ChineseRemainder
- To: mathgroup at smc.vnet.net
- Subject: [mg56971] ChineseRemainder
- From: "Dana DeLouis" <delouis at bellsouth.net>
- Date: Wed, 11 May 2005 05:25:04 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
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)"
- Follow-Ups:
- Re: ChineseRemainder
- From: Chris Chiasson <chris.chiasson@gmail.com>
- Re: ChineseRemainder