I am given the following function as a brain teaser:

def t(x, y, z):
    if x <= z:
        return y
        return t(t(x  - 1 , y  , z ) , t(y  - 1 , z  , x ) , t (z  - 1 , x  , y ))

the task is to remove the recursion in the else-branch. the only allowed expressions are of type arithmetic (+,-,*,/,%), logical (&&,||,!), relational (<,<=,==,>=,>,!=) and if-else.

My first approach was to order the input in lexicographical order and trying to deduce a relationship which seemed to look like like this initially:

    if y > x
        return x
        return z

but that does not work, a counterexample is t(20, 7, 18)=8. So the underlying relationship seems to be a bit more complex. I have also tried to formally infer the solution by differentiating the cases for the '>'-relation between the variables but I did not get far with that either.


Here is an equivalent way to calculate $t(x,y,z)$.

We distinguish between three cases:

  1. $x \leq z$: the answer is $y$.

  2. $x = z + 1$: the answer is $z$ if $y \leq x$, and $x$ otherwise.

  3. $x-z \geq 2$: the answer is $z$ (if $x-z$ is odd) or $\min(y+1,z)$ (if $x-z$ is even) if $y \leq x+1$, and $x$ (if $y-x$ is even) or $z+1$ (if $y-x$ is odd) otherwise.

You should be able to prove this characterization by induction. The second case is identical to the third case, except that the condition $y \leq x$ is replaced by the condition $y \leq x+1$.

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  • $\begingroup$ yes thats it, thanks! tbh, I don't even know how to approach multidimensional induction (I was not even sure if it is applicable to this question), but I will try to figure it out by reading some sources. How long did it take you to get to the solution? $\endgroup$ – Doflaminhgo Nov 19 '19 at 9:58
  • 1
    $\begingroup$ It took about an hour. I looked at the sequences $t(x,0,z),t(x,1,z),\ldots$ for various small values of $x,z$, and the pattern was obvious. $\endgroup$ – Yuval Filmus Nov 19 '19 at 11:28
  • $\begingroup$ okay, based on your last comment I managed to find a slightly more verbose solution compared to yours, which should be equivalent (no min-function, but more if-elses). I honestly don't see how it's obvious though, it took me a couple hours and I probably could not even have done it, if I didn't glance at your solution beforehand (the odd-even-relations were difficult for me to spot). So thanks again! $\endgroup$ – Doflaminhgo Nov 19 '19 at 22:06

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