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add simulator to explore the trade-offs for hashed-based digital signatures

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Zooko O'Whielacronx 2010-08-18 20:06:30 -07:00
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#!python
# MIT or any Tahoe license.
# range of hash output lengths
range_L_hash = [128]
lg_M = 53 # lg(required number of signatures before losing security)
limit_bytes = 480000 # limit on signature length
limit_cost = 500 # limit on Mcycles_Sig + weight_ver*Mcycles_ver
weight_ver = 1 # how important verification cost is relative to signature cost
# (note: setting this too high will just exclude useful candidates)
L_block = 512 # bitlength of hash input blocks
L_pad = 64 # bitlength of hash padding overhead (for M-D hashes)
L_label = 80 # bitlength of hash position label
L_prf = 256 # bitlength of hash output when used as a PRF
cycles_per_byte = 15.8 # cost of hash
Mcycles_per_block = cycles_per_byte * L_block / (8 * 1000000.0)
from math import floor, ceil, log, log1p, pow, e, sqrt
from sys import stderr
from gc import collect
def lg(x):
return log(x, 2)
def ln(x):
return log(x, e)
def ceil_log(x, B):
return int(ceil(log(x, B)))
def ceil_div(x, y):
return int(ceil(float(x) / float(y)))
def floor_div(x, y):
return int(floor(float(x) / float(y)))
# number of compression function evaluations to hash k hash-outputs
# we assume that there is a label in each block
def compressions(k):
return ceil_div(k + L_pad, L_block - L_label)
# sum of power series sum([pow(p, i) for i in range(n)])
def sum_powers(p, n):
if p == 1: return n
return (pow(p, n) - 1)/(p - 1)
def make_candidate(B, K, K1, K2, q, T, T_min, L_hash, lg_N, sig_bytes, c_sign, c_ver, c_ver_pm):
Mcycles_sign = c_sign * Mcycles_per_block
Mcycles_ver = c_ver * Mcycles_per_block
Mcycles_ver_pm = c_ver_pm * Mcycles_per_block
cost = Mcycles_sign + weight_ver*Mcycles_ver
if sig_bytes >= limit_bytes or cost > limit_cost:
return []
return [{
'B': B, 'K': K, 'K1': K1, 'K2': K2, 'q': q, 'T': T,
'T_min': T_min,
'L_hash': L_hash,
'lg_N': lg_N,
'sig_bytes': sig_bytes,
'c_sign': c_sign,
'Mcycles_sign': Mcycles_sign,
'c_ver': c_ver,
'c_ver_pm': c_ver_pm,
'Mcycles_ver': Mcycles_ver,
'Mcycles_ver_pm': Mcycles_ver_pm,
'cost': cost,
}]
# K1 = size of root Merkle tree
# K = size of middle Merkle trees
# K2 = size of leaf Merkle trees
# q = number of revealed private keys per signed message
# Winternitz with B < 4 is never optimal. For example, going from B=4 to B=2 halves the
# chain depth, but that is cancelled out by doubling (roughly) the number of digits.
range_B = xrange(4, 33)
M = pow(2, lg_M)
def calculate(K, K1, K2, q_max, L_hash, trees):
candidates = []
lg_K = lg(K)
lg_K1 = lg(K1)
lg_K2 = lg(K2)
# We want the optimal combination of q and T. That takes too much time and memory
# to search for directly, so we start by calculating the lowest possible value of T
# for any q. Then for potential values of T, we calculate the smallest q such that we
# will have at least L_hash bits of security against forgery using revealed private keys
# (i.e. this method of forgery is no easier than finding a hash preimage), provided
# that fewer than 2^lg_S_min messages are signed.
# min height of certification tree (excluding root and bottom layer)
T_min = ceil_div(lg_M - lg_K1, lg_K)
last_q = None
for T in xrange(T_min, T_min+21):
# lg(total number of leaf private keys)
lg_S = lg_K1 + lg_K*T
lg_N = lg_S + lg_K2
# Suppose that m signatures have been made. The number of times X that a given bucket has
# been chosen follows a binomial distribution B(m, p) where p = 1/S and S is the number of
# buckets. I.e. Pr(X = x) = C(m, x) * p^x * (1-p)^(m-x).
#
# If an attacker picks a random seed and message that falls into a bucket that has been
# chosen x times, then at most q*x private values in that bucket have been revealed, so
# (ignoring the possibility of guessing private keys, which is negligable) the attacker's
# success probability for a forgery using the revealed values is at most min(1, q*x / K2)^q.
#
# Let j = floor(K2/q). Conditioning on x, we have
#
# Pr(forgery) = sum_{x = 0..j}(Pr(X = x) * (q*x / K2)^q) + Pr(x > j)
# = sum_{x = 1..j}(Pr(X = x) * (q*x / K2)^q) + Pr(x > j)
#
# We lose nothing by approximating (q*x / K2)^q as 1 for x > 4, i.e. ignoring the resistance
# of the HORS scheme to forgery when a bucket has been chosen 5 or more times.
#
# Pr(forgery) < sum_{x = 1..4}(Pr(X = x) * (q*x / K2)^q) + Pr(x > 4)
#
# where Pr(x > 4) = 1 - sum_{x = 0..4}(Pr(X = x))
#
# We use log arithmetic here because values very close to 1 cannot be represented accurately
# in floating point, but their logarithms can (provided we use appropriate functions such as
# log1p).
lg_p = -lg_S
lg_1_p = log1p(-pow(2, lg_p))/ln(2) # lg(1-p), computed accurately
j = 5
lg_px = [lg_1_p * M]*j
# We approximate lg(M-x) as lg(M)
lg_px_step = lg_M + lg_p - lg_1_p
for x in xrange(1, j):
lg_px[x] = lg_px[x-1] - lg(x) + lg_px_step
def find_min_q():
for q in xrange(1, q_max+1):
lg_q = lg(q)
lg_pforge = [lg_px[x] + (lg_q*x - lg_K2)*q for x in xrange(1, j)]
if max(lg_pforge) < -L_hash + lg(j) and lg_px[j-1] + 1.0 < -L_hash:
#print "K = %d, K1 = %d, K2 = %d, L_hash = %d, lg_K2 = %.3f, q = %d, lg_pforge_1 = %.3f, lg_pforge_2 = %.3f, lg_pforge_3 = %.3f" \
# % (K, K1, K2, L_hash, lg_K2, q, lg_pforge_1, lg_pforge_2, lg_pforge_3)
return q
return None
q = find_min_q()
if q is None or q == last_q:
# if q hasn't decreased, this will be strictly worse than the previous candidate
continue
last_q = q
# number of compressions to compute the Merkle hashes
(h_M, c_M, _) = trees[K]
(h_M1, c_M1, _) = trees[K1]
(h_M2, c_M2, (dau, tri)) = trees[K2]
# B = generalized Winternitz base
for B in range_B:
# n is the number of digits needed to sign the message representative and checksum.
# The representation is base-B, except that we allow the most significant digit
# to be up to 2B-1.
n_L = ceil_div(L_hash-1, lg(B))
firstL_max = floor_div(pow(2, L_hash)-1, pow(B, n_L-1))
C_max = firstL_max + (n_L-1)*(B-1)
n_C = ceil_log(ceil_div(C_max, 2), B)
n = n_L + n_C
firstC_max = floor_div(C_max, pow(B, n_C-1))
# Total depth of Winternitz hash chains. The chains for the most significant
# digit of the message representative and of the checksum may be a different
# length to those for the other digits.
c_D = (n-2)*(B-1) + firstL_max + firstC_max
# number of compressions to hash a Winternitz public key
c_W = compressions(n*L_hash + L_label)
# bitlength of a single Winternitz signature and authentication path
L_MW = (n + h_M ) * L_hash
L_MW1 = (n + h_M1) * L_hash
# bitlength of the HORS signature and authentication paths
# For all but one of the q authentication paths, one of the sibling elements in
# another path is made redundant where they intersect. This cancels out the hash
# that would otherwise be needed at the bottom of the path, making the total
# length of the signature q*h_M2 + 1 hashes, rather than q*(h_M2 + 1).
L_leaf = (q*h_M2 + 1) * L_hash
# length of the overall GMSS+HORS signature and seeds
sig_bytes = ceil_div(L_MW1 + T*L_MW + L_leaf + L_prf + ceil(lg_N), 8)
c_MW = K *(c_D + c_W) + c_M + ceil_div(K *n*L_hash, L_prf)
c_MW1 = K1*(c_D + c_W) + c_M1 + ceil_div(K1*n*L_hash, L_prf)
# For simplicity, c_sign and c_ver don't take into account compressions saved
# as a result of intersecting authentication paths in the HORS signature, so
# are slight overestimates.
c_sign = c_MW1 + T*c_MW + q*(c_M2 + 1) + ceil_div(K2*L_hash, L_prf)
# *expected* number of compressions to verify a signature
c_ver = c_D/2.0 + c_W + c_M1 + T*(c_D/2.0 + c_W + c_M) + q*(c_M2 + 1)
c_ver_pm = (1 + T)*c_D/2.0
candidates += make_candidate(B, K, K1, K2, q, T, T_min, L_hash, lg_N, sig_bytes, c_sign, c_ver, c_ver_pm)
return candidates
def search():
for L_hash in range_L_hash:
print >>stderr, "collecting... \r",
collect()
print >>stderr, "precomputing... \r",
"""
# d/dq (lg(q+1) + L_hash/q) = 1/(ln(2)*(q+1)) - L_hash/q^2
# Therefore lg(q+1) + L_hash/q is at a minimum when 1/(ln(2)*(q+1)) = L_hash/q^2.
# Let alpha = L_hash*ln(2), then from the quadratic formula, the integer q that
# minimizes lg(q+1) + L_hash/q is the floor or ceiling of (alpha + sqrt(alpha^2 - 4*alpha))/2.
# (We don't want the other solution near 0.)
alpha = floor(L_hash*ln(2)) # float
q = floor((alpha + sqrt(alpha*(alpha-4)))/2)
if lg(q+2) + L_hash/(q+1) < lg(q+1) + L_hash/q:
q += 1
lg_S_margin = lg(q+1) + L_hash/q
q_max = int(q)
q = floor(L_hash*ln(2)) # float
if lg(q+1) + L_hash/(q+1) < lg(q) + L_hash/q:
q += 1
lg_S_margin = lg(q) + L_hash/q
q_max = int(q)
"""
q_max = 4000
# find optimal Merkle tree shapes for this L_hash and each K
trees = {}
K_max = 50
c2 = compressions(2*L_hash + L_label)
c3 = compressions(3*L_hash + L_label)
for dau in xrange(0, 10):
a = pow(2, dau)
for tri in xrange(0, ceil_log(30-dau, 3)):
x = int(a*pow(3, tri))
h = dau + 2*tri
c_x = int(sum_powers(2, dau)*c2 + a*sum_powers(3, tri)*c3)
for y in xrange(1, x+1):
if tri > 0:
# If the bottom level has arity 3, then for every 2 nodes by which the tree is
# imperfect, we can save c3 compressions by pruning 3 leaves back to their parent.
# If the tree is imperfect by an odd number of nodes, we can prune one extra leaf,
# possibly saving a compression if c2 < c3.
c_y = c_x - floor_div(x-y, 2)*c3 - ((x-y) % 2)*(c3-c2)
else:
# If the bottom level has arity 2, then for each node by which the tree is
# imperfect, we can save c2 compressions by pruning 2 leaves back to their parent.
c_y = c_x - (x-y)*c2
if y not in trees or (h, c_y, (dau, tri)) < trees[y]:
trees[y] = (h, c_y, (dau, tri))
#for x in xrange(1, K_max+1):
# print x, trees[x]
candidates = []
progress = 0
fuzz = 0
complete = (K_max-1)*(2200-200)/100
for K in xrange(2, K_max+1):
for K2 in xrange(200, 2200, 100):
for K1 in xrange(max(2, K-fuzz), min(K_max, K+fuzz)+1):
candidates += calculate(K, K1, K2, q_max, L_hash, trees)
progress += 1
print >>stderr, "searching: %3d %% \r" % (100.0 * progress / complete,),
print >>stderr, "filtering... \r",
step = 2.0
bins = {}
limit = floor_div(limit_cost, step)
for bin in xrange(0, limit+2):
bins[bin] = []
for c in candidates:
bin = floor_div(c['cost'], step)
bins[bin] += [c]
del candidates
# For each in a range of signing times, find the best candidate.
best = []
for bin in xrange(0, limit):
candidates = bins[bin] + bins[bin+1] + bins[bin+2]
if len(candidates) > 0:
best += [min(candidates, key=lambda c: c['sig_bytes'])]
def format_candidate(candidate):
return ("%(B)3d %(K)3d %(K1)3d %(K2)5d %(q)4d %(T)4d "
"%(L_hash)4d %(lg_N)5.1f %(sig_bytes)7d "
"%(c_sign)7d (%(Mcycles_sign)7.2f) "
"%(c_ver)7d +/-%(c_ver_pm)5d (%(Mcycles_ver)5.2f +/-%(Mcycles_ver_pm)5.2f) "
) % candidate
print >>stderr, " \r",
if len(best) > 0:
print " B K K1 K2 q T L_hash lg_N sig_bytes c_sign (Mcycles) c_ver ( Mcycles )"
print "---- ---- ---- ------ ---- ---- ------ ------ --------- ------------------ --------------------------------"
best.sort(key=lambda c: (c['sig_bytes'], c['cost']))
last_sign = None
last_ver = None
for c in best:
if last_sign is None or c['c_sign'] < last_sign or c['c_ver'] < last_ver:
print format_candidate(c)
last_sign = c['c_sign']
last_ver = c['c_ver']
print
else:
print "No candidates found for L_hash = %d or higher." % (L_hash)
return
del bins
del best
print "Maximum signature size: %d bytes" % (limit_bytes,)
print "Maximum (signing + %d*verification) cost: %.1f Mcycles" % (weight_ver, limit_cost)
print "Hash parameters: %d-bit blocks with %d-bit padding and %d-bit labels, %.2f cycles per byte" \
% (L_block, L_pad, L_label, cycles_per_byte)
print "PRF output size: %d bits" % (L_prf,)
print "Security level given by L_hash is maintained for up to 2^%d signatures.\n" % (lg_M,)
search()