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. If- kis specified, then it returns the combinatorial class of combinations of- msetof size- k.- 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 of- range(n)of size- kwhen 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 of- range(n)of size- kwhere- kis the length of- comb.- 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