def tu_decomposition(s, v, verbose=False):
"""Using the knowledge that v is a subspace of Z_s of dimension n, a
TU-decomposition (as defined in [Perrin17]) of s is performed and T
and U are returned.
"""
N = int(log(len(s), 2))
# building B
basis = extract_basis(v[0], N)
t = len(basis)
basis = complete_basis(basis, N)
B = Matrix(GF(2), N, N, [tobin(x, N) for x in reversed(basis)])
if verbose:
print "B= (rank={})\n{}".format(B.rank(), B.str())
# building A
basis = complete_basis(extract_basis(v[1], N), N)
A = Matrix(GF(2), N, N, [tobin(x, N) for x in reversed(basis)])
if verbose:
print "A= (rank={})\n{}".format(A.rank(), A.str())
# building linear equivalent s_prime
s_prime = [
apply_bin_mat(s[apply_bin_mat(x, A.inverse())], B)
for x in xrange(0, 2**N)
]
# TU decomposition of s'
T, U = get_tu_open(s_prime, t)
if verbose:
print "T=["
for i in xrange(0, 2**(N - t)):
print " {} {}".format(T[i], is_permutation(T[i]))
print "]\nU=["
for i in xrange(0, 2**t):
print " {} {}".format(U[i], is_permutation(U[i]))
print "]"
return T, U
def get_ccz_equivalent_permutation_cartesian(s, v0, v1, verbose=False):
"""Takes as input two vector spaces v0 and v1 of the set of zeroes in
the LAT of s, each being the cartesian product of two spaces, and
returns a permutation CCZ equivalent to s obtained using these CCZ
spaces.
"""
N = int(log(len(s), 2))
# building B
e1 = extract_basis(v0[0], N)
t = len(e1)
e2 = extract_basis(v1[0], N)
B = Matrix(GF(2), N, N, [tobin(x, N) for x in e1 + e2])
if verbose:
print "B= (rank={})\n{}".format(B.rank(), B.str())
# building A
e1 = extract_basis(v0[1], N)
e2 = extract_basis(v1[1], N)
A = Matrix(GF(2), N, N, [tobin(x, N) for x in e1 + e2])
if verbose:
print "A= (rank={})\n{}".format(A.rank(), A.str())
# building linear equivalent s_prime
s_prime = [
apply_bin_mat(s[apply_bin_mat(x, A.inverse())], B)
for x in xrange(0, 2**N)
]
# TU decomposition of s'
T_inv, U = get_tu_closed(s_prime, t)
if verbose:
print "T_inv=["
for i in xrange(0, 2**t):
print " {} {}".format(T_inv[i], is_permutation(T_inv[i]))
print "]\nU=["
for i in xrange(0, 2**t):
print " {} {}".format(U[i], is_permutation(U[i]))
print "]"
# TU-unfolding
T = [inverse(row) for row in T_inv]
result = [0 for x in xrange(0, 2**N)]
for l in xrange(0, 2**t):
for r in xrange(0, 2**(N - t)):
x = (l << (N - t)) | r
y_r = T[r][l]
y_l = U[y_r][r]
result[x] = (y_l << (t)) | y_r
return result
def get_ccz_equivalent_permutation(l, v0, v1, verbose=False):
N = int(log(len(l), 2))
mask = sum(int(1 << i) for i in xrange(0, N))
L = v0 + v1
L = Matrix(GF(2), 2 * N, 2 * N, [tobin(x, 2 * N) for x in L]).transpose()
if verbose:
print "\n\nrank={}\n{}".format(L.rank(), L.str())
new_l = [[0 for b in xrange(0, 2**N)] for a in xrange(0, 2**N)]
for a, b in itertools.product(xrange(0, 2**N), xrange(0, 2**N)):
v = apply_bin_mat((a << N) | b, L)
a_prime, b_prime = v >> N, v & mask
new_l[a][b] = l[a_prime][b_prime]
return invert_lat(new_l)
