Your problem can be solved using constraint programming. In python this can be implemented using the python-constraint library.
First, we need a way to check if two slots are consistent with each other. this is a function that returns true if two slots share a tag and their rimeframes overlap. In python this can be implemented using the following function
def checkNoOverlap(slot1, slot2):
shareTags = False
for tag in slot1['tags']:
if tag in slot2['tags']:
shareTags = True
break
if not shareTags: return True
return not (slot2['begin'] <= slot1['begin'] <= slot2['end'] or
slot2['begin'] <= slot1['end'] <= slot2['end'])
I was not sure whether you wanted the tags to be completely the same (like {foo: bar} equals {foo: bar}) or only the keys (like {foo: bar} equals {foo: qux}), but you can change that in the function above.
Consistency check
We can use the python-constraint module for the two kinds of functionality you requested.
The second functionality is the easiest. To implement this, we can use the function isConsistent(set) which takes a list of slots in the provided data structure as input. The function will then feed all the slots to python-constraint and will check if the list of slots is consistent (no 2 slots that shouldn't overlap, overlap) and return the consistency.
def isConsistent(set):
#initialize python-constraint context
problem = Problem()
#add all slots the context as variables with a singleton domain
for i in range(len(set)):
problem.addVariable(i, [set[i]])
#add a constraint for each possible pair of slots
for i in range(len(set)):
for j in range(len(set)):
#we don't want slots to be checked against themselves
if i == j:
continue
#this constraint uses the checkNoOverlap function
problem.addConstraint(lambda a,b: checkNoOverlap(a, b), (i, j))
# getSolutions returns all the possible combinations of domain elements
# because all domains are singleton, this either returns a list with length 1 (consistent) or 0 (inconsistent)
return not len(problem.getSolutions()) == 0
This function can be called whenever a user wants to add a reservation slot. The input slot can be added to the list of already existing slots and the consistency can be checked. If it is consistent, the new slot an be reserverd. Else, the new slot overlaps and should be rejected.
Finding available slots
This problem is a bit trickier. We can use the same functionality as above with a few significant changes. Instead of adding the new slot together with the existing slot, we now want to add all possible slots to the already existing slots. We can then check the consistency of all those possible slots with the reserved slots and ask the constraint system for the combinations that are consistent.
Because the number of possible slots would be infinite if we didn't put any restrictions on it, we first need to declare some parameters for the program:
MIN = 149780000 #available time slots can never start earlier then this time
MAX = 149790000 #available time slots can never start later then this time
GRANULARITY = 1*60 #possible time slots are always at least one minut different from each other
We can now continue to the main function. It looks a lot like the consistency check, but instead of the new slot from the user, we now add a variable to discover all available slots.
def availableSlots(tags, set):
#same as above
problem = Problem()
for i in range(len(set)):
problem.addVariable(i, [set[i]])
#add an extra variable for the available slot is added, with a domain of all possible slots
problem.addVariable(len(set), generatePossibleSlots(MIN, MAX, GRANULARITY, tags))
for i in range(len(set) +1):
for j in range(len(set) +1):
if i == j:
continue
problem.addConstraint(lambda a, b: checkNoOverlap(a, b), (i, j))
#extract the available time slots from the solution for clean output
return filterAvailableSlots(problem.getSolutions())
I use some helper functions to keep the code cleaner. They are included here.
def filterAvailableSlots(possibleCombinations):
result = []
for slots in possibleCombinations:
for key, slot in slots.items():
if slot['type'] == 'available':
result.append(slot)
return result
def generatePossibleSlots(min, max, granularity, tags):
possibilities = []
for i in range(min, max - 1, granularity):
for j in range(i + 1, max, granularity):
possibleSlot = {
'type': 'available',
'begin': i,
'end': j,
'tags': tags
}
possibilities.append(possibleSlot)
return tuple(possibilities)
You can now use the function getAvailableSlots(tags, set) with the tags for which you want the available slots and a set of already reserved slots. Note that this function really return all the consistent possible slots, so no effort is done to find the one of maximum lenght or for other optimalizations.
Hope this helps! (I got it to work as you described in my pycharm)
