Known algorithm for efficiently distributing items and satisfying minima? - algorithm

Question

For the following problem I'm wondering if there is a known algorithm already as I don't want to reinvent the wheel.
In this case it's about hotel rooms, but I think that is rather irrelevant:
name | max guests | min guests
1p | 1 | 1
2p | 2 | 2
3p | 3 | 2
4p | 4 | 3
I'm trying to distribute a certain amount of guests over available rooms, but the distribution has to satisfy the 'min guests' criteria of the rooms. Also, the rooms need to be used as efficiently as possible.
Let's take 7 guests for example. I wouldn't want this combination:
3 x 3p ( 1 x 3 guests, 2 x 2 guests )
.. this would satisfy the minimum criteria, but would be inefficient. Rather I'm looking for combinations such as:
1 x 3p and 1 x 4p
3 x 2p and 1 x 1p
etc...
I would think this is a familiar problem. Is there any known algorithm already to solve this problem?
To clarify:
By efficient I mean, distribute guests in such a way that rooms are filled up as much as possible (guests preferences are of secondary concern here, and are not important for the algorithm I'm looking for).
I do want all permutations that satisfy this efficiency criteria though. So in above example 7 x 1p would be fine as well.
So in summary:
Is there a known algorithm that is able to distribute items as efficiently as possible over slots with a min and max capacity, always satisfying the min criteria and trying to satisfy the max criteria as much as possible.

Answer 1

You need to use dynamic programming, define a cost function, and try to fit people in possible rooms having a cost function as small as possible.
Your cost function can be something like :
Sum of vacancy in rooms + number of rooms
It can be a bit similar to the least rageness problem : Word wrap to X lines instead of maximum width (Least raggedness)
You fit people in room, as you fit words in line.
The constraints are the vacancies in the rooms instead of being the length of the lines. (infinite cost if you don't fullfil the constraints)
and the recursion relation is pretty much the same .
Hope it helps

Answer 2

$sql = "SELECT *
FROM rooms
WHERE min_guests <= [$num_of_guests]
ORDER BY max_guests DESC
LIMIT [$num_of_guests]";
$query = $this->db->query($sql);
$remaining_guests = $num_of_guests;
$rooms = array();
$filled= false;
foreach($query->result() as $row)
{
if(!$filled)
{
$rooms[] = $row;
$remaining_guests -= $row->max_guests;
if(remaining_guests <= 0)
{
$filled = true;
break;
}
}
}
Recursive function:
public function getRoomsForNumberOfGuests($number)
{
$sql = "SELECT *
FROM rooms
WHERE min_guests <= $number
ORDER BY max_guests DESC
LIMIT 1";
$query = $this->db->query($sql);
$remaining_guests = $number;
$rooms = array();
foreach($query->result() as $row)
{
$rooms[] = $row;
$remaining_guests -= $row->max_guests;
if($remaining_guests > 0)
{
$rooms = array_merge($this->getRoomsForNumberOfGuests($remaining_guests), $rooms);
}
}
return $rooms;
}
Would something like this work for ya? Not sure what language your in?

Answer 3

For efficint = minimum rooms used, perhaps this would work. To minimise the number of rooms used you want to put max guests in the large rooms.
So sort the rooms in descending order of max guests, then allocate guests to them in that order, placing max guests in each room in turn. Try to place all remaining guests is any remaining room that will accept that many min guests; if that is impossible, back-track and try again. When back tracking, hold back the room with the smallest min guests. Held back rooms are not allocated guests in the max guests phase.
EDIT
As Ricky Bobby pointed out, this does not work as such, because of the difficulty of the back-tracking. I'm keeping this answer for now, more as a warning than as a suggestion :-)

Answer 4

This can be done as a fairly straightforward modification of the recursive algorithm to enumerate integer partitions. In Python:
import collections
RoomType = collections.namedtuple("RoomType", ("name", "max_guests", "min_guests"))
def room_combinations(count, room_types, partial_combo=()):
if count == 0:
yield partial_combo
elif room_types and count > 0:
room = room_types[0]
for guests in range(room.min_guests, room.max_guests + 1):
yield from room_combinations(
count - guests, room_types, partial_combo + ((room.name, guests),)
)
yield from room_combinations(count, room_types[1:], partial_combo)
for combo in room_combinations(
7,
[
RoomType("1p", 1, 1),
RoomType("2p", 2, 2),
RoomType("3p", 3, 2),
RoomType("4p", 4, 3),
],
):
print(combo)

Answer 5

we should select a list of room (maybe more than one time for each room) in a way that sum of the min value of selected room get equal or smaller than the number of guests and sum of the max value get equal or bigger than the number of guests.
I defined cost as total free space in selected rooms. (cost = max - min)
We can check for the answer with this code and find all possible combinations with the minimum cost. (c# code)
class FindRooms
{
// input
int numberOfGuest = 7; // your goal
List<Room> rooms;
// output
int cost = -1; // total free space in rooms.
List<String> options; // list of possible combinations for the best cost
private void solve()
{
// fill rooms data
// fill numberOfGuest
// run this function to find the answer
addMoreRoom("", 0, 0);
// cost and options are ready to use
}
// this function add room to the list recursively
private void addMoreRoom(String selectedRooms, int minPerson, int maxPerson)
{
// check is it acceptable
if (minPerson <= numberOfGuest && maxPerson >= numberOfGuest)
{
// check is it better than or equal to previous result
if(maxPerson - minPerson == cost)
{
options.Add(selectedRooms);
}
else if (maxPerson - minPerson < cost)
{
cost = maxPerson - minPerson;
options.Clear();
options.Add(selectedRooms);
}
}
// check if too many room selected
if (minPerson > numberOfGuest)
return;
// add more room recursively
foreach (Room room in rooms)
{
// add room and min and max space to current state and check
addMoreRoom(selectedRooms + "," + room, minPerson + room.min, maxPerson + room.max);
}
}
public class Room
{
public String name;
public int min;
public int max;
}
}