{{{ #!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() }}}