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126 lines
3.9 KiB
Python
126 lines
3.9 KiB
Python
# Generate test cases for a bignum implementation.
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import sys
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# integer square roots
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def sqrt(n):
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d = long(n)
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a = 0L
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# b must start off as a power of 4 at least as large as n
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ndigits = len(hex(long(n)))
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b = 1L << (ndigits*4)
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while 1:
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a = a >> 1
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di = 2*a + b
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if di <= d:
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d = d - di
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a = a + b
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b = b >> 2
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if b == 0: break
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return a
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# continued fraction convergents of a rational
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def confrac(n, d):
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coeffs = [(1,0),(0,1)]
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while d != 0:
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i = n / d
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n, d = d, n % d
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coeffs.append((coeffs[-2][0]-i*coeffs[-1][0],
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coeffs[-2][1]-i*coeffs[-1][1]))
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return coeffs
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def findprod(target, dir = +1, ratio=(1,1)):
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# Return two numbers whose product is as close as we can get to
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# 'target', with any deviation having the sign of 'dir', and in
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# the same approximate ratio as 'ratio'.
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r = sqrt(target * ratio[0] * ratio[1])
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a = r / ratio[1]
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b = r / ratio[0]
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if a*b * dir < target * dir:
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a = a + 1
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b = b + 1
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assert a*b * dir >= target * dir
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best = (a,b,a*b)
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while 1:
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improved = 0
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a, b = best[:2]
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coeffs = confrac(a, b)
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for c in coeffs:
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# a*c[0]+b*c[1] is as close as we can get it to zero. So
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# if we replace a and b with a+c[1] and b+c[0], then that
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# will be added to our product, along with c[0]*c[1].
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da, db = c[1], c[0]
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# Flip signs as appropriate.
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if (a+da) * (b+db) * dir < target * dir:
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da, db = -da, -db
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# Multiply up. We want to get as close as we can to a
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# solution of the quadratic equation in n
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#
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# (a + n da) (b + n db) = target
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# => n^2 da db + n (b da + a db) + (a b - target) = 0
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A,B,C = da*db, b*da+a*db, a*b-target
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discrim = B^2-4*A*C
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if discrim > 0 and A != 0:
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root = sqrt(discrim)
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vals = []
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vals.append((-B + root) / (2*A))
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vals.append((-B - root) / (2*A))
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if root * root != discrim:
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root = root + 1
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vals.append((-B + root) / (2*A))
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vals.append((-B - root) / (2*A))
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for n in vals:
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ap = a + da*n
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bp = b + db*n
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pp = ap*bp
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if pp * dir >= target * dir and pp * dir < best[2]*dir:
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best = (ap, bp, pp)
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improved = 1
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if not improved:
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break
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return best
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def hexstr(n):
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s = hex(n)
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if s[:2] == "0x": s = s[2:]
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if s[-1:] == "L": s = s[:-1]
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return s
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# Tests of multiplication which exercise the propagation of the last
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# carry to the very top of the number.
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for i in range(1,4200):
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a, b, p = findprod((1<<i)+1, +1, (i, i*i+1))
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print "mul", hexstr(a), hexstr(b), hexstr(p)
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a, b, p = findprod((1<<i)+1, +1, (i, i+1))
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print "mul", hexstr(a), hexstr(b), hexstr(p)
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# Simple tests of modmul.
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for ai in range(20, 200, 60):
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a = sqrt(3<<(2*ai-1))
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for bi in range(20, 200, 60):
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b = sqrt(5<<(2*bi-1))
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for m in range(20, 600, 32):
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m = sqrt(2**(m+1))
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print "modmul", hexstr(a), hexstr(b), hexstr(m), hexstr((a*b) % m)
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# Simple tests of modpow.
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for i in range(64, 4097, 63):
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modulus = sqrt(1<<(2*i-1)) | 1
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base = sqrt(3*modulus*modulus) % modulus
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expt = sqrt(modulus*modulus*2/5)
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print "pow", hexstr(base), hexstr(expt), hexstr(modulus), hexstr(pow(base, expt, modulus))
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if i <= 1024:
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# Test even moduli, which can't be done by Montgomery.
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modulus = modulus - 1
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print "pow", hexstr(base), hexstr(expt), hexstr(modulus), hexstr(pow(base, expt, modulus))
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print "pow", hexstr(i), hexstr(expt), hexstr(modulus), hexstr(pow(i, expt, modulus))
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