$L_1 =(a^nb^n)$ and $L_2 =(a^nb^{2n})$. Is $L_1 cup L_2$ DCFL?
1
I think that since $a^nb^n$ is not regular (applied pumping lemma), so is $L_2$.
Therefore, $L_1 cup L_2$ is not cfL.
Is that correct?
context-free-grammar regular-expressions
asked Dec 11 '14 at 7:52
Sachin DivakarSachin Divakar
1076
add a comment |
1
I think that since $a^nb^n$ is not regular (applied pumping lemma), so is $L_2$.
Therefore, $L_1 cup L_2$ is not cfL.
Is that correct?
context-free-grammar regular-expressions
asked Dec 11 '14 at 7:52
Sachin DivakarSachin Divakar
1076
Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
– Steven Stadnicki
Dec 11 '14 at 8:47
Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
– Sachin Divakar
Dec 11 '14 at 8:59
add a comment |
1
1
1
I think that since $a^nb^n$ is not regular (applied pumping lemma), so is $L_2$.
Therefore, $L_1 cup L_2$ is not cfL.
Is that correct?
context-free-grammar regular-expressions
asked Dec 11 '14 at 7:52
Sachin DivakarSachin Divakar
1076
I think that since $a^nb^n$ is not regular (applied pumping lemma), so is $L_2$.
Therefore, $L_1 cup L_2$ is not cfL.
Is that correct?
context-free-grammar regular-expressions
context-free-grammar regular-expressions
asked Dec 11 '14 at 7:52
Sachin DivakarSachin Divakar
1076
asked Dec 11 '14 at 7:52
Sachin DivakarSachin Divakar
1076
edited Dec 11 '14 at 8:36
Sachin Divakar
asked Dec 11 '14 at 7:52
Sachin DivakarSachin Divakar
1076
asked Dec 11 '14 at 7:52
Sachin DivakarSachin Divakar
1076
asked Dec 11 '14 at 7:52
Sachin DivakarSachin Divakar
1076
1076
Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
– Steven Stadnicki
Dec 11 '14 at 8:47
Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
– Sachin Divakar
Dec 11 '14 at 8:59
add a comment |
Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
– Steven Stadnicki
Dec 11 '14 at 8:47
Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
– Sachin Divakar
Dec 11 '14 at 8:59
Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
– Steven Stadnicki
Dec 11 '14 at 8:47
Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
– Steven Stadnicki
Dec 11 '14 at 8:47
Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
– Sachin Divakar
Dec 11 '14 at 8:59
Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
– Sachin Divakar
Dec 11 '14 at 8:59
add a comment |
1 Answer
1
active
oldest
votes
0
Hint:
$L_1$ is produced by
S := $epsilon$
| aSb
$L_2$ is produced by
S := $epsilon$
| aSbb
(assuming $a^a$ is a typo, meaning $a^n$)
answered Dec 11 '14 at 8:09
RonaldRonald
673510
yes $a^n$ but is it dCFL
– Sachin Divakar
Dec 11 '14 at 8:32
1
Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
– Ronald
Dec 11 '14 at 10:42
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',
autoActivateHeartbeat: false,
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
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1063132%2fl-1-anbn-and-l-2-anb2n-is-l-1-cup-l-2-dcfl%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
0
Hint:
$L_1$ is produced by
S := $epsilon$
| aSb
$L_2$ is produced by
S := $epsilon$
| aSbb
(assuming $a^a$ is a typo, meaning $a^n$)
answered Dec 11 '14 at 8:09
RonaldRonald
673510
yes $a^n$ but is it dCFL
– Sachin Divakar
Dec 11 '14 at 8:32
1
Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
– Ronald
Dec 11 '14 at 10:42
add a comment |
0
Hint:
$L_1$ is produced by
S := $epsilon$
| aSb
$L_2$ is produced by
S := $epsilon$
| aSbb
(assuming $a^a$ is a typo, meaning $a^n$)
answered Dec 11 '14 at 8:09
RonaldRonald
673510
yes $a^n$ but is it dCFL
– Sachin Divakar
Dec 11 '14 at 8:32
1
Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
– Ronald
Dec 11 '14 at 10:42
add a comment |
0
0
0
Hint:
$L_1$ is produced by
S := $epsilon$
| aSb
$L_2$ is produced by
S := $epsilon$
| aSbb
(assuming $a^a$ is a typo, meaning $a^n$)
answered Dec 11 '14 at 8:09
RonaldRonald
673510
Hint:
$L_1$ is produced by
S := $epsilon$
| aSb
$L_2$ is produced by
S := $epsilon$
| aSbb
(assuming $a^a$ is a typo, meaning $a^n$)
answered Dec 11 '14 at 8:09
RonaldRonald
673510
answered Dec 11 '14 at 8:09
RonaldRonald
673510
answered Dec 11 '14 at 8:09
RonaldRonald
673510
answered Dec 11 '14 at 8:09
RonaldRonald
673510
673510
yes $a^n$ but is it dCFL
– Sachin Divakar
Dec 11 '14 at 8:32
1
Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
– Ronald
Dec 11 '14 at 10:42
add a comment |
yes $a^n$ but is it dCFL
– Sachin Divakar
Dec 11 '14 at 8:32
1
Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
– Ronald
Dec 11 '14 at 10:42
yes $a^n$ but is it dCFL
– Sachin Divakar
Dec 11 '14 at 8:32
yes $a^n$ but is it dCFL
– Sachin Divakar
Dec 11 '14 at 8:32
1
1
Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
– Ronald
Dec 11 '14 at 10:42
Both $L_1$ and $L_2$ are DCFL. Their union isn't because it cannot be deterministically decided to process single b's or pairs of b's after a sequence of a's.
– Ronald
Dec 11 '14 at 10:42
add a comment |
draft saved
draft discarded
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.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1063132%2fl-1-anbn-and-l-2-anb2n-is-l-1-cup-l-2-dcfl%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
Being CFL is entirely different from being a DCFL and both are entirely different from being regular. Which do you actually want?
– Steven Stadnicki
Dec 11 '14 at 8:47
Is it DCFL or not/am i understanding correctly being DCFL means it can be accepted by a deterministic PDA
– Sachin Divakar
Dec 11 '14 at 8:59