How to show that the inverse of a primitive recursive permutation of $\mathbb{N}$ is not necessarily a primitive recursive function?
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1$\begingroup$ Does it necessarily have to be a bijection? The inverse of the Ackermann function is known to be primitive recursive (since Ackermann has a primitive recursive graph). (I guess maybe you can pad this out into a primitive recursive bijection too). $\endgroup$– Anupam DasCommented May 9 at 14:15
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$\begingroup$ @AnupamDas: yes. It is easy to show that recursive permutations form a group under composition. But primitive recursive permutations do not; that is why I would like to show (or see a proof...) that the inverse of a primitive recursive permutation might be non-primitive recursive. Thank you for pointing out that property of Ackermann function. $\endgroup$– ijonCommented May 10 at 9:30
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4$\begingroup$ I had not seen this before, but a quick search led to descriptions of (essentially the same) counterexamples here (math.stackexchange.com/questions/2975305/…) and here (ncatlab.org/nlab/show/partial+recursive+function#inverse). $\endgroup$– Noam ZeilbergerCommented May 10 at 20:11
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$\begingroup$ @NoamZeilberger Thanks! $\endgroup$– ijonCommented May 14 at 20:32
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1 Answer
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Given a set $X\not\ni 0$, let $\pi_X$ be the permutation of $\mathbb{N}$ defined as follows: $$\pi_X(x)=\begin{cases} x+1 & \mbox{ if }x\not\in X,\\ \max(\{y\in X: y<x\}\cup \{0\}) & \mbox{ if }x\in X.\\ \end{cases}$$
This basically breaks $\mathbb{N}$ into finite loops. Now if the characteristic function of $X$ is primitive recursive, so is the function $\pi_X$. However, if $X$ is extremely sparse then $\pi_X^{-1}$ will fail to be primitive recursive since it will have to grow extremely quickly.
In particular, if we let $X$ be the range of the Ackermann function, we get $\pi_X$ primitive recursive but $\pi_X^{-1}$ not primitive recursive.
EDIT: I just learned of an old strengthening of the above, due to Cannonito and Finkelstein ("On primitive recursive permutations and their inverses"). The argument is short enough to reproduce here.
Theorem (Cannonito/Finkelstein, improving on Kuznecov/Kent/J. Robinson): Let $\mathscr{E}^i$ be the $i$th level of the Grzegorczyk hierarchy. Then there are $A,C\in\mathscr{E}^3$ such that, for every total recursive unary $f$, there is a permutation $B\in\mathscr{E}^3$ with $$f=AB^{-1}C.$$ In particular, taking $f$ "wild" requires $B^{-1}$ to be "wild" as well.
Proof. Letting $e$ be a fixed index for $f$, we define the required functions as follows:
$C\simeq \lambda x.2x$.
$A$ is an appropriately universal function $U$, such that whenever $y$ codes a halting computation of $\varphi_e(x)$ then $U(y)=\varphi_e(x)$. (It's not obvious that this can be done in $\mathscr{E}^3$, but it is true.)
Finally, to get $B$ we consider the set $E$ of numbers which code the (halting) computation of $f$ (via the fixed index $e$) on some input. The characteristic function of $E$ is in $\mathscr{E}^3$, and so the permutation sending $y$ to $2x$ if $y\in E$ and $y$ codes the behavior of $f(x)$, and sending $y$ to $$\max\{0,2\left(\sum_{k\le y}\chi_E(k)\right)-1\}$$ otherwise. This permutation is our $B$.
Cannonito/Finkelstein go on to derive some neat group-theoretic consequences of the above.
