How does one use the Nerode-Myhill theorem to prove that a language is regular?How I can find all equivalence...
How to know if I am a 'Real Developer'
Identical projects by students at two different colleges: still plagiarism?
Badly designed reimbursement form. What does that say about the company?
Found a major flaw in paper from home university – to which I would like to return
Does an increasing sequence of reals converge if the difference of consecutive terms approaches zero?
Why don't hotels offer (at least) 1 kitchen bookable by any guest?
How can a kingdom keep the secret of a missing monarchy from the public?
Does a star need to be inside a galaxy?
Which was the first story to feature helmets which reads your mind to control a machine?
Negotiating 1-year delay to my Assistant Professor Offer
How does one use the Nerode-Myhill theorem to prove that a language is regular?
Why would you use 2 alternate layout buttons instead of 1, when only one can be selected at once
Is layered encryption more secure than long passwords?
Bitcoin automatically diverted to bech32 address
Why is Shelob considered evil?
Why are energy weapons seen as more acceptable in children's shows than guns that fire bullets?
How to use the viewer node?
How can guns be countered by melee combat without raw-ability or exceptional explanations?
How to not forget my phone in the bathroom?
Why does finding small effects in large studies indicate publication bias?
Was Opportunity's last message to Earth "My battery is low and it's getting dark"?
multiple price sets?
What have we got?
Why there is no EEPROM in STM32F4 MCUs
How does one use the Nerode-Myhill theorem to prove that a language is regular?
How I can find all equivalence classes by Myhill-Nerode?finding separating words (Nerode)How to prove that a language is not regular?Undergrad resources for identifying regular languages with Myhill-Nerode matricesunion of two equivalence classes (Myhill–Nerode theorem)A Myhill-Nerode type characterization of the regular languages using fooling sets?Non-regularity of the set of primes in unary encoding using Myhill-NerodeHow do I show that an equivalence class of a language containing an empty string is infiniteShow that language generated by grammar is regularHow do you prove that the set of decimal representation of the 4 divisble natural numbers is regular?Proving that L is not regular by showing that $equiv_L$ has infinite indexHow I can find all equivalence classes by Myhill-Nerode?
$begingroup$
Showing that a language is not regular is straight-forward, because all one needs to do is find an infinite set of inputs which has an injective mapping to the set of equivalence classes which compose that language.
How does one show that the set of equivalence classes of $L$ is finite? For instance, how would one show that the simple language $L = {s: |s| equiv 0 mod 2}$ has a finite number of equivalence classes?
I think that showing that there is a surjective mapping is not sufficient, because the image may still be of infinite size.
complexity-theory computability regular-languages
asked 4 hours ago
AleksandrAleksandr
153
$endgroup$
add a comment |
$begingroup$
Showing that a language is not regular is straight-forward, because all one needs to do is find an infinite set of inputs which has an injective mapping to the set of equivalence classes which compose that language.
How does one show that the set of equivalence classes of $L$ is finite? For instance, how would one show that the simple language $L = {s: |s| equiv 0 mod 2}$ has a finite number of equivalence classes?
I think that showing that there is a surjective mapping is not sufficient, because the image may still be of infinite size.
complexity-theory computability regular-languages
asked 4 hours ago
AleksandrAleksandr
153
$endgroup$
add a comment |
$begingroup$
Showing that a language is not regular is straight-forward, because all one needs to do is find an infinite set of inputs which has an injective mapping to the set of equivalence classes which compose that language.
How does one show that the set of equivalence classes of $L$ is finite? For instance, how would one show that the simple language $L = {s: |s| equiv 0 mod 2}$ has a finite number of equivalence classes?
I think that showing that there is a surjective mapping is not sufficient, because the image may still be of infinite size.
complexity-theory computability regular-languages
asked 4 hours ago
AleksandrAleksandr
153
$endgroup$
Showing that a language is not regular is straight-forward, because all one needs to do is find an infinite set of inputs which has an injective mapping to the set of equivalence classes which compose that language.
How does one show that the set of equivalence classes of $L$ is finite? For instance, how would one show that the simple language $L = {s: |s| equiv 0 mod 2}$ has a finite number of equivalence classes?
I think that showing that there is a surjective mapping is not sufficient, because the image may still be of infinite size.
complexity-theory computability regular-languages
complexity-theory computability regular-languages
asked 4 hours ago
AleksandrAleksandr
153
asked 4 hours ago
AleksandrAleksandr
153
asked 4 hours ago
AleksandrAleksandr
153
asked 4 hours ago
AleksandrAleksandr
153
asked 4 hours ago
AleksandrAleksandr
153
153
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is just as straightforward to show a language is regular using Nerod-Myhill theorem as to show a language is not regular using that theorem since Nerode-Myhill theorem characterizes when a language is a regular.
For example, let us take the simple language $L={s: |s|equiv 0 text{ (mod 2)}}$. There are two equivalent classes of $L$, assuming $a$ is a letter in the alphabet.
- $left[epsilonright]_{equiv_{L}}=left{s: |s|equiv 0text{ (mod 2)}right}$
- $left[aright]_{equiv_{L}}=left{s: |s|equiv 1text{ (mod 2)}right}$
It should be easy for you to show that the above two classes are well-defined equivalence classes of $L$. There are no other equivalence class of $L$, since every word belongs to one of two classes. If a word is of even length, it belongs to class $left[epsilonright]_{equiv_{L}}$; otherwise, it belongs to class $left[aright]_{equiv_{L}}$.
Hence, there are two equivalence classes of $L$ in total. That is, the number of equivalence classes of $L$ is finite. According to the Nerode-Myhill theorem, $L$ must be regular.
I think that showing that there is a surjective mapping is not sufficient, because the image may still be of infinite size.
I am not sure which surjective mapping you are talking about.
As you can see from the example above, it is enough to show that the union of the finitely many equivalence classes you have found contains all words. Why? If we have another equivalence class $E,$ let $win E$. Then $w$ must belong to $F$, one of the equivalence classes you have found. Since $E$ and $F$ share one word, they are the same equivalence class. That is, $E$ has been found.
You may want to check a few related questions and answers such as finding separating words (Nerode) and how I can find all equivalence classes by Myhill-Nerode?.
answered 3 hours ago
Apass.JackApass.Jack
10.9k1939
$endgroup$
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f104571%2fhow-does-one-use-the-nerode-myhill-theorem-to-prove-that-a-language-is-regular%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
$begingroup$
It is just as straightforward to show a language is regular using Nerod-Myhill theorem as to show a language is not regular using that theorem since Nerode-Myhill theorem characterizes when a language is a regular.
For example, let us take the simple language $L={s: |s|equiv 0 text{ (mod 2)}}$. There are two equivalent classes of $L$, assuming $a$ is a letter in the alphabet.
- $left[epsilonright]_{equiv_{L}}=left{s: |s|equiv 0text{ (mod 2)}right}$
- $left[aright]_{equiv_{L}}=left{s: |s|equiv 1text{ (mod 2)}right}$
It should be easy for you to show that the above two classes are well-defined equivalence classes of $L$. There are no other equivalence class of $L$, since every word belongs to one of two classes. If a word is of even length, it belongs to class $left[epsilonright]_{equiv_{L}}$; otherwise, it belongs to class $left[aright]_{equiv_{L}}$.
Hence, there are two equivalence classes of $L$ in total. That is, the number of equivalence classes of $L$ is finite. According to the Nerode-Myhill theorem, $L$ must be regular.
I think that showing that there is a surjective mapping is not sufficient, because the image may still be of infinite size.
I am not sure which surjective mapping you are talking about.
As you can see from the example above, it is enough to show that the union of the finitely many equivalence classes you have found contains all words. Why? If we have another equivalence class $E,$ let $win E$. Then $w$ must belong to $F$, one of the equivalence classes you have found. Since $E$ and $F$ share one word, they are the same equivalence class. That is, $E$ has been found.
You may want to check a few related questions and answers such as finding separating words (Nerode) and how I can find all equivalence classes by Myhill-Nerode?.
answered 3 hours ago
Apass.JackApass.Jack
10.9k1939
$endgroup$
add a comment |
$begingroup$
It is just as straightforward to show a language is regular using Nerod-Myhill theorem as to show a language is not regular using that theorem since Nerode-Myhill theorem characterizes when a language is a regular.
For example, let us take the simple language $L={s: |s|equiv 0 text{ (mod 2)}}$. There are two equivalent classes of $L$, assuming $a$ is a letter in the alphabet.
- $left[epsilonright]_{equiv_{L}}=left{s: |s|equiv 0text{ (mod 2)}right}$
- $left[aright]_{equiv_{L}}=left{s: |s|equiv 1text{ (mod 2)}right}$
It should be easy for you to show that the above two classes are well-defined equivalence classes of $L$. There are no other equivalence class of $L$, since every word belongs to one of two classes. If a word is of even length, it belongs to class $left[epsilonright]_{equiv_{L}}$; otherwise, it belongs to class $left[aright]_{equiv_{L}}$.
Hence, there are two equivalence classes of $L$ in total. That is, the number of equivalence classes of $L$ is finite. According to the Nerode-Myhill theorem, $L$ must be regular.
I think that showing that there is a surjective mapping is not sufficient, because the image may still be of infinite size.
I am not sure which surjective mapping you are talking about.
As you can see from the example above, it is enough to show that the union of the finitely many equivalence classes you have found contains all words. Why? If we have another equivalence class $E,$ let $win E$. Then $w$ must belong to $F$, one of the equivalence classes you have found. Since $E$ and $F$ share one word, they are the same equivalence class. That is, $E$ has been found.
You may want to check a few related questions and answers such as finding separating words (Nerode) and how I can find all equivalence classes by Myhill-Nerode?.
answered 3 hours ago
Apass.JackApass.Jack
10.9k1939
$endgroup$
add a comment |
$begingroup$
It is just as straightforward to show a language is regular using Nerod-Myhill theorem as to show a language is not regular using that theorem since Nerode-Myhill theorem characterizes when a language is a regular.
For example, let us take the simple language $L={s: |s|equiv 0 text{ (mod 2)}}$. There are two equivalent classes of $L$, assuming $a$ is a letter in the alphabet.
- $left[epsilonright]_{equiv_{L}}=left{s: |s|equiv 0text{ (mod 2)}right}$
- $left[aright]_{equiv_{L}}=left{s: |s|equiv 1text{ (mod 2)}right}$
It should be easy for you to show that the above two classes are well-defined equivalence classes of $L$. There are no other equivalence class of $L$, since every word belongs to one of two classes. If a word is of even length, it belongs to class $left[epsilonright]_{equiv_{L}}$; otherwise, it belongs to class $left[aright]_{equiv_{L}}$.
Hence, there are two equivalence classes of $L$ in total. That is, the number of equivalence classes of $L$ is finite. According to the Nerode-Myhill theorem, $L$ must be regular.
I think that showing that there is a surjective mapping is not sufficient, because the image may still be of infinite size.
I am not sure which surjective mapping you are talking about.
As you can see from the example above, it is enough to show that the union of the finitely many equivalence classes you have found contains all words. Why? If we have another equivalence class $E,$ let $win E$. Then $w$ must belong to $F$, one of the equivalence classes you have found. Since $E$ and $F$ share one word, they are the same equivalence class. That is, $E$ has been found.
You may want to check a few related questions and answers such as finding separating words (Nerode) and how I can find all equivalence classes by Myhill-Nerode?.
answered 3 hours ago
Apass.JackApass.Jack
10.9k1939
$endgroup$
It is just as straightforward to show a language is regular using Nerod-Myhill theorem as to show a language is not regular using that theorem since Nerode-Myhill theorem characterizes when a language is a regular.
For example, let us take the simple language $L={s: |s|equiv 0 text{ (mod 2)}}$. There are two equivalent classes of $L$, assuming $a$ is a letter in the alphabet.
- $left[epsilonright]_{equiv_{L}}=left{s: |s|equiv 0text{ (mod 2)}right}$
- $left[aright]_{equiv_{L}}=left{s: |s|equiv 1text{ (mod 2)}right}$
It should be easy for you to show that the above two classes are well-defined equivalence classes of $L$. There are no other equivalence class of $L$, since every word belongs to one of two classes. If a word is of even length, it belongs to class $left[epsilonright]_{equiv_{L}}$; otherwise, it belongs to class $left[aright]_{equiv_{L}}$.
Hence, there are two equivalence classes of $L$ in total. That is, the number of equivalence classes of $L$ is finite. According to the Nerode-Myhill theorem, $L$ must be regular.
I think that showing that there is a surjective mapping is not sufficient, because the image may still be of infinite size.
I am not sure which surjective mapping you are talking about.
As you can see from the example above, it is enough to show that the union of the finitely many equivalence classes you have found contains all words. Why? If we have another equivalence class $E,$ let $win E$. Then $w$ must belong to $F$, one of the equivalence classes you have found. Since $E$ and $F$ share one word, they are the same equivalence class. That is, $E$ has been found.
You may want to check a few related questions and answers such as finding separating words (Nerode) and how I can find all equivalence classes by Myhill-Nerode?.
answered 3 hours ago
Apass.JackApass.Jack
10.9k1939
edited 2 hours ago
answered 3 hours ago
Apass.JackApass.Jack
10.9k1939
answered 3 hours ago
Apass.JackApass.Jack
10.9k1939
answered 3 hours ago
Apass.JackApass.Jack
10.9k1939
10.9k1939
add a comment |
add a comment |
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.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f104571%2fhow-does-one-use-the-nerode-myhill-theorem-to-prove-that-a-language-is-regular%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
