20091122, 18:44  #1 
Dec 2008
7^{2}·17 Posts 
Residue Number System
I was wondering what the residue number representation of 1 was? (Or for negative numbers in general)
Last fiddled with by flouran on 20091122 at 18:47 
20091122, 19:01  #2 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10257_{8} Posts 
I'm guessing you mean Prime95's residue thing at the end of an LL test?
I believe it would loop around to 2^p2, (as the mod is 2^p1) which would be (in binary) 111...111110, which would be (in hex) [some number 1F]...FFFE, which would make a residue of 0xFFFFFFFFFFFFFFFE (think that's the right length). So in general: for negative numbers x, it's the residue for 2^p1+x. Last fiddled with by MiniGeek on 20091122 at 19:05 
20091122, 21:22  #3 
"William"
May 2003
New Haven
23×103 Posts 
"1" means that number which, when added to 1, results in zero. For example, in base 12 "clock arithmetic", "1" is another name for 11. In "modulo p" arithmetic, it is another name for p1.

20091122, 22:34  #4 
Dec 2008
7^{2}·17 Posts 
No. Not what I was asking.
Last fiddled with by flouran on 20091122 at 23:30 
20091122, 22:42  #5  
Dec 2008
341_{16} Posts 
Quote:
Wikipedia gives a somewhat decent account of what RNS is: http://en.wikipedia.org/wiki/Residue_number_system Thus, for example, 0 in RNS would be (0,0,0). 1 in RNS is (1,1,1). 2 in RNS is (0,2,2). 3 in RNS is (1,0,3). 41 in RNS is (1,2,6). I think perhaps 1 would be (1,1,1). But that seems wrong Last fiddled with by flouran on 20091122 at 22:44 

20091122, 22:47  #6 
Jun 2008
2^{3}·3^{2} Posts 
How do you figure that this is wrong? (1,1,1) corresponds to (1,2,6) mod (2,3,7).

20091122, 22:50  #7  
Dec 2008
7^{2}×17 Posts 
Quote:
Nice to see that I was correct Last fiddled with by flouran on 20091122 at 23:30 

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