FoVeOOS'11 Competition: challenge 2, in Java
Maximum of a tree of integers, in Java
Authors: Claude Marché
Topics: Trees
References: The COST FoVeOOS'11 Competition
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//@+ TerminationPolicy = user
/*@ axiomatic integer_max {
@ logic integer max(integer x, integer y);
@ axiom max_is_ge : \forall integer x y; max(x,y) >= x && max(x,y) >= y;
@ axiom max_is_some : \forall integer x y; max(x,y) == x || max(x,y) == y;
@ }
@*/
class Int {
//@ ensures \result == max(x,y);
public static int max(int x, int y);
}
/*@ axiomatic Mem {
@ predicate mem{L}(int x, Tree t);
@ axiom mem_null{L}: \forall int x; ! mem(x,null);
@ axiom mem_root{L}: \forall Tree t; t != null ==>
@ mem(t.value,t);
@ axiom mem_root_eq{L}: \forall int x, Tree t; t != null ==>
@ x==t.value ==> mem(x,t);
@ axiom mem_left{L}: \forall int x, Tree t; t != null ==>
@ mem(x,t.left) ==> mem(x,t);
@ axiom mem_right{L}: \forall int x, Tree t; t != null ==>
@ mem(x,t.right) ==> mem(x,t);
@ axiom mem_inversion{L}: \forall int x, Tree t;
@ mem(x,t) ==> t != null &&
@ (x==t.value || mem(x,t.left) || mem(x,t.right));
@ }
@*/
/* attempt to prove termination, not succesful yet */
/* axiomatic Finite {
@ predicate has_size{L}(Tree t, integer s);
@ axiom has_size_null{L}: has_size(null,0);
@ axiom has_size_non_null{L}: \forall Tree t; t != null ==>
@ \forall integer s1 s2;
@ has_size(t.left,s1) && has_size(t.right,s2) ==>
@ has_size(t,s1+s2+1) ;
@ axiom has_size_inversion{L}: \forall Tree t, integer s;
@ has_size(t,s) ==>
@ (t == null && s == 0) ||
@ (t != null && \exists integer s1 s2;
@ has_size(t.left,s1) && has_size(t.right,s2) &&
@ 0 <= s1 && 0 <= s2 && s == s1+s2+1) ;
@ predicate size_decreases{L}(Tree t1, Tree t2) =
@ \exists integer s1 s2; has_size(t1,s1) && has_size(t2,s2) && s1 > s2;
@ }
@*/
class Tree {
int value;
Tree left;
Tree right;
/*@ // requires \exists integer s; has_size(this,s);
@ // decreases this for size_decreases;
@ ensures mem(\result,this) &&
@ \forall int x; mem(x,this) ==> \result >= x;
@*/
int tree_max() {
int m = value;
if (left != null) m = Int.max(m,left.tree_max());
if (right != null) m = Int.max(m,right.tree_max());
return m;
}
}