How to program a schedule [closed]

Question

I have to build a program that schedules based on certain rules. I'm not sure how to explain it, so let me give you an example..

You have Five People A,B,C,D,E. And you Have another set of people S1 S2 S3 S4 S5 S6 S7.

If A B C D and E are available every hour from 9 to 5, and S1 S2 S3 S4 S5 S6 and S7 have a list of 3 people they want to see from {A,B,C,D,E}

That's my problem and I'm not sure where to begin...

Thanks for your help!

Answer 1

Here's some python code that will do the trick. You will want to update VISITOR_PEOPLE. And if some people get to schedule before others, you'll need to reorder VISITOR_IDS.

Edit: I added some more code to account for the fact that people can't be in a different place at the same time. You might want to make that more efficient (i.e. don't try to schedule a time that will not work). I'll let you figure that out though ;)

import sys

HOURS = ['9:00AM', '10:00AM', '11:00AM', '12:00PM', '1:00PM', '2:00PM', '3:00PM', '4:00PM']
PEOPLE_IDS = ['A', 'B', 'C', 'D', 'E']
VISITOR_IDS = ['S1', 'S2', 'S3', 'S4', 'S5', 'S6', 'S7']
VISITOR_PEOPLE = {'S1': ['A', 'B', 'C'], 
                  'S2': ['A', 'D', 'E'], 
                  'S3': ['B', 'E', 'D'], 
                  'S4': ['D', 'E', 'A'], 
                  'S5': ['C', 'D', 'E'], 
                  'S6': ['A', 'D', 'C'], 
                  'S7': ['B', 'C', 'D']
                  }

def main():
    people = {}
    for id in PEOPLE_IDS:
        people[id] = Person(id)
    visitors = {}
    for id in VISITOR_IDS:
        visitors[id] = Visitor(id, VISITOR_PEOPLE[id], people)
    for v in visitors.values():
        v.printSchedule()

class Person:
    def __init__(self, id):
        self.id = id
        self.schedule = [False]*8  # False = free, True = busy
    def scheduleTime(self):
        # schedules next available hour and returns that hour
        for i in range(len(self.schedule)):
            if not self.schedule[i]:
                self.schedule[i] = True
                return HOURS[i]
        return 'unavailable'
    def unscheduleTime(self, index):
        self.schedule[index] = False

class Visitor:
    def __init__(self, id, people_requests, people):
        self.id = id
        self.schedule = {} # {person_id: hour}
        for p in people_requests:
            bad_times = set()  # times that Visitor is busy
            time = people[p].scheduleTime()
            while time in self.schedule.values():  # keep scheduling a time until you get one that works for both the Visitor and Person
                bad_times.add(time)
                time = people[p].scheduleTime()
            self.schedule[p] = time
            for t in bad_times:  # unschedule bad_times from Person
                people[p].unscheduleTime(HOURS.index(t))
    def printSchedule(self):
        print 'Schedule for %s [Person (time)]:' % self.id
        for p,t in self.schedule.items():
            print '    %s (%s)' % (p,t)

if __name__ == '__main__':
    sys.exit(main())

Answer 2

Here's one approach:

Start with S1, assign him the three people he wants at 9am, then go to S2 and try to schedule his meeting at 9am. Continue until you have a conflict, then move that meeting to 10am. Return to 9am for the next one. If there is a conflict at 10 also, move to 11, etc.

Once the program tries to schedule a meeting after hours, you'll know you've hit a case where all the meetings aren't possible in a single day.