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Do there exist two computable functions, a and b, which can construct every computable function by a finite serie of a's and b's which is function composed? Fx. let's take the serie, a,b,a,b,b,a,a,a , which function composed is the function, a∘b∘a∘b∘b∘a∘a∘a ( =a(b(a(b(b(a(a(a(x)))))))) ), this function is the function described by the serie, a,b,a,b,b,a,a,a. And I want to know if every program can be described, by such serie.

If such functions exist, can you tell a example of a and b?

Thanks.

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  • $\begingroup$ What do you think? What have you tried, and where did you get stuck? In what context did you run into this problem, or what's the practical motivation? $\endgroup$ – D.W. Jun 4 '14 at 16:00
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If such functions existed, they would constitute a computable enumeration of all computable functions, which is impossible for the following reason. Suppose you had a computable enumeration $f_i$ of all computable functions. The function $g\colon i \mapsto f_i(i) + 1$ is then computable, but by definition $g \neq f_i$ for all $i$.

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  • $\begingroup$ What if the serie is infinite, is it then possible to have a and b? or if the series n'th element is calculated by a function which is described by a finite serie of a's and b's? $\endgroup$ – Mathematica Jun 4 '14 at 15:08
  • $\begingroup$ @Mathematica If the series is infinite then I'm not sure how you'd evaluate the expression. Regarding your second question, I'll let you think about it. $\endgroup$ – Yuval Filmus Jun 5 '14 at 3:19
  • $\begingroup$ +1, diagonal argument a-la Turing, however what about the functions add and multiply, dont they produce all (numeric) computable functions (including exponentiation)? $\endgroup$ – Nikos M. Jun 5 '14 at 21:00
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    $\begingroup$ @NikosM. Perhaps they produce all primitive recursive functions, but certainly not all recursive functions, since there is no effective enumeration of all recursive functions. $\endgroup$ – Yuval Filmus Jun 6 '14 at 0:41

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