Hmm, upon further inspection your algorithm is the following: 
sub divide {
  my $mid = @_/2;
  $ITER++;
  return if @_ == 1;
  divide(@_[0 .. $mid - 1]);
  divide(@_[$mid .. $#_]);
}

for (10 .. 30) {
  $ITER = 0;
  divide(1 .. $_);
  printf "%2d  %2d\n", $_, $ITER;
}

[download]
The order of this is N, really. Specifically, it's 2*N-1. T(N) is the time it takes to call divide() for a list of N values: 
T(1) = 1
T(2) = 1 + 2*T(1)
T(4) = 1 + 2*T(2)
    ...
T(N) = 1 + 2*T(N/2)

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We can then expand backwards from the generality: 
T(N) = 1 + 2*T(N/2)
     = 1 + 2*(1 + 2*T(N/4))
     = 1 + 2 + 4*T(N/4)
-->  = 3 + 4*T(N/4)
     = 3 + 4*(1 + 2*T(N/8))
     = 3 + 4 + 8*T(N/8)
-->  = 7 + 8*T(N/8)
     = 7 + 8*(1 + 2*T(N/16))
     = 7 + 8 + 16*T(N/16)
-->  = 15 + 16*T(N/16)

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The general equation here is T(N) = 2^k - 1 + 2^k * T(N / 2^k). We will make the following substitution: k = log(N) (base 2). We now have: 
T(N) = 2^k - 1 + 2^k * T(N / 2^k)
     = 2^(log N) - 1 + 2^(log N) * T(N / 2^log(N))
     = N - 1 + N * T(N/N)
     = N - 1 + N * 1
     = 2*N - 1

[download]


$_="goto+F.print+chop;\n=yhpaj";F1:eval

In reply to RE: RE: Re: finding min and max of array recursivly by japhy
in thread finding min and max of array recursivly by mbond

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