## Coincidence count

Topics: Array Data Structure

Tools: Why3

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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 array.Array
use ref.Refint
use set.Fsetint
use set.FsetComprehension

function setof (a: array 'a) : set 'a =
map (fun x -> a[x]) (interval 0 (length a))

function drop (a: array 'a) (n: int) : set 'a =
map (fun x -> a[x]) (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
```