
min_max(+Goal, ?Template, ?Solution, ?C)

   Find the solution of Goal that minimizes the maximum of elements of C,
and unify the minimized Template with Solution.



Arguments
   +Goal               A callable term.
   ?Template           A term containing all or some of Goal's variables
   ?Solution           Term to be unified with the minimized Template
   ?C                  A linear term or a list of linear terms.

Type
   library(fd)

Description
   If C is a linear term, a solution of the goal Goal is found that
   minimizes the value of C. If C is a list of linear terms, the returned
   solution minimizes the maximum value of terms in the list.  The solution
   is found using the branch and bound method; as soon as a partial
   solution is found that is worse than a previous solution, the search is
   abandoned and a new solution is searched for.  Every time a new better
   solution is found, the event 280 is raised, its default handler prints
   the current cost.


   Solutions will be unified with a copy of Template where the variables
   are replaced with their minimized values. Typically, the Template will
   contain all or a subset of Goal's variables.


   min_max/2 can be written in terms of min_max/4 as follows:



	min_max(Goal, Cost) :-
	    min_max(Goal, Goal, Goal, Cost).



Modules
   This predicate is sensitive to its module context (tool predicate, see @/2).

Fail Conditions
      Fails if there is no solution to Goal.



Resatisfiable
      No.

Examples
   
    % Find the minimal C and bind X to the corresponding value
    [eclipse]: X::1..3, C #= 3-X, min_max(indomain(X), X, X, C).
    Found a solution with cost 2
    Found a solution with cost 1
    Found a solution with cost 0
    X = 3
    C = 0
    yes.

    % Find the minimal C and don't bind anything
    [eclipse]: X::1..3, C #= 3-X, min_max(indomain(X), [], [], C).
    Found a solution with cost 2
    Found a solution with cost 1
    Found a solution with cost 0
    X = X{[1..3]}
    C = C{[0..2]}

    Delayed goals:
	    -3 + X{[1..3]} + C{[0..2]}#=0
    yes.

    % Find the minimal C and return it in MinC. Don't bind X or C.
    [eclipse]: X::1..3, C #= 3-X, min_max(indomain(X), C, MinC, C).
    Found a solution with cost 2
    Found a solution with cost 1
    Found a solution with cost 0
    X = X{[1..3]}
    MinC = 0
    C = C{[0..2]}

    Delayed goals:
	    -3 + X{[1..3]} + C{[0..2]}#=0
    yes.






See Also
   min_max / 2, min_max / 5, min_max / 6, min_max / 8, minimize / 2, minimize / 4, minimize / 5, minimize / 6, minimize / 8
