Re: ChineseRemainder
- To: mathgroup at smc.vnet.net
- Subject: [mg56995] Re: ChineseRemainder
- From: dh <dh at metrohm.ch>
- Date: Thu, 12 May 2005 02:32:35 -0400 (EDT)
- References: <d5ski6$nvj$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
HiDana, see below Sincerely, Daniel Dana DeLouis 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 is correct. By definition of ChineseRemainder: r == list2 mod list1. Therefore Mathematica says 164 == 0 mod 4 and because 0 == 4 mod 4 this also means that 164 == 4 mod 4 > > 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)" >