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Create truth table of two events

Answered
Huib Meulenbelt
Edited
  • Investigator
  • Gurobi-versary
  • Conversationalist

Hi all,

 

Let's assume there are two possible events in a fight that lasts r rounds between blue and red:

  • fight lasts over 1.5 rounds
  • red has more significant strikes in round 4

The corresponding truth table looks as follows. 

The events are compatible in every scenario except the scenario where the fight doesn't last at least 1.5 rounds. Logically, if the fight doesn't last more than 1.5 rounds, the fourth round never takes place.

I write the following model:

from gurobipy import Model, GRB


r_max = 6

r_p_min = 1.5
r_p_max = 6
r_q_min = 4
r_q_max = 4

m = Model()

p = m.addVar(vtype=GRB.BINARY)
q = m.addVar(vtype=GRB.BINARY)
r = m.addVar(vtype=GRB.INTEGER, lb=1, ub=r_max)

m.addGenConstrIndicator(p, True, r_p_min <= r)
m.addGenConstrIndicator(p, True, r <= r_p_max)

m.addGenConstrIndicator(q, True, r_q_min <= r)
m.addGenConstrIndicator(q, True, r <= r_q_max)

How should the constraints look like for the scenarios where p = 0 and/or q = 0?

The model should generate the following solutions

r = 1: p = 0 and q = 0
r = 2: p = 1 and q = 0
r = 3: p = 1 and q = 0
r = 4: p = 1 and q = 1
r = 5: p = 1 and q = 0
r = 6: p = 1 and q = 0

 

import itertools

truth_table = dict.fromkeys(itertools.product([1, 0], repeat=2), 0)

for s in range(m.SolCount):
    m.setParam(GRB.Param.SolutionNumber, s)
    truth_table[p.Xn, q.Xn] = 1

print(truth_table)
>> {(1, 1): 1, (1, 0): 1, (0, 1): 0, (0, 0): 1}

 

Thank you

 

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Comments

4 comments

Date Votes
  • Maliheh Aramon
    • Gurobi Staff Gurobi Staff

    Hi Huib, 

    How should the constraints look like for the scenarios where p = 0 and/or q = 0?

    Do you want to model the constraints below?

    • If p = 0, then q = 0 and r = 1

    If this is the case, you should add the following to your implementation:

    m.addGenConstrIndicator(p, False, q == 0)
m.addGenConstrIndicator(p, False, r == 1)


    Is this what you are looking for?

    Best regards,
    Maliheh

    0
  • Huib Meulenbelt
    • Investigator
    • Gurobi-versary
    • Conversationalist

    Hi,

    There are two scenarios for `p`

    If p = 1` then r_p_min ≤ r ≤ r_p_max  

    I created this constraint by writing:

    m.addGenConstrIndicator(p, True, r_p_min <= r)
m.addGenConstrIndicator(p, True, r <= r_p_max)

    Now, I want to model the constraints for the case where `p = 0`

    If p = 0` then r_p_min > r  OR  r > r_p_max  

    I tried the following, but this doesn't work

    m.addGenConstrIndicator(p, False, r_p_min > r)
m.addGenConstrIndicator(p, False, r > r_p_max)

    Please advice

     

    0
  • Maliheh Aramon
    • Gurobi Staff Gurobi Staff
    Edited

    I am still struggling to understand what p and q really mean in your optimization problem. It is hard for me to comprehend your problem as an optimization problem. 

    Just focusing on this constraint:

    If p = 0` then r_p_min > r  OR  r > r_p_max  

    Assuming \(r\), \(r_{p, min}\), and \(r_{p, max}\) are integer, one approach is to define two binary variables \(y_1\) and \(y_2\) and \(M\) as a sufficiently large value. Then write:

    \[\begin{align} r \leq (r_{p, min} - 1) + M (1 - y_1) & \notag \\ r  \geq  (r_{p, max} + 1) - M (1- y_2) & \notag \\ y_1 + y_2 \geq 1 - p & \notag \\ y_1 + y_2 \leq 1 + p \end{align}\] 

    If \(p = 0\), the last two constraints ensure \(y_1 + y_2 = 1\) meaning that exactly one of the \(y\)s equals 1. 

    • If \(y_1 = 1, y_2=0\), then \( r \leq (r_{p, min} - 1\)) is enforced and the constraint \(r  \geq  (r_{p, max} + 1) - M\) is redundant.
    • If \(y_1 = 0, y_2=1\), then \( r \leq (r_{p, min} -1) + M\) is redundant and \(r  \geq  (r_{p, max} + 1) \) is enforced. 

    Instead of the last two constraints, you can also use Gurobi's indicator constraint and implement \(\text{if}~ p = 0, \text{then}~ y_1 + y_2 = 1\).
     

    Gurobi has an AI tool called Gurobi AI Modeling Assistant that you might find useful when exploring how to model your problem.

    Best regards,

    Maliheh

    1
  • Huib Meulenbelt
    Edited
    • Investigator
    • Gurobi-versary
    • Conversationalist

    Hi Maliheh,

    Thank you for your solution. I am getting now the output that I expect

     

    import itertools
from gurobipy import Model, GRB


truth_table = dict.fromkeys(itertools.product([1, 0], repeat=2), 0)


r_max = 6

r_p_min = 1.5
r_p_max = 6
r_q_min = 4
r_q_max = 4

M = 10000

model = Model()

p = model.addVar(vtype=GRB.BINARY)
q = model.addVar(vtype=GRB.BINARY)
r = model.addVar(vtype=GRB.INTEGER, lb=0, ub=r_max)

i = model.addVar(vtype=GRB.BINARY)
j = model.addVar(vtype=GRB.BINARY)
n = model.addVar(vtype=GRB.BINARY)
m = model.addVar(vtype=GRB.BINARY)

model.addConstr(r <= (r_p_min - 1) + M * (1 - i))
model.addConstr(r >= (r_p_max + 1) - M * (1 - j))
model.addGenConstrIndicator(p, False, i + j == 1)
model.addGenConstrIndicator(p, True, r_p_min <= r)
model.addGenConstrIndicator(p, True, r <= r_p_max)

model.addConstr(r <= (r_q_min - 1) + M * (1 - n))
model.addConstr(r >= (r_q_max + 1) - M * (1 - m))
model.addGenConstrIndicator(q, False, n + m == 1)
model.addGenConstrIndicator(q, True, r_q_min <= r)
model.addGenConstrIndicator(q, True, r <= r_q_max)

model.Params.PoolSearchMode = 2
model.optimize()

for s in range(model.SolCount):
    model.setParam(GRB.Param.SolutionNumber, s)
    truth_table[p.Xn, q.Xn] = 1
    print(f'r: {r.Xn}, p: {p.Xn}, q: {q.Xn}')

print(truth_table)
    0

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