Complexity of the Kitten Adoption problem

Question

This came up while I was trying to answer this question on Wiring Length Minimization. I was going to call this the "polygamous marriage" problem, but the internet, so kittens. Yay!

Suppose we have $M$ kittens that need to be adopted by $N$ people, $M > N$. For each kitten, $i$ and each person $j$ there is a cost $c_{ij}$. We would like to minimize the total cost of getting all the kittens adopted. There is also a set of constraints: each person $j$ is able to adopt no more than $u_j$ kittens.

Without the constraints the problem is easy; each kitten $i$ goes with the person $j$ for which $c_{ij}$ is minimal. With the constraints is there an efficient algorithm for this problem or is it NP-hard?

Answer 1

This is the min-cost max-flow problem.

Consider a graph $G=(A\cup B\cup \{s,t\},E)$, where $A$ is the set of kittens, $B$ is the set of people.

Let $C:E\to \mathbb{R}^+$ be the capacity of the edges, and $c:E\to \mathbb{R^+}$ be the cost of an edge. We make sure that

1. There is an edge between $a_i,b_j$, where $a_i\in A$ and $b_j\in B$, and $C(a_i,b_j)=1$, $c(a_i,b_j)=c_{i,j}$.
2. There is an edge between $s$ and $a_i\in A$, and $C(s,a_i)=1$, $c(s,a_i)=0$.
3. There is an edge between $b_j\in B$ and $t$, and $C(b_j,t)=u_j$, $c(b_j,t)=0$.

If the max flow is $M$, then we know there exist a solution. You can construct a min-cost solution from the min-cost max-flow.

Answer 2

This is the minimum weight perfect matching problem which is polynomial. Consider the complete bipartite graph $(L , R , E)$, in which $L$ contains a node $l_i$ for each kitten $i$, $R$ consists of $u_j$ copies of the node $r_{j}$ for each person $j$, and edges $e_{ij} \in E$ between $l_i$ and each copy of $r_{j}$, with weights $c_{ij}$.

We know that $|L| \leq |R|$, otherwise not all kittens can be assigned to persons.

Since perfect matching must match all the nodes, we need to add dummy nodes to $L$ (to get $|L| = |R|$) and connect them with zero weight edges to all the nodes in $R$.

Answer 3

Possibly of interest is the observation that one can reduce Partition to a variant of this problem. Given is an instance of Partition with elements $\{x_1,\dots,x_q\}$ with $q$ even from which we have to choose a subset $S \subseteq \{1,\dots,q\}$ with $|S|=q/2$ such that $\sum_{i\in S} x_i = \sum_{i\not\in S} x_i = K$. (Note that the requirement to choose exactly half the elements is not the usual form but this form is still NP-hard.) Let each kitten be an element of the set; let there be two people; let the weights be $c_{i1} = x_i$ and $c_{i2} = -x_i$; let $u_1 = u_2 = q/2$. Then this instance of Kitten Adoption has a maximum of $0$ iff the instance of Partition has a solution.

If the costs in Kitten Adoption are meant to always be positive then one can just add a large enough constant $C$ to all weights to ensure they are the right sign, when the required threshold is $Cq$ instead of 0.

I am not sure what this says about the complexity of the original problem, but given the often-seen "one of minimize/maximize is NP-hard, the other is in P" setup for combinatorial optimization problems, I would start looking for an efficient algorithm.