def __new__(cls, *args):
from diofant.matrices.immutable import ImmutableMatrix
args = map(sympify, args)
mat = ImmutableMatrix(*args)
obj = Basic.__new__(cls, mat)
return obj
def __new__(cls, mat):
mat = _sympify(mat)
if not mat.is_Matrix:
raise TypeError("mat should be a matrix")
if not mat.is_square:
raise ShapeError("Inverse of non-square matrix %s" % mat)
return Basic.__new__(cls, mat)
def __new__(cls, subset, superset):
"""
Default constructor.
It takes the subset and its superset as its parameters.
Examples
========
>>> from diofant.combinatorics.subsets import Subset
>>> a = Subset(['c', 'd'], ['a', 'b', 'c', 'd'])
>>> a.subset
['c', 'd']
>>> a.superset
['a', 'b', 'c', 'd']
>>> a.size
2
"""
if len(subset) > len(superset):
raise ValueError('Invalid arguments have been provided. The superset must be larger than the subset.')
for elem in subset:
if elem not in superset:
raise ValueError('The superset provided is invalid as it does not contain the element %i' % elem)
obj = Basic.__new__(cls)
obj._subset = subset
obj._superset = superset
return obj
def __new__(cls, *args, **options):
# (Try to) sympify args first
newargs = []
for ec in args:
# ec could be a ExprCondPair or a tuple
pair = ExprCondPair(*getattr(ec, 'args', ec))
cond = pair.cond
if cond == false:
continue
if not isinstance(cond, (bool, Relational, Boolean)):
raise TypeError(
"Cond %s is of type %s, but must be a Relational,"
" Boolean, or a built-in bool." % (cond, type(cond)))
newargs.append(pair)
if cond == S.true:
break
if options.pop('evaluate', True):
r = cls.eval(*newargs)
else:
r = None
if r is None:
return Basic.__new__(cls, *newargs, **options)
else:
return r
def _new(cls, *args, **kwargs):
if len(args) == 1 and isinstance(args[0], ImmutableMatrix):
return args[0]
rows, cols, flat_list = cls._handle_creation_inputs(*args, **kwargs)
rows = Integer(rows)
cols = Integer(cols)
mat = Tuple(*flat_list)
return Basic.__new__(cls, rows, cols, mat)
def __new__(cls, *args, **kwargs):
args = list(map(sympify, args))
check = kwargs.get('check', True)
obj = Basic.__new__(cls, *args)
if check:
validate(*args)
return obj
def __new__(cls, partition, integer=None):
"""
Generates a new IntegerPartition object from a list or dictionary.
The partition can be given as a list of positive integers or a
dictionary of (integer, multiplicity) items. If the partition is
preceeded by an integer an error will be raised if the partition
does not sum to that given integer.
Examples
========
>>> from diofant.combinatorics.partitions import IntegerPartition
>>> a = IntegerPartition([5, 4, 3, 1, 1])
>>> a
IntegerPartition(14, (5, 4, 3, 1, 1))
>>> print(a)
[5, 4, 3, 1, 1]
>>> IntegerPartition({1:3, 2:1})
IntegerPartition(5, (2, 1, 1, 1))
If the value that the partion should sum to is given first, a check
will be made to see n error will be raised if there is a discrepancy:
>>> IntegerPartition(10, [5, 4, 3, 1])
Traceback (most recent call last):
...
ValueError: The partition is not valid
"""
if integer is not None:
integer, partition = partition, integer
if isinstance(partition, (dict, Dict)):
_ = []
for k, v in sorted(partition.items(), reverse=True):
if not v:
continue
k, v = as_int(k), as_int(v)
_.extend([k]*v)
partition = tuple(_)
else:
partition = tuple(sorted(map(as_int, partition), reverse=True))
sum_ok = False
if integer is None:
integer = sum(partition)
sum_ok = True
else:
integer = as_int(integer)
if not sum_ok and sum(partition) != integer:
raise ValueError("Partition did not add to %s" % integer)
if any(i < 1 for i in partition):
raise ValueError("The summands must all be positive.")
obj = Basic.__new__(cls, integer, partition)
obj.partition = list(partition)
obj.integer = integer
return obj
def __new__(cls, *args, **kwargs):
check = kwargs.get('check', True)
args = list(map(sympify, args))
obj = Basic.__new__(cls, *args)
factor, matrices = obj.as_coeff_matrices()
if check:
validate(*matrices)
return obj
def _new(cls, *args, **kwargs):
s = MutableSparseMatrix(*args)
rows = Integer(s.rows)
cols = Integer(s.cols)
mat = Dict(s._smat)
obj = Basic.__new__(cls, rows, cols, mat)
obj.rows = s.rows
obj.cols = s.cols
obj._smat = s._smat
return obj
def __new__(cls, *args, **kw_args):
"""The constructor for the Prufer object.
Examples
========
>>> from diofant.combinatorics.prufer import Prufer
A Prufer object can be constructed from a list of edges:
>>> a = Prufer([[0, 1], [0, 2], [0, 3]])
>>> a.prufer_repr
[0, 0]
If the number of nodes is given, no checking of the nodes will
be performed; it will be assumed that nodes 0 through n - 1 are
present:
>>> Prufer([[0, 1], [0, 2], [0, 3]], 4)
Prufer([[0, 1], [0, 2], [0, 3]], 4)
A Prufer object can be constructed from a Prufer sequence:
>>> b = Prufer([1, 3])
>>> b.tree_repr
[[0, 1], [1, 3], [2, 3]]
"""
ret_obj = Basic.__new__(cls, *args, **kw_args)
args = [list(args[0])]
if args[0] and iterable(args[0][0]):
if not args[0][0]:
raise ValueError(
'Prufer expects at least one edge in the tree.')
if len(args) > 1:
nnodes = args[1]
else:
nodes = set(flatten(args[0]))
nnodes = max(nodes) + 1
if nnodes != len(nodes):
missing = set(range(nnodes)) - nodes
if len(missing) == 1:
msg = 'Node %s is missing.' % missing.pop()
else:
msg = 'Nodes %s are missing.' % list(sorted(missing))
raise ValueError(msg)
ret_obj._tree_repr = [list(i) for i in args[0]]
ret_obj._nodes = nnodes
else:
ret_obj._prufer_repr = args[0]
ret_obj._nodes = len(ret_obj._prufer_repr) + 2
return ret_obj
def __new__(cls, n, *args, **kw_args):
"""
Default constructor.
It takes a single argument ``n`` which gives the dimension of the Gray
code. The starting Gray code string (``start``) or the starting ``rank``
may also be given; the default is to start at rank = 0 ('0...0').
Examples
========
>>> from diofant.combinatorics.graycode import GrayCode
>>> a = GrayCode(3)
>>> a
GrayCode(3)
>>> a.n
3
>>> a = GrayCode(3, start='100')
>>> a.current
'100'
>>> a = GrayCode(4, rank=4)
>>> a.current
'0110'
>>> a.rank
4
"""
if n < 1 or int(n) != n:
raise ValueError(
'Gray code dimension must be a positive integer, not %i' % n)
n = int(n)
args = (n, ) + args
obj = Basic.__new__(cls, *args)
if 'start' in kw_args:
obj._current = kw_args["start"]
if len(obj._current) > n:
raise ValueError('Gray code start has length %i but '
'should not be greater than %i' %
(len(obj._current), n))
elif 'rank' in kw_args:
if int(kw_args["rank"]) != kw_args["rank"]:
raise ValueError('Gray code rank must be a positive integer, '
'not %i' % kw_args["rank"])
obj._rank = int(kw_args["rank"]) % obj.selections
obj._current = obj.unrank(n, obj._rank)
return obj
def __new__(cls, *mats):
return Basic.__new__(BlockDiagMatrix, *mats)
def __new__(cls, corners, faces=[], pgroup=[]):
"""
The constructor of the Polyhedron group object.
It takes up to three parameters: the corners, faces, and
allowed transformations.
The corners/vertices are entered as a list of arbitrary
expressions that are used to identify each vertex.
The faces are entered as a list of tuples of indices; a tuple
of indices identifies the vertices which define the face. They
should be entered in a cw or ccw order; they will be standardized
by reversal and rotation to be give the lowest lexical ordering.
If no faces are given then no edges will be computed.
>>> from diofant.combinatorics.polyhedron import Polyhedron
>>> Polyhedron(list('abc'), [(1, 2, 0)]).faces
{(0, 1, 2)}
>>> Polyhedron(list('abc'), [(1, 0, 2)]).faces
{(0, 1, 2)}
The allowed transformations are entered as allowable permutations
of the vertices for the polyhedron. Instance of Permutations
(as with faces) should refer to the supplied vertices by index.
These permutation are stored as a PermutationGroup.
Examples
========
>>> from diofant.combinatorics.permutations import Permutation
>>> Permutation.print_cyclic = False
>>> from diofant.abc import w, x, y, z
Here we construct the Polyhedron object for a tetrahedron.
>>> corners = [w, x, y, z]
>>> faces = [(0, 1, 2), (0, 2, 3), (0, 3, 1), (1, 2, 3)]
Next, allowed transformations of the polyhedron must be given. This
is given as permutations of vertices.
Although the vertices of a tetrahedron can be numbered in 24 (4!)
different ways, there are only 12 different orientations for a
physical tetrahedron. The following permutations, applied once or
twice, will generate all 12 of the orientations. (The identity
permutation, Permutation(range(4)), is not included since it does
not change the orientation of the vertices.)
>>> pgroup = [Permutation([[0, 1, 2], [3]]), \
Permutation([[0, 1, 3], [2]]), \
Permutation([[0, 2, 3], [1]]), \
Permutation([[1, 2, 3], [0]]), \
Permutation([[0, 1], [2, 3]]), \
Permutation([[0, 2], [1, 3]]), \
Permutation([[0, 3], [1, 2]])]
The Polyhedron is now constructed and demonstrated:
>>> tetra = Polyhedron(corners, faces, pgroup)
>>> tetra.size
4
>>> tetra.edges
{(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)}
>>> tetra.corners
(w, x, y, z)
It can be rotated with an arbitrary permutation of vertices, e.g.
the following permutation is not in the pgroup:
>>> tetra.rotate(Permutation([0, 1, 3, 2]))
>>> tetra.corners
(w, x, z, y)
An allowed permutation of the vertices can be constructed by
repeatedly applying permutations from the pgroup to the vertices.
Here is a demonstration that applying p and p**2 for every p in
pgroup generates all the orientations of a tetrahedron and no others:
>>> all = ( (w, x, y, z), \
(x, y, w, z), \
(y, w, x, z), \
(w, z, x, y), \
(z, w, y, x), \
(w, y, z, x), \
(y, z, w, x), \
(x, z, y, w), \
(z, y, x, w), \
(y, x, z, w), \
(x, w, z, y), \
(z, x, w, y) )
>>> got = []
>>> for p in (pgroup + [p**2 for p in pgroup]):
... h = Polyhedron(corners)
... h.rotate(p)
... got.append(h.corners)
...
>>> set(got) == set(all)
True
The make_perm method of a PermutationGroup will randomly pick
permutations, multiply them together, and return the permutation that
can be applied to the polyhedron to give the orientation produced
by those individual permutations.
Here, 3 permutations are used:
>>> tetra.pgroup.make_perm(3) # doctest: +SKIP
Permutation([0, 3, 1, 2])
To select the permutations that should be used, supply a list
of indices to the permutations in pgroup in the order they should
be applied:
>>> use = [0, 0, 2]
>>> p002 = tetra.pgroup.make_perm(3, use)
>>> p002
Permutation([1, 0, 3, 2])
Apply them one at a time:
>>> tetra.reset()
>>> for i in use:
... tetra.rotate(pgroup[i])
...
>>> tetra.vertices
(x, w, z, y)
>>> sequentially = tetra.vertices
Apply the composite permutation:
>>> tetra.reset()
>>> tetra.rotate(p002)
>>> tetra.corners
(x, w, z, y)
>>> tetra.corners in all and tetra.corners == sequentially
True
Notes
=====
Defining permutation groups
---------------------------
It is not necessary to enter any permutations, nor is necessary to
enter a complete set of transformations. In fact, for a polyhedron,
all configurations can be constructed from just two permutations.
For example, the orientations of a tetrahedron can be generated from
an axis passing through a vertex and face and another axis passing
through a different vertex or from an axis passing through the
midpoints of two edges opposite of each other.
For simplicity of presentation, consider a square --
not a cube -- with vertices 1, 2, 3, and 4:
1-----2 We could think of axes of rotation being:
| | 1) through the face
| | 2) from midpoint 1-2 to 3-4 or 1-3 to 2-4
3-----4 3) lines 1-4 or 2-3
To determine how to write the permutations, imagine 4 cameras,
one at each corner, labeled A-D:
A B A B
1-----2 1-----3 vertex index:
| | | | 1 0
| | | | 2 1
3-----4 2-----4 3 2
C D C D 4 3
original after rotation
along 1-4
A diagonal and a face axis will be chosen for the "permutation group"
from which any orientation can be constructed.
>>> pgroup = []
Imagine a clockwise rotation when viewing 1-4 from camera A. The new
orientation is (in camera-order): 1, 3, 2, 4 so the permutation is
given using the *indices* of the vertices as:
>>> pgroup.append(Permutation((0, 2, 1, 3)))
Now imagine rotating clockwise when looking down an axis entering the
center of the square as viewed. The new camera-order would be
3, 1, 4, 2 so the permutation is (using indices):
>>> pgroup.append(Permutation((2, 0, 3, 1)))
The square can now be constructed:
** use real-world labels for the vertices, entering them in
camera order
** for the faces we use zero-based indices of the vertices
in *edge-order* as the face is traversed; neither the
direction nor the starting point matter -- the faces are
only used to define edges (if so desired).
>>> square = Polyhedron((1, 2, 3, 4), [(0, 1, 3, 2)], pgroup)
To rotate the square with a single permutation we can do:
>>> square.rotate(square.pgroup[0])
>>> square.corners
(1, 3, 2, 4)
To use more than one permutation (or to use one permutation more
than once) it is more convenient to use the make_perm method:
>>> p011 = square.pgroup.make_perm([0, 1, 1]) # diag flip + 2 rotations
>>> square.reset() # return to initial orientation
>>> square.rotate(p011)
>>> square.corners
(4, 2, 3, 1)
Thinking outside the box
------------------------
Although the Polyhedron object has a direct physical meaning, it
actually has broader application. In the most general sense it is
just a decorated PermutationGroup, allowing one to connect the
permutations to something physical. For example, a Rubik's cube is
not a proper polyhedron, but the Polyhedron class can be used to
represent it in a way that helps to visualize the Rubik's cube.
>>> from diofant.utilities.iterables import flatten, unflatten
>>> from diofant import symbols, sstr
>>> from diofant.combinatorics import RubikGroup
>>> facelets = flatten([symbols(s+'1:5') for s in 'UFRBLD'])
>>> def show():
... pairs = unflatten(r2.corners, 2)
... print(sstr(pairs[::2]))
... print(sstr(pairs[1::2]))
...
>>> r2 = Polyhedron(facelets, pgroup=RubikGroup(2))
>>> show()
[(U1, U2), (F1, F2), (R1, R2), (B1, B2), (L1, L2), (D1, D2)]
[(U3, U4), (F3, F4), (R3, R4), (B3, B4), (L3, L4), (D3, D4)]
>>> r2.rotate(0) # cw rotation of F
>>> show()
[(U1, U2), (F3, F1), (U3, R2), (B1, B2), (L1, D1), (R3, R1)]
[(L4, L2), (F4, F2), (U4, R4), (B3, B4), (L3, D2), (D3, D4)]
Predefined Polyhedra
====================
For convenience, the vertices and faces are defined for the following
standard solids along with a permutation group for transformations.
When the polyhedron is oriented as indicated below, the vertices in
a given horizontal plane are numbered in ccw direction, starting from
the vertex that will give the lowest indices in a given face. (In the
net of the vertices, indices preceded by "-" indicate replication of
the lhs index in the net.)
tetrahedron, tetrahedron_faces
------------------------------
4 vertices (vertex up) net:
0 0-0
1 2 3-1
4 faces:
(0,1,2) (0,2,3) (0,3,1) (1,2,3)
cube, cube_faces
----------------
8 vertices (face up) net:
0 1 2 3-0
4 5 6 7-4
6 faces:
(0,1,2,3)
(0,1,5,4) (1,2,6,5) (2,3,7,6) (0,3,7,4)
(4,5,6,7)
octahedron, octahedron_faces
----------------------------
6 vertices (vertex up) net:
0 0 0-0
1 2 3 4-1
5 5 5-5
8 faces:
(0,1,2) (0,2,3) (0,3,4) (0,1,4)
(1,2,5) (2,3,5) (3,4,5) (1,4,5)
dodecahedron, dodecahedron_faces
--------------------------------
20 vertices (vertex up) net:
0 1 2 3 4 -0
5 6 7 8 9 -5
14 10 11 12 13-14
15 16 17 18 19-15
12 faces:
(0,1,2,3,4)
(0,1,6,10,5) (1,2,7,11,6) (2,3,8,12,7) (3,4,9,13,8) (0,4,9,14,5)
(5,10,16,15,14) (
6,10,16,17,11) (7,11,17,18,12) (8,12,18,19,13) (9,13,19,15,14)
(15,16,17,18,19)
icosahedron, icosahedron_faces
------------------------------
12 vertices (face up) net:
0 0 0 0 -0
1 2 3 4 5 -1
6 7 8 9 10 -6
11 11 11 11 -11
20 faces:
(0,1,2) (0,2,3) (0,3,4) (0,4,5) (0,1,5)
(1,2,6) (2,3,7) (3,4,8) (4,5,9) (1,5,10)
(2,6,7) (3,7,8) (4,8,9) (5,9,10) (1,6,10)
(6,7,11,) (7,8,11) (8,9,11) (9,10,11) (6,10,11)
>>> from diofant.combinatorics.polyhedron import cube
>>> cube.edges
{(0, 1), (0, 3), (0, 4), ..., (4, 7), (5, 6), (6, 7)}
If you want to use letters or other names for the corners you
can still use the pre-calculated faces:
>>> corners = list('abcdefgh')
>>> Polyhedron(corners, cube.faces).corners
(a, b, c, d, e, f, g, h)
References
==========
[1] www.ocf.berkeley.edu/~wwu/articles/platonicsolids.pdf
"""
faces = [minlex(f, directed=False, is_set=True) for f in faces]
corners, faces, pgroup = args = \
[Tuple(*a) for a in (corners, faces, pgroup)]
obj = Basic.__new__(cls, *args)
obj._corners = tuple(corners) # in order given
obj._faces = FiniteSet(*faces)
if pgroup and pgroup[0].size != len(corners):
raise ValueError("Permutation size unequal to number of corners.")
# use the identity permutation if none are given
obj._pgroup = PermutationGroup((
pgroup or [Perm(range(len(corners)))] ))
return obj
def __new__(cls, name, antisym, **kwargs):
obj = Basic.__new__(cls, name, antisym, **kwargs)
obj.name = name
obj.antisym = antisym
return obj
def __new__(cls, *args, **kwargs):
args = map(sympify, args)
return Basic.__new__(cls, *args, **kwargs)
def __new__(cls, name, n, m):
n, m = sympify(n), sympify(m)
obj = Basic.__new__(cls, name, n, m)
return obj