# Construct a problem from scratch with variables of various
# types. Adds indicator constraints and shows how to retrieve
# such data once it has been added to the problem using the
# API functions.
#
# (C) Fair Isaac Corp., 1983-2024
from __future__ import print_function
import xpress as xp
N = 40
S = range(N)
m = xp.problem("test restriction")
m.controls.miprelstop = 0
#
# All variables used in this example
#
v1 = m.addVariable(lb=0, ub=10, threshold=5, vartype=xp.continuous)
v2 = m.addVariable(lb=1, ub=7, threshold=5, vartype=xp.continuous)
v3 = m.addVariable(lb=5, ub=10, threshold=7, vartype=xp.semicontinuous)
v4 = m.addVariable(lb=1, ub=7, threshold=3, vartype=xp.semiinteger)
vb = m.addVariable(vartype=xp.integer, lb=0, ub=1)
y = [m.addVariable(name="y{0}".format(i), lb=0, ub=2*N) for i in S]
cc = xp.constraint(body=v1 - v2, lb=2, ub=15)
cc0 = xp.constraint(body=v1 + v2, lb=2, ub=15)
# Adds both y, a vector (list) of variables, and v1 and v2, two scalar
# variables.
m.addConstraint(cc)
# Indices of variables can be retrieved both using their name and
# their Python object.
print("index of y[0] from name: ", m.getIndexFromName(xp.names_column, "y0"))
print("index of y[0]: ", m.getIndex(y[0]))
# Indicator constraints consist of a tuple with a condition on a
# binary variable and a constraint).
ind1 = (vb == 1, v1 + v2 >= 6)
ind2 = (vb == 1, v1 + v3 >= 7)
# Adds the first indicator constraint
m.addIndicator(ind1)
# Adds another indicator constraint and the second one defined above
m.addIndicator((vb == 1, v1 + v3 <= 10), ind2)
print("get index: var v1 -->", m.getIndex(v1), "; con cc -->", m.getIndex(cc))
ii_inds = []
ii_comps = []
m.getindicators(ii_inds, ii_comps, 1, 3)
print("getind: ", ii_inds, ii_comps)
# objective overwritten at each setObjective()
m.setObjective(xp.Sum([i*y[i] for i in S]))
m.optimize()
# Retrieve a solution: first declare an empty string, then call the
# getmipsol() function to fill it up.
mipsol = []
m.getmipsol(mipsol)
s1 = m.getSolution(v1, v2, y[10:30]) # get a subset of the solutions
s2 = m.getSolution(S) # can get it with indices as well
print("v1: ", m.getSolution(v1),
", v2: ", m.getSolution(v2),
"; sol vector: ", m.getSolution(),
"; obj: ", m.getObjVal(),
sep="") # default separator between strings is " "
# Adds yet another constraint to the problem and saves it, then
# removes an SOS and saves another version
m.addConstraint((1.25 * v1 - 2.5*v2 + 4.3) * (3.1 * v2 - 2 * v1 - 5.2)
+ 72.5 * v1**2 + 73 * v2**2 <= 1950)
m.write("restriction", "lp")
m.optimize()
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