0

Using the definition of the Ackermann function on page 247 of this paper, (sidenote, great paper): $$ \alpha(k,m,n) = \begin{cases} m+n, & k=1 \\ m, & n=1 \\ \alpha(k-1,m,\alpha(k,m,n-1)),& \text{otherwise} \end{cases} $$

Is there some function $f$ such that $f(k,m,n)$ returns the number of times $\alpha$ is used to calculate $\alpha(k,m,n)$?

Some trivial values would be $f(1,m,n)=f(k,m,1)=1$, and $f(2,2,2)=2$, found manually by counting the number of $\alpha$s in the expanded form of $\alpha$.

Note: There is neither a tag for "ackermann-function" or "recursion", I lack the privilege to create these but I feel they would have some limited use.

Simply Beautiful Art
  • 76,603
user189728
  • 670
  • 1
    Your question doesn't make a whole lot of sense in the vocabulary of mathematics. Does there exist a function? Sure. You have just defined such a function.

    You mean is there a nice neat closed form? How many times does a recursive algorithm call upon itself might be a good CS question. I think your question might get more play on cs.stackexchange.com but I am not so sure of the community there and how they would respond.

     – Mason Jul 15 '18 at 04:40
  • 1
    @Mason I initially intended to ask this on stackoverflow, but I felt it wasn't CS-related enough – user189728 Jul 15 '18 at 14:55
  • Oh no. Ackermann Function is right out of the CS Camp. They should be very interested: Check out: https://cs.stackexchange.com/questions/47227/time-complexity-of-ackermanns-function. The problem will be that you are going to have a more precise question to make much progress. – Mason Jul 15 '18 at 17:58
  • @Mason is there a simple way to migrate questions? – user189728 Jul 15 '18 at 17:59
  • Look it up on the Meta and post the link here. I dunno know. but I need to. – Mason Jul 15 '18 at 17:59
  • Sorry. I misread your comment. You are correct: StackOverflow is not the right choice. Maybe cs.se after you shop your question. – Mason Jul 15 '18 at 18:01
  • Not sure what you meant about [tag:ackermann-function] and [tag:recursion] since they already exist. I have [edit]ed your question to include these tags. – Simply Beautiful Art Aug 07 '18 at 13:29
  • How are you getting $f(2,2,2)=2$? I'm getting $3$, since $\alpha(2,2,2)=\alpha(1,2,\alpha(2,2,1))=4$ with $3$ instances of $\alpha$. – Simply Beautiful Art Aug 10 '18 at 12:53
  • Also, would $$f(k,m,n)=\begin{cases}1,&k=1\text{ or }n=1\1+f(k,m,n-1)+f(k-1,m,\alpha(k,m,n-1)),&\rm else\end{cases}$$ work as an answer? – Simply Beautiful Art Aug 10 '18 at 12:55
  • Note that $m$ does not change over the iterations, so much of the same analysis applies. – Simply Beautiful Art Apr 04 '20 at 01:25

0 Answers0