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Are regular expressions a*a and aa* equivalent?


How to convert finite automata to regular expressions?Finding simpler equivalent regular expressionsRegular expressions and semi-linear sets∅-free regular expressions?Why are regular expressions defined with union, concatenation and star operations?How to generate regular expression for language of regular expressions?How to read regular expressions?Regular expression for even/odd string on alphabetDo all Regular Expressions describe Regular Languages?How to prove that $$x$$ is a regular language if $x$ is derived from $L=w$ by substituting substrings?













1












$begingroup$


Let $Sigma = a, b$ an alphabet. Are the regular expressions $a^*a$ and $aa^*$ over $Sigma$ equivalent? Even though concatenation is not commutative, in this case it seems like the statement is correct, but I am not sure.










share|cite|improve this question











$endgroup$











  • $begingroup$
    They are equivalent because both accept exactly "a string of at least one a", but for educational purposes it would be interesting to see a formal proof, which I can't produce.
    $endgroup$
    – Albert Hendriks
    13 hours ago










  • $begingroup$
    me neither, that's why I am not sure
    $endgroup$
    – Yamahari
    13 hours ago










  • $begingroup$
    You can prove it directly on the regular expressions: Let $w in L(aa^ast)$, i.e. $w = a cdot u$ with $u in L(a^ast)$, i.e. $u = a^k$. for some $k in mathbbN$. Then $w = a^k+1 = a^k a in L(a^ast a)$. The other direction is analogous. However, for these simple expression you might also want to compute the corresponding NFAs and transform them to the minimal DFAs. If these DFAs coincide the regexes have also to be equivalent.
    $endgroup$
    – ttnick
    12 hours ago











  • $begingroup$
    @ttnick Converting to an NFA and then determinizing is much more work than just reasoning about what strings are matched.
    $endgroup$
    – David Richerby
    12 hours ago










  • $begingroup$
    ok this question confused me quite a bit since I read the title first and thought the "?" was part of the regex
    $endgroup$
    – Esben Skov Pedersen
    9 hours ago















1












$begingroup$


Let $Sigma = a, b$ an alphabet. Are the regular expressions $a^*a$ and $aa^*$ over $Sigma$ equivalent? Even though concatenation is not commutative, in this case it seems like the statement is correct, but I am not sure.










share|cite|improve this question











$endgroup$











  • $begingroup$
    They are equivalent because both accept exactly "a string of at least one a", but for educational purposes it would be interesting to see a formal proof, which I can't produce.
    $endgroup$
    – Albert Hendriks
    13 hours ago










  • $begingroup$
    me neither, that's why I am not sure
    $endgroup$
    – Yamahari
    13 hours ago










  • $begingroup$
    You can prove it directly on the regular expressions: Let $w in L(aa^ast)$, i.e. $w = a cdot u$ with $u in L(a^ast)$, i.e. $u = a^k$. for some $k in mathbbN$. Then $w = a^k+1 = a^k a in L(a^ast a)$. The other direction is analogous. However, for these simple expression you might also want to compute the corresponding NFAs and transform them to the minimal DFAs. If these DFAs coincide the regexes have also to be equivalent.
    $endgroup$
    – ttnick
    12 hours ago











  • $begingroup$
    @ttnick Converting to an NFA and then determinizing is much more work than just reasoning about what strings are matched.
    $endgroup$
    – David Richerby
    12 hours ago










  • $begingroup$
    ok this question confused me quite a bit since I read the title first and thought the "?" was part of the regex
    $endgroup$
    – Esben Skov Pedersen
    9 hours ago













1












1








1


1



$begingroup$


Let $Sigma = a, b$ an alphabet. Are the regular expressions $a^*a$ and $aa^*$ over $Sigma$ equivalent? Even though concatenation is not commutative, in this case it seems like the statement is correct, but I am not sure.










share|cite|improve this question











$endgroup$




Let $Sigma = a, b$ an alphabet. Are the regular expressions $a^*a$ and $aa^*$ over $Sigma$ equivalent? Even though concatenation is not commutative, in this case it seems like the statement is correct, but I am not sure.







regular-expressions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









Apass.Jack

12.9k1939




12.9k1939










asked 13 hours ago









YamahariYamahari

465




465











  • $begingroup$
    They are equivalent because both accept exactly "a string of at least one a", but for educational purposes it would be interesting to see a formal proof, which I can't produce.
    $endgroup$
    – Albert Hendriks
    13 hours ago










  • $begingroup$
    me neither, that's why I am not sure
    $endgroup$
    – Yamahari
    13 hours ago










  • $begingroup$
    You can prove it directly on the regular expressions: Let $w in L(aa^ast)$, i.e. $w = a cdot u$ with $u in L(a^ast)$, i.e. $u = a^k$. for some $k in mathbbN$. Then $w = a^k+1 = a^k a in L(a^ast a)$. The other direction is analogous. However, for these simple expression you might also want to compute the corresponding NFAs and transform them to the minimal DFAs. If these DFAs coincide the regexes have also to be equivalent.
    $endgroup$
    – ttnick
    12 hours ago











  • $begingroup$
    @ttnick Converting to an NFA and then determinizing is much more work than just reasoning about what strings are matched.
    $endgroup$
    – David Richerby
    12 hours ago










  • $begingroup$
    ok this question confused me quite a bit since I read the title first and thought the "?" was part of the regex
    $endgroup$
    – Esben Skov Pedersen
    9 hours ago
















  • $begingroup$
    They are equivalent because both accept exactly "a string of at least one a", but for educational purposes it would be interesting to see a formal proof, which I can't produce.
    $endgroup$
    – Albert Hendriks
    13 hours ago










  • $begingroup$
    me neither, that's why I am not sure
    $endgroup$
    – Yamahari
    13 hours ago










  • $begingroup$
    You can prove it directly on the regular expressions: Let $w in L(aa^ast)$, i.e. $w = a cdot u$ with $u in L(a^ast)$, i.e. $u = a^k$. for some $k in mathbbN$. Then $w = a^k+1 = a^k a in L(a^ast a)$. The other direction is analogous. However, for these simple expression you might also want to compute the corresponding NFAs and transform them to the minimal DFAs. If these DFAs coincide the regexes have also to be equivalent.
    $endgroup$
    – ttnick
    12 hours ago











  • $begingroup$
    @ttnick Converting to an NFA and then determinizing is much more work than just reasoning about what strings are matched.
    $endgroup$
    – David Richerby
    12 hours ago










  • $begingroup$
    ok this question confused me quite a bit since I read the title first and thought the "?" was part of the regex
    $endgroup$
    – Esben Skov Pedersen
    9 hours ago















$begingroup$
They are equivalent because both accept exactly "a string of at least one a", but for educational purposes it would be interesting to see a formal proof, which I can't produce.
$endgroup$
– Albert Hendriks
13 hours ago




$begingroup$
They are equivalent because both accept exactly "a string of at least one a", but for educational purposes it would be interesting to see a formal proof, which I can't produce.
$endgroup$
– Albert Hendriks
13 hours ago












$begingroup$
me neither, that's why I am not sure
$endgroup$
– Yamahari
13 hours ago




$begingroup$
me neither, that's why I am not sure
$endgroup$
– Yamahari
13 hours ago












$begingroup$
You can prove it directly on the regular expressions: Let $w in L(aa^ast)$, i.e. $w = a cdot u$ with $u in L(a^ast)$, i.e. $u = a^k$. for some $k in mathbbN$. Then $w = a^k+1 = a^k a in L(a^ast a)$. The other direction is analogous. However, for these simple expression you might also want to compute the corresponding NFAs and transform them to the minimal DFAs. If these DFAs coincide the regexes have also to be equivalent.
$endgroup$
– ttnick
12 hours ago





$begingroup$
You can prove it directly on the regular expressions: Let $w in L(aa^ast)$, i.e. $w = a cdot u$ with $u in L(a^ast)$, i.e. $u = a^k$. for some $k in mathbbN$. Then $w = a^k+1 = a^k a in L(a^ast a)$. The other direction is analogous. However, for these simple expression you might also want to compute the corresponding NFAs and transform them to the minimal DFAs. If these DFAs coincide the regexes have also to be equivalent.
$endgroup$
– ttnick
12 hours ago













$begingroup$
@ttnick Converting to an NFA and then determinizing is much more work than just reasoning about what strings are matched.
$endgroup$
– David Richerby
12 hours ago




$begingroup$
@ttnick Converting to an NFA and then determinizing is much more work than just reasoning about what strings are matched.
$endgroup$
– David Richerby
12 hours ago












$begingroup$
ok this question confused me quite a bit since I read the title first and thought the "?" was part of the regex
$endgroup$
– Esben Skov Pedersen
9 hours ago




$begingroup$
ok this question confused me quite a bit since I read the title first and thought the "?" was part of the regex
$endgroup$
– Esben Skov Pedersen
9 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

Yes, they're equivalent. Informally, it's clear that "any number of $a$s followed by one more" is the same thing as "a $a$ followed by any number more." However, there was a request in the comments for something more formal so here goes...



If $R$ and $S$ are regular expressions, then the concatenation $RS$ matches any string $w=w_1dots w_ell$ (where the $w_i$ are the individual characters) such that, for some $d$, $w_1dots w_d$ matches $R$ and $w_d+1dots w_ell$ matches $S$. Here, $d=0$ and $d=ell$ mean that we've split $w$ into $varepsilon$ and $w$ and vice-versa.



So, consider $R=a^*$ and $S=a$. Then $RS$ matches any string $w=w_1dots w_ell$ such that, for some $d$, $w_1dots w_d$ matches $a^*$ and $w_d+1dots w_ell$ matches $a$. We must have $d=ell-1$ and $w_ell=a$ because only the string $a$ matches $a$, and it has length $1$. And we must have $w_1=dots=w_ell-1=a$ because that is the only length-$(ell-1)$ string that matches $a^*$. So $w=a^ell$ for some $ellgeq 1$.



Considering $R=a$ and $S=a^*$, an almost identical argument shows that any string that matches $aa^*$ must also be $a^ell$ for some $ellgeq 1$, so the two regular expressions do indeed match the same language.



At an intermediate level of formality, you can argue that $a^*$ matches any string $a^ell$ for $ellgeq 0$, and $a$ matches the single string $a=a^1$. So $a^*a$ matches any string $a^ell a$ for $ellgeq 0$, i.e., any string $a^ell+1$ for $ellgeq 0$, i.e., any string $a^ell$ for $ell>0$. Similarly, $aa^*$ matches any string $aa^ell=a^1+ell$ for $ellgeq 0$, i.e., $a^ell$ for $ell>0$.






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    4












    $begingroup$

    Yes, they're equivalent. Informally, it's clear that "any number of $a$s followed by one more" is the same thing as "a $a$ followed by any number more." However, there was a request in the comments for something more formal so here goes...



    If $R$ and $S$ are regular expressions, then the concatenation $RS$ matches any string $w=w_1dots w_ell$ (where the $w_i$ are the individual characters) such that, for some $d$, $w_1dots w_d$ matches $R$ and $w_d+1dots w_ell$ matches $S$. Here, $d=0$ and $d=ell$ mean that we've split $w$ into $varepsilon$ and $w$ and vice-versa.



    So, consider $R=a^*$ and $S=a$. Then $RS$ matches any string $w=w_1dots w_ell$ such that, for some $d$, $w_1dots w_d$ matches $a^*$ and $w_d+1dots w_ell$ matches $a$. We must have $d=ell-1$ and $w_ell=a$ because only the string $a$ matches $a$, and it has length $1$. And we must have $w_1=dots=w_ell-1=a$ because that is the only length-$(ell-1)$ string that matches $a^*$. So $w=a^ell$ for some $ellgeq 1$.



    Considering $R=a$ and $S=a^*$, an almost identical argument shows that any string that matches $aa^*$ must also be $a^ell$ for some $ellgeq 1$, so the two regular expressions do indeed match the same language.



    At an intermediate level of formality, you can argue that $a^*$ matches any string $a^ell$ for $ellgeq 0$, and $a$ matches the single string $a=a^1$. So $a^*a$ matches any string $a^ell a$ for $ellgeq 0$, i.e., any string $a^ell+1$ for $ellgeq 0$, i.e., any string $a^ell$ for $ell>0$. Similarly, $aa^*$ matches any string $aa^ell=a^1+ell$ for $ellgeq 0$, i.e., $a^ell$ for $ell>0$.






    share|cite|improve this answer









    $endgroup$

















      4












      $begingroup$

      Yes, they're equivalent. Informally, it's clear that "any number of $a$s followed by one more" is the same thing as "a $a$ followed by any number more." However, there was a request in the comments for something more formal so here goes...



      If $R$ and $S$ are regular expressions, then the concatenation $RS$ matches any string $w=w_1dots w_ell$ (where the $w_i$ are the individual characters) such that, for some $d$, $w_1dots w_d$ matches $R$ and $w_d+1dots w_ell$ matches $S$. Here, $d=0$ and $d=ell$ mean that we've split $w$ into $varepsilon$ and $w$ and vice-versa.



      So, consider $R=a^*$ and $S=a$. Then $RS$ matches any string $w=w_1dots w_ell$ such that, for some $d$, $w_1dots w_d$ matches $a^*$ and $w_d+1dots w_ell$ matches $a$. We must have $d=ell-1$ and $w_ell=a$ because only the string $a$ matches $a$, and it has length $1$. And we must have $w_1=dots=w_ell-1=a$ because that is the only length-$(ell-1)$ string that matches $a^*$. So $w=a^ell$ for some $ellgeq 1$.



      Considering $R=a$ and $S=a^*$, an almost identical argument shows that any string that matches $aa^*$ must also be $a^ell$ for some $ellgeq 1$, so the two regular expressions do indeed match the same language.



      At an intermediate level of formality, you can argue that $a^*$ matches any string $a^ell$ for $ellgeq 0$, and $a$ matches the single string $a=a^1$. So $a^*a$ matches any string $a^ell a$ for $ellgeq 0$, i.e., any string $a^ell+1$ for $ellgeq 0$, i.e., any string $a^ell$ for $ell>0$. Similarly, $aa^*$ matches any string $aa^ell=a^1+ell$ for $ellgeq 0$, i.e., $a^ell$ for $ell>0$.






      share|cite|improve this answer









      $endgroup$















        4












        4








        4





        $begingroup$

        Yes, they're equivalent. Informally, it's clear that "any number of $a$s followed by one more" is the same thing as "a $a$ followed by any number more." However, there was a request in the comments for something more formal so here goes...



        If $R$ and $S$ are regular expressions, then the concatenation $RS$ matches any string $w=w_1dots w_ell$ (where the $w_i$ are the individual characters) such that, for some $d$, $w_1dots w_d$ matches $R$ and $w_d+1dots w_ell$ matches $S$. Here, $d=0$ and $d=ell$ mean that we've split $w$ into $varepsilon$ and $w$ and vice-versa.



        So, consider $R=a^*$ and $S=a$. Then $RS$ matches any string $w=w_1dots w_ell$ such that, for some $d$, $w_1dots w_d$ matches $a^*$ and $w_d+1dots w_ell$ matches $a$. We must have $d=ell-1$ and $w_ell=a$ because only the string $a$ matches $a$, and it has length $1$. And we must have $w_1=dots=w_ell-1=a$ because that is the only length-$(ell-1)$ string that matches $a^*$. So $w=a^ell$ for some $ellgeq 1$.



        Considering $R=a$ and $S=a^*$, an almost identical argument shows that any string that matches $aa^*$ must also be $a^ell$ for some $ellgeq 1$, so the two regular expressions do indeed match the same language.



        At an intermediate level of formality, you can argue that $a^*$ matches any string $a^ell$ for $ellgeq 0$, and $a$ matches the single string $a=a^1$. So $a^*a$ matches any string $a^ell a$ for $ellgeq 0$, i.e., any string $a^ell+1$ for $ellgeq 0$, i.e., any string $a^ell$ for $ell>0$. Similarly, $aa^*$ matches any string $aa^ell=a^1+ell$ for $ellgeq 0$, i.e., $a^ell$ for $ell>0$.






        share|cite|improve this answer









        $endgroup$



        Yes, they're equivalent. Informally, it's clear that "any number of $a$s followed by one more" is the same thing as "a $a$ followed by any number more." However, there was a request in the comments for something more formal so here goes...



        If $R$ and $S$ are regular expressions, then the concatenation $RS$ matches any string $w=w_1dots w_ell$ (where the $w_i$ are the individual characters) such that, for some $d$, $w_1dots w_d$ matches $R$ and $w_d+1dots w_ell$ matches $S$. Here, $d=0$ and $d=ell$ mean that we've split $w$ into $varepsilon$ and $w$ and vice-versa.



        So, consider $R=a^*$ and $S=a$. Then $RS$ matches any string $w=w_1dots w_ell$ such that, for some $d$, $w_1dots w_d$ matches $a^*$ and $w_d+1dots w_ell$ matches $a$. We must have $d=ell-1$ and $w_ell=a$ because only the string $a$ matches $a$, and it has length $1$. And we must have $w_1=dots=w_ell-1=a$ because that is the only length-$(ell-1)$ string that matches $a^*$. So $w=a^ell$ for some $ellgeq 1$.



        Considering $R=a$ and $S=a^*$, an almost identical argument shows that any string that matches $aa^*$ must also be $a^ell$ for some $ellgeq 1$, so the two regular expressions do indeed match the same language.



        At an intermediate level of formality, you can argue that $a^*$ matches any string $a^ell$ for $ellgeq 0$, and $a$ matches the single string $a=a^1$. So $a^*a$ matches any string $a^ell a$ for $ellgeq 0$, i.e., any string $a^ell+1$ for $ellgeq 0$, i.e., any string $a^ell$ for $ell>0$. Similarly, $aa^*$ matches any string $aa^ell=a^1+ell$ for $ellgeq 0$, i.e., $a^ell$ for $ell>0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 12 hours ago









        David RicherbyDavid Richerby

        68.5k15103194




        68.5k15103194



























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