예제 #1
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    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
예제 #2
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 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)
예제 #3
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    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
예제 #4
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    def __new__(cls, *args):
        from diofant.matrices.immutable import ImmutableMatrix
        args = map(sympify, args)
        mat = ImmutableMatrix(*args)

        obj = Basic.__new__(cls, mat)
        return obj
예제 #5
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    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
예제 #6
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 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)
예제 #7
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    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
예제 #8
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    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
예제 #9
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 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
예제 #10
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    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
예제 #11
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def test_sympyissue_6100():
    assert x**1.0 != x
    assert x != x**1.0
    assert true != x**1.0
    assert x**1.0 is not True
    assert x is not True
    assert x*y != (x*y)**1.0
    assert (x**1.0)**1.0 != x
    assert (x**1.0)**2.0 == x**2
    b = Basic()
    assert Pow(b, 1.0, evaluate=False) != b
    # if the following gets distributed as a Mul (x**1.0*y**1.0 then
    # __eq__ methods could be added to Symbol and Pow to detect the
    # power-of-1.0 case.
    assert isinstance((x*y)**1.0, Pow)
예제 #12
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def test_Idx_construction():
    i, a, b = symbols('i a b', integer=True)
    assert Idx(i) != Idx(i, 1)
    assert Idx(i, a) == Idx(i, (0, a - 1))
    assert Idx(i, oo) == Idx(i, (0, oo))

    pytest.raises(TypeError, lambda: Idx(x))
    pytest.raises(TypeError, lambda: Idx(0.5))
    pytest.raises(TypeError, lambda: Idx(i, x))
    pytest.raises(TypeError, lambda: Idx(i, 0.5))
    pytest.raises(TypeError, lambda: Idx(i, (x, 5)))
    pytest.raises(TypeError, lambda: Idx(i, (2, x)))
    pytest.raises(TypeError, lambda: Idx(i, (2, 3.5)))
    pytest.raises(ValueError, lambda: Idx(i, (1, 2, 3)))
    pytest.raises(TypeError, lambda: Idx(i, Basic()))
예제 #13
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    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
예제 #14
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def test_sympyissue_6100():
    x = Symbol('x')
    y = Symbol('y')
    assert x**1.0 == x
    assert x == x**1.0
    assert S.true != x**1.0
    assert x**1.0 is not True
    assert x is not True
    assert x*y == (x*y)**1.0
    assert (x**1.0)**1.0 == x
    assert (x**1.0)**2.0 == x**2
    b = Basic()
    assert Pow(b, 1.0, evaluate=False) == b
    # if the following gets distributed as a Mul (x**1.0*y**1.0 then
    # __eq__ methods could be added to Symbol and Pow to detect the
    # power-of-1.0 case.
    assert ((x*y)**1.0).func is Pow
예제 #15
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def test_sort():
    assert sort(str)(Basic(3, 1, 2)) == Basic(1, 2, 3)
예제 #16
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 def __new__(cls, *mats):
     return Basic.__new__(BlockDiagMatrix, *mats)
예제 #17
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    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
예제 #18
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 def __new__(cls, *args, **kwargs):
     args = map(sympify, args)
     return Basic.__new__(cls, *args, **kwargs)
예제 #19
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 def __new__(cls, name, antisym, **kwargs):
     obj = Basic.__new__(cls, name, antisym, **kwargs)
     obj.name = name
     obj.antisym = antisym
     return obj
예제 #20
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def test_flatten():
    assert flatten(Basic(1, 2, Basic(3, 4))) == Basic(1, 2, 3, 4)
예제 #21
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def test_rm_id():
    rmzeros = rm_id(lambda x: x == 0)
    assert rmzeros(Basic(0, 1)) == Basic(1)
    assert rmzeros(Basic(0, 0)) == Basic(0)
    assert rmzeros(Basic(2, 1)) == Basic(2, 1)
예제 #22
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 def __new__(cls, name, n, m):
     n, m = sympify(n), sympify(m)
     obj = Basic.__new__(cls, name, n, m)
     return obj
예제 #23
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def test_unpack():
    assert unpack(Basic(2)) == 2
    assert unpack(Basic(2, 3)) == Basic(2, 3)