Objective Landscapes for Constraint Programming

Abstract

This paper presents the concept of objective landscape in the context of Constraint Programming. An objective landscape is a light-weight structure providing some information on the relation between decision variables and objective values, that can be quickly computed once and for all at the beginning of the resolution and is used to guide the search. It is particularly useful on decision variables with large domains and with a continuous semantics, which is typically the case for time or resource quantity variables in scheduling problems. This concept was recently implemented in the automatic search of CP Optimizer and resulted in an average speed-up of about 50% on scheduling problems with up to almost 2 orders of magnitude for some applications.

Appendix: Proof of Proposition 3

Let $v \in D_i$, we want to prove that $L_i(v) = f^*_i(v)$. We distinguish 3 cases depending on the position of v with respect to $X^L_i(Z^L)$ and $X^U_i(Z^L)$.

Case 1: $v \in [X^L_i(Z^L),X^U_i(Z^L)]$

By definition of the landscape function we have $L_i(v)=Z^L$.

We first note that by definition of $Z^L$, for every objective value $z < Z^L$ as the propagation of $f(x)\le z$ fails, it means that $Z^L$ is a lower bound on $Z^*=\min _{x \in D} f(x)$.

As by definition of function $f^*_i$ we have $\forall v \in D_i, Z^* \le f^*_i(v)$. We can deduce: $L_i(v) ( = Z^L ) \le f^*_i(v)$.

We will now use the bound-consistency property of propagation to show that $f^*_i(X^L_i(Z^L)) \le Z^L$ and $f^*_i(X^U_i(Z^L)) \le Z^L$. By definition $X^L_i(Z^L)$ is the minimal value in the domain of $x_i$ after propagation of $f(x)\le Z^L$. Because propagation is bound-consistency on $x_i$, it means that there exist some x with $x_i=X^L_i(Z^L)$ such that $f(x) \le Z^L$ (otherwise, this value would have been filtered from the domain). Furthermore, by definition of function $f^*_i$, we have $f^*_i(X^L_i(Z^L)) \le f(x)$, thus $f^*_i(X^L_i(Z^L)) \le Z^L$. The proof that $f^*_i(X^U_i(Z^L)) \le Z^L$ is symmetrical. Finally, the quasiconvexity of $f^*_i$ implies that for all $v \in [X^L_i(Z^L),X^U_i(Z^L)]$, $f^*_i(v) \le Z^L (= L_i(v))$.

Case 2: $v < X^L_i(Z^L)$

By definition of the landscape we have $L_i(v)= \min {\{z | X^L_i(z) \le v\}}$.

We first prove that $f^*_i(v) \ge L_i(v)$. By contradiction, suppose that $f^*_i(v) < L_i(v)$ and let $z' = f^*_i(v)$. This would mean that there exist x with $x_i=v$ such that $f(x) = z'$. For such a $z'$ we would have $X^L_i(z') \le v$ because x supports objective value $z'$ and thus value v cannot be removed from the domain of $x_i$. Such a value $z' < L_i(v)$ such that $X^L_i(z') \le v$ contradict the fact $L_i(v)$ is the smallest z satisfying $X^L_i(z) \le v$. This proves $f^*_i(v) \ge L_i(v)$.

For proving that $f^*_i(v) \le L_i(v)$, we distinguish 2 sub-cases depending on whether or not there exist an objective value z such that $X^L_i(z) = v$.

Case 2.1: There exist a value z such that $X^L_i(z) = v$ so that $L_i(v)=z$.

By definition of $X^L_i$, this means that v is the minimal value in the domain of $x_i$ after propagating $f(x) \le z$. Because propagation is bound-consistent on $x_i$, it means that there exist some x with $x_i=v$ such that $f(x) \le z$ (otherwise, this value would have been filtered from the domain). Furthermore, by definition of function $f^*_i$, we have $f^*_i(v) \le f(x)$, thus $f^*_i(v) \le z (=L_i(v))$.

Case 2.2: There does not exist any value z such that $X^L_i(z) = v$.

This is the case of a discontinuity of function $X^L_i(z)$. Let $v^+ = X^L_i(L_i(v))$ denote the value of $X^L_i$ at the discontinuity. By definition of $L_i$, we have $L_i(v)=L_i(v^+)$. Furthermore, as $v^+$ satisfies case 2.1 above, we also know that $f^*_i(v^+) \le L_i(v^+)$. We have seen in case 1 that $f^*_i(X^L_i(Z^L)) \le Z^L$ and, of course, $Z^L \le L_i(v^+)$. To summarize: $v \in [X^L_i(Z^L), v^+]$, $f^*_i(X^L_i(Z^L)) \le L_i(v^+)$ and $f(v^+) \le L_i(v^+)$. Thus, by quasiconvexity of $f^*_i$: $f(v) \le L_i(v^+) (=L_i(v))$.

Case 3: $v > X^U_i(Z^L)$

This case is the symmetrical of Case 2.    $\square$