ORTools CP-SAT Solver. Constraint to require two lists of variables to be drawn from the same set of integers

Question

I have two lists of variables M and T. I would like to create a constraint where the set of unique values between M and T are identical.

From the solution variables I would like:

set(T) == set(M) -> True

So far, I have tried creating a matrix of the differences between each element of M and T

diffs = M[:, None] - T

Then building a constraint that the product of each element in individual row and every individual column of diffs is 0. This should ensure that each element of M has an element in T that it is equal to and vice versa.

for m in range(num_groups):
    model.AddMultiplicationEquality(0, diffs[m, :])
for t in range(num_groups):
    model.AddMultiplicationEquality(0, diffs[:, t])

Upon solving this model with no objective or other constraints I receive an invalid status. I am new to ORTools.

Full Program

ortools==9.3.10497

from ortools.sat.python import cp_model
import numpy as np

model = cp_model.CpModel()
num_groups = 4
STATUS  = ['UNKNOWN', 'MODEL_INVALID', 'FEASIBLE', 'INFEASIBLE', 'OPTIMAL']
M = []
for m in range(num_groups):
    M.append(model.NewIntVar(lb=0, ub=num_groups, name=f'M_{m}'))
T = []
for t in range(num_groups):
    T.append(model.NewIntVar(lb=0, ub=num_groups, name=f'T_{t}'))
M = np.array(M)
T = np.array(T)


diffs = M[:, None] - T
for m in range(num_groups):
    model.AddMultiplicationEquality(0, diffs[m, :])
for tt in range(num_groups):
    model.AddMultiplicationEquality(0, diffs[:, tt])

solver = cp_model.CpSolver()
status = solver.Solve(model)
print(STATUS[status])

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community, Apr 22 '22 at 01:58
It's perfectly clear what you're asking to those who use OR-Tools. Would it be possible to formulate your problem so that you don't need two lists, but use the variables from the first list for both contexts? That would automatically enforce your requirement without any additional overhead. — Christopher Hamkins, Apr 22 '22 at 07:19

Laurent Perron · Accepted Answer · 2022-04-28T20:58:17.573

This is catastrophic. Multiplications are very slow.

just create one Boolean variable per value, one Boolean variable per value and per variable.

Now, you want to link everything together.

for a variable x in either set and a value v:

    assign = {}
    assign[x, v] = model.NewBoolVar(f'assign_{x}_{v}')
    model.Add(x == v).OnlyEnforceIf(assign[x, v])
    model.Add(x != v).OnlyEnforceIf(assign[x, v].Not())

for a value v:

    appears = {}
    appears[v] = model.NewBoolVar(f'appears_{v}')

    # If v does not appear, no variable can be assigned its value.
    for x in T U M:
        model.AddImplication(appears[v].Not(), assign[x, v].Not())

    # If v appears, at least one variable in both T and M must be assigned to v.
    model.AddBoolOr(assign[x, v] for x in T).OnlyEnforceIf(appears[v])
    model.AddBoolOr(assign[x, v] for x in M).OnlyEnforceIf(appears[v])

Thank you for your help. This solved my problem and helped me get a better handle on the design patterns using OR Tools. Only comment for future readers is within the T U M for loop, the second argument of AddImplication should be `assign[x, v].Not()` — Patrick Bray, Apr 28 '22 at 19:18

ORTools CP-SAT Solver. Constraint to require two lists of variables to be drawn from the same set of integers

1 Answers1