
minimize(+Goal, ?Template, ?Solution, ?C, +Lower, +Upper, +Percent, +Timeout)

   Find the solution of Goal that minimizes the maximum of elements of C,
within the bounds set by Lower,Upper and Percent in time not longer than
Timeout.



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.
   +Lower              An integer.
   +Upper              An integer.
   +Percent            An integer.
   +Timeout            A number.

Type
   library(fd)

Description
   This is the most general version of the minimize predicate with all
   options.


   A solution of the goal Goal is found that minimizes the value of C. 
   The solution is found using the branch and bound method.  Whenever
   a better solution is found, the upper cost bound is tightened and
   the search for a better solution continues.


   The starting assumption is that the value to minimize is less than
   Upper and that any value less than Lower can be considered as a
   final solution.  Moreover, solutions whose minimized values are
   closer than Percent % are considered equal.  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.


   If Timeout is not zero, the predicate will stop after Timeout seconds
   and report the best solution it has found so far.  Calls with specified
   Timeout cannot be nested.


   All other variants of minimize can be written in terms of minimize/9, eg.



	minimize(Goal, Cost) :-
	    minint(Min), maxint(Max),
	    minimize(Goal, Goal, Goal, Cost, Min, Max, 0, 0).



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, minimize(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, minimize(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, minimize(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 / 4, min_max / 5, min_max / 6, min_max / 8, minimize / 2, minimize / 4, minimize / 5, minimize / 6, minimize_bound_check / 0
