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Re: Bug ???

• To: mathgroup at smc.vnet.net
• Subject: [mg81813] Re: [mg81805] Bug ???
• From: Murray Eisenberg <murray at math.umass.edu>
• Date: Thu, 4 Oct 2007 04:20:05 -0400 (EDT)
• Organization: Mathematics & Statistics, Univ. of Mass./Amherst
• References: <200710020928.FAA28247@smc.vnet.net> <200710030635.CAA29450@smc.vnet.net> <200710031034.GAA27140@smc.vnet.net>
• Reply-to: murray at math.umass.edu

If I understand correctly what you're trying to do, the reason the 
results are different is that you're asking for, and getting different 
results.

It's often a lot simpler to "think whole list at once" rather than to 
think of an iterative process.  And to see what's happening, form the 
whole list and only then select from it the ones you want.  Moreover, 
let's try a smaller upper limit, say 10, instead of your 1000 so as to 
avoid a huge amount of output.

First, here's the entire table of pairs:

   Table[{n^2-(Floor[n^2 (2/3)])^2,n^2}, {n,1,10}]
{{1,1},{0,4},{-27,9},{-84,16},{-231,25},{-540,36},{-975,49},{-1700,64},{-2835,81},{-4256,100}}

Your "not working" code does the same thing as this:

   Select[Table[{n^2-(Floor[n^2(2/3)])^2,n^2},{n,1,10}],First[#]==0&]
{{0, 4}}

But your second, "working", code does the same thing as this:

   Select[Table[{n^2- (Floor[n^2 (2/3)])^2,n^2}, {n,1,10}], First[#]<1&]
{{0,4},{-27,9},{-84,16},{-231,25},{-540,36},{-975,49},{-1700,64},{-2835,81},{-4256,100}}

Quite evidently now, the two conditions (==0 vs. <1) are quite different 
in their effects.

By the way, the "think whole list at once" method is faster for your 
upper limit of 1000:

   First@Timing[Select[Table[{n^2-(Floor[n^2 (2/3)])^2, n^2},{n,1,1000}],
        First[#]<1 &];]
0.016

   First@Timing[
a={};Do[k=n^2-(Floor[n^2(2/3)])^2;If[k<1,AppendTo[a,{k,n^2}]],{n,1,1000}];a;]
0.062

(These were timed with a fresh kernel.)

The timing advantage for the whole-list method by a factor of roughly 4 
persists for a larger upper limit than 10000.  For example, for an upper 
limit of 10000, the whole-list method has timing 0.282 whereas the 
iterative method has timing 1.687, roughly a factor of 5 to 6.

Of course if you really want to do the calculation for a VERY large 
value of the upper limit, eventually memory space will become more 
significant and you'll likely need to use the iterative method.


Artur wrote:
> Why doesn't work:
> a={};Do[k=n^2-(Floor[n^2(2/3)])^2;If[k=0,
> AppendTo[a,{k,n^2}]],{n,1,1000}];a

> but working:

> a={};Do[k=n^2-(Floor[n^2(2/3)])^2;If[k<1,
> AppendTo[a,{k,n^2}]],{n,1,1000}];a
> 
> Artur Jasinski, Poland


-- 
Murray Eisenberg                     murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305

• References:
  • help with polynomial solutions
    • From: "Jules P. Aronson" <aronson@nlm.nih.gov>
  • Re: help with polynomial solutions
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
  • Bug ???
    • From: Artur <grafix@csl.pl>