$A,C$ are recursive and $Acdot B = C$, Is $B$ is recursive?











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Past year paper question.



Prove or disprove: Suppose $A,C$ are recursive and $Acdot B = C$ (The dot denotes concatenation). Then $B$ is recursive.



I think it is false.



Let $B$ be any non-recursive language containing the empty string $epsilon$.



Let $A=C=Sigma^*$.



Then $A cdot B = C$.



Question is how can I construct $B$?










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  • Why do you mean by constructing a non-recursive language?
    – J.-E. Pin
    Nov 21 at 8:35










  • I mean is there a concrete example of a non-recursive language, that contains the empty string.
    – eatfood
    Nov 21 at 8:57















up vote
0
down vote

favorite












Past year paper question.



Prove or disprove: Suppose $A,C$ are recursive and $Acdot B = C$ (The dot denotes concatenation). Then $B$ is recursive.



I think it is false.



Let $B$ be any non-recursive language containing the empty string $epsilon$.



Let $A=C=Sigma^*$.



Then $A cdot B = C$.



Question is how can I construct $B$?










share|cite|improve this question






















  • Why do you mean by constructing a non-recursive language?
    – J.-E. Pin
    Nov 21 at 8:35










  • I mean is there a concrete example of a non-recursive language, that contains the empty string.
    – eatfood
    Nov 21 at 8:57













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Past year paper question.



Prove or disprove: Suppose $A,C$ are recursive and $Acdot B = C$ (The dot denotes concatenation). Then $B$ is recursive.



I think it is false.



Let $B$ be any non-recursive language containing the empty string $epsilon$.



Let $A=C=Sigma^*$.



Then $A cdot B = C$.



Question is how can I construct $B$?










share|cite|improve this question













Past year paper question.



Prove or disprove: Suppose $A,C$ are recursive and $Acdot B = C$ (The dot denotes concatenation). Then $B$ is recursive.



I think it is false.



Let $B$ be any non-recursive language containing the empty string $epsilon$.



Let $A=C=Sigma^*$.



Then $A cdot B = C$.



Question is how can I construct $B$?







formal-languages






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asked Nov 21 at 8:29









eatfood

1827




1827












  • Why do you mean by constructing a non-recursive language?
    – J.-E. Pin
    Nov 21 at 8:35










  • I mean is there a concrete example of a non-recursive language, that contains the empty string.
    – eatfood
    Nov 21 at 8:57


















  • Why do you mean by constructing a non-recursive language?
    – J.-E. Pin
    Nov 21 at 8:35










  • I mean is there a concrete example of a non-recursive language, that contains the empty string.
    – eatfood
    Nov 21 at 8:57
















Why do you mean by constructing a non-recursive language?
– J.-E. Pin
Nov 21 at 8:35




Why do you mean by constructing a non-recursive language?
– J.-E. Pin
Nov 21 at 8:35












I mean is there a concrete example of a non-recursive language, that contains the empty string.
– eatfood
Nov 21 at 8:57




I mean is there a concrete example of a non-recursive language, that contains the empty string.
– eatfood
Nov 21 at 8:57










1 Answer
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1
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Probably you can just suppose that there exists a non-recursive language that contains the empty string, and nobody will demand a proof of this.



Adding or removing a finite number of strings will not change the recursiveness or non-recursiveness of a language. In particular, adding the empty string will not change the complexity (take a TM for the language, add by a non-deterministic choice the possibility to accept the empty string instead of starting the original computation; even if the original TM rejects the empty string, the new one will accept it). So if you are allowed to suppose without proof that there exist non-recursive languages, then you can just take any of them and add the empty string and it will still be non-recursive.






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    1 Answer
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    Probably you can just suppose that there exists a non-recursive language that contains the empty string, and nobody will demand a proof of this.



    Adding or removing a finite number of strings will not change the recursiveness or non-recursiveness of a language. In particular, adding the empty string will not change the complexity (take a TM for the language, add by a non-deterministic choice the possibility to accept the empty string instead of starting the original computation; even if the original TM rejects the empty string, the new one will accept it). So if you are allowed to suppose without proof that there exist non-recursive languages, then you can just take any of them and add the empty string and it will still be non-recursive.






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      up vote
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      down vote













      Probably you can just suppose that there exists a non-recursive language that contains the empty string, and nobody will demand a proof of this.



      Adding or removing a finite number of strings will not change the recursiveness or non-recursiveness of a language. In particular, adding the empty string will not change the complexity (take a TM for the language, add by a non-deterministic choice the possibility to accept the empty string instead of starting the original computation; even if the original TM rejects the empty string, the new one will accept it). So if you are allowed to suppose without proof that there exist non-recursive languages, then you can just take any of them and add the empty string and it will still be non-recursive.






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        Probably you can just suppose that there exists a non-recursive language that contains the empty string, and nobody will demand a proof of this.



        Adding or removing a finite number of strings will not change the recursiveness or non-recursiveness of a language. In particular, adding the empty string will not change the complexity (take a TM for the language, add by a non-deterministic choice the possibility to accept the empty string instead of starting the original computation; even if the original TM rejects the empty string, the new one will accept it). So if you are allowed to suppose without proof that there exist non-recursive languages, then you can just take any of them and add the empty string and it will still be non-recursive.






        share|cite|improve this answer












        Probably you can just suppose that there exists a non-recursive language that contains the empty string, and nobody will demand a proof of this.



        Adding or removing a finite number of strings will not change the recursiveness or non-recursiveness of a language. In particular, adding the empty string will not change the complexity (take a TM for the language, add by a non-deterministic choice the possibility to accept the empty string instead of starting the original computation; even if the original TM rejects the empty string, the new one will accept it). So if you are allowed to suppose without proof that there exist non-recursive languages, then you can just take any of them and add the empty string and it will still be non-recursive.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 9:49









        Peter Leupold

        56826




        56826






























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