Show that the following language is context-free/not context free by expressing the language as the union of...
up vote
1
down vote
favorite
I want to show that the language $L = $ {$a^mba^nba^p:m=n $ or $n = p$ or $m = p$} is either context-free or not context free by expressing the language as a union of three other languages $L_1$, $L_2$, and $L_3$.
By knowing if these three other languages are context-free/not context-free, I hope to indicate whether the language $L$ is context-free/not context-free.
Can anyone help me pick these three other languages that when unioned together form $L$?
combinatorics formal-languages regular-language context-free-grammar
asked Nov 19 at 18:06
etnie1031
133
add a comment |
up vote
1
down vote
favorite
I want to show that the language $L = $ {$a^mba^nba^p:m=n $ or $n = p$ or $m = p$} is either context-free or not context free by expressing the language as a union of three other languages $L_1$, $L_2$, and $L_3$.
By knowing if these three other languages are context-free/not context-free, I hope to indicate whether the language $L$ is context-free/not context-free.
Can anyone help me pick these three other languages that when unioned together form $L$?
combinatorics formal-languages regular-language context-free-grammar
asked Nov 19 at 18:06
etnie1031
133
Doesn't the use of "or" in the language description to join three possible conditions give you a clue?
– rici
Nov 19 at 18:24
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to show that the language $L = $ {$a^mba^nba^p:m=n $ or $n = p$ or $m = p$} is either context-free or not context free by expressing the language as a union of three other languages $L_1$, $L_2$, and $L_3$.
By knowing if these three other languages are context-free/not context-free, I hope to indicate whether the language $L$ is context-free/not context-free.
Can anyone help me pick these three other languages that when unioned together form $L$?
combinatorics formal-languages regular-language context-free-grammar
asked Nov 19 at 18:06
etnie1031
133
I want to show that the language $L = $ {$a^mba^nba^p:m=n $ or $n = p$ or $m = p$} is either context-free or not context free by expressing the language as a union of three other languages $L_1$, $L_2$, and $L_3$.
By knowing if these three other languages are context-free/not context-free, I hope to indicate whether the language $L$ is context-free/not context-free.
Can anyone help me pick these three other languages that when unioned together form $L$?
combinatorics formal-languages regular-language context-free-grammar
combinatorics formal-languages regular-language context-free-grammar
asked Nov 19 at 18:06
etnie1031
133
asked Nov 19 at 18:06
etnie1031
133
asked Nov 19 at 18:06
etnie1031
133
asked Nov 19 at 18:06
etnie1031
133
asked Nov 19 at 18:06
etnie1031
133
133
Doesn't the use of "or" in the language description to join three possible conditions give you a clue?
– rici
Nov 19 at 18:24
add a comment |
Doesn't the use of "or" in the language description to join three possible conditions give you a clue?
– rici
Nov 19 at 18:24
Doesn't the use of "or" in the language description to join three possible conditions give you a clue?
– rici
Nov 19 at 18:24
Doesn't the use of "or" in the language description to join three possible conditions give you a clue?
– rici
Nov 19 at 18:24
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Let $L_1={a^mba^nba^p:m=n}$. Consider the following grammar that generates this language with starting symbol $S$: $$
Slongrightarrow Cbtext{ }|text{ }Sa \
C longrightarrow btext{ }|text{ }aCa
$$
This grammar is context-free, so $L_1$ is context-free. A similar grammar will show that $L_2={a^mba^nba^p:n=p}$ is context-free.
Let $L_3={a^mba^nba^p:m=p}$. Consider the following grammar that generates this language with starting symbol $S$:
$$
Slongrightarrow bCbtext{ }|text{ }aSa \
C longrightarrow epsilontext{ }|text{ }aC
$$
This grammar is context-free, so $L_3$ is context-free. Since $L=L_1cup L_2cup L_3$, then $L$ is context-free.
answered Nov 19 at 19:06
Joey Kilpatrick
1,183422
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: "69"
};
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',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: 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%2fmath.stackexchange.com%2fquestions%2f3005281%2fshow-that-the-following-language-is-context-free-not-context-free-by-expressing%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
up vote
0
down vote
Let $L_1={a^mba^nba^p:m=n}$. Consider the following grammar that generates this language with starting symbol $S$: $$
Slongrightarrow Cbtext{ }|text{ }Sa \
C longrightarrow btext{ }|text{ }aCa
$$
This grammar is context-free, so $L_1$ is context-free. A similar grammar will show that $L_2={a^mba^nba^p:n=p}$ is context-free.
Let $L_3={a^mba^nba^p:m=p}$. Consider the following grammar that generates this language with starting symbol $S$:
$$
Slongrightarrow bCbtext{ }|text{ }aSa \
C longrightarrow epsilontext{ }|text{ }aC
$$
This grammar is context-free, so $L_3$ is context-free. Since $L=L_1cup L_2cup L_3$, then $L$ is context-free.
answered Nov 19 at 19:06
Joey Kilpatrick
1,183422
add a comment |
up vote
0
down vote
Let $L_1={a^mba^nba^p:m=n}$. Consider the following grammar that generates this language with starting symbol $S$: $$
Slongrightarrow Cbtext{ }|text{ }Sa \
C longrightarrow btext{ }|text{ }aCa
$$
This grammar is context-free, so $L_1$ is context-free. A similar grammar will show that $L_2={a^mba^nba^p:n=p}$ is context-free.
Let $L_3={a^mba^nba^p:m=p}$. Consider the following grammar that generates this language with starting symbol $S$:
$$
Slongrightarrow bCbtext{ }|text{ }aSa \
C longrightarrow epsilontext{ }|text{ }aC
$$
This grammar is context-free, so $L_3$ is context-free. Since $L=L_1cup L_2cup L_3$, then $L$ is context-free.
answered Nov 19 at 19:06
Joey Kilpatrick
1,183422
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $L_1={a^mba^nba^p:m=n}$. Consider the following grammar that generates this language with starting symbol $S$: $$
Slongrightarrow Cbtext{ }|text{ }Sa \
C longrightarrow btext{ }|text{ }aCa
$$
This grammar is context-free, so $L_1$ is context-free. A similar grammar will show that $L_2={a^mba^nba^p:n=p}$ is context-free.
Let $L_3={a^mba^nba^p:m=p}$. Consider the following grammar that generates this language with starting symbol $S$:
$$
Slongrightarrow bCbtext{ }|text{ }aSa \
C longrightarrow epsilontext{ }|text{ }aC
$$
This grammar is context-free, so $L_3$ is context-free. Since $L=L_1cup L_2cup L_3$, then $L$ is context-free.
answered Nov 19 at 19:06
Joey Kilpatrick
1,183422
Let $L_1={a^mba^nba^p:m=n}$. Consider the following grammar that generates this language with starting symbol $S$: $$
Slongrightarrow Cbtext{ }|text{ }Sa \
C longrightarrow btext{ }|text{ }aCa
$$
This grammar is context-free, so $L_1$ is context-free. A similar grammar will show that $L_2={a^mba^nba^p:n=p}$ is context-free.
Let $L_3={a^mba^nba^p:m=p}$. Consider the following grammar that generates this language with starting symbol $S$:
$$
Slongrightarrow bCbtext{ }|text{ }aSa \
C longrightarrow epsilontext{ }|text{ }aC
$$
This grammar is context-free, so $L_3$ is context-free. Since $L=L_1cup L_2cup L_3$, then $L$ is context-free.
answered Nov 19 at 19:06
Joey Kilpatrick
1,183422
answered Nov 19 at 19:06
Joey Kilpatrick
1,183422
answered Nov 19 at 19:06
Joey Kilpatrick
1,183422
answered Nov 19 at 19:06
Joey Kilpatrick
1,183422
1,183422
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2fmath.stackexchange.com%2fquestions%2f3005281%2fshow-that-the-following-language-is-context-free-not-context-free-by-expressing%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
Doesn't the use of "or" in the language description to join three possible conditions give you a clue?
– rici
Nov 19 at 18:24