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MathGroup Archive 2005

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Re: ChineseRemainder

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56993] Re: [mg56971] ChineseRemainder
  • From: Chris Chiasson <chris.chiasson at gmail.com>
  • Date: Thu, 12 May 2005 02:32:32 -0400 (EDT)
  • References: <200505110925.FAA24094@smc.vnet.net>
  • Reply-to: Chris Chiasson <chris.chiasson at gmail.com>
  • Sender: owner-wri-mathgroup at wolfram.com

Mod[4,4] is zero.

I think the routine should return a null set.


On 5/11/05, Dana DeLouis <delouis at bellsouth.net> wrote:
> Hello.  I have a question on the function "ChineseRemainder."  Could anyone
> offer an explanation on the following behavior?  Thanks.
> 
> Here's the package...
> 
> Needs["NumberTheory`NumberTheoryFunctions`"]
> 
> The following small example has no solution, and returns the null set, which
> is ok...
> 
> list1={2,3,4};list2={9,4,8};
> 
> r=ChineseRemainder[list1,list2]
> 
> {}
> 
> The following returns a solution of 94.
> 
> list1={2,3,4};list2={4,7,9};
> 
> r=ChineseRemainder[list1,list2]
> 
> 94
> 
> According to Help on this function, we can test the solution with the
> following, and it correctly returns list1.
> 
> Mod[r,list2]
> 
> {2,3,4}
> 
> These similar numbers return a different solution.
> 
> list1={2,3,4};list2={9,7,4};
> 
> r=ChineseRemainder[list1,list2]
> 
> 164
> 
> However, it does not correctly return list1.  (According to help)
> 
> Mod[r,list2]
> 
> {2,3,0}
> 
> This happens often in my program, and I'm having a tough time trusting the
> solution.  Does anyone familiar with this have any insight?  Thanks.
> 
> I note that Mathematica's built-in ChineseRemainder function returns the
> same answer, so I'm sure Mathematica is doing it correctly.  I guess I don't
> understand why Mod[r,list2] will often not return list1 (According to help).
> 
> Reduce`RChineseRemainder[list1, list2]
> 
> 164
> 
> Thank you for any insight.
> Dana
> 
> $Version
>  "5.1 for Microsoft Windows (January 27, 2005)"
> 
> 


-- 
Chris Chiasson
http://chrischiasson.com/
1 (810) 265-3161


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