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Some tips and tricks for the CP-SAT solver.

Resources:

  • https://github.com/or-tools/awesome_or-tools
  • https://stackoverflow.com/questions/tagged/or-tools

Solver Parameters

By default, OR-tools will try to use all available cores but you can set num_search_workers manually.

solver.parameters.num_search_workers = 8

If the program silently crashes you can enable logging to get more info.

solver.parameters.log_search_progress = True

You can also see which worker is better for your problem and improve the single core performance by setting the same parameters.

Search for all optimal solutions

You can’t enumerate all solutions if you have an objective, so you have to do it in two steps:

# Get the optimal objective value
model.Maximize(objective)
solver.Solve(model)

# Set the objective to a fixed value
# use round() instead of int()
model.Add(objective == round(solver.ObjectiveValue()))
model.Proto().ClearField('objective')

# Search for all solutions
solver.parameters.enumerate_all_solutions = True
solver.Solve(model, cp_model.VarArraySolutionPrinter([x, y, z]))

Non-contiguous Intvar

Alternatives to model.NewIntVar(lb, ub, name), solving might be faster:

# List of values
model.NewIntVarFromDomain(
    cp_model.Domain.FromValues([1, 3, 4, 6]), 'x'
)

# List of intervals
model.NewIntVarFromDomain(
    cp_model.Domain.FromIntervals([[1, 2], [4, 6]]), 'x'
)

# Exclude [-1, 1]
model.NewIntVarFromDomain(
    cp_model.Domain.FromIntervals([[cp_model.INT_MIN, -2], [2, cp_model.INT_MAX]]), 'x'
)

# Constant
model.NewConstant(154)

Domains

Working with Domains gives a lot of flexibility and should be faster than creating multiple constraints.

domain = cp_model.Domain(0, 10)
domain = cp_model.Domain.FromValues([1, 3, 4, 6])
domain = cp_model.Domain.FromIntervals([[1, 2], [4, 6]])
domain = cp_model.Domain.FromIntervals([
    [cp_model.INT_MIN, -2],
    [2, cp_model.INT_MAX]
])

model.AddLinearExpressionInDomain(sum(my_vars), domain).OnlyEnforceIf(b)

Iff, equivalence, boolean product

p <=> (x and y)

# (x and y) => p, rewrite as not(x and y) or p
model.AddBoolOr([x.Not(), y.Not(), p])

# p => (x and y)
model.AddImplication(p, x)
model.AddImplication(p, y)

Boolean sum, boolean or

p <=> (x or y)

# p => (x or y), rewrite as not(p) or x or y
model.AddBoolOr([x, y, p.Not()])

# (x or y) => p
model.AddImplication(x, p)
model.AddImplication(y, p)

Note: see De Morgan’s laws

Boolean Implications

# a => b (both booleans)
model.AddImplication(a, b)

# a <=> b (use only one of them if you can)
model.Add(a == b)

# a and b and c => d
model.Add(d == 1).OnlyEnforceIf([a, b, c])
or
model.AddBoolOr([a.Not(), b.Not(), c.Not(), d])

If-Then-Else

Using intermediate boolean variables.

b = model.NewBoolVar('b')

# Implement b == (x >= 5).
model.Add(x >= 5).OnlyEnforceIf(b)
model.Add(x < 5).OnlyEnforceIf(b.Not())

Solution hint / Warm start

It may speed up the search.

num_vals = 3
x = model.NewIntVar(0, num_vals - 1, 'x')
y = model.NewIntVar(0, num_vals - 1, 'y')
z = model.NewIntVar(0, num_vals - 1, 'z')

model.Add(x != y)

model.Maximize(x + 2 * y + 3 * z)

# Solution hinting: x <- 1, y <- 2
model.AddHint(x, 1)
model.AddHint(y, 2)

Storing Multi-index variables

I recommend using dictionary comprehensions:

employee_works_day = {
    (e, day): model.NewBoolVar(f"{e} works {day}")
    for e in employees
    for day in days
}

Variable product (Non linear)

You have to create an intermediate variable:

x = model.NewIntvar(0, 8, 'x')
y = model.NewIntvar(0, 5, 'y')
xy = model.NewIntVar(0, 8*5, 'xy')

model.AddMultiplicationEquality(xy, [x, y])

Black box function (Non linear)

As your last resort you can precalculate all the posible values and use an element constraint.

from ortools.sat.python import cp_model

model = cp_model.CpModel()
x = model.NewIntVar(0, 10, "")
two_to_the_x = model.NewIntVar(1, 2 ** 10, "")

precalculated = [2 ** i for i in range(11)]
model.AddElement(x, precalculated, two_to_the_x)

# test
model.Add(x == 3)

solver = cp_model.CpSolver()
solver.Solve(model)

print("x", solver.Value(x))
print("2**x", solver.Value(two_to_the_x))

Load model Proto

This allows you to use different machines.

from google.protobuf import text_format
from ortools.sat.python import cp_model

model = cp_model.CpModel()
a = model.NewIntVar(0, 10, "a")
b = model.NewIntVar(0, 10, "b")
model.Maximize(a + b)

new_model = cp_model.CpModel()
text_format.Parse(str(model), new_model.Proto())

solver = cp_model.CpSolver()
status = solver.Solve(new_model)

print(solver.StatusName(status))
print(solver.ObjectiveValue())

Circuit constraint (ordering)

Ordering the numbers from 1 to 10 so that we maximize the distance between between numbers:

from itertools import permutations
from ortools.sat.python import cp_model

model = cp_model.CpModel()
solver = cp_model.CpSolver()

literals = {}
# An arc is just a (int, int, BoolVar) tuple
all_arcs = []
nodes = range(1, 11)

for i in nodes:
    # We use 0 as a dummy nodes as we don't have an actual circuit
    literals[0, i] = model.NewBoolVar(f"0 -> {i}")  # start arc
    literals[i, 0] = model.NewBoolVar(f"{i} -> 0")  # end arc
    all_arcs.append([0, i, literals[0, i]])
    all_arcs.append([i, 0, literals[i, 0]])

for i, j in permutations(nodes, 2):
    # this booleans will be true if the arc is present
    literals[i, j] = model.NewBoolVar(f"{i} -> {j}")
    all_arcs.append([i, j, literals[i, j]])
# to make an arc optional, add the [i, i, True] loop

model.AddCircuit(all_arcs)
model.Maximize(sum(literals[i, j] * abs(i - j) for i, j in permutations(nodes, 2)))
solver.Solve(model)

node = 0
print(node, end="")
while True:
    for i in nodes:
        if i != node and solver.Value(literals[node, i]):
            print(f" -> {i}", end="")
            node = i
            break
    else:
        break
print(" -> 0")

Fairness, distribute items evenly

  • Maximize the minimum value
model = cp_model.CpModel()
n_tasks = 100
n_employees = 3

min_tasks = model.NewIntVar(0, n_tasks, "")
employee_tasks = [model.NewIntVar(0, n_tasks, "") for _ in range(n_employees)]
model.Add(sum(employee_tasks) == n_tasks)

model.AddMinEquality(min_tasks, employee_tasks)

model.Maximize(min_tasks)
solver = cp_model.CpSolver()
status = solver.Solve(model)

print([solver.Value(n) for n in employee_tasks])
  • Minimize delta to the average value
model = cp_model.CpModel()
n_tasks = 100
n_employees = 3
avg = n_tasks // n_employees

delta = model.NewIntVar(0, n_tasks, "")
employee_tasks = [model.NewIntVar(0, n_tasks, "") for _ in range(n_employees)]
model.Add(sum(employee_tasks) == n_tasks)

for i in range(n_employees):
    model.Add(employee_tasks[i] <= avg + delta)
    model.Add(employee_tasks[i] >= avg - delta)

model.Minimize(delta)
solver = cp_model.CpSolver()
status = solver.Solve(model)

print([solver.Value(n) for n in employee_tasks])

Multiobjective optimization

Two ways to achieve that:

  • Add a weight to each objective
  • Solve with the first objective, constraint the objective with the solution, hint and solve with the new objective.
from ortools.sat.python import cp_model

model = cp_model.CpModel()
solver = cp_model.CpSolver()
x = model.NewIntVar(0, 10, "x")
y = model.NewIntVar(0, 10, "y")

# Maximize x
model.Maximize(x)
solver.Solve(model)
print("x", solver.Value(x))
print("y", solver.Value(y))
print()

# Hint (speed up solving)
model.AddHint(x, solver.Value(x))
model.AddHint(y, solver.Value(y))

# Maximize y (and constraint prev objective)
model.Add(x == round(solver.ObjectiveValue()))  # use <= or >= if not optimal
model.Maximize(y)

solver.Solve(model)
print("x", solver.Value(x))
print("y", solver.Value(y))

Soft constraints

https://stackoverflow.com/a/66377562

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