from collections import defaultdict, OrderedDict from itertools import ( combinations, combinations_with_replacement, permutations, product, product as cartes ) import random from operator import gt from sympy.core import Basic # this is the logical location of these functions from sympy.core.compatibility import (as_int, is_sequence, iterable, ordered) from sympy.core.compatibility import default_sort_key # noqa: F401 import sympy from sympy.utilities.enumerative import ( multiset_partitions_taocp, list_visitor, MultisetPartitionTraverser) 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): """ Recursively denest iterable containers. >>> from sympy.utilities.iterables 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.core import Basic >>> class MyOp(Basic): ... pass ... >>> flatten([MyOp(1, MyOp(2, 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: reducible = lambda x: is_sequence(x, set) else: reducible = lambda x: 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[k] = [] for hi in how: if type(hi) is int: rv[k].extend(seq[i: i + hi]) i += hi else: n = sum(flatten(hi)) hi_type = type(hi) rv[k].append(hi_type(reshape(seq[i: i + n], hi)[0])) i += n rv[k] = container(rv[k]) return type(seq)(rv) def group(seq, multiple=True): """ Splits a sequence into a list of lists of equal, adjacent elements. Examples ======== >>> from sympy.utilities.iterables 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 not seq: return [] current, groups = [seq[0]], [] for elem in seq[1:]: if elem == current[-1]: current.append(elem) else: groups.append(current) current = [elem] groups.append(current) if multiple: return groups for i, current in enumerate(groups): groups[i] = (current[0], len(current)) return groups 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 """ rv = defaultdict(int) for s in seq: rv[s] += 1 return dict(rv) def postorder_traversal(node, keys=None): """ Do a postorder traversal of a tree. This generator recursively yields nodes that it has visited in a postorder fashion. That is, it descends through the tree depth-first to yield all of a node's children's postorder traversal before yielding the node itself. Parameters ========== node : sympy expression The expression to traverse. keys : (default None) sort key(s) The key(s) used to sort args of Basic objects. When None, args of Basic objects are processed in arbitrary order. If key is defined, it will be passed along to ordered() as the only key(s) to use to sort the arguments; if ``key`` is simply True then the default keys of ``ordered`` will be used (node count and default_sort_key). Yields ====== subtree : sympy expression All of the subtrees in the tree. Examples ======== >>> from sympy.utilities.iterables import postorder_traversal >>> from sympy.abc import w, x, y, z The nodes are returned in the order that they are encountered unless key is given; simply passing key=True will guarantee that the traversal is unique. >>> list(postorder_traversal(w + (x + y)*z)) # doctest: +SKIP [z, y, x, x + y, z*(x + y), w, w + z*(x + y)] >>> list(postorder_traversal(w + (x + y)*z, keys=True)) [w, z, x, y, x + y, z*(x + y), w + z*(x + y)] """ if isinstance(node, Basic): args = node.args if keys: if keys != True: args = ordered(args, keys, default=False) else: args = ordered(args) for arg in args: yield from postorder_traversal(arg, keys) elif iterable(node): for item in node: yield from postorder_traversal(item, keys) yield node def interactive_traversal(expr): """Traverse a tree asking a user which branch to choose. """ from sympy.printing import pprint RED, BRED = '\033[0;31m', '\033[1;31m' GREEN, BGREEN = '\033[0;32m', '\033[1;32m' YELLOW, BYELLOW = '\033[0;33m', '\033[1;33m' # noqa BLUE, BBLUE = '\033[0;34m', '\033[1;34m' # noqa MAGENTA, BMAGENTA = '\033[0;35m', '\033[1;35m'# noqa CYAN, BCYAN = '\033[0;36m', '\033[1;36m' # noqa END = '\033[0m' def cprint(*args): print("".join(map(str, args)) + END) def _interactive_traversal(expr, stage): if stage > 0: print() cprint("Current expression (stage ", BYELLOW, stage, END, "):") print(BCYAN) pprint(expr) print(END) if isinstance(expr, Basic): if expr.is_Add: args = expr.as_ordered_terms() elif expr.is_Mul: args = expr.as_ordered_factors() else: args = expr.args elif hasattr(expr, "__iter__"): args = list(expr) else: return expr n_args = len(args) if not n_args: return expr for i, arg in enumerate(args): cprint(GREEN, "[", BGREEN, i, GREEN, "] ", BLUE, type(arg), END) pprint(arg) print() if n_args == 1: choices = '0' else: choices = '0-%d' % (n_args - 1) try: choice = input("Your choice [%s,f,l,r,d,?]: " % choices) except EOFError: result = expr print() else: if choice == '?': cprint(RED, "%s - select subexpression with the given index" % choices) cprint(RED, "f - select the first subexpression") cprint(RED, "l - select the last subexpression") cprint(RED, "r - select a random subexpression") cprint(RED, "d - done\n") result = _interactive_traversal(expr, stage) elif choice in ['d', '']: result = expr elif choice == 'f': result = _interactive_traversal(args[0], stage + 1) elif choice == 'l': result = _interactive_traversal(args[-1], stage + 1) elif choice == 'r': result = _interactive_traversal(random.choice(args), stage + 1) else: try: choice = int(choice) except ValueError: cprint(BRED, "Choice must be a number in %s range\n" % choices) result = _interactive_traversal(expr, stage) else: if choice < 0 or choice >= n_args: cprint(BRED, "Choice must be in %s range\n" % choices) result = _interactive_traversal(expr, stage) else: result = _interactive_traversal(args[choice], stage + 1) return result return _interactive_traversal(expr, 0) 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(list(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 a generator of 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.utilities.iterables 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 yield from permutations(seq, n) else: if n == 0: yield () else: yield from 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.utilities.iterables 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: for k in range(len(seq) + 1): yield from subsets(seq, k, repetition) else: if not repetition: yield from combinations(seq, k) else: yield from 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 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 any(not s for s in 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 any(not s for s in 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: key = lambda value: 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 } 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 } 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/Connected_component_(graph_theory) .. [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 ''' if key is None: key = sympy.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) """ if g is None: if type(m) is dict: if n > sum(m.values()): return g = [[k, m[k]] for k in ordered(m)] else: m = list(m) if n > len(m): 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 if sum(v for k, v in g) < 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 """ if g is None: if type(m) is dict: 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 type(vector) is 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 partions 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 type(multiset) is 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: 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: http://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 pow = int(2**(ceil(log(n, 2)))) sum = 0 partition = [] while pow: if sum + pow <= n: partition.append(pow) sum += pow pow >>= 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.utilities.iterables import has_dups >>> from sympy import 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 uniq = set() return any(True for s in seq if s in uniq or uniq.add(s)) def has_variety(seq): """Return True if there are any different elements in ``seq``. Examples ======== >>> from sympy.utilities.iterables 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] 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] http://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 generate_derangements(perm): """ Routine to generate unique derangements. TODO: This will be rewritten to use the ECO operator approach once the permutations branch is in master. 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, 3])) [[1, 0, 3, 2], [1, 2, 3, 0], [1, 3, 0, 2], [2, 0, 3, 1], \ [2, 3, 0, 1], [2, 3, 1, 0], [3, 0, 1, 2], [3, 2, 0, 1], \ [3, 2, 1, 0]] >>> list(generate_derangements([0, 1, 1])) [] See Also ======== sympy.functions.combinatorial.factorials.subfactorial """ for p in multiset_permutations(perm): if not any(i == j for i, j in zip(perm, p)): yield 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] http://mathworld.wolfram.com/Necklace.html """ return uniq(minlex(i, directed=not free) for i in variations(list(range(k)), n, repetition=True)) 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): """ Return the rotation of the sequence in which the lexically smallest elements appear first, e.g. `cba ->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 -> 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') """ if key is None: key = sympy.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 kbins(l, k, ordered=None): """ Return sequence ``l`` partitioned into ``k`` bins. Examples ======== >>> from __future__ import print_function 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 """ def partition(lista, bins): # EnricoGiampieri's partition generator from # https://stackoverflow.com/questions/13131491/ # partition-n-items-into-k-bins-in-python-lazily if len(lista) == 1 or bins == 1: yield [lista] elif len(lista) > 1 and bins > 1: for i in range(1, len(lista)): for part in partition(lista[i:], bins - 1): if len([lista[:i]] + part) == bins: yield [lista[:i]] + part if ordered is None: yield from partition(l, k) elif ordered == 11: for pl in multiset_permutations(l): pl = list(pl) yield from partition(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 cartes(*[(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.utilities.iterables 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/2/library/itertools.html#recipes roundrobin('ABC', 'D', 'EF') --> A D E B F C Recipe credited to George Sakkis """ import itertools nexts = itertools.cycle(iter(it).__next__ for it in iterables) pending = len(iterables) while pending: try: for next in nexts: yield next() except StopIteration: pending -= 1 nexts = itertools.cycle(itertools.islice(nexts, pending))