from collections import Counter, defaultdict, OrderedDict from itertools import ( chain, combinations, combinations_with_replacement, cycle, islice, permutations, product, groupby ) # For backwards compatibility from itertools import product as cartes # noqa: F401 from operator import gt # this is the logical location of these functions from sympy.utilities.enumerative import ( multiset_partitions_taocp, list_visitor, MultisetPartitionTraverser) from sympy.utilities.misc import as_int from sympy.utilities.decorator import deprecated def is_palindromic(s, i=0, j=None): """ Return True if the sequence is the same from left to right as it is from right to left in the whole sequence (default) or in the Python slice ``s[i: j]``; else False. Examples ======== >>> from sympy.utilities.iterables import is_palindromic >>> is_palindromic([1, 0, 1]) True >>> is_palindromic('abcbb') False >>> is_palindromic('abcbb', 1) False Normal Python slicing is performed in place so there is no need to create a slice of the sequence for testing: >>> is_palindromic('abcbb', 1, -1) True >>> is_palindromic('abcbb', -4, -1) True See Also ======== sympy.ntheory.digits.is_palindromic: tests integers """ i, j, _ = slice(i, j).indices(len(s)) m = (j - i)//2 # if length is odd, middle element will be ignored return all(s[i + k] == s[j - 1 - k] for k in range(m)) def flatten(iterable, levels=None, cls=None): # noqa: F811 """ Recursively denest iterable containers. >>> from sympy import flatten >>> flatten([1, 2, 3]) [1, 2, 3] >>> flatten([1, 2, [3]]) [1, 2, 3] >>> flatten([1, [2, 3], [4, 5]]) [1, 2, 3, 4, 5] >>> flatten([1.0, 2, (1, None)]) [1.0, 2, 1, None] If you want to denest only a specified number of levels of nested containers, then set ``levels`` flag to the desired number of levels:: >>> ls = [[(-2, -1), (1, 2)], [(0, 0)]] >>> flatten(ls, levels=1) [(-2, -1), (1, 2), (0, 0)] If cls argument is specified, it will only flatten instances of that class, for example: >>> from sympy import Basic, S >>> class MyOp(Basic): ... pass ... >>> flatten([MyOp(S(1), MyOp(S(2), S(3)))], cls=MyOp) [1, 2, 3] adapted from https://kogs-www.informatik.uni-hamburg.de/~meine/python_tricks """ from sympy.tensor.array import NDimArray if levels is not None: if not levels: return iterable elif levels > 0: levels -= 1 else: raise ValueError( "expected non-negative number of levels, got %s" % levels) if cls is None: def reducible(x): return is_sequence(x, set) else: def reducible(x): return isinstance(x, cls) result = [] for el in iterable: if reducible(el): if hasattr(el, 'args') and not isinstance(el, NDimArray): el = el.args result.extend(flatten(el, levels=levels, cls=cls)) else: result.append(el) return result def unflatten(iter, n=2): """Group ``iter`` into tuples of length ``n``. Raise an error if the length of ``iter`` is not a multiple of ``n``. """ if n < 1 or len(iter) % n: raise ValueError('iter length is not a multiple of %i' % n) return list(zip(*(iter[i::n] for i in range(n)))) def reshape(seq, how): """Reshape the sequence according to the template in ``how``. Examples ======== >>> from sympy.utilities import reshape >>> seq = list(range(1, 9)) >>> reshape(seq, [4]) # lists of 4 [[1, 2, 3, 4], [5, 6, 7, 8]] >>> reshape(seq, (4,)) # tuples of 4 [(1, 2, 3, 4), (5, 6, 7, 8)] >>> reshape(seq, (2, 2)) # tuples of 4 [(1, 2, 3, 4), (5, 6, 7, 8)] >>> reshape(seq, (2, [2])) # (i, i, [i, i]) [(1, 2, [3, 4]), (5, 6, [7, 8])] >>> reshape(seq, ((2,), [2])) # etc.... [((1, 2), [3, 4]), ((5, 6), [7, 8])] >>> reshape(seq, (1, [2], 1)) [(1, [2, 3], 4), (5, [6, 7], 8)] >>> reshape(tuple(seq), ([[1], 1, (2,)],)) (([[1], 2, (3, 4)],), ([[5], 6, (7, 8)],)) >>> reshape(tuple(seq), ([1], 1, (2,))) (([1], 2, (3, 4)), ([5], 6, (7, 8))) >>> reshape(list(range(12)), [2, [3], {2}, (1, (3,), 1)]) [[0, 1, [2, 3, 4], {5, 6}, (7, (8, 9, 10), 11)]] """ m = sum(flatten(how)) n, rem = divmod(len(seq), m) if m < 0 or rem: raise ValueError('template must sum to positive number ' 'that divides the length of the sequence') i = 0 container = type(how) rv = [None]*n for k in range(len(rv)): _rv = [] for hi in how: if isinstance(hi, int): _rv.extend(seq[i: i + hi]) i += hi else: n = sum(flatten(hi)) hi_type = type(hi) _rv.append(hi_type(reshape(seq[i: i + n], hi)[0])) i += n rv[k] = container(_rv) return type(seq)(rv) def group(seq, multiple=True): """ Splits a sequence into a list of lists of equal, adjacent elements. Examples ======== >>> from sympy import group >>> group([1, 1, 1, 2, 2, 3]) [[1, 1, 1], [2, 2], [3]] >>> group([1, 1, 1, 2, 2, 3], multiple=False) [(1, 3), (2, 2), (3, 1)] >>> group([1, 1, 3, 2, 2, 1], multiple=False) [(1, 2), (3, 1), (2, 2), (1, 1)] See Also ======== multiset """ if multiple: return [(list(g)) for _, g in groupby(seq)] return [(k, len(list(g))) for k, g in groupby(seq)] def _iproduct2(iterable1, iterable2): '''Cartesian product of two possibly infinite iterables''' it1 = iter(iterable1) it2 = iter(iterable2) elems1 = [] elems2 = [] sentinel = object() def append(it, elems): e = next(it, sentinel) if e is not sentinel: elems.append(e) n = 0 append(it1, elems1) append(it2, elems2) while n <= len(elems1) + len(elems2): for m in range(n-len(elems1)+1, len(elems2)): yield (elems1[n-m], elems2[m]) n += 1 append(it1, elems1) append(it2, elems2) def iproduct(*iterables): ''' Cartesian product of iterables. Generator of the Cartesian product of iterables. This is analogous to itertools.product except that it works with infinite iterables and will yield any item from the infinite product eventually. Examples ======== >>> from sympy.utilities.iterables import iproduct >>> sorted(iproduct([1,2], [3,4])) [(1, 3), (1, 4), (2, 3), (2, 4)] With an infinite iterator: >>> from sympy import S >>> (3,) in iproduct(S.Integers) True >>> (3, 4) in iproduct(S.Integers, S.Integers) True .. seealso:: `itertools.product `_ ''' if len(iterables) == 0: yield () return elif len(iterables) == 1: for e in iterables[0]: yield (e,) elif len(iterables) == 2: yield from _iproduct2(*iterables) else: first, others = iterables[0], iterables[1:] for ef, eo in _iproduct2(first, iproduct(*others)): yield (ef,) + eo def multiset(seq): """Return the hashable sequence in multiset form with values being the multiplicity of the item in the sequence. Examples ======== >>> from sympy.utilities.iterables import multiset >>> multiset('mississippi') {'i': 4, 'm': 1, 'p': 2, 's': 4} See Also ======== group """ return dict(Counter(seq).items()) def ibin(n, bits=None, str=False): """Return a list of length ``bits`` corresponding to the binary value of ``n`` with small bits to the right (last). If bits is omitted, the length will be the number required to represent ``n``. If the bits are desired in reversed order, use the ``[::-1]`` slice of the returned list. If a sequence of all bits-length lists starting from ``[0, 0,..., 0]`` through ``[1, 1, ..., 1]`` are desired, pass a non-integer for bits, e.g. ``'all'``. If the bit *string* is desired pass ``str=True``. Examples ======== >>> from sympy.utilities.iterables import ibin >>> ibin(2) [1, 0] >>> ibin(2, 4) [0, 0, 1, 0] If all lists corresponding to 0 to 2**n - 1, pass a non-integer for bits: >>> bits = 2 >>> for i in ibin(2, 'all'): ... print(i) (0, 0) (0, 1) (1, 0) (1, 1) If a bit string is desired of a given length, use str=True: >>> n = 123 >>> bits = 10 >>> ibin(n, bits, str=True) '0001111011' >>> ibin(n, bits, str=True)[::-1] # small bits left '1101111000' >>> list(ibin(3, 'all', str=True)) ['000', '001', '010', '011', '100', '101', '110', '111'] """ if n < 0: raise ValueError("negative numbers are not allowed") n = as_int(n) if bits is None: bits = 0 else: try: bits = as_int(bits) except ValueError: bits = -1 else: if n.bit_length() > bits: raise ValueError( "`bits` must be >= {}".format(n.bit_length())) if not str: if bits >= 0: return [1 if i == "1" else 0 for i in bin(n)[2:].rjust(bits, "0")] else: return variations(range(2), n, repetition=True) else: if bits >= 0: return bin(n)[2:].rjust(bits, "0") else: return (bin(i)[2:].rjust(n, "0") for i in range(2**n)) def variations(seq, n, repetition=False): r"""Returns an iterator over the n-sized variations of ``seq`` (size N). ``repetition`` controls whether items in ``seq`` can appear more than once; Examples ======== ``variations(seq, n)`` will return `\frac{N!}{(N - n)!}` permutations without repetition of ``seq``'s elements: >>> from sympy import variations >>> list(variations([1, 2], 2)) [(1, 2), (2, 1)] ``variations(seq, n, True)`` will return the `N^n` permutations obtained by allowing repetition of elements: >>> list(variations([1, 2], 2, repetition=True)) [(1, 1), (1, 2), (2, 1), (2, 2)] If you ask for more items than are in the set you get the empty set unless you allow repetitions: >>> list(variations([0, 1], 3, repetition=False)) [] >>> list(variations([0, 1], 3, repetition=True))[:4] [(0, 0, 0), (0, 0, 1), (0, 1, 0), (0, 1, 1)] .. seealso:: `itertools.permutations `_, `itertools.product `_ """ if not repetition: seq = tuple(seq) if len(seq) < n: return iter(()) # 0 length iterator return permutations(seq, n) else: if n == 0: return iter(((),)) # yields 1 empty tuple else: return product(seq, repeat=n) def subsets(seq, k=None, repetition=False): r"""Generates all `k`-subsets (combinations) from an `n`-element set, ``seq``. A `k`-subset of an `n`-element set is any subset of length exactly `k`. The number of `k`-subsets of an `n`-element set is given by ``binomial(n, k)``, whereas there are `2^n` subsets all together. If `k` is ``None`` then all `2^n` subsets will be returned from shortest to longest. Examples ======== >>> from sympy import subsets ``subsets(seq, k)`` will return the `\frac{n!}{k!(n - k)!}` `k`-subsets (combinations) without repetition, i.e. once an item has been removed, it can no longer be "taken": >>> list(subsets([1, 2], 2)) [(1, 2)] >>> list(subsets([1, 2])) [(), (1,), (2,), (1, 2)] >>> list(subsets([1, 2, 3], 2)) [(1, 2), (1, 3), (2, 3)] ``subsets(seq, k, repetition=True)`` will return the `\frac{(n - 1 + k)!}{k!(n - 1)!}` combinations *with* repetition: >>> list(subsets([1, 2], 2, repetition=True)) [(1, 1), (1, 2), (2, 2)] If you ask for more items than are in the set you get the empty set unless you allow repetitions: >>> list(subsets([0, 1], 3, repetition=False)) [] >>> list(subsets([0, 1], 3, repetition=True)) [(0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1)] """ if k is None: if not repetition: return chain.from_iterable((combinations(seq, k) for k in range(len(seq) + 1))) else: return chain.from_iterable((combinations_with_replacement(seq, k) for k in range(len(seq) + 1))) else: if not repetition: return combinations(seq, k) else: return combinations_with_replacement(seq, k) def filter_symbols(iterator, exclude): """ Only yield elements from `iterator` that do not occur in `exclude`. Parameters ========== iterator : iterable iterator to take elements from exclude : iterable elements to exclude Returns ======= iterator : iterator filtered iterator """ exclude = set(exclude) for s in iterator: if s not in exclude: yield s def numbered_symbols(prefix='x', cls=None, start=0, exclude=(), *args, **assumptions): """ Generate an infinite stream of Symbols consisting of a prefix and increasing subscripts provided that they do not occur in ``exclude``. Parameters ========== prefix : str, optional The prefix to use. By default, this function will generate symbols of the form "x0", "x1", etc. cls : class, optional The class to use. By default, it uses ``Symbol``, but you can also use ``Wild`` or ``Dummy``. start : int, optional The start number. By default, it is 0. Returns ======= sym : Symbol The subscripted symbols. """ exclude = set(exclude or []) if cls is None: # We can't just make the default cls=Symbol because it isn't # imported yet. from sympy.core import Symbol cls = Symbol while True: name = '%s%s' % (prefix, start) s = cls(name, *args, **assumptions) if s not in exclude: yield s start += 1 def capture(func): """Return the printed output of func(). ``func`` should be a function without arguments that produces output with print statements. >>> from sympy.utilities.iterables import capture >>> from sympy import pprint >>> from sympy.abc import x >>> def foo(): ... print('hello world!') ... >>> 'hello' in capture(foo) # foo, not foo() True >>> capture(lambda: pprint(2/x)) '2\\n-\\nx\\n' """ from io import StringIO import sys stdout = sys.stdout sys.stdout = file = StringIO() try: func() finally: sys.stdout = stdout return file.getvalue() def sift(seq, keyfunc, binary=False): """ Sift the sequence, ``seq`` according to ``keyfunc``. Returns ======= When ``binary`` is ``False`` (default), the output is a dictionary where elements of ``seq`` are stored in a list keyed to the value of keyfunc for that element. If ``binary`` is True then a tuple with lists ``T`` and ``F`` are returned where ``T`` is a list containing elements of seq for which ``keyfunc`` was ``True`` and ``F`` containing those elements for which ``keyfunc`` was ``False``; a ValueError is raised if the ``keyfunc`` is not binary. Examples ======== >>> from sympy.utilities import sift >>> from sympy.abc import x, y >>> from sympy import sqrt, exp, pi, Tuple >>> sift(range(5), lambda x: x % 2) {0: [0, 2, 4], 1: [1, 3]} sift() returns a defaultdict() object, so any key that has no matches will give []. >>> sift([x], lambda x: x.is_commutative) {True: [x]} >>> _[False] [] Sometimes you will not know how many keys you will get: >>> sift([sqrt(x), exp(x), (y**x)**2], ... lambda x: x.as_base_exp()[0]) {E: [exp(x)], x: [sqrt(x)], y: [y**(2*x)]} Sometimes you expect the results to be binary; the results can be unpacked by setting ``binary`` to True: >>> sift(range(4), lambda x: x % 2, binary=True) ([1, 3], [0, 2]) >>> sift(Tuple(1, pi), lambda x: x.is_rational, binary=True) ([1], [pi]) A ValueError is raised if the predicate was not actually binary (which is a good test for the logic where sifting is used and binary results were expected): >>> unknown = exp(1) - pi # the rationality of this is unknown >>> args = Tuple(1, pi, unknown) >>> sift(args, lambda x: x.is_rational, binary=True) Traceback (most recent call last): ... ValueError: keyfunc gave non-binary output The non-binary sifting shows that there were 3 keys generated: >>> set(sift(args, lambda x: x.is_rational).keys()) {None, False, True} If you need to sort the sifted items it might be better to use ``ordered`` which can economically apply multiple sort keys to a sequence while sorting. See Also ======== ordered """ if not binary: m = defaultdict(list) for i in seq: m[keyfunc(i)].append(i) return m sift = F, T = [], [] for i in seq: try: sift[keyfunc(i)].append(i) except (IndexError, TypeError): raise ValueError('keyfunc gave non-binary output') return T, F def take(iter, n): """Return ``n`` items from ``iter`` iterator. """ return [ value for _, value in zip(range(n), iter) ] def dict_merge(*dicts): """Merge dictionaries into a single dictionary. """ merged = {} for dict in dicts: merged.update(dict) return merged def common_prefix(*seqs): """Return the subsequence that is a common start of sequences in ``seqs``. >>> from sympy.utilities.iterables import common_prefix >>> common_prefix(list(range(3))) [0, 1, 2] >>> common_prefix(list(range(3)), list(range(4))) [0, 1, 2] >>> common_prefix([1, 2, 3], [1, 2, 5]) [1, 2] >>> common_prefix([1, 2, 3], [1, 3, 5]) [1] """ if not all(seqs): return [] elif len(seqs) == 1: return seqs[0] i = 0 for i in range(min(len(s) for s in seqs)): if not all(seqs[j][i] == seqs[0][i] for j in range(len(seqs))): break else: i += 1 return seqs[0][:i] def common_suffix(*seqs): """Return the subsequence that is a common ending of sequences in ``seqs``. >>> from sympy.utilities.iterables import common_suffix >>> common_suffix(list(range(3))) [0, 1, 2] >>> common_suffix(list(range(3)), list(range(4))) [] >>> common_suffix([1, 2, 3], [9, 2, 3]) [2, 3] >>> common_suffix([1, 2, 3], [9, 7, 3]) [3] """ if not all(seqs): return [] elif len(seqs) == 1: return seqs[0] i = 0 for i in range(-1, -min(len(s) for s in seqs) - 1, -1): if not all(seqs[j][i] == seqs[0][i] for j in range(len(seqs))): break else: i -= 1 if i == -1: return [] else: return seqs[0][i + 1:] def prefixes(seq): """ Generate all prefixes of a sequence. Examples ======== >>> from sympy.utilities.iterables import prefixes >>> list(prefixes([1,2,3,4])) [[1], [1, 2], [1, 2, 3], [1, 2, 3, 4]] """ n = len(seq) for i in range(n): yield seq[:i + 1] def postfixes(seq): """ Generate all postfixes of a sequence. Examples ======== >>> from sympy.utilities.iterables import postfixes >>> list(postfixes([1,2,3,4])) [[4], [3, 4], [2, 3, 4], [1, 2, 3, 4]] """ n = len(seq) for i in range(n): yield seq[n - i - 1:] def topological_sort(graph, key=None): r""" Topological sort of graph's vertices. Parameters ========== graph : tuple[list, list[tuple[T, T]] A tuple consisting of a list of vertices and a list of edges of a graph to be sorted topologically. key : callable[T] (optional) Ordering key for vertices on the same level. By default the natural (e.g. lexicographic) ordering is used (in this case the base type must implement ordering relations). Examples ======== Consider a graph:: +---+ +---+ +---+ | 7 |\ | 5 | | 3 | +---+ \ +---+ +---+ | _\___/ ____ _/ | | / \___/ \ / | V V V V | +----+ +---+ | | 11 | | 8 | | +----+ +---+ | | | \____ ___/ _ | | \ \ / / \ | V \ V V / V V +---+ \ +---+ | +----+ | 2 | | | 9 | | | 10 | +---+ | +---+ | +----+ \________/ where vertices are integers. This graph can be encoded using elementary Python's data structures as follows:: >>> V = [2, 3, 5, 7, 8, 9, 10, 11] >>> E = [(7, 11), (7, 8), (5, 11), (3, 8), (3, 10), ... (11, 2), (11, 9), (11, 10), (8, 9)] To compute a topological sort for graph ``(V, E)`` issue:: >>> from sympy.utilities.iterables import topological_sort >>> topological_sort((V, E)) [3, 5, 7, 8, 11, 2, 9, 10] If specific tie breaking approach is needed, use ``key`` parameter:: >>> topological_sort((V, E), key=lambda v: -v) [7, 5, 11, 3, 10, 8, 9, 2] Only acyclic graphs can be sorted. If the input graph has a cycle, then ``ValueError`` will be raised:: >>> topological_sort((V, E + [(10, 7)])) Traceback (most recent call last): ... ValueError: cycle detected References ========== .. [1] https://en.wikipedia.org/wiki/Topological_sorting """ V, E = graph L = [] S = set(V) E = list(E) for v, u in E: S.discard(u) if key is None: def key(value): return value S = sorted(S, key=key, reverse=True) while S: node = S.pop() L.append(node) for u, v in list(E): if u == node: E.remove((u, v)) for _u, _v in E: if v == _v: break else: kv = key(v) for i, s in enumerate(S): ks = key(s) if kv > ks: S.insert(i, v) break else: S.append(v) if E: raise ValueError("cycle detected") else: return L def strongly_connected_components(G): r""" Strongly connected components of a directed graph in reverse topological order. Parameters ========== graph : tuple[list, list[tuple[T, T]] A tuple consisting of a list of vertices and a list of edges of a graph whose strongly connected components are to be found. Examples ======== Consider a directed graph (in dot notation):: digraph { A -> B A -> C B -> C C -> B B -> D } .. graphviz:: digraph { A -> B A -> C B -> C C -> B B -> D } where vertices are the letters A, B, C and D. This graph can be encoded using Python's elementary data structures as follows:: >>> V = ['A', 'B', 'C', 'D'] >>> E = [('A', 'B'), ('A', 'C'), ('B', 'C'), ('C', 'B'), ('B', 'D')] The strongly connected components of this graph can be computed as >>> from sympy.utilities.iterables import strongly_connected_components >>> strongly_connected_components((V, E)) [['D'], ['B', 'C'], ['A']] This also gives the components in reverse topological order. Since the subgraph containing B and C has a cycle they must be together in a strongly connected component. A and D are connected to the rest of the graph but not in a cyclic manner so they appear as their own strongly connected components. Notes ===== The vertices of the graph must be hashable for the data structures used. If the vertices are unhashable replace them with integer indices. This function uses Tarjan's algorithm to compute the strongly connected components in `O(|V|+|E|)` (linear) time. References ========== .. [1] https://en.wikipedia.org/wiki/Strongly_connected_component .. [2] https://en.wikipedia.org/wiki/Tarjan%27s_strongly_connected_components_algorithm See Also ======== sympy.utilities.iterables.connected_components """ # Map from a vertex to its neighbours V, E = G Gmap = {vi: [] for vi in V} for v1, v2 in E: Gmap[v1].append(v2) return _strongly_connected_components(V, Gmap) def _strongly_connected_components(V, Gmap): """More efficient internal routine for strongly_connected_components""" # # Here V is an iterable of vertices and Gmap is a dict mapping each vertex # to a list of neighbours e.g.: # # V = [0, 1, 2, 3] # Gmap = {0: [2, 3], 1: [0]} # # For a large graph these data structures can often be created more # efficiently then those expected by strongly_connected_components() which # in this case would be # # V = [0, 1, 2, 3] # Gmap = [(0, 2), (0, 3), (1, 0)] # # XXX: Maybe this should be the recommended function to use instead... # # Non-recursive Tarjan's algorithm: lowlink = {} indices = {} stack = OrderedDict() callstack = [] components = [] nomore = object() def start(v): index = len(stack) indices[v] = lowlink[v] = index stack[v] = None callstack.append((v, iter(Gmap[v]))) def finish(v1): # Finished a component? if lowlink[v1] == indices[v1]: component = [stack.popitem()[0]] while component[-1] is not v1: component.append(stack.popitem()[0]) components.append(component[::-1]) v2, _ = callstack.pop() if callstack: v1, _ = callstack[-1] lowlink[v1] = min(lowlink[v1], lowlink[v2]) for v in V: if v in indices: continue start(v) while callstack: v1, it1 = callstack[-1] v2 = next(it1, nomore) # Finished children of v1? if v2 is nomore: finish(v1) # Recurse on v2 elif v2 not in indices: start(v2) elif v2 in stack: lowlink[v1] = min(lowlink[v1], indices[v2]) # Reverse topological sort order: return components def connected_components(G): r""" Connected components of an undirected graph or weakly connected components of a directed graph. Parameters ========== graph : tuple[list, list[tuple[T, T]] A tuple consisting of a list of vertices and a list of edges of a graph whose connected components are to be found. Examples ======== Given an undirected graph:: graph { A -- B C -- D } .. graphviz:: graph { A -- B C -- D } We can find the connected components using this function if we include each edge in both directions:: >>> from sympy.utilities.iterables import connected_components >>> V = ['A', 'B', 'C', 'D'] >>> E = [('A', 'B'), ('B', 'A'), ('C', 'D'), ('D', 'C')] >>> connected_components((V, E)) [['A', 'B'], ['C', 'D']] The weakly connected components of a directed graph can found the same way. Notes ===== The vertices of the graph must be hashable for the data structures used. If the vertices are unhashable replace them with integer indices. This function uses Tarjan's algorithm to compute the connected components in `O(|V|+|E|)` (linear) time. References ========== .. [1] https://en.wikipedia.org/wiki/Component_%28graph_theory%29 .. [2] https://en.wikipedia.org/wiki/Tarjan%27s_strongly_connected_components_algorithm See Also ======== sympy.utilities.iterables.strongly_connected_components """ # Duplicate edges both ways so that the graph is effectively undirected # and return the strongly connected components: V, E = G E_undirected = [] for v1, v2 in E: E_undirected.extend([(v1, v2), (v2, v1)]) return strongly_connected_components((V, E_undirected)) def rotate_left(x, y): """ Left rotates a list x by the number of steps specified in y. Examples ======== >>> from sympy.utilities.iterables import rotate_left >>> a = [0, 1, 2] >>> rotate_left(a, 1) [1, 2, 0] """ if len(x) == 0: return [] y = y % len(x) return x[y:] + x[:y] def rotate_right(x, y): """ Right rotates a list x by the number of steps specified in y. Examples ======== >>> from sympy.utilities.iterables import rotate_right >>> a = [0, 1, 2] >>> rotate_right(a, 1) [2, 0, 1] """ if len(x) == 0: return [] y = len(x) - y % len(x) return x[y:] + x[:y] def least_rotation(x, key=None): ''' Returns the number of steps of left rotation required to obtain lexicographically minimal string/list/tuple, etc. Examples ======== >>> from sympy.utilities.iterables import least_rotation, rotate_left >>> a = [3, 1, 5, 1, 2] >>> least_rotation(a) 3 >>> rotate_left(a, _) [1, 2, 3, 1, 5] References ========== .. [1] https://en.wikipedia.org/wiki/Lexicographically_minimal_string_rotation ''' from sympy.functions.elementary.miscellaneous import Id if key is None: key = Id S = x + x # Concatenate string to it self to avoid modular arithmetic f = [-1] * len(S) # Failure function k = 0 # Least rotation of string found so far for j in range(1,len(S)): sj = S[j] i = f[j-k-1] while i != -1 and sj != S[k+i+1]: if key(sj) < key(S[k+i+1]): k = j-i-1 i = f[i] if sj != S[k+i+1]: if key(sj) < key(S[k]): k = j f[j-k] = -1 else: f[j-k] = i+1 return k def multiset_combinations(m, n, g=None): """ Return the unique combinations of size ``n`` from multiset ``m``. Examples ======== >>> from sympy.utilities.iterables import multiset_combinations >>> from itertools import combinations >>> [''.join(i) for i in multiset_combinations('baby', 3)] ['abb', 'aby', 'bby'] >>> def count(f, s): return len(list(f(s, 3))) The number of combinations depends on the number of letters; the number of unique combinations depends on how the letters are repeated. >>> s1 = 'abracadabra' >>> s2 = 'banana tree' >>> count(combinations, s1), count(multiset_combinations, s1) (165, 23) >>> count(combinations, s2), count(multiset_combinations, s2) (165, 54) """ from sympy.core.sorting import ordered if g is None: if isinstance(m, dict): if any(as_int(v) < 0 for v in m.values()): raise ValueError('counts cannot be negative') N = sum(m.values()) if n > N: return g = [[k, m[k]] for k in ordered(m)] else: m = list(m) N = len(m) if n > N: return try: m = multiset(m) g = [(k, m[k]) for k in ordered(m)] except TypeError: m = list(ordered(m)) g = [list(i) for i in group(m, multiple=False)] del m else: # not checking counts since g is intended for internal use N = sum(v for k, v in g) if n > N or not n: yield [] else: for i, (k, v) in enumerate(g): if v >= n: yield [k]*n v = n - 1 for v in range(min(n, v), 0, -1): for j in multiset_combinations(None, n - v, g[i + 1:]): rv = [k]*v + j if len(rv) == n: yield rv def multiset_permutations(m, size=None, g=None): """ Return the unique permutations of multiset ``m``. Examples ======== >>> from sympy.utilities.iterables import multiset_permutations >>> from sympy import factorial >>> [''.join(i) for i in multiset_permutations('aab')] ['aab', 'aba', 'baa'] >>> factorial(len('banana')) 720 >>> len(list(multiset_permutations('banana'))) 60 """ from sympy.core.sorting import ordered if g is None: if isinstance(m, dict): if any(as_int(v) < 0 for v in m.values()): raise ValueError('counts cannot be negative') g = [[k, m[k]] for k in ordered(m)] else: m = list(ordered(m)) g = [list(i) for i in group(m, multiple=False)] del m do = [gi for gi in g if gi[1] > 0] SUM = sum([gi[1] for gi in do]) if not do or size is not None and (size > SUM or size < 1): if not do and size is None or size == 0: yield [] return elif size == 1: for k, v in do: yield [k] elif len(do) == 1: k, v = do[0] v = v if size is None else (size if size <= v else 0) yield [k for i in range(v)] elif all(v == 1 for k, v in do): for p in permutations([k for k, v in do], size): yield list(p) else: size = size if size is not None else SUM for i, (k, v) in enumerate(do): do[i][1] -= 1 for j in multiset_permutations(None, size - 1, do): if j: yield [k] + j do[i][1] += 1 def _partition(seq, vector, m=None): """ Return the partition of seq as specified by the partition vector. Examples ======== >>> from sympy.utilities.iterables import _partition >>> _partition('abcde', [1, 0, 1, 2, 0]) [['b', 'e'], ['a', 'c'], ['d']] Specifying the number of bins in the partition is optional: >>> _partition('abcde', [1, 0, 1, 2, 0], 3) [['b', 'e'], ['a', 'c'], ['d']] The output of _set_partitions can be passed as follows: >>> output = (3, [1, 0, 1, 2, 0]) >>> _partition('abcde', *output) [['b', 'e'], ['a', 'c'], ['d']] See Also ======== combinatorics.partitions.Partition.from_rgs """ if m is None: m = max(vector) + 1 elif isinstance(vector, int): # entered as m, vector vector, m = m, vector p = [[] for i in range(m)] for i, v in enumerate(vector): p[v].append(seq[i]) return p def _set_partitions(n): """Cycle through all partitions of n elements, yielding the current number of partitions, ``m``, and a mutable list, ``q`` such that ``element[i]`` is in part ``q[i]`` of the partition. NOTE: ``q`` is modified in place and generally should not be changed between function calls. Examples ======== >>> from sympy.utilities.iterables import _set_partitions, _partition >>> for m, q in _set_partitions(3): ... print('%s %s %s' % (m, q, _partition('abc', q, m))) 1 [0, 0, 0] [['a', 'b', 'c']] 2 [0, 0, 1] [['a', 'b'], ['c']] 2 [0, 1, 0] [['a', 'c'], ['b']] 2 [0, 1, 1] [['a'], ['b', 'c']] 3 [0, 1, 2] [['a'], ['b'], ['c']] Notes ===== This algorithm is similar to, and solves the same problem as, Algorithm 7.2.1.5H, from volume 4A of Knuth's The Art of Computer Programming. Knuth uses the term "restricted growth string" where this code refers to a "partition vector". In each case, the meaning is the same: the value in the ith element of the vector specifies to which part the ith set element is to be assigned. At the lowest level, this code implements an n-digit big-endian counter (stored in the array q) which is incremented (with carries) to get the next partition in the sequence. A special twist is that a digit is constrained to be at most one greater than the maximum of all the digits to the left of it. The array p maintains this maximum, so that the code can efficiently decide when a digit can be incremented in place or whether it needs to be reset to 0 and trigger a carry to the next digit. The enumeration starts with all the digits 0 (which corresponds to all the set elements being assigned to the same 0th part), and ends with 0123...n, which corresponds to each set element being assigned to a different, singleton, part. This routine was rewritten to use 0-based lists while trying to preserve the beauty and efficiency of the original algorithm. References ========== .. [1] Nijenhuis, Albert and Wilf, Herbert. (1978) Combinatorial Algorithms, 2nd Ed, p 91, algorithm "nexequ". Available online from https://www.math.upenn.edu/~wilf/website/CombAlgDownld.html (viewed November 17, 2012). """ p = [0]*n q = [0]*n nc = 1 yield nc, q while nc != n: m = n while 1: m -= 1 i = q[m] if p[i] != 1: break q[m] = 0 i += 1 q[m] = i m += 1 nc += m - n p[0] += n - m if i == nc: p[nc] = 0 nc += 1 p[i - 1] -= 1 p[i] += 1 yield nc, q def multiset_partitions(multiset, m=None): """ Return unique partitions of the given multiset (in list form). If ``m`` is None, all multisets will be returned, otherwise only partitions with ``m`` parts will be returned. If ``multiset`` is an integer, a range [0, 1, ..., multiset - 1] will be supplied. Examples ======== >>> from sympy.utilities.iterables import multiset_partitions >>> list(multiset_partitions([1, 2, 3, 4], 2)) [[[1, 2, 3], [4]], [[1, 2, 4], [3]], [[1, 2], [3, 4]], [[1, 3, 4], [2]], [[1, 3], [2, 4]], [[1, 4], [2, 3]], [[1], [2, 3, 4]]] >>> list(multiset_partitions([1, 2, 3, 4], 1)) [[[1, 2, 3, 4]]] Only unique partitions are returned and these will be returned in a canonical order regardless of the order of the input: >>> a = [1, 2, 2, 1] >>> ans = list(multiset_partitions(a, 2)) >>> a.sort() >>> list(multiset_partitions(a, 2)) == ans True >>> a = range(3, 1, -1) >>> (list(multiset_partitions(a)) == ... list(multiset_partitions(sorted(a)))) True If m is omitted then all partitions will be returned: >>> list(multiset_partitions([1, 1, 2])) [[[1, 1, 2]], [[1, 1], [2]], [[1, 2], [1]], [[1], [1], [2]]] >>> list(multiset_partitions([1]*3)) [[[1, 1, 1]], [[1], [1, 1]], [[1], [1], [1]]] Counting ======== The number of partitions of a set is given by the bell number: >>> from sympy import bell >>> len(list(multiset_partitions(5))) == bell(5) == 52 True The number of partitions of length k from a set of size n is given by the Stirling Number of the 2nd kind: >>> from sympy.functions.combinatorial.numbers import stirling >>> stirling(5, 2) == len(list(multiset_partitions(5, 2))) == 15 True These comments on counting apply to *sets*, not multisets. Notes ===== When all the elements are the same in the multiset, the order of the returned partitions is determined by the ``partitions`` routine. If one is counting partitions then it is better to use the ``nT`` function. See Also ======== partitions sympy.combinatorics.partitions.Partition sympy.combinatorics.partitions.IntegerPartition sympy.functions.combinatorial.numbers.nT """ # This function looks at the supplied input and dispatches to # several special-case routines as they apply. if isinstance(multiset, int): n = multiset if m and m > n: return multiset = list(range(n)) if m == 1: yield [multiset[:]] return # If m is not None, it can sometimes be faster to use # MultisetPartitionTraverser.enum_range() even for inputs # which are sets. Since the _set_partitions code is quite # fast, this is only advantageous when the overall set # partitions outnumber those with the desired number of parts # by a large factor. (At least 60.) Such a switch is not # currently implemented. for nc, q in _set_partitions(n): if m is None or nc == m: rv = [[] for i in range(nc)] for i in range(n): rv[q[i]].append(multiset[i]) yield rv return if len(multiset) == 1 and isinstance(multiset, str): multiset = [multiset] if not has_variety(multiset): # Only one component, repeated n times. The resulting # partitions correspond to partitions of integer n. n = len(multiset) if m and m > n: return if m == 1: yield [multiset[:]] return x = multiset[:1] for size, p in partitions(n, m, size=True): if m is None or size == m: rv = [] for k in sorted(p): rv.extend([x*k]*p[k]) yield rv else: from sympy.core.sorting import ordered multiset = list(ordered(multiset)) n = len(multiset) if m and m > n: return if m == 1: yield [multiset[:]] return # Split the information of the multiset into two lists - # one of the elements themselves, and one (of the same length) # giving the number of repeats for the corresponding element. elements, multiplicities = zip(*group(multiset, False)) if len(elements) < len(multiset): # General case - multiset with more than one distinct element # and at least one element repeated more than once. if m: mpt = MultisetPartitionTraverser() for state in mpt.enum_range(multiplicities, m-1, m): yield list_visitor(state, elements) else: for state in multiset_partitions_taocp(multiplicities): yield list_visitor(state, elements) else: # Set partitions case - no repeated elements. Pretty much # same as int argument case above, with same possible, but # currently unimplemented optimization for some cases when # m is not None for nc, q in _set_partitions(n): if m is None or nc == m: rv = [[] for i in range(nc)] for i in range(n): rv[q[i]].append(i) yield [[multiset[j] for j in i] for i in rv] def partitions(n, m=None, k=None, size=False): """Generate all partitions of positive integer, n. Parameters ========== m : integer (default gives partitions of all sizes) limits number of parts in partition (mnemonic: m, maximum parts) k : integer (default gives partitions number from 1 through n) limits the numbers that are kept in the partition (mnemonic: k, keys) size : bool (default False, only partition is returned) when ``True`` then (M, P) is returned where M is the sum of the multiplicities and P is the generated partition. Each partition is represented as a dictionary, mapping an integer to the number of copies of that integer in the partition. For example, the first partition of 4 returned is {4: 1}, "4: one of them". Examples ======== >>> from sympy.utilities.iterables import partitions The numbers appearing in the partition (the key of the returned dict) are limited with k: >>> for p in partitions(6, k=2): # doctest: +SKIP ... print(p) {2: 3} {1: 2, 2: 2} {1: 4, 2: 1} {1: 6} The maximum number of parts in the partition (the sum of the values in the returned dict) are limited with m (default value, None, gives partitions from 1 through n): >>> for p in partitions(6, m=2): # doctest: +SKIP ... print(p) ... {6: 1} {1: 1, 5: 1} {2: 1, 4: 1} {3: 2} References ========== .. [1] modified from Tim Peter's version to allow for k and m values: https://code.activestate.com/recipes/218332-generator-for-integer-partitions/ See Also ======== sympy.combinatorics.partitions.Partition sympy.combinatorics.partitions.IntegerPartition """ if (n <= 0 or m is not None and m < 1 or k is not None and k < 1 or m and k and m*k < n): # the empty set is the only way to handle these inputs # and returning {} to represent it is consistent with # the counting convention, e.g. nT(0) == 1. if size: yield 0, {} else: yield {} return if m is None: m = n else: m = min(m, n) k = min(k or n, n) n, m, k = as_int(n), as_int(m), as_int(k) q, r = divmod(n, k) ms = {k: q} keys = [k] # ms.keys(), from largest to smallest if r: ms[r] = 1 keys.append(r) room = m - q - bool(r) if size: yield sum(ms.values()), ms.copy() else: yield ms.copy() while keys != [1]: # Reuse any 1's. if keys[-1] == 1: del keys[-1] reuse = ms.pop(1) room += reuse else: reuse = 0 while 1: # Let i be the smallest key larger than 1. Reuse one # instance of i. i = keys[-1] newcount = ms[i] = ms[i] - 1 reuse += i if newcount == 0: del keys[-1], ms[i] room += 1 # Break the remainder into pieces of size i-1. i -= 1 q, r = divmod(reuse, i) need = q + bool(r) if need > room: if not keys: return continue ms[i] = q keys.append(i) if r: ms[r] = 1 keys.append(r) break room -= need if size: yield sum(ms.values()), ms.copy() else: yield ms.copy() def ordered_partitions(n, m=None, sort=True): """Generates ordered partitions of integer ``n``. Parameters ========== m : integer (default None) The default value gives partitions of all sizes else only those with size m. In addition, if ``m`` is not None then partitions are generated *in place* (see examples). sort : bool (default True) Controls whether partitions are returned in sorted order when ``m`` is not None; when False, the partitions are returned as fast as possible with elements sorted, but when m|n the partitions will not be in ascending lexicographical order. Examples ======== >>> from sympy.utilities.iterables import ordered_partitions All partitions of 5 in ascending lexicographical: >>> for p in ordered_partitions(5): ... print(p) [1, 1, 1, 1, 1] [1, 1, 1, 2] [1, 1, 3] [1, 2, 2] [1, 4] [2, 3] [5] Only partitions of 5 with two parts: >>> for p in ordered_partitions(5, 2): ... print(p) [1, 4] [2, 3] When ``m`` is given, a given list objects will be used more than once for speed reasons so you will not see the correct partitions unless you make a copy of each as it is generated: >>> [p for p in ordered_partitions(7, 3)] [[1, 1, 1], [1, 1, 1], [1, 1, 1], [2, 2, 2]] >>> [list(p) for p in ordered_partitions(7, 3)] [[1, 1, 5], [1, 2, 4], [1, 3, 3], [2, 2, 3]] When ``n`` is a multiple of ``m``, the elements are still sorted but the partitions themselves will be *unordered* if sort is False; the default is to return them in ascending lexicographical order. >>> for p in ordered_partitions(6, 2): ... print(p) [1, 5] [2, 4] [3, 3] But if speed is more important than ordering, sort can be set to False: >>> for p in ordered_partitions(6, 2, sort=False): ... print(p) [1, 5] [3, 3] [2, 4] References ========== .. [1] Generating Integer Partitions, [online], Available: https://jeromekelleher.net/generating-integer-partitions.html .. [2] Jerome Kelleher and Barry O'Sullivan, "Generating All Partitions: A Comparison Of Two Encodings", [online], Available: https://arxiv.org/pdf/0909.2331v2.pdf """ if n < 1 or m is not None and m < 1: # the empty set is the only way to handle these inputs # and returning {} to represent it is consistent with # the counting convention, e.g. nT(0) == 1. yield [] return if m is None: # The list `a`'s leading elements contain the partition in which # y is the biggest element and x is either the same as y or the # 2nd largest element; v and w are adjacent element indices # to which x and y are being assigned, respectively. a = [1]*n y = -1 v = n while v > 0: v -= 1 x = a[v] + 1 while y >= 2 * x: a[v] = x y -= x v += 1 w = v + 1 while x <= y: a[v] = x a[w] = y yield a[:w + 1] x += 1 y -= 1 a[v] = x + y y = a[v] - 1 yield a[:w] elif m == 1: yield [n] elif n == m: yield [1]*n else: # recursively generate partitions of size m for b in range(1, n//m + 1): a = [b]*m x = n - b*m if not x: if sort: yield a elif not sort and x <= m: for ax in ordered_partitions(x, sort=False): mi = len(ax) a[-mi:] = [i + b for i in ax] yield a a[-mi:] = [b]*mi else: for mi in range(1, m): for ax in ordered_partitions(x, mi, sort=True): a[-mi:] = [i + b for i in ax] yield a a[-mi:] = [b]*mi def binary_partitions(n): """ Generates the binary partition of n. A binary partition consists only of numbers that are powers of two. Each step reduces a `2^{k+1}` to `2^k` and `2^k`. Thus 16 is converted to 8 and 8. Examples ======== >>> from sympy.utilities.iterables import binary_partitions >>> for i in binary_partitions(5): ... print(i) ... [4, 1] [2, 2, 1] [2, 1, 1, 1] [1, 1, 1, 1, 1] References ========== .. [1] TAOCP 4, section 7.2.1.5, problem 64 """ from math import ceil, log power = int(2**(ceil(log(n, 2)))) acc = 0 partition = [] while power: if acc + power <= n: partition.append(power) acc += power power >>= 1 last_num = len(partition) - 1 - (n & 1) while last_num >= 0: yield partition if partition[last_num] == 2: partition[last_num] = 1 partition.append(1) last_num -= 1 continue partition.append(1) partition[last_num] >>= 1 x = partition[last_num + 1] = partition[last_num] last_num += 1 while x > 1: if x <= len(partition) - last_num - 1: del partition[-x + 1:] last_num += 1 partition[last_num] = x else: x >>= 1 yield [1]*n def has_dups(seq): """Return True if there are any duplicate elements in ``seq``. Examples ======== >>> from sympy import has_dups, Dict, Set >>> has_dups((1, 2, 1)) True >>> has_dups(range(3)) False >>> all(has_dups(c) is False for c in (set(), Set(), dict(), Dict())) True """ from sympy.core.containers import Dict from sympy.sets.sets import Set if isinstance(seq, (dict, set, Dict, Set)): return False unique = set() try: return any(True for s in seq if s in unique or unique.add(s)) except TypeError: return len(seq) != len(list(uniq(seq))) def has_variety(seq): """Return True if there are any different elements in ``seq``. Examples ======== >>> from sympy import has_variety >>> has_variety((1, 2, 1)) True >>> has_variety((1, 1, 1)) False """ for i, s in enumerate(seq): if i == 0: sentinel = s else: if s != sentinel: return True return False def uniq(seq, result=None): """ Yield unique elements from ``seq`` as an iterator. The second parameter ``result`` is used internally; it is not necessary to pass anything for this. Note: changing the sequence during iteration will raise a RuntimeError if the size of the sequence is known; if you pass an iterator and advance the iterator you will change the output of this routine but there will be no warning. Examples ======== >>> from sympy.utilities.iterables import uniq >>> dat = [1, 4, 1, 5, 4, 2, 1, 2] >>> type(uniq(dat)) in (list, tuple) False >>> list(uniq(dat)) [1, 4, 5, 2] >>> list(uniq(x for x in dat)) [1, 4, 5, 2] >>> list(uniq([[1], [2, 1], [1]])) [[1], [2, 1]] """ try: n = len(seq) except TypeError: n = None def check(): # check that size of seq did not change during iteration; # if n == None the object won't support size changing, e.g. # an iterator can't be changed if n is not None and len(seq) != n: raise RuntimeError('sequence changed size during iteration') try: seen = set() result = result or [] for i, s in enumerate(seq): if not (s in seen or seen.add(s)): yield s check() except TypeError: if s not in result: yield s check() result.append(s) if hasattr(seq, '__getitem__'): yield from uniq(seq[i + 1:], result) else: yield from uniq(seq, result) def generate_bell(n): """Return permutations of [0, 1, ..., n - 1] such that each permutation differs from the last by the exchange of a single pair of neighbors. The ``n!`` permutations are returned as an iterator. In order to obtain the next permutation from a random starting permutation, use the ``next_trotterjohnson`` method of the Permutation class (which generates the same sequence in a different manner). Examples ======== >>> from itertools import permutations >>> from sympy.utilities.iterables import generate_bell >>> from sympy import zeros, Matrix This is the sort of permutation used in the ringing of physical bells, and does not produce permutations in lexicographical order. Rather, the permutations differ from each other by exactly one inversion, and the position at which the swapping occurs varies periodically in a simple fashion. Consider the first few permutations of 4 elements generated by ``permutations`` and ``generate_bell``: >>> list(permutations(range(4)))[:5] [(0, 1, 2, 3), (0, 1, 3, 2), (0, 2, 1, 3), (0, 2, 3, 1), (0, 3, 1, 2)] >>> list(generate_bell(4))[:5] [(0, 1, 2, 3), (0, 1, 3, 2), (0, 3, 1, 2), (3, 0, 1, 2), (3, 0, 2, 1)] Notice how the 2nd and 3rd lexicographical permutations have 3 elements out of place whereas each "bell" permutation always has only two elements out of place relative to the previous permutation (and so the signature (+/-1) of a permutation is opposite of the signature of the previous permutation). How the position of inversion varies across the elements can be seen by tracing out where the largest number appears in the permutations: >>> m = zeros(4, 24) >>> for i, p in enumerate(generate_bell(4)): ... m[:, i] = Matrix([j - 3 for j in list(p)]) # make largest zero >>> m.print_nonzero('X') [XXX XXXXXX XXXXXX XXX] [XX XX XXXX XX XXXX XX XX] [X XXXX XX XXXX XX XXXX X] [ XXXXXX XXXXXX XXXXXX ] See Also ======== sympy.combinatorics.permutations.Permutation.next_trotterjohnson References ========== .. [1] https://en.wikipedia.org/wiki/Method_ringing .. [2] https://stackoverflow.com/questions/4856615/recursive-permutation/4857018 .. [3] https://web.archive.org/web/20160313023044/http://programminggeeks.com/bell-algorithm-for-permutation/ .. [4] https://en.wikipedia.org/wiki/Steinhaus%E2%80%93Johnson%E2%80%93Trotter_algorithm .. [5] Generating involutions, derangements, and relatives by ECO Vincent Vajnovszki, DMTCS vol 1 issue 12, 2010 """ n = as_int(n) if n < 1: raise ValueError('n must be a positive integer') if n == 1: yield (0,) elif n == 2: yield (0, 1) yield (1, 0) elif n == 3: yield from [(0, 1, 2), (0, 2, 1), (2, 0, 1), (2, 1, 0), (1, 2, 0), (1, 0, 2)] else: m = n - 1 op = [0] + [-1]*m l = list(range(n)) while True: yield tuple(l) # find biggest element with op big = None, -1 # idx, value for i in range(n): if op[i] and l[i] > big[1]: big = i, l[i] i, _ = big if i is None: break # there are no ops left # swap it with neighbor in the indicated direction j = i + op[i] l[i], l[j] = l[j], l[i] op[i], op[j] = op[j], op[i] # if it landed at the end or if the neighbor in the same # direction is bigger then turn off op if j == 0 or j == m or l[j + op[j]] > l[j]: op[j] = 0 # any element bigger to the left gets +1 op for i in range(j): if l[i] > l[j]: op[i] = 1 # any element bigger to the right gets -1 op for i in range(j + 1, n): if l[i] > l[j]: op[i] = -1 def generate_involutions(n): """ Generates involutions. An involution is a permutation that when multiplied by itself equals the identity permutation. In this implementation the involutions are generated using Fixed Points. Alternatively, an involution can be considered as a permutation that does not contain any cycles with a length that is greater than two. Examples ======== >>> from sympy.utilities.iterables import generate_involutions >>> list(generate_involutions(3)) [(0, 1, 2), (0, 2, 1), (1, 0, 2), (2, 1, 0)] >>> len(list(generate_involutions(4))) 10 References ========== .. [1] https://mathworld.wolfram.com/PermutationInvolution.html """ idx = list(range(n)) for p in permutations(idx): for i in idx: if p[p[i]] != i: break else: yield p def multiset_derangements(s): """Generate derangements of the elements of s *in place*. Examples ======== >>> from sympy.utilities.iterables import multiset_derangements, uniq Because the derangements of multisets (not sets) are generated in place, copies of the return value must be made if a collection of derangements is desired or else all values will be the same: >>> list(uniq([i for i in multiset_derangements('1233')])) [[None, None, None, None]] >>> [i.copy() for i in multiset_derangements('1233')] [['3', '3', '1', '2'], ['3', '3', '2', '1']] >>> [''.join(i) for i in multiset_derangements('1233')] ['3312', '3321'] """ from sympy.core.sorting import ordered # create multiset dictionary of hashable elements or else # remap elements to integers try: ms = multiset(s) except TypeError: # give each element a canonical integer value key = dict(enumerate(ordered(uniq(s)))) h = [] for si in s: for k in key: if key[k] == si: h.append(k) break for i in multiset_derangements(h): yield [key[j] for j in i] return mx = max(ms.values()) # max repetition of any element n = len(s) # the number of elements ## special cases # 1) one element has more than half the total cardinality of s: no # derangements are possible. if mx*2 > n: return # 2) all elements appear once: singletons if len(ms) == n: yield from _set_derangements(s) return # find the first element that is repeated the most to place # in the following two special cases where the selection # is unambiguous: either there are two elements with multiplicity # of mx or else there is only one with multiplicity mx for M in ms: if ms[M] == mx: break inonM = [i for i in range(n) if s[i] != M] # location of non-M iM = [i for i in range(n) if s[i] == M] # locations of M rv = [None]*n # 3) half are the same if 2*mx == n: # M goes into non-M locations for i in inonM: rv[i] = M # permutations of non-M go to M locations for p in multiset_permutations([s[i] for i in inonM]): for i, pi in zip(iM, p): rv[i] = pi yield rv # clean-up (and encourages proper use of routine) rv[:] = [None]*n return # 4) single repeat covers all but 1 of the non-repeats: # if there is one repeat then the multiset of the values # of ms would be {mx: 1, 1: n - mx}, i.e. there would # be n - mx + 1 values with the condition that n - 2*mx = 1 if n - 2*mx == 1 and len(ms.values()) == n - mx + 1: for i, i1 in enumerate(inonM): ifill = inonM[:i] + inonM[i+1:] for j in ifill: rv[j] = M for p in permutations([s[j] for j in ifill]): rv[i1] = s[i1] for j, pi in zip(iM, p): rv[j] = pi k = i1 for j in iM: rv[j], rv[k] = rv[k], rv[j] yield rv k = j # clean-up (and encourages proper use of routine) rv[:] = [None]*n return ## general case is handled with 3 helpers: # 1) `finish_derangements` will place the last two elements # which have arbitrary multiplicities, e.g. for multiset # {c: 3, a: 2, b: 2}, the last two elements are a and b # 2) `iopen` will tell where a given element can be placed # 3) `do` will recursively place elements into subsets of # valid locations def finish_derangements(): """Place the last two elements into the partially completed derangement, and yield the results. """ a = take[1][0] # penultimate element a_ct = take[1][1] b = take[0][0] # last element to be placed b_ct = take[0][1] # split the indexes of the not-already-assigned elements of rv into # three categories forced_a = [] # positions which must have an a forced_b = [] # positions which must have a b open_free = [] # positions which could take either for i in range(len(s)): if rv[i] is None: if s[i] == a: forced_b.append(i) elif s[i] == b: forced_a.append(i) else: open_free.append(i) if len(forced_a) > a_ct or len(forced_b) > b_ct: # No derangement possible return for i in forced_a: rv[i] = a for i in forced_b: rv[i] = b for a_place in combinations(open_free, a_ct - len(forced_a)): for a_pos in a_place: rv[a_pos] = a for i in open_free: if rv[i] is None: # anything not in the subset is set to b rv[i] = b yield rv # Clean up/undo the final placements for i in open_free: rv[i] = None # additional cleanup - clear forced_a, forced_b for i in forced_a: rv[i] = None for i in forced_b: rv[i] = None def iopen(v): # return indices at which element v can be placed in rv: # locations which are not already occupied if that location # does not already contain v in the same location of s return [i for i in range(n) if rv[i] is None and s[i] != v] def do(j): if j == 1: # handle the last two elements (regardless of multiplicity) # with a special method yield from finish_derangements() else: # place the mx elements of M into a subset of places # into which it can be replaced M, mx = take[j] for i in combinations(iopen(M), mx): # place M for ii in i: rv[ii] = M # recursively place the next element yield from do(j - 1) # mark positions where M was placed as once again # open for placement of other elements for ii in i: rv[ii] = None # process elements in order of canonically decreasing multiplicity take = sorted(ms.items(), key=lambda x:(x[1], x[0])) yield from do(len(take) - 1) rv[:] = [None]*n def random_derangement(t, choice=None, strict=True): """Return a list of elements in which none are in the same positions as they were originally. If an element fills more than half of the positions then an error will be raised since no derangement is possible. To obtain a derangement of as many items as possible--with some of the most numerous remaining in their original positions--pass `strict=False`. To produce a pseudorandom derangment, pass a pseudorandom selector like `choice` (see below). Examples ======== >>> from sympy.utilities.iterables import random_derangement >>> t = 'SymPy: a CAS in pure Python' >>> d = random_derangement(t) >>> all(i != j for i, j in zip(d, t)) True A predictable result can be obtained by using a pseudorandom generator for the choice: >>> from sympy.core.random import seed, choice as c >>> seed(1) >>> d = [''.join(random_derangement(t, c)) for i in range(5)] >>> assert len(set(d)) != 1 # we got different values By reseeding, the same sequence can be obtained: >>> seed(1) >>> d2 = [''.join(random_derangement(t, c)) for i in range(5)] >>> assert d == d2 """ if choice is None: import secrets choice = secrets.choice def shuffle(rv): '''Knuth shuffle''' for i in range(len(rv) - 1, 0, -1): x = choice(rv[:i + 1]) j = rv.index(x) rv[i], rv[j] = rv[j], rv[i] def pick(rv, n): '''shuffle rv and return the first n values ''' shuffle(rv) return rv[:n] ms = multiset(t) tot = len(t) ms = sorted(ms.items(), key=lambda x: x[1]) # if there are not enough spaces for the most # plentiful element to move to then some of them # will have to stay in place M, mx = ms[-1] n = len(t) xs = 2*mx - tot if xs > 0: if strict: raise ValueError('no derangement possible') opts = [i for (i, c) in enumerate(t) if c == ms[-1][0]] pick(opts, xs) stay = sorted(opts[:xs]) rv = list(t) for i in reversed(stay): rv.pop(i) rv = random_derangement(rv, choice) for i in stay: rv.insert(i, ms[-1][0]) return ''.join(rv) if type(t) is str else rv # the normal derangement calculated from here if n == len(ms): # approx 1/3 will succeed rv = list(t) while True: shuffle(rv) if all(i != j for i,j in zip(rv, t)): break else: # general case rv = [None]*n while True: j = 0 while j > -len(ms): # do most numerous first j -= 1 e, c = ms[j] opts = [i for i in range(n) if rv[i] is None and t[i] != e] if len(opts) < c: for i in range(n): rv[i] = None break # try again pick(opts, c) for i in range(c): rv[opts[i]] = e else: return rv return rv def _set_derangements(s): """ yield derangements of items in ``s`` which are assumed to contain no repeated elements """ if len(s) < 2: return if len(s) == 2: yield [s[1], s[0]] return if len(s) == 3: yield [s[1], s[2], s[0]] yield [s[2], s[0], s[1]] return for p in permutations(s): if not any(i == j for i, j in zip(p, s)): yield list(p) def generate_derangements(s): """ Return unique derangements of the elements of iterable ``s``. Examples ======== >>> from sympy.utilities.iterables import generate_derangements >>> list(generate_derangements([0, 1, 2])) [[1, 2, 0], [2, 0, 1]] >>> list(generate_derangements([0, 1, 2, 2])) [[2, 2, 0, 1], [2, 2, 1, 0]] >>> list(generate_derangements([0, 1, 1])) [] See Also ======== sympy.functions.combinatorial.factorials.subfactorial """ if not has_dups(s): yield from _set_derangements(s) else: for p in multiset_derangements(s): yield list(p) def necklaces(n, k, free=False): """ A routine to generate necklaces that may (free=True) or may not (free=False) be turned over to be viewed. The "necklaces" returned are comprised of ``n`` integers (beads) with ``k`` different values (colors). Only unique necklaces are returned. Examples ======== >>> from sympy.utilities.iterables import necklaces, bracelets >>> def show(s, i): ... return ''.join(s[j] for j in i) The "unrestricted necklace" is sometimes also referred to as a "bracelet" (an object that can be turned over, a sequence that can be reversed) and the term "necklace" is used to imply a sequence that cannot be reversed. So ACB == ABC for a bracelet (rotate and reverse) while the two are different for a necklace since rotation alone cannot make the two sequences the same. (mnemonic: Bracelets can be viewed Backwards, but Not Necklaces.) >>> B = [show('ABC', i) for i in bracelets(3, 3)] >>> N = [show('ABC', i) for i in necklaces(3, 3)] >>> set(N) - set(B) {'ACB'} >>> list(necklaces(4, 2)) [(0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 1), (1, 1, 1, 1)] >>> [show('.o', i) for i in bracelets(4, 2)] ['....', '...o', '..oo', '.o.o', '.ooo', 'oooo'] References ========== .. [1] https://mathworld.wolfram.com/Necklace.html .. [2] Frank Ruskey, Carla Savage, and Terry Min Yih Wang, Generating necklaces, Journal of Algorithms 13 (1992), 414-430; https://doi.org/10.1016/0196-6774(92)90047-G """ # The FKM algorithm if k == 0 and n > 0: return a = [0]*n yield tuple(a) if n == 0: return while True: i = n - 1 while a[i] == k - 1: i -= 1 if i == -1: return a[i] += 1 for j in range(n - i - 1): a[j + i + 1] = a[j] if n % (i + 1) == 0 and (not free or all(a <= a[j::-1] + a[-1:j:-1] for j in range(n - 1))): # No need to test j = n - 1. yield tuple(a) def bracelets(n, k): """Wrapper to necklaces to return a free (unrestricted) necklace.""" return necklaces(n, k, free=True) def generate_oriented_forest(n): """ This algorithm generates oriented forests. An oriented graph is a directed graph having no symmetric pair of directed edges. A forest is an acyclic graph, i.e., it has no cycles. A forest can also be described as a disjoint union of trees, which are graphs in which any two vertices are connected by exactly one simple path. Examples ======== >>> from sympy.utilities.iterables import generate_oriented_forest >>> list(generate_oriented_forest(4)) [[0, 1, 2, 3], [0, 1, 2, 2], [0, 1, 2, 1], [0, 1, 2, 0], \ [0, 1, 1, 1], [0, 1, 1, 0], [0, 1, 0, 1], [0, 1, 0, 0], [0, 0, 0, 0]] References ========== .. [1] T. Beyer and S.M. Hedetniemi: constant time generation of rooted trees, SIAM J. Computing Vol. 9, No. 4, November 1980 .. [2] https://stackoverflow.com/questions/1633833/oriented-forest-taocp-algorithm-in-python """ P = list(range(-1, n)) while True: yield P[1:] if P[n] > 0: P[n] = P[P[n]] else: for p in range(n - 1, 0, -1): if P[p] != 0: target = P[p] - 1 for q in range(p - 1, 0, -1): if P[q] == target: break offset = p - q for i in range(p, n + 1): P[i] = P[i - offset] break else: break def minlex(seq, directed=True, key=None): r""" Return the rotation of the sequence in which the lexically smallest elements appear first, e.g. `cba \rightarrow acb`. The sequence returned is a tuple, unless the input sequence is a string in which case a string is returned. If ``directed`` is False then the smaller of the sequence and the reversed sequence is returned, e.g. `cba \rightarrow abc`. If ``key`` is not None then it is used to extract a comparison key from each element in iterable. Examples ======== >>> from sympy.combinatorics.polyhedron import minlex >>> minlex((1, 2, 0)) (0, 1, 2) >>> minlex((1, 0, 2)) (0, 2, 1) >>> minlex((1, 0, 2), directed=False) (0, 1, 2) >>> minlex('11010011000', directed=True) '00011010011' >>> minlex('11010011000', directed=False) '00011001011' >>> minlex(('bb', 'aaa', 'c', 'a')) ('a', 'bb', 'aaa', 'c') >>> minlex(('bb', 'aaa', 'c', 'a'), key=len) ('c', 'a', 'bb', 'aaa') """ from sympy.functions.elementary.miscellaneous import Id if key is None: key = Id best = rotate_left(seq, least_rotation(seq, key=key)) if not directed: rseq = seq[::-1] rbest = rotate_left(rseq, least_rotation(rseq, key=key)) best = min(best, rbest, key=key) # Convert to tuple, unless we started with a string. return tuple(best) if not isinstance(seq, str) else best def runs(seq, op=gt): """Group the sequence into lists in which successive elements all compare the same with the comparison operator, ``op``: op(seq[i + 1], seq[i]) is True from all elements in a run. Examples ======== >>> from sympy.utilities.iterables import runs >>> from operator import ge >>> runs([0, 1, 2, 2, 1, 4, 3, 2, 2]) [[0, 1, 2], [2], [1, 4], [3], [2], [2]] >>> runs([0, 1, 2, 2, 1, 4, 3, 2, 2], op=ge) [[0, 1, 2, 2], [1, 4], [3], [2, 2]] """ cycles = [] seq = iter(seq) try: run = [next(seq)] except StopIteration: return [] while True: try: ei = next(seq) except StopIteration: break if op(ei, run[-1]): run.append(ei) continue else: cycles.append(run) run = [ei] if run: cycles.append(run) return cycles def sequence_partitions(l, n, /): r"""Returns the partition of sequence $l$ into $n$ bins Explanation =========== Given the sequence $l_1 \cdots l_m \in V^+$ where $V^+$ is the Kleene plus of $V$ The set of $n$ partitions of $l$ is defined as: .. math:: \{(s_1, \cdots, s_n) | s_1 \in V^+, \cdots, s_n \in V^+, s_1 \cdots s_n = l_1 \cdots l_m\} Parameters ========== l : Sequence[T] A nonempty sequence of any Python objects n : int A positive integer Yields ====== out : list[Sequence[T]] A list of sequences with concatenation equals $l$. This should conform with the type of $l$. Examples ======== >>> from sympy.utilities.iterables import sequence_partitions >>> for out in sequence_partitions([1, 2, 3, 4], 2): ... print(out) [[1], [2, 3, 4]] [[1, 2], [3, 4]] [[1, 2, 3], [4]] Notes ===== This is modified version of EnricoGiampieri's partition generator from https://stackoverflow.com/questions/13131491/partition-n-items-into-k-bins-in-python-lazily See Also ======== sequence_partitions_empty """ # Asserting l is nonempty is done only for sanity check if n == 1 and l: yield [l] return for i in range(1, len(l)): for part in sequence_partitions(l[i:], n - 1): yield [l[:i]] + part def sequence_partitions_empty(l, n, /): r"""Returns the partition of sequence $l$ into $n$ bins with empty sequence Explanation =========== Given the sequence $l_1 \cdots l_m \in V^*$ where $V^*$ is the Kleene star of $V$ The set of $n$ partitions of $l$ is defined as: .. math:: \{(s_1, \cdots, s_n) | s_1 \in V^*, \cdots, s_n \in V^*, s_1 \cdots s_n = l_1 \cdots l_m\} There are more combinations than :func:`sequence_partitions` because empty sequence can fill everywhere, so we try to provide different utility for this. Parameters ========== l : Sequence[T] A sequence of any Python objects (can be possibly empty) n : int A positive integer Yields ====== out : list[Sequence[T]] A list of sequences with concatenation equals $l$. This should conform with the type of $l$. Examples ======== >>> from sympy.utilities.iterables import sequence_partitions_empty >>> for out in sequence_partitions_empty([1, 2, 3, 4], 2): ... print(out) [[], [1, 2, 3, 4]] [[1], [2, 3, 4]] [[1, 2], [3, 4]] [[1, 2, 3], [4]] [[1, 2, 3, 4], []] See Also ======== sequence_partitions """ if n < 1: return if n == 1: yield [l] return for i in range(0, len(l) + 1): for part in sequence_partitions_empty(l[i:], n - 1): yield [l[:i]] + part def kbins(l, k, ordered=None): """ Return sequence ``l`` partitioned into ``k`` bins. Examples ======== The default is to give the items in the same order, but grouped into k partitions without any reordering: >>> from sympy.utilities.iterables import kbins >>> for p in kbins(list(range(5)), 2): ... print(p) ... [[0], [1, 2, 3, 4]] [[0, 1], [2, 3, 4]] [[0, 1, 2], [3, 4]] [[0, 1, 2, 3], [4]] The ``ordered`` flag is either None (to give the simple partition of the elements) or is a 2 digit integer indicating whether the order of the bins and the order of the items in the bins matters. Given:: A = [[0], [1, 2]] B = [[1, 2], [0]] C = [[2, 1], [0]] D = [[0], [2, 1]] the following values for ``ordered`` have the shown meanings:: 00 means A == B == C == D 01 means A == B 10 means A == D 11 means A == A >>> for ordered_flag in [None, 0, 1, 10, 11]: ... print('ordered = %s' % ordered_flag) ... for p in kbins(list(range(3)), 2, ordered=ordered_flag): ... print(' %s' % p) ... ordered = None [[0], [1, 2]] [[0, 1], [2]] ordered = 0 [[0, 1], [2]] [[0, 2], [1]] [[0], [1, 2]] ordered = 1 [[0], [1, 2]] [[0], [2, 1]] [[1], [0, 2]] [[1], [2, 0]] [[2], [0, 1]] [[2], [1, 0]] ordered = 10 [[0, 1], [2]] [[2], [0, 1]] [[0, 2], [1]] [[1], [0, 2]] [[0], [1, 2]] [[1, 2], [0]] ordered = 11 [[0], [1, 2]] [[0, 1], [2]] [[0], [2, 1]] [[0, 2], [1]] [[1], [0, 2]] [[1, 0], [2]] [[1], [2, 0]] [[1, 2], [0]] [[2], [0, 1]] [[2, 0], [1]] [[2], [1, 0]] [[2, 1], [0]] See Also ======== partitions, multiset_partitions """ if ordered is None: yield from sequence_partitions(l, k) elif ordered == 11: for pl in multiset_permutations(l): pl = list(pl) yield from sequence_partitions(pl, k) elif ordered == 00: yield from multiset_partitions(l, k) elif ordered == 10: for p in multiset_partitions(l, k): for perm in permutations(p): yield list(perm) elif ordered == 1: for kgot, p in partitions(len(l), k, size=True): if kgot != k: continue for li in multiset_permutations(l): rv = [] i = j = 0 li = list(li) for size, multiplicity in sorted(p.items()): for m in range(multiplicity): j = i + size rv.append(li[i: j]) i = j yield rv else: raise ValueError( 'ordered must be one of 00, 01, 10 or 11, not %s' % ordered) def permute_signs(t): """Return iterator in which the signs of non-zero elements of t are permuted. Examples ======== >>> from sympy.utilities.iterables import permute_signs >>> list(permute_signs((0, 1, 2))) [(0, 1, 2), (0, -1, 2), (0, 1, -2), (0, -1, -2)] """ for signs in product(*[(1, -1)]*(len(t) - t.count(0))): signs = list(signs) yield type(t)([i*signs.pop() if i else i for i in t]) def signed_permutations(t): """Return iterator in which the signs of non-zero elements of t and the order of the elements are permuted. Examples ======== >>> from sympy.utilities.iterables import signed_permutations >>> list(signed_permutations((0, 1, 2))) [(0, 1, 2), (0, -1, 2), (0, 1, -2), (0, -1, -2), (0, 2, 1), (0, -2, 1), (0, 2, -1), (0, -2, -1), (1, 0, 2), (-1, 0, 2), (1, 0, -2), (-1, 0, -2), (1, 2, 0), (-1, 2, 0), (1, -2, 0), (-1, -2, 0), (2, 0, 1), (-2, 0, 1), (2, 0, -1), (-2, 0, -1), (2, 1, 0), (-2, 1, 0), (2, -1, 0), (-2, -1, 0)] """ return (type(t)(i) for j in permutations(t) for i in permute_signs(j)) def rotations(s, dir=1): """Return a generator giving the items in s as list where each subsequent list has the items rotated to the left (default) or right (``dir=-1``) relative to the previous list. Examples ======== >>> from sympy import rotations >>> list(rotations([1,2,3])) [[1, 2, 3], [2, 3, 1], [3, 1, 2]] >>> list(rotations([1,2,3], -1)) [[1, 2, 3], [3, 1, 2], [2, 3, 1]] """ seq = list(s) for i in range(len(seq)): yield seq seq = rotate_left(seq, dir) def roundrobin(*iterables): """roundrobin recipe taken from itertools documentation: https://docs.python.org/3/library/itertools.html#itertools-recipes roundrobin('ABC', 'D', 'EF') --> A D E B F C Recipe credited to George Sakkis """ nexts = cycle(iter(it).__next__ for it in iterables) pending = len(iterables) while pending: try: for nxt in nexts: yield nxt() except StopIteration: pending -= 1 nexts = cycle(islice(nexts, pending)) class NotIterable: """ Use this as mixin when creating a class which is not supposed to return true when iterable() is called on its instances because calling list() on the instance, for example, would result in an infinite loop. """ pass def iterable(i, exclude=(str, dict, NotIterable)): """ Return a boolean indicating whether ``i`` is SymPy iterable. True also indicates that the iterator is finite, e.g. you can call list(...) on the instance. When SymPy is working with iterables, it is almost always assuming that the iterable is not a string or a mapping, so those are excluded by default. If you want a pure Python definition, make exclude=None. To exclude multiple items, pass them as a tuple. You can also set the _iterable attribute to True or False on your class, which will override the checks here, including the exclude test. As a rule of thumb, some SymPy functions use this to check if they should recursively map over an object. If an object is technically iterable in the Python sense but does not desire this behavior (e.g., because its iteration is not finite, or because iteration might induce an unwanted computation), it should disable it by setting the _iterable attribute to False. See also: is_sequence Examples ======== >>> from sympy.utilities.iterables import iterable >>> from sympy import Tuple >>> things = [[1], (1,), set([1]), Tuple(1), (j for j in [1, 2]), {1:2}, '1', 1] >>> for i in things: ... print('%s %s' % (iterable(i), type(i))) True <... 'list'> True <... 'tuple'> True <... 'set'> True True <... 'generator'> False <... 'dict'> False <... 'str'> False <... 'int'> >>> iterable({}, exclude=None) True >>> iterable({}, exclude=str) True >>> iterable("no", exclude=str) False """ if hasattr(i, '_iterable'): return i._iterable try: iter(i) except TypeError: return False if exclude: return not isinstance(i, exclude) return True def is_sequence(i, include=None): """ Return a boolean indicating whether ``i`` is a sequence in the SymPy sense. If anything that fails the test below should be included as being a sequence for your application, set 'include' to that object's type; multiple types should be passed as a tuple of types. Note: although generators can generate a sequence, they often need special handling to make sure their elements are captured before the generator is exhausted, so these are not included by default in the definition of a sequence. See also: iterable Examples ======== >>> from sympy.utilities.iterables import is_sequence >>> from types import GeneratorType >>> is_sequence([]) True >>> is_sequence(set()) False >>> is_sequence('abc') False >>> is_sequence('abc', include=str) True >>> generator = (c for c in 'abc') >>> is_sequence(generator) False >>> is_sequence(generator, include=(str, GeneratorType)) True """ return (hasattr(i, '__getitem__') and iterable(i) or bool(include) and isinstance(i, include)) @deprecated( """ Using postorder_traversal from the sympy.utilities.iterables submodule is deprecated. Instead, use postorder_traversal from the top-level sympy namespace, like sympy.postorder_traversal """, deprecated_since_version="1.10", active_deprecations_target="deprecated-traversal-functions-moved") def postorder_traversal(node, keys=None): from sympy.core.traversal import postorder_traversal as _postorder_traversal return _postorder_traversal(node, keys=keys) @deprecated( """ Using interactive_traversal from the sympy.utilities.iterables submodule is deprecated. Instead, use interactive_traversal from the top-level sympy namespace, like sympy.interactive_traversal """, deprecated_since_version="1.10", active_deprecations_target="deprecated-traversal-functions-moved") def interactive_traversal(expr): from sympy.interactive.traversal import interactive_traversal as _interactive_traversal return _interactive_traversal(expr) @deprecated( """ Importing default_sort_key from sympy.utilities.iterables is deprecated. Use from sympy import default_sort_key instead. """, deprecated_since_version="1.10", active_deprecations_target="deprecated-sympy-core-compatibility", ) def default_sort_key(*args, **kwargs): from sympy import default_sort_key as _default_sort_key return _default_sort_key(*args, **kwargs) @deprecated( """ Importing default_sort_key from sympy.utilities.iterables is deprecated. Use from sympy import default_sort_key instead. """, deprecated_since_version="1.10", active_deprecations_target="deprecated-sympy-core-compatibility", ) def ordered(*args, **kwargs): from sympy import ordered as _ordered return _ordered(*args, **kwargs)