Pulp Python problem setting constraints when summing values in a column

Hi this is my first question here so go easy on me if I format things incorrectly.

I’m trying to model a table where each value is either 1 or 0.
I’d like to determine whether the sum of a column is 0 or not 0, then check how many columns are > 0.
The underlying problem I’m trying to solve is appointment scheduling, where each column represent one appointment. I’ve simplified it here as in the original I’m using a dataframe to match clinician competencies to patient needs (each row is a patient need). My problem started when I tried to ensure all variables could only be equal to 1 if in one if they were in one of 2 columns, hence my simplified code here to try to work out where I am going wrong.

I’ve set up a pulp variable dictionary with ROWS and COLS as the keys, and value == 0 or 1.

In the problem definition I’m trying to assign a value of 1 to the column sum if sum of the row values in the column is >= 1 and 0 otherwise, then summing the total. This should allow me to set the total number of columns that sum to >= 1, for example only 2 columns are represented by non zero variables.

In the code below my aim is for the total sum of all variables to be minimised BUT there should be 2 columns that contain a variable 1 i.e. 2 columns sum to >=1.

Thanks in advance.

import pulp as Pulp
ROWS = range(1, 6)
COLS = range(1,5)

prob = Pulp.LpProblem("Fewestcolumns", Pulp.LpMinimize)
choices = Pulp.LpVariable.dicts("Choice", (ROWS, COLS), cat="Integer", lowBound=0, upBound=1)
prob += Pulp.lpSum([choices[row][col] for row in ROWS for col in COLS])
prob += Pulp.lpSum([1 if Pulp.lpSum([choices[row][col] for row in ROWS]) >= 1 else 0 for col in COLS]) == 2



prob.solve()

print("Status:", Pulp.LpStatus[prob.status])
for v in prob.variables():
    print(v.name, "=", v.varValue)`

My results:

C:UsersxxxComputingLinearProgrammingScriptspython.exe C:/Users/xxx/Computing/LinearProgramming/LinearProgTest.py
Welcome to the CBC MILP Solver 
Version: 2.10.3 
Build Date: Dec 15 2019 

command line - C:UsersxxxxComputingLinearProgramminglibsite-packagespulpsolverdircbcwin64cbc.exe C:UserssimonAppDataLocalTemp4f8ff67726844bde8abe98316b6338c4-pulp.mps timeMode elapsed branch printingOptions all solution C:UserssimonAppDataLocalTemp4f8ff67726844bde8abe98316b6338c4-pulp.sol (default strategy 1)
At line 2 NAME          MODEL
At line 3 ROWS
At line 6 COLUMNS
At line 67 RHS
At line 69 BOUNDS
At line 90 ENDATA
Problem MODEL has 1 rows, 20 columns and 0 elements
Coin0008I MODEL read with 0 errors
Option for timeMode changed from cpu to elapsed
Problem is infeasible - 0.00 seconds
Option for printingOptions changed from normal to all
Total time (CPU seconds):       0.01   (Wallclock seconds):       0.01

Status: Infeasible
Choice_1_1 = 0.0
Choice_1_2 = 0.0
Choice_1_3 = 0.0
Choice_1_4 = 0.0
Choice_2_1 = 0.0
Choice_2_2 = 0.0
Choice_2_3 = 0.0
Choice_2_4 = 0.0
Choice_3_1 = 0.0
Choice_3_2 = 0.0
Choice_3_3 = 0.0
Choice_3_4 = 0.0
Choice_4_1 = 0.0
Choice_4_2 = 0.0
Choice_4_3 = 0.0
Choice_4_4 = 0.0
Choice_5_1 = 0.0
Choice_5_2 = 0.0
Choice_5_3 = 0.0
Choice_5_4 = 0.0

Process finished with exit code 0

I was expecting a list of variables a bit like this, with a possible solution:

Status: Optimal
Choice_1_1 = 1.0
Choice_1_2 = 1.0
Choice_1_3 = 0.0
Choice_1_4 = 0.0
Choice_2_1 = 0.0
Choice_2_2 = 0.0
Choice_2_3 = 0.0
Choice_2_4 = 0.0
Choice_3_1 = 0.0
Choice_3_2 = 0.0
Choice_3_3 = 0.0
Choice_3_4 = 0.0
Choice_4_1 = 0.0
Choice_4_2 = 0.0
Choice_4_3 = 0.0
Choice_4_4 = 0.0
Choice_5_1 = 0.0
Choice_5_2 = 0.0
Choice_5_3 = 0.0
Choice_5_4 = 0.0

Edits:
Many thanks AirSquid for pointing me in the right direction. I’m still struggling with big M constraints.

I tried this:

import pulp as Pulp
ROWS = range(1, 6)
COLS = range(1,5)

prob = Pulp.LpProblem("Fewestcolumns", Pulp.LpMaximize)
choices = Pulp.LpVariable.dicts("Choice", (ROWS, COLS), cat="Integer", lowBound=0, upBound=1)
used = Pulp.LpVariable.dicts("used", COLS, cat="Binary")
b = Pulp.LpVariable.dicts("b", COLS, cat="Binary")

prob += Pulp.lpSum([choices[row][col] for row in ROWS for col in COLS])
for rows, items in choices.items():
    prob += Pulp.lpSum(cols for cols in items.values()) == 1

M = 20
for col in COLS:
    prob += b[col] >= (Pulp.lpSum([choices[row][col] for row in ROWS]) - 1) / M
    prob += used[col] >= M * (b[col] - 1)

prob += Pulp.lpSum([used[col] for col in COLS]) == 2
prob.solve()

print("Status:", Pulp.LpStatus[prob.status])
for v in prob.variables():
    print(v.name, "=", v.varValue)

I got the following results:

 Result - Optimal solution found

Objective value:                5.00000000
Enumerated nodes:               0
Total iterations:               0
Time (CPU seconds):             0.00
Time (Wallclock seconds):       0.00

Option for printingOptions changed from normal to all
Total time (CPU seconds):       0.01   (Wallclock seconds):       0.02

Status: Optimal
Choice_1_1 = 0.0
Choice_1_2 = 0.0
Choice_1_3 = 0.0
Choice_1_4 = 1.0
Choice_2_1 = 0.0
Choice_2_2 = 0.0
Choice_2_3 = 0.0
Choice_2_4 = 1.0
Choice_3_1 = 0.0
Choice_3_2 = 0.0
Choice_3_3 = 0.0
Choice_3_4 = 1.0
Choice_4_1 = 0.0
Choice_4_2 = 0.0
Choice_4_3 = 0.0
Choice_4_4 = 1.0
Choice_5_1 = 0.0
Choice_5_2 = 0.0
Choice_5_3 = 0.0
Choice_5_4 = 1.0
b_1 = 1.0
b_2 = 1.0
b_3 = 1.0
b_4 = 1.0
used_1 = 1.0
used_2 = 1.0
used_3 = 0.0
used_4 = 0.0

Process finished with exit code 0

Not sure what I did wrong – I was hoping for some 1.0s in columns that aren’t column 4. Any more hints please?