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Re: Fibonacci golf with one state variable

by dragonchild (Archbishop)
on Oct 07, 2008 at 14:25 UTC ( #715797=note: print w/ replies, xml ) Need Help??


in reply to Fibonacci golf with one state variable

How about a tail-recursive version with no state variables?

perl -E 'sub f{say$_[0];goto&f($_[1],$_[0]+$_[1])}f(1,1)'
And I'm sure that could be reduced further. And, no, the stack doesn't count as a state variable. :-p

My criteria for good software:
  1. Does it work?
  2. Can someone else come in, make a change, and be reasonably certain no bugs were introduced?


Comment on Re: Fibonacci golf with one state variable
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Re^2: Fibonacci golf with one state variable
by aufflick (Deacon) on Oct 08, 2008 at 02:05 UTC

    Sure, of course...

    I'm wondering, though, why the goto&f? It works perfectly well as:

    perl -E 'sub f{say$_[0];f($_[1],$_[0]+$_[1])}f(1,1)' |head -10

    If you want to use the & form that saves on stack operations by not creating a new @_, you could do this (paying careful attention to the right to left execution):

    perl -E 'sub f{say$_[0];$_[1]=$_[1]+shift;&f}f(1,1)' |head -10

    Which is exactly the same length, but less operations. Now if only Perl auto-optimised tail recursion...

    It's a shame that unshift doesn't default to @_ like shift otherwise I could save one character by switching the order of the values on the stack:

    perl -E 'sub f{say$_[1];unshift@_,$_[0]+pop;&f}f(1,1)' |head -10

    Update: I just read the perldoc for goto and I have to say I didn't know that goto &NAME form, so thanks for the learning exercise!. Like my form it won't result in recursive stack blowout. Without studying the internals too carefully I don't know which would result in less operations. Maybe a benchmark will show.

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