31 KiB
The safegcd implementation in libsecp256k1 explained
This document explains the modular inverse implementation in the src/modinv*.h files. It is based
on the paper
"Fast constant-time gcd computation and modular inversion"
by Daniel J. Bernstein and Bo-Yin Yang. The references below are for the Date: 2019.04.13 version.
The actual implementation is in C of course, but for demonstration purposes Python3 is used here. Most implementation aspects and optimizations are explained, except those that depend on the specific number representation used in the C code.
1. Computing the Greatest Common Divisor (GCD) using divsteps
The algorithm from the paper (section 11), at a very high level, is this:
def gcd(f, g):
"""Compute the GCD of an odd integer f and another integer g."""
assert f & 1 # require f to be odd
delta = 1 # additional state variable
while g != 0:
assert f & 1 # f will be odd in every iteration
if delta > 0 and g & 1:
delta, f, g = 1 - delta, g, (g - f) // 2
elif g & 1:
delta, f, g = 1 + delta, f, (g + f) // 2
else:
delta, f, g = 1 + delta, f, (g ) // 2
return abs(f)
It computes the greatest common divisor of an odd integer f and any integer g. Its inner loop keeps rewriting the variables f and g alongside a state variable δ that starts at 1, until g=0 is reached. At that point, |f| gives the GCD. Each of the transitions in the loop is called a "division step" (referred to as divstep in what follows).
For example, gcd(21, 14) would be computed as:
- Start with δ=1 f=21 g=14
- Take the third branch: δ=2 f=21 g=7
- Take the first branch: δ=-1 f=7 g=-7
- Take the second branch: δ=0 f=7 g=0
- The answer |f| = 7.
Why it works:
- Divsteps can be decomposed into two steps (see paragraph 8.2 in the paper):
- (a) If g is odd, replace (f,g) with (g,g-f) or (f,g+f), resulting in an even g.
- (b) Replace (f,g) with (f,g/2) (where g is guaranteed to be even).
- Neither of those two operations change the GCD:
- For (a), assume gcd(f,g)=c, then it must be the case that f=a c and g=b c for some integers a and b. As (g,g-f)=(b c,(b-a)c) and (f,f+g)=(a c,(a+b)c), the result clearly still has common factor c. Reasoning in the other direction shows that no common factor can be added by doing so either.
- For (b), we know that f is odd, so gcd(f,g) clearly has no factor 2, and we can remove it from g.
- The algorithm will eventually converge to g=0. This is proven in the paper (see theorem G.3).
- It follows that eventually we find a final value f' for which gcd(f,g) = gcd(f',0). As the gcd of f' and 0 is |f'| by definition, that is our answer.
Compared to more traditional GCD algorithms, this one has the property of only ever looking at the low-order bits of the variables to decide the next steps, and being easy to make constant-time (in more low-level languages than Python). The δ parameter is necessary to guide the algorithm towards shrinking the numbers' magnitudes without explicitly needing to look at high order bits.
Properties that will become important later:
- Performing more divsteps than needed is not a problem, as f does not change anymore after g=0.
- Only even numbers are divided by 2. This means that when reasoning about it algebraically we do not need to worry about rounding.
- At every point during the algorithm's execution the next N steps only depend on the bottom N bits of f and g, and on δ.
2. From GCDs to modular inverses
We want an algorithm to compute the inverse a of x modulo M, i.e. the number a such that a x=1 mod M. This inverse only exists if the GCD of x and M is 1, but that is always the case if M is prime and 0 < x < M. In what follows, assume that the modular inverse exists. It turns out this inverse can be computed as a side effect of computing the GCD by keeping track of how the internal variables can be written as linear combinations of the inputs at every step (see the extended Euclidean algorithm). Since the GCD is 1, such an algorithm will compute numbers a and b such that a x + b M = 1*. Taking that expression mod M gives a x mod M = 1, and we see that a is the modular inverse of x mod M.
A similar approach can be used to calculate modular inverses using the divsteps-based GCD algorithm shown above, if the modulus M is odd. To do so, compute gcd(f=M,g=x), while keeping track of extra variables d and e, for which at every step d = f/x (mod M) and e = g/x (mod M). f/x here means the number which multiplied with x gives f mod M. As f and g are initialized to M and x respectively, d and e just start off being 0 (M/x mod M = 0/x mod M = 0) and 1 (x/x mod M = 1).
def div2(M, x):
"""Helper routine to compute x/2 mod M (where M is odd)."""
assert M & 1
if x & 1: # If x is odd, make it even by adding M.
x += M
# x must be even now, so a clean division by 2 is possible.
return x // 2
def modinv(M, x):
"""Compute the inverse of x mod M (given that it exists, and M is odd)."""
assert M & 1
delta, f, g, d, e = 1, M, x, 0, 1
while g != 0:
# Note that while division by two for f and g is only ever done on even inputs, this is
# not true for d and e, so we need the div2 helper function.
if delta > 0 and g & 1:
delta, f, g, d, e = 1 - delta, g, (g - f) // 2, e, div2(M, e - d)
elif g & 1:
delta, f, g, d, e = 1 + delta, f, (g + f) // 2, d, div2(M, e + d)
else:
delta, f, g, d, e = 1 + delta, f, (g ) // 2, d, div2(M, e )
# Verify that the invariants d=f/x mod M, e=g/x mod M are maintained.
assert f % M == (d * x) % M
assert g % M == (e * x) % M
assert f == 1 or f == -1 # |f| is the GCD, it must be 1
# Because of invariant d = f/x (mod M), 1/x = d/f (mod M). As |f|=1, d/f = d*f.
return (d * f) % M
Also note that this approach to track d and e throughout the computation to determine the inverse is different from the paper. There (see paragraph 12.1 in the paper) a transition matrix for the entire computation is determined (see section 3 below) and the inverse is computed from that. The approach here avoids the need for 2x2 matrix multiplications of various sizes, and appears to be faster at the level of optimization we're able to do in C.
3. Batching multiple divsteps
Every divstep can be expressed as a matrix multiplication, applying a transition matrix (1/2 t) to both vectors [f, g] and [d, e] (see paragraph 8.1 in the paper):
t = [ u, v ]
[ q, r ]
[ out_f ] = (1/2 * t) * [ in_f ]
[ out_g ] = [ in_g ]
[ out_d ] = (1/2 * t) * [ in_d ] (mod M)
[ out_e ] [ in_e ]
where (u, v, q, r) is (0, 2, -1, 1), (2, 0, 1, 1), or (2, 0, 0, 1), depending on which branch is taken. As above, the resulting f and g are always integers.
Performing multiple divsteps corresponds to a multiplication with the product of all the individual divsteps' transition matrices. As each transition matrix consists of integers divided by 2, the product of these matrices will consist of integers divided by 2N (see also theorem 9.2 in the paper). These divisions are expensive when updating d and e, so we delay them: we compute the integer coefficients of the combined transition matrix scaled by 2N, and do one division by 2N as a final step:
def divsteps_n_matrix(delta, f, g):
"""Compute delta and transition matrix t after N divsteps (multiplied by 2^N)."""
u, v, q, r = 1, 0, 0, 1 # start with identity matrix
for _ in range(N):
if delta > 0 and g & 1:
delta, f, g, u, v, q, r = 1 - delta, g, (g - f) // 2, 2*q, 2*r, q-u, r-v
elif g & 1:
delta, f, g, u, v, q, r = 1 + delta, f, (g + f) // 2, 2*u, 2*v, q+u, r+v
else:
delta, f, g, u, v, q, r = 1 + delta, f, (g ) // 2, 2*u, 2*v, q , r
return delta, (u, v, q, r)
As the branches in the divsteps are completely determined by the bottom N bits of f and g, this function to compute the transition matrix only needs to see those bottom bits. Furthermore all intermediate results and outputs fit in (N+1)-bit numbers (unsigned for f and g; signed for u, v, q, and r) (see also paragraph 8.3 in the paper). This means that an implementation using 64-bit integers could set N=62 and compute the full transition matrix for 62 steps at once without any big integer arithmetic at all. This is the reason why this algorithm is efficient: it only needs to update the full-size f, g, d, and e numbers once every N steps.
We still need functions to compute:
[ out_f ] = (1/2^N * [ u, v ]) * [ in_f ]
[ out_g ] ( [ q, r ]) [ in_g ]
[ out_d ] = (1/2^N * [ u, v ]) * [ in_d ] (mod M)
[ out_e ] ( [ q, r ]) [ in_e ]
Because the divsteps transformation only ever divides even numbers by two, the result of t [f,g] is always even. When t is a composition of N divsteps, it follows that the resulting f and g will be multiple of 2N, and division by 2N is simply shifting them down:
def update_fg(f, g, t):
"""Multiply matrix t/2^N with [f, g]."""
u, v, q, r = t
cf, cg = u*f + v*g, q*f + r*g
# (t / 2^N) should cleanly apply to [f,g] so the result of t*[f,g] should have N zero
# bottom bits.
assert cf % 2**N == 0
assert cg % 2**N == 0
return cf >> N, cg >> N
The same is not true for d and e, and we need an equivalent of the div2 function for division by 2N mod M.
This is easy if we have precomputed 1/M mod 2N (which always exists for odd M):
def div2n(M, Mi, x):
"""Compute x/2^N mod M, given Mi = 1/M mod 2^N."""
assert (M * Mi) % 2**N == 1
# Find a factor m such that m*M has the same bottom N bits as x. We want:
# (m * M) mod 2^N = x mod 2^N
# <=> m mod 2^N = (x / M) mod 2^N
# <=> m mod 2^N = (x * Mi) mod 2^N
m = (Mi * x) % 2**N
# Subtract that multiple from x, cancelling its bottom N bits.
x -= m * M
# Now a clean division by 2^N is possible.
assert x % 2**N == 0
return (x >> N) % M
def update_de(d, e, t, M, Mi):
"""Multiply matrix t/2^N with [d, e], modulo M."""
u, v, q, r = t
cd, ce = u*d + v*e, q*d + r*e
return div2n(M, Mi, cd), div2n(M, Mi, ce)
With all of those, we can write a version of modinv that performs N divsteps at once:
def modinv(M, Mi, x):
"""Compute the modular inverse of x mod M, given Mi=1/M mod 2^N."""
assert M & 1
delta, f, g, d, e = 1, M, x, 0, 1
while g != 0:
# Compute the delta and transition matrix t for the next N divsteps (this only needs
# (N+1)-bit signed integer arithmetic).
delta, t = divsteps_n_matrix(delta, f % 2**N, g % 2**N)
# Apply the transition matrix t to [f, g]:
f, g = update_fg(f, g, t)
# Apply the transition matrix t to [d, e]:
d, e = update_de(d, e, t, M, Mi)
return (d * f) % M
This means that in practice we'll always perform a multiple of N divsteps. This is not a problem because once g=0, further divsteps do not affect f, g, d, or e anymore (only δ keeps increasing). For variable time code such excess iterations will be mostly optimized away in later sections.
4. Avoiding modulus operations
So far, there are two places where we compute a remainder of big numbers modulo M: at the end of
div2n in every update_de, and at the very end of modinv after potentially negating d due to the
sign of f. These are relatively expensive operations when done generically.
To deal with the modulus operation in div2n, we simply stop requiring d and e to be in range
[0,M) all the time. Let's start by inlining div2n into update_de, and dropping the modulus
operation at the end:
def update_de(d, e, t, M, Mi):
"""Multiply matrix t/2^N with [d, e] mod M, given Mi=1/M mod 2^N."""
u, v, q, r = t
cd, ce = u*d + v*e, q*d + r*e
# Cancel out bottom N bits of cd and ce.
md = -((Mi * cd) % 2**N)
me = -((Mi * ce) % 2**N)
cd += md * M
ce += me * M
# And cleanly divide by 2**N.
return cd >> N, ce >> N
Let's look at bounds on the ranges of these numbers. It can be shown that |u|+|v| and |q|+|r|
never exceed 2N (see paragraph 8.3 in the paper), and thus a multiplication with t will have
outputs whose absolute values are at most 2N times the maximum absolute input value. In case the
inputs d and e are in (-M,M), which is certainly true for the initial values d=0 and e=1 assuming
M > 1, the multiplication results in numbers in range (-2NM,2NM). Subtracting less than 2N
times M to cancel out N bits brings that up to (-2N+1M,2NM), and
dividing by 2N at the end takes it to (-2M,M). Another application of update_de would take that
to (-3M,2M), and so forth. This progressive expansion of the variables' ranges can be
counteracted by incrementing d and e by M whenever they're negative:
...
if d < 0:
d += M
if e < 0:
e += M
cd, ce = u*d + v*e, q*d + r*e
# Cancel out bottom N bits of cd and ce.
...
With inputs in (-2M,M), they will first be shifted into range (-M,M), which means that the
output will again be in (-2M,M), and this remains the case regardless of how many update_de
invocations there are. In what follows, we will try to make this more efficient.
Note that increasing d by M is equal to incrementing cd by u M and ce by q M. Similarly, increasing e by M is equal to incrementing cd by v M and ce by r M. So we could instead write:
...
cd, ce = u*d + v*e, q*d + r*e
# Perform the equivalent of incrementing d, e by M when they're negative.
if d < 0:
cd += u*M
ce += q*M
if e < 0:
cd += v*M
ce += r*M
# Cancel out bottom N bits of cd and ce.
md = -((Mi * cd) % 2**N)
me = -((Mi * ce) % 2**N)
cd += md * M
ce += me * M
...
Now note that we have two steps of corrections to cd and ce that add multiples of M: this increment, and the decrement that cancels out bottom bits. The second one depends on the first one, but they can still be efficiently combined by only computing the bottom bits of cd and ce at first, and using that to compute the final md, me values:
def update_de(d, e, t, M, Mi):
"""Multiply matrix t/2^N with [d, e], modulo M."""
u, v, q, r = t
md, me = 0, 0
# Compute what multiples of M to add to cd and ce.
if d < 0:
md += u
me += q
if e < 0:
md += v
me += r
# Compute bottom N bits of t*[d,e] + M*[md,me].
cd, ce = (u*d + v*e + md*M) % 2**N, (q*d + r*e + me*M) % 2**N
# Correct md and me such that the bottom N bits of t*[d,e] + M*[md,me] are zero.
md -= (Mi * cd) % 2**N
me -= (Mi * ce) % 2**N
# Do the full computation.
cd, ce = u*d + v*e + md*M, q*d + r*e + me*M
# And cleanly divide by 2**N.
return cd >> N, ce >> N
One last optimization: we can avoid the md M and me M multiplications in the bottom bits of cd and ce by moving them to the md and me correction:
...
# Compute bottom N bits of t*[d,e].
cd, ce = (u*d + v*e) % 2**N, (q*d + r*e) % 2**N
# Correct md and me such that the bottom N bits of t*[d,e]+M*[md,me] are zero.
# Note that this is not the same as {md = (-Mi * cd) % 2**N} etc. That would also result in N
# zero bottom bits, but isn't guaranteed to be a reduction of [0,2^N) compared to the
# previous md and me values, and thus would violate our bounds analysis.
md -= (Mi*cd + md) % 2**N
me -= (Mi*ce + me) % 2**N
...
The resulting function takes d and e in range (-2M,M) as inputs, and outputs values in the same
range. That also means that the d value at the end of modinv will be in that range, while we want
a result in [0,M). To do that, we need a normalization function. It's easy to integrate the
conditional negation of d (based on the sign of f) into it as well:
def normalize(sign, v, M):
"""Compute sign*v mod M, where v is in range (-2*M,M); output in [0,M)."""
assert sign == 1 or sign == -1
# v in (-2*M,M)
if v < 0:
v += M
# v in (-M,M). Now multiply v with sign (which can only be 1 or -1).
if sign == -1:
v = -v
# v in (-M,M)
if v < 0:
v += M
# v in [0,M)
return v
And calling it in modinv is simply:
...
return normalize(f, d, M)
5. Constant-time operation
The primary selling point of the algorithm is fast constant-time operation. What code flow still depends on the input data so far?
- the number of iterations of the while g ≠ 0 loop in
modinv - the branches inside
divsteps_n_matrix - the sign checks in
update_de - the sign checks in
normalize
To make the while loop in modinv constant time it can be replaced with a constant number of
iterations. The paper proves (Theorem 11.2) that 741 divsteps are sufficient for any 256-bit
inputs, and safegcd-bounds shows that the slightly better bound 724 is
sufficient even. Given that every loop iteration performs N divsteps, it will run a total of
⌈724/N⌉ times.
To deal with the branches in divsteps_n_matrix we will replace them with constant-time bitwise
operations (and hope the C compiler isn't smart enough to turn them back into branches; see
valgrind_ctime_test.c for automated tests that this isn't the case). To do so, observe that a
divstep can be written instead as (compare to the inner loop of gcd in section 1).
x = -f if delta > 0 else f # set x equal to (input) -f or f
if g & 1:
g += x # set g to (input) g-f or g+f
if delta > 0:
delta = -delta
f += g # set f to (input) g (note that g was set to g-f before)
delta += 1
g >>= 1
To convert the above to bitwise operations, we rely on a trick to negate conditionally: per the definition of negative numbers in two's complement, (-v == ~v + 1) holds for every number v. As -1 in two's complement is all 1 bits, bitflipping can be expressed as xor with -1. It follows that -v == (v ^ -1) - (-1). Thus, if we have a variable c that takes on values 0 or -1, then (v ^ c) - c is v if c=0 and -v if c=-1.
Using this we can write:
x = -f if delta > 0 else f
in constant-time form as:
c1 = (-delta) >> 63
# Conditionally negate f based on c1:
x = (f ^ c1) - c1
To use that trick, we need a helper mask variable c1 that resolves the condition δ>0 to -1
(if true) or 0 (if false). We compute c1 using right shifting, which is equivalent to dividing by
the specified power of 2 and rounding down (in Python, and also in C under the assumption of a typical two's complement system; see
assumptions.h for tests that this is the case). Right shifting by 63 thus maps all
numbers in range [-263,0) to -1, and numbers in range [0,263) to 0.
Using the facts that x&0=0 and x&(-1)=x (on two's complement systems again), we can write:
if g & 1:
g += x
as:
# Compute c2=0 if g is even and c2=-1 if g is odd.
c2 = -(g & 1)
# This masks out x if g is even, and leaves x be if g is odd.
g += x & c2
Using the conditional negation trick again we can write:
if g & 1:
if delta > 0:
delta = -delta
as:
# Compute c3=-1 if g is odd and delta>0, and 0 otherwise.
c3 = c1 & c2
# Conditionally negate delta based on c3:
delta = (delta ^ c3) - c3
Finally:
if g & 1:
if delta > 0:
f += g
becomes:
f += g & c3
It turns out that this can be implemented more efficiently by applying the substitution η=-δ. In this representation, negating δ corresponds to negating η, and incrementing δ corresponds to decrementing η. This allows us to remove the negation in the c1 computation:
# Compute a mask c1 for eta < 0, and compute the conditional negation x of f:
c1 = eta >> 63
x = (f ^ c1) - c1
# Compute a mask c2 for odd g, and conditionally add x to g:
c2 = -(g & 1)
g += x & c2
# Compute a mask c for (eta < 0) and odd (input) g, and use it to conditionally negate eta,
# and add g to f:
c3 = c1 & c2
eta = (eta ^ c3) - c3
f += g & c3
# Incrementing delta corresponds to decrementing eta.
eta -= 1
g >>= 1
A variant of divsteps with better worst-case performance can be used instead: starting δ at 1/2 instead of 1. This reduces the worst case number of iterations to 590 for 256-bit inputs (which can be shown using convex hull analysis). In this case, the substitution ζ=-(δ+1/2) is used instead to keep the variable integral. Incrementing δ by 1 still translates to decrementing ζ by 1, but negating δ now corresponds to going from ζ to -(ζ+1), or ~ζ. Doing that conditionally based on c3 is simply:
...
c3 = c1 & c2
zeta ^= c3
...
By replacing the loop in divsteps_n_matrix with a variant of the divstep code above (extended to
also apply all f operations to u, v and all g operations to q, r), a constant-time version of
divsteps_n_matrix is obtained. The full code will be in section 7.
These bit fiddling tricks can also be used to make the conditional negations and additions in
update_de and normalize constant-time.
6. Variable-time optimizations
In section 5, we modified the divsteps_n_matrix function (and a few others) to be constant time.
Constant time operations are only necessary when computing modular inverses of secret data. In
other cases, it slows down calculations unnecessarily. In this section, we will construct a
faster non-constant time divsteps_n_matrix function.
To do so, first consider yet another way of writing the inner loop of divstep operations in
gcd from section 1. This decomposition is also explained in the paper in section 8.2. We use
the original version with initial δ=1 and η=-δ here.
for _ in range(N):
if g & 1 and eta < 0:
eta, f, g = -eta, g, -f
if g & 1:
g += f
eta -= 1
g >>= 1
Whenever g is even, the loop only shifts g down and decreases η. When g ends in multiple zero
bits, these iterations can be consolidated into one step. This requires counting the bottom zero
bits efficiently, which is possible on most platforms; it is abstracted here as the function
count_trailing_zeros.
def count_trailing_zeros(v):
"""
When v is zero, consider all N zero bits as "trailing".
For a non-zero value v, find z such that v=(d<<z) for some odd d.
"""
if v == 0:
return N
else:
return (v & -v).bit_length() - 1
i = N # divsteps left to do
while True:
# Get rid of all bottom zeros at once. In the first iteration, g may be odd and the following
# lines have no effect (until "if eta < 0").
zeros = min(i, count_trailing_zeros(g))
eta -= zeros
g >>= zeros
i -= zeros
if i == 0:
break
# We know g is odd now
if eta < 0:
eta, f, g = -eta, g, -f
g += f
# g is even now, and the eta decrement and g shift will happen in the next loop.
We can now remove multiple bottom 0 bits from g at once, but still need a full iteration whenever there is a bottom 1 bit. In what follows, we will get rid of multiple 1 bits simultaneously as well.
Observe that as long as η ≥ 0, the loop does not modify f. Instead, it cancels out bottom bits of g and shifts them out, and decreases η and i accordingly - interrupting only when η becomes negative, or when i reaches 0. Combined, this is equivalent to adding a multiple of f to g to cancel out multiple bottom bits, and then shifting them out.
It is easy to find what that multiple is: we want a number w such that g+w f has a few bottom
zero bits. If that number of bits is L, we want g+w f mod 2L = 0, or w = -g/f mod 2L. Since f
is odd, such a w exists for any L. L cannot be more than i steps (as we'd finish the loop before
doing more) or more than η+1 steps (as we'd run eta, f, g = -eta, g, -f at that point), but
apart from that, we're only limited by the complexity of computing w.
This code demonstrates how to cancel up to 4 bits per step:
NEGINV16 = [15, 5, 3, 9, 7, 13, 11, 1] # NEGINV16[n//2] = (-n)^-1 mod 16, for odd n
i = N
while True:
zeros = min(i, count_trailing_zeros(g))
eta -= zeros
g >>= zeros
i -= zeros
if i == 0:
break
# We know g is odd now
if eta < 0:
eta, f, g = -eta, g, -f
# Compute limit on number of bits to cancel
limit = min(min(eta + 1, i), 4)
# Compute w = -g/f mod 2**limit, using the table value for -1/f mod 2**4. Note that f is
# always odd, so its inverse modulo a power of two always exists.
w = (g * NEGINV16[(f & 15) // 2]) % (2**limit)
# As w = -g/f mod (2**limit), g+w*f mod 2**limit = 0 mod 2**limit.
g += w * f
assert g % (2**limit) == 0
# The next iteration will now shift out at least limit bottom zero bits from g.
By using a bigger table more bits can be cancelled at once. The table can also be implemented as a formula. Several formulas are known for computing modular inverses modulo powers of two; some can be found in Hacker's Delight second edition by Henry S. Warren, Jr. pages 245-247. Here we need the negated modular inverse, which is a simple transformation of those:
- Instead of a 3-bit table:
- -f or f ^ 6
- Instead of a 4-bit table:
- 1 - f(f + 1)
- -(f + (((f + 1) & 4) << 1))
- For larger tables the following technique can be used: if w=-1/f mod 2L, then w(w f+2) is
-1/f mod 22L. This allows extending the previous formulas (or tables). In particular we
have this 6-bit function (based on the 3-bit function above):
- f(f2 - 2)
This loop, again extended to also handle u, v, q, and r alongside f and g, placed in
divsteps_n_matrix, gives a significantly faster, but non-constant time version.
7. Final Python version
All together we need the following functions:
- A way to compute the transition matrix in constant time, using the
divsteps_n_matrixfunction from section 2, but with its loop replaced by a variant of the constant-time divstep from section 5, extended to handle u, v, q, r:
def divsteps_n_matrix(zeta, f, g):
"""Compute zeta and transition matrix t after N divsteps (multiplied by 2^N)."""
u, v, q, r = 1, 0, 0, 1 # start with identity matrix
for _ in range(N):
c1 = zeta >> 63
# Compute x, y, z as conditionally-negated versions of f, u, v.
x, y, z = (f ^ c1) - c1, (u ^ c1) - c1, (v ^ c1) - c1
c2 = -(g & 1)
# Conditionally add x, y, z to g, q, r.
g, q, r = g + (x & c2), q + (y & c2), r + (z & c2)
c1 &= c2 # reusing c1 here for the earlier c3 variable
zeta = (zeta ^ c1) - 1 # inlining the unconditional zeta decrement here
# Conditionally add g, q, r to f, u, v.
f, u, v = f + (g & c1), u + (q & c1), v + (r & c1)
# When shifting g down, don't shift q, r, as we construct a transition matrix multiplied
# by 2^N. Instead, shift f's coefficients u and v up.
g, u, v = g >> 1, u << 1, v << 1
return zeta, (u, v, q, r)
- The functions to update f and g, and d and e, from section 2 and section 4, with the constant-time
changes to
update_defrom section 5:
def update_fg(f, g, t):
"""Multiply matrix t/2^N with [f, g]."""
u, v, q, r = t
cf, cg = u*f + v*g, q*f + r*g
return cf >> N, cg >> N
def update_de(d, e, t, M, Mi):
"""Multiply matrix t/2^N with [d, e], modulo M."""
u, v, q, r = t
d_sign, e_sign = d >> 257, e >> 257
md, me = (u & d_sign) + (v & e_sign), (q & d_sign) + (r & e_sign)
cd, ce = (u*d + v*e) % 2**N, (q*d + r*e) % 2**N
md -= (Mi*cd + md) % 2**N
me -= (Mi*ce + me) % 2**N
cd, ce = u*d + v*e + M*md, q*d + r*e + M*me
return cd >> N, ce >> N
- The
normalizefunction from section 4, made constant time as well:
def normalize(sign, v, M):
"""Compute sign*v mod M, where v in (-2*M,M); output in [0,M)."""
v_sign = v >> 257
# Conditionally add M to v.
v += M & v_sign
c = (sign - 1) >> 1
# Conditionally negate v.
v = (v ^ c) - c
v_sign = v >> 257
# Conditionally add M to v again.
v += M & v_sign
return v
- And finally the
modinvfunction too, adapted to use ζ instead of δ, and using the fixed iteration count from section 5:
def modinv(M, Mi, x):
"""Compute the modular inverse of x mod M, given Mi=1/M mod 2^N."""
zeta, f, g, d, e = -1, M, x, 0, 1
for _ in range((590 + N - 1) // N):
zeta, t = divsteps_n_matrix(zeta, f % 2**N, g % 2**N)
f, g = update_fg(f, g, t)
d, e = update_de(d, e, t, M, Mi)
return normalize(f, d, M)
- To get a variable time version, replace the
divsteps_n_matrixfunction with one that uses the divsteps loop from section 5, and amodinvversion that calls it without the fixed iteration count:
NEGINV16 = [15, 5, 3, 9, 7, 13, 11, 1] # NEGINV16[n//2] = (-n)^-1 mod 16, for odd n
def divsteps_n_matrix_var(eta, f, g):
"""Compute eta and transition matrix t after N divsteps (multiplied by 2^N)."""
u, v, q, r = 1, 0, 0, 1
i = N
while True:
zeros = min(i, count_trailing_zeros(g))
eta, i = eta - zeros, i - zeros
g, u, v = g >> zeros, u << zeros, v << zeros
if i == 0:
break
if eta < 0:
eta, f, u, v, g, q, r = -eta, g, q, r, -f, -u, -v
limit = min(min(eta + 1, i), 4)
w = (g * NEGINV16[(f & 15) // 2]) % (2**limit)
g, q, r = g + w*f, q + w*u, r + w*v
return eta, (u, v, q, r)
def modinv_var(M, Mi, x):
"""Compute the modular inverse of x mod M, given Mi = 1/M mod 2^N."""
eta, f, g, d, e = -1, M, x, 0, 1
while g != 0:
eta, t = divsteps_n_matrix_var(eta, f % 2**N, g % 2**N)
f, g = update_fg(f, g, t)
d, e = update_de(d, e, t, M, Mi)
return normalize(f, d, Mi)