FPTAS definition

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I read that a fully polynomial time approximation scheme (FPTAS) has time complexity polynomial in the input size and also polynomial in $1/epsilon$. However, this leaves some ambiguity.



For example, $T(n,1/epsilon)=min(n^1/epsilon,(1/epsilon)^n)$ is both polynomial in $n$ (when holding $1/epsilon$ constant) and polynomial in $1/epsilon$ (when holding $n$ constant). But $T$ is not polynomial in $n/epsilon$ or $(n+1/epsilon)$ because $sqrt x^sqrt x$ and $(x/2)^x/2$ aren't bounded by polynomials.



Does this mean an algorithm with running time $T$ isn't an FTPAS, and perhaps a better definition would be that it has to be polynomial in $n+1/epsilon$?




algorithms complexity-theory terminology




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    up vote
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    down vote

    favorite












    I read that a fully polynomial time approximation scheme (FPTAS) has time complexity polynomial in the input size and also polynomial in $1/epsilon$. However, this leaves some ambiguity.



    For example, $T(n,1/epsilon)=min(n^1/epsilon,(1/epsilon)^n)$ is both polynomial in $n$ (when holding $1/epsilon$ constant) and polynomial in $1/epsilon$ (when holding $n$ constant). But $T$ is not polynomial in $n/epsilon$ or $(n+1/epsilon)$ because $sqrt x^sqrt x$ and $(x/2)^x/2$ aren't bounded by polynomials.



    Does this mean an algorithm with running time $T$ isn't an FTPAS, and perhaps a better definition would be that it has to be polynomial in $n+1/epsilon$?




    algorithms complexity-theory terminology




    share|cite|improve this question





















      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I read that a fully polynomial time approximation scheme (FPTAS) has time complexity polynomial in the input size and also polynomial in $1/epsilon$. However, this leaves some ambiguity.



      For example, $T(n,1/epsilon)=min(n^1/epsilon,(1/epsilon)^n)$ is both polynomial in $n$ (when holding $1/epsilon$ constant) and polynomial in $1/epsilon$ (when holding $n$ constant). But $T$ is not polynomial in $n/epsilon$ or $(n+1/epsilon)$ because $sqrt x^sqrt x$ and $(x/2)^x/2$ aren't bounded by polynomials.



      Does this mean an algorithm with running time $T$ isn't an FTPAS, and perhaps a better definition would be that it has to be polynomial in $n+1/epsilon$?




      algorithms complexity-theory terminology




      share|cite|improve this question











      I read that a fully polynomial time approximation scheme (FPTAS) has time complexity polynomial in the input size and also polynomial in $1/epsilon$. However, this leaves some ambiguity.



      For example, $T(n,1/epsilon)=min(n^1/epsilon,(1/epsilon)^n)$ is both polynomial in $n$ (when holding $1/epsilon$ constant) and polynomial in $1/epsilon$ (when holding $n$ constant). But $T$ is not polynomial in $n/epsilon$ or $(n+1/epsilon)$ because $sqrt x^sqrt x$ and $(x/2)^x/2$ aren't bounded by polynomials.



      Does this mean an algorithm with running time $T$ isn't an FTPAS, and perhaps a better definition would be that it has to be polynomial in $n+1/epsilon$?





      algorithms complexity-theory terminology





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      asked 2 days ago









      Akababa

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          The standard definition is incredibly ambiguous, since it is not clear what "polynomial in $n$ and in $1/epsilon$" means. Judging from examples, it means $O(n^C(1/epsilon)^D)$ for some constants $C,D$. Without loss of generality, $C=D$, and then we get the simpler definition $O((n/epsilon)^C)$.






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          • Thanks for clarifying. Although might $O((n+1/epsilon)^C)$ be more precise because when $epsilon=n$ is very large, $n/epsilon=1$ but it should still take at least $O(n)$ to find a feasible solution?
            – Akababa
            2 days ago










          • The big O holds in the limit $ntoinfty$ and $epsilonto0$. We can assume that $epsilon < 1 < n$, say.
            – Yuval Filmus
            2 days ago










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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          The standard definition is incredibly ambiguous, since it is not clear what "polynomial in $n$ and in $1/epsilon$" means. Judging from examples, it means $O(n^C(1/epsilon)^D)$ for some constants $C,D$. Without loss of generality, $C=D$, and then we get the simpler definition $O((n/epsilon)^C)$.






          share|cite|improve this answer





















          • Thanks for clarifying. Although might $O((n+1/epsilon)^C)$ be more precise because when $epsilon=n$ is very large, $n/epsilon=1$ but it should still take at least $O(n)$ to find a feasible solution?
            – Akababa
            2 days ago










          • The big O holds in the limit $ntoinfty$ and $epsilonto0$. We can assume that $epsilon < 1 < n$, say.
            – Yuval Filmus
            2 days ago














          up vote
          2
          down vote



          accepted










          The standard definition is incredibly ambiguous, since it is not clear what "polynomial in $n$ and in $1/epsilon$" means. Judging from examples, it means $O(n^C(1/epsilon)^D)$ for some constants $C,D$. Without loss of generality, $C=D$, and then we get the simpler definition $O((n/epsilon)^C)$.






          share|cite|improve this answer





















          • Thanks for clarifying. Although might $O((n+1/epsilon)^C)$ be more precise because when $epsilon=n$ is very large, $n/epsilon=1$ but it should still take at least $O(n)$ to find a feasible solution?
            – Akababa
            2 days ago










          • The big O holds in the limit $ntoinfty$ and $epsilonto0$. We can assume that $epsilon < 1 < n$, say.
            – Yuval Filmus
            2 days ago












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          The standard definition is incredibly ambiguous, since it is not clear what "polynomial in $n$ and in $1/epsilon$" means. Judging from examples, it means $O(n^C(1/epsilon)^D)$ for some constants $C,D$. Without loss of generality, $C=D$, and then we get the simpler definition $O((n/epsilon)^C)$.






          share|cite|improve this answer













          The standard definition is incredibly ambiguous, since it is not clear what "polynomial in $n$ and in $1/epsilon$" means. Judging from examples, it means $O(n^C(1/epsilon)^D)$ for some constants $C,D$. Without loss of generality, $C=D$, and then we get the simpler definition $O((n/epsilon)^C)$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered 2 days ago









          Yuval Filmus

          179k12167327




          179k12167327











          • Thanks for clarifying. Although might $O((n+1/epsilon)^C)$ be more precise because when $epsilon=n$ is very large, $n/epsilon=1$ but it should still take at least $O(n)$ to find a feasible solution?
            – Akababa
            2 days ago










          • The big O holds in the limit $ntoinfty$ and $epsilonto0$. We can assume that $epsilon < 1 < n$, say.
            – Yuval Filmus
            2 days ago
















          • Thanks for clarifying. Although might $O((n+1/epsilon)^C)$ be more precise because when $epsilon=n$ is very large, $n/epsilon=1$ but it should still take at least $O(n)$ to find a feasible solution?
            – Akababa
            2 days ago










          • The big O holds in the limit $ntoinfty$ and $epsilonto0$. We can assume that $epsilon < 1 < n$, say.
            – Yuval Filmus
            2 days ago















          Thanks for clarifying. Although might $O((n+1/epsilon)^C)$ be more precise because when $epsilon=n$ is very large, $n/epsilon=1$ but it should still take at least $O(n)$ to find a feasible solution?
          – Akababa
          2 days ago




          Thanks for clarifying. Although might $O((n+1/epsilon)^C)$ be more precise because when $epsilon=n$ is very large, $n/epsilon=1$ but it should still take at least $O(n)$ to find a feasible solution?
          – Akababa
          2 days ago












          The big O holds in the limit $ntoinfty$ and $epsilonto0$. We can assume that $epsilon < 1 < n$, say.
          – Yuval Filmus
          2 days ago




          The big O holds in the limit $ntoinfty$ and $epsilonto0$. We can assume that $epsilon < 1 < n$, say.
          – Yuval Filmus
          2 days ago












           

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