Interval tree with added dimension of subset matching?

Question

This is an algorithmic question about a somewhat complex problem. The foundation is this:

A scheduling system based on available slots and reserved slot

Answer 1

Here's a solution, I'll include all the code below.

1. Create a table of slots, and a table of reservations

2. Create a matrix of reservations x slots

which is populated by true or false values based on whether that reservation-slot combination are possible

3. Figure out the best mapping that allows for the most Reservation-Slot Combinations

Note: my current solution scales poorly with very large arrays as it involves looping through all possible permutations of a list with size = number of slots. I've posted another question to see if anyone can find a better way of doing this. However, this solution is accurate and can be optimized

Python Code Source

Part 1

from IPython.display import display
import pandas as pd
import datetime

available_data = [
    ['SlotA', datetime.time(11, 0, 0), datetime.time(12, 30, 0), set(list('ABD'))],
    ['SlotB',datetime.time(12, 0, 0), datetime.time(13, 30, 0), set(list('C'))],
    ['SlotC',datetime.time(12, 0, 0), datetime.time(13, 30, 0), set(list('ABCD'))],
    ['SlotD',datetime.time(12, 0, 0), datetime.time(13, 30, 0), set(list('AD'))],
]

reservation_data = [
    ['ReservationA', datetime.time(11, 15, 0), datetime.time(12, 15, 0), set(list('AD'))],
    ['ReservationB', datetime.time(11, 15, 0), datetime.time(12, 15, 0), set(list('A'))],
    ['ReservationC', datetime.time(12, 0, 0), datetime.time(12, 15, 0), set(list('C'))],
    ['ReservationD', datetime.time(12, 0, 0), datetime.time(12, 15, 0), set(list('C'))],
    ['ReservationE', datetime.time(12, 0, 0), datetime.time(12, 15, 0), set(list('D'))]
]

reservations = pd.DataFrame(data=reservation_data, columns=['reservations', 'begin', 'end', 'tags']).set_index('reservations')
slots = pd.DataFrame(data=available_data, columns=['slots', 'begin', 'end', 'tags']).set_index('slots')

display(slots)
display(reservations)

Part 2

def is_possible_combination(r):
    return (r['begin'] >= slots['begin']) & (r['end'] <= slots['end']) & (r['tags'] <= slots['tags'])

solution_matrix = reservations.apply(is_possible_combination, axis=1).astype(int)
display(solution_matrix)

Part 3

import numpy as np
from itertools import permutations

# add dummy columns to make the matrix square if it is not
sqr_matrix = solution_matrix
if sqr_matrix.shape[0] > sqr_matrix.shape[1]:
    # uhoh, there are more reservations than slots... this can't be good
    for i in range(sqr_matrix.shape[0] - sqr_matrix.shape[1]):
        sqr_matrix.loc[:,'FakeSlot' + str(i)] = [1] * sqr_matrix.shape[0]
elif sqr_matrix.shape[0] < sqr_matrix.shape[1]:
    # there are more slots than customers, why doesn't anyone like us?
    for i in range(sqr_matrix.shape[0] - sqr_matrix.shape[1]):
        sqr_matrix.loc['FakeCustomer' + str(i)] = [1] * sqr_matrix.shape[1]

# we only want the values now
A = solution_matrix.values.astype(int)

# make an identity matrix (the perfect map)
imatrix = np.diag([1]*A.shape[0])

# randomly swap columns on the identity matrix until they match. 
n = A.shape[0]

# this will hold the map that works the best
best_map_so_far = np.zeros([1,1])

for column_order in permutations(range(n)):
    # this is an identity matrix with the columns swapped according to the permutation
    imatrix = np.zeros(A.shape)
    for row, column in enumerate(column_order):
        imatrix[row,column] = 1

    # is this map better than the previous best?
    if sum(sum(imatrix * A)) > sum(sum(best_map_so_far)):
        best_map_so_far = imatrix

    # could it be? a perfect map??
    if sum(sum(imatrix * A)) == n:
        break

if sum(sum(imatrix * A)) != n:
    print('a perfect map was not found')

output = pd.DataFrame(A*imatrix, columns=solution_matrix.columns, index=solution_matrix.index, dtype=int)
display(output)