Go Back > Great Internet Mersenne Prime Search > Math > Number Theory Discussion Group

Thread Tools
Old 2020-10-02, 20:22   #1
bhelmes's Avatar
Mar 2016

7·41 Posts
Default calculation of modulo Mp

A peaceful and pleasant day for you,

I do not understand how the calculation modulo a Mersenne prime is made:
"Arithmetic modulo a Mersenne number is particularly efficient on a binary computer"

Perhaps this is an interesting question also for others.

Greetings from the modulo operator

bhelmes is offline   Reply With Quote
Old 2020-10-02, 21:43   #2
Batalov's Avatar
Mar 2008

2×4,591 Posts

Read first ?

Originally Posted by
...and it performs the mod 2P-1 step for free.
In very simple terms:
Fact 1. When doing mod operations, (a * b) (mod Mp) = a (mod Mp) * b (mod Mp). Consequence: you never need to keep "the real value"; always keep only the value (mod Mp). This means that it will never be more than p bits long.

Fact 2. if we multiply one value by another and both are less than p bits long, you will get a result that is < 2p bits long.

Fact 3 (assuming mult or square is already done; there was no question about that part).
Mod operation becomes this: cut the bits above p-th bit. Slide them down, align with lower part. Add. Only if 1 bit carry sticks out above p bits, cut it again and add 1 to lower part. Done!

The DWT actually doesn't need this operation literally. But even if it did, this is how cheap computationally it would have been. Essentially - there is no division. Division by 2p-1 is luckily this elegant and easy.
Batalov is offline   Reply With Quote
Old 2020-10-03, 07:18   #3
Romulan Interpreter
LaurV's Avatar
Jun 2011

100010111110012 Posts

Probably easier for him, you have a number a which you want to reduce (mod m), so you are looking for a number b such as a=b (mod m). From the definition of the modulus, if a=b (mod m), this means that there is a number k, such as a=k*m+b. Now, if m=2^p-1, then you have a=k*(2^p-1)+b, or other way written, a=k*(2^p)-k+b. When you represent this is binary, on 2p bits, the first most-significant p bits (MSB) of the representation contain the k value (because multiplying with 2^p just moves k, p bits to the left), and the last p LSB bits (least significant bits) contain the value b-k. If you add the two halves together, you get b. That's all**.

(**Edit: except, sometimes you may get it all 1, by addition, and then the result is zero, or you may need to repeat the procedure once, but those "trifles" won't be discussed here.)

Edit 2, as I have some more time, and there is no reply yet, say you want to reduce 107 (mod 31), now 107 is 3*32+11, i.e. 3*2^5+11, therefore when you represent it in binary on 10 bits (5+5) you have 00011 01011. The MSB contains 3, and the LSB contain 11 (decimal, sorry for the confusion, I will "go advanced" now to change the font for the binary numbers here, and color all of them nicely). This is 3*32+11=3*(31+1)+11=3*31+3+11. The 3*31 part won't matter for the modulus. So, all you have to do, is to add the two halves to get 11+3=14. Which is the right result of 107 (mod 31).

Last fiddled with by LaurV on 2020-10-05 at 07:56
LaurV is offline   Reply With Quote
Old 2020-10-03, 08:49   #4
paulunderwood's Avatar
Sep 2002
Database er0rr

5·701 Posts

2^p - 1 == 0 mod Mp
2^p == 1 mod Mp
k*2^p == k mod Mp
That is k shifted to the left p-bits is equivalent to k.

If a and b each have no more than p bits then a*b is at most 2*p-1 bits, you can just add the number composed of the top most bits (above p-1 counting from 0) to the number composed of the bottom p bits, to get the result mod Mp.

Last fiddled with by paulunderwood on 2020-10-03 at 09:35
paulunderwood is offline   Reply With Quote
Old 2020-10-03, 17:03   #5
bhelmes's Avatar
Mar 2016

7×41 Posts

Most people know the 9-rule in the decimalsystem
which based on the fact that 10 = 1 mod 9

For Mersenne numbers it is the same reflection
that 2^p = 1 mod 2^p -1

Thanks for all replies,

now we know one advantage from the Mersenne numbers.

Last fiddled with by bhelmes on 2020-10-03 at 17:04
bhelmes is offline   Reply With Quote

Thread Tools

Similar Threads
Thread Thread Starter Forum Replies Last Post
Factorial modulo a prime axn Computer Science & Computational Number Theory 66 2011-09-01 21:55
modulo operation for polynomials? smslca Math 3 2011-04-18 17:18
Order of 3 modulo a Mersenne prime T.Rex Math 7 2009-03-13 10:46
N! modulo for large N Cyclamen Persicum Math 2 2003-12-10 10:52
The modulo operation, how is it computed? eepiccolo Math 7 2003-01-08 03:07

All times are UTC. The time now is 22:00.

Tue Dec 1 22:00:04 UTC 2020 up 82 days, 19:11, 1 user, load averages: 1.55, 1.83, 1.99

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.