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### Re^3: Non-recursive Ackermann

by ikegami (Pope)
 on Feb 04, 2011 at 16:24 UTC ( #886243=note: print w/replies, xml ) Need Help??

in reply to Re^2: Non-recursive Ackermann

[ Just a few rushed little comments ]

Thanks!

we don't know the range.

with the minimal effort required for each step

I reached the same conclusions.

it requires that it has multiple, independant, recursive calls at each level.

That's why I used fib at first and called Ackermann a whole other ball game.

The interesting thing now is: how many recursive algorithms fit that pattern?

Any divide and conquer algorithm, for starters. That's where the dependency graph came in. I wanted to see if there was a split in it that could be exploited.

wonder why you do a loop initialisation rather than

I didn't know what I was going to end up with when I started. The code got refactored multiple times. I didn't micro-optimise.

(Upd: Put differently, the snippet is a theoretical implementation, not a practical one. Work still needs to be done to make it practical. Well, as practical as Ackermann can be. )

why for(;;) not while( 1 )?

It's what the code I learned from used. I've never used while (1). I like for (;;) cause I read it as "for ever". Trivia: Since the entire bound check of while (CONSTANT) gets optimised away, and since "while" and C-style "for" use the same opcodes, while (1) and for (;;) are practically identical internally.

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 [Eily]: you could tie a variable into not having the same value each time, if you like to make people who try to debug your code facepalm [Corion]: perl -wle 'package o; use overload q("") => sub {warn "str"; ""}, bool => sub{warn "bool"; 1}; package main; my \$o={}; bless \$o => o; print "Yay" if (\$o && !length(\$o))' [Corion]: But people writing such code should document the objects they construct and why it makes sense for an object to be invisible as string while being true in a boolean context [hippo]: That's equal parts clever and horrendous. [Eily]: the overload version wouldn't return true with "\$x" && !length \$x though, I guess [hippo]: The more I look at this code, the more \$x is a plain old scalar and the more this condition will never be true. I'm calling it a bug at this point. [hippo]: Thanks for your input which has soothed my sanity (a little) [Corion]: Eily: Sure - if you force both things into stringy things, then you break that magic. But that would also mean that you changed the expression, as now \$x = 0.00 will be true instead of false as it were before [Corion]: Ah no, at least in my feeble experiments that doesn't change the meaning [Corion]: We sell sanity in small packages ;)

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