def test_mul():
a, b = [0, 2, 1, 3], [0, 1, 3, 2]
assert _af_rmul(a, b) == [0, 2, 3, 1]
assert _af_rmuln(a, b, list(range(4))) == [0, 2, 3, 1]
assert rmul(Permutation(a), Permutation(b)).array_form == [0, 2, 3, 1]
a = Permutation([0, 2, 1, 3])
b = (0, 1, 3, 2)
c = (3, 1, 2, 0)
assert Permutation.rmul(a, b, c) == Permutation([1, 2, 3, 0])
assert Permutation.rmul(a, c) == Permutation([3, 2, 1, 0])
pytest.raises(TypeError, lambda: Permutation.rmul(b, c))
n = 6
m = 8
a = [Permutation.unrank_nonlex(n, i).array_form for i in range(m)]
h = list(range(n))
for i in range(m):
h = _af_rmul(h, a[i])
h2 = _af_rmuln(*a[:i + 1])
assert h == h2
def test_mul():
a, b = [0, 2, 1, 3], [0, 1, 3, 2]
assert _af_rmul(a, b) == [0, 2, 3, 1]
assert _af_rmuln(a, b, list(range(4))) == [0, 2, 3, 1]
assert rmul(Permutation(a), Permutation(b)).array_form == [0, 2, 3, 1]
a = Permutation([0, 2, 1, 3])
b = (0, 1, 3, 2)
c = (3, 1, 2, 0)
assert Permutation.rmul(a, b, c) == Permutation([1, 2, 3, 0])
assert Permutation.rmul(a, c) == Permutation([3, 2, 1, 0])
pytest.raises(TypeError, lambda: Permutation.rmul(b, c))
n = 6
m = 8
a = [Permutation.unrank_nonlex(n, i).array_form for i in range(m)]
h = list(range(n))
for i in range(m):
h = _af_rmul(h, a[i])
h2 = _af_rmuln(*a[:i + 1])
assert h == h2
def _strip_af(h, base, orbits, transversals, j):
"""
optimized _strip, with h, transversals and result in array form
if the stripped elements is the identity, it returns False, base_len + 1
j h[base[i]] == base[i] for i <= j
"""
base_len = len(base)
for i in range(j + 1, base_len):
beta = h[base[i]]
if beta == base[i]:
continue
if beta not in orbits[i]:
return h, i + 1
u = transversals[i][beta]
if h == u:
return False, base_len + 1
h = _af_rmul(_af_invert(u), h)
return h, base_len + 1
def canonical_free(base, gens, g, num_free):
"""
canonicalization of a tensor with respect to free indices
choosing the minimum with respect to lexicographical ordering
in the free indices
``base``, ``gens`` BSGS for slot permutation group
``g`` permutation representing the tensor
``num_free`` number of free indices
The indices must be ordered with first the free indices
see explanation in double_coset_can_rep
The algorithm is a variation of the one given in [2].
Examples
========
>>> from diofant.combinatorics import Permutation
>>> from diofant.combinatorics.tensor_can import canonical_free
>>> gens = [[1, 0, 2, 3, 5, 4], [2, 3, 0, 1, 4, 5],[0, 1, 3, 2, 5, 4]]
>>> gens = [Permutation(h) for h in gens]
>>> base = [0, 2]
>>> g = Permutation([2, 1, 0, 3, 4, 5])
>>> canonical_free(base, gens, g, 4)
[0, 3, 1, 2, 5, 4]
Consider the product of Riemann tensors
``T = R^{a}_{d0}^{d1,d2}*R_{d2,d1}^{d0,b}``
The order of the indices is ``[a,b,d0,-d0,d1,-d1,d2,-d2]``
The permutation corresponding to the tensor is
``g = [0,3,4,6,7,5,2,1,8,9]``
In particular ``a`` is position ``0``, ``b`` is in position ``9``.
Use the slot symmetries to get `T` is a form which is the minimal
in lexicographic order in the free indices ``a`` and ``b``, e.g.
``-R^{a}_{d0}^{d1,d2}*R^{b,d0}_{d2,d1}`` corresponding to
``[0, 3, 4, 6, 1, 2, 7, 5, 9, 8]``
>>> from diofant.combinatorics.tensor_can import riemann_bsgs, tensor_gens
>>> base, gens = riemann_bsgs
>>> size, sbase, sgens = tensor_gens(base, gens, [[], []], 0)
>>> g = Permutation([0, 3, 4, 6, 7, 5, 2, 1, 8, 9])
>>> canonical_free(sbase, [Permutation(h) for h in sgens], g, 2)
[0, 3, 4, 6, 1, 2, 7, 5, 9, 8]
"""
g = g.array_form
size = len(g)
if not base:
return g[:]
transversals = get_transversals(base, gens)
m = len(base)
for x in sorted(g[:-2]):
if x not in base:
base.append(x)
h = g
for i, transv in enumerate(transversals):
b = base[i]
h_i = [size] * num_free
# find the element s in transversals[i] such that
# _af_rmul(h, s) has its free elements with the lowest position in h
s = None
for sk in transv.values():
h1 = _af_rmul(h, sk)
hi = [h1.index(ix) for ix in range(num_free)]
if hi < h_i:
h_i = hi
s = sk
if s:
h = _af_rmul(h, s)
return h
def double_coset_can_rep(dummies, sym, b_S, sgens, S_transversals, g):
"""
Butler-Portugal algorithm for tensor canonicalization with dummy indices
dummies
list of lists of dummy indices,
one list for each type of index;
the dummy indices are put in order contravariant, covariant
[d0, -d0, d1, -d1, ...].
sym
list of the symmetries of the index metric for each type.
possible symmetries of the metrics
* 0 symmetric
* 1 antisymmetric
* None no symmetry
b_S
base of a minimal slot symmetry BSGS.
sgens
generators of the slot symmetry BSGS.
S_transversals
transversals for the slot BSGS.
g
permutation representing the tensor.
Return 0 if the tensor is zero, else return the array form of
the permutation representing the canonical form of the tensor.
A tensor with dummy indices can be represented in a number
of equivalent ways which typically grows exponentially with
the number of indices. To be able to establish if two tensors
with many indices are equal becomes computationally very slow
in absence of an efficient algorithm.
The Butler-Portugal algorithm [3] is an efficient algorithm to
put tensors in canonical form, solving the above problem.
Portugal observed that a tensor can be represented by a permutation,
and that the class of tensors equivalent to it under slot and dummy
symmetries is equivalent to the double coset `D*g*S`
(Note: in this documentation we use the conventions for multiplication
of permutations p, q with (p*q)(i) = p[q[i]] which is opposite
to the one used in the Permutation class)
Using the algorithm by Butler to find a representative of the
double coset one can find a canonical form for the tensor.
To see this correspondence,
let `g` be a permutation in array form; a tensor with indices `ind`
(the indices including both the contravariant and the covariant ones)
can be written as
`t = T(ind[g[0],..., ind[g[n-1]])`,
where `n= len(ind)`;
`g` has size `n + 2`, the last two indices for the sign of the tensor
(trick introduced in [4]).
A slot symmetry transformation `s` is a permutation acting on the slots
`t -> T(ind[(g*s)[0]],..., ind[(g*s)[n-1]])`
A dummy symmetry transformation acts on `ind`
`t -> T(ind[(d*g)[0]],..., ind[(d*g)[n-1]])`
Being interested only in the transformations of the tensor under
these symmetries, one can represent the tensor by `g`, which transforms
as
`g -> d*g*s`, so it belongs to the coset `D*g*S`.
Let us explain the conventions by an example.
Given a tensor `T^{d3 d2 d1}{}_{d1 d2 d3}` with the slot symmetries
`T^{a0 a1 a2 a3 a4 a5} = -T^{a2 a1 a0 a3 a4 a5}`
`T^{a0 a1 a2 a3 a4 a5} = -T^{a4 a1 a2 a3 a0 a5}`
and symmetric metric, find the tensor equivalent to it which
is the lowest under the ordering of indices:
lexicographic ordering `d1, d2, d3` then and contravariant index
before covariant index; that is the canonical form of the tensor.
The canonical form is `-T^{d1 d2 d3}{}_{d1 d2 d3}`
obtained using `T^{a0 a1 a2 a3 a4 a5} = -T^{a2 a1 a0 a3 a4 a5}`.
To convert this problem in the input for this function,
use the following labelling of the index names
(- for covariant for short) `d1, -d1, d2, -d2, d3, -d3`
`T^{d3 d2 d1}{}_{d1 d2 d3}` corresponds to `g = [4,2,0,1,3,5,6,7]`
where the last two indices are for the sign
`sgens = [Permutation(0,2)(6,7), Permutation(0,4)(6,7)]`
sgens[0] is the slot symmetry `-(0,2)`
`T^{a0 a1 a2 a3 a4 a5} = -T^{a2 a1 a0 a3 a4 a5}`
sgens[1] is the slot symmetry `-(0,4)`
`T^{a0 a1 a2 a3 a4 a5} = -T^{a4 a1 a2 a3 a0 a5}`
The dummy symmetry group D is generated by the strong base generators
`[(0,1),(2,3),(4,5),(0,1)(2,3),(2,3)(4,5)]`
The dummy symmetry acts from the left
`d = [1,0,2,3,4,5,6,7]` exchange `d1 -> -d1`
`T^{d3 d2 d1}{}_{d1 d2 d3} == T^{d3 d2}{}_{d1}{}^{d1}{}_{d2 d3}`
`g=[4,2,0,1,3,5,6,7] -> [4,2,1,0,3,5,6,7] = _af_rmul(d, g)`
which differs from `_af_rmul(g, d)`.
The slot symmetry acts from the right
`s = [2,1,0,3,4,5,7,6]` exchanges slots 0 and 2 and changes sign
`T^{d3 d2 d1}{}_{d1 d2 d3} == -T^{d1 d2 d3}{}_{d1 d2 d3}`
`g=[4,2,0,1,3,5,6,7] -> [0,2,4,1,3,5,7,6] = _af_rmul(g, s)`
Example in which the tensor is zero, same slot symmetries as above:
`T^{d3}{}_{d1,d2}{}^{d1}{}_{d3}{}^{d2}`
`= -T^{d3}{}_{d1,d3}{}^{d1}{}_{d2}{}^{d2}` under slot symmetry `-(2,4)`;
`= T_{d3 d1}{}^{d3}{}^{d1}{}_{d2}{}^{d2}` under slot symmetry `-(0,2)`;
`= T^{d3}{}_{d1 d3}{}^{d1}{}_{d2}{}^{d2}` symmetric metric;
`= 0` since two of these lines have tensors differ only for the sign.
The double coset D*g*S consists of permutations `h = d*g*s` corresponding
to equivalent tensors; if there are two `h` which are the same apart
from the sign, return zero; otherwise
choose as representative the tensor with indices
ordered lexicographically according to `[d1, -d1, d2, -d2, d3, -d3]`
that is `rep = min(D*g*S) = min([d*g*s for d in D for s in S])`
The indices are fixed one by one; first choose the lowest index
for slot 0, then the lowest remaining index for slot 1, etc.
Doing this one obtains a chain of stabilizers
`S -> S_{b0} -> S_{b0,b1} -> ...` and
`D -> D_{p0} -> D_{p0,p1} -> ...`
where `[b0, b1, ...] = range(b)` is a base of the symmetric group;
the strong base `b_S` of S is an ordered sublist of it;
therefore it is sufficient to compute once the
strong base generators of S using the Schreier-Sims algorithm;
the stabilizers of the strong base generators are the
strong base generators of the stabilizer subgroup.
`dbase = [p0,p1,...]` is not in general in lexicographic order,
so that one must recompute the strong base generators each time;
however this is trivial, there is no need to use the Schreier-Sims
algorithm for D.
The algorithm keeps a TAB of elements `(s_i, d_i, h_i)`
where `h_i = d_i*g*s_i` satisfying `h_i[j] = p_j` for `0 <= j < i`
starting from `s_0 = id, d_0 = id, h_0 = g`.
The equations `h_0[0] = p_0, h_1[1] = p_1,...` are solved in this order,
choosing each time the lowest possible value of p_i
For `j < i`
`d_i*g*s_i*S_{b_0,...,b_{i-1}}*b_j = D_{p_0,...,p_{i-1}}*p_j`
so that for dx in `D_{p_0,...,p_{i-1}}` and sx in
`S_{base[0],...,base[i-1]}` one has `dx*d_i*g*s_i*sx*b_j = p_j`
Search for dx, sx such that this equation holds for `j = i`;
it can be written as `s_i*sx*b_j = J, dx*d_i*g*J = p_j`
`sx*b_j = s_i**-1*J; sx = trace(s_i**-1, S_{b_0,...,b_{i-1}})`
`dx**-1*p_j = d_i*g*J; dx = trace(d_i*g*J, D_{p_0,...,p_{i-1}})`
`s_{i+1} = s_i*trace(s_i**-1*J, S_{b_0,...,b_{i-1}})`
`d_{i+1} = trace(d_i*g*J, D_{p_0,...,p_{i-1}})**-1*d_i`
`h_{i+1}*b_i = d_{i+1}*g*s_{i+1}*b_i = p_i`
`h_n*b_j = p_j` for all j, so that `h_n` is the solution.
Add the found `(s, d, h)` to TAB1.
At the end of the iteration sort TAB1 with respect to the `h`;
if there are two consecutive `h` in TAB1 which differ only for the
sign, the tensor is zero, so return 0;
if there are two consecutive `h` which are equal, keep only one.
Then stabilize the slot generators under `i` and the dummy generators
under `p_i`.
Assign `TAB = TAB1` at the end of the iteration step.
At the end `TAB` contains a unique `(s, d, h)`, since all the slots
of the tensor `h` have been fixed to have the minimum value according
to the symmetries. The algorithm returns `h`.
It is important that the slot BSGS has lexicographic minimal base,
otherwise there is an `i` which does not belong to the slot base
for which `p_i` is fixed by the dummy symmetry only, while `i`
is not invariant from the slot stabilizer, so `p_i` is not in
general the minimal value.
This algorithm differs slightly from the original algorithm [3]:
the canonical form is minimal lexicographically, and
the BSGS has minimal base under lexicographic order.
Equal tensors `h` are eliminated from TAB.
Examples
========
>>> from diofant.combinatorics.permutations import Permutation
>>> from diofant.combinatorics.perm_groups import PermutationGroup
>>> from diofant.combinatorics.tensor_can import double_coset_can_rep, get_transversals
>>> gens = [Permutation(x) for x in [[2, 1, 0, 3, 4, 5, 7, 6], [4, 1, 2, 3, 0, 5, 7, 6]]]
>>> base = [0, 2]
>>> g = Permutation([4, 2, 0, 1, 3, 5, 6, 7])
>>> transversals = get_transversals(base, gens)
>>> double_coset_can_rep([list(range(6))], [0], base, gens, transversals, g)
[0, 1, 2, 3, 4, 5, 7, 6]
>>> g = Permutation([4, 1, 3, 0, 5, 2, 6, 7])
>>> double_coset_can_rep([list(range(6))], [0], base, gens, transversals, g)
0
"""
size = g.size
g = g.array_form
num_dummies = size - 2
indices = list(range(num_dummies))
all_metrics_with_sym = all([_ is not None for _ in sym])
num_types = len(sym)
dumx = dummies[:]
dumx_flat = []
for dx in dumx:
dumx_flat.extend(dx)
b_S = b_S[:]
sgensx = [h._array_form for h in sgens]
if b_S:
S_transversals = transversal2coset(size, b_S, S_transversals)
# strong generating set for D
dsgsx = []
for i in range(num_types):
dsgsx.extend(dummy_sgs(dumx[i], sym[i], num_dummies))
ginv = _af_invert(g)
idn = list(range(size))
# TAB = list of entries (s, d, h) where h = _af_rmuln(d,g,s)
# for short, in the following d*g*s means _af_rmuln(d,g,s)
TAB = [(idn, idn, g)]
for i in range(size - 2):
b = i
testb = b in b_S and sgensx
if testb:
sgensx1 = [_af_new(_) for _ in sgensx]
deltab = _orbit(size, sgensx1, b)
else:
deltab = {b}
# p1 = min(IMAGES) = min(Union D_p*h*deltab for h in TAB)
if all_metrics_with_sym:
md = _min_dummies(dumx, sym, indices)
else:
md = [
min(_orbit(size, [_af_new(ddx) for ddx in dsgsx], ii))
for ii in range(size - 2)
]
p_i = min([min([md[h[x]] for x in deltab]) for s, d, h in TAB])
dsgsx1 = [_af_new(_) for _ in dsgsx]
Dxtrav = _orbit_transversal(size, dsgsx1, p_i, False, af=True) \
if dsgsx else None
if Dxtrav:
Dxtrav = [_af_invert(x) for x in Dxtrav]
# compute the orbit of p_i
for ii in range(num_types):
if p_i in dumx[ii]:
# the orbit is made by all the indices in dum[ii]
if sym[ii] is not None:
deltap = dumx[ii]
else:
# the orbit is made by all the even indices if p_i
# is even, by all the odd indices if p_i is odd
p_i_index = dumx[ii].index(p_i) % 2
deltap = dumx[ii][p_i_index::2]
break
else:
deltap = [p_i]
TAB1 = []
nTAB = len(TAB)
while TAB:
s, d, h = TAB.pop()
if min([md[h[x]] for x in deltab]) != p_i:
continue
deltab1 = [x for x in deltab if md[h[x]] == p_i]
# NEXT = s*deltab1 intersection (d*g)**-1*deltap
dg = _af_rmul(d, g)
dginv = _af_invert(dg)
sdeltab = [s[x] for x in deltab1]
gdeltap = [dginv[x] for x in deltap]
NEXT = [x for x in sdeltab if x in gdeltap]
# d, s satisfy
# d*g*s*base[i-1] = p_{i-1}; using the stabilizers
# d*g*s*S_{base[0],...,base[i-1]}*base[i-1] =
# D_{p_0,...,p_{i-1}}*p_{i-1}
# so that to find d1, s1 satisfying d1*g*s1*b = p_i
# one can look for dx in D_{p_0,...,p_{i-1}} and
# sx in S_{base[0],...,base[i-1]}
# d1 = dx*d; s1 = s*sx
# d1*g*s1*b = dx*d*g*s*sx*b = p_i
for j in NEXT:
if testb:
# solve s1*b = j with s1 = s*sx for some element sx
# of the stabilizer of ..., base[i-1]
# sx*b = s**-1*j; sx = _trace_S(s, j,...)
# s1 = s*trace_S(s**-1*j,...)
s1 = _trace_S(s, j, b, S_transversals)
if not s1:
continue
else:
s1 = [s[ix] for ix in s1]
else:
s1 = s
assert s1[b] == j
# solve d1*g*j = p_i with d1 = dx*d for some element dg
# of the stabilizer of ..., p_{i-1}
# dx**-1*p_i = d*g*j; dx**-1 = trace_D(d*g*j,...)
# d1 = trace_D(d*g*j,...)**-1*d
# to save an inversion in the inner loop; notice we did
# Dxtrav = [perm_af_invert(x) for x in Dxtrav] out of the loop
if Dxtrav:
d1 = _trace_D(dg[j], p_i, Dxtrav)
if not d1:
continue
else:
if p_i != dg[j]:
continue
d1 = idn
assert d1[dg[j]] == p_i
d1 = [d1[ix] for ix in d]
h1 = [d1[g[ix]] for ix in s1]
assert h1[b] == p_i
TAB1.append((s1, d1, h1))
# if TAB contains equal permutations, keep only one of them;
# if TAB contains equal permutations up to the sign, return 0
TAB1.sort(key=lambda x: x[-1])
nTAB1 = len(TAB1)
prev = [0] * size
while TAB1:
s, d, h = TAB1.pop()
if h[:-2] == prev[:-2]:
if h[-1] != prev[-1]:
return 0
else:
TAB.append((s, d, h))
prev = h
# stabilize the SGS
sgensx = [h for h in sgensx if h[b] == b]
if b in b_S:
b_S.remove(b)
_dumx_remove(dumx, dumx_flat, p_i)
dsgsx = []
for i in range(num_types):
dsgsx.extend(dummy_sgs(dumx[i], sym[i], num_dummies))
return TAB[0][-1]
def test_Permutation():
# don't auto fill 0
pytest.raises(ValueError, lambda: Permutation([1]))
p = Permutation([0, 1, 2, 3])
# call as bijective
assert [p(i) for i in range(p.size)] == list(p)
# call as operator
assert p(list(range(p.size))) == list(p)
# call as function
assert list(p(1, 2)) == [0, 2, 1, 3]
# conversion to list
assert list(p) == list(range(4))
assert Permutation(size=4) == Permutation(3)
assert Permutation(Permutation(3), size=5) == Permutation(4)
# cycle form with size
assert Permutation([[1, 2]], size=4) == Permutation([[1, 2], [0], [3]])
# random generation
assert Permutation.random(2) in (Permutation([1, 0]), Permutation([0, 1]))
p = Permutation([2, 5, 1, 6, 3, 0, 4])
q = Permutation([[1], [0, 3, 5, 6, 2, 4]])
assert len({p, p}) == 1
r = Permutation([1, 3, 2, 0, 4, 6, 5])
ans = Permutation(_af_rmuln(*[w.array_form for w in (p, q, r)])).array_form
assert rmul(p, q, r).array_form == ans
# make sure no other permutation of p, q, r could have given
# that answer
for a, b, c in permutations((p, q, r)):
if (a, b, c) == (p, q, r):
continue
assert rmul(a, b, c).array_form != ans
assert p.support() == list(range(7))
assert q.support() == [0, 2, 3, 4, 5, 6]
assert Permutation(p.cyclic_form).array_form == p.array_form
assert p.cardinality == 5040
assert q.cardinality == 5040
assert q.cycles == 2
assert rmul(q, p) == Permutation([4, 6, 1, 2, 5, 3, 0])
assert rmul(p, q) == Permutation([6, 5, 3, 0, 2, 4, 1])
assert _af_rmul(p.array_form, q.array_form) == \
[6, 5, 3, 0, 2, 4, 1]
assert rmul(Permutation([[1, 2, 3], [0, 4]]),
Permutation([[1, 2, 4], [0], [3]])).cyclic_form == \
[[0, 4, 2], [1, 3]]
assert q.array_form == [3, 1, 4, 5, 0, 6, 2]
assert q.cyclic_form == [[0, 3, 5, 6, 2, 4]]
assert q.full_cyclic_form == [[0, 3, 5, 6, 2, 4], [1]]
assert p.cyclic_form == [[0, 2, 1, 5], [3, 6, 4]]
t = p.transpositions()
assert t == [(0, 5), (0, 1), (0, 2), (3, 4), (3, 6)]
assert Permutation.rmul(*[Permutation(Cycle(*ti)) for ti in (t)])
assert Permutation([1, 0]).transpositions() == [(0, 1)]
assert p**13 == p
assert q**0 == Permutation(list(range(q.size)))
assert q**-2 == ~q**2
assert q**2 == Permutation([5, 1, 0, 6, 3, 2, 4])
assert q**3 == q**2 * q
assert q**4 == q**2 * q**2
a = Permutation(1, 3)
b = Permutation(2, 0, 3)
I = Permutation(3)
assert ~a == a**-1
assert a * ~a == I
assert a * b**-1 == a * ~b
ans = Permutation(0, 5, 3, 1, 6)(2, 4)
assert (p + q.rank()).rank() == ans.rank()
assert (p + q.rank())._rank == ans.rank()
assert (q + p.rank()).rank() == ans.rank()
pytest.raises(TypeError, lambda: p + Permutation(list(range(10))))
assert (p - q.rank()).rank() == Permutation(0, 6, 3, 1, 2, 5, 4).rank()
assert p.rank() - q.rank() < 0 # for coverage: make sure mod is used
assert (q - p.rank()).rank() == Permutation(1, 4, 6, 2)(3, 5).rank()
assert p * q == Permutation(_af_rmuln(*[list(w) for w in (q, p)]))
assert p * Permutation([]) == p
assert Permutation([]) * p == p
assert p * Permutation([[0, 1]]) == Permutation([2, 5, 0, 6, 3, 1, 4])
assert Permutation([[0, 1]]) * p == Permutation([5, 2, 1, 6, 3, 0, 4])
pq = p ^ q
assert pq == Permutation([5, 6, 0, 4, 1, 2, 3])
assert pq == rmul(q, p, ~q)
qp = q ^ p
assert qp == Permutation([4, 3, 6, 2, 1, 5, 0])
assert qp == rmul(p, q, ~p)
pytest.raises(ValueError, lambda: p ^ Permutation([]))
assert p.commutator(q) == Permutation(0, 1, 3, 4, 6, 5, 2)
assert q.commutator(p) == Permutation(0, 2, 5, 6, 4, 3, 1)
assert p.commutator(q) == ~q.commutator(p)
pytest.raises(ValueError, lambda: p.commutator(Permutation([])))
assert len(p.atoms()) == 7
assert q.atoms() == {0, 1, 2, 3, 4, 5, 6}
assert p.inversion_vector() == [2, 4, 1, 3, 1, 0]
assert q.inversion_vector() == [3, 1, 2, 2, 0, 1]
assert Permutation.from_inversion_vector(p.inversion_vector()) == p
assert Permutation.from_inversion_vector(q.inversion_vector()).array_form\
== q.array_form
pytest.raises(ValueError,
lambda: Permutation.from_inversion_vector([0, 2]))
assert Permutation([i for i in range(500, -1, -1)]).inversions() == 125250
s = Permutation([0, 4, 1, 3, 2])
assert s.parity() == 0
_ = s.cyclic_form # needed to create a value for _cyclic_form
assert len(s._cyclic_form) != s.size and s.parity() == 0
assert not s.is_odd
assert s.is_even
assert Permutation([0, 1, 4, 3, 2]).parity() == 1
assert _af_parity([0, 4, 1, 3, 2]) == 0
assert _af_parity([0, 1, 4, 3, 2]) == 1
s = Permutation([0])
assert s.is_Singleton
assert Permutation([]).is_Empty
r = Permutation([3, 2, 1, 0])
assert (r**2).is_Identity
assert rmul(~p, p).is_Identity
assert (~p)**13 == Permutation([5, 2, 0, 4, 6, 1, 3])
assert ~(r**2).is_Identity
assert p.max() == 6
assert p.min() == 0
q = Permutation([[6], [5], [0, 1, 2, 3, 4]])
assert q.max() == 4
assert q.min() == 0
p = Permutation([1, 5, 2, 0, 3, 6, 4])
q = Permutation([[1, 2, 3, 5, 6], [0, 4]])
assert p.ascents() == [0, 3, 4]
assert q.ascents() == [1, 2, 4]
assert r.ascents() == []
assert p.descents() == [1, 2, 5]
assert q.descents() == [0, 3, 5]
assert Permutation(r.descents()).is_Identity
assert p.inversions() == 7
# test the merge-sort with a longer permutation
big = list(p) + list(range(p.max() + 1, p.max() + 130))
assert Permutation(big).inversions() == 7
assert p.signature() == -1
assert q.inversions() == 11
assert q.signature() == -1
assert rmul(p, ~p).inversions() == 0
assert rmul(p, ~p).signature() == 1
assert p.order() == 6
assert q.order() == 10
assert (p**(p.order())).is_Identity
assert p.length() == 6
assert q.length() == 7
assert r.length() == 4
assert p.runs() == [[1, 5], [2], [0, 3, 6], [4]]
assert q.runs() == [[4], [2, 3, 5], [0, 6], [1]]
assert r.runs() == [[3], [2], [1], [0]]
assert p.index() == 8
assert q.index() == 8
assert r.index() == 3
assert p.get_precedence_distance(q) == q.get_precedence_distance(p)
assert p.get_adjacency_distance(q) == p.get_adjacency_distance(q)
assert p.get_positional_distance(q) == p.get_positional_distance(q)
p = Permutation([0, 1, 2, 3])
q = Permutation([3, 2, 1, 0])
assert p.get_precedence_distance(q) == 6
assert p.get_adjacency_distance(q) == 3
assert p.get_positional_distance(q) == 8
p = Permutation([0, 3, 1, 2, 4])
q = Permutation.josephus(4, 5, 2)
assert p.get_adjacency_distance(q) == 3
pytest.raises(ValueError,
lambda: p.get_adjacency_distance(Permutation([])))
pytest.raises(ValueError,
lambda: p.get_positional_distance(Permutation([])))
pytest.raises(ValueError,
lambda: p.get_precedence_distance(Permutation([])))
a = [Permutation.unrank_nonlex(4, i) for i in range(5)]
iden = Permutation([0, 1, 2, 3])
for i in range(5):
for j in range(i + 1, 5):
assert a[i].commutes_with(a[j]) == \
(rmul(a[i], a[j]) == rmul(a[j], a[i]))
if a[i].commutes_with(a[j]):
assert a[i].commutator(a[j]) == iden
assert a[j].commutator(a[i]) == iden
a = Permutation(3)
b = Permutation(0, 6, 3)(1, 2)
assert a.cycle_structure == {1: 4}
assert b.cycle_structure == {2: 1, 3: 1, 1: 2}
def test_Permutation():
# don't auto fill 0
pytest.raises(ValueError, lambda: Permutation([1]))
p = Permutation([0, 1, 2, 3])
# call as bijective
assert [p(i) for i in range(p.size)] == list(p)
# call as operator
assert p(list(range(p.size))) == list(p)
# call as function
assert list(p(1, 2)) == [0, 2, 1, 3]
# conversion to list
assert list(p) == list(range(4))
assert Permutation(size=4) == Permutation(3)
assert Permutation(Permutation(3), size=5) == Permutation(4)
# cycle form with size
assert Permutation([[1, 2]], size=4) == Permutation([[1, 2], [0], [3]])
# random generation
assert Permutation.random(2) in (Permutation([1, 0]), Permutation([0, 1]))
p = Permutation([2, 5, 1, 6, 3, 0, 4])
q = Permutation([[1], [0, 3, 5, 6, 2, 4]])
assert len({p, p}) == 1
r = Permutation([1, 3, 2, 0, 4, 6, 5])
ans = Permutation(_af_rmuln(*[w.array_form for w in (p, q, r)])).array_form
assert rmul(p, q, r).array_form == ans
# make sure no other permutation of p, q, r could have given
# that answer
for a, b, c in permutations((p, q, r)):
if (a, b, c) == (p, q, r):
continue
assert rmul(a, b, c).array_form != ans
assert p.support() == list(range(7))
assert q.support() == [0, 2, 3, 4, 5, 6]
assert Permutation(p.cyclic_form).array_form == p.array_form
assert p.cardinality == 5040
assert q.cardinality == 5040
assert q.cycles == 2
assert rmul(q, p) == Permutation([4, 6, 1, 2, 5, 3, 0])
assert rmul(p, q) == Permutation([6, 5, 3, 0, 2, 4, 1])
assert _af_rmul(p.array_form, q.array_form) == \
[6, 5, 3, 0, 2, 4, 1]
assert rmul(Permutation([[1, 2, 3], [0, 4]]),
Permutation([[1, 2, 4], [0], [3]])).cyclic_form == \
[[0, 4, 2], [1, 3]]
assert q.array_form == [3, 1, 4, 5, 0, 6, 2]
assert q.cyclic_form == [[0, 3, 5, 6, 2, 4]]
assert q.full_cyclic_form == [[0, 3, 5, 6, 2, 4], [1]]
assert p.cyclic_form == [[0, 2, 1, 5], [3, 6, 4]]
t = p.transpositions()
assert t == [(0, 5), (0, 1), (0, 2), (3, 4), (3, 6)]
assert Permutation.rmul(*[Permutation(Cycle(*ti)) for ti in t])
assert Permutation([1, 0]).transpositions() == [(0, 1)]
assert p**13 == p
assert q**0 == Permutation(list(range(q.size)))
assert q**-2 == ~q**2
assert q**2 == Permutation([5, 1, 0, 6, 3, 2, 4])
assert q**3 == q**2*q
assert q**4 == q**2*q**2
a = Permutation(1, 3)
b = Permutation(2, 0, 3)
I = Permutation(3)
assert ~a == a**-1
assert a*~a == I
assert a*b**-1 == a*~b
ans = Permutation(0, 5, 3, 1, 6)(2, 4)
assert (p + q.rank()).rank() == ans.rank()
assert (p + q.rank())._rank == ans.rank()
assert (q + p.rank()).rank() == ans.rank()
pytest.raises(TypeError, lambda: p + Permutation(list(range(10))))
assert (p - q.rank()).rank() == Permutation(0, 6, 3, 1, 2, 5, 4).rank()
assert p.rank() - q.rank() < 0 # for coverage: make sure mod is used
assert (q - p.rank()).rank() == Permutation(1, 4, 6, 2)(3, 5).rank()
assert p*q == Permutation(_af_rmuln(*[list(w) for w in (q, p)]))
assert p*Permutation([]) == p
assert Permutation([])*p == p
assert p*Permutation([[0, 1]]) == Permutation([2, 5, 0, 6, 3, 1, 4])
assert Permutation([[0, 1]])*p == Permutation([5, 2, 1, 6, 3, 0, 4])
pq = p ^ q
assert pq == Permutation([5, 6, 0, 4, 1, 2, 3])
assert pq == rmul(q, p, ~q)
qp = q ^ p
assert qp == Permutation([4, 3, 6, 2, 1, 5, 0])
assert qp == rmul(p, q, ~p)
pytest.raises(ValueError, lambda: p ^ Permutation([]))
assert p.commutator(q) == Permutation(0, 1, 3, 4, 6, 5, 2)
assert q.commutator(p) == Permutation(0, 2, 5, 6, 4, 3, 1)
assert p.commutator(q) == ~q.commutator(p)
pytest.raises(ValueError, lambda: p.commutator(Permutation([])))
assert len(p.atoms()) == 7
assert q.atoms() == {0, 1, 2, 3, 4, 5, 6}
assert p.inversion_vector() == [2, 4, 1, 3, 1, 0]
assert q.inversion_vector() == [3, 1, 2, 2, 0, 1]
assert Permutation.from_inversion_vector(p.inversion_vector()) == p
assert Permutation.from_inversion_vector(q.inversion_vector()).array_form\
== q.array_form
pytest.raises(ValueError, lambda: Permutation.from_inversion_vector([0, 2]))
assert Permutation([i for i in range(500, -1, -1)]).inversions() == 125250
s = Permutation([0, 4, 1, 3, 2])
assert s.parity() == 0
s.cyclic_form # needed to create a value for _cyclic_form
assert len(s._cyclic_form) != s.size and s.parity() == 0
assert not s.is_odd
assert s.is_even
assert Permutation([0, 1, 4, 3, 2]).parity() == 1
assert _af_parity([0, 4, 1, 3, 2]) == 0
assert _af_parity([0, 1, 4, 3, 2]) == 1
s = Permutation([0])
assert s.is_Singleton
assert Permutation([]).is_Empty
r = Permutation([3, 2, 1, 0])
assert (r**2).is_Identity
assert rmul(~p, p).is_Identity
assert (~p)**13 == Permutation([5, 2, 0, 4, 6, 1, 3])
assert ~(r**2).is_Identity
assert p.max() == 6
assert p.min() == 0
q = Permutation([[6], [5], [0, 1, 2, 3, 4]])
assert q.max() == 4
assert q.min() == 0
p = Permutation([1, 5, 2, 0, 3, 6, 4])
q = Permutation([[1, 2, 3, 5, 6], [0, 4]])
assert p.ascents() == [0, 3, 4]
assert q.ascents() == [1, 2, 4]
assert r.ascents() == []
assert p.descents() == [1, 2, 5]
assert q.descents() == [0, 3, 5]
assert Permutation(r.descents()).is_Identity
assert p.inversions() == 7
# test the merge-sort with a longer permutation
big = list(p) + list(range(p.max() + 1, p.max() + 130))
assert Permutation(big).inversions() == 7
assert p.signature() == -1
assert q.inversions() == 11
assert q.signature() == -1
assert rmul(p, ~p).inversions() == 0
assert rmul(p, ~p).signature() == 1
assert p.order() == 6
assert q.order() == 10
assert (p**(p.order())).is_Identity
assert p.length() == 6
assert q.length() == 7
assert r.length() == 4
assert p.runs() == [[1, 5], [2], [0, 3, 6], [4]]
assert q.runs() == [[4], [2, 3, 5], [0, 6], [1]]
assert r.runs() == [[3], [2], [1], [0]]
assert p.index() == 8
assert q.index() == 8
assert r.index() == 3
assert p.get_precedence_distance(q) == q.get_precedence_distance(p)
assert p.get_adjacency_distance(q) == p.get_adjacency_distance(q)
assert p.get_positional_distance(q) == p.get_positional_distance(q)
p = Permutation([0, 1, 2, 3])
q = Permutation([3, 2, 1, 0])
assert p.get_precedence_distance(q) == 6
assert p.get_adjacency_distance(q) == 3
assert p.get_positional_distance(q) == 8
p = Permutation([0, 3, 1, 2, 4])
q = Permutation.josephus(4, 5, 2)
assert p.get_adjacency_distance(q) == 3
pytest.raises(ValueError, lambda: p.get_adjacency_distance(Permutation([])))
pytest.raises(ValueError, lambda: p.get_positional_distance(Permutation([])))
pytest.raises(ValueError, lambda: p.get_precedence_distance(Permutation([])))
a = [Permutation.unrank_nonlex(4, i) for i in range(5)]
iden = Permutation([0, 1, 2, 3])
for i in range(5):
for j in range(i + 1, 5):
assert a[i].commutes_with(a[j]) == \
(rmul(a[i], a[j]) == rmul(a[j], a[i]))
if a[i].commutes_with(a[j]):
assert a[i].commutator(a[j]) == iden
assert a[j].commutator(a[i]) == iden
a = Permutation(3)
b = Permutation(0, 6, 3)(1, 2)
assert a.cycle_structure == {1: 4}
assert b.cycle_structure == {2: 1, 3: 1, 1: 2}
def _strip(g, base, orbits, transversals):
"""
Attempt to decompose a permutation using a (possibly partial) BSGS
structure.
This is done by treating the sequence ``base`` as an actual base, and
the orbits ``orbits`` and transversals ``transversals`` as basic orbits and
transversals relative to it.
This process is called "sifting". A sift is unsuccessful when a certain
orbit element is not found or when after the sift the decomposition
doesn't end with the identity element.
The argument ``transversals`` is a list of dictionaries that provides
transversal elements for the orbits ``orbits``.
Parameters
==========
``g`` - permutation to be decomposed
``base`` - sequence of points
``orbits`` - a list in which the ``i``-th entry is an orbit of ``base[i]``
under some subgroup of the pointwise stabilizer of `
`base[0], base[1], ..., base[i - 1]``. The groups themselves are implicit
in this function since the only information we need is encoded in the orbits
and transversals
``transversals`` - a list of orbit transversals associated with the orbits
``orbits``.
Examples
========
>>> from diofant.combinatorics import Permutation
>>> Permutation.print_cyclic = True
>>> from diofant.combinatorics.named_groups import SymmetricGroup
>>> from diofant.combinatorics.permutations import Permutation
>>> from diofant.combinatorics.util import _strip
>>> S = SymmetricGroup(5)
>>> S.schreier_sims()
>>> g = Permutation([0, 2, 3, 1, 4])
>>> _strip(g, S.base, S.basic_orbits, S.basic_transversals)
(Permutation(4), 5)
Notes
=====
The algorithm is described in [1],pp.89-90. The reason for returning
both the current state of the element being decomposed and the level
at which the sifting ends is that they provide important information for
the randomized version of the Schreier-Sims algorithm.
References
==========
[1] Holt, D., Eick, B., O'Brien, E.
"Handbook of computational group theory"
See Also
========
diofant.combinatorics.perm_groups.PermutationGroup.schreier_sims
diofant.combinatorics.perm_groups.PermutationGroup.schreier_sims_random
"""
h = g._array_form
base_len = len(base)
for i in range(base_len):
beta = h[base[i]]
if beta == base[i]:
continue
if beta not in orbits[i]:
return _af_new(h), i + 1
u = transversals[i][beta]._array_form
h = _af_rmul(_af_invert(u), h)
return _af_new(h), base_len + 1
def canonicalize_naive(g, dummies, sym, *v):
"""
Canonicalize tensor formed by tensors of the different types
g permutation representing the tensor
dummies list of dummy indices
msym symmetry of the metric
v is a list of (base_i, gens_i, n_i, sym_i) for tensors of type `i`
base_i, gens_i BSGS for tensors of this type
n_i number ot tensors of type `i`
sym_i symmetry under exchange of two component tensors of type `i`
None no symmetry
0 commuting
1 anticommuting
Return 0 if the tensor is zero, else return the array form of
the permutation representing the canonical form of the tensor.
Examples
========
>>> from diofant.combinatorics.testutil import canonicalize_naive
>>> from diofant.combinatorics.tensor_can import get_symmetric_group_sgs
>>> from diofant.combinatorics import Permutation, PermutationGroup
>>> g = Permutation([1, 3, 2, 0, 4, 5])
>>> base2, gens2 = get_symmetric_group_sgs(2)
>>> canonicalize_naive(g, [2, 3], 0, (base2, gens2, 2, 0))
[0, 2, 1, 3, 4, 5]
"""
from diofant.combinatorics.perm_groups import PermutationGroup
from diofant.combinatorics.tensor_can import gens_products, dummy_sgs
from diofant.combinatorics.permutations import Permutation, _af_rmul
v1 = []
for i in range(len(v)):
base_i, gens_i, n_i, sym_i = v[i]
v1.append((base_i, gens_i, [[]]*n_i, sym_i))
size, sbase, sgens = gens_products(*v1)
dgens = dummy_sgs(dummies, sym, size-2)
if isinstance(sym, int):
num_types = 1
dummies = [dummies]
sym = [sym]
else:
num_types = len(sym)
dgens = []
for i in range(num_types):
dgens.extend(dummy_sgs(dummies[i], sym[i], size - 2))
S = PermutationGroup(sgens)
D = PermutationGroup([Permutation(x) for x in dgens])
dlist = list(D.generate(af=True))
g = g.array_form
st = set()
for s in S.generate(af=True):
h = _af_rmul(g, s)
for d in dlist:
q = tuple(_af_rmul(d, h))
st.add(q)
a = list(st)
a.sort()
prev = (0,)*size
for h in a:
if h[:-2] == prev[:-2]:
if h[-1] != prev[-1]:
return 0
prev = h
return list(a[0])