Poly1305 is a Message Authentication Code—a cryptographic primitive for authenticating a message with a shared secret key, like HMAC.
Although it's really a fraction of the complexity of e.g. elliptic curves, most of the implementations I've read look decidedly like magic, mysteriously multiplying values by enchanted constants, and shuffling bits like The Sorcerer's Apprentice in Fantasia. Even the paper does not explain why and how its design decisions lead to faster code!
Still, after reverse-engineering what the implementations were doing, I grew convinced that cryptography code could be perfectly understandable if only we commented it.
I set out to prove this, and the code below is the Poly1305 implementation that came out. It should hopefully be readable with just an understanding of what MACs are for, a beginner level of Go, and high school math (like exponents and moduli). It aims to document both what it is doing and how, so it should not require any knowledge of Poly1305 specifically.
It also happens to be 75% faster than the existing Go code it is meant to replace. The amd64 assembly implementation in golang.org/x/crypto/poly1305 is only 30-60% faster than this code, which provides some timid hope for my dream of reducing the assembly in the Go crypto standard libraries year over year.
(If the layout below does not work for you, here's the same code on Gitiles.)
// Copyright 2019 The Go Authors. All rights reserved.
// Use of this source code is governed by a BSD-style
// license that can be found at https://golang.org/LICENSE.
package poly1305
import (
"encoding/binary"
"math/bits"
)
// Poly1305 [RFC 7539] is a relatively simple MAC algorithm: the authentication
// tag for a 64 bytes message is approximately
//
// s + m[0:16] * r⁴ + m[16:32] * r³ + m[32:48] * r² + m[48:64] * r mod 2¹³⁰ - 5
//
// for some secret r and s. It can be computed sequentially like
//
// for len(msg) > 0:
// h += read(msg, 16)
// h *= r
// h %= 2¹³⁰ - 5
// return h + s
//
// All the complexity is about doing performant constant-time math on numbers
// larger than any available numeric type.
// MAC is an io.Writer computing a Poly1305 authentication tag of the data
// written to it. After Sum has been called, Write must not be used anymore.
type MAC struct {
macState
buffer [TagSize]byte
offset int
}
// TagSize is the size, in bytes, of a poly1305 authenticator.
const TagSize = 16
// New returns a new MAC for a single-use key.
func New(key *[32]byte) *MAC {
h := &MAC{}
initialize(key, &h.r, &h.s)
return h
}
// Write splits the incoming message into TagSize chunks, and passes them to
// update. It buffers incomplete chunks.
func (h *MAC) Write(p []byte) (int, error) {
if h.offset > 0 {
n := copy(h.buffer[h.offset:], p)
if h.offset+n < TagSize {
h.offset += n
return len(p), nil
}
p = p[n:]
h.offset = 0
update(&h.macState, h.buffer[:])
}
if n := len(p) - (len(p) % TagSize); n > 0 {
update(&h.macState, p[:n])
p = p[n:]
}
if len(p) > 0 {
h.offset += copy(h.buffer[h.offset:], p)
}
return len(p), nil
}
// Sum flushes the last incomplete chunk from the buffer, if any, and generates
// the MAC output. It does not modify the MAC's state, in order to allow for
// multiple calls to Sum, even if no Write should be performed after Sum.
func (h *MAC) Sum(out *[TagSize]byte) {
state := h.macState
if h.offset > 0 {
update(&state, h.buffer[:h.offset])
}
finalize(out, &state.h, &state.s)
}
// macState holds numbers in saturated 64-bit little-endian limbs. That is,
// the value of [x0, x1, x2] is x[0] + x[1] * 2⁶⁴ + x[2] * 2¹²⁸.
type macState struct {
// h is the main accumulator. It is to be interpreted modulo 2¹³⁰ - 5, but
// can grow larger during and after rounds.
h [3]uint64
// r and s are the private key components.
r [2]uint64
s [2]uint64
}
// [rMask0, rMask1] is the specified Poly1305 clamping mask in little-endian. It
// clears some bits of the secret coefficient to make it possible to implement
// multiplication more efficiently.
const (
rMask0 = 0x0FFFFFFC0FFFFFFF
rMask1 = 0x0FFFFFFC0FFFFFFC
)
func initialize(key *[32]byte, r, s *[2]uint64) {
r[0] = binary.LittleEndian.Uint64(key[0:8]) & rMask0
r[1] = binary.LittleEndian.Uint64(key[8:16]) & rMask1
s[0] = binary.LittleEndian.Uint64(key[16:24])
s[1] = binary.LittleEndian.Uint64(key[24:32])
}
// uint128 holds a 128-bit number as two 64-bit limbs, for use with the
// bits.Mul64 and bits.Add64 intrinsics.
type uint128 struct {
lo, hi uint64
}
func mul64(a, b uint64) uint128 {
hi, lo := bits.Mul64(a, b)
return uint128{lo, hi}
}
func add128(a, b uint128) uint128 {
lo, c := bits.Add64(a.lo, b.lo, 0)
hi, c := bits.Add64(a.hi, b.hi, c)
if c != 0 {
panic("poly1305: unexpected overflow")
}
return uint128{lo, hi}
}
func shiftRightBy2(a uint128) uint128 {
a.lo = a.lo>>2 | (a.hi&3)<<62
a.hi = a.hi >> 2
return a
}
// update absorbs msg into the state.h accumulator. For each chunk of 128
// bits of message, it computes
//
// h₊ = (h + m) * r mod 2¹³⁰ - 5
//
// If the msg length is not a multiple of TagSize, it assumes the last
// incomplete chunk is the final one.
func update(state *macState, msg []byte) {
h0, h1, h2 := state.h[0], state.h[1], state.h[2]
r0, r1 := state.r[0], state.r[1]
for len(msg) > 0 {
var c uint64
// For the first step, h + m, we use a chain of bits.Add64 intrinsics.
// The resulting value of h might exceed 2¹³⁰ - 5, but will be partially
// reduced at the end of the multiplication below.
//
// The spec requires us to set a bit just above the message size, not to
// hide leading zeroes. For full chunks, that's 1 << 128, so we can just
// add 1 to the most significant (2¹²⁸) limb, h2.
if len(msg) >= TagSize {
h0, c = bits.Add64(h0, binary.LittleEndian.Uint64(msg[0:8]), 0)
h1, c = bits.Add64(h1, binary.LittleEndian.Uint64(msg[8:16]), c)
h2 += c + 1
msg = msg[TagSize:]
} else {
var buf [TagSize]byte
copy(buf[:], msg)
buf[len(msg)] = 1
h0, c = bits.Add64(h0, binary.LittleEndian.Uint64(buf[0:8]), 0)
h1, c = bits.Add64(h1, binary.LittleEndian.Uint64(buf[8:16]), c)
h2 += c
msg = nil
}
// Multiplication of big number limbs is similar to elementary school
// columnar multiplication. Instead of digits, there are 64-bit limbs.
//
// We are multiplying a 3 limbs number, h, by a 2 limbs number, r.
//
// h2 h1 h0 x
// r1 r0 =
// ----------------
// h2r0 h1r0 h0r0 <-- individual 128-bit products
// + h2r1 h1r1 h0r1
// ------------------------
// m3 m2 m1 m0 <-- result in 128-bit overlapping limbs
// ------------------------
// m3.hi m2.hi m1.hi m0.hi <-- carry propagation
// + m3.lo m2.lo m1.lo m0.lo
// -------------------------------
// t4 t3 t2 t1 t0 <-- final result in 64-bit limbs
//
// The main difference from pen-and-paper multiplication is that we do
// carry propagation in a separate step, as if we wrote two digit sums
// at first (the 128-bit limbs), and then carried the tens all at once.
h0r0 := mul64(h0, r0)
h1r0 := mul64(h1, r0)
h2r0 := mul64(h2, r0)
h0r1 := mul64(h0, r1)
h1r1 := mul64(h1, r1)
h2r1 := mul64(h2, r1)
// Since h2 is known to be at most 5 (see below), and r0 and r1 have their
// top 4 bits cleared by the mask, we know that their product is not going
// to overflow 64 bits, so we can ignore the high part of the products.
//
// This also means that the product doesn't have a fifth limb (t4).
if h2r0.hi != 0 {
panic("poly1305: unexpected overflow")
}
if h2r1.hi != 0 {
panic("poly1305: unexpected overflow")
}
m0 := h0r0
m1 := add128(h1r0, h0r1) // These two additions don't overflow thanks again
m2 := add128(h2r0, h1r1) // to the 4 masked bits at the top of r0 and r1.
m3 := h2r1
t0 := m0.lo
t1, c := bits.Add64(m1.lo, m0.hi, 0)
t2, c := bits.Add64(m2.lo, m1.hi, c)
t3, _ := bits.Add64(m3.lo, m2.hi, c)
// Now we have the result as 4 64-bit limbs, and we need to reduce it
// modulo 2¹³⁰ - 5. The special shape of this Crandall prime lets us do
// a cheap partial reduction according to the reduction identity
//
// c * 2¹³⁰ + n = c * 5 + n mod 2¹³⁰ - 5
//
// because 2¹³⁰ = 5 mod 2¹³⁰ - 5. Partial reduction since the result is
// likely to be larger than 2¹³⁰ - 5, but still small enough to fit the
// assumptions we make about h in the rest of the code.
//
// See also https://speakerdeck.com/gtank/engineering-prime-numbers?slide=23
// We split the final result at the 2¹³⁰ mark into h and cc, the carry.
// Note that the carry bits are effectively shifted left by 2, in other
// words, cc = c * 4 for the c in the reduction identity.
h0, h1, h2 = t0, t1, t2&maskLow2Bits
cc := uint128{t2 & maskNotLow2Bits, t3}
// To add c * 5 to h, we first add cc = c * 4, and then add (cc >> 2) = c.
h0, c = bits.Add64(h0, cc.lo, 0)
h1, c = bits.Add64(h1, cc.hi, c)
h2 += c
cc = shiftRightBy2(cc)
h0, c = bits.Add64(h0, cc.lo, 0)
h1, c = bits.Add64(h1, cc.hi, c)
h2 += c
// h2 is at most 3 + 1 + 1 = 5, making the whole of h at most
//
// 5 * 2¹²⁸ + (2¹²⁸ - 1) = 6 * 2¹²⁸ - 1
}
state.h[0], state.h[1], state.h[2] = h0, h1, h2
}
const (
maskLow2Bits uint64 = 0x0000000000000003
maskNotLow2Bits uint64 = ^maskLow2Bits
)
// select64 returns x if v == 1 and y if v == 0, in constant time.
func select64(v, x, y uint64) uint64 { return ^(v-1)&x | (v-1)&y }
// [p0, p1, p2] is 2¹³⁰ - 5 in little endian order.
const (
p0 = 0xFFFFFFFFFFFFFFFB
p1 = 0xFFFFFFFFFFFFFFFF
p2 = 0x0000000000000003
)
// finalize completes the modular reduction of h and computes
//
// out = h + s mod 2¹²⁸
//
func finalize(out *[TagSize]byte, h *[3]uint64, s *[2]uint64) {
h0, h1, h2 := h[0], h[1], h[2]
// After the partial reduction in update, h might be more than 2¹³⁰ - 5, but
// will be less than 2 * (2¹³⁰ - 5). To complete the reduction in constant
// time, we compute t = h - (2¹³⁰ - 5), and select h as the result if the
// subtraction underflows, and t otherwise.
t0, b := bits.Sub64(h0, p0, 0)
t1, b := bits.Sub64(h1, p1, b)
_, b = bits.Sub64(h2, p2, b)
// h = h if h < p else h - p
h0 = select64(b, h0, t0)
h1 = select64(b, h1, t1)
// Finally, we compute the last Poly1305 step
//
// tag = h + s mod 2¹²⁸
//
// by just doing a wide addition with the 128 low bits of h and discarding
// the overflow.
h0, c := bits.Add64(h0, s[0], 0)
h1, _ = bits.Add64(h1, s[1], c)
binary.LittleEndian.PutUint64(out[0:8], h0)
binary.LittleEndian.PutUint64(out[8:16], h1)
}
If you made it this far, consider following me on Twitter!
To prove that crypto code can be understandable, I gave my best shot at writing a readable Poly1305 implementation. It tries to explain both what it’s doing and how. (It’s also 75% faster than the current one.) https://t.co/4eKIw42uUU
— Filippo Valsorda (@FiloSottile) April 2, 2019