Dealing with big numbers in Haskell

Question

I'm trying to implement the Miller test in Haskell (Not Miller-Rabin.) I'm dealing with big numbers, and in particular I need to exponentiate big numbers and take the modulus of a large number mod another large number.

Are there any standard functions for doing this? The normal expt function ^ tells me I run out of memory before it computes a result. For example, I'd like to do:

(mod (8888^38071670985) 9746347772161)

I could implement my own algorithms, but it'd be nice if these already exist.

http://stackoverflow.com/questions/1184296/why-can-haskell-handle-very-large-numbers-easily ..., your exponent is extremely large... however.... — Kristopher Micinski, Mar 25 '12 at 21:01
NVM about implementing my own. I looked at the Haskell implementations of these algorithms. They're exactly how I would have implemented them. — afkbowflexin, Mar 25 '12 at 21:02
Maybe try to estimate how much memory you would need for storing your exponent. That is enormous number you want to store. — Trismegistos, Mar 26 '12 at 19:07

score 6 · Accepted Answer · edited Mar 25 '12 at 21:34

6

There is modular exponentiation (and much more) in the arithmoi package.

Since I wrote it, I'd be interested to hear if you find it useful and what could be improved.

If you try to compute

(mod (8888^38071670985) 9746347772161)

as it stands, the intermediate result 8888^38071670985 would occupy roughly 5*10¹¹ bits, about 60GB. Even if you have so much RAM, that is close to (maybe slightly above) the limits of GMP (the size field in the GMP integers is four bytes).

So you also have to reduce the intermediate results during the calculation. Not only does that let the computation fit into memory without problems, but it's also faster since the involved numbers remain fairly small.

edited Mar 25 '12 at 21:34

ehird

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answered Mar 25 '12 at 21:05

Daniel Fischer

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This worked brilliantly. I should have done this from the beginning. I don't know why I didn't. I've been staring at the algorithm for a while now anyways. For whatever reason, I kept trying to do the expt and the mod separately and didn't put it together to use properties of congruences to reduce the problem. – afkbowflexin Mar 25 '12 at 21:45

score 1 · Answer 2 · answered Mar 25 '12 at 21:26

1

An approximation to your number before taking modulo is

  10^log(8888^38071670985)
= 10^(38071670985 * log(8888))
= 10^(1.5 * 10^11)

In other words it has around 1.5 * 10^11 digits. It would need around

1.5 * 10^11 / log(2) / 8 / (2^30) = 58GB

of memory just to represent.

So starting with this may not be the best idea. Does the library have to support calculation with this large numbers?

answered Mar 25 '12 at 21:26

Irfy

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Modular exponentiation does not require keeping the "number before taking modulo" – user102008 Mar 26 '12 at 02:45

Dealing with big numbers in Haskell

2 Answers2