On the core and nucleolus of directed acyclic graph games

Abstract

We introduce directed acyclic graph (DAG) games, a generalization of standard tree games, to study cost sharing on networks. This structure has not been previously analyzed from a cooperative game theoretic perspective. Every monotonic and subadditive cost game—including monotonic minimum cost spanning tree games—can be modeled as a DAG-game. We provide an efficiently verifiable condition satisfied by a large class of directed acyclic graphs that is sufficient for the balancedness of the associated DAG-game. We introduce a network canonization process and prove various structural results for the core of canonized DAG-games. In particular, we characterize classes of coalitions that have a constant payoff in the core. In addition, we identify a subset of the coalitions that is sufficient to determine the core. This result also guarantees that the nucleolus can be found in polynomial time for a large class of DAG-games.

Appendix

Theorem 2

Let N be a player set and ${\hat{c}}:2^N \rightarrow \mathbb {R}$ a monotonic, subadditive cost function. There exists a DAG-network $\mathcal {D}=(G(V,A),\delta )$ and a residency mapping $\mathcal {R}:N\rightarrow V$ such that

$$\begin{aligned} c_{(\mathcal {D},\mathcal {R})}(S)={\hat{c}}(S) \quad \forall S \subseteq N. \end{aligned}$$

Proof

We will use the characteristic DAG-representation of ${\hat{c}}$. We have to show that the cost of the cheapest trunk of any coalition S equals to ${\hat{c}}(S)$. This is trivially true for the singleton coalitions. Each player—as a singleton—will use its direct connection to the root as any other route could be only more expensive due to the monotonicity of ${\hat{c}}$. Now let S be an arbitrary non-singleton coalition. We may suppose that $\delta (a_S)>0$, otherwise both ${\hat{c}}(S)$ and $c_{(\mathcal {D},\mathcal {R})}(S)$ are trivially zero. Let $T'$ be the trunk whose arc set consists of $a_S$ and the zero arcs $\{a^i_S \ | \ i \in S\}$ and whose node set consists of the nodes spanning these arcs. We will prove $T' \in T_S$. Let $T \in T_S$ and suppose that $\{a_{S_1},\dots ,a_{S_k}\}$ are the non-zero arcs of T. If $S=S_{\ell }$ for some $\ell \in \{1,\dots ,k\}$, then by deleting the $a_{S_j}$, $j\ne \ell$ arcs and the zero arcs that enter $\mathbf {p}_{S_j}$, $j\ne \ell$ we can obtain $T'$. This would imply $T' \in T_S$ as the cost could only decrease by deleting these arcs. If $S\subset S_{\ell }$ for some $\ell \in \{1,\dots ,k\}$, then again we can obtain a weakly cheaper trunk by connecting the members of S via $a_S$ and deleting all the other $a_{S_{\ell }}$ arcs. Thus $S_{\ell } \subset S$ for all $\ell \in \{1,\dots ,k\}$. Although a player may be connected to more than one $\mathbf {p}_{S_{\ell }}$ node, in real he only needs one of the $a_{S_{\ell }}$ arcs to reach the root. Let us assign the players of S to one of the $a_{S_{\ell }}$ arcs. Let us denote by $S^{\ell } \subset S$ those players of S that were assigned to $S_{\ell }$. Note that the $\{S^{\ell } \ | \ {\ell }\in \{1,\dots ,k\}, S^{\ell }\ne \emptyset \}$ coalitions comprise a partition of S. If $S^{\ell }=\emptyset$ for some $\ell \in \{1,\dots ,k\}$ then $a_{S_{\ell }}$ and the entering zero arcs of $\mathbf {p}_{S_{\ell }}$ can be deleted. If $\emptyset \ne S^{\ell } \subset S_{\ell }$ then the players of $S^{\ell }$ can be reassigned to $a_{S^{\ell }}$, that is, we can delete $a_{S_{\ell }}$ and the entering zero arcs of $\mathbf {p}_{S_{\ell }}$ and construct $a_{S^{\ell }}$ and the the entering zero arcs of $\mathbf {p}_{S^{\ell }}$ instead. Due to the monotonicity, the cost can only decrease this way, while all the players of S are still able to reach the root. Let us perform this transformation for all the $S_{\ell }$ coalitions. We obtain a trunk $\hat{T} \in T_S$ such that

$$\begin{aligned} A(\hat{T})=\{a_{S^{\ell }} \ | \ \ell \in \{1,\dots ,k\}, S^{\ell }\ne \emptyset \}\cup \{a^i_{S^{\ell }} \ | \ \ell \in \{1,\dots ,k\},S^{\ell }\ne \emptyset ,i \in S\}. \end{aligned}$$

By the subadditivity of ${{\hat{c}}}$

$$\begin{aligned} {\hat{c}}(S)&\le \sum _{\ell \in \{1,\dots ,k\},S^{\ell }\ne \emptyset }{\hat{c}}(S^{\ell }) \\ \delta (a_S)&\le \sum _{\ell \in \{1,\dots ,k\},S^{\ell }\ne \emptyset }\delta (a_{S^{\ell }}) \end{aligned}$$

Thus, $T'$ is as least as cheap as $\hat{T}$, from which $T' \in T_S$ follows. $\square$

Lemma 1

The node set of every trunk that corresponds to a coalition $S\subset N$ can be obtained by deleting some branches from V. The removed branches can be chosen in such way that each of them originates from a passage. Formally for any $S \subset N$ there exists $Q_1,\dots , Q_k \subset V$ and $\mathbf {p}_1,\dots , \mathbf {p}_k \in V$ such that

$$\begin{aligned} V(T_S)=V{\setminus }\cup _{j=1}^kB_{\mathbf {p}_j}^{Q_j}, \end{aligned}$$

where $\mathbf {p}_j$ is a passage for all $j \in \{1,2,\dots ,k\}$.

Proof

Any trunk T has a representation where V(T) is obtained by removing branches from V (any single node is a branch in itself if we trim all its children). The only thing we need to prove is that these branches can be picked in such way that each of them originates from a passage. Let $\{\mathbf {p}_1,\dots ,\mathbf {p}_{k}\} \subset V{\setminus } V(T_S)$ denote those passages that connect to $V(T_S)$ from the outside, i.e. for which $\pi (\mathbf {p}_j)\in V(T_S)$ for all $j=1,\dots ,k$. Due to the definition of $T_S$ there exists at least one such passage. Note that any entering zero arc is included in the trunk of S even if no player of S resides there, due to the definition of $T_S$. If we remove all the $B_{\mathbf {p}_1},\dots ,B_{\mathbf {p}_{k}}$ branches from V it can happen that we removed some nodes in $V(T_S)$ as well i.e. $V{\setminus } (\cup _{j=1}^{k}B_{\mathbf {p}_j}) \subset V(T_S)$. In order to retain all the nodes of $V(T_S)$ we trim the $B_{\mathbf {p}_j}$ branches where they intersect with $V(T_S)$. Let $Q_j = V(T_S) \cap B_{\mathbf {p}_j}$ then $B_{\mathbf {p}_j}^{Q_j}$ is a proper branch for any j and $V(T_S) =V{\setminus } (\cup _{j=1}^{k}B_{\mathbf {p}_j}^{Q_j})$. $\square$

Lemma 4

If $B_{\mathbf {p}}^Q$ is a building block, then $x(N(B_{\mathbf {p}}^Q))= \tau (B_{\mathbf {p}}^Q,N)$ for any core allocation x.

Proof

Since $\pi (\mathbf {p})$ is free, it is a junction and $x(N(\pi (\mathbf {p})))=0$. We know from Theorem 3 that $exc(N(V{\setminus } B_{\pi (\mathbf {p})}),x)=0$ for any core allocation x. It follows that $exc(N((V{\setminus } B_{\pi (\mathbf {p})}) \cup \{\pi (\mathbf {p})\}),x)=0$ is also true. With a similar argument as in Lemma 3 it can be shown that $x(N(B_{\mathbf {p}}^Q))\le \tau (B_{\mathbf {p}}^Q,N)$.

Each node of Q has (at least) two semi-arc-disjoint paths that leads to the root. As $B_{\mathbf {p}}^Q$ does not contain a free node one of these paths for each node by-passes $B_{\mathbf {p}}^Q$. We prove this by contradiction. Let $\mathbf {q} \in Q$ an arbitrary free node. Suppose there exists two semi-arc-disjoint paths in $T_N$, $P_1$ and $P_2$ that leads from $\mathbf {q}$ to the root and crosses $B_{\mathbf {p}}^Q$. Let $\mathbf {q}_1 \in B_{\mathbf {p}}^Q \cap V(P_1)$ be such that there exist no other $\mathbf {q}' \in B_{\mathbf {p}}^Q \cap V(P_1)$ such that $\mathbf {q}' \prec \mathbf {q}_1$. Similarly let $\mathbf {q}_2$ be the node closest to the root that is an element of both $B_{\mathbf {p}}^Q$ and $P_2$. As $\mathbf {q}_1$ and $\mathbf {q}_2$ lie on semi-arc-disjoint paths, one of them—say $\mathbf {q}_1$—is not $\mathbf {p}$. Thus the $P_1$ path leaves the $B_{\mathbf {p}}^Q$ node set at $\mathbf {q}_1$ on a zero-arc. There leads a path in $T_N$ from $\mathbf {q}_1$ to $\pi (\mathbf {p})$ through $B_{\mathbf {p}}^{Q}$ that is arc-disjoint of $P_1$. As $\pi (\mathbf {p})$ is free there leads two semi-arc-disjoint paths $P_3$ and $P_4$ from $\pi (\mathbf {p})$ to the root. Without loss of generality we may assume that $P_1$ intersects with $P_3$ first (or at the same time as it intersects with $P_4$). Let us denote this node by $\mathbf {q}^*$. Note that if $\mathbf {q}^*$ is a common node of $P_3$ and $P_4$ it is a junction, otherwise the two paths would not be semi-arc-disjoint. Let $P_A$ be the path that starts from $\mathbf {q}_1$, follows $P_1$ till $\mathbf {q}^*$, then reaches the root following $P_3$. Let $P_B$ be the path that originates at $\mathbf {q}_1$,reaches $\pi (\mathbf {p})$ using only $T_N$-arcs and nodes from $B_{\mathbf {p}}^Q$, and goes to the root following $P_4$. By construction $P_A$ and $P_B$ are semi-arc-disjoint, thus $\mathbf {q}_1$ is free, which contradicts the assumption that $B_{\mathbf {p}}^Q$ is a building block.

It follows that there exists a path in $T_N$ for every $\mathbf {q} \in Q$ that leads to the root, that does not pass through any node of $B_{\mathbf {p}}^Q$. A straightforward consequence is that $B_{\mathbf {p}}^Q$ is a proper branch and every node in $V {\setminus } B_{\mathbf {p}}^Q$ can reach the root by using only $T_N$-arcs. Note that there is no zero-arc that leaves $B_{\mathbf {p}}^Q$ and enters in $V{\setminus } B_{\mathbf {p}}^Q$, otherwise $B_{\mathbf {p}}^Q$ would contain a free node. Thus the node set $V {\setminus } B_{\mathbf {p}}^Q$ corresponds to a trunk, namely to $T_{N(V {\setminus } B_{\mathbf {p}}^Q)}$. Finally, for any core allocation x

$$\begin{aligned} c_{\mathcal {D}}(N)-x(N)&=[c_{\mathcal {D}}(N(V{\setminus } B_{\mathbf {p}}^Q))-x(N(V{\setminus } B_{\mathbf {p}}^Q))]+[\tau (B_{\mathbf {p}}^Q,N)-x(N(B_{\mathbf {p}}^Q))] \\ 0&=[exc(N(V{\setminus } B_{\mathbf {p}}^Q)),x)]+[\tau (B_{\mathbf {p}}^Q,N)-x(N(B_{\mathbf {p}}^Q))] \end{aligned}$$

Both expressions in the square brackets are non-negative, thus $x(N(B_{\mathbf {p}}^Q))= \tau (B_{\mathbf {p}}^Q,N)$. $\square$

Lemma 5

Let $\cup _{j=1}^k B_{\mathbf {p}_j}^{F_j}$ be a union of branches such that $\mathbf {p}_j$ is a passage, $\pi (\mathbf {p}_j) \in F$ and $F_j \subset F$ for $j=1,\dots ,k$. Then $\cup _{j=1}^k B_{\mathbf {p}_j}^{F_j}$ can be decomposed into a disjoint union of building blocks and free nodes.

Proof

The proof proceeds by induction on the number of nodes. If $\cup _{j=1}^k B_{\mathbf {p}_j}^{F_j}$ consist of a single node, then $k=1$ and $B_{\mathbf {p}_1}^{F_1}$ must be a building block. Now suppose the lemma is true for node sets with less than l nodes and let $|\cup _{j=1}^k B_{\mathbf {p}_j}^{F_j}|=l$. Let $B_{\mathbf {p}_1}^Q$ be a branch where $Q=B_{\mathbf {p}_1} \cap F$ and let $B_{\mathbf {p}_1}^{Q'}$ be the standard form of this branch. Note that $B_{\mathbf {p}_1}^{Q'}$ is a building block and it is a subset of $B_{\mathbf {p}_1}^{F_1}$. Let us delete $B_{\mathbf {p}_1}^{Q'}$ from $B_{\mathbf {p}_1}^{F_1}$. If $Q'\cap B_{\mathbf {p}_1}^{F_1}$ is not empty we delete those nodes too (these are free as all the nodes of $Q'$ are free). If some descendant of a node in $Q'$ is a junction then it is free therefore it can be deleted too. If we deleted all the free nodes in this way and there are still some nodes in $B_{\mathbf {p}_1}^{F_1}$ then those must be passages. Let us denote these by $\mathbf {p}_1',\dots ,\mathbf {p}_K'$. Note that $\pi (\mathbf {p}_1'),\dots ,\pi (\mathbf {p}_K') \in F$. Hence the remaining nodes can be written as $\cup _{i=1}^K B_{\mathbf {p}_i'}^{F_1}\cup _{j=2}^k B_{\mathbf {p}_j}^{F_j}$. By reindexing $\mathbf {p}_i'$ we are done as $|\cup _{i=1}^K B_{\mathbf {p}_i'}^{F_1}\cup _{j=2}^k B_{\mathbf {p}_j}^{F_j}|<l$. $\square$