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in reply to Re: Speed/Efficiency tweaks for a fannkuch benchmark script?
in thread Speed/Efficiency tweaks for a fannkuch benchmark script?

Hmm, the only problem with this is that my optimised routine gives the wrong answer. The bug is that, if the largest element is at the beginning or the smallest is at the end of the list, my test will skip it when it shouldn't. So the condition really needs to be something like: 
            my ($n, $first, $last) = ($#$copy, @$copy[0,-1]);
            if ( ( $first == $n+1 || $first != max(@$copy[0..($first -
+ 1)]) )
              && ( $last == 1 || $last != min(@$copy[($last - 1)..$n])
+ ))
            {
                ...

[download]
Now it's a little faster than the original on my laptop, and a little slower on the Linux box.

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Re^3: Speed/Efficiency tweaks for a fannkuch benchmark script?
by thundergnat (Deacon) on Dec 01, 2005 at 19:20 UTC

    Yes, but... The only case where it would yeild a wrong answer is when $n == 2. So just check for that.

    if ((max( @$copy[ 0 .. ( $copy->[0] - 1 ) ] ) != $copy->[0]
   && min( @$copy[ ( $copy->[-1] - 1 ) .. $#$copy ] ) != $copy->[-1])
   || @$copy == 2 )
   {

    [download]

    Or am I missing something?

[reply]
[d/l]
      It also fails for $n == 3, where the answers should be 231 and 312, each of which take 2 steps.

      So just do || @$copy <= 3, one might say; and indeed that would give the right answers. But how do we know it would give the right answers, except by comparing the output with the unoptimized version? Is it possible to prove that this is safe?

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[d/l]
[select]

        Ah. Yes, ok I see your point now. I might be able to make a case for n being in the first position being safe to ignore, but how to prove when 1 is in the last...

[reply]

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