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382 lines
12 KiB
Python
382 lines
12 KiB
Python
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# --- ELLIPTIC CURVE MATH ------------------------------------------------------
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#
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# curve definition: y^2 = x^3 - p*x - q
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# over finite field: Z/nZ* (prime residue classes modulo a prime number n)
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#
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#
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# COPYRIGHT (c) 2010 by Toni Mattis <solaris@live.de>
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# ------------------------------------------------------------------------------
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'''
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Module for elliptic curve arithmetic over a prime field GF(n).
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E(GF(n)) takes the form y**2 == x**3 - p*x - q (mod n) for a prime n.
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0. Structures used by this module
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PARAMETERS and SCALARS are non-negative (long) integers.
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A POINT (x, y), usually denoted p1, p2, ...
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is a pair of (long) integers where 0 <= x < n and 0 <= y < n
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A POINT in PROJECTIVE COORDINATES, usually denoted jp1, jp2, ...
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takes the form (X, Y, Z, Z**2, Z**3) where x = X / Z**2
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and y = Y / z**3. This form is called Jacobian coordinates.
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The NEUTRAL element "0" or "O" is represented by None
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in both coordinate systems.
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1. Basic Functions
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euclid() Is the Extended Euclidean Algorithm.
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inv() Computes the multiplicative inversion modulo n.
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curve_q() Finds the curve parameter q (mod n)
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when p and a point are given.
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element() Tests whether a point (x, y) is on the curve.
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2. Point transformations
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to_projective() Converts a point (x, y) to projective coordinates.
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from_projective() Converts a point from projective coordinates
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to (x, y) using the transformation described above.
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neg() Computes the inverse point -P in both coordinate
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systems.
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3. Slow point arithmetic
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These algorithms make use of basic geometry and modular arithmetic
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thus being suitable for small numbers and academic study.
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add() Computes the sum of two (x, y)-points
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mul() Perform scalar multiplication using "double & add"
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4. Fast point arithmetic
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These algorithms make use of projective coordinates, signed binary
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expansion and a JSP-like approach (joint sparse form).
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The following functions consume and return projective coordinates:
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addf() Optimized point addition.
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doublef() Optimized point doubling.
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mulf() Highly optimized scalar multiplication.
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muladdf() Highly optimized addition of two products.
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The following functions use the optimized ones above but consume
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and output (x, y)-coordinates for a more convenient usage:
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mulp() Encapsulates mulf()
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muladdp() Encapsulates muladdf()
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For single additions add() is generally faster than an encapsulation of
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addf() which would involve expensive coordinate transformations.
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Hence there is no addp() and doublep().
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'''
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# BASIC MATH -------------------------------------------------------------------
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def euclid(a, b):
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'''Solve x*a + y*b = ggt(a, b) and return (x, y, ggt(a, b))'''
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# Non-recursive approach hence suitable for large numbers
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x = yy = 0
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y = xx = 1
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while b:
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q = a // b
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a, b = b, a % b
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x, xx = xx - q * x, x
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y, yy = yy - q * y, y
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return xx, yy, a
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def inv(a, n):
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'''Perform inversion 1/a modulo n. a and n should be COPRIME.'''
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# coprimality is not checked here in favour of performance
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i = euclid(a, n)[0]
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while i < 0:
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i += n
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return i
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def curve_q(x, y, p, n):
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'''Find curve parameter q mod n having point (x, y) and parameter p'''
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return ((x * x - p) * x - y * y) % n
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def element(point, p, q, n):
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'''Test, whether the given point is on the curve (p, q, n)'''
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if point:
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x, y = point
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return (x * x * x - p * x - q) % n == (y * y) % n
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else:
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return True
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def to_projective(p):
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'''Transform point p given as (x, y) to projective coordinates'''
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if p:
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return (p[0], p[1], 1, 1, 1)
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else:
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return None # Identity point (0)
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def from_projective(jp, n):
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'''Transform a point from projective coordinates to (x, y) mod n'''
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if jp:
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return (jp[0] * inv(jp[3], n)) % n, (jp[1] * inv(jp[4], n)) % n
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else:
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return None # Identity point (0)
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def neg(p, n):
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'''Compute the inverse point to p in any coordinate system'''
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return (p[0], (n - p[1]) % n) + p[2:] if p else None
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# POINT ADDITION ---------------------------------------------------------------
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# addition of points in y**2 = x**3 - p*x - q over <Z/nZ*; +>
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def add(p, q, n, p1, p2):
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'''Add points p1 and p2 over curve (p, q, n)'''
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if p1 and p2:
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x1, y1 = p1
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x2, y2 = p2
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if (x1 - x2) % n:
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s = ((y1 - y2) * inv(x1 - x2, n)) % n # slope
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x = (s * s - x1 - x2) % n # intersection with curve
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return (x, n - (y1 + s * (x - x1)) % n)
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else:
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if (y1 + y2) % n: # slope s calculated by derivation
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s = ((3 * x1 * x1 - p) * inv(2 * y1, n)) % n
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x = (s * s - 2 * x1) % n # intersection with curve
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return (x, n - (y1 + s * (x - x1)) % n)
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else:
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return None
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else: # either p1 is not none -> ret. p1, otherwiese p2, which may be
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return p1 if p1 else p2 # none too.
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# faster addition: redundancy in projective coordinates eliminates
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# expensive inversions mod n.
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def addf(p, q, n, jp1, jp2):
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'''Add jp1 and jp2 in projective (jacobian) coordinates.'''
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if jp1 and jp2:
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x1, y1, z1, z1s, z1c = jp1
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x2, y2, z2, z2s, z2c = jp2
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s1 = (y1 * z2c) % n
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s2 = (y2 * z1c) % n
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u1 = (x1 * z2s) % n
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u2 = (x2 * z1s) % n
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if (u1 - u2) % n:
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h = (u2 - u1) % n
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r = (s2 - s1) % n
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hs = (h * h) % n
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hc = (hs * h) % n
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x3 = (-hc - 2 * u1 * hs + r * r) % n
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y3 = (-s1 * hc + r * (u1 * hs - x3)) % n
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z3 = (z1 * z2 * h) % n
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z3s = (z3 * z3) % n
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z3c = (z3s * z3) % n
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return (x3, y3, z3, z3s, z3c)
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else:
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if (s1 + s2) % n:
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return doublef(p, q, n, jp1)
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else:
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return None
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else:
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return jp1 if jp1 else jp2
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# explicit point doubling using redundant coordinates
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def doublef(p, q, n, jp):
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'''Double jp in projective (jacobian) coordinates'''
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if not jp:
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return None
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x1, y1, z1, z1p2, z1p3 = jp
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y1p2 = (y1 * y1) % n
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a = (4 * x1 * y1p2) % n
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b = (3 * x1 * x1 - p * z1p3 * z1) % n
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x3 = (b * b - 2 * a) % n
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y3 = (b * (a - x3) - 8 * y1p2 * y1p2) % n
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z3 = (2 * y1 * z1) % n
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z3p2 = (z3 * z3) % n
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return x3, y3, z3, z3p2, (z3p2 * z3) % n
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# SCALAR MULTIPLICATION --------------------------------------------------------
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# scalar multiplication p1 * c = p1 + p1 + ... + p1 (c times) in O(log(n))
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def mul(p, q, n, p1, c):
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'''multiply point p1 by scalar c over curve (p, q, n)'''
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res = None
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while c > 0:
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if c & 1:
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res = add(p, q, n, res, p1)
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c >>= 1 # c = c / 2
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p1 = add(p, q, n, p1, p1) # p1 = p1 * 2
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return res
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# this method allows _signed_bin() to choose between 1 and -1. It will select
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# the sign which leaves the higher number of zeroes in the binary
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# representation (the higher GDB).
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def _gbd(n):
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'''Compute second greatest base-2 divisor'''
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i = 1
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if n <= 0: return 0
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while not n % i:
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i <<= 1
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return i >> 2
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# This method transforms n into a binary representation having signed bits.
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# A signed binary expansion contains more zero-bits hence reducing the number
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# of additions required by a multiplication algorithm.
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#
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# Example: 15 ( 0b1111 ) can be written as 16 - 1, resulting in (1,0,0,0,-1)
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# and saving 2 additions. Subtraction can be performed as
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# efficiently as addition.
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def _signed_bin(n):
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'''Transform n into an optimized signed binary representation'''
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r = []
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while n > 1:
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if n & 1:
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cp = _gbd(n + 1)
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cn = _gbd(n - 1)
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if cp > cn: # -1 leaves more zeroes -> subtract -1 (= +1)
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r.append(-1)
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n += 1
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else: # +1 leaves more zeroes -> subtract +1 (= -1)
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r.append(+1)
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n -= 1
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else:
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r.append(0) # be glad about one more zero
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n >>= 1
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r.append(n)
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return r[::-1]
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# This multiplication algorithm combines signed binary expansion and
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# fast addition using projective coordinates resulting in 5 to 10 times
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# faster multiplication.
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def mulf(p, q, n, jp1, c):
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'''Multiply point jp1 by c in projective coordinates'''
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sb = _signed_bin(c)
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res = None
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jp0 = neg(jp1, n) # additive inverse of jp1 to be used fot bit -1
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for s in sb:
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res = doublef(p, q, n, res)
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if s:
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res = addf(p, q, n, res, jp1) if s > 0 else \
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addf(p, q, n, res, jp0)
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return res
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# Encapsulates mulf() in order to enable flat coordinates (x, y)
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def mulp(p, q, n, p1, c):
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'''Multiply point p by c using fast multiplication'''
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return from_projective(mulf(p, q, n, to_projective(p1), c), n)
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# Sum of two products using Shamir's trick and signed binary expansion
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def muladdf(p, q, n, jp1, c1, jp2, c2):
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'''Efficiently compute c1 * jp1 + c2 * jp2 in projective coordinates'''
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s1 = _signed_bin(c1)
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s2 = _signed_bin(c2)
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diff = len(s2) - len(s1)
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if diff > 0:
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s1 = [0] * diff + s1
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elif diff < 0:
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s2 = [0] * -diff + s2
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jp1p2 = addf(p, q, n, jp1, jp2)
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jp1n2 = addf(p, q, n, jp1, neg(jp2, n))
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precomp = ((None, jp2, neg(jp2, n)),
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(jp1, jp1p2, jp1n2),
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(neg(jp1, n), neg(jp1n2, n), neg(jp1p2, n)))
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res = None
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for i, j in zip(s1, s2):
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res = doublef(p, q, n, res)
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if i or j:
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res = addf(p, q, n, res, precomp[i][j])
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return res
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# Encapsulate muladdf()
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def muladdp(p, q, n, p1, c1, p2, c2):
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'''Efficiently compute c1 * p1 + c2 * p2 in (x, y)-coordinates'''
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return from_projective(muladdf(p, q, n,
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to_projective(p1), c1,
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to_projective(p2), c2), n)
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# POINT COMPRESSION ------------------------------------------------------------
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# Compute the square root modulo n
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# Determine the sign-bit of a point allowing to reconstruct y-coordinates
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# when x and the sign-bit are given:
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def sign_bit(p1):
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'''Return the signedness of a point p1'''
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return p1[1] % 2 if p1 else 0
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# Reconstruct the y-coordinate when curve parameters, x and the sign-bit of
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# the y coordinate are given:
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def y_from_x(x, p, q, n, sign):
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'''Return the y coordinate over curve (p, q, n) for given (x, sign)'''
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# optimized form of (x**3 - p*x - q) % n
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a = (((x * x) % n - p) * x - q) % n
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if __name__ == "__main__":
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import rsa
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import time
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t = time.time()
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n = rsa.get_prime(256/8, 20)
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tp = time.time() - t
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p = rsa.random.randint(1, n)
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p1 = (rsa.random.randint(1, n), rsa.random.randint(1, n))
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q = curve_q(p1[0], p1[1], p, n)
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r1 = rsa.random.randint(1,n)
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r2 = rsa.random.randint(1,n)
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q1 = mulp(p, q, n, p1, r1)
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q2 = mulp(p, q, n, p1, r2)
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s1 = mulp(p, q, n, q1, r2)
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s2 = mulp(p, q, n, q2, r1)
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s1 == s2
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tt = time.time() - t
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def test(tcount, bits = 256):
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n = rsa.get_prime(bits/8, 20)
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p = rsa.random.randint(1, n)
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p1 = (rsa.random.randint(1, n), rsa.random.randint(1, n))
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q = curve_q(p1[0], p1[1], p, n)
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p2 = mulp(p, q, n, p1, rsa.random.randint(1, n))
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c1 = [rsa.random.randint(1, n) for i in xrange(tcount)]
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c2 = [rsa.random.randint(1, n) for i in xrange(tcount)]
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c = zip(c1, c2)
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t = time.time()
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for i, j in c:
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from_projective(addf(p, q, n,
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mulf(p, q, n, to_projective(p1), i),
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mulf(p, q, n, to_projective(p2), j)), n)
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t1 = time.time() - t
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t = time.time()
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for i, j in c:
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muladdp(p, q, n, p1, i, p2, j)
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t2 = time.time() - t
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return tcount, t1, t2
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