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Make bignum.py self-contained, by importing versions of the two

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functions I was depending on from my personal Python maths utility
module.

[originally from svn r9104]
This commit is contained in:
Simon Tatham 2011-02-22 00:06:12 +00:00
parent 77180221bd
commit 9d4005e5c1
1 changed files with 33 additions and 11 deletions
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44
testdata/bignum.py vendored
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@ -1,14 +1,40 @@
# Generate test cases for a bignum implementation. # Generate test cases for a bignum implementation.
import sys import sys
import mathlib
# integer square roots
def sqrt(n):
d = long(n)
a = 0L
# b must start off as a power of 4 at least as large as n
ndigits = len(hex(long(n)))
b = 1L << (ndigits*4)
while 1:
a = a >> 1
di = 2*a + b
if di <= d:
d = d - di
a = a + b
b = b >> 2
if b == 0: break
return a
# continued fraction convergents of a rational
def confrac(n, d):
coeffs = [(1,0),(0,1)]
while d != 0:
i = n / d
n, d = d, n % d
coeffs.append((coeffs[-2][0]-i*coeffs[-1][0],
coeffs[-2][1]-i*coeffs[-1][1]))
return coeffs
def findprod(target, dir = +1, ratio=(1,1)): def findprod(target, dir = +1, ratio=(1,1)):
# Return two numbers whose product is as close as we can get to # Return two numbers whose product is as close as we can get to
# 'target', with any deviation having the sign of 'dir', and in # 'target', with any deviation having the sign of 'dir', and in
# the same approximate ratio as 'ratio'. # the same approximate ratio as 'ratio'.
r = mathlib.sqrt(target * ratio[0] * ratio[1]) r = sqrt(target * ratio[0] * ratio[1])
a = r / ratio[1] a = r / ratio[1]
b = r / ratio[0] b = r / ratio[0]
if a*b * dir < target * dir: if a*b * dir < target * dir:
@ -22,11 +48,7 @@ def findprod(target, dir = +1, ratio=(1,1)):
improved = 0 improved = 0
a, b = best[:2] a, b = best[:2]
terms = mathlib.confracr(a, b, output=None) coeffs = confrac(a, b)
coeffs = [(1,0),(0,1)]
for t in terms:
coeffs.append((coeffs[-2][0]-t*coeffs[-1][0],
coeffs[-2][1]-t*coeffs[-1][1]))
for c in coeffs: for c in coeffs:
# a*c[0]+b*c[1] is as close as we can get it to zero. So # a*c[0]+b*c[1] is as close as we can get it to zero. So
# if we replace a and b with a+c[1] and b+c[0], then that # if we replace a and b with a+c[1] and b+c[0], then that
@ -45,7 +67,7 @@ def findprod(target, dir = +1, ratio=(1,1)):
A,B,C = da*db, b*da+a*db, a*b-target A,B,C = da*db, b*da+a*db, a*b-target
discrim = B^2-4*A*C discrim = B^2-4*A*C
if discrim > 0 and A != 0: if discrim > 0 and A != 0:
root = mathlib.sqrt(discrim) root = sqrt(discrim)
vals = [] vals = []
vals.append((-B + root) / (2*A)) vals.append((-B + root) / (2*A))
vals.append((-B - root) / (2*A)) vals.append((-B - root) / (2*A))
@ -83,9 +105,9 @@ for i in range(1,4200):
# Simple tests of modpow. # Simple tests of modpow.
for i in range(64, 4097, 63): for i in range(64, 4097, 63):
modulus = mathlib.sqrt(1<<(2*i-1)) | 1 modulus = sqrt(1<<(2*i-1)) | 1
base = mathlib.sqrt(3*modulus*modulus) % modulus base = sqrt(3*modulus*modulus) % modulus
expt = mathlib.sqrt(modulus*modulus*2/5) expt = sqrt(modulus*modulus*2/5)
print "pow", hexstr(base), hexstr(expt), hexstr(modulus), hexstr(pow(base, expt, modulus)) print "pow", hexstr(base), hexstr(expt), hexstr(modulus), hexstr(pow(base, expt, modulus))
if i <= 1024: if i <= 1024:
# Test even moduli, which can't be done by Montgomery. # Test even moduli, which can't be done by Montgomery.