Time complexity of a language whose alphabet has a single symbol

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Consider a language $L$ such that $L subseteq Sigma^*$, where the cardinality of $Sigma$ is $1$ (i.e. the alphabet has only one symbol). E.g. $L subseteq a^*$.



Can anything be said about the time complexity of $L$? Is it the case that $L in P$?



I can't come up with any example of a language with a single symbol that is not in $P$.






time-complexity decision-problem







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    up vote
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    down vote

    favorite












    Consider a language $L$ such that $L subseteq Sigma^*$, where the cardinality of $Sigma$ is $1$ (i.e. the alphabet has only one symbol). E.g. $L subseteq a^*$.



    Can anything be said about the time complexity of $L$? Is it the case that $L in P$?



    I can't come up with any example of a language with a single symbol that is not in $P$.






    time-complexity decision-problem







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Consider a language $L$ such that $L subseteq Sigma^*$, where the cardinality of $Sigma$ is $1$ (i.e. the alphabet has only one symbol). E.g. $L subseteq a^*$.



      Can anything be said about the time complexity of $L$? Is it the case that $L in P$?



      I can't come up with any example of a language with a single symbol that is not in $P$.






      time-complexity decision-problem







      share|cite|improve this question













      Consider a language $L$ such that $L subseteq Sigma^*$, where the cardinality of $Sigma$ is $1$ (i.e. the alphabet has only one symbol). E.g. $L subseteq a^*$.



      Can anything be said about the time complexity of $L$? Is it the case that $L in P$?



      I can't come up with any example of a language with a single symbol that is not in $P$.






      time-complexity decision-problem




      time-complexity decision-problem






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      share|cite|improve this question











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      asked 5 hours ago









      denidare

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          The language $a^n : text the $n$th Turing machine halts on the empty string $ is undecidable.



          In fact, a random unary language is undecidable almost surely, for the simple reason that there are only countably many computable languages.






          share|cite|improve this answer




















          • Thanks, this makes sense. Would anything change if we restricted $L$ to be decidable?
            – denidare
            5 hours ago










          • Perhaps you can search a bit on the Internet. You might find things like "if there exists a unary language that is NP-complete, then P = NP" and other interesting observations.
            – Pål GD
            4 hours ago






          • 1




            Not really. You can ask whether the $n$th Turing machine halts within $T(n)$ steps on the empty string, which should be hard to determine given much less than $T(n)$ time.
            – Yuval Filmus
            4 hours ago










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          up vote
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          The language $a^n : text the $n$th Turing machine halts on the empty string $ is undecidable.



          In fact, a random unary language is undecidable almost surely, for the simple reason that there are only countably many computable languages.






          share|cite|improve this answer




















          • Thanks, this makes sense. Would anything change if we restricted $L$ to be decidable?
            – denidare
            5 hours ago










          • Perhaps you can search a bit on the Internet. You might find things like "if there exists a unary language that is NP-complete, then P = NP" and other interesting observations.
            – Pål GD
            4 hours ago






          • 1




            Not really. You can ask whether the $n$th Turing machine halts within $T(n)$ steps on the empty string, which should be hard to determine given much less than $T(n)$ time.
            – Yuval Filmus
            4 hours ago














          up vote
          3
          down vote













          The language $a^n : text the $n$th Turing machine halts on the empty string $ is undecidable.



          In fact, a random unary language is undecidable almost surely, for the simple reason that there are only countably many computable languages.






          share|cite|improve this answer




















          • Thanks, this makes sense. Would anything change if we restricted $L$ to be decidable?
            – denidare
            5 hours ago










          • Perhaps you can search a bit on the Internet. You might find things like "if there exists a unary language that is NP-complete, then P = NP" and other interesting observations.
            – Pål GD
            4 hours ago






          • 1




            Not really. You can ask whether the $n$th Turing machine halts within $T(n)$ steps on the empty string, which should be hard to determine given much less than $T(n)$ time.
            – Yuval Filmus
            4 hours ago












          up vote
          3
          down vote










          up vote
          3
          down vote









          The language $a^n : text the $n$th Turing machine halts on the empty string $ is undecidable.



          In fact, a random unary language is undecidable almost surely, for the simple reason that there are only countably many computable languages.






          share|cite|improve this answer












          The language $a^n : text the $n$th Turing machine halts on the empty string $ is undecidable.



          In fact, a random unary language is undecidable almost surely, for the simple reason that there are only countably many computable languages.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 5 hours ago









          Yuval Filmus

          185k12176337




          185k12176337











          • Thanks, this makes sense. Would anything change if we restricted $L$ to be decidable?
            – denidare
            5 hours ago










          • Perhaps you can search a bit on the Internet. You might find things like "if there exists a unary language that is NP-complete, then P = NP" and other interesting observations.
            – Pål GD
            4 hours ago






          • 1




            Not really. You can ask whether the $n$th Turing machine halts within $T(n)$ steps on the empty string, which should be hard to determine given much less than $T(n)$ time.
            – Yuval Filmus
            4 hours ago
















          • Thanks, this makes sense. Would anything change if we restricted $L$ to be decidable?
            – denidare
            5 hours ago










          • Perhaps you can search a bit on the Internet. You might find things like "if there exists a unary language that is NP-complete, then P = NP" and other interesting observations.
            – Pål GD
            4 hours ago






          • 1




            Not really. You can ask whether the $n$th Turing machine halts within $T(n)$ steps on the empty string, which should be hard to determine given much less than $T(n)$ time.
            – Yuval Filmus
            4 hours ago















          Thanks, this makes sense. Would anything change if we restricted $L$ to be decidable?
          – denidare
          5 hours ago




          Thanks, this makes sense. Would anything change if we restricted $L$ to be decidable?
          – denidare
          5 hours ago












          Perhaps you can search a bit on the Internet. You might find things like "if there exists a unary language that is NP-complete, then P = NP" and other interesting observations.
          – Pål GD
          4 hours ago




          Perhaps you can search a bit on the Internet. You might find things like "if there exists a unary language that is NP-complete, then P = NP" and other interesting observations.
          – Pål GD
          4 hours ago




          1




          1




          Not really. You can ask whether the $n$th Turing machine halts within $T(n)$ steps on the empty string, which should be hard to determine given much less than $T(n)$ time.
          – Yuval Filmus
          4 hours ago




          Not really. You can ask whether the $n$th Turing machine halts within $T(n)$ steps on the empty string, which should be hard to determine given much less than $T(n)$ time.
          – Yuval Filmus
          4 hours ago

















           

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