Combinations¶
AUTHORS:
Mike Hansen (2007): initial implementation
Vincent Delecroix (2011): cleaning, bug corrections, doctests
Antoine Genitrini (2020) : new implementation of the lexicographic unranking of combinations
- class sage.combinat.combination.ChooseNK(mset, k)[source]¶
Bases:
Combinations_setk
- sage.combinat.combination.Combinations(mset, k=None)[source]¶
Return the combinatorial class of combinations of the multiset
mset. Ifkis specified, then it returns the combinatorial class of combinations ofmsetof sizek.A combination of a multiset \(M\) is an unordered selection of \(k\) objects of \(M\), where every object can appear at most as many times as it appears in \(M\).
The combinatorial classes correctly handle the cases where
msethas duplicate elements.EXAMPLES:
sage: C = Combinations(range(4)); C Combinations of [0, 1, 2, 3] sage: C.list() [[], [0], [1], [2], [3], [0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3], [0, 1, 2], [0, 1, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]] sage: C.cardinality() 16
>>> from sage.all import * >>> C = Combinations(range(Integer(4))); C Combinations of [0, 1, 2, 3] >>> C.list() [[], [0], [1], [2], [3], [0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3], [0, 1, 2], [0, 1, 3], [0, 2, 3], [1, 2, 3], [0, 1, 2, 3]] >>> C.cardinality() 16
sage: C2 = Combinations(range(4),2); C2 Combinations of [0, 1, 2, 3] of length 2 sage: C2.list() [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]] sage: C2.cardinality() 6
>>> from sage.all import * >>> C2 = Combinations(range(Integer(4)),Integer(2)); C2 Combinations of [0, 1, 2, 3] of length 2 >>> C2.list() [[0, 1], [0, 2], [0, 3], [1, 2], [1, 3], [2, 3]] >>> C2.cardinality() 6
sage: Combinations([1,2,2,3]).list() [[], [1], [2], [3], [1, 2], [1, 3], [2, 2], [2, 3], [1, 2, 2], [1, 2, 3], [2, 2, 3], [1, 2, 2, 3]]
>>> from sage.all import * >>> Combinations([Integer(1),Integer(2),Integer(2),Integer(3)]).list() [[], [1], [2], [3], [1, 2], [1, 3], [2, 2], [2, 3], [1, 2, 2], [1, 2, 3], [2, 2, 3], [1, 2, 2, 3]]
sage: Combinations([1,2,3], 2).list() [[1, 2], [1, 3], [2, 3]]
>>> from sage.all import * >>> Combinations([Integer(1),Integer(2),Integer(3)], Integer(2)).list() [[1, 2], [1, 3], [2, 3]]
sage: mset = [1,1,2,3,4,4,5] sage: Combinations(mset,2).list() [[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 3], [2, 4], [2, 5], [3, 4], [3, 5], [4, 4], [4, 5]]
>>> from sage.all import * >>> mset = [Integer(1),Integer(1),Integer(2),Integer(3),Integer(4),Integer(4),Integer(5)] >>> Combinations(mset,Integer(2)).list() [[1, 1], [1, 2], [1, 3], [1, 4], [1, 5], [2, 3], [2, 4], [2, 5], [3, 4], [3, 5], [4, 4], [4, 5]]
sage: mset = ["d","a","v","i","d"] sage: Combinations(mset,3).list() [['d', 'd', 'a'], ['d', 'd', 'v'], ['d', 'd', 'i'], ['d', 'a', 'v'], ['d', 'a', 'i'], ['d', 'v', 'i'], ['a', 'v', 'i']]
>>> from sage.all import * >>> mset = ["d","a","v","i","d"] >>> Combinations(mset,Integer(3)).list() [['d', 'd', 'a'], ['d', 'd', 'v'], ['d', 'd', 'i'], ['d', 'a', 'v'], ['d', 'a', 'i'], ['d', 'v', 'i'], ['a', 'v', 'i']]
sage: X = Combinations([1,2,3,4,5],3) sage: [x for x in X] [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]
>>> from sage.all import * >>> X = Combinations([Integer(1),Integer(2),Integer(3),Integer(4),Integer(5)],Integer(3)) >>> [x for x in X] [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]
It is possible to take combinations of Sage objects:
sage: Combinations([vector([1,1]), vector([2,2]), vector([3,3])], 2).list() # needs sage.modules [[(1, 1), (2, 2)], [(1, 1), (3, 3)], [(2, 2), (3, 3)]]
>>> from sage.all import * >>> Combinations([vector([Integer(1),Integer(1)]), vector([Integer(2),Integer(2)]), vector([Integer(3),Integer(3)])], Integer(2)).list() # needs sage.modules [[(1, 1), (2, 2)], [(1, 1), (3, 3)], [(2, 2), (3, 3)]]
- class sage.combinat.combination.Combinations_msetk(mset, k)[source]¶
Bases:
Parent- cardinality()[source]¶
Return the size of combinations(mset, k).
IMPLEMENTATION: Wraps GAP’s NrCombinations.
EXAMPLES:
sage: mset = [1,1,2,3,4,4,5] sage: Combinations(mset,2).cardinality() # needs sage.libs.gap 12
>>> from sage.all import * >>> mset = [Integer(1),Integer(1),Integer(2),Integer(3),Integer(4),Integer(4),Integer(5)] >>> Combinations(mset,Integer(2)).cardinality() # needs sage.libs.gap 12
- class sage.combinat.combination.Combinations_set(mset)[source]¶
Bases:
Combinations_mset- cardinality()[source]¶
Return the size of Combinations(set).
EXAMPLES:
sage: Combinations(range(16000)).cardinality() == 2^16000 True
>>> from sage.all import * >>> Combinations(range(Integer(16000))).cardinality() == Integer(2)**Integer(16000) True
- class sage.combinat.combination.Combinations_setk(mset, k)[source]¶
Bases:
Combinations_msetk- cardinality()[source]¶
Return the size of combinations(set, k).
EXAMPLES:
sage: Combinations(range(16000), 5).cardinality() 8732673194560003200
>>> from sage.all import * >>> Combinations(range(Integer(16000)), Integer(5)).cardinality() 8732673194560003200
- list()[source]¶
EXAMPLES:
sage: Combinations([1,2,3,4,5],3).list() [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]
>>> from sage.all import * >>> Combinations([Integer(1),Integer(2),Integer(3),Integer(4),Integer(5)],Integer(3)).list() [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]
- sage.combinat.combination.from_rank(r, n, k)[source]¶
Return the combination of rank
rin the subsets ofrange(n)of sizekwhen listed in lexicographic order.The algorithm used is based on factoradics and presented in [DGH2020]. It is there compared to the other from the literature.
EXAMPLES:
sage: import sage.combinat.combination as combination sage: combination.from_rank(0,3,0) () sage: combination.from_rank(0,3,1) (0,) sage: combination.from_rank(1,3,1) (1,) sage: combination.from_rank(2,3,1) (2,) sage: combination.from_rank(0,3,2) (0, 1) sage: combination.from_rank(1,3,2) (0, 2) sage: combination.from_rank(2,3,2) (1, 2) sage: combination.from_rank(0,3,3) (0, 1, 2)
>>> from sage.all import * >>> import sage.combinat.combination as combination >>> combination.from_rank(Integer(0),Integer(3),Integer(0)) () >>> combination.from_rank(Integer(0),Integer(3),Integer(1)) (0,) >>> combination.from_rank(Integer(1),Integer(3),Integer(1)) (1,) >>> combination.from_rank(Integer(2),Integer(3),Integer(1)) (2,) >>> combination.from_rank(Integer(0),Integer(3),Integer(2)) (0, 1) >>> combination.from_rank(Integer(1),Integer(3),Integer(2)) (0, 2) >>> combination.from_rank(Integer(2),Integer(3),Integer(2)) (1, 2) >>> combination.from_rank(Integer(0),Integer(3),Integer(3)) (0, 1, 2)
- sage.combinat.combination.rank(comb, n, check=True)[source]¶
Return the rank of
combin the subsets ofrange(n)of sizekwherekis the length ofcomb.The algorithm used is based on combinadics and James McCaffrey’s MSDN article. See: Wikipedia article Combinadic.
EXAMPLES:
sage: import sage.combinat.combination as combination sage: combination.rank((), 3) 0 sage: combination.rank((0,), 3) 0 sage: combination.rank((1,), 3) 1 sage: combination.rank((2,), 3) 2 sage: combination.rank((0,1), 3) 0 sage: combination.rank((0,2), 3) 1 sage: combination.rank((1,2), 3) 2 sage: combination.rank((0,1,2), 3) 0 sage: combination.rank((0,1,2,3), 3) Traceback (most recent call last): ... ValueError: len(comb) must be <= n sage: combination.rank((0,0), 2) Traceback (most recent call last): ... ValueError: comb must be a subword of (0,1,...,n) sage: combination.rank([1,2], 3) 2 sage: combination.rank([0,1,2], 3) 0
>>> from sage.all import * >>> import sage.combinat.combination as combination >>> combination.rank((), Integer(3)) 0 >>> combination.rank((Integer(0),), Integer(3)) 0 >>> combination.rank((Integer(1),), Integer(3)) 1 >>> combination.rank((Integer(2),), Integer(3)) 2 >>> combination.rank((Integer(0),Integer(1)), Integer(3)) 0 >>> combination.rank((Integer(0),Integer(2)), Integer(3)) 1 >>> combination.rank((Integer(1),Integer(2)), Integer(3)) 2 >>> combination.rank((Integer(0),Integer(1),Integer(2)), Integer(3)) 0 >>> combination.rank((Integer(0),Integer(1),Integer(2),Integer(3)), Integer(3)) Traceback (most recent call last): ... ValueError: len(comb) must be <= n >>> combination.rank((Integer(0),Integer(0)), Integer(2)) Traceback (most recent call last): ... ValueError: comb must be a subword of (0,1,...,n) >>> combination.rank([Integer(1),Integer(2)], Integer(3)) 2 >>> combination.rank([Integer(0),Integer(1),Integer(2)], Integer(3)) 0