Scheduling - exactly n staff or none each shift?

[Danny Cheng]

I am trying to solve a shift scheduling problem, where each shift should have exactly n staff, or have none at all. Not all the shifts need to be filled. Is this possible? I am looking at the nurse scheduling example and that is a great starting point, but would appreciate any suggestions on what other examples I should study up on.

[Mario Leung]

Hi, 

if you follow the nurse rostering sample (https://developers.google.com/optimization/scheduling/employee_scheduling), you can simply add variables to record the number of staff handling the specific shifts, and only allow the number to be equal to n or 0. Take a look at the sample pseudo-code below
...
for (int s : all_shifts) {

  for (int d : all_days) {

     IntVar num_nurse_allowed = cp_model.NewIntVar...

    std::vector<BoolVar> num_nurse_worked;  

    for (int n : all_nurses) {
      auto key = std::make_tuple(n, d, s);
       num_nurse_worked.push_back(shifts[key]);
    }

    cp_model.AddEquality( num_nurse_allowed, LinearExpr::Sum(num_nurse_worked));

    cp_model.Add( (num_nurse_allowed == 0) or (num_nurse_allowed == n));

  }

...

[Danny Cheng]

thanks, that seems rather straight forward, will take a closer look at this tomorrow.

[Laurent Perron]

It does not work. 

cp_model.Add( (num_nurse_allowed == 0) or (num_nurse_allowed == n));

is not interpreted correctly. what is 'or' ?

I would suggest 

  for (int d : all_days) {

    BoolVar nurses_working = cp_model.NewBoolVar...

    std::vector<BoolVar> num_nurse_worked;  

    for (int n : all_nurses) {
       auto key = std::make_tuple(n, d, s);
       num_nurse_worked.push_back(shifts[key]);
    }

    cp_model.AddEquality( nurses_working * n, LinearExpr::Sum(num_nurse_worked));

  }

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