# Integer division, step 0: no remainder

Exciting times in SBCL-land! Not only will Google Summer of Code support two students to work on SBCL (one will improve our support for correct Unicode manipulation, and the other our strength reduction for integer division), but we also sprouted a new Linux/ARM port! As Christophe points out, this a nice coincidence: (most?) ARM chips lack hardware integer division units. I find the integer division project even more interesting because I believe we can cover all three standard division operators (floor, truncate, and ceiling) with an unified code generator.

I first looked into integer division by constants four years ago, and I was immediately struck by the ad hoc treatment of the transformation: I have yet to find a paper that summarises and relates algorithms that are currently in use. Worse, the pseudocode tends to assume fixed-width integer, which drowns the interesting logic in bignum-management noise. Back when I had free time, I uploaded an early draft of what may become a more enlightening introduction to the topic. My goal was to unite all the simplification algorithms I’d seen and to generalise them to SBCL’s needs: our optimisers benefit from precise integer range derivation, and codegen ought to deal with tagged fixnums. The draft should take shape as the GSoC project progresses.

There is one widespread – but very specialised – integer division algorithm that does not fit in the draft: multiplication by modular inverses. I’m guessing it’s common because it’s the first thing that comes to mind when we say division-by-multiplication. The transformation is also so specialised that I find it’s usually mentioned in contexts where it wouldn’t work. Still, it’s a nice application of algebra and the coefficients are simple enough to generate at runtime (even in C or assembly language), so here goes.

# Multiplicative inverses for integer division

Let $$a$$ and $$m$$ be naturals. The multiplicative inverse of $$a$$ modulo $$m$$ is a natural $$b$$ such that $$a \times b \equiv 1 \mod m$$. Machine arithmetic is naturally modular (e.g., mod $$m = 2\sp{32}$$). This seems perfect!

There are a couple issues here:

1. we have to find the modular inverse;
2. the modular inverse only exists if $$\mathop{gcd}(a, m) = 1$$;
3. multiplicative inversion and integer division only coincide when the remainder is zero.

For a concrete example of the third issue, consider $$11$$, the multiplicative inverse of $$3 \mod 16$$: $$3\times 11 = 33 \equiv 1 \mod 16$$ and $$6 \times 11 = 66 \equiv 2 \mod 16$$. However, $$4 \times 11 = 44 \equiv 12 \mod 16$$, and $$12$$ is nowhere close to $$4 \div 3$$.

This post addresses the first two points. There is no workaround for the last one.

We can generate a modular inverse with the extended Euclidean algorithm. Wikipedia shows the iterative version, which I can never remember, so I’ll instead construct the simple recursive one.

We already assume that $\mathop{gcd}(a, b) = 1$ and we wish to find $$x, y$$ such that $ax + by = 1.$ Bézout’s identity guarantees that such coefficients exist.

Things are simpler if we assume that $$a < b$$ (they can only be equal if $$a = b = 1$$, and that case is both annoying and uninteresting).

If $$a = 1$$, $$a + b0 = 1$$.

Otherwise, let $$q = \lfloor b/a\rfloor$$ and $$r = b - qa$$. $\mathop{gcd}(a, r) = \mathop{gcd}(a, b) = 1,$ and, given $ax’ + ry’ = 1,$ we can revert our change to find $ax’ + (b - qa)y’ = a(x’ - qy’) + by’ = 1.$

We’re working in modular arithmetic, so we can sprinkle mod m without changing the result. In C, this will naturally happen for unsigned integers, via overflows. In CL, we can still force modular reduction, just to convince ourselves that we don’t need bignums.

And a quick sanity check:

CL-USER> (loop for m from 2 upto (ash 1 10)
do (loop for i from 1 below m
when (= 1 (gcd i m))
do (inverse i m)))
NIL ; no assertion failure


The second issue is that the multiplicative inverse only exists if our divisor and our modulo (e.g., $$2\sp{32}$$) are coprime. The good news is that $$\mathop{gcd}(a, 2\sp{w})$$ can only be a power of two. We only have to factor our divisor $$a = 2\sp{s} v$$, and find $$i$$, the multiplicative inverse of $$v$$. Division by $$a$$ is then a right shift by $$s$$ and a multiplication by $$i$$.

And now, a final round of tests:

CL-USER> (defun test-divisor (d m)
(let ((divisor (divisor d m)))
(loop for i upfrom 0
for j from 0 by d below m
do (assert (= (funcall divisor j) i)))))
TEST-DIVISOR
CL-USER> (loop for width from 1 upto 20
for mod = (ash 1 width)
do (loop for x from 1 below mod
do (test-divisor x mod)))
NIL


A simple transformation from integer division to shift and multiplication… that works only in very specific conditions.

# What are modular inverses good for, then?

I’ve only seen this transformation used for pointer subtractions in C-like languages: machines count in chars and programs in whatever the pointers point to. Pointer arithmetic is only defined within the same array, so the compiler can assume that the distance between the two pointers is a multiple of the object size.

The following program is deep in undefined behaviour, for example.

pkhuong:tmp pkhuong \$ clang foo.c && ./a.out
-2635249153387078801 0


What I find interesting is that, if we pay attention to the correctness analysis, it’s clear that general div-by-mul transformations benefit from known common factors between the divisor and the dividend. In the extreme case, when the dividend is always a multiple of the divisor, we can convert the division to a single double-wide multiplication, without any shift or additional multi-word arithmetic. On architectures with fast multipliers or ones that let us compute the high half of product without the low part, the general case (coupled with a tight analysis) may be marginally quicker than this specialised transformation. Yet, both GCC and clang convert pointer subtractions to shifts and multiplications by modular inverses.

In the end multiplicative inverses seem mostly useful as a red herring, and as a minimal-complexity low hanging fruit. The only reason I use them is that it’s easy to generate the coefficients in C, which is helpful when allocation sizes are determined at runtime.

May 11th, 2014