Secret santa algorithm - algorithm

Question

Every Christmas we draw names for gift exchanges in my family. This usually involves mulitple redraws until no one has pulled their spouse. So this year I coded up my own name drawing app that takes in a bunch of names, a bunch of disallowed pairings, and sends off an email to everyone with their chosen giftee.
Right now, the algorithm works like this (in pseudocode):
function DrawNames(list allPeople, map disallowedPairs) returns map
// Make a list of potential candidates
foreach person in allPeople
person.potentialGiftees = People
person.potentialGiftees.Remove(person)
foreach pair in disallowedPairs
if pair.first = person
person.Remove(pair.second)
// Loop through everyone and draw names
while allPeople.count > 0
currentPerson = allPeople.findPersonWithLeastPotentialGiftees
giftee = pickRandomPersonFrom(currentPerson.potentialGiftees)
matches[currentPerson] = giftee
allPeople.Remove(currentPerson)
foreach person in allPeople
person.RemoveIfExists(giftee)
return matches
Does anyone who knows more about graph theory know some kind of algorithm that would work better here? For my purposes, this works, but I'm curious.
EDIT: Since the emails went out a while ago, and I'm just hoping to learn something I'll rephrase this as a graph theory question. I'm not so interested in the special cases where the exclusions are all pairs (as in spouses not getting each other). I'm more interested in the cases where there are enough exclusions that finding any solution becomes the hard part. My algorithm above is just a simple greedy algorithm that I'm not sure would succeed in all cases.
Starting with a complete directed graph and a list of vertex pairs. For each vertex pair, remove the edge from the first vertex to the second.
The goal is to get a graph where each vertex has one edge coming in, and one edge leaving.

Answer 1

Just make a graph with edges connecting two people if they are allowed to share gifts and then use a perfect matching algorithm. (Look for "Paths, Trees, and Flowers" for the (clever) algorithm)

Answer 2

I was just doing this myself, in the end the algorithm I used doesn't exactly model drawing names out of a hat, but it's pretty damn close. Basically shuffle the list, and then pair each person with the next person in the list. The only difference with drawing names out of a hat is that you get one cycle instead of potentially getting mini subgroups of people who only exchange gifts with each other. If anything that might be a feature.
Implementation in Python:
import random
from collections import deque
def pairup(people):
""" Given a list of people, assign each one a secret santa partner
from the list and return the pairings as a dict. Implemented to always
create a perfect cycle"""
random.shuffle(people)
partners = deque(people)
partners.rotate()
return dict(zip(people,partners))

Answer 3

I wouldn't use disallowed pairings, since that greatly increases the complexity of the problem. Just enter everyone's name and address into a list. Create a copy of the list and keep shuffling it until the addresses in each position of the two lists don't match. This will ensure that no one gets themselves, or their spouse.
As a bonus, if you want to do this secret-ballot-style, print envelopes from the first list and names from the second list. Don't peek while stuffing the envelopes. (Or you could just automate emailing everyone thier pick.)
There are even more solutions to this problem on this thread.

Answer 4

Hmmm. I took a course in graph theory, but simpler is to just randomly permute your list, pair each consecutive group, then swap any element that is disallowed with another. Since there's no disallowed person in any given pair, the swap will always succeed if you don't allow swaps with the group selected. Your algorithm is too complex.

Answer 5

Create a graph where each edge is "giftability" Vertices that represent Spouses will NOT be adjacent. Select an edge at random (that is a gift assignment). Delete all edges coming from the gifter and all edges going to the receiver and repeat.

Answer 6

There is a concept in graph theory called a Hamiltonian Circuit that describes the "goal" you describe. One tip for anybody who finds this is to tell users which "seed" was used to generate the graph. This way if you have to re-generate the graph you can. The "seed" is also useful if you have to add or remove a person. In that case simply choose a new "seed" and generate a new graph, making sure to tell participants which "seed" is the current/latest one.

Answer 7

I just created a web app that will do exactly this - http://www.secretsantaswap.com/
My algorithm allows for subgroups. It's not pretty, but it works.
Operates as follows:
1. assign a unique identifier to all participants, remember which subgroup they're in
2. duplicate and shuffle that list (the targets)
3. create an array of the number of participants in each subgroup
4. duplicate array from [3] for targets
5. create a new array to hold the final matches
6. iterate through participants assigning the first target that doesn't match any of the following criteria:
A. participant == target
B. participant.Subgroup == target.Subgroup
C. choosing the target will cause a subgroup to fail in the future (e.g. subgroup 1 must always have at least as many non-subgroup 1 targets remaining as participants subgroup 1 participants remaining)
D. participant(n+1) == target (n +1)
If we assign the target we also decrement the arrays from 3 and 4
So, not pretty (at all) but it works. Hope it helps,
Dan Carlson

Answer 8

Here a simple implementation in java for the secret santa problem.
public static void main(String[] args) {
ArrayList<String> donor = new ArrayList<String>();
donor.add("Micha");
donor.add("Christoph");
donor.add("Benj");
donor.add("Andi");
donor.add("Test");
ArrayList<String> receiver = (ArrayList<String>) donor.clone();
Collections.shuffle(donor);
for (int i = 0; i < donor.size(); i++) {
Collections.shuffle(receiver);
int target = 0;
if(receiver.get(target).equals(donor.get(i))){
target++;
}
System.out.println(donor.get(i) + " => " + receiver.get(target));
receiver.remove(receiver.get(target));
}
}

Answer 9

Python solution here.
Given a sequence of (person, tags), where tags is itself a (possibly empty) sequence of strings, my algorithm suggests a chain of persons where each person gives a present to the next in the chain (the last person obviously is paired with the first one).
The tags exist so that the persons can be grouped and every time the next person is chosen from the group most dis-joined to the last person chosen. The initial person is chosen by an empty set of tags, so it will be picked from the longest group.
So, given an input sequence of:
example_sequence= [
("person1", ("male", "company1")),
("person2", ("female", "company2")),
("person3", ("male", "company1")),
("husband1", ("male", "company2", "marriage1")),
("wife1", ("female", "company1", "marriage1")),
("husband2", ("male", "company3", "marriage2")),
("wife2", ("female", "company2", "marriage2")),
]
a suggestion is:
['person1 [male,company1]',
'person2 [female,company2]',
'person3 [male,company1]',
'wife2 [female,marriage2,company2]',
'husband1 [male,marriage1,company2]',
'husband2 [male,marriage2,company3]',
'wife1 [female,marriage1,company1]']
Of course, if all persons have no tags (e.g. an empty tuple) then there is only one group to choose from.
There isn't always an optimal solution (think an input sequence of 10 women and 2 men, their genre being the only tag given), but it does a good work as much as it can.
Py2/3 compatible.
import random, collections
class Statistics(object):
def __init__(self):
self.tags = collections.defaultdict(int)
def account(self, tags):
for tag in tags:
self.tags[tag] += 1
def tags_value(self, tags):
return sum(1./self.tags[tag] for tag in tags)
def most_disjoined(self, tags, groups):
return max(
groups.items(),
key=lambda kv: (
-self.tags_value(kv[0] & tags),
len(kv[1]),
self.tags_value(tags - kv[0]) - self.tags_value(kv[0] - tags),
)
)
def secret_santa(people_and_their_tags):
"""Secret santa algorithm.
The lottery function expects a sequence of:
(name, tags)
For example:
[
("person1", ("male", "company1")),
("person2", ("female", "company2")),
("person3", ("male", "company1")),
("husband1", ("male", "company2", "marriage1")),
("wife1", ("female", "company1", "marriage1")),
("husband2", ("male", "company3", "marriage2")),
("wife2", ("female", "company2", "marriage2")),
]
husband1 is married to wife1 as seen by the common marriage1 tag
person1, person3 and wife1 work at the same company.
…
The algorithm will try to match people with the least common characteristics
between them, to maximize entrop— ehm, mingling!
Have fun."""
# let's split the persons into groups
groups = collections.defaultdict(list)
stats = Statistics()
for person, tags in people_and_their_tags:
tags = frozenset(tag.lower() for tag in tags)
stats.account(tags)
person= "%s [%s]" % (person, ",".join(tags))
groups[tags].append(person)
# shuffle all lists
for group in groups.values():
random.shuffle(group)
output_chain = []
prev_tags = frozenset()
while 1:
next_tags, next_group = stats.most_disjoined(prev_tags, groups)
output_chain.append(next_group.pop())
if not next_group: # it just got empty
del groups[next_tags]
if not groups: break
prev_tags = next_tags
return output_chain
if __name__ == "__main__":
example_sequence = [
("person1", ("male", "company1")),
("person2", ("female", "company2")),
("person3", ("male", "company1")),
("husband1", ("male", "company2", "marriage1")),
("wife1", ("female", "company1", "marriage1")),
("husband2", ("male", "company3", "marriage2")),
("wife2", ("female", "company2", "marriage2")),
]
print("suggested chain (each person gives present to next person)")
import pprint
pprint.pprint(secret_santa(example_sequence))