(** Coincidence count Exercise proposed by Rustan Leino at Dagstuhl seminar 14171, April 2014 You are given two sequences of integers, sorted in increasing order and without duplicate elements, and you count the number of elements that appear in both sequences (in linear time and constant space). Authors: Jean-Christophe FilliÃ¢tre (CNRS) Andrei Paskevich (UniversitÃ© Paris Sud) *) module CoincidenceCount use import array.Array use import ref.Refint use import set.Fsetint use import set.FsetComprehension function setof (a: array 'a) : set 'a = map (get a) (interval 0 (length a)) function drop (a: array 'a) (n: int) : set 'a = map (get a) (interval n (length a)) lemma drop_left: forall a: array 'a, n: int. 0 <= n < length a -> drop a n == add a[n] (drop a (n+1)) predicate increasing (a: array int) = forall i j: int. 0 <= i < j < length a -> a[i] < a[j] function cc (a b: array int) : int = cardinal (inter (setof a) (setof b)) lemma not_mem_inter_r: forall a: array int, i: int, s: set int. 0 <= i < length a -> not (mem a[i] s) -> inter (drop a i) s == inter (drop a (i+1)) s lemma not_mem_inter_l: forall a: array int, i: int, s: set int. 0 <= i < length a -> not (mem a[i] s) -> inter s (drop a i) == inter s (drop a (i+1)) let coincidence_count (a b: array int) : int requires { increasing a } requires { increasing b } ensures { result = cc a b } = let i = ref 0 in let j = ref 0 in let c = ref 0 in while !i < length a && !j < length b do invariant { 0 <= !i <= length a } invariant { 0 <= !j <= length b } invariant { !c + cardinal (inter (drop a !i) (drop b !j)) = cc a b } variant { length a + length b - !i - !j } if a[!i] < b[!j] then incr i else if a[!i] > b[!j] then incr j else begin assert { inter (drop a !i) (drop b !j) == add a[!i] (inter (drop a (!i+1)) (drop b (!j+1))) }; assert { not (mem a[!i] (drop a (!i+1))) }; incr i; incr j; incr c end done; !c end