Graphical house allocation with identical valuations

Abstract

The classical house allocation problem involves assigning n houses (or items) to n agents according to their preferences. A key criterion in such problems is satisfying some fairness constraints such as envy-freeness. We consider a generalization of this problem, called Graphical House Allocation, wherein the agents are placed along the vertices of a graph (corresponding to a social network), and each agent can only experience envy towards its neighbors. Our goal is to minimize the aggregate envy among the agents as a natural fairness objective, i.e., the sum of the envy value over all edges in a social graph. We focus on graphical house allocation with identical valuations. When agents have identical and evenly-spaced valuations, our problem reduces to the well-studied Minimum Linear Arrangement. For identical valuations with possibly uneven spacing, we show a number of deep and surprising ways in which our setting is a departure from this classical problem. More broadly, we contribute several structural and computational results for various classes of graphs, including NP-hardness results for disjoint unions of paths, cycles, stars, cliques, and complete bipartite graphs; we also obtain fixed-parameter tractable (and, in some cases, polynomial-time) algorithms for paths, cycles, stars, cliques, complete bipartite graphs, and their disjoint unions. Additionally, a conceptual contribution of our work is the formulation of a structural property for disconnected graphs that we call splittability, which results in efficient parameterized algorithms for finding optimal allocations.

Appendices

Appendix A: Distinct valuations

Lemma A.1

Given any instance (N, H, G, v) of Graphical House Allocation, there exists a valuation function $v'$ such that $v'$ gives each house a distinct value, and any optimal allocation under $v'$ is also optimal under v.

Proof

Let $\delta > 0$ be the smallest nonzero envy difference between two allocations of H to G under the valuation v, and let $\gamma > 0$ be the smallest nonzero difference between the values of two houses. If either $\delta$ or $\gamma$ are not well-defined, then all allocations have the same optimal envy, and we can define any arbitrary one-to-one function $v'$ to satisfy the lemma. So assume both $\delta$ and $\gamma$ are well-defined and positive. Define $\epsilon = \min \{\delta /2, \gamma \}$. We will show that there is a one-to-one valuation function $v'$, such that for any allocation $\pi$, the total envy under $v'$ differs from the total envy under v by at most an additive term of $\epsilon$. For $h_k \in H$, define

$$\begin{aligned} v'(h_k) := v(h_k) + \frac{\epsilon }{n^22^k}. \end{aligned}$$

It is easy to see that this function is one-to-one by the definition of $\epsilon$. For any allocation $\pi$ on G, consider the envy between agents i and j. If $\pi (i) = h_k$ and $\pi (j) = h_\ell$, we have, using the triangle inequality,

$$\begin{aligned} \big |v'(\pi (i)) - v'(\pi (j))\big |&= \bigg |v(\pi (i)) - v(\pi (j)) + \frac{\epsilon }{n^22^k} - \frac{\epsilon }{n^22^\ell }\bigg | \\&\le \big |v(\pi (i)) - v(\pi (j))\big | + \frac{\epsilon }{n^2}\bigg |\frac{1}{2^k} - \frac{1}{2^\ell }\bigg |\\&< \big |v(\pi (i)) - v(\pi (j))\big | + \frac{\epsilon }{n^2}. \end{aligned}$$

We also similarly have

$$\begin{aligned} \big |v'(\pi (i)) - v'(\pi (j))\big |&= \bigg |v(\pi (i)) - v(\pi (j)) + \frac{\epsilon }{n^22^k} - \frac{\epsilon }{n^22^\ell }\bigg | \\&\ge \big |v(\pi (i)) - v(\pi (j))\big | - \frac{\epsilon }{n^2}\bigg |\frac{1}{2^k} - \frac{1}{2^\ell }\bigg |\\&> \big |v(\pi (i)) - v(\pi (j))\big | - \frac{\epsilon }{n^2}. \end{aligned}$$

Summing over the at most $n^2$ edges of G, we have $\textsf{Envy}_v(G, \pi ) - \epsilon< \textsf{Envy}_{v'}(G, \pi ) < \textsf{Envy}_v(G, \pi ) + \epsilon$, as desired, where the subscripts v and $v'$ denote the valuation functions being used in each case.

For any allocation $\pi ^{*}$ which minimizes envy under $v'$, if we compare against another allocation $\pi '$ such that $\pi ^{*}$ and $\pi '$ have different total envies under v, we see that

$$\begin{aligned} \textsf{Envy}_v(G, \pi ^{*}) - \epsilon< \textsf{Envy}_{v'}(G, \pi ^{*}) \le \textsf{Envy}_{v'}(G, \pi ') < \textsf{Envy}_v(G, \pi ') + \epsilon . \end{aligned}$$

By the definition of $\epsilon = \min \{\delta /2, \gamma \}$, we can infer that if $\pi ^{*}$ is optimal under $v'$, then it must be optimal under v as well. If $\pi ^{*}$ is not optimal under v, then there will be an allocation $\pi '$ which is optimal under v that violates the inequality above; that is, we will have $\textsf{Envy}_v(G, \pi ') \le \textsf{Envy}_v(G, \pi ^{*}) - 2\epsilon$ by the definition of $\epsilon$. $\square$

Appendix B: Technical Proofs from Section 4

Theorem 4.6

When G is the graph $K_{r, s}$ ($r > s$), the minimum envy allocation $\pi ^*$ has the following property:

• (1) If $r - s =: 2m$ is even, then the first and last m houses are allocated to the larger part, and for all $i \in [s]$, the houses $h_{m + 2i - 1}$ and $h_{m +2i}$ are allocated to different parts.

• (2) If $r - s =: 2m + 1$ is odd, then the first m and last $m + 1$ houses are allocated to the larger part. For all $i \in [s]$, the houses $h_{m + 2i - 1}$ and $h_{m+ 2i}$ are allocated to the larger and smaller parts respectively.

Moreover, all allocations which satisfy this property have the same (optimal) envy.

Proof

This proof is very similar to that of Theorem 4.5. Again, for notational ease, let the graph have bipartition (L, R), with $|L| = r > s = |R|$. We refer to the properties in the theorem statement when $r - s$ is even and odd as the optimal even property and the optimal odd property respectively. This proof will also use the notation $n^{<}_{L, \pi }(x)$, $n^{>}_{L, \pi }(x), n^{<}_{R, \pi }(x)$ and $n^{>}_{R, \pi }(x)$ defined in Definition 2.3.

Case 1 $r - s$ is even. We split the proof into two claims.

Claim B.1

Any optimal allocation allocates the first m houses to agents in L.

Proof of Claim B.1

Assume for contradiction that this is not true. That is, there is an optimal allocation $\pi$ such that:

$$\begin{aligned}&\pi (h_j) \in L \text { for all } j \in [k] \text { for some } 0 \le k < m, \\&\pi (h_{k+j}) \in R \text { for all } j \in [l] \text { for some } l > 0, \\&\pi (h_{k+l+1}) \in L. \end{aligned}$$

Create an allocation $\pi '$ from $\pi$ by swapping $h_{k+l}$ and $h_{k+l+1}$. We can now compare the aggregate envy of $\pi$ and $\pi '$ using arguments similar to those in Theorem 4.5.

$$\begin{aligned}&\textsf{Envy}(\pi ', G) - \textsf{Envy}(\pi , G) \\&\quad = [n^{<}_{L, \pi }(v(h_{k+l+1})) - n^{>}_{L, \pi }(v(h_{k+l+1}))](v(h_{k+l+1}) - v(h_{k+l})) \\&\qquad + [n^{>}_{R, \pi }(v(h_{k+l})) - n^{<}_{R, \pi }(v(h_{k+l}))](v(h_{k+l+1}) - v(h_{k+l})) \\&\quad = (v(h_{k+l+1}) - v(h_{k+l})) \\&\qquad [n^{<}_{L, \pi }(v(h_{k+l+1})) - n^{>}_{L, \pi }(v(h_{k+l+1})) + n^{>}_{R, \pi }(v(h_{k+l})) - n^{<}_{R, \pi }(v(h_{k+l}))] \\&\quad = [k - (r - (k+1)) + (s - l) - (l-1)](v(h_{2i}) - v(h_{2i-1})) \\&\quad = [2k - (r -s) + 2 - 2l](v(h_{2i}) - v(h_{2i-1})) \\&\quad <0. \end{aligned}$$

The last inequality follows from the fact that $l \ge 1$ and $k < m = (r - s)/2$. This contradicts the optimality of $\pi$. $\square$

Claim B.2

In any optimal allocation, for any $i \in [s]$, $h_{m+ 2i-1}$ and $h_{m+ 2i}$ cannot be allocated to the same part.

Proof of Claim B.2

Assume for contradiction that this is not true. Let $\pi$ be an optimal allocation that satisfies Claim B.1 but not Claim B.2. Choose j as the least i such that $h_{m+ 2i-1}$ and $h_{m + 2i}$ are allocated to the same part, say L. Let $\{h_{m+ 2j-1}, h_{m + 2j}, \ldots , h_{m + 2j+k}\}$ be a set of houses allocated to agents in L such that $h_{m + 2j+k+1}$ is allocated to some agent in R ($k \ge 0$). Create an allocation $\pi '$ from $\pi$ by swapping $h_{m + 2j+k}$ and $h_{m + 2j+k+1}$. We can compare the envy between $\pi '$ and $\pi$.

$$\begin{aligned}&\textsf{Envy}(\pi ', G) - \textsf{Envy}(\pi , G) \\&\quad = [n^{>}_{L, \pi }(v(h_{m+2j+k})) - n^{<}_{L, \pi }(v(h_{m+2j+k}))](v(h_{m+2j+k+1}) - v(h_{m+2j+k})) \\&\qquad + [n^{<}_{R, \pi }(v(h_{m+2j+k+1})) - n^{>}_{R, \pi }(v(h_{m+2j+k+1}))](v(h_{m+2j+k+1}) - v(h_{m+2j+k})) \\&\quad = (v(h_{m+2j+k+1}) - v(h_{m+2j+k})) [n^{>}_{L, \pi }(v(h_{m+2j+k})) - n^{<}_{L, \pi }(v(h_{m+2j+k})) \\&\qquad + n^{<}_{R, \pi }(v(h_{m+2j+k+1})) - n^{>}_{R, \pi }(v(h_{m+2j+k+1}))] \\&\quad = (v(h_{m+2j+k+1}) - v(h_{m+2j+k})) [(r - (m + k+2 + j-1)) \\&\qquad - (m + k+1+j-1) + (j-1) - (s - j)] \\&\quad =[2j - 2(k+j) - 2](v(h_{m+2j+k+1}) - v(h_{m+2j+k})) \\&\quad = [-2k - 2](v(h_{m+2j+k+1}) - v(h_{m+2j+k})) \\&\quad <0. \end{aligned}$$

The final inequality holds since $k \ge 0$. Again, we contradict the optimality of $\pi$. $\square$

Claim B.2 also implies that none of the final $m = (r - s)/2$ houses are allocated to agents in R; this is because all agents in R have already been assigned houses by Claim B.2. We can therefore conclude that these houses must be allocated to agents in L in any optimal allocation.

To show that any allocation that satisfies the optimal even property has the same aggregate envy, we use a swapping based argument similar to Theorem 4.5. Let $\pi$ be any allocation that satisfies the optimal even property. Pick an arbitrary $i \in [s]$ and let $\pi '$ be the allocation that results from swapping $h_{m + 2i - 1}$ and $h_{m + 2i}$ in $\pi$. Assume that $h_{m + 2i - 1}$ is allocated to L in $\pi$. The proof for R flows similarly. Let us compare the envy of the two allocations.

$$\begin{aligned}&\textsf{Envy}(\pi ', G) - \textsf{Envy}(\pi , G) \\&\quad = [n^{>}_{L, \pi }(v(h_{m+2i-1})) - n^{<}_{L, \pi }(v(h_{m + 2i-1}))](v(h_{m + 2i}) - v(h_{m + 2i-1})) \\&\qquad + [n^{<}_{R, \pi }(v(h_{m + 2i})) - n^{>}_{R, \pi }(v(h_{m + 2i}))](v(h_{m + 2i}) - v(h_{m + 2i-1})) \\&\quad = (v(h_{m + 2i}) - v(h_{m + 2i-1})) \\&\qquad [n^{>}_{L, \pi }(v(h_{m + 2i-1})) - n^{<}_{L, \pi }(v(h_{m + 2i-1})) + n^{<}_{R, \pi }(v(h_{m + 2i})) - n^{>}_{R, \pi }(v(h_{m + 2i}))] \\& \quad = [(r- (i+m)) - (m + i-1) + (i-1) - (s - i)](v(h_{m + 2i}) - v(h_{m + 2i-1})) \\& \quad =0. \end{aligned}$$

Case 2 $r - s$ is odd. This is, unsurprisingly, very similar to the previous case. We similarly split the proof into two claims.

Claim B.3

Any optimal allocation allocates the first m houses to agents in L.

The proof of this claim is exactly the same as the proof to the Claim B.1. The key difference in this case is that $m = (r-s-1)/2$ but this does not affect the proof as we can still use the inequality $k < (r-s)/2$ since $k < m$. So we move on to the second claim.

Claim B.4

In any optimal allocation, for any $i \in [s]$, $h_{m+ 2i-1}$ is allocated to some agent in L and $h_{m+ 2i}$ is allocated to some agent in R.

Proof

This proof is again very similar to Claim B.2. However, there are some subtle differences.

Assume for contradiction that the claim is not true. Let $\pi$ be an optimal allocation that satisfies Claim B.3 but not Claim B.4. Choose j as the least i where the claim is violated. That is, either $h_{m + 2j-1}$ is allocated to R or $h_{m + 2j}$ is allocated to L. In this proof, we assume the latter has occured. The proof for the former is very similar. In other words, both $h_{m+ 2j-1}$ and $h_{m+ 2j}$ are allocated to some agents in L. Let $h_{m+ 2j-1}, h_{m + 2j}, \ldots , h_{m + 2j+k}$ be a set of houses allocated to agents in L such that $h_{m + 2j+k+1}$ is allocated to some agent in R. Let $\pi '$ be the allocation that results from swapping $h_{m + 2j+k}$ and $h_{m + 2j+k+1}$. We can compare the envy between $\pi '$ and $\pi$:

$$\begin{aligned}&\textsf{Envy}(\pi ', G) - \textsf{Envy}(\pi , G) \\&\quad = [n^{>}_{L, \pi }(v(h_{m +2j+k})) - n^{<}_{L, \pi }(v(h_{m+2j+k}))](v(h_{m+2j+k+1}) - v(h_{m+2j+k})) \\&\qquad + [n^{<}_{R, \pi }(v(h_{m+2j+k+1})) - n^{>}_{R, \pi }(v(h_{m+2j+k+1}))](v(h_{m+2j+k+1}) - v(h_{m+2j+k})) \\&\quad = (v(h_{m+2j+k+1}) - v(h_{m+2j+k})) [n^{>}_{L, \pi }(v(h_{m+2j+k})) - n^{<}_{L, \pi }(v(h_{m+2j+k})) \\&\qquad + n^{<}_{R, \pi }(v(h_{m+2j+k+1})) - n^{>}_{R, \pi }(v(h_{m+2j+k+1}))] \\&\quad = (v(h_{m+2j+k+1}) - v(h_{m+2j+k})) \\&\qquad [(r - (m + k+2 + j-1)) - (m + k+1+j-1) + (j-1) - (s - j)] \\&\quad =[2j - 2(k+j) - 1](v(h_{m+2j+k+1}) - v(h_{m+2j+k})) \\&\quad = [-2k - 1](v(h_{m+2j+k+1}) - v(h_{m+2j+k})) \\&\quad <0. \end{aligned}$$

The final inequality holds since $k \ge 0$. The optimality of $\pi$ has been contradicted. $\square$

Claim B.4 also implies that none of the final $m+1$ houses are allocated to agents in R. We can therefore conclude that these houses must be allocated to agents in L in any optimal allocation.

Note that the optimal odd property specifies exactly which houses must be allocated to L and R in any optimal allocation. Any two allocations which satisfy the optimal odd property can only differ over which agents in L and R houses are allocated to and not which houses are allocated to L and R. It is easy to see that this difference cannot lead to a difference in envy over the complete bipartite graph. $\square$

Appendix C: Technical Proofs from Section 5

Theorem 5.15

Let G be a disjoint union of cliques with arbitrary sizes, $K_{n_1} + \cdots + K_{n_r}$, where $n_1 \ge \ldots \ge n_r$. Then, G is splittable (but not necessarily strongly splittable if the $n_i$’s are not all equal). In particular, for all $1 \le i < j \le r$, in every optimal allocation, $K_{n_i}$ splits $K_{n_j}$.

Proof

Let $\pi$ be any minimum envy allocation. Assume for contradiction that there exist two cliques (say K and $K'$) such that $|K| > |K'|$ and K does not receive a contiguous set of valuations with respect to the houses in $K \cup K'$. The case where $|K| = |K'|$ has been shown in Theorem 5.13. Let the houses in $K \cup K'$ have values $\{a_1, a_2, \ldots , a_{|K \cup K'|}\}$ such that $a_1< a_2< \ldots < a_{|K \cup K'|}$. Since each house has a unique value, we refer to houses using their values for the rest of this proof.

By our assumptions, the houses allocated to K must be split. Therefore there must be some houses in $K'$ that are better than the houses allocated to some nodes in K and worse than houses allocated to other nodes in K. This can be formalized as follows

$$\begin{aligned}&\pi (a_j) \in K' \text { for all } j \in [\ell ] \text { and some } \ell \ge 0 \\&\pi (a_{l+j}) \in K \text { for all } j \in [m] \text { and some } m> 0 \\&\pi (a_{l+ m + j}) \in K' \text { for all } j \in [k] \text { and some } k > 0 \\&\pi (a_{l+m+k+1}) \in K \end{aligned}$$

We will frequently use the notation $n_{K, \pi }^{<}(x)$ and $n_{K, \pi }^{>}(x)$ (defined in Definition 2.3) for each clique K.

Construct the allocation $\pi '$ starting at $\pi$ and swapping the houses $a_{l+m+k}$ and $a_{l+m+k+1}$. For any node in K whose value is less than $a_{l+m+k+1}$ under $\pi$, the total envy between them and their neighbors increases by $a_{l+m+k+1} - a_{l+m+k}$ in $\pi '$. For any node in K whose value is greater than $a_{l+m+k+1}$ under $\pi$, the total envy between them and their neighbors decreases by $a_{l+m+k+1} - a_{l+m+k}$ in $\pi '$. We can show something similar for $K'$. This gives us the total change in envy as

$$\begin{aligned}&\textsf{Envy}(\pi ', G) - \textsf{Envy}(\pi , G) \\&\quad = \textsf{Envy}(\pi ', K \cup K') - \textsf{Envy}(\pi , K \cup K') \\&\quad = \left[ n^{<}_{K', \pi }(a_{l+m+k}) - n^{>}_{K', \pi }(a_{l+m+k}) \right] \left( a_{l+m+k+1} - a_{l+m+k} \right) \\&\qquad + \left[ n^{>}_{K, \pi }(a_{l+m+k+1}) - n^{<}_{K, \pi }(a_{l+m+l+1}) \right] \left( a_{l+m+k+1} - a_{l+m+k} \right) \\&\quad = \left( a_{l+m+k+1} - a_{l+m+k} \right) \\&\qquad \left[ n^{<}_{K', \pi }(a_{l+m+k}) - n^{>}_{K', \pi }(a_{l+m+k}) + n^{>}_{K, \pi }(a_{l+m+k+1}) - n^{<}_{K, \pi }(a_{l+m+l+1}) \right] \\&\quad = \left[ (l+k-1) - (|K'| - l - k) + (|K| - (m+1)) - m \right] (a_{l+m+k+1} - a_{l+m+k}) \\&\quad = \left[ |K| - |K'| + 2(l+k) - 2m - 2 \right] (a_{l+m+k+1} - a_{l+m+k}) \end{aligned}$$

Note that due to the optimality of $\pi$, we must have $\textsf{Envy}(\pi ', G) - \textsf{Envy}(\pi , G) \ge 0$. Since $a_{l+m+k+1} - a_{l+m+k} > 0$ by construction, this implies $|K| - |K'| + 2(l+k) - 2m - 2 \ge 0$. Removing the $-2$, we get $|K| - |K'| + 2(l+k) - 2m > 0$. This gives us the following observation.

Observation C.1

$|K'| - |K| - 2(l+k) + 2m < 0$

Construct another allocation $\pi ''$ as follows: start at $\pi$ and for every $j \in [\min \{m, k\}]$, swap $a_{l+m + 1 - j}$ with $a_{l+m+j}$. In each swap, we swap one house in K with one house in $K'$. Using a similar argument, we can compare the total envy of $\pi ''$ and $\pi$.

$$\begin{aligned}&\textsf{Envy}(\pi '', G) - \textsf{Envy}(\pi , G) \nonumber \\&\quad = \textsf{Envy}(\pi '', K \cup K') - \textsf{Envy}(\pi , K \cup K') \nonumber \\&\quad = \left[ n^{<}_{K, \pi }(a_{l+m+1 - \min \{m, k\}}) - n^{>}_{K, \pi }(a_{l+m}) + n^{>}_{K', \pi }(a_{l+m+\min \{m, k\}}) - n^{<}_{K', \pi }(a_{l+m+1}) \right] \nonumber \\&\qquad \left[ \sum _{j \in [\min \{m, k\}]}(a_{l+m+j} - a_{l+m+1-j}) \right] \nonumber \\&\quad = \left[ (m-\min \{m, k\}) - (|K| - m) + (|K'| - (l+\min \{k,m\})) - l \right] \nonumber \\&\qquad \left[ \sum _{j \in [\min \{m, k\}]}(a_{l+m+j} - a_{l+m+1-j}) \right] \nonumber \\&\quad = \left[ |K'| - |K| + 2m - 2(\min \{m, k\} + l) \right] \nonumber \\&\qquad \left[ \sum _{j \in [\min \{m, k\}]}(a_{l+m+j} - a_{l+m+1-j}) \right] \end{aligned}$$

(C1)

Note that the second term is always strictly positive since $a_{l+m+j} > a_{l+m+1-j}$ for all $j \in \min \{m,k\}$. If we show that the first term $|K'| - |K| + 2m - 2(\min \{m, k\} + l)$ is negative, we contradict the optimality of $\pi$. We have two possible cases.

Case 1 $k \le m$. In this case, (C1) reduces to

$$\begin{aligned}&\textsf{Envy}(\pi '', G) - \textsf{Envy}(\pi , G) \nonumber \\&\quad = \left[ |K'| - |K| + 2m - 2(k + l) \right] \left[ \sum _{j \in [\min \{m, k\}]}(a_{l+m+j} - a_{l+m+1-j})\right] \end{aligned}$$

From Observation C.1, the first term is negative.

Case 2 $k > m$. In this case, (C1) reduces to

$$\begin{aligned}&\textsf{Envy}(\pi '', G) - \textsf{Envy}(\pi , G) \\&\quad = \left[ |K'| - |K| + 2m - 2(m + l) \right] \left[ \sum _{j \in [\min \{m, k\}]}(a_{l+m+j} - a_{l+m+1-j}) \right] \\&\quad = \left[ |K'| - |K| - 2l \right] \left[ \sum _{j \in [\min \{m, k\}]}(a_{l+m+j} - a_{l+m+1-j}) \right] \end{aligned}$$

Since $|K| > |K'|$ and $l \ge 0$, the first term is negative.

To conclude, it cannot be the case that the houses in K are split. $\square$

Theorem 5.17

If $G = K_{r, s}$ for any $r, s \in {\mathbb {N}}$, then $G + G$ is strongly splittable.

Proof

Let $(V = L \cup R, E)$ and $(V' = L' \cup R', E')$ be the set of vertices and edges of each copy of G. There exists a bijective mapping $\tau : V \mapsto V'$ such that for every node $v \in V$, $\tau (v) \in L'$ if and only if $v \in L$.

Let $\pi$ be any allocation on $G+G$, we show that if $\pi$ does not allocate contiguous intervals to each component, we can create a better allocation $\pi '$.

Let $a_1< a_2 < \ldots a_{r+s}$ be the values allocated to the nodes in V and $b_1< b_2 < \ldots b_{r+s}$ be the values allocated to the nodes in $V'$ in some optimal allocation $\pi$. We rearrange the goods allocated to $V'$ such that if node $v \in V$ receives $a_i$, then node $\tau (v)$ receives $b_{r+s - i}$. If the allocation of a values to V is optimal, then from our characterization of bipartite graphs (Theorem 4.6), we know that this allocation of b houses to $V'$ is optimal as well.

If each component is not allocated a contiguous interval, the least valued $r+s$ houses must have some a values and some b values. Let’s call the least valued $r+s$ houses $H'$ and let’s say there are k $a_i$’s in $H'$. Therefore $H'$ contains $a_1, a_2, \ldots , a_k$ and $b_1, b_2, \ldots , b_{r+s-k}$.

We create a new allocation $\pi '$ starting at $\pi$ and for all $i \in [k]$, we swap $a_i$ with $b_{r+s-i}$. Note that for each house among the least-valued $r + s$ houses, if $a_i$ is allocated to $v \in V$, we swap the houses given to v and $\tau (v)$, thereby creating $\pi '$ from $\pi$.

Let us now compute the change in envy between $\pi '$ and $\pi$. We do this by showing that, for every edge $(u, v) \in E$, the total sum of the envies along the edges (u, v) and $(\tau (u), \tau (v))$ decreases. Before we go into the math, note that if $(u, v) \in E$, then $(\tau (u), \tau (v)) \in E'$ by our definition of $\tau$.

Case 1 u and v are unaffected by the swap. Then $\tau (u)$ and $\tau (v)$ are unaffected as well. Therefore the total envy along these two edges does not change.

Case 2 u and v are both affected by the swap. Then, $\textit{envy}_{\pi '}(u,v) = \textit{envy}_{\pi }(\tau (u),\tau (v))$ and $\textit{envy}_{\pi }(u,v) = \textit{envy}_{\pi '}(\tau (u),\tau (v))$. Therefore, the total envy along these two edges does not change.

Case 3 Only u is affected by the swap. This means $\tau (v)$ is not affected by the swap. The total envy along these two edges under $\pi$ is

$$\begin{aligned} \textit{envy}_{\pi }(u,v) + \textit{envy}_{\pi }(\tau (u),\tau (v)) = (a_j - a_i) + (b_{r+s-i} - b_{r+s-j}) \end{aligned}$$

where $j> k > i$. This can be re-written as

$$\begin{aligned} \textit{envy}_{\pi }(u,v) + \textit{envy}_{\pi }(\tau (u),\tau (v))&= 2\min \{a_j, b_{r+s -i}\} + |a_j - b_{r+s-i}| \\&\quad - 2\max \{a_i, b_{r+s -j}\} + |a_i - b_{r+s-j}| \end{aligned}$$

The total envy along these two edges under $\pi '$ is

$$\begin{aligned} \textit{envy}_{\pi '}(u,v) + \textit{envy}_{\pi '}(\tau (u),\tau (v)) = |a_j - b_{r+s-i}| + |a_i - b_{r+s-j}| \end{aligned}$$

The change in envy is

$$\begin{aligned} 2\max \{a_i, b_{r+s -j}\} - 2\min \{a_j, b_{r+s -i}\} < 0 \end{aligned}$$

The inequality holds since $j> k > i$.

When $k \ge 1$, at least one edge belongs to Case 3 and so the total envy of $\pi '$ is strictly less than the total envy of $\pi$. $\square$

Theorem 5.19

Let G be a disjoint union of symmetric complete bipartite graphs $K_{n_1, n_1} + K_{n_2, n_2} + \cdots + K_{n_{\ell }, n_{\ell }}$, where $n_1 \ge n_2 \ge \ldots \ge n_{\ell }$. Then G is splittable (but not necessarily strongly splittable if $n_1 > n_\ell$) and the order of splittability is $K_{n_1, n_1}, \ldots , K_{n_{\ell }, n_{\ell }}$.

Proof

We prove complete symmetric bipartite graphs are not strongly splittable in Proposition 5.20, so we focus on proving splittability here. Consider two complete bipartite graphs $G_1 = K_{r, r}$ and $G_2 = K_{s, s}$ such that $r < s$. Assume houses with values $a_1, \ldots a_{2r + 2s}$ such that $a_1< \ldots < a_{2r+2s}$ are allocated to these two graphs. Since house values are unique, we will say the value $a_i$ is allocated to a node j if the unique house with value $a_i$ is allocated to the node j.

We need to show that $K_{s, s}$ is allocated a contiguous interval of values in at least one optimal allocation. Assume for contradiction that this is not true. Let $\pi$ be an allocation where

$$\begin{aligned}&a_1, \ldots , a_{\ell _1} \text { is allocated to }G_1\text { for some }\ell _1 \ge 0 \\&a_{\ell _1 + 1}, \ldots , a_{\ell _1 + \ell _2} \text { is allocated to }G_2\text { for some }\ell _2> 0 \\&a_{\ell _1+ \ell _2 + 1}, \ldots , a_{\ell _1 + \ell _2 + \ell _3} \text { is allocated to }G_1\text { for some }\ell _3> 0 \\&a_{\ell _1+\ell _2 + \ell _3+ 1} \text { is allocated to }G_2\text { for some }\ell _3 > 0 \end{aligned}$$

Since we assumed no optimal allocation gives a contiguous set of values to $G_2$, all optimal allocations must have the above structure for some $\ell _1, \ell _2$ and $\ell _3$. If there are multiple optimal allocations, pick one such that $\ell _1$ is maximized. Break any further ties by picking one such that $\ell _2$ is maximized. Finally, break ties by ensuring $\ell _1 + \ell _2 + \ell _3$ is minimized. If there are still multiple envy minimizing allocations, pick one arbitrarily.

Since $G_1$ and $G_2$ are complete bipartite graphs, we refer to the nodes in the ‘left’ part of $G_1$ and $G_2$ using $L_1$ and $L_2$ respectively. Similarly, we refer to the ‘right’ part of nodes using $R_1$ and $R_2$. Since we assume $\pi$ is optimal, the allocations to $G_1$ and $G_2$ must satisfy the structural properties from Theorem 4.5. Specifically, if the values $b_1, \ldots , b_{2y}$ are allocated to $G_i$ for some $i \in [2]$, we assume $b_1, b_3, b_5, \ldots , b_{2y - 1}$ are allocated to $L_i$.

Swap $a_{\ell _1 + \ell _2 + \ell _3}$ with $a_{\ell _1 + \ell _2 + \ell _3 + 1}$ in $\pi$ to create a new allocation $\pi '$. Let us compare the envies of $\pi$ and $\pi '$. Observe that

$$\begin{aligned}&\textsf{Envy}(\pi ', G_1 + G_2) - \textsf{Envy}(\pi , G_1 + G_2) \nonumber \\&\quad = (a_{\ell _1 + \ell _2 + \ell _3 + 1} - a_{\ell _1 + \ell _2 + \ell _3}) \times \nonumber \\&\qquad \bigg [ \big \lceil \frac{n^{<}_{G_1}(a_{\ell _1 + \ell _2 + \ell _3})}{2} \big \rceil - \left( r - \big \lceil \frac{n^{<}_{G_1}(a_{\ell _1 + \ell _2 + \ell _3})}{2} \big \rceil \right) \nonumber \\&\qquad - \big \lceil \frac{n^{<}_{G_2}(a_{\ell _1 + \ell _2 + \ell _3 + 1})}{2} \big \rceil + \left( s -\big \lceil \frac{n^{<}_{G_2}(a_{\ell _1 + \ell _2 + \ell _3 + 1})}{2} \big \rceil \right) \bigg ] \nonumber \\&\quad = (a_{\ell _1 + \ell _2 + \ell _3 + 1} - a_{\ell _1 + \ell _2 + \ell _3}) \times \left[ 2\big \lceil \frac{\ell _1 + \ell _3 - 1}{2} \big \rceil - 2 \big \lceil \frac{\ell _2}{2} \big \rceil + s - r \right] \end{aligned}$$

(C2)

where $n_{G_i, \pi }^{<}(x)$ and $n_{G_i, \pi }^{>}(x)$ are defined according to Definition 2.3.

Fig. 13

Measuring the value $\textsf{Envy}(\pi ', G_1 + G_2) - \textsf{Envy}(\pi , G_1 + G_2)$. We assume values are allocated in increasing order from the top to the bottom with least valued nodes at the top of the graph and the highest valued nodes at the bottom of the graph. Only the edges which see a change in envy are drawn. The exact change in envy for the edges in $K_{r, r}$ is described. A similar argument can be used to measure the exact change in envy in $K_{s, s}$

An explanation for how this expression is computed is presented in Fig. 13. Note that (C2) must be strictly positive by our choice of optimal allocation — $\pi '$ either has a bigger $\ell _2$ or has a smaller $\ell _1 + \ell _2 + \ell _3$ than $\pi$. The first term in (C2) is always positive, the second term only contains integers, so it must be lower bounded by 1. This gives us the following observation:

Observation C.2

$2\big \lceil \frac{\ell _1 + \ell _3 - 1}{2} \big \rceil - 2 \big \lceil \frac{\ell _2}{2} \big \rceil + s - r \ge 1$.

Let us now construct a third allocation $\pi ''$ from $\pi$ by swapping $\{a_{\ell _1 + \ell _2 - \min \{\ell _2, \ell _3\} + 1}, \ldots , a_{\ell _1 + \ell _2}\}$ from $G_2$ with $\{a_{\ell _1 + \ell _2 + 1}, \ldots , a_{\ell _1 + \ell _2 + \min \{\ell _2, \ell _3\}}\}$ from $G_1$. When we swap these two sets, we ensure we swap them in order. That is,

$$\begin{aligned}&a_{\ell _1 + \ell _2 - \min \{\ell _2, \ell _3\} + 1} \text { is swapped with } a_{\ell _1 + \ell _2 + 1}, \\&a_{\ell _1 + \ell _2 - \min \{\ell _2, \ell _3\} + 2} \text { is swapped with } a_{\ell _1 + \ell _2 + 2}, \end{aligned}$$

and so on. Note that we swap exactly $\min \{\ell _2, \ell _3\}$ values and with this careful swap, the edges between the values in each of these sets is preserved. That is, an edge between $a_{\ell _1 + \ell _2 + 1}$ and $a_{\ell _1 + \ell _2 + 2}$ exists in $\pi ''$ if and only if it exists in $\pi$. Using an argument similar to Fig. 13, we can find the difference in envy between $\pi ''$ and $\pi$ as:

$$\begin{aligned}&\textsf{Envy}(\pi '', G_1 + G_2) - \textsf{Envy}(\pi , G_1 + G_2) \nonumber \\&\quad = c_1 \Bigg [ \big \lceil \frac{n^{<}_{G_2}(a_{\ell _1 + \ell _2 - \min \{\ell _2, \ell _3\} + 1})}{2} \big \rceil - \left( s - \left\lfloor \frac{\min \{\ell _2, \ell _3\}}{2}\right\rfloor - \big \lceil \frac{n^{<}_{G_2}(a_{\ell _1 + \ell _2 - \min \{\ell _2, \ell _3\} + 1})}{2} \big \rceil \right) \nonumber \\&\qquad - \big \lceil \frac{n^{<}_{G_1}(a_{\ell _1 + \ell _2 + 1})}{2} \big \rceil + \left( r - \left\lfloor \frac{\min \{\ell _2, \ell _3\}}{2}\right\rfloor - \big \lceil \frac{n^{<}_{G_1}(a_{\ell _1 + \ell _2 + 1})}{2} \big \rceil \right) \Bigg ] \nonumber \\&\qquad + c_2 \Bigg [ \left\lfloor \frac{n^{<}_{G_2}(a_{\ell _1 + \ell _2 - \min \{\ell _2, \ell _3\} + 1})}{2} \right\rfloor - \left( s - \big \lceil \frac{\min \{\ell _2, \ell _3\}}{2}\big \rceil - \left\lfloor \frac{n^{<}_{G_2}(a_{\ell _1 + \ell _2 - \min \{\ell _2, \ell _3\} + 1})}{2} \right\rfloor \right) \nonumber \\&\qquad - \left\lfloor \frac{n^{<}_{G_1}(a_{\ell _1 + \ell _2 + 1})}{2} \right\rfloor + \left( r - \big \lceil \frac{\min \{\ell _2, \ell _3\}}{2}\big \rceil - \left\lfloor \frac{n^{<}_{G_1}(a_{\ell _1 + \ell _2 + 1})}{2} \right\rfloor \right) \Bigg ] \nonumber \\&\qquad \text {where } c_1 = \sum _{j = 0}^{\big \lceil \frac{\min \{\ell _2, \ell _3\}}{2}\big \rceil - 1} \left( a_{\ell _1 + \ell _2 + 2j + 1} - a_{\ell _1 + \ell _2 - \min \{\ell _2, \ell _3\} + 2j + 1}\right) \nonumber \\&\qquad \text { and } c_2 = \sum _{j = 0}^{\left\lfloor \frac{\min \{\ell _2, \ell _3\}}{2}\right\rfloor - 1} \left( a_{\ell _1 + \ell _2 + 2j + 2} - a_{\ell _1 + \ell _2 - \min \{\ell _2, \ell _3\} + 2j + 2}\right) . \end{aligned}$$

The only thing to keep in mind about $c_1$ and $c_2$ are that they are positive constants. The above expression can be simplified as

$$\begin{aligned}&\textsf{Envy}(\pi '', G_1 + G_2) - \textsf{Envy}(\pi , G_1 + G_2) \nonumber \\&\quad = c_1 \left[ 2\big \lceil \frac{\ell _2 - \min \{\ell _2, \ell _3\}}{2} \big \rceil - 2\big \lceil \frac{\ell _1}{2} \big \rceil + r - s \right] \nonumber \\&\qquad + c_2 \left[ 2\left\lfloor \frac{\ell _2 - \min \{\ell _2, \ell _3\}}{2} \right\rfloor - 2\left\lfloor \frac{\ell _1}{2} \right\rfloor + r - s \right] \nonumber \\&\quad \le (c_1+ c_2) \nonumber \\&\qquad \left[ \max \left\{ 2\big \lceil \frac{\ell _2 - \min \{\ell _2, \ell _3\}}{2} \big \rceil - 2\big \lceil \frac{\ell _1}{2} \big \rceil , 2\left\lfloor \frac{\ell _2 - \min \{\ell _2, \ell _3\}}{2} \right\rfloor - 2\left\lfloor \frac{\ell _1}{2} \right\rfloor \right\} + r - s \right] \end{aligned}$$

(C3)

Again, (C3) must be strictly positive due to our choice of optimal allocation. $c_1$ and $c_2$ are positive constants, so this comes down to the second term. Note immediately that the second term cannot be positive if $\ell _2 \le \ell _3$. Therefore, we can assume $\ell _2 > \ell _3$, and using the fact that all the terms inside the second term are integers, we can make the following observation:

Observation C.3

$\max \left\{ 2\big \lceil \frac{\ell _2 - \ell _3}{2} \big \rceil - 2\big \lceil \frac{\ell _1}{2} \big \rceil , 2\left\lfloor \frac{\ell _2 - \ell _3}{2} \right\rfloor - 2\left\lfloor \frac{\ell _1}{2} \right\rfloor \right\} + r - s \ge 1$.

Adding up Observations C.2 and C.3, we get

$$\begin{aligned}&\max \bigg \{2\big \lceil \frac{\ell _2 - \ell _3}{2} \big \rceil - 2\big \lceil \frac{\ell _1}{2} \big \rceil + 2\big \lceil \frac{\ell _1 + \ell _3 - 1}{2} \big \rceil - 2 \big \lceil \frac{\ell _2}{2} \big \rceil , \\&\qquad 2\left\lfloor \frac{\ell _2 - \ell _3}{2} \right\rfloor - 2\left\lfloor \frac{\ell _1}{2} \right\rfloor + 2\big \lceil \frac{\ell _1 + \ell _3 - 1}{2} \big \rceil - 2 \big \lceil \frac{\ell _2}{2} \big \rceil \bigg \} \ge 2 \end{aligned}$$

It is easy to verify that the left hand side in the above inequality is upper bounded at 1; if there were no ceilings or floors, the left hand side would equal $-1$. The ceilings and floors, adversarially set, can only increase this value by 2. Therefore, the above expression can never be true and we have arrived at a glorious contradiction. $\square$

Proposition 5.23

There exists a 3-regular unsplittable graph.

Proof

The following lemma will prove to be useful.

Lemma C.4

For any wheel $W_{2t + 1}$, and any two non-rim vertices $u_1, u_2 \in V(W_{2t + 1})$, there are three $u_1$-$u_2$ paths that are disjoint except at the endpoints.

Proof

The shortest path $P_0$ along the outer cycle is one path from $u_1$ to $u_2$. Call the remainder of the outer cycle the “longer $u_1$-$u_2$ path”. Now, consider the path $P_1$ going from $u_1$ to its mate along its diagonal, and then to $u_2$ along the longer $u_1$-$u_2$ path. Also consider the path $P_2$ that takes $u_1$ to the mate of $u_2$ along the cycle on the longer $u_1$-$u_2$ path, and then across to $u_2$ on the diagonal. Note that $P_0$, $P_1$, and $P_2$ are all internally disjoint paths on this graph from $u_1$ to $u_2$. See Fig. 14 for an illustration. $\square$

Fig. 14

Illustrative example of three disjoint paths between non-rim vertices $u_1$ and $u_2$, drawn here in three different colors: red, blue, and green

We continue with the proof. Consider the instance shown in Fig. 11. Call the inter-cluster gaps $I_1$, $I_2$, and $I_3$ respectively. By analyzing the size of any minimum cut in the given graph with exactly 401 vertices on one side, we can easily show that every allocation will need to have at least one edge go over $I_1$ (since there is no way to put 401 vertices of the graph without having at least one edge across the cut). Using a similar argument on minimum cuts with exactly 702 (resp. 903) vertices on one side, we can also show that at least two (resp. one) edges must go over $I_2$ (resp. $I_3$) in every allocation. So, the optimal envy must be at least $|I_1| + 2|I_2| + |I_3|$. Furthermore, this is realizable by the obvious allocation that maps the cluster sizes to the corresponding wheels. Therefore, any allocation that puts more than one edge on either $I_1$ or $I_3$, or more than two edges on $I_2$, must be strictly suboptimal.

Consider any optimal allocation. We first claim that $W_{101}$ must be entirely inside the fourth cluster. Otherwise, some other wheel $W'$ has its vertices appearing in the last cluster. If only the rim of $W'$ appears in the last cluster, then its two neighbors in $W'$ both appear in other clusters, so that $I_3$ has at least two edges passing over it, contradiction. So some non-rim vertex of $W'$ appears in the fourth cluster. The fourth cluster is not enough to fit all of $W'$, and so some non-rim vertex from $W'$ appears in a different cluster as well. By Lemma C.4, this requires at least three edges over $I_3$, contradiction. Therefore, $W_{101}$ fits snugly inside the fourth cluster.

We now claim that $W_{201}$ must be entirely inside the third cluster. Otherwise, either $W_{301}$ or $W_{401}$ has some presence in the third cluster, say $W_{301}$. If this is a non-rim vertex, then again by Lemma C.4, we must have at least three edges over $I_2$, contradiction. So at best, the third cluster can have a rim vertex from $W_{301}$. This vertex’s neighbors in $W_{301}$ must be on either the first or second cluster, accounting for two edges above the interval $I_2$. But then, the third cluster must have some vertex from the bicycle $B_{401, 201}$, but also does not have enough space to fit the entire bicycle. Hence, there must also be at least one edge over the interval $I_2$ from the bicycle $B_{401, 201}$, accounting for a total of three or more edges over $I_2$, contradiction. A similar argument holds when $W_{401}$ has some presence in the third cluster.

Finally, we claim that the copy of $W_{401}$ must be entirely inside the first cluster. Otherwise, there is at least one vertex from $W_{301}$ in the first cluster, and therefore at least one vertex from $W_{401}$ in the second cluster. Of course, the second cluster cannot fit in at least 100 vertices from $W_{401}$, and so there is at least one non-rim $W_{401}$-vertex in the second cluster (otherwise its two neighbors correspond to two edges over $I_1$, contradiction), and at least one non-rim $W_{401}$ vertex in the first cluster, which by Lemma C.4 is a contradiction. $\square$