## Stream: general

### Topic: long multiplication in Lean

#### Kevin Buzzard (Jan 12 2019 at 12:59):

nat is great for proving things, but is computationally inefficient because it uses O(n) memory to store the natural number n:

#reduce 10000 * 10000 -- deep recursion error

pos_num is a computationally efficient implementation of the positive integers in Lean.

inductive pos_num : Type
| one  : pos_num
| bit1 : pos_num → pos_num
| bit0 : pos_num → pos_num


Now we only need O(log n) memory to store n. But I must be using it incorrectly because I see performance issues:

import data.num.basic

-- takes a few seconds on my machine
#reduce (10000 : pos_num) * 10000 -- binary repn of 10^8

-- (deterministic) timeout
#reduce (1000000 : pos_num) * 1000000


Why are these things not instantaneous, like they would be in a computer algebra system? Lean has clearly solved these problems somehow, because computationally efficient types are presumably at the root of why these proofs work:

import tactic.norm_num

-- all this is immediate
example : (10000 : ℕ) * 10000 = 100000000 := by norm_num
example : (1000000 : ℕ) * 1000000 = 1000000000000 := by norm_num


Oh! tactic.norm_num does not even seem to import data.num.basic! So it must be using something else. The norm_num.lean file looks much less scary to me than it did a year ago, but I still can't see how it is doing multiplication (maybe because there is no 50 line module docstring ;-) )

#### Mario Carneiro (Jan 12 2019 at 13:12):

The norm_num interactive tactic actually delegates addition, subtraction and multiplication to tactic.norm_num in core

#### Kevin Buzzard (Jan 12 2019 at 13:18):

Oh! And am I right in thinking that this is written in C++?

#### Kevin Buzzard (Jan 12 2019 at 13:18):

meta constant norm_num : expr → tactic (expr × expr)

meta constants are in some sense invisible to me.

#### Rob Lewis (Jan 12 2019 at 13:27):

Yes, the core norm_num is in C++. For addition and multiplication it works basically like a special purpose simplifier: bit0 a + bit0 b simplifies to bit0 (a + b), etc. But it does this in the context of proving an equality, so it really changes bit0 a + bit0 b = c to bit0 (a + b) = c where c is in normal form. This makes it easy to reduce - and / to + and *.

#### Mario Carneiro (Jan 12 2019 at 14:45):

The reason why #reduce (10000 : pos_num) is slow is because the parser produces bit0 and bit1 applications, which could be defined in terms of the constructors of the pos_num type but instead uses the fixed definition _root_.bit0 which is defined using self-addition. As a result the bit0 operation is linear time instead of O(1), and a full numeral should be O(n^2) to evaluate (rather than O(n)). But it is still much better than the exponential time implementation for nat. This could be solved if there was a typeclass like has_bit providing the bit operations directly

#### Mario Carneiro (Jan 12 2019 at 14:49):

It looks like the equation compiler definitions of pos_num.add and pos_num.succ are also significant factors. If I define the functions using pos_num.rec or the induction tactic then it goes much faster

Last updated: May 08 2021 at 04:14 UTC