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Re: Pi day

  • To: mathgroup at
  • Subject: [mg108458] Re: Pi day
  • From: Ray Koopman <koopman at>
  • Date: Thu, 18 Mar 2010 04:34:12 -0500 (EST)
  • References: <hnnk1f$cfe$>

On Mar 16, 2:46 am, Ray Koopman <koop... at> wrote:
> The code I gave can certainly be speeded up, but right now I'm more
> concerned about the algorithm. 'hi' is meant to be the smallest value
> of the form s[[i]]/s[[j]] that is greater than Pi, and 'lo' is meant
> to be the largest value of the form s[[i]]/s[[j]] that is less than
> Pi. 'hi' is OK, but I wonder about 'lo'.
> Specifically, potential values of 'lo' with j > k are never checked,
> where k is the largest value of j for which s[[-1]]/s[[j]] > Pi. When
> the While[j < k, ...] loop exits, we have j = k, s[[i]]/s[[j]] > Pi,
> and s[[i-1]]/s[[j]] < Pi.  s[[-1]]/s[[k+1]] < Pi, but might it not
> also be > the current 'lo'?

One more check is necessary. If j > k then s[[i]]/s[[j]] < Pi for
all i, so we need to check the largest possible new value for 'lo'.

Here is the revised code, compiled.

hiloPi = Compile[{}, Module[{s,i,j,k,hi,lo,ijhi,ijlo},
s = (Permutations@Range@9).(10^Range[8,0,-1]);
i = j = 1; While[s[[i]]/s[[j]] < Pi, i++];
hi = s[[ i ]]/s[[j]]; ijhi = { i ,j};
lo = s[[i-1]]/s[[j]]; ijlo = {i-1,j};
k = Length@s; While[s[[-1]]/s[[k]] < Pi, k--];
While[j < k, j++; While[s[[i]]/s[[j]] < Pi, i++];
  If[s[[ i ]]/s[[j]] < hi, hi = s[[ i ]]/s[[j]]; ijhi = { i ,j}];
  If[s[[i-1]]/s[[j]] > lo, lo = s[[i-1]]/s[[j]]; ijlo = {i-1,j}]];
If[s[[-1]]/s[[k+1]] > lo, lo = s[[-1]]/s[[k+1]]; ijlo = {-1,k+1}];


{1.23 Second,{{429751836,136794258},{467895213,148935672}}}

{1.0185541299279066*^-10, -8.499689840846258*^-11}

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