

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.

automata regular-languages finite-automata

share|cite|improve this question

edited 2 hours ago

Yuval Filmus

194k14183347

asked 2 hours ago

truvakingtruvaking

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.

$endgroup$

share|cite|improve this question

edited 2 hours ago

Yuval Filmus

194k14183347

asked 2 hours ago

truvakingtruvaking

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.

share|cite|improve this question

edited 2 hours ago

Yuval Filmus

194k14183347

asked 2 hours ago

truvakingtruvaking

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.

edited 2 hours ago

Yuval Filmus

194k14183347

edited 2 hours ago

Yuval Filmus

194k14183347

asked 2 hours ago

truvakingtruvaking

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.

asked 2 hours ago

truvakingtruvaking

284

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

answered 2 hours ago

Yuval FilmusYuval Filmus

194k14183347

$endgroup$

add a comment | 

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

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

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

Name

Email

Required, but never shown

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

answered 2 hours ago

Yuval FilmusYuval Filmus

194k14183347

$endgroup$

add a comment | 

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

answered 2 hours ago

Yuval FilmusYuval Filmus

194k14183347

$endgroup$

add a comment | 

$begingroup$

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

answered 2 hours ago

Yuval FilmusYuval Filmus

194k14183347

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Yuval FilmusYuval Filmus

194k14183347

answered 2 hours ago

Yuval FilmusYuval Filmus

194k14183347

answered 2 hours ago

Yuval FilmusYuval Filmus

194k14183347