Subsets satisfying a hereditary property¶

sage.combinat.subsets_hereditary.
subsets_with_hereditary_property
(f, X, max_obstruction_size=None, ncpus=1)¶ Return all subsets \(S\) of \(X\) such that \(f(S)\) is true.
The boolean function \(f\) must be decreasing, i.e. \(f(S)\Rightarrow f(S')\) if \(S'\subseteq S\).
This function is implemented to call \(f\) as few times as possible. More precisely, \(f\) will be called on all sets \(S\) such that \(f(S)\) is true, as well as on all inclusionwise minimal sets \(S\) such that \(f(S)\) is false.
The problem that this function answers is also known as the learning problem on monotone boolean functions, or as computing the set of winning coalitions in a simple game.
INPUT:
f
– a boolean function which takes as input a list of elements fromX
.X
– a list/iterable.max_obstruction_size
(integer) – if you know that there is a \(k\) such that \(f(S)\) is true if and only if \(f(S')\) is true for all \(S'\subseteq S\) with \(S'\leq k\), setmax_obstruction_size=k
. It may dramatically decrease the number of calls to \(f\). Set toNone
by default, meaning \(k=X\).ncpus
– number of cpus to use for this computation. Note that changing the value from \(1\) (default) to anything different enables parallel computations which can have a cost by itself, so it is not necessarily a good move. In some cases, however, it is a great move. Set toNone
to automatically detect and use the maximum number of cpus available.Note
Parallel computations are performed through the
parallel()
decorator. See its documentation for more information, in particular with respect to the memory context.
EXAMPLES:
Sets whose elements all have the same remainder mod 2:
sage: from sage.combinat.subsets_hereditary import subsets_with_hereditary_property sage: f = lambda x: (not x) or all(xx%2 == x[0]%2 for xx in x) sage: list(subsets_with_hereditary_property(f,range(4))) [[], [0], [1], [2], [3], [0, 2], [1, 3]]
Same, on two threads:
sage: sorted(subsets_with_hereditary_property(f,range(4),ncpus=2)) [[], [0], [0, 2], [1], [1, 3], [2], [3]]
One can use this function to compute the independent sets of a graph. We know, however, that in this case the maximum obstructions are the edges, and have size 2. We can thus set
max_obstruction_size=2
, which reduces the number of calls to \(f\) from 91 to 56:sage: num_calls=0 sage: g = graphs.PetersenGraph() sage: def is_independent_set(S): ....: global num_calls ....: num_calls+=1 ....: return g.subgraph(S).size()==0 sage: l1=list(subsets_with_hereditary_property(is_independent_set,g.vertices())) sage: num_calls 91 sage: num_calls=0 sage: l2=list(subsets_with_hereditary_property(is_independent_set,g.vertices(),max_obstruction_size=2)) sage: num_calls 56 sage: l1==l2 True