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In the 1965 article "On the computational complexity of algorithms" by Hartmanis and Stearns, the authors conjecture that if a real-time Turing Machine computes the real number $r$ in, for example, base 10, then $r$ is either a rational number or a transcendental number.

Is there a computable transcendental number that is not computable by a real-time Turing machine in, for example, base 10?

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  • $\begingroup$ If I understand your question correctly, Chaitin's constants are examples of such numbers: They are transcendental and not computable at all. $\endgroup$
    – Bruno
    Commented Dec 29, 2014 at 9:12
  • $\begingroup$ @Bruno,but Chaitin's constants is not computable, or semicomputable, so it is not the numbers that is computable transcendental number and is not computable by a real-time Turing machine. $\endgroup$ Commented Dec 29, 2014 at 11:25
  • $\begingroup$ My mistake, I didn't notice that you asked for a computable number... $\endgroup$
    – Bruno
    Commented Dec 29, 2014 at 16:59

1 Answer 1

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Let $L$ be an EXPTIME-complete language, and let $r \in (0,1)$ be the corresponding real. Clearly $r$ is computable. The number $r$ cannot be algebraic since the $n$th bit of an algebraic number can be computed in time $n^{O(1)}$ (Datta and Pratap). Since the $n$th bit of any number computable by a real-time Turing machine can be computed in time $O(n)$, $r$ cannot be computed by a real-time Turing machine.

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  • $\begingroup$ Excellent, But I have to think about it carefully. And I have just found that the Datta and Pratap is a paper that is just published recently. $\endgroup$ Commented Dec 29, 2014 at 11:57
  • $\begingroup$ Presumably it had been known that the binary expansion of algebraic numbers can be computed in polynomial time. Their paper is just the first one I could find, and it actually proves stronger results. $\endgroup$ Commented Dec 29, 2014 at 13:06
  • $\begingroup$ Yes, I have conjectured for a long time that the binary expansion of algebraic numbers can be computed in polynomial time,but have not found any proof of that, thank you again for your answer and the referenced paper $\endgroup$ Commented Dec 30, 2014 at 0:32

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