# An application of the Xpress callbacks to a Benders decomposition algorithm.
#
# Solves the problem:
# min c1*x + c2*y
# st  A1*x + A2*y <=b (m constraints)
# x binary n1-dimensional vector
# y >=0 continuous n2-dimensional vector
#
# NOTE: This example is for illustration/tutorial purposes only. There is no
# benefit in solving the generic problem defined below using Benders
# decomposition.
# Courtesy of Georgios Patsakis (UC Berkeley, Amazon) and Richard L.-Y. Chen (Amazon).
#
# (C) 2019-2025 Fair Isaac Corporation

# Import necessary packages.
import xpress as xp
import sys

# Initialize problem parameters.
c1 = [1,6,5,7]                    # n1 x 1
c2 = [9,3,0,2,3]                  # n2 x 1
b  = [-3,-4,1,4,5]                # m  x 1
A1 = [[0, -2, 3, 2],
      [-5, 0, -3, 1],
      [1, 0, 4, -2],
      [0, -3, 4, -1],
      [-5, -4, 3, 0]]
A2 = [[3, 4, 2, 0, -5],
      [0, 2, 3, -2, 1],
      [2, 0, 1, -3, -5],
      [-5, 3, -2, -3, 0],
      [-2, 3, -1, 2, -4]]
m  = len(b)
n1 = len(c1)
n2 = len(c2)

# Absolute accuracy for optimal objective.
ObjAbsAccuracy=0.00001

# SOLVE ORIGINAL PROBLEM WITHOUT DECOMPOSING
# Define Problem.
p = xp.problem()

# Define Variables.
x = [p.addVariable(vartype=xp.binary) for _ in range(n1)]
y = [p.addVariable() for _ in range(n2)] # positive real variable

# Define Constraints.
constr = [xp.Sum(A1[ii][jj]*x[jj] for jj in range(n1)) + \
          xp.Sum(A2[ii][jj]*y[jj] for jj in range(n2))   \
                        <= b[ii] for ii in range(m)]
p.addConstraint(constr)

# Define Objective.
p.setObjective(xp.Sum(c1[jj]*x[jj] for jj in range(n1)) + \
               xp.Sum(c2[jj]*y[jj] for jj in range(n2)) , \
               sense=xp.minimize)

# Solve and retrieve solution.
p.optimize()

if p.attributes.solvestatus != xp.SolveStatus.COMPLETED or \
   p.attributes.solstatus != xp.SolStatus.OPTIMAL:
    raise RuntimeError('Problem could not be solved to MIP optimality')

yopt = p.getSolution(y)
print("The optimal objective is:", p.attributes.objval)
print("The first stage variables are:", p.getSolution(x))
print("The second stage variables are:", yopt)
print("The objective of the second stage is:",
       sum(c2[jj]*yopt[jj] for jj in range(n2)))

# SOLVE DECOMPOSED PROBLEM
# We define the following functions:
# - subproblem: function that solves the subproblem for a given first stage
#   variable xhat. If the subproblem has an optimal solution, it returns the
#   the optimal dual multiplier associated with the non anticipativaty
#   constraint, the optimal objective, and the label "Optimal". If the
#   subproblem is infeasible, it returns the dual ray of unboundedness and
#   the label "Infeasible".
#
# - integer_callback: An integer callback is triggered every time an integer
#   solution is found. The function integer_callback is an input to the
#   addPreIntsolCallback function and has a specified input - output format based
#   on the Xpress documentation. It takes as input the main optimization
#   problem name, a user data structure, a variable that indicates whether
#   the integer solution comes from a heuristic or from a node relaxation,
#   and the proposed update to the upper bound if this integer solution is
#   accepted as feasible. The inputs are populated by the solver when the
#   callback is triggered. The function returns whether the integer solution
#   found should be accepted as feasible and the proposed update to the upper
#   bound if the solution is feasible (which could be the same as the one
#   proposed by the solver as part of the input).

# Subproblem Solver Function.
def subproblem(xhat):
    # Input:
    # - xhat: n1x1a array is the first stage variable,
    # passed to the subproblem.
    #
    # Output:
    # return (lamb: n1x1 array, beta: scalar, flag:string)
    # - If the subproblem is Infeasible, the flag is 'Infeasible', and
    #   lamb (n1x1) and beta (1x1) are the components of the dual ray of
    #   unboundedness necessary to write a feasibility cut of the form:
    #       sum(x[i]*lamb[i] for all i) +beta >= 0
    # - If the subproblem is Optimal, the flag is 'Optimal', beta (1x1) is the
    #   optimal objective, and lamb (n1x1) is the optimal dual multiplier
    #   associated with the non anticipativaty constraint, so that the
    #   optimality cut will have the form:
    #       theta >= sum(x[i]*lamb[i] for all i) + beta

    # IMPORTANT NOTE: The subproblem needs to ensure tha upper bounds or non
    # zero lower bounds should ONLY BE DEFINED THROUGH CONSTRAINTS, not as part
    # of the variable definition. This is to ensure the feasibility cuts are
    # correct.
    # Side Note: It is a good practice to ensure that the subproblem is always
    # feasible through modeling, by say adding slacks to all the constraints,
    # hence avoiding feasibility cuts completely. This is because feasibility
    # cuts tend to be extremely slow. In this code we implemented feasibility
    # cuts for reasons of completeness.

    # Define and Solve Subproblem.

    # Initialize Problem.
    r=xp.problem()

    # Define Variables.
    y=[r.addVariable() for _ in range(n2)] # Positive real variable - second stage.
    z=[r.addVariable(lb=-xp.infinity) for _ in range(n1)] # Dummy copy variable,
                                           # must have the exact shape as xhat.
    epsilon=r.addVariable(lb=-xp.infinity) # Dummy variable to extract part of the
                                           # infeasibility ray (if infeasible).

    # Define Constraints.
    # Dummy cosntraint for optimality/feasibility cuts.
    # Note: make sure the index of the variable and of the position in the
    # constraint array is the same (or has a known mapping).
    dummy1= [z[i]==xhat[i] for i in range(n1)]
    # Dummy constraint for feasibility cut.
    dummy2= epsilon==1
    # Second stage constraints, where RHS b has been multiplied by epsilon.
    constr=[xp.Sum(A1[ii][jj]*z[jj] for jj in range(n1)) + \
            xp.Sum(A2[ii][jj]*y[jj] for jj in range(n2))   \
            - epsilon*b[ii]<=0 for ii in range(m)]
    r.addConstraint(constr,dummy1,dummy2)

    # Define Objective.
    r.setObjective(xp.Sum(c2[jj]*y[jj] for jj in range(n2)),
            sense=xp.minimize)

    # Disable presolve. The only reason to do this is because in the case of
    # an infeasible problem, the presolve might identify infeasibility of the
    # problem without running simplex. Because of that, no dual infeasibility
    # certificate will be found, hence a dual ray of unboundedness will not be
    # returned, which means that a feasibility cut can not be formed. In
    # practice, we should avoid infeasibility cuts altogether through modeling
    # and have presolve on for a better performance in the subproblem. If an
    # infeasibility cut can not be avoided, formulating the dual explicitly is
    # the only option.
    r.controls.presolve = 0

    # Silence output.
    r.controls.outputlog = 0

    # Solve optimization.
    r.optimize()

    # Find the indices of constraint dummy1, which will be used to access the
    # dual multipliers corresponding to the entries of xhat.
    xind1=[dummy1[ii].index for ii in range(n1)]

    if r.attributes.solvestatus != xp.SolveStatus.COMPLETED:
        print("ERROR: Subproblem was not solved. Terminating.")
        sys.exit()

    # Take cases depending on subproblem status.
    if r.attributes.solstatus == xp.SolStatus.OPTIMAL:
        # Optimal subproblem.
        print("Optimal Subproblem")
        # Retrieve optimal dual multiplier and objective and return.
        dualmult=r.getDuals()
        lamb=[dualmult[ii] for ii in xind1]
        beta=r.attributes.objval
        return(lamb,beta,'Optimal')

    elif r.attributes.solstatus == xp.SolStatus.INFEAS:
        # Infeasible subproblem.
        print("Infeasible Subproblem")
        if not r.hasdualray ():
            print ("Could not retrieve a dual ray, return no good cut instead:")
            # This is just a cheat if the subproblem is found infeasible and
            # the optimizer fails to find a dual ray. Since all the first stage
            # variables are binary, we add a No-Good-Cut to exclude the point
            # xhat instead of an infeasibility cut.
            xhatones=set(ii for ii in range(n1) if xhat[ii]>=0.5)
            lamb=[2*xhat[ii]-1 for ii in range(n1)]
            beta=-sum(xhat)+1
        else:
            # Extract the dual ray.
            dray = r.getDualRay()
            print ("Dual Ray:", dray)
            # Extract the part of the dual ray corresponding to the first stage
            # variables xhat.
            lamb=[dray[ii] for ii in xind1]
            # Extract the constant from the dual ray entry of constraint dummy 2.
            beta=dray[dummy2.index]
        return(lamb,beta,'Infeasible')
    else:
        print("ERROR: Subproblem not optimal or infeasible. Terminating.")
        sys.exit()

# Integer Callback Function
def integer_callback(p, data, soltype, cutoff):
    # Input:
    # NOTE: the input is populated by the solver when an integer solution is
    # found.
    # - p: an xp.problem() (the main problem from which the callbacks were
    # triggered)
    # - data: structure that is passed to the callback from the
    #   problem, and is specified in the addPreIntsolCallback function. No
    #   data is needed here, so the addPreIntsolCallback() call uses None.
    # - soltype: 0 if the integer solution was found by a node relaxation,
    # and not 0 if the integer solution was found by a heuristic
    # - cutoff: the integer solution found by the solver is associated with an
    # upper bound, if this integer solution is feasible to full problem. cutoff
    # is the upper bound (slightly decreased) associated with that integer
    # solution. If the user decides that the integer solution is indeed feasible,
    # the user can modify that cutoff, if the default solver value misrepresents
    # the upper bound, and return a new cutoff (newcutoff)
    #
    # Output:
    # return (isreject, newcutoff)
    # - isreject: True to reject integer solution, False to accept it
    # - newcutoff: in case isreject is False, possible update to the cutoff.

    print("Entering Integer Callback.")
    if soltype ==0:
        # In this case the integer solution was found by a node relaxation of
        # the B&B tree, so we accept it as is. The node callback has taken
        # care of ensuring the solution is feasible to the subproblem.
        print("Integer solution from optimal node relaxation.")
        print("Solution accepted. Cutoff:", cutoff)
        return(False,None)
    else:
        print("Integer solution found from heuristic. Checking subproblem.")
        # In any other case, we need to solve the subproblem.

        # Retrieve xhat and thetahat from the heuristic solution.
        xhat = p.getCallbackSolution([x[ii] for ii in range(n1)])
        thetahat = p.getCallbackSolution(theta)
        print("Solution tested x=",xhat, "and theta=",thetahat)
        print("Cutoff is:",cutoff)

        # Solve the subproblem.
        (dual_mult,opt,status)=subproblem(xhat)

        if status=='Infeasible':
            # In the case of an infeasible subproblem, reject the solution found.
            return (True,None)
        elif status=='Optimal':
            if thetahat>=opt-ObjAbsAccuracy:
                # In this case the vector (xhat,thetahat) is feasible, so accept
                # it and accept the cutoff.
                print("Accepting pair x=",xhat,",theta=",thetahat )
                print("Accept new cutoff:",cutoff)
                return(False,None)
            else:
                # In this case the solution (xhat,thetahat) is infeasible so
                # reject it.
                return (True,None)
        else:
            print("We shouldn't reach this point.")
            print("Rejecting pair x=",xhat,",theta=",thetahat )
            return(True,None)

# Node Callback Function
def node_callback(p, data):
    # Input:
    # Note: the input is populated by the solver when the relaxation of a node
    # is solved
    # - p: the xp.problem corresponding to the main problem, from which the
    # callback is triggered.
    # - data: structure that is passed to the callback from the
    #   problem, and is specified in the addOptnodeCallback function. No
    #   data is needed here, so the addOptnodeCallback() call uses None.
    #
    # Output:
    # return 0

    print("We entered a node callback")

    # Obtain node relaxation solution.
    xind = [x[ii].index for ii in range(n1)]
    thetaind = [theta.index]
    xhat = p.getCallbackSolution([x[ii] for ii in range(n1)])
    thetahat = p.getCallbackSolution(theta)
    print("Solution tested x=",xhat, "and theta=",thetahat)

    # Now let's see if the solution satisfies the subproblem induced constraints.
    # Solve the subproblem.
    (dual_mult,opt,status)=subproblem(xhat)
    if status=='Infeasible':
        # Create feasibility cut to add to node.
        coefficients=dual_mult
        rhs=-opt
        print("Add cut. Cut stats:")
        print("coefficients:",coefficients)
        print("<=rhs:",rhs)
        print("column indices:",xind)
        # Add the cut (this will reject the point).
        p.addCuts([1],['L'],[rhs],[0,len(xhat)],xind,coefficients)
        print("Rejecting pair x=",xhat,",theta=",thetahat, "with a cut." )
        return 0
    elif status=='Optimal':
        if thetahat>=opt-ObjAbsAccuracy:
            # In this case the vector (xhat,thetahat) is feasible, so accept.
            print("Accepting pair x=",xhat,",theta=",thetahat )
            print("An integer callback will be triggered for that pair later.")
            return 0
        else:
            # In this case the solution (xhat,thetahat) is infeasible so reject
            # it and add an optimality cut.
            coefficients=[1]+[-dual_mult[ii] for ii in range(len(dual_mult))]
            rhs=opt-sum(dual_mult[ii]*xhat[ii] for ii in range(len(dual_mult)))
            print("Store cut. Cut stats:")
            print("coefficients:",coefficients)
            print(">=rhs:",rhs)
            print("column indices:",thetaind+xind)
            # Add the cut (this will reject the point).
            p.addCuts([1],['G'],[rhs],[0,len(xhat)+1],thetaind+xind,coefficients)
            print("Rejecting pair x=",xhat,",theta=",thetahat,"with a cut." )
            return 0
    else:
        print("ERROR: Subproblem not optimal or infeasible. Terminating.")
        sys.exit()

# Main Problem.

# Define main problem.
p=xp.problem()

# Define first stage variables.
x=[p.addVariable(vartype=xp.binary) for _ in range(n1)]
theta=p.addVariable() # (positive) second stage cost - change the default lower
                      # bound if second stage cost may not be positive.

# Define objective.
p.setObjective(xp.Sum(c1[jj]*x[jj] for jj in range(n1)) +  theta,
               sense=xp.minimize)

# Add Integer callback. i.e., when an integer solution is found, the function
# integer_callback will be called by the solver.
p.addPreIntsolCallback(integer_callback, None, 0)

# Add node_callback. i.e., when a node relaxation is solved, the function
# node_callback will be called by the solver.
p.addOptnodeCallback(node_callback, None, 0)

# Deactivate presolve. This is to ensure that the solver will not eliminate
# variables, so the indices we retrieve for the variables and the cuts we add
# correspond to/contain existing optimization variables. Presolve can be
# activated, but additional steps are required for that (presolving the cuts
# the same way as the rows of the matrix).
p.setControl({"presolve":0,"mippresolve":0,"symmetry":0})

# Solve the main problem.
p.optimize()

if p.attributes.solvestatus == xp.SolveStatus.COMPLETED and \
   p.attributes.solstatus == xp.SolStatus.OPTIMAL:

    # Print results.
    print("The optimal objective is:",p.attributes.objval)
    print("The first stage variables are:",p.getSolution(x))
else:
    print("Could not solve the problem")