http://alicebob.cryptoland.net/understanding-the-montgomery-reduction-algorithm/ Fri, 06 Jan 2012 10:22:56 +0000 hourly 1 http://wordpress.org/?v=3.3.1 http://alicebob.cryptoland.net/understanding-the-montgomery-reduction-algorithm/#comment-320 Conrado Fri, 06 Jan 2012 10:22:56 +0000 http://alicebob.cryptoland.net/?p=220#comment-320 Oops. Thanks! Oops. Thanks!

]]> http://alicebob.cryptoland.net/understanding-the-montgomery-reduction-algorithm/#comment-319 Martin Fri, 06 Jan 2012 08:48:47 +0000 http://alicebob.cryptoland.net/?p=220#comment-319 x * p[0] mod B = 1 not x * p[0] mod B = 0. x * p[0] mod B = 1
not x * p[0] mod B = 0.

]]> http://alicebob.cryptoland.net/understanding-the-montgomery-reduction-algorithm/#comment-318 Conrado Tue, 03 Jan 2012 01:19:10 +0000 http://alicebob.cryptoland.net/?p=220#comment-318 Sure! I am assuming that you are storing your numbers in byte vectors, therefore p[0] is the lower byte of the prime modulus. Then, 1/p[0] mod B is the inverse, modulo B, of p[0]; i.e. the number x such that x * p[0] mod B = 1. In order to compute it, you can use the <a href="http://alicebob.cryptoland.net/understanding-the-extended-euclidian-algorithm/" rel="nofollow">extended Euclidean algorithm</a>; but in your case you can simply compute x * p[0] mod B for all bytes x until the result is one. Sure! I am assuming that you are storing your numbers in byte vectors, therefore p[0] is the lower byte of the prime modulus. Then, 1/p[0] mod B is the inverse, modulo B, of p[0]; i.e. the number x such that x * p[0] mod B = 1. In order to compute it, you can use the extended Euclidean algorithm; but in your case you can simply compute x * p[0] mod B for all bytes x until the result is one.

]]> http://alicebob.cryptoland.net/understanding-the-montgomery-reduction-algorithm/#comment-317 Daniel Mon, 02 Jan 2012 23:48:18 +0000 http://alicebob.cryptoland.net/?p=220#comment-317 Can you go more in depth on "the magic number" section? I'm a little confused on how to calculate -(1/p[0]) mod B. I've got a little toy implementation that uses B=256. Can you go more in depth on “the magic number” section? I’m a little confused on how to calculate -(1/p[0]) mod B. I’ve got a little toy implementation that uses B=256.

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