Skip to main content

is_lex_leader

Function is_lex_leader 

Source
pub fn is_lex_leader(group: &[Perm], a: &[bool]) -> bool
Expand description

Is a the lexicographic leader of its orbit — a ≤_lex a∘g for every g? The semantic canonical test; the CNF predicate lex_leader_sbp accepts exactly these assignments.