Commit e747314a authored by MARCHE Claude's avatar MARCHE Claude
Browse files

LCP: improved doc + separate module for test harness

parent 7cb45e8d
{h <code>}
{1 VerifyThis@fm2012 competition, Challenge 1: Longest Repeated Substring}
The following is the original description of the verification task,
reproduced verbatim. A <a
implementation in Java</a> (based on code by Robert Sedgewick and
Kevin Wayne) was also given. The Why3 implementation below follows the
Java implementation as faithfully as possible.
Longest Common Prefix (LCP) - 45 minutes
......@@ -21,22 +31,16 @@ Pseudocode:
int lcp(int[] a, int x, int y) {
int l = 0;
while (x+l<a.length && y+l<a.length && a[x+l]==a[y+l]) {
while (x+l&lt;a.length && y+l&lt;a.length && a[x+l]==a[y+l]) {
return l;
......@@ -61,7 +65,6 @@ sorted suffix array is:
......@@ -71,10 +74,7 @@ comparison on arrays, a sorting routine, and LCP.
The client code ( uses these to solve the LRS problem. Verify
that it does so correctly.
(Based on code by Robert Sedgewick and Kevin Wayne.)
{h </code>}
......@@ -95,12 +95,13 @@ use import array.Array
at respective positions [x] and [y] in array [a] are identical. In
other words, the array parts a[x..x+l-1] and a[y..y+l-1] are equal
predicate is_common_prefix (a:array int) (x y:int) (l:int) =
predicate is_common_prefix (a:array int) (x y l:int) =
0 <= l /\ x+l <= a.length /\ y+l <= a.length /\
(forall i:int. 0 <= i < l -> a[x+i] = a[y+i])
(* helps the proof of [lcp] (but not mandatory) and needed later for [le_trans] *)
lemma not_common_prefix_if_last_different:
(** This lemma helps for the proof of [lcp] (but is not mandatory) and
is needed later (for [le_trans]) *)
lemma not_common_prefix_if_last_char_are_different:
forall a:array int, x y:int, l:int.
0 <= l /\ x+l < a.length /\ y+l < a.length /\ a[x+l] <> a[y+l] ->
not is_common_prefix a x y (l+1)
......@@ -108,13 +109,13 @@ lemma not_common_prefix_if_last_different:
(** [is_longest_common_prefix a x y l] is true when [l] is the maximal
length such that prefixes at positions [x] and [y] in array [a]
are identical. *)
predicate is_longest_common_prefix (a:array int) (x y:int) (l:int) =
predicate is_longest_common_prefix (a:array int) (x y l:int) =
is_common_prefix a x y l /\
forall m:int. l < m -> not (is_common_prefix a x y m)
(* helps proving [lcp] (but not mandatory), and needed
for proving [lcp_sym] and [le_trans] lemmas, and the post of [compare]
function in the absurd case *)
(** This lemma helps for proving [lcp] (but again is not mandatory),
and is needed for proving [lcp_sym] and [le_trans] lemmas, and the
post-condition of [compare] function in the "absurd" case *)
lemma longest_common_prefix_succ:
forall a:array int, x y l:int.
is_common_prefix a x y l /\ not (is_common_prefix a x y (l+1)) ->
......@@ -136,7 +137,16 @@ let lcp (a:array int) (x y:int) : int
assert { not is_common_prefix a x y (!l+1) };
module LCP_test
(** test harness for lcp *)
use import array.Array
use import LCP
let test () =
let arr = Array.make 4 0 in
arr[0]<-1; arr[1]<-2; arr[2]<-2; arr[3]<-5;
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