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diff --git a/src/secp256k1/doc/safegcd_implementation.md b/src/secp256k1/doc/safegcd_implementation.md new file mode 100644 index 0000000000..3ae556f9a7 --- /dev/null +++ b/src/secp256k1/doc/safegcd_implementation.md @@ -0,0 +1,765 @@ +# 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"](https://gcd.cr.yp.to/papers.html#safegcd) +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: + +```python +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](https://en.wikipedia.org/wiki/Euclidean_algorithm), 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](https://en.wikipedia.org/wiki/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*). + +```python +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 *2<sup>N</sup>* (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 *2<sup>N</sup>*, and +do one division by *2<sup>N</sup>* as a final step: + +```python +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 *2<sup>N</sup>*, and division by *2<sup>N</sup>* is simply shifting them down: + +```python +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 *2<sup>N</sup> mod M*. +This is easy if we have precomputed *1/M mod 2<sup>N</sup>* (which always exists for odd *M*): + +```python +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: + +```python3 +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: + +```python +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 *2<sup>N</sup>* (see paragraph 8.3 in the paper), and thus a multiplication with *t* will have +outputs whose absolute values are at most *2<sup>N</sup>* 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 *(-2<sup>N</sup>M,2<sup>N</sup>M)*. Subtracting less than *2<sup>N</sup>* +times *M* to cancel out *N* bits brings that up to *(-2<sup>N+1</sup>M,2<sup>N</sup>M)*, and +dividing by *2<sup>N</sup>* 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: + +```python + ... + 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: + +```python + ... + 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: + +```python +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: + +```python + ... + # 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: + +```python +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: + +```python + ... + 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](https://github.com/sipa/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). + +```python + 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: + +```python + x = -f if delta > 0 else f +``` + +in constant-time form as: + +```python + 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 *[-2<sup>63</sup>,0)* to *-1*, and numbers in range *[0,2<sup>63</sup>)* to *0*. + +Using the facts that *x&0=0* and *x&(-1)=x* (on two's complement systems again), we can write: + +```python + if g & 1: + g += x +``` + +as: + +```python + # 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: + +```python + if g & 1: + if delta > 0: + delta = -delta +``` + +as: + +```python + # 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: + +```python + if g & 1: + if delta > 0: + f += g +``` + +becomes: + +```python + 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: + +```python + # 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: + +```python + ... + 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. + +```python +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`. + +```python +def count_trailing_zeros(v): + """For a non-zero value v, find z such that v=(d<<z) for some odd d.""" + 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 2<sup>L</sup> = 0*, or *w = -g/f mod 2<sup>L</sup>*. 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: + +```python +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 2<sup>L</sup>*, then *w(w f+2)* is + *-1/f mod 2<sup>2L</sup>*. This allows extending the previous formulas (or tables). In particular we + have this 6-bit function (based on the 3-bit function above): + - *f(f<sup>2</sup> - 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_matrix` function + 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*: + +```python +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_de` from section 5: + +```python +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 `normalize` function from section 4, made constant time as well: + +```python +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 `modinv` function too, adapted to use *ζ* instead of *δ*, and using the fixed + iteration count from section 5: + +```python +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_matrix` function with one that uses the + divsteps loop from section 5, and a `modinv` version that calls it without the fixed iteration + count: + +```python +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) +``` |