prove or disprove if the following language is regular language
for $A,Bsubseteq{0,1}^*$ regular languages
prove or disprove that $L_2$ is a regular language:
$L_2 = {x in A | exists y in B : |x|_1 =|y|_1 }$
$|x|_1$ means the number of appearances of 1 in the word.
if $L_2$ regular language demonstrate his automata, else disprove with pump lemma or with clousre properties.
I think $L_2$ is a regular language because I can choose $A=B$ for example and I'll get regular language or for example $B={0,1}^*$.
I am not sure how to demonstrate his automata.
formal-languages automata
asked Nov 24 at 22:09
UltimateMath
746
add a comment |
for $A,Bsubseteq{0,1}^*$ regular languages
prove or disprove that $L_2$ is a regular language:
$L_2 = {x in A | exists y in B : |x|_1 =|y|_1 }$
$|x|_1$ means the number of appearances of 1 in the word.
if $L_2$ regular language demonstrate his automata, else disprove with pump lemma or with clousre properties.
I think $L_2$ is a regular language because I can choose $A=B$ for example and I'll get regular language or for example $B={0,1}^*$.
I am not sure how to demonstrate his automata.
formal-languages automata
asked Nov 24 at 22:09
UltimateMath
746
Are $A$ and $B$ regular languages? Also, I assume prove that "$L_2$ is a formal language" is a typo, since this would be trivial and doesn't match the title. Should it say "regular language"?
– Joey Kilpatrick
Nov 25 at 4:02
@JoeyKilpatrick you are right, edited.
– UltimateMath
Nov 25 at 7:43
Let $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$, then $L_2 = A cap B'$, can you find an automatum for $B'$?
– Alex Vong
Nov 25 at 8:01
add a comment |
for $A,Bsubseteq{0,1}^*$ regular languages
prove or disprove that $L_2$ is a regular language:
$L_2 = {x in A | exists y in B : |x|_1 =|y|_1 }$
$|x|_1$ means the number of appearances of 1 in the word.
if $L_2$ regular language demonstrate his automata, else disprove with pump lemma or with clousre properties.
I think $L_2$ is a regular language because I can choose $A=B$ for example and I'll get regular language or for example $B={0,1}^*$.
I am not sure how to demonstrate his automata.
formal-languages automata
asked Nov 24 at 22:09
UltimateMath
746
for $A,Bsubseteq{0,1}^*$ regular languages
prove or disprove that $L_2$ is a regular language:
$L_2 = {x in A | exists y in B : |x|_1 =|y|_1 }$
$|x|_1$ means the number of appearances of 1 in the word.
if $L_2$ regular language demonstrate his automata, else disprove with pump lemma or with clousre properties.
I think $L_2$ is a regular language because I can choose $A=B$ for example and I'll get regular language or for example $B={0,1}^*$.
I am not sure how to demonstrate his automata.
formal-languages automata
formal-languages automata
asked Nov 24 at 22:09
UltimateMath
746
asked Nov 24 at 22:09
UltimateMath
746
edited Nov 25 at 7:40
asked Nov 24 at 22:09
UltimateMath
746
asked Nov 24 at 22:09
UltimateMath
746
asked Nov 24 at 22:09
UltimateMath
746
746
Are $A$ and $B$ regular languages? Also, I assume prove that "$L_2$ is a formal language" is a typo, since this would be trivial and doesn't match the title. Should it say "regular language"?
– Joey Kilpatrick
Nov 25 at 4:02
@JoeyKilpatrick you are right, edited.
– UltimateMath
Nov 25 at 7:43
Let $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$, then $L_2 = A cap B'$, can you find an automatum for $B'$?
– Alex Vong
Nov 25 at 8:01
add a comment |
Are $A$ and $B$ regular languages? Also, I assume prove that "$L_2$ is a formal language" is a typo, since this would be trivial and doesn't match the title. Should it say "regular language"?
– Joey Kilpatrick
Nov 25 at 4:02
@JoeyKilpatrick you are right, edited.
– UltimateMath
Nov 25 at 7:43
Let $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$, then $L_2 = A cap B'$, can you find an automatum for $B'$?
– Alex Vong
Nov 25 at 8:01
Are $A$ and $B$ regular languages? Also, I assume prove that "$L_2$ is a formal language" is a typo, since this would be trivial and doesn't match the title. Should it say "regular language"?
– Joey Kilpatrick
Nov 25 at 4:02
Are $A$ and $B$ regular languages? Also, I assume prove that "$L_2$ is a formal language" is a typo, since this would be trivial and doesn't match the title. Should it say "regular language"?
– Joey Kilpatrick
Nov 25 at 4:02
@JoeyKilpatrick you are right, edited.
– UltimateMath
Nov 25 at 7:43
@JoeyKilpatrick you are right, edited.
– UltimateMath
Nov 25 at 7:43
Let $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$, then $L_2 = A cap B'$, can you find an automatum for $B'$?
– Alex Vong
Nov 25 at 8:01
Let $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$, then $L_2 = A cap B'$, can you find an automatum for $B'$?
– Alex Vong
Nov 25 at 8:01
add a comment |
2 Answers
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I am not entire sure if my approach work, but here we go:
Firstly, we observe that we can write $L_2 = A cap B'$ where $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$. Since we can create a FSA for the intersection of two regular languages, it suffices to find a FSA for $B'$.
Before proceeding, consider a more concrete description of $B'$. Note that we can write $$B' = {0^{alpha_0} 1 0^{alpha_1} 1 0^{alpha_2} dots 1 0^{alpha_{|y|_1}} mid alpha_j in mathbb{N}, y in B}$$ Intuitively, $B'$ is constructed by taking all strings in $B$, chops away all $0$'s, and adding back any number of $0$'s in between.
Now given a FSA $M$ for $B$, we can construct a corresponding FSA $M'$ for $B'$ by the following transformation:
Case 1:
should be converted to
Case 2:
Nothing needs to be changed.
Case 3:
should be converted to
Case 4:
Nothing needs to be changed.
This completes the FSA.
answered Nov 25 at 12:05
Alex Vong
1,227819
add a comment |
Let $pi: {0,1}^* to 1^*$ be the monoid morphism defined by $pi(u) = |u|_1$. Since regular languages are closed under morphisms and inverses of morphisms, $R =pi(B)$ is regular and $L_2 = A cap pi^{-1}(R)$ is regular.
answered Nov 25 at 11:23
J.-E. Pin
18.3k21754
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
I am not entire sure if my approach work, but here we go:
Firstly, we observe that we can write $L_2 = A cap B'$ where $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$. Since we can create a FSA for the intersection of two regular languages, it suffices to find a FSA for $B'$.
Before proceeding, consider a more concrete description of $B'$. Note that we can write $$B' = {0^{alpha_0} 1 0^{alpha_1} 1 0^{alpha_2} dots 1 0^{alpha_{|y|_1}} mid alpha_j in mathbb{N}, y in B}$$ Intuitively, $B'$ is constructed by taking all strings in $B$, chops away all $0$'s, and adding back any number of $0$'s in between.
Now given a FSA $M$ for $B$, we can construct a corresponding FSA $M'$ for $B'$ by the following transformation:
Case 1:
should be converted to
Case 2:
Nothing needs to be changed.
Case 3:
should be converted to
Case 4:
Nothing needs to be changed.
This completes the FSA.
answered Nov 25 at 12:05
Alex Vong
1,227819
add a comment |
I am not entire sure if my approach work, but here we go:
Firstly, we observe that we can write $L_2 = A cap B'$ where $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$. Since we can create a FSA for the intersection of two regular languages, it suffices to find a FSA for $B'$.
Before proceeding, consider a more concrete description of $B'$. Note that we can write $$B' = {0^{alpha_0} 1 0^{alpha_1} 1 0^{alpha_2} dots 1 0^{alpha_{|y|_1}} mid alpha_j in mathbb{N}, y in B}$$ Intuitively, $B'$ is constructed by taking all strings in $B$, chops away all $0$'s, and adding back any number of $0$'s in between.
Now given a FSA $M$ for $B$, we can construct a corresponding FSA $M'$ for $B'$ by the following transformation:
Case 1:
should be converted to
Case 2:
Nothing needs to be changed.
Case 3:
should be converted to
Case 4:
Nothing needs to be changed.
This completes the FSA.
answered Nov 25 at 12:05
Alex Vong
1,227819
add a comment |
I am not entire sure if my approach work, but here we go:
Firstly, we observe that we can write $L_2 = A cap B'$ where $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$. Since we can create a FSA for the intersection of two regular languages, it suffices to find a FSA for $B'$.
Before proceeding, consider a more concrete description of $B'$. Note that we can write $$B' = {0^{alpha_0} 1 0^{alpha_1} 1 0^{alpha_2} dots 1 0^{alpha_{|y|_1}} mid alpha_j in mathbb{N}, y in B}$$ Intuitively, $B'$ is constructed by taking all strings in $B$, chops away all $0$'s, and adding back any number of $0$'s in between.
Now given a FSA $M$ for $B$, we can construct a corresponding FSA $M'$ for $B'$ by the following transformation:
Case 1:
should be converted to
Case 2:
Nothing needs to be changed.
Case 3:
should be converted to
Case 4:
Nothing needs to be changed.
This completes the FSA.
answered Nov 25 at 12:05
Alex Vong
1,227819
I am not entire sure if my approach work, but here we go:
Firstly, we observe that we can write $L_2 = A cap B'$ where $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$. Since we can create a FSA for the intersection of two regular languages, it suffices to find a FSA for $B'$.
Before proceeding, consider a more concrete description of $B'$. Note that we can write $$B' = {0^{alpha_0} 1 0^{alpha_1} 1 0^{alpha_2} dots 1 0^{alpha_{|y|_1}} mid alpha_j in mathbb{N}, y in B}$$ Intuitively, $B'$ is constructed by taking all strings in $B$, chops away all $0$'s, and adding back any number of $0$'s in between.
Now given a FSA $M$ for $B$, we can construct a corresponding FSA $M'$ for $B'$ by the following transformation:
Case 1:
should be converted to
Case 2:
Nothing needs to be changed.
Case 3:
should be converted to
Case 4:
Nothing needs to be changed.
This completes the FSA.
answered Nov 25 at 12:05
Alex Vong
1,227819
answered Nov 25 at 12:05
Alex Vong
1,227819
answered Nov 25 at 12:05
Alex Vong
1,227819
answered Nov 25 at 12:05
Alex Vong
1,227819
1,227819
add a comment |
add a comment |
Let $pi: {0,1}^* to 1^*$ be the monoid morphism defined by $pi(u) = |u|_1$. Since regular languages are closed under morphisms and inverses of morphisms, $R =pi(B)$ is regular and $L_2 = A cap pi^{-1}(R)$ is regular.
answered Nov 25 at 11:23
J.-E. Pin
18.3k21754
add a comment |
Let $pi: {0,1}^* to 1^*$ be the monoid morphism defined by $pi(u) = |u|_1$. Since regular languages are closed under morphisms and inverses of morphisms, $R =pi(B)$ is regular and $L_2 = A cap pi^{-1}(R)$ is regular.
answered Nov 25 at 11:23
J.-E. Pin
18.3k21754
add a comment |
Let $pi: {0,1}^* to 1^*$ be the monoid morphism defined by $pi(u) = |u|_1$. Since regular languages are closed under morphisms and inverses of morphisms, $R =pi(B)$ is regular and $L_2 = A cap pi^{-1}(R)$ is regular.
answered Nov 25 at 11:23
J.-E. Pin
18.3k21754
Let $pi: {0,1}^* to 1^*$ be the monoid morphism defined by $pi(u) = |u|_1$. Since regular languages are closed under morphisms and inverses of morphisms, $R =pi(B)$ is regular and $L_2 = A cap pi^{-1}(R)$ is regular.
answered Nov 25 at 11:23
J.-E. Pin
18.3k21754
edited Nov 25 at 12:22
answered Nov 25 at 11:23
J.-E. Pin
18.3k21754
answered Nov 25 at 11:23
J.-E. Pin
18.3k21754
answered Nov 25 at 11:23
J.-E. Pin
18.3k21754
18.3k21754
add a comment |
add a comment |
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Are $A$ and $B$ regular languages? Also, I assume prove that "$L_2$ is a formal language" is a typo, since this would be trivial and doesn't match the title. Should it say "regular language"?
– Joey Kilpatrick
Nov 25 at 4:02
@JoeyKilpatrick you are right, edited.
– UltimateMath
Nov 25 at 7:43
Let $B' = {s in Sigma^* mid |s|_1 = |y|_1 text{ for some } y in B}$, then $L_2 = A cap B'$, can you find an automatum for $B'$?
– Alex Vong
Nov 25 at 8:01