On the computation of probabilistic coalition structures

Abstract

In Coalition Structure Generation (CSG), one seeks to form a partition of a given set of agents into coalitions such that the sum of the values of each coalition is maximized. This paper introduces a model for Probabilistic CSG (PCSG), which extends the standard CSG model to account for the stochastic nature of the environment, i.e., when some of the agents considered at start may be finally defective. In PCSG, the goal is to maximize the expected utility of a coalition structure. We show that the problem is ${\mathsf{NP}}^{\mathsf {PP}}$-hard in the general case, but remains in ${\mathsf{NP}}$ for two natural subclasses of PCSG instances, when the characteristic function that gives the utility of every coalition is represented using a marginal contribution network (MC-net). Two encoding schemes are presented for these subclasses and empirical results are reported, showing that computing a coalition structure with maximal expected utility can be done efficiently for PCSG instances of reasonable size. This is an extended and revised version of the paper entitled “Probabilistic Coalition Structure Generation” published in the proceedings of KR’18, pages 663–664 [33].

Appendix: Proofs of propositions

Proposition 2

Let $\langle A, f\rangle$ be a CFG. Then every optimistic CFG extension of $\langle A, f\rangle$ is equivalent to $\langle A, f\rangle$.

Proof

Let $\langle A, f\rangle$ be a CFG and let $\langle A, g, p_\top \rangle$ be any optimistic CFG extension of $\langle A, f\rangle$. By definition of a CFG extension (cf. Definition 3), we have that $g(C, \omega _{A}) = f(C)$ for each coalition $C \subseteq A$. So by Eq. 2, for each coalition structure $CS \in \varPi _A$, we get that $G(CS, \omega _{A}) = \sum _{C \in CS}{g(C, \omega _{A})} = \sum _{C \in CS}{f(C)} = F(CS)$. Hence, by definition of $p_\top$ and $U(CS)$, we get that $U(CS) = \sum _{\omega _{P} \in \varOmega _A}p(\omega _{P}) \cdot G(CS, \omega _{P}) = G(CS, \omega _{A}) = F(CS)$. This concludes the proof. $\square$

Proposition 3

Let $\langle A, g, p\rangle$ be a PCFG.

• (i) If $g$ is subadditive, then the coalition structure $CS = \{\{a_i\} \mid a_i \in A\}$ is an optimal one.

• (ii) If $g$ is superadditive, then the coalition structure $CS = \{A\}$ is an optimal one.

Proof

We provide the proof only for (i), as (ii) can be proved in a similar manner. Let $\langle A, g, p\rangle$ be a PCFG, and assume that $g$ is subadditive. Let $CS = \{\{a_i\} \mid a_i \in A\}$ and $CS^{\prime} \in \varPi _A$, and let us prove that $U(CS^{\prime}) \le U(CS)$. By definition of $U(CS)$ and $U(CS^{\prime})$ (cf. Equation 3), we have that

$$\begin{aligned} U(CS^{\prime}) = \sum _{\omega _{P} \in \varOmega _A}{p(\omega _{P}) \cdot \sum _{C \in CS}{g(C, \omega _{P})}} \end{aligned}$$

and

$$\begin{aligned} U(CS) = \sum _{\omega _{P} \in \varOmega _A}{p(\omega _{P}) \cdot \sum _{a \in A}{g(\{a\}, \omega _{P})}}. \end{aligned}$$

So we just need to show that $\sum _{C \in CS}{g(C, \omega _{P})} \le \sum _{a \in A}{g(\{a\}, \omega _{P})}$. Yet $g$ is subadditive, so

$$\begin{aligned} \sum _{C \in CS}{g(C, \omega _{P})} \le \sum _{a \in A}{g(\{a\}, \omega _{P})}, \end{aligned}$$

from which we can conclude that

$$\begin{aligned} U(CS^{\prime}) \le U(CS). \end{aligned}$$

$\square$

Proposition 4

Given a PCFG $\langle A, g, p\rangle$, for every coalition structure $CS \subseteq \varPi _A$, we have that

$$\begin{aligned} U(CS) = \sum _{C \in CS}{u(C)}. \end{aligned}$$

Proof

We have that:

$$\begin{aligned} \begin{array}{ll} U(CS) & = \sum _{P \subseteq A}{p(\omega _{P}) \cdot G(CS, \omega _{P})}\\ & = \sum _{P \subseteq A}{p(\omega _{P}) \cdot \sum _{C \in CS}{g(C, \omega _{P})}}\\ & = \sum _{C \in CS}\sum _{P \subseteq A}p(\omega _{P}) \cdot g(C, \omega _{P})\\ & = \sum _{C \in CS}{u(C)}. \end{array} \end{aligned}$$

This concludes the proof. $\square$

Proposition 5

Given a PCFG $\langle A, g, p\rangle$, for every coalition $C \subseteq A$, we have that

$$\begin{aligned} u(C) = \sum _{P \subseteq C}{p(\langle P, C\setminus P\rangle )} \cdot {g(C, \omega _{P})}. \end{aligned}$$

The proof uses the following lemma:

Lemma 1

Let $\langle A, g, p\rangle$ be a PCFG, let $\langle Q, R\rangle$ be an event from ${\mathcal{E}}_A$ and $C \subseteq A$ such that $C \cap (Q \cup R) = \emptyset$. We have that $\langle Q, R\rangle = \bigcup _{P \subseteq C}{\langle Q \cup P, R \cup (C \setminus P)\rangle }$.

Proof

We prove it by induction on the size of C. The result trivially holds if $|C| = 0$, i.e., if $C = \emptyset$. Assume the result holds for $|C| = k$, for some k such that $0 \le k \le n - 1 - |Q \cup R|$. So:

$$\begin{aligned} \langle Q, R\rangle = \bigcup _{P \subseteq C}{\langle Q \cup P, R \cup (C \setminus P)\rangle }. \end{aligned}$$

Let $a_i \in A \setminus (Q \cup R)$. We know that $\langle \{a_i\}, \emptyset \rangle \cup \langle \emptyset , \{a_i\}\rangle = \varOmega _A$. Thus:

$$\begin{aligned} \begin{array}{ll} \langle Q, R\rangle & = \bigcup _{P \subseteq C}{\langle Q \cup P, R \cup (C \setminus P)\rangle }\\ & = \bigcup _{P \subseteq C}{\langle Q \cup P, R \cup (C \setminus P)\rangle } \cap (\langle \{a_i\}, \emptyset \rangle \cup \langle \emptyset , \{a_i\}\rangle )\\ & = (\bigcup _{P \subseteq C}{\langle Q \cup P, R \cup (C \setminus P)\rangle \cap \langle \{a_i\}, \emptyset \rangle }) \cup \\ & \quad (\bigcup _{P \subseteq C}{\langle Q \cup P, R \cup (C \setminus P)\rangle \cap \langle \emptyset , \{a_i\}\rangle })\\ & = (\bigcup _{P \subseteq C}{\langle Q \cup P \cup \{a_i\}, R \cup (C \setminus P)\rangle }) \cup \\ & \quad (\bigcup _{P \subseteq C}{\langle Q \cup P, R \cup (C \setminus P) \cup \{a_i\}\rangle })\\ & = \bigcup _{P \subseteq C \cup \{a_i\}}{\langle Q \cup P, R \cup (C\setminus P)\rangle }. \end{array} \end{aligned}$$

We have just proved that the result holds for any C, $|C| = k + 1$, which concludes the proof. $\square$

We can now prove Proposition 5:

Proof of Proposition 5

By definition (cf. Eq. 4), we have that:

$$\begin{aligned} u(C) = \sum _{P \subseteq A}{p(\omega _{P}) \cdot g(C, \omega _{P})}. \end{aligned}$$

For any subset C of A, every subset P of A is by construction the union of a subset $P_C$ of C with a subset $P_{\overline{C}}$ of $\overline{C}$, such that $P_C \cap P_{\overline{C}} = \emptyset$. Accordingly,

$$\begin{aligned} u(C) = \sum _{P_C \subseteq C, P_{\overline{C}} \subseteq \overline{C}}{p(\langle P_C \cup P_{\overline{C}}, \overline{P_C \cup P_{\overline{C}}}\rangle ) \cdot } g(C, \omega _{P_C \cup P_{\overline{C}}}). \end{aligned}$$

Yet from Equation 1, we know that $g(C, \omega _{P_C \cup P_{\overline{C}}}) = g(C, \omega _{P_C})$. So we can rewrite $u(C)$ as:

$$\begin{aligned} u(C) = \sum _{P_C \subseteq C} g(C, \omega _{P_C}) \cdot \sum _{P_{\overline{C}} \subseteq \overline{C}}{p(\langle P_C \cup P_{\overline{C}}, \overline{P_C \cup P_{\overline{C}}}\rangle )}. \end{aligned}$$

Yet for any $P_C \subseteq C$ and any $P_{\overline{C}} \subseteq \overline{C}$, we have that $\overline{P_C \cup P_{\overline{C}}} = (C \setminus P_C) \cup (\overline{C} \setminus P_{\overline{C}})$, so the event $\langle P_C \cup P_{\overline{C}}, \overline{P_C \cup P_{\overline{C}}}\rangle$ corresponds to the event $\langle P_C \cup P_{\overline{C}}, (C \setminus P_C) \cup (\overline{C} \setminus P_{\overline{C}})\rangle$. And by Lemma 1, for any $P_C \subseteq C$, we have that $\bigcup _{P_{\overline{C}} \subseteq \overline{C}}{\langle P_C \cup P_{\overline{C}}, (C \setminus P_C) \cup (\overline{C} \setminus P_{\overline{C}})\rangle } = \langle P_C, C \setminus P_C\rangle$. That is to say, for any $P_C \subseteq C$, we get that $\bigcup _{P_{\overline{C}} \subseteq \overline{C}}{\langle P_C \cup P_{\overline{C}}, \overline{P_C \cup P_{\overline{C}}}\rangle } = \langle P_C, C \setminus P_C\rangle$. Hence, $\sum _{P_{\overline{C}} \subseteq \overline{C}}{p(\langle P_C \cup P_{\overline{C}}, \overline{P_C \cup P_{\overline{C}}}\rangle )} = p(\langle P_C, C \setminus P_C\rangle )$. Therefore, we can rewrite $u(C)$ as follows:

$$\begin{aligned} u(C) = \sum _{P_C \subseteq C} p(\langle P_C, C\setminus P_C\rangle ) \cdot g(C, \omega _{P_C}), \end{aligned}$$

which, using Equation 1 again, can be written equivalently as:

$$\begin{aligned} u(C) = \sum _{P \subseteq C} p(\langle P, C\setminus P\rangle ) \cdot g(C, \omega _{P}). \end{aligned}$$

This concludes the proof. $\square$

Proposition 6

Let $\langle A, g^{f}_{cau}, p\rangle$ be a cautious PCFG. For any coalition $C \subseteq A$, we have that

$$\begin{aligned} u_{cau}(C) = p(\langle C, \emptyset \rangle ) \cdot {g^{f}_{cau}(C, \omega _{C})}. \end{aligned}$$

Proof

According to Corollary 1, for any coalition structure $CS \in \varPi _A$, we have that:

$$\begin{aligned} U_{cau}(CS) = \sum _{C \in CS}{\sum _{P \subseteq C}{p(\langle P, C\setminus P\rangle ) \cdot g^{f}_{cau}(C, \omega _{P})}}. \end{aligned}$$

Yet by definition of $g^{f}_{cau}$, for any given coalition $C \subseteq A$ and any $P \subseteq C$, we have that $g^{f}_{cau}(C, \omega _{P}) = 0$ whenever $C \not \subseteq P$. Hence, in the equation, for any coalition C one can restrict ourselves to events $\langle P, C \setminus P\rangle$ where $P = C$, that corresponds to the single event $\langle C, \emptyset \rangle$. Therefore, $U_{cau}(CS)$ is simplified as $U_{cau}(CS) = \sum _{C \in CS}p(\langle C, \emptyset \rangle ) \cdot g^{f}_{cau}(C, \omega _{C})$. This concludes the proof. $\square$

Proposition 7

${\mathbf{DP}}$-${\mathbf{CS}}$ is ${\mathsf {PP}}$-hard and ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}^{{\mathsf {PP}}}$-hard. Hardness results hold for flexible PCFGs, and even when the set of events $\{\langle \{a_i\}, \emptyset \rangle \mid a_i \in A\}$ are pairwise independent.

Proof

1. We prove that ${\mathbf{DP}}$-${\mathbf{CS}}$ is ${\mathsf {PP}}$-hard for flexible PCFGs and when all events from $\{\langle \{a_i\}, \emptyset \rangle \mid a_i \in A\}$ are pairwise independent by considering a reduction in polynomial time from the ${\mathsf {PP}}$-complete problem $MAJSAT$: given a propositional formula $\varphi$ in Conjunctive Normal Form (CNF), does the majority of assignments satisfy $\varphi$? Consider such a formula $\varphi$ defined over the set of propositional variables $X = \{x_1, \ldots , x_n\}$. Since $\varphi$ is in CNF, it can be viewed as a set of clauses $\{cl_i \mid cl_i \in \varphi \}$ interpreted conjunctively, where each clause is a disjunction of literals over X. Now, let us associate with $\varphi$ the flexible PCFG $\langle A, g^{f}_{fle}, p\rangle$ and the coalition structure CS, where $A = \{a_1, \ldots , a_n\}$, $CS_* = \{A\}$, all events from $\{\langle \{a_i\}, \emptyset \rangle \mid a_i \in A\}$ are pairwise independent, $p(\langle \{a_i\}, \emptyset \rangle ) = 0.5$ for each $a_i \in A$, and $g^{f}_{fle}$be characterized by a function $f: 2^A \mapsto \mathbb {N}$ defined for each coalition $C \subseteq A$ as

$f(C) = 1$ if for every clause $cl_i \in \varphi$, there is a literal $l_j \in cl_i$ such that ($a_j \in C$ if and only if $l_j$ is a positive literal); and $f(C) = 0$ in the remaining cases.

Note that each value $f(C)$ is not explicitely represented for each coalition $C \subseteq A$ and each outcome $\omega _{P} \in \varOmega _A$ in a table of exponential size, since each such value is completely characterized in polynomial time by $\varphi$ given C. Then we recall that $g^{f}_{fle}$ is characterized for each coalition $C \subseteq A$ and each outcome $\omega _{P} \in \varOmega _A$ as $g^{f}_{fle}(C, \omega _{P}) = f(C \cap P)$.

Let us show that the majority of propositional assignments satisfies $\varphi$ if and only if $U(CS_*) \ge 0.5$ (i.e., $k = 0.5$). Consider the one-to-one correspondence between the set of all outcomes $\varOmega _A$ and the set of all propositional assignments associating each outcome $\omega _{P} \in \varOmega _A$ with the propositional assignment $I_P$ defined for every variable $x_i \in X$ as $I_P(x_i) = 1$ if and only if $a_i \in P$. Then it can be seen by definition of $f$ that for each outcome $\omega _{P} \in \varOmega _A$, $f(P) = 1$ if $I_P$ satisfies $\varphi$ (denoted by $I_P \models \varphi$), otherwise $f(P) = 0$. Now, by definition of $p$, for each outcome $\omega _{P} \in \varOmega _A$, we have that $p(\omega _{P}) = 0.5^n$. Hence,

$$\begin{aligned} \begin{array}{ll} U_{fle}(CS_*) & = \sum _{\omega _{P} \in \varOmega _A}{p(\omega _{P}) \cdot G_{fle}(CS_*, \omega _{P})}\\ & = \sum _{\omega _{P} \in \varOmega _A}{p(\omega _{P}) \cdot G_{fle}(\{A\}, \omega _{P})}\\ & = \sum _{\omega _{P} \in \varOmega _A}{p(\omega _{P}) \cdot g^{f}_{fle}(A, \omega _{P})}\\ & = \sum _{\omega _{P} \in \varOmega _A}{0.5^n \cdot f(P)}\\ & = 0.5^n \cdot \sum _{\omega _{P} \in \varOmega _A}{f(P)}\\ & = 0.5^n \cdot |\{I_P \mid I_P \models \varphi \}|. \end{array} \end{aligned}$$

The majority of propositional assignments satisfies $\varphi$ if and only if $|\{I_P \mid I_P \models \varphi \}| \ge 2^{n-1}$ if and only if $U_{fle}(CS_*) \ge 0.5^n \cdot 2^{n-1} = 0.5$. Hence, ${\mathbf{DP}}$-${\mathbf{CS}}$ is ${\mathsf {PP}}$-hard.

2. We prove that ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}^{{\mathsf {PP}}}$-hard for flexible PCFGs and when all events from $\{\langle \{a_i\}, \emptyset \rangle \mid a_i \in A\}$ are pairwise independent by considering a reduction in polynomial time from the ${\mathsf{NP}}^{{\mathsf {PP}}}$-complete problem $E$-$MAJSAT$: given a propositional formula $\varphi$ in Conjunctive Normal Form (CNF), defined over $X \cup Y$ where X and Y are two disjoint sets of propositional variables, does there exist an assignment $\gamma _X$ over X such that the majority of assignments over Y satisfies $\varphi _{\mid \gamma _X}$ (i.e., $\varphi$ conditioned on $\gamma _X$)? Consider such a formula $\varphi$ defined over the set of propositional variables $X \cup Y$, where $X = \{x_1, \ldots , x_n\}$ and $Y = \{y_1, \ldots , y_n\}$. Without loss of generality, we assume that $n \ge 2$, for technical reasons that will be of use in the latter part of the proof. Since $\varphi$ is in CNF, it can be viewed as a set of clauses $\{cl_i \mid cl_i \in \varphi \}$ interpreted conjunctively, where each clause is a disjunction of literals over $X \cup Y$.

Now, let us associate with $\varphi$ the flexible PCFG $\langle A, g^{f}_{fle}, p\rangle$, where $A = \{new, a_1, a^{\prime}_1, b_1, b^{\prime}_1, \ldots , a_n, a^{\prime}_n, b_n, b^{\prime}_n\}$, all events from $\{\langle \{new\},$ $\emptyset \rangle \} \cup$ $\bigcup \{\langle \{a_i\}, \emptyset \rangle ,$ $\langle \{a^{\prime}_i\}, \emptyset \rangle , \langle \{b_i\}, \emptyset \rangle , \langle \{b^{\prime}_i\}, \emptyset \rangle \mid i \in \{1, \ldots , n\}\}$ are pairwise independent, $p(\langle \{new\}, \emptyset \rangle ) = 1$, and $p(\langle \{a_i\}, \emptyset \rangle ) = p(\langle \{a^{\prime}_i\}, \emptyset \rangle ) = p(\langle \{b_i\}, \emptyset \rangle ) = p(\langle \{b^{\prime}_i\}, \emptyset \rangle ) = 0.5$ for each $i \in \{1, \ldots ,$ $n\}$. The function $g^{f}_{fle}$ is characterized by a function $f: 2^A \mapsto \mathbb {N}$; yet before describing how $f$ is defined, let us introduce some preliminary notions on coalitions from A that will be useful in the proof.

A set $C \subseteq A$ is said to be canonical if the following conditions are jointly satisfied:

• 1. $new \in C$;

• 2. for each $i \in \{1, \ldots , n\}$, $\{a_i, a^{\prime}_i\} \subseteq C$ or $\{b_i, b^{\prime}_i\} \subseteq C$;

• 3. for each $i \in \{1, \ldots , n\}$, $\{a_i, a^{\prime}_i\} \cap C = \emptyset$ or $\{b_i, b^{\prime}_i\} \cap C = \emptyset$.

For instance, for $n = 4$, $C_1 = \{new, a_1, a^{\prime}_1, b_2, b^{\prime}_2, b_3, b^{\prime}_3,$ $a_4, a^{\prime}_4\}$ is a canonical coalition. Note that a canonical coalition always contains $2n + 1$ elements, and that there are exactly $2^n$ canonical coalitions.

A set $C \subseteq A$ is said to be sub-canonical if the following conditions are jointly satisfied:

• 1. C is a proper subset of a canonical coalition;

• 2. $new \in C$;

• 3. for each $i \in \{1, \ldots , n\}$, $a_i \in C$ if and only if $a^{\prime}_i \notin C$;

• 4. for each $i \in \{1, \ldots , n\}$, $b_i \in C$ if and only if $b^{\prime}_i \notin C$.

Note that when C is sub-canonical, there is exactly one canonical coalition $C^{\prime}$ such that $C \subseteq C$, and we also say that C is a sub-canonical set w.r.t. $C^{\prime}$. For instance, for $n = 4$, $C_2 = \{new, a_1, b^{\prime}_2, b_3, a_4\}$ is a sub-canonical coalition of $C_1$. Additionally, one can remark that a sub-canonical coalition C always contains $n + 1$ elements, that for each $i \in \{1, \ldots , n\}$, exactly one element from $\{a_i, a^{\prime}_i, b_i, b^{\prime}_i\}$ belongs to a sub-canonical coalition C, and that for any given canonical coalition $C^{\prime}$, there are exactly $2^n$ sub-canonical coalitions w.r.t. $C^{\prime}$.

Lastly, let us define the function $\alpha$ associating any literal $l_i$ over $X \cup Y$ with a pair of agents from A, defined as:

$$\begin{aligned} \alpha (l_i) = \left\{ \begin{array}{ll} \{a_i, a^{\prime}_i\} &\quad {\text{if}}\,l_i\,{\text{is a positive literal over}}\,X,\\ \{b_i, b^{\prime}_i\} &\quad{\text{if}}\,\,l_i\,{\text{ is a negative literal over}}\,X,\\ \{a_i, b_i\} &\quad{\text{if}}\,\,l_i\,{\text{is a positive literal over}}\,Y,\\ \{a^{\prime}_i, b^{\prime}_i\} &\quad{\text{if}}\,\,l_i\,\,{\text{is a negative literal over}}\,Y. \end{array}\right. \end{aligned}$$

We are now ready to define $f$, for each coalition $C \subseteq A$, as:

$$\begin{aligned} f(C) = \left\{ \begin{array}{ll} 2^{5n} &\quad{\text{if}}\,\,C\,{\text{is canonical}},\\ 2^{n+1} &\quad{\text{if}}\,\,C\,{\text{is sub-canonical and for each clause}}\,cl_i \in \varphi ,\\ &\quad {\text{there is a literal}}\, l_j \in cl_i\,{\text{such that}}\,\alpha (l_j) \cap C \ne \emptyset ,\\ 0 &\quad {\text{in the remaining cases}}. \end{array}\right. \end{aligned}$$

Then we recall that $g^{f}_{fle}$ is characterized for each coalition $C \subseteq A$ and each outcome $\omega _{P} \in \varOmega _A$ as $g^{f}_{fle}(C, \omega _{P}) = f(C \cap P)$.

We intend now to prove that there is an assignment $\gamma _X$ over X such that the majority of assignments over Y satisfies $\varphi _{\mid \gamma _X}$ if and only if there exists a coalition structure CS such that $U_{fle}(CS) \ge 2^{3n} + 1$ according to the flexible PCFG $\langle A, g^{f}_{fle}, p\rangle$.

(Only if part) Assume that there is an assignment $\gamma _X$ over X such that the majority of assignments over Y satisfies $\varphi _{\mid \gamma _X}$. We associate with $\gamma _X$ the coalition structure $CS_{\gamma _X} = \{C_{\gamma _X}, A \setminus C_{\gamma _X}\}$, where $C_{\gamma _X}$ is the canonical coalition defined as:

$$\begin{aligned} \begin{array}{ll} C_{\gamma _X} &= \bigcup \{\{a_i, a^{\prime}_i\} \mid \gamma _X(x_i) = 1, i \in \{1, \ldots , n\}\}\\ & \quad \cup \bigcup \{\{b_i, b^{\prime}_i\} \mid \gamma _X(x_i) = 0, i \in \{1, \ldots , n\}\}\\ &\quad \cup \{new\}. \end{array} \end{aligned}$$

Let us show that $U_{fle}(CS_{\gamma _X}) \ge 2^{3n} + 1$. From Proposition 4 we know that

$$\begin{aligned} U_{fle}(CS_{\gamma _X}) = u_{fle}(C_{\gamma _X}) + u_{fle}(A \setminus C_{\gamma _X}). \end{aligned}$$

So it is enough to prove that $u_{fle}(C_{\gamma _X}) \ge 2^{3n} + 1$. Let us first recall that according to Proposition 5, $u_{fle}(C_{\gamma _X})$ is defined as:

$$\begin{aligned} u_{fle}(C_{\gamma _X}) = \sum _{P \subseteq C_{\gamma _X}}{p(\langle P, C_{\gamma _X}\setminus P\rangle )} \cdot g^{f}_{fle}(C_{\gamma _X}, \omega _{P}). \end{aligned}$$

Equivalently,

$$\begin{aligned} u_{fle}(C_{\gamma _X}) = \sum _{P \subseteq C_{\gamma _X}}{p(\langle P, C_{\gamma _X}\setminus P\rangle )} \cdot f(P). \end{aligned}$$

Yet we know by definition of $f$ that $f(P) = 2^{5n}$ when P is canonical, that is – under the condition that $P \subseteq C_{\gamma _X}$ – exactly the case when $P = C_{\gamma _X}$. On the other hand, since $p(\langle \{new\}, \emptyset \rangle ) = 1$, since $p(\langle \{a_i\}, \emptyset \rangle ) = p(\langle \{a^{\prime}_i\}, \emptyset \rangle ) = p(\langle \{b_i\}, \emptyset \rangle ) = p(\langle \{b^{\prime}_i\}, \emptyset \rangle ) = 0.5$ for each $i \in \{1, \ldots , n\}$, since all these events are pairwise independent and since $C_{\gamma _X}$ contains $2n + 1$ elements, we have that $p(\langle C_{\gamma _X}, \emptyset \rangle ) = 0.5^{2n}$. Thus $p(\langle C_{\gamma _X}, \emptyset \rangle ) \cdot f(C_{\gamma _X}) = 0.5^{2n} \cdot 2^{5n} = 2^{3n}$. Hence,

$$\begin{aligned} u_{fle}(C_{\gamma _X}) = 2^{3n} + \delta , \end{aligned}$$

where $\delta = \sum _{P \subsetneq C_{\gamma _X}}{p(\langle P, C_{\gamma _X}\setminus P\rangle )} \cdot f(P)$. Since we need to show that $u_{fle}(C_{\gamma _X}) \ge 2^{3n} + 1$, what remains to be shown is that $\delta \ge 1$.

Now, let us consider the set ${\mathcal{E}}_A^{C_{\gamma _X}} \subseteq {\mathcal{E}}_A$ defined as the set of events $\langle Q, R\rangle$ satisfying the following set of conditions:

• (i) Q is a sub-canonical set w.r.t. $C_{\gamma _X}$;

• (ii) $R = C_{\gamma _X} \setminus Q$.

Note that for each $\langle Q, R\rangle \in {\mathcal{E}}_A^{C_{\gamma _X}}$, $Q \cup R = C_{\gamma _X}$, $|Q| = n+1$ and $|R| = n$, and that $|{\mathcal{E}}_A^{C_{\gamma _X}}| = 2^n$.

By construction of ${\mathcal{E}}_A^{C_{\gamma _X}}$, we have that ${\mathcal{E}}_A^{C_{\gamma _X}} \subseteq \{\langle Q, C_{\gamma _X}\setminus Q\rangle \mid Q \subsetneq C_{\gamma _X}\}$. Thus:

$$\begin{aligned} \delta \ge \sum _{\langle Q, R\rangle \in {\mathcal{E}}_A^{C_{\gamma _X}}}{p(\langle Q, R\rangle ) \cdot f(Q)}. \end{aligned}$$

Now, since $p(\langle \{new\}, \emptyset \rangle ) = 1$, since $p(\langle \{a_i\}, \emptyset \rangle ) = p(\langle \{a^{\prime}_i\}, \emptyset \rangle ) = p(\langle \{b_i\}, \emptyset \rangle ) = p(\langle \{b^{\prime}_i\}, \emptyset \rangle ) = 0.5$ for each $i \in \{1, \ldots , n\}$, and since all these events are pairwise independent, we get for each $\langle Q, R\rangle \in {\mathcal{E}}_A^{C_{\gamma _X}}$ that:

$$\begin{aligned} \begin{array}{ll} p(\langle Q, R\rangle ) & = \prod _{a \in Q}{p(\langle \{a\}, \emptyset \rangle )} \cdot \prod _{a \in R}{p(\langle \emptyset , \{a\}\rangle )}\\ & = 1 \cdot 0.5^n \cdot (1-0.5)^n\\ & = 0.5^{2n}. \end{array} \end{aligned}$$

So we get that

$$\begin{aligned} \delta \ge \sum _{\langle Q, R\rangle \in {\mathcal{E}}_A^{C_{\gamma _X}}}{0.5^{2n} \cdot f(Q)}. \end{aligned}$$

Equivalently,

$$\begin{aligned} \delta \ge 0.5^{2n} \cdot \sum _{\langle Q, R\rangle \in {\mathcal{E}}_A^{C_{\gamma _X}}}{f(Q)}. \end{aligned}$$

Now, let us build a one-to-one correspondence $\beta$ between the set ${\mathcal{E}}_A^{C_{\gamma _X}}$ and the set of propositional assignments over Y as follows. For each event $\langle Q, R\rangle \in {\mathcal{E}}_A^{C_{\gamma _X}}$, we set $\beta (\langle Q, R\rangle )$ to be the propositional assignment $\gamma _Y$ over Y defined for each $y_i \in Y$ as $\gamma _Y(y_i) = 1$ if and only if $\{a_i, b_i\} \cap Q \ne \emptyset$ (i.e., $\gamma _Y(y_i) = 0$ when $\{a^{\prime}_i, b^{\prime}_i\} \cap Q \ne \emptyset$).

It can be easily verified by construction of $C_{\gamma _X}$ and ${\mathcal{E}}_A^{C_{\gamma _X}}$ that for every propositional assignment $\gamma _Y$ over Y that satisfies $\varphi _{\mid \gamma _X}$, the event $\langle Q, R\rangle \in {\mathcal{E}}_A^{C_{\gamma _X}}$ defined as $\langle Q, R\rangle = \beta ^{-1}(\gamma _Y)$ meets the conditions given in the definition of $f$ to satisfy $f(Q) = 2^{n+1}$. Since a majority of propositional assignments over Y satisfies $\varphi _{\mid \gamma _X}$, this means that $f(Q) = 2^{n+1}$ for a majority of events $\langle Q, R\rangle$ from ${\mathcal{E}}_A^{C_{\gamma _X}}$. Yet we know that $|{\mathcal{E}}_A^{C_{\gamma _X}}| = 2^n$, so this means that $f(Q) = 2^{n+1}$ for at least $2^{n-1}$ events $\langle Q, R\rangle$ from ${\mathcal{E}}_A^{C_{\gamma _X}}$. Hence,

$$\begin{aligned} \sum _{\langle Q, R\rangle \in {\mathcal{E}}_A^{C_{\gamma _X}}}{f(Q) \ge 2^{n-1} \cdot 2^{n+1} = 2^{2n}}. \end{aligned}$$

We got that $\delta \ge 0.5^{2n} \cdot 2^{2n}$, i.e., $\delta \ge 1$, thus $u_{fle}(C_{\gamma _X}) \ge 2^{3n} + 1$, so $U_{fle}(CS_{\gamma _X}) \ge 2^{3n} + 1$. This concludes the (only if) part of the proof.

(If part) Assume that there exists a coalition structure CS such that $U_{fle}(CS) \ge 2^{3n} + 1$ according to the flexible PCFG $\langle A, g^{f}_{fle}, p\rangle$. From Proposition 4, we know that $U_{fle}(CS)$ is defined as:

$$\begin{aligned} U_{fle}(CS) = \sum _{C \in CS}{u_{fle}(C)}, \end{aligned}$$

where

$$\begin{aligned} u_{fle}(C) = \sum _{\omega _{P} \in \varOmega _A}{p(\omega _{P}) \cdot g^{f}_{fle}(C, \omega _{P})}, \end{aligned}$$

or equivalently, where

$$\begin{aligned} u_{fle}(C) = \sum _{\omega _{P} \in \varOmega _A}{p(\omega _{P}) \cdot f(C \cap P)}, \end{aligned}$$

Yet for every coalition C such that $new \notin C$, we know from the definition of $f$ that $f(C \cap P) = 0$ for every $\omega _{P} \in \varOmega _A$; and thus for each such coalition C, $u_{fle}(C) = 0$. This means that CS contains a coalition $C^*$ such that $u_{fle}(C^*) \ge 2^{3n} + 1$.

Let us show that such coalition $C^*$ is canonical. Let us recall from Proposition 5 that $u_{fle}(C^*)$ is characterized as follows:

$$\begin{aligned} u_{fle}(C^*) = \sum _{P \subseteq C^*}{p(\langle P, C^*\setminus P\rangle )} \cdot g(C^*, \omega _{P}). \end{aligned}$$

Equivalently,

$$\begin{aligned} u_{fle}(C^*) = \sum _{P \subseteq C^*}{p(\langle P, C^*\setminus P\rangle )} \cdot f(P). \end{aligned}$$

Toward a contradiction, assume that $C^*$ is not canonical. We fall into two cases:

• Case 1: $C^*$ is a strict superset of a canonical coalition C. Then $u_{fle}(C^*)$ can be written as $u_{fle}(C^*) = \delta _1 + \delta _2$, where

  $$\begin{aligned} \delta _1 = p(\langle C, C^*\setminus C\rangle ) \cdot f(C), \end{aligned}$$

  and

  $$\begin{aligned} \delta _2 = \sum _{P \subseteq C^*, P \ne C}{p(\langle P, C^*\setminus P\rangle )} \cdot f(P). \end{aligned}$$

  We first compute $\delta _1$. On the one hand, $p(\langle \{new\}, \emptyset \rangle ) = 1$, $p(\langle \{a_i\}, \emptyset \rangle ) = p(\langle \{a^{\prime}_i\}, \emptyset \rangle ) = p(\langle \{b_i\}, \emptyset \rangle ) = p(\langle \{b^{\prime}_i\}, \emptyset \rangle ) = 0.5$ for each $i \in \{1, \ldots , n\}$, and all these events are pairwise independent, so we have that $p(\langle C, C^*\setminus C\rangle ) = 0.5^{|C^*|-1}$. On the other hand, according to the definition of $f$, we have that $f(C) = 2^{5n}$ since C is canonical. Hence, $\delta _1 = 0.5^{|C^*|-1} \cdot 2^{5n}$, that is, $\delta _1 = 2^{5n-|C^*|+1}$.

  Let us now provide an upper bound for $\delta _2$. On the one hand, we know that for each $P \subseteq C^*$ such that $P \ne C$, $p(\langle P, C^*\setminus P\rangle ) = 0.5^{|C^*|-1}$; and $P = C^* \cap P$ is not canonical, so according to the definition of $f$, $f(P) \le 2^{n+1}$. Thus for each $P \subseteq C^*$ such that $P \ne C$,

  $$\begin{aligned} p(\langle P, C^*\setminus P\rangle ) \cdot f(P) \le 0.5^{|C^*|-1} \cdot 2^{n+1}. \end{aligned}$$

  Hence,

  $$\begin{aligned} \delta _2 \le \sum _{P \subseteq C^*, P \ne C}{0.5^{|C^*|-1} \cdot 2^{n+1}}, \end{aligned}$$

  so

  $$\begin{aligned} \delta _2 < 2^{|C^*|} \cdot 0.5^{|C^*|-1} \cdot 2^{n+1}, \end{aligned}$$

  that is, $\delta _2 < 2^{n+2}$.

  We got that $\delta _1 + \delta _2 < 2^{5n-|C^*|+1} + 2^{n+2}$. Yet since C is canonical, $|C| = 2n+1$. Since $C^*$ is a strict superset of C, we get that $|C^*| \ge 2n+2$. So $\delta _1 + \delta _2 < 2^{3n-1} + 2^{n+2}$. As we initially assumed at the beginning of this proof and without loss of generality that $n \ge 2$, we get that $2^{n+2} < 2^{3n-1}$, so $\delta _1 + \delta _2 < 2^{3n-1} \cdot 2^{3n-1}$, thus $\delta _1 + \delta _2 < 2^{3n}$, and so $u_{fle}(C^*) < 2^{3n}$. This contradicts $u_{fle}(C^*) \ge 2^{3n} + 1$.

• Case 2: $C^*$ is not a strict superset of a canonical coalition C. Since $C^*$ is assumed not to be canonical, we know that no subset of $C^*$ is canonical. So for each $P \subseteq C^*$, according to the definition of $f$, $f(P) \le 2^{n+1}$; and we know that $p(\langle P, C^*\setminus P\rangle ) = 0.5^{|C^*|-1}$. Thus

  $$\begin{aligned} u_{fle}(C^*) \le \sum _{P \subseteq C^*}{0.5^{|C^*|-1} \cdot 2^{n+1}}, \end{aligned}$$

  that is,

  $$\begin{aligned} u_{fle}(C^*) \le 2^{|C^*|} \cdot 0.5^{|C^*|-1} \cdot 2^{n+1}, \end{aligned}$$

  or equivalently,

  $$\begin{aligned} u_{fle}(C^*) \le 2^n. \end{aligned}$$

  This contradicts $u_{fle}(C^*) \ge 2^{3n} + 1$.

Both cases lead to a contradiction, so we know that $C^*$ is canonical.

Now, since $C^*$ is canonical, we have that $f(C^*) = 2^{5n}$ by definition of $f$. And since $|C^*| = 2n+1$, we have that $p(\langle C^*, \emptyset \rangle ) = 0.5^{2n}$. Thus

$$\begin{aligned} p(\langle C^*, \emptyset \rangle ) \cdot f(C^*) = 0.5^{2n} \cdot 2^{5n} = 2^{3n}. \end{aligned}$$

This means that $u_{fle}(C^*)$ can be written as

$$\begin{aligned} u_{fle}(C^*) = 2^{3n} + \delta _3, \end{aligned}$$

where

$$\begin{aligned} \delta _3 = \sum _{P \subseteq C^*, P \ne C^*}{p(\langle P, C^*\setminus P\rangle )} \cdot f(P). \end{aligned}$$

Since we know that $u_{fle}(C^*) \ge 2^{3n} + 1$, we get that $\delta _3 \ge 1$.

Now, obviously enough, $\delta _3$ can be written as $\delta _3 = \delta _3^1 + \delta _3^2$, where

$$\begin{aligned} \begin{array}{ll} \delta _3^1 & = \sum _{\begin{array}{c} P \subseteq C^*, P \ne C^*,\\ \langle P, C^*\setminus P\rangle \in {\mathcal{E}}_A^{C^*} \end{array}}{p(\langle P, C^*\setminus P\rangle )} \cdot f(P)\\ & = \sum _{\langle P, C^*\setminus P\rangle \in {\mathcal{E}}_A^{C^*}}{p(\langle P, C^*\setminus P\rangle )} \cdot f(P), \end{array} \end{aligned}$$

and

$$\begin{aligned} \delta _3^2 = \sum _{\begin{array}{c} P \subseteq C^*, P \ne C^*,\\ \langle P, C^*\setminus P\rangle \notin {\mathcal{E}}_A^{C^*} \end{array}}{p(\langle P, C^*\setminus P\rangle )} \cdot f(P), \end{aligned}$$

where the set ${\mathcal{E}}_A^{C^*} \subseteq {\mathcal{E}}_A$ is defined as the set of events $\langle Q, R\rangle$ satisfying the following set of conditions:

• (i) Q is a sub-canonical set w.r.t. $C^*$;

• (ii) $R = C^* \setminus P$.

Yet one can see that for every $P \subseteq C^*$ such that $P \ne C^*$, if $\langle P, C^*\setminus P\rangle \notin {\mathcal{E}}_A^{C^*}$ then the set $P = C^* \cap P$ is neither canonical nor sub-canonical, which means that $f(P) = 0$ according to the definition of $f$. Hence, $\delta _3^2 = 0$. Since we know that $\delta _3 \ge 1$ and $\delta _3 = \delta _3^1 + \delta _3^2$, we get that $\delta _3^1 \ge 1$. Yet for each event $\langle P, C^*\setminus P\rangle \in {\mathcal{E}}_A^{C^*}$, $p(\langle P, C^*\setminus P\rangle ) = 0.5^{2n}$ ($C^*$ is canonical, so $|C^*| = 2n+1$). Thus

$$\begin{aligned} \delta _3^1 = \sum _{\langle P, C^*\setminus P\rangle \in {\mathcal{E}}_A^{C^*}}{0.5^{2n} \cdot f(P)}. \end{aligned}$$

Let us prove that there is a majority of events $\langle P, C^*\setminus P\rangle$ from ${\mathcal{E}}_A^{C^*}$ such that $f(P) = 2^{n+1}$.

We know that for each $\langle P, C^*\setminus P\rangle \in {\mathcal{E}}_A^{C^*}$, $P = C^* \cap P$ is not canonical, so by definition of $f$, we know that $f(P) \le 2^{n+1}$. So what we want to show is that there is a majority of events $\langle P, C^*\setminus P\rangle$ from ${\mathcal{E}}_A^{C^*}$ that satisfy $f(P) \ge 2^{n+1}$. Assume toward a contradiction that this is not the case.

According to the definition of $f$, this means that there is a strict majority of events $\langle P, C^*\setminus P\rangle$ from ${\mathcal{E}}_A^{C^*}$ that satisfy $f(P) = 0$. Since $|{\mathcal{E}}_A^{C^*}| = 2^n$ ($C^*$ is canonical and thus it admits $2^n$ sub-canonical subsets), this means that:

$$\begin{aligned} \delta _3^1 < 0.5 \cdot 2^n \cdot 0.5^{2n} \cdot 2^{n+1}. \end{aligned}$$

We get that $\delta _3^1 < 1$. This contradicts $\delta _3^1 \ge 1$. Therefore, there is a majority of events $\langle P, C^*\setminus P\rangle$ from ${\mathcal{E}}_A^{C^*}$ such that $f(P) = 2^{n+1}$.

Now, let us associate with $C^*$ the propositional assignment $\gamma _X^{C^*}$ over X defined for every $x_i \in X$ as $\gamma _X^{C^*}(x_i) = 1$ if and only if $\{a_i, a^{\prime}_i\} \subseteq C^*$ (recall that since $C^*$ is canonical, we set $\gamma _X^{C^*}(x_i) = 0$ when $\{a_i, a^{\prime}_i\} \cap C^* = \emptyset$ and $\{b_i, b^{\prime}_i\} \subseteq C^*$). And let us build a one-to-one correspondence $\beta$ between the set of events ${\mathcal{E}}_A^{C^*}$ and the set of propositional assignments over Y as follows. For each event $\langle P, C^*\setminus P\rangle \in {\mathcal{E}}_A^{C^*}$, we set $\beta (\langle P, C^*\setminus P\rangle )$ to be the propositional assignment $\gamma _Y$ over Y defined for each $y_i \in Y$ as $\gamma _Y(y_i) = 1$ if and only if $\{a_i, b_i\} \cap P \ne \emptyset$ (i.e., $\gamma _Y(y_i) = 0$ when $\{a^{\prime}_i, b^{\prime}_i\} \cap P \ne \emptyset$).

Now, consider any event $\langle P, C^*\setminus P\rangle$ from ${\mathcal{E}}_A^{C^*}$ such that $f(P) = 2^{n+1}$. By definition of $f$, this means that for each clause $cl_i \in \varphi$, there is a literal $l_j \in cl_i$ such that $\alpha (l_j) \cap P \ne \emptyset$. Then it can be directly verified by construction of $\gamma _X^{C^*}$ and $\beta (\langle P, C^*\setminus P\rangle )$ that the interpretation $\gamma _X^{C^*} \cup \beta (\langle P, C^*\setminus P\rangle )$ satisfies each clause $cl_i \in \varphi$. Since we have proved that there is a majority of events $\langle P, C^*\setminus P\rangle$ from ${\mathcal{E}}_A^{C^*}$ such that $f(P) = 2^{n+1}$, this means that there is a majority of assignments over Y that satisfies $\varphi _{\mid \gamma _X^{C^*}}$. So we have shown that there is an assignment $\gamma _X$ over X such that the majority of assignments over Y satisfies $\varphi _{\mid \gamma _X^{C^*}}$. This concludes the (if) part of the proof.

We have proved that there is an assignment $\gamma _X$ over X such that the majority of assignments over Y satisfies $\varphi _{\mid \gamma _X}$ if and only if there exists a coalition structure CS such that $U_{fle}(CS) \ge 2^{3n} + 1$ according to the PCFG $\langle A, g^{f}_{fle}, p\rangle$ Therefore, ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}^{{\mathsf {PP}}}$-hard. $\square$

Proposition 8

${\mathbf{DP}}$-${\mathbf{CS}}$ is in ${\mathsf{P}}$ and ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}$-complete for cautious PCFGs.

Proof

1. Let us first prove that ${\mathbf{DP}}$-${\mathbf{CS}}$ is in ${\mathsf{P}}$ for cautious PCFGs. From Proposition 6, we have for any coalition $C \subseteq A$ that

$$\begin{aligned} u_{cau}(C) = p(\langle C, \emptyset \rangle ) \cdot g^{f}_{cau}(C, \omega _{C}), \end{aligned}$$

yet $p(\langle C, \emptyset \rangle )$ and $g^{f}_{cau}(C, \omega _{P})$ are computed in polynomial time, so $u_{cau}(C)$ is computed in polynomial time. Yet from Proposition 4, for every coalition structure $CS \in \varPi _A$,

$$\begin{aligned} U_{cau}(CS) = \sum _{C \in CS}{u_{cau}(C)}, \end{aligned}$$

which is then be computed in polynomial time. Therefore, ${\mathbf{DP}}$-${\mathbf{CS}}$ is in ${\mathsf{P}}$ for cautious PCFGs.

2. Let us now prove that ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}$-complete for cautious PCFGs. ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is in ${\mathsf{NP}}$ for cautious PCFGs, since the fact that a (polynomial-size) coalition structure $CS \in \varPi _A$ can be guessed in polynomial time and checked in ${\mathsf{P}}$ according to point 1 of this proof.

To show that ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}$-hard for cautious PCFGs, it is enough to remark that the standard CSG problem is ${\mathsf{NP}}$-hard, and that from Proposition 2, any CFG $\langle A, f\rangle$ is equivalent to its optimistic cautious CFG extension $\langle A, g^{f}_{cau}, p_\top \rangle$, i.e., where for each outcome $\omega _{P} \in \varOmega _A$, $p_\top (\omega _{P}) = 1$ if $P = A$, and $p_\top (\omega _{P}) = 0$ otherwise.

Therefore, ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}$-complete for cautious PCFGs. $\square$

Proposition 9

For any $CS \in \varPi _A$,

$$\begin{aligned} U_{cau}(CS) = \sum _{r_i \in {\mathcal{R}}^*}{w_i \cdot p(\langle CS(i), \emptyset \rangle )}, \end{aligned}$$

where ${\mathcal{R}}^*$ is the set of rules $r_i$ activated by a coalition $CS(i) \in CS$.

Proof

We know from Proposition 4 that for each $CS \in \varPi _A$,

$$\begin{aligned} U_{cau}(CS) = \sum _{C \in CS}{u_{cau}(C)}, \end{aligned}$$

and from Proposition 6 that for each $C \in CS$,

$$\begin{aligned} u_{cau}(C) = p(\langle C, \emptyset \rangle ) \cdot g^{f}_{cau}(C, \omega _{C}). \end{aligned}$$

Yet $g^{f}_{cau}$ is characterized by $f$, which itself is represented by an MC-net R. So by definition of $f$ and $g^{f}_{cau}$, we get that

$$\begin{aligned} g^{f}_{cau}(C, \langle C, \emptyset \rangle ) = f(C) = \sum _{r_i \in {\mathcal{R}}^*_C} w_i, \end{aligned}$$

where $R^*_C$ is the set of rules from R that are activated by C.

Hence,

$$\begin{aligned} u_{cau}(C) = p(\langle C, \emptyset \rangle ) \cdot \sum _{r_i \in {\mathcal{R}}^*_C} w_i. \end{aligned}$$

And since $p(\langle C, \emptyset \rangle )$ is independent of $r_i$ in the above equation, we get that

$$\begin{aligned} u_{cau}(C) = \sum _{r_i \in {\mathcal{R}}^*_C}{w_i \cdot p(\langle C, \emptyset \rangle )}. \end{aligned}$$

Hence, for any $CS \in \varPi _A$,

$$\begin{aligned} U_{cau}(CS) = \sum _{C \in CS}{\sum _{r_i \in {\mathcal{R}}^*_C}{w_i \cdot p(\langle C, \emptyset \rangle )}}. \end{aligned}$$

Yet we know that each rule $r_i$ from ${\mathcal{R}}$ is activated by at most one coalition from CS. This means that for all coalitions $C, C^{\prime}$, $C \ne C^{\prime}$, we have that ${\mathcal{R}}^*_C \cap {\mathcal{R}}^*_{C^{\prime}} = \emptyset$. Therefore,

$$\begin{aligned} U_{cau}(CS) = \sum _{r_i \in {\mathcal{R}}^*}{w_i \cdot p(\langle CS(i), \emptyset \rangle )}, \end{aligned}$$

where ${\mathcal{R}}^*$ is the set of rules $r_i$ activated by a coalition $CS(i) \in CS$.

This concludes the proof. $\square$

Proposition 10

For any $CS \in \varPi _A$,

$$\begin{aligned} U_{fle}(CS) = \sum _{r_i \in {\mathcal{R}}^+}{w_i \cdot p(\langle \gamma _i^+, CS(i) \cap \gamma _i^-\rangle )}, \end{aligned}$$

where ${\mathcal{R}}^+$ is the set of rules $r_i$ partially activated by a coalition $CS(i) \in CS$.

Proof

We know from Proposition 4 that for each $CS \in \varPi _A$,

$$\begin{aligned} U_{fle}(CS) = \sum _{C \in CS}{u_{fle}(C)}, \end{aligned}$$

and that for each $C \in CS$,

$$\begin{aligned} u_{fle}(C) = \sum _{P \subseteq C}{p(\langle P, C\setminus P\rangle )} \cdot g^{f}_{fle}(C, \omega _{P}). \end{aligned}$$

Yet $g^{f}_{fle}$ is characterized by $f$, which itself is represented by an MC-net R. So by definition of $f$ and $g^{f}_{fle}$, we get for each $P \subseteq C$ that

$$\begin{aligned} g^{f}_{fle}(C, \langle P, C\setminus P\rangle ) = f(C \cap P) = f(P) = \sum _{r_i \in {\mathcal{R}}^*_P} w_i, \end{aligned}$$

where ${\mathcal{R}}^*_P$ is the set of rules from ${\mathcal{R}}$ that are activated by P. Hence,

$$\begin{aligned} u_{fle}(C) = \sum _{P \subseteq C}{p(\langle P, C\setminus P\rangle )} \cdot \sum _{r_i \in {\mathcal{R}}^*_P} w_i. \end{aligned}$$

And since $p(\langle P, C\setminus P\rangle )$ is independent of $r_i$ in the above equation, we get that

$$\begin{aligned} u_{fle}(C) = \sum _{P \subseteq C}{\sum _{r_i \in {\mathcal{R}}^*_P} w_i \cdot p(\langle P, C\setminus P\rangle )}. \end{aligned}$$

Equivalently, we can write

$$\begin{aligned} u_{fle}(C) = \sum _{r_i \in {\mathcal{R}}}\sum _{P \subseteq C}{w_i(P) \cdot p(\langle P, C\setminus P\rangle )}, \end{aligned}$$

where $w_i(P) = w_i$ if $r_i \in {\mathcal{R}}^*_P$, otherwise $w_i(P) = 0$.

Yet for all $P \subseteq C$, we have that $r_i \in {\mathcal{R}}^*_P$ precisely for those events $\langle P, C\setminus P\rangle$ where $\gamma _i^+ \subseteq P$ and $\gamma _i^- \cap C \subseteq C\setminus P$, i.e., $w_i(P) = w_i$ when $\gamma _i^+ \subseteq P$ and $\gamma _i^- \cap C \subseteq C\setminus P$, otherwise $w_i(P) = 0$. Hence,

$$\begin{aligned} u_{fle}(C) = \sum _{r_i \in {\mathcal{R}}, \gamma _i^+ \subseteq C}w_i \cdot p(\langle \gamma _i^+, \gamma _i^- \cap C\rangle ). \end{aligned}$$

Equivalently,

$$\begin{aligned} u_{fle}(C) = \sum _{r_i \in {\mathcal{R}}^+_C}w_i \cdot p(\langle \gamma _i^+, \gamma _i^- \cap C\rangle ), \end{aligned}$$

where ${\mathcal{R}}^+_C$ is the set of rules $r_i$ partially activated by C. Hence, for any $CS \in \varPi _A$,

$$\begin{aligned} U_{fle}(CS) = \sum _{C \in CS}{\sum _{r_i \in {\mathcal{R}}^+_C}w_i \cdot p(\langle \gamma _i^+, \gamma _i^- \cap C\rangle )}. \end{aligned}$$

Yet we know that each rule $r_i$ from ${\mathcal{R}}$ is partially activated by at most one coalition from CS. This means that for all coalitions $C, C^{\prime}$, $C \ne C^{\prime}$, we have that ${\mathcal{R}}^+_C \cap {\mathcal{R}}^+_{C^{\prime}} = \emptyset$. Therefore, for any $CS \in \varPi _A$,

$$\begin{aligned} U_{fle}(CS) = \sum _{r_i \in {\mathcal{R}}^+}{w_i \cdot p(\langle \gamma _i^+, CS(i) \cap \gamma _i^-\rangle )}, \end{aligned}$$

where ${\mathcal{R}}^+$ is the set of rules $r_i$ partially activated by a coalition $CS(i) \in CS$.

This concludes the proof. $\square$

Proposition 11

${\mathbf{DP}}$-${\mathbf{CS}}$ is in ${\mathsf{P}}$ and ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}$-complete for MC-net based flexible PCFGs.

Proof

1. From Proposition 6, we have for any $CS \in \varPi _A$ that

$$\begin{aligned} U_{fle}(CS) = \sum _{r_i \in R^+}{w_i \cdot p(\langle \gamma _i^+, CS(i) \cap \gamma _i^-\rangle )}, \end{aligned}$$

where ${\mathcal{R}}^+$ is the set of rules $r_i$ partially activated by a coalition $CS(i) \in CS$.

Yet $p(\langle Q, R\rangle )$ is computed in polynomial time for any event $\langle Q, R\rangle$, so $\langle \gamma _i^+, CS(i) \cap \gamma _i^-\rangle$ is computed in polynomial time for any rule $r_i \in {\mathcal{R}}$ and the coalition CS(i) that partially activated $r_i$ in any $CS \in \varPi _A$. Hence, for any $CS \in \varPi _A$, $U_{fle}(CS)$ is computed in polynomial time in the size of $|{\mathcal{R}}|$. Therefore, ${\mathbf{DP}}$-${\mathbf{CS}}$ is in ${\mathsf{P}}$ for MC-net based flexible PCFGs.

2. ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is in ${\mathsf{NP}}$, since the fact that a (polynomial-size) coalition structure $CS \in \varPi _A$ can be guessed in polynomial time and checked in ${\mathsf{P}}$ according to point 1 of this proof.

To show that ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}$-hard for cautious PCFGs, it is enough to remark that the standard CSG problem is ${\mathsf{NP}}$-hard even when f is represented as an MC-net [23], and that from Proposition 2, any MC-net based CFG $\langle A, f\rangle$ is equivalent to its optimistic flexible MC-net based CFG extension $\langle A, g^{f}_{fle}, p_\top \rangle$, i.e., where for each outcome $\omega _{P} \in \varOmega _A$, $p_\top (\omega _{P}) = 1$ if $P = A$, and $p_\top (\omega _{P}) = 0$ otherwise.

Therefore, ${\mathbf{DP}}$-$\exists {\mathbf{CS}}$ is ${\mathsf{NP}}$-complete for MC-net based flexible PCFGs. $\square$