An extended knowledge compilation map for conditional preference statements-based and generalized additive utilities-based languages

Abstract

Conditional preference statements have been used to compactly represent preferences over combinatorial domains. They are at the core of CP-nets and their generalizations, and lexicographic preference trees. Several works have addressed the complexity of some queries (optimization, dominance in particular). We extend in this paper some of these results, and study other queries which have not been addressed so far, like equivalence, and transformations, like conditioning and variable elimination, thereby contributing to a knowledge compilation map for languages based on conditional preference statements. We also study the expressiveness and complexity of queries and transformations for generalized additive utilities.

Appendices

A Proofs for Section 3(Languages)

Proposition 1

Let $\varphi$ be an LP-tree over $\mathcal X$, then $\succeq _\varphi$ as defined above is a partial order. Furthermore, $\succeq _\varphi$ is a linear order if and only if 1) every attribute appears on every branch and 2) every preference rule specifies a linear order.

Proof

By definition, ${\succeq } _{\varphi }$ is reflexive. The fact that it is antisymmetric follows from the antisymmetry of the local preference relations in the conditional preference tables. For transitivity, the proof given by [8] is for a restricted family of LP-trees, so we recast it here for our more general family of LP-trees. Suppose that $o \succeq _{\varphi } o' \succeq _{\varphi } o''$ and o, $o'$, $o''$ are distinct. There must be a node N at which $\{o,o'\}$ is decided, let W be the set of attributes that labels N, then $o[\textsf {Anc}(N)] = o'[\textsf {Anc}(N)]$, and there is one rule ${\alpha } : {\ge }$ such that $o[\textsf {NonInst}(N)] = o'[\textsf {NonInst}(N)] \models {\alpha }$ and $o[W] \ge o'[W]$. Similarly, let $N'$ be the node at which $\{o',o''\}$ is decided, let $W'$ be the set of attributes that labels $N'$, then $o[\textsf {Anc}(N')] =o'[\textsf {Anc}(N')]$, and there is one rule ${\alpha } ' : {\ge } '$ s.t. $o'[\textsf {NonInst}(N')] = o^{\prime \prime }[\textsf {NonInst}(N')] \models {\alpha }'$ and $o'[W'] \ge ' o^{\prime \prime }[W^{\prime }]$.

If $N = N^{\prime }$, then $o[\textsf {Anc}(N)] =o^{\prime }[\textsf {Anc}(N)]= o^{\prime \prime }[\textsf {Anc}(N)]$, and $o[W]> o^{\prime }[W] > o^{\prime \prime }[W]$ since ${\ge }$ is antisymmetric, thus $o[W] > o^{\prime \prime }[W]$ because ${\ge } = {\ge }^{\prime }$ is also transitive, hence N decides $\{o,o^{\prime \prime }\}$ and $o {\succeq } _{\varphi } o^{\prime \prime }$.

If $N {\ne } N'$, note that both nodes are in the unique branch in ${\varphi }$ that corresponds to $o^{\prime }$, so one of N, $N^{\prime }$ must be above the other. Suppose that N is above $N^{\prime }$, then, it must be the case that $o^{\prime }[W] = o^{\prime \prime }[W]$, and $o[W] \ne o^{\prime }[W]$, thus N decides $\{o,o^{\prime \prime }\}$; moreover, since $\textsf {NonInst}(N) \subseteq \textsf {NonInst}(N^{\prime })$, $o[\textsf {NonInst}(N)] = o^{\prime }[\textsf {NonInst}(N)] = o^{\prime \prime }[\textsf {NonInst}(N)] \models {\alpha }$, and $o[W] \ge o^{\prime }[W] = o^{\prime \prime }[W]$; hence $o \succeq _{\varphi }o^{\prime \prime }$. The case where $N^{\prime }$ is above N is similar.

For the second part of the proposition, suppose first that every attribute appears on every branch and that every preference rule specifies a linear order: we will show that ${\succeq }_{\varphi }$ is antisymmetric and connex. For antisymmetry, consider distinct alternatives $o, o' {\in } \underline{\mathcal{X}}$: because every attribute appears on every branch, there must be a node N, labelled with some $W \subseteq \mathcal X$, that decides $\{o,o^{\prime }\}$, and a unique rule ${\alpha }: {\ge }$ at N such that $o[\textsf {NonInst}(N)] = o'[\textsf {NonInst}(N)] \models {\alpha }$; ${\ge }$ must be a linear order over W, so either $o[W] > o'[W]$ and $o \succ _{\varphi } o'$, or $o'[W] > o[W]$ and $o' \succ _{\varphi } o$: $\succeq _{\varphi }$ is connex and antisymmetric. For the converse, assuming that either there is some branch where some attribute does not appear, or that there is some rule at some node that does not define a linear order, it is not difficult to define two distinct alternatives that cannot be compared with $\succeq _{\varphi }$.

B Proofs for Section 4 (Expressiveness)

Proposition 3

$\textsf {CP} = {\bigcup } _{k \in {\mathbb {N}}}\, \textsf {k-CP}$ and, for every $k \in {\mathbb {N}}$, $k \ge 2$:

$$\begin{aligned}&\textsf {CP}{\wedge } \,{\sqsubseteq \!\!\!\sqsupseteq }\,\textsf {CP}{\ntriangleright } \,{\sqsubseteq \!\!\!\sqsupseteq }\,\textsf {CP}{\wedge } {\ntriangleright } \,{\sqsubseteq \!\!\!\sqsupseteq }\, \textsf {CP} \,{\sqsupset }\, \textsf {k-CP} \,{\sqsubseteq \!\!\!\sqsupseteq }\, \textsf {k-CP}{\wedge } \,{\sqsupset }\, \textsf {k-CP}{\ntriangleright } \,{\sqsubseteq \!\!\!\sqsupseteq }\, \textsf {k-CP}{\wedge } {\ntriangleright }\\ &\;\;\;\qquad \,\,\textsf {k-CP} \,{\sqsupset }\, \textsf {(k-1)-CP}\quad \text { and }\quad \textsf {k-CP}{\wedge } {\ntriangleright } \,{\sqsupset }\, \textsf {(k-1)-CP}{\wedge } {\ntriangleright } \,{\sqsupset }\, \textsf {CPnet}. \end{aligned}$$

Proof

That $\textsf {CP} {\sqsubseteq \!\!\!\sqsupseteq }\textsf {CP}{\wedge }{\sqsubseteq \!\!\!\sqsupseteq }\textsf {CP}{\ntriangleright }{\sqsubseteq \!\!\!\sqsupseteq }\textsf {CP}{\wedge } {\ntriangleright }$ follows from Property 2.

By definition $\textsf {CP} {\supset } \textsf {1-CP} {\supset } \textsf {1-CP}{\ntriangleright } {\supset } \textsf {1-CP}{\wedge }{\ntriangleright }$ and $\textsf {1-CP} {\supset } \textsf {1-CP}{\wedge } {\supset } \textsf {1-CP}{\wedge }{\ntriangleright }$, thus $\textsf {CP} {\sqsupseteq }\textsf {1-CP} {\sqsupseteq }\textsf {1-CP}{\ntriangleright } {\sqsupseteq }\textsf {1-CP}{\wedge }{\ntriangleright }$ and $\textsf {1-CP} {\sqsupseteq }\textsf {1-CP}{\wedge } {\sqsupseteq }\textsf {1-CP}{\wedge }{\ntriangleright }$. Restricting to conjunction of literals does not induce a loss in expressiveness because, given a statement ${\alpha } | V : x {\ge } x'$, it is possible to compute a DNF logically equivalent to ${\alpha }$, and then consider a set of statements, each statement having one disjunct of the DNF as conditioning part. Example 3 prove that $\textsf {CP} \sqsupset \textsf {1-CP}$. Example 4 proves that $\textsf {1-CP}{\sqsupset }\textsf {1-CP}{\ntriangleright }$, it can be generalized to prove that $\textsf {k-CP}{\wedge } {\sqsupset }\textsf {k-CP}{\ntriangleright }$ by considering k binary attributes $A{_1} , {\ldots } , A{_k}$ instead of A, and the preorder $a{_1} {\ldots } a{_k} b {\succ } a{_1} {\ldots } a{_k} \bar{b} {\succ } \bar{a}{_1} {\ldots } \bar{a}{_k} b {\succ } \bar{a}{_1} {\ldots } \bar{a}{_k} \bar{b}$, which can be represented in $\textsf {k-CP}{\wedge }$ but not in $\textsf {k-CP}{\ntriangleright }$.

To prove that $\textsf {k-CP} \,{\sqsupset }\, \textsf {(k-1)-CP}$, simply consider k binary attributes $A{_1} , {\ldots } , A{_k}$ and the preorder that contains a single pair: $a{_1} {\ldots } a{_k} {\succ } \bar{a}{_1} {\ldots } \bar{a}{_k}$, it can be represented in $\textsf {k-CP}$ with the single statement $a{_1} {\ldots } a{_k} {\ge } \bar{a}{_1} {\ldots } \bar{a}{_k}$, but not in $\textsf {(k-1)-CP}$. Note that this statement is in $\textsf {k-CP}{\wedge } {\ntriangleright }$, so it proves that $\textsf {k-CP}{\wedge } {\ntriangleright }\,{\sqsupset }\, \textsf {(k-1)-CP}{\wedge } {\ntriangleright }$. The fact that follows from the “completeness” condition in the definition of CP-nets: in a CP-net, every attribute must have some local preference rules associated to it, whereas a formula in $\textsf {1-CP}{\wedge } {\ntriangleright }$ may consist of one rule only.

Proposition 4

$\textsf {LPT} = {\bigcup } _{k{\in } {\mathbb {N}} } \textsf {k-LPT}$ and, for every $k \in {\mathbb {N}}$:

$$\begin{aligned}&\textsf {CP} \; {{\sqsupset }}\; \textsf {LPT} \;{\sqsubseteq \!\!\!\sqsupseteq }\; \textsf {LPT}{\wedge } \;{\sqsupset }\; \textsf {k-LPT} \;{\sqsubseteq \!\!\!\sqsupseteq }\; \textsf {k-LPT}{\wedge } \;{\sqsupset }\; \textsf {(k-1)-LPT}\\ &\;\;\qquad \textsf {k-CP} \;{\sqsupset }\; \textsf {k-LPT} \;{\sqsupset }\; \textsf {k-LPT}^{\textsf {lin}} \;{\sqsupset }\; \textsf {(k-1)-LPT}^{\textsf {lin}}. \end{aligned}$$

Proof

LP trees can only represent antisymmetric preorders, so $\textsf {LPT}$ is strictly less expressive than $\textsf {CP}$. That $\textsf {k-LPT}{\wedge } {\sqsupseteq }\textsf {k-LPT}$ follows from the fact that the condition of every CP-statement in the set of CP-statements that correspond to some $\textsf {k-LPT}$ can be represented with a DNF. To see that for every $k {\ge } 1$, consider some k-LP-tree ${\varphi }$ with k attributes $X{_1} , {\ldots } , X{_k}$ at the root, and a linear order with $x{_1} {\ldots } x{_k}$ as top element, and $\bar{x}{_1} {\ldots } \bar{x}{_k}$ as second best element: then $o {\succeq } _{\varphi } o'$ for every pair of alternatives $(o,o')$ such that $o[X{_1} {\ldots } X{_k} ] = x{_1} {\ldots } x{_k}$ and $o'[X{_1} {\ldots } X{_k} ] = \bar{x}{_1} {\ldots } \bar{x}{_k}$; no LP-tree in $\textsf {(k-1)-LPT}$ can represent that. Note that we can choose ${\varphi }$ to be linear, so that proves that too.

To show that , consider an LP-tree with attributes $X{_1} {\ldots } X{_k}$ at the root, with $|{\underline{\{X_{1} {\ldots } X_{k} \}}}|$ children, where every child is labelled with binary attribute Y, and at least two children order y and $\bar{y}$ differently: no linear k-LP tree can represent the same order.

That $\textsf {k-CP} \;{\sqsupseteq }\; \textsf {k-LPT}$ follows from the remark below Proposition 1 that describes a set of CP-statements equivalent to a given LP-tree ${\varphi }$: it is not difficult to check that if every node in ${\varphi }$ has at most k attributes, then the corresponding CP-statements are all in $\textsf {k-CP}$. To prove that , consider a CP-net ${\varphi }$ over $k+1$ binary attributes $X{_1} , {\ldots } , X_{k+1}$, with $x_i {\ge } \bar{x}_i$ for every $1 {\le } i {\le } k+1$ (thus the CP-net has no edge): clearly ${\varphi } {\in } \textsf {1-CP} {\subseteq } \textsf {k-CP}$. Consider now some LP-tree ${\psi }$ with $j {\le } k$ attributes at the root; w.l.o.g. we can assume that these attributes are $X{_1} , {\ldots } , X_j$; then the CPT at the root of ${\psi }$ must contain the preorder over $X{_1} , {\ldots } , X_j$ defined by the set of CP-statements $\{ x_i {\ge } \bar{x}_i | 1 {\le } i {\le } j \}$. But then $x{_1} {\ldots } x_j\bar{x}_{j+1}{\ldots } \bar{x}_{k+1} {\succ } _{\psi } \bar{x}{_1} {\ldots } \bar{x}_j x_{j+1}{\ldots } x_{k+1}$, whereas $x{_1} {\ldots } x_j\bar{x}_{j+1}{\ldots } \bar{x}_{k+1} {\not\succ } _{\varphi } \bar{x}{_1} {\ldots } \bar{x}_j x_{j+1}{\ldots } x_{k+1}$. Since $\textsf {LPT} {\sqsupseteq }\textsf {CP} {\sqsupset }\textsf {k-CP}$ and $\textsf {k-CP} \;{\sqsupseteq }\; \textsf {k-LPT}$, we can conclude too that $\textsf {LPT} {\sqsupset }\textsf {k-LPT}$.

Proposition 6

For every $k \in {\mathbb {N}}$: $\textsf {GAI}_{k+1} {\sqsupset }\textsf {GAI}_k$, and $(k-1)-\textsf {CP} \not\sqsupset \textsf {GAI}_{k}$.

Proof

We first prove the first statement. $\textsf {GAI}_{k+1} {\sqsupseteq }\textsf {GAI}_k$ is clear, since $\textsf {GAI}_{k+1} \supseteq \textsf {GAI}_k$, so it suffices to show that the increase in expressiveness is strict.

Fix the set of attributes $\mathcal X = \{X_1, \ldots , X_{k+1}\}$ and set as the domain of each attribute $\underline{X_i}=\{0,1\}$. For every set $A\subseteq \mathcal X$, define the indicator function $I_A(X_1, \ldots , X_{k+1})$ as the function that, given as input an alternative $o\in \underline{\mathcal{X}}$, returns 1 if for all $X\in A$ we have $o(X)=1$ and 0 otherwise. Set $\varphi = \{I_\mathcal{X}\}$, then we have that $g_\varphi (o) = I_\mathcal{X}(o)$. For every A define $o_A$ to be the alternative that is 1 on exactly the attributes in A. Then $g_\varphi$ induces the total preorder $\succeq$ in which $o_\mathcal{X}$ strictly dominates all other alternatives, whereas for all pairs $o, o'$ both different from $o_\mathcal{X}$ we have $o \sim o'$.

Clearly, $\succeq$ is expressed in $\textsf {GAI}_{k+1}$. We claim that it cannot be expressed in $\textsf {GAI}_{k}$. To this end, assume that this were wrong, then there is a set $\varphi = \{g_{Z_1}, \dots ,g_{Z_m}\}$ of real valued functions bearing on strict subsets $Z_i$ of $\mathcal X$ such that $g_\varphi$ induces the order $\succeq$ on $\underline{\mathcal{X}}$. Without loss of generality, assume that for every $A\subset \mathcal X$ the set $\mathcal X$ contains exactly one function $g_A$. It will be convenient to represent $g_{A}$ as a weighted sum of indicator functions.

We use the following representation result for functions from $\{0,1\}^{\ell }\rightarrow \mathbb {R}$ whose proof can e.g. be found in [22, Section 13.2].

Lemma 1

For every function $f:\{0,1\}^{\ell } \rightarrow \mathbb {R}$ with $\ell \in \mathbb {N}$ and in variables $x_1', \ldots , x_\ell '$, there are coefficients $c_A\in \mathbb {R}$ for $A\subseteq \{x_1', \ldots , x_\ell '\}$ such that

$$\begin{aligned} f(x_1', \ldots , x_\ell ') = \sum _{A\subseteq \{x_1', \ldots , x_\ell '\}} c_A I_A(x_1', \ldots , x_\ell '). \end{aligned}$$

Proof

Applying this to the utility functions, it follows directly that, for every $A\subset \mathcal X$, there are coefficients $\lambda _{A,B}$ for $B\subseteq A$ such that for all alternatives $o\in \underline{\mathcal{X}}$ we have

$$\begin{aligned} g_A(o) = \sum _{B\subseteq A} \lambda _{A,B} I_{B}(o). \end{aligned}$$

We get by summing the $g_A$ that there are coefficients $\lambda _B$ such that for all $o\in \underline{\mathcal{X}}$

$$\begin{aligned} g_\varphi (o) = \sum _{B\subset \mathcal X} \lambda _B I_{B}(o). \end{aligned}$$

(1)

By subtracting values in some of the $g_A$, we may assume w.l.o.g. that $g_\varphi (o)= 0$ for all $o\ne o_\mathcal{X}$.

We claim that, for all $B\subset \mathcal X$, we have $\lambda _B = 0$. We show this by induction on the size of B. For $B=\emptyset$, we have with (1) that $0 = g(o_\emptyset ) = \lambda _{\emptyset } I_\emptyset (o_\emptyset ) = \lambda _\emptyset$. For non-empty $B\subset \mathcal X$, we have $g_\varphi (o_B) = \sum _{C\subset \mathcal X} \lambda _C I_{C}(o_B) = \sum _{C\subseteq B} \lambda _C I_{C}(o_B)$. However, by the induction hypothesis, we know that for $C\subset B$ we have $\lambda _C = 0$, so $0= \lambda _B I_{B}(o_B) = \lambda _B$.

Plugging the $\lambda _B$ into (1), we get that $g_\varphi (o_\mathcal{X}) = 0$ which contradicts the assumption that in $\succeq$ the alternative $o_\mathcal{X}$ strictly dominates all others.

For the second statement, consider k binary attributes $A_1, \ldots , A_k$ such that $a_1\ldots a_k \succ \bar{a}_1\ldots \bar{a}_k$. Extend this to an arbitrary complete preference relation such that for all other alternatives o we have $o \succ a_1\ldots a_k$. Clearly, any such order can be expressed as a $\textsf {GAI}_{k}$ by simply giving all alternatives o a utility that yields this order in a single k-ary function $g_{\{A_1, \ldots , A_k\}}$. We claim that this order cannot be expressed by a $(k-1)-\textsf {CP}$. Assume this were false, so there is a set of preference statements defining the order and in which the set of swapped attributes never contains more than $k-1$ attributes. In particular, there is such a statement $\alpha \mid V: w \ge w'$ that sanctions $a_1\ldots a_k \succ \bar{a}_1\ldots \bar{a}_k$ (this comparison cannot be obtained by transitivity, since all other alternatives have a utility that is strictly greater than that of $a_1\ldots a_k$). By assumption w cannot contain all attributes, so there is one attribute, say w.l.o.g. $A_1$ that does not appear in w. If $A_1$ is not in V, then, by definition, applying the statement cannot swap the value of $A_1$, so it cannot justify $a_1\ldots a_k \succ \bar{a}_1\ldots \bar{a}_k$. So $A_1$ must appear. Then $A_1 \notin \textsf {Var}(\alpha )$, thus the statement also sanctions $a_1a_2\ldots a_k \succ a_1\bar{a}_2\ldots \bar{a}_k$ which contradicts the order we want to define. So as we claimed, $\succ$ is not defined by any $(k-1)-\textsf {CP}$.

C Proofs for Section 5 (Succinctness)

Proposition 9

Any complete LPT can be transformed in polytime and space into an equivalent GAI.

Proof

A complete LP tree ${\varphi }$ induces a linear order over $\underline{\mathcal{X}}$, thus we can define the rank of alternative o w.r.t. ${\succeq } _{\varphi }$: $\textsf {rank}({\varphi } ,o) = 1 +$ the number of alternatives strictly preferred to o, so that the most preferred alternative has rank 1, the least preferred has rank $|{\underline{\mathcal{X}}}|$:

$$\begin{aligned} \textsf {rank}({\varphi } ,o) = 1+|{\{o' \in \underline{\mathcal{X}}\mid o' {\succ } _{\varphi } o\}}|. \end{aligned}$$

Fargier et al. [28] explain how $\textsf {rank}({\varphi } ,o)$ can be decomposed as a weighted sum of “local” ranks associated to the nodes of ${\varphi }$:

$$\begin{aligned} \textsf {rank}({\varphi } ,o) = 1 + \sum _{N {\in }\textsf {nodes}({\varphi } )}^{{\alpha } :{\ge } {\in } \textsf {CPT}(N)} \llbracket o[\textsf {Inst}(N)] \!=\! \textsf {inst}(N) {\wedge } o \models {\alpha } \rrbracket \\ \nonumber {\times } \big ( r({\ge } ,o[\textsf {Var}(N)]) - 1\big ) {\times } |{\underline{\textsf {Desc}(N)}}| \end{aligned}$$

where :

• $\textsf {nodes}({\varphi } )$ denotes the set of nodes of ${\varphi }$;

• $\llbracket o[\textsf {Inst}(N)] = \textsf {inst}(N) {\wedge } o \models {\alpha } \rrbracket$ is an indicator function, that equals 1 when the condition $o[\textsf {Inst}(N)] = \textsf {inst}(N) {\wedge } o \models {\alpha }$ is true; that is, when N is on the branch of ${\varphi }$ that corresponds to o, and ${\alpha } :{\ge }$ is the rule that orders at N alternatives that have same values as o for the attributes in the ancestor nodes of N; and equals 0 otherwise;

• $r({\ge } ,o[\textsf {Var}(N)])$ denotes the rank in $\underline{\textsf {Var}(N)}$ with respect to ${\ge }$ of the instantiation given by o to $\textsf {Var}(N)$; so that $r({\ge } ,o[\textsf {Var}(N)]) - 1$ is the number of subtrees rooted at children of N that are less preferred than o at N;

• $\textsf {Desc}(N) = \mathcal X - (\textsf {Anc}(N){\cup } \textsf {Var}(N))$ is the set of attributes that appear below N in that branch, so that $|{\underline{\textsf {Desc}(N)}}|$ is the number of instantiations that are “contained" in every subtree of ${\varphi }$ rooted at any one child of N.

Thus we can define, for every node N of ${\varphi }$, and every rule ${\alpha } :\,\ge \, \in \textsf {CPT}(N)$, a sub-utility $u_{N,{\alpha }}$ as follows:

$$ \begin{aligned} u_{N,{\alpha } }(o) = \left\{ \begin{array}{l} \big ( r({\ge } ,o[\textsf {Var}(N)]) \!-\! 1\big ) \times |{\underline{\textsf {Desc}(N)}}|\, \text { if }\, o[\textsf {Inst}(N)] = \textsf {inst}(N) \& o \models {\alpha } \\ 0 \text { otherwise } \end{array}\right. \end{aligned}$$

and define a utility $u_{\varphi }$ that orders the alternatives as ${\varphi }$ as follows:

$$\begin{aligned} u_{\varphi } = - \sum _{N {\in } \textsf {nodes}({\varphi } )}^{{\alpha } :{\ge } {\in } \textsf {CPT}(N)} u_{N,{\alpha } } \end{aligned}$$

The number of non-null entries in the table of every $u_{N,{\alpha } }$ is equal to $\underline{\textsf {Var}(N)} - 1$, which also corresponds to the space needed to represent the linear order ${\ge }$ of the rule ${\alpha } :{\ge }$. Assuming that $\textsf {Var}(N)$ (resp. $\textsf {Desc}(N)$) contains p (resp. q) attributes, the largest entry cannot be larger than $d^{p+q} \le d^n$, where d is the size of the largest attribute domain, so the number of digits needed for representing the non-null values is polynomial in n and d. Thus the size of the representation of $u_{\varphi }$ is polynomial in the size of ${\varphi }$.

D Proofs for Section 6 (Queries)

Proposition 11

${\succ }$ -comparison and ${\bowtie }$ -comparison are $\textsf {NP}$-hard for the language of acyclic CP-nets, and tractable for polytree CP-nets.

Proposition 11 is proved using a result about the ordering query introduced in [10]: it is a particular case of the top-p that is recalled in Section 6.4.

Ordering Given $S \subseteq \underline{\mathcal{X}}$ with $|S| = 2$, and ${\varphi }$, return some $o \in S$ such that $o' \not\succ _{\varphi } o$, where $o'$ is the other element of S.

Note that when S contains exactly two elements, at least one of them is not strictly dominated by the other; it the two elements in S are incomparable, then the ordering query may return any one of them.

Proof

Note that for ayclic CP-nets (and thus for polytree CP-nets), ${\succeq }$ -comparison and ${\succ }$ -comparison are “almost" equivalent, in the sense that for different alternatives o and $o'$, $o \succ _{\varphi } o'$ iff $o \succeq _{\varphi } o'$ (because acyclic CP-nets are consistent). In particular, ${\succeq }$ -comparison can be reduced to ${\succ }$ -comparison for consistent languages, thus ${\succ }$ -comparison is $\textsf {NP}$ hard for acyclic CP-nets because ${\succeq }$ -comparison is hard for this language [10, Theorems 15, 16].

${\succ }$ -comparison can also be reduced, still for languages that guarantee consistency, to ${\bowtie }$ -comparison: consider alternatives $o \ne o'$, in order to check if $o \succ o'$ we can ask if $o \bowtie o'$: if the answer is “yes”, then $o \not\succ o'$; if the answer is “no”, ask the ordering query for $S = \{o,o'\}$: the answer must be, in polynomial time for acyclic CP-nets [10, Theorem 5], that $o \not\succ o'$ or $o' \not\succ o$: if the answer is that $o \not\succ o'$, it answers the initial query; if the answer is that $o' \not\succ o$, since we know that $o \succ o'$ or $o' \succ o$ because o and $o'$ are not incomparable and $\succeq$ is antisymmetric, it must be the case that $o \succ o'$.

Finally, ${\succeq }$ -comparison is tractable for polytree CP-nets, and two calls of this query at most can answer ${\succ }$ -comparison and ${\bowtie }$ -comparison.

Proposition 12

equivalence is $\textsf {coNP}$-hard for 1-, and for 1-$LPT{\wedge }$, both restricted to binary attributes.

Given a propositional language $\mathcal P$ we define $\mathcal P ^{{\vee }}$ to be the set of finite disjunctions of formulas in $\mathcal P$, and:

$\textsf {1-CP} {\ntriangleright } \mathcal P$ is $\textsf {1-CP}{\ntriangleright }$ restricted to those statements such that the condition is in $\mathcal P$ $\textsf {1-LPT} \mathcal P$ is $\textsf {1-LPT}$ restricted to those LP-trees such that the condition of every rule is in $\mathcal P$.

The proof of the proposition is based on the following lemma, which formalizes the intuition suggested by Example 9.

Lemma 2

Given a propositional language $\mathcal {P}$ closed for conjunction, equivalence for $\mathcal P ^{{\vee } }$ (in the sense of propositional logic), reduces to equivalence for $\textsf {1-CP}{\ntriangleright }\mathcal P$ restricted to acyclic formulas, and to equivalence for $\textsf {1-LPT}\mathcal P$.

Proof

Consider two formulas $\alpha = \bigvee _{i\in I} \alpha _i$ and $\alpha ' = \bigvee _{i\in I'}\alpha '_i$ over a set $\mathcal X$ of binary attributes, with all $\alpha _i$’s and $\alpha '_i$’s in $\mathcal P$; take some binary attribute $X \notin \mathcal X$, with values x and $\bar{x}$, and let $\varphi = \{\alpha _i : x \ge \bar{x} | i \in I\}$ and $\varphi ' = \{ \alpha '_i : x \ge \bar{x} | i \in I'\}$. Note that $\varphi , \varphi ' \in \textsf {1-CP}{\ntriangleright } \mathcal P$, that they are acyclic, and that they can be computed in time polynomial in $|\alpha |+|\alpha '|$. Then $\varphi ^* = \{(ox,o\bar{x}) | o \in \underline{\mathcal{X}}, o \models {\alpha } \}$ and for every $o_1 , o_2 \in \underline{\mathcal{X}}$, for every $x_1 , x_2 \in \underline{X}$, $o_1 x_1 \succeq _\varphi o_2 x_2$ if and only if $o_1 = o_2$, $x_1 = x$, $x_2 = \bar{x}$ and $o_1 \models {\alpha }$; similarly, $o_1 x_1 \succeq _{\varphi '} o_2 x_2$ if and only if $o_1 = o_2$, $x_1 = x$, $x_2 = \bar{x}$ and $o_1 \models {\alpha } '$. Thus $\alpha \equiv \alpha '$ if and only if for every $o \in \underline{\mathcal{X}}$, $o \models \alpha \Leftrightarrow o \models \alpha '$, iff for every $o \in \underline{\mathcal{X}}$, $ox \succeq _\varphi o\bar{x} \Leftrightarrow ox \succeq _{\varphi '} o\bar{x}$, if and only if $\varphi \equiv \varphi '$.

Similarly, we can define two linear 1-LP-trees $\psi$ and $\psi '$ as follows: the top $| \mathcal X |$ nodes are labelled with attributes from $\mathcal X$, in any order and with no rule; then there is one node labelled with X, and the same preference rules as above.

Proposition 14

${\succeq }$ -cut counting and ${\succ }$ -cut counting are $\#\textsf {P}$-hard for CP-nets and acyclic CP-nets.

Proof

Remember that a vertex cover in a graph $G=(V,E)$ is a set $S\subseteq V$ such that for each edge $uv\in E$ we have $u\in S$ or $v\in S$. The problem $\#\textsf {VertexCover}$ is, given a graph G, to count its vertex covers. $\#\textsf {VertexCover}$ is well-known to be $\#\textsf {P}$-hard [47], so we will use a reduction from $\#\textsf {VertexCover}$ to $\succ$-cut-counting to establish the claim.

So let $G=(V,E)$ be a graph. For every vertex $v\in V$ we introduce an attribute $V_v$ and for every edge $e= uv\in E$ we introduce an attribute $E_{uv}$. Note that for convenience we denote $E_{uv}$ also by $E_e$ sometimes. Finally, we introduce attributes $D_i$ for $i\in [|V|+|E|+1]$. The attributes $V_v$ have no parents. Let $e_1, \ldots , e_m$ be an order of the edges in E where $e_i = u_iv_i$. For $i>1$ the attribute $E_{e_i}$ has the parents $V_{u_i}, V_{v_i}, E_{e_{i-1}}$. The attribute $E_{e_1}$ has parents $V_{u_1}, V_{v_1}$. Finally, the attributes $D_j$ all have the single parent $E_{e_m}$.

We next describe the CPTs for all attributes: all attributes have values in $\{0,1\}$. All $V_{v}$ have the order $1 \ge 0$. For all $D_i$, we have that if $E_{e_m}$ has value 0 then the order is $0 \ge 1$ and if $E_{e_m}$ has value 1, then $1\ge 0$. For $E_{e_1}$ we have the order $1\ge 0$ if and only if at least one of $V_{u_1}, V_{v_1}$ has value 1 and the order $0\ge 1$ otherwise. Finally, for $i>1$, we have the order $1\ge 0$ if and only if $E_{i-1}$ has value 1 and at least one of $V_{u_i}, V_{v_i}$ has value 1. Otherwise $E_{e_i}$ has the order $0\ge 1$. Call the resulting CP-net $\varphi$.

Note that one can easily see that no attribute in an increasing flip** sequence can ever be flipped back to 0 from 0: for the attributes $V_v$ this is immediate. For the $E_{e_j}$ it follows with an easy induction and the fact that it is true for the $V_v$. For the $D_j$ finally it follows from the fact that $E_{e_m}$ can never flip back to 0.

Let o be the assignment that assigns 0 to all attributes. Let $o'$ be an assignment such that $o'$ is reachable from o by an increasing flip** sequence, or equivalently $o' \succeq _\varphi o$. We claim that if $E_{e_m}$ has value 1 in $o'$, then $S := \{v \in V\mid o'(V_v) = 1\}$ is a vertex cover of G. To see this, first observe that in fact all $E_{e_j}$ must take the value 1 in $o'$: to flip $E_{e_j}$ to 1, we must have flipped $E_{j-1}$ before (if it exists) and since we can never flip back to 0, $E_{j-1}$ must take 1 in $o'$. But then when we flipped $E_{e_j}$ to 1, at least one of $V_{v_j}, V_{u_j}$ must have had value 1 and since we cannot flip it back, in $o'$ one of $V_{v_j}, V_{u_j}$ must have value 1. So for every $e_j$ we have that one of $V_{v_j}, V_{u_j}$ must have value 1 which proves that S is a vertex cover as claimed.

Now for $S\subseteq V$, define $o_S$ to be the assignment that assigns 1 to $V_v$ if and only if $v\in S$, assigns 1 to all $E_{e_i}$ and assigns 0 to all $D_j$. We claim that $o_S \succeq _\varphi o$ if and only if S is a vertex cover of G. First note that if S is a vertex cover, we can flip all $V_v$ accordingly and then iteratively flip all $E_{e_j}$ to reach $o_S$. The other direction is clear from what we saw above, observing that $E_{e_m}$ takes value 1 in $o'$.

Observe that for every $o_S$, where S is a vertex cover, we can flip an arbitrary subset of the $D_j$ to 1 to reach an assignment $o'\succeq _\varphi o_S \succeq _\varphi o$. Note that for different vertex covers $S_1, S_2$, there is no such $o' \succeq o_{S_1}$ and $o' \succeq o_{S_2}$ since $o_{S_1}$ and $o_{S_2}$ differ on the $V_v$ and in the construction of the $o'$ from the $o_S$ we do not change those. It follows that

$$\begin{aligned}&\{o' \mid o' \succeq _\varphi o'', o''(E_{e_m}) = 1\} =\\ &\qquad \qquad \qquad \qquad \qquad \qquad \bigcup _{S \text { vertex cover of } G} \{o' \mid o'\succeq _\varphi o_S, \forall V_v: o_S(V_v) = o'(V_v)\} \end{aligned}$$

and the union is disjoint. Now for every vertex cover S of G, we have

$$\begin{aligned} |\{o'' \mid o''\succeq _\varphi o_S, \forall V_v: o_S(V_v) = o''(V_v)\}| = 2^{|V|+|E|+1}. \end{aligned}$$

Let s be the number of vertex covers of G. It follows that

$$\begin{aligned} |\{o' \mid o'\succeq _\varphi o\}|&= |\{o' \mid o' \succeq _\varphi o, o'(E_{e_m}) = 0\}| + |\{o' \mid o'' \succeq _\varphi o, o''(E_{e_m}) = 1\}|\\ &= |\{o' \mid o' \succeq _\varphi o, o'(E_{e_m}) = 0\}| + s 2^{|V|+|E|+1}. \end{aligned}$$

Now since in no $o'$ with $o'(E_{e_m}) = 0$ any of the $D_j$ can be flipped to 1 in any increasing flip** sequence, we have

$$\begin{aligned} |\{o' \mid o' \succeq _\varphi o, o'(E_{e_m}) = 0\}| < 2^{|V|+|E|}, \end{aligned}$$

since such $o'$ have only $|V|+|E|-1$ attributes with domain $\{0,1\}$ which are not forced to be constant 0. Consequently, s can be inferred from $|\{o' \mid o'\succeq _\varphi o\}|$ by a single integer division which completes the reduction.

This proves that ${\succeq }$ -cut counting for acyclic CP-nets is as hard as #VertexCover; this holds for ${\succ }$ -cut counting since in the case of acyclic CP-nets, $\succeq$ is antisymmetric. And this hardness result extends to the larger class of CP-nets.

Proposition 15

${\succ }$ -cut counting is $\#\textsf {P}$-complete for GAI, GAI$_k$ and GAI$_1$.

For containment in $\#\textsf {P}$, observe that all elements in $\textsf {CUT}^{\succ ,o}(\varphi )$ have polynomial size, so we can easily guess them and compare in polynomial time to o since ${\succ }$ -comparison can be solved in polynomial time for GAI.

For hardness, we reduce from the problem #SubsetSum which is, given a set $S=\{s_1, \ldots , s_n\}$ of positive integers and an additional integer k, to count the number of subsets of S that sum up to k. #SubsetSum is well-known to be $\#\textsf {P}$-complete, see e.g. [27]. It will be convenient to work with a slight variant which we call #SubsetSum$_>$ and which is, given the same type of input as for #SubsetSum, to count the number of subsets of S which sum up to a value greater than k. There is an easy oracle reduction from #SubsetSum to #SubsetSum$_>$: given an input S, k, call an oracle for #SubsetSum$_>$ on the two inputs $S, k-1$ and S, k. Then the answer to the #SubsetSum instance is the difference of the answers of the oracle calls. It follows that #SubsetSum$_>$ is $\#\textsf {P}$-hard.

We now reduce #SubsetSum$_>$ to ${\succ }$ -cut counting for GAI$_1$. So let $S= \{s_1, \ldots , s_n\}$ and k be an instance of #SubsetSum$_>$. We construct n functions $g_i(X_i)$ for $i=1, \ldots , n$ where $\underline{X_i} = \{0,1\}$. We set $g_i(0)=0$ and $g_i(1)=s_i$. Moreover, we add a function $g_Y(Y)$ where $\underline{Y}= \{0,1\}$ and $g_Y(0) = 1$ and $g_Y(1) = k$. Set $\varphi = \{g_1, \ldots , g_n, g_Y\}$ and $\mathcal X = \{X_1, \ldots , X_n, Y\}$. This completes the construction of the GAI. Call the induced relation $\succeq$.

To complete the reduction, let $o^*\in \underline{\mathcal{X}}$ be the alternative that sets Y to 1 and all other attributes to 0. Then $g_\varphi (o^*) = k$. Moreover, for $o\in \underline{\mathcal{X}}$, we have that $g_\varphi (o)> k$ if and only if $o(Y) = 1$ and there is an $i\in [n]$ such that $o(X_i)=1$—i.e. the set $\{i\in [n]\mid o(X_i)=1\}$ is non-empty—, or $o(Y) = 0$ and $\sum _{i\in [n]} g_i(o[X_i]) = \sum _{i\in [n]:o(X_i)=1} s_i > k$. Note that there are $2^n-1$ alternatives of the former type, corresponding to the non-empty subsets of [n], so the number of subsets of S that sum up to values greater than k is $|\textsf {CUT}^{\succ , o^*}(\varphi )| - 2^n+1$. Thus, one oracle call to ${\succ }$ -cut counting allows solving #SubsetSum$_>$ in polynomial time which completes the reduction.

Proposition 16

${\succeq }$ -cut extraction is tractable for CP, and ${\succ }$ -cut extraction is tractable for acyclic CP-nets. ${\succ }$ -cut counting and ${\succ }$ -cut extraction are PSPACE-hard for 1-CP${\ntriangleright }{\wedge }$. ${\succ }$ -cut extraction, ${\succeq }$ -cut extraction and ${\succ }$ -cut counting are tractable for LP-trees.

Proof

${\succeq }$ -cut extraction is easy for CP: given o and $\varphi$, in order to return an element of $\textsf {CUT}^{\succeq ,o}(\varphi )$, it is sufficient to find one statement in $\varphi$ which sanctions an improving swap for o. For acyclic CP-nets (and more generally for any language that guarantees consistency), $\succ$ is the asymmetric part of $\succeq$, thus ${\succ }$ -cut extraction is equivalent to ${\succeq }$ -cut extraction and is tractable.

Note that alternative o is undominated iff CUT$^{\succ ,o}(\varphi ) = \emptyset$, iff $|\textsf {CUT}^{\succ ,o}(\varphi )| = 0$; therefore, ${\succ }$ -cut counting and ${\succ }$ -cut extraction are at least as hard as undominated check, they are therefore PSPACE-hard for 1-CP${\ntriangleright }{\wedge }$.

Finally, $\succ$ -cut extraction, $\succeq$ -cut extraction and ${\succ }$ -cut counting are tractable for LP-trees: for LP-tree $\varphi$, given o, in order to find some $o'$ such that $o' \succ _\varphi o$ (resp. $o' \succeq _\varphi o$), it is possible to traverse the tree, starting at the root, guided by the values assigned by o, until reaching a node where the value(s) assigned by o for the attributes at that node is/are strictly dominated (resp. dominated) by other values at that node. Also, when going down $\varphi$ in the branch that corresponds to o, it is possible, at each node N encountered, labelled with T, to count the number of $t'$ in $\underline{T}$ such that $t > o[t']$ (according to the preference rule $\beta :\ge ^{\beta }$ at N such that $o \models \beta$), and to multiply this number by the sizes of the domains of the attributes that have not been encountered yet; adding these sums of products along the branch will give the number of alternatives o such that $o' \succ _\varphi o$.

Proposition 17

${\succeq }$ -cut extraction, $\succ$ -cut extraction, undominated check, and undominated extract are tractable for GAI$_1$.

Given a GAI $\varphi$, we can simply choose the values for the attributes in such a way that the utilities are maximized. Since the utilities are unary, this leads to a consistent and thus also maximal alternative $o^*$. For ${\succeq }$ -cut extraction, $o^*$ is always a valid output, so it solves the problem independent of the additional input alternative o. For $\succ$ -cut extraction, we check if $g_\varphi (o^*)> g_\varphi (o)$. If so, we return $o^*$ again. Otherwise, due to the maximality of $o^*$, we have $g_\varphi (o^*)= g_\varphi (o)$ and thus there is no alternative strictly dominating o and thus no valid output.

For undominated check we have that o is undominated if and only if $g_{\varphi }(o) = g_{\varphi }(o^*)$ which we can check efficiently. Finally, for undominated extract we can simply return $o^*$.

Proposition 18

${\succeq }$ -cut extraction and $\succ$ -cut extraction are $\textsf {NP}$-complete for GAI$_k$ for $k\ge 2$ and GAI. undominated check is $\textsf {coNP}$-complete and undominated extract is $\textsf {NP}$-hard for GAI$_k$ for $k\ge 2$ and for GAI.

Proof

Containment in $\textsf {NP}$, resp. $\textsf {coNP}$, is easy to see in all cases since alternatives can be compared efficiently

We show hardness for all problems by reduction from 3-Coloring which is, given a simple, undirected graph $G=(V,E)$, to decide if there is an assignment $c:V\rightarrow \{r,g,b\}$ such that for all edges $uv\in E$ we have $c(v)\ne c(u)$. The map** c is called a coloring and it is said to be valid if it satisfies the condition on the edges. 3-Coloring is well-known to be $\textsf {NP}$-complete, see e.g. [43, Theorem 9.8].

We use the same construction of a GAI$_2$ $\varphi _G$ from a graph G for all problems. So let a graph G be given in which w.l.o.g. every vertex has at least two neighbors (vertices with fewer than two neighbors can iteratively be deleted without changing the answer to the 3-Coloring question). We also assume that G is connected; if it is not, we can connect the different connected components iteratively by adding edges without changing the answer to the 3-Coloring question. We construct a GAI$_2$ representation as follows: for every vertex $v\in V$, we introduce an attribute $X_v$ with domain $\underline{X_v} = \{r,g,b,d\}$. For every edge $e=uv$, we construct a utility function $g_{uv}$ in the variables $X_u, X_v$ and which takes value 1 on inputs rb, rg, br, bg, gr, gb, dd and 0 on all other inputs. Setting $\varphi _G = \{g_{uv}\mid uv\in E\}$ completes the construction of the GAI $\varphi _G$. Let $\succeq$ be the order that $\varphi _G$ induces.

We first show hardness for ${\succeq }$ -cut extraction. To this end, let $o_d$ be the alternative in which all attributes take value d. Then all $g_{uv}$ evaluate to 1 on $o_d$, so $g_{\varphi _G}(o_d) = |E|$. Now consider $o \in \textsf {CUT}^{\succeq ,o_d}$. Assume first that some attribute of o takes value d. Since not all attributes can take value d and G is connected, there must be an edge uv such that $d = o(u) \ne o(v)$. Then $g_{uv}(o)=0$ and $g_{\varphi _G}(o)< |E| = g_{\varphi _G}(o_d)$, so $o \notin \textsf {CUT}^{\succeq ,o_d}$ which contradicts the choice of o. Consequently, we must have that all $X_v$ takes values in $\{r,g,b\}$ in o. Moreover, for all $uv\in E$, we must have that $o(X_u)\ne o(X_v)$. Thus, setting $c(v) = o(X_v)$ for all $v\in V$ yields a valid coloring of G. So if there is an element in $\textsf {CUT}^{\succeq ,o_d}$, the graph G is 3-colorable. The other way round, if G has a valid 3-coloring c, then defining o for all $X_v$ by $o(X_v) = c(v)$ yields an alternative in $\textsf {CUT}^{\succeq ,o_d}$. This shows $\textsf {NP}$-hardness of ${\succeq }$ -cut extraction for GAI$_2$ and thus for all GAI$_k$ with $k\ge 2$ and GAI.

The reasoning for $\succ$ -cut extraction is similar. The only difference is that for one arbitrary edge uv we set $g_{uv}(d,d)$ to 0. Call the resulting GAI $\varphi _G'$. We have $g_{\varphi _G'}(o_d)= |E|-1$. The rest of the reduction and the argument for completeness is exactly as that for ${\succeq }$ -cut extraction.

For undominated check, observe that $o_d$ is dominating for $\varphi _G'$ if and only if there is no alternative o with $g_{\varphi _G'}(o_d) < g_{\varphi _G'}(o)$. Reasoning as above, this is exactly the case if and only if G has no valid 3-coloring. Thus undominated check is $\textsf {coNP}$-hard.

Finally, to show hardness of undominated extract, multiply all utility values in $g_G$ by 2. Then, for one arbitrary edge uv set $g_{uv}(d,d)$ to 1. Call the resulting GAI $\varphi _{G}''$. Then we have $g_{\varphi _G''}(o_d)= 2|E|-1$. Moreover, for all alternatives o encoding a valid 3-coloring, we have $g_{\varphi _G''}(o)= 2|E|$. Finally, for all other alternatives o, we have $g_{\varphi _G''}(o)\le 2|E|-2$. So in any case an undominated alternative is either a valid 3-coloring of the graph G or $o_d$, hence $o_d$ is undominated if and only if G is not 3-colorable which shows that undominated extract is $\textsf {NP}$-hard.

E Proofs for Section 7 (Transformations)

Proposition 19

Given a preorder $\succeq$ over $\underline{\mathcal{X}}$, given $V \subseteq \mathcal X$, let $U \subseteq \mathcal X \setminus V$. If $v,v' \in \underline{V}$ and $v \succeq ^{\downarrow V}_{\text {w.opt.}} v'$, then there is some $u \in \underline{U}$ such that for no $u' \in \underline{U}$ it holds that $u'v' \succ uv$.

Proof

Assume that $v \succeq ^{\downarrow V}_{\text {w.opt.}} v'$, and let $u \in \underline{U}$ be such that for no other $u_1 \in \underline{U}$ it is the case that $u_1 v \succ uv$: such a u must exist because $\underline{U}$ is finite; if there is some completion $u'$ of $v'$ such that $u'v' \succ uv$, then, since $v \succeq ^{\downarrow V}_{\text {w.opt.}} v'$, there must be some $u_1 \in \underline{U}$ such that $u_1 v \succeq u'v'$, but then $u_1 v \succ uv$, which is a contradiction.

Proposition 20

All four projections defined above are equivalent for the 1-GAI language and the language that contains complete LP-trees of $\textsf {1-LPT}^{\textsf {lin}}$, and can be computed in polynomial time.

Proof

Let $\varphi \in \textsf {1-GAI}$, let X be any attribute, $\varphi$ is defined by a function of the form $g(o) = g_X(o[X]) + \sum _{Y \ne X} g_Y(o[Y])$. Let $\succeq$ denote the associated weak order over the set of alternatives. Let $\psi$ be defined by $h(o) = \sum _{Y \ne X} g_Y(o[Y])$. We show that $o \succeq _{\text {low}} o'$ iff $o \succeq _{\text {up}} o'$ iff $o \succeq _{\text {w.opt.}} o'$ iff $o \succeq _{\text {s.opt}} o'$ iff $o \succeq _\psi o'$:

$$\begin{aligned} o \succeq _{\psi } o'{} & {} \Leftrightarrow h(o) \ge h(o') \\{} & {} \Leftrightarrow ( \forall x \in \underline{X} : g_X(x) + h(o) \ge g_X(x) + h(o') ) \Leftrightarrow o \succeq _{\text {low}} o' \\{} & {} \Leftrightarrow ( \exists x \in \underline{X} : g_X(x) + h(o) \ge g_X(x) + h(o') ) \Leftrightarrow o \succeq _{\text {up}} o' \\{} & {} \Leftrightarrow ( \forall x' \in \underline{X} \exists x \in \underline{X} : g_X(x) + h(o) \ge g_X(x') + h(o') ) (\text {take } x=x') \\{} & {} \Leftrightarrow o \succeq _{\text {w.opt.}} o' \\{} & {} \Leftrightarrow ( \forall x' \in \underline{X} : g_X(\underset{x \in \underline{X}}{\textsf {argmax}} ~g_X(x)) + h(o) \ge g_X(x') + h(o') ) \\{} & {} \Leftrightarrow o \succeq _{\text {s.opt.}} o' \end{aligned}$$

Suppose now that $\varphi$ is a complete, linear LP-tree in 1-LPT over $\mathcal X$, and let $X \in \mathcal X$. Let $\psi$ be the LP-tree defined by removing node X, redirecting the parent of X to the unique child of X when X is an internal node of $\varphi$. Consider alternatives $o, o' \in \underline{\mathcal{X} \setminus \mathcal{X}}$, let $x \in \underline{X}$. Let Y be the attribute that decides the pair $\{o,o'\}$ in $\psi$ then $o \succeq _\psi o'$ iff $o[Y] > o'[Y]$ in $\textsf {CPT}(Y)$. Suppose first that Y is an ancestor of X in $\varphi$, then $\{ox,o'x'\}$ is decided at Y in $\varphi$ forall $x,x \in \underline{X}$, thus $o[Y] > o'[Y] \Rightarrow ( \forall x,x' \in \underline{X} : ox \succeq _\varphi o'x' ) \Rightarrow o \succeq _\pi o'$ for all four projections; and $o \succeq _\pi o' \Rightarrow o \succeq _\psi o'$ for any of the four projections. Suppose now that X is an ancestor of Y in $\varphi$, then for all $x, x' \in \underline{X}$ 1) the pair $\{ox,o'x\}$ is decided at Y, whereas 2) pair $\{ox,o'x'\}$ with $x \ne x'$ is decided at X. From 1) it follows that $o \succeq _\psi o' \Leftrightarrow o \succeq _{\text {low}} o' \Leftrightarrow o \succeq _{\text {up}} o'$; moreover, let $x_0$ be the optimal value for $\underline{X}$ in $\textsf {CPT}(X)$ in $\varphi$ (which exists and is unique because $\varphi$ is a complete LP-tree, thus the linear order over $\underline{X}$ in $\textsf {CPT}(X)$ is a linear order), then $ox_0 \succeq _\varphi o'x'$ for all $x' \in \underline{X}$, thus $o \succeq _{\text {s.opt.}} o'$; [5] mention that $o \succeq _{\text {s.opt}} \Rightarrow o \succeq _{\text {w.opt}} o'$; lastly, if $o \succeq _{\text {w.opt}} o'$ then there exists x such that $ox \succeq _\varphi o'x_0$, and $ox_0 \succeq _\varphi ox$, thus $ox_0 \succeq _\varphi o'x_0$, thus $o[Y] > o'[Y]$.

Proposition 21

If conditioning can be done in polynomial time for language $\mathcal L$ but the extraction of an undominated alternative is $\textsf {NP}$-hard, then the strong optimistic projections cannot be computed in polynomial time for $\mathcal L$ (unless $\textsf {P} = \textsf {NP}$).

Proof

We assume that for any preorder expressed in the language $\mathcal L$, any strong optimistic projection leads to a preorder that again can be expressed in $\mathcal L$. If this is not true, then the statement of the theorem is trivially true, even without the assumption $\textsf {P} \ne \textsf {NP}$.

We give an algorithm that, given a preorder $\succeq$ encoded in $\mathcal L$, computes an undominated alternative o in polynomial time, assuming polynomial time algorithms for conditioning and computation of strong optimistic projection. The algorithm considers the attributes of $\succeq$ in sequence, say from $V_1$ to $V_n$. The value $v_1$ of $V_1$ is obtained by projecting $\succeq$ onto $V_1$ - then an undominated value $v_1$ is chosen for $V_1$; indeed, $v_1 \succeq ^{\downarrow \{V1\}}_{\text {s.opt.}} v_1'$ means that there exist an assignment v of $\{V_2, \dots , V_n\}$ such that $v_1. v \succeq v_1'. v'$ for all $v'$ – $V_1 = v_1$ in one of the non dominated solutions. Then the original formula is conditioned: value $v_1$ is assigned to $V_1$ and the procedure is repeated for the next variable - and this until all the variables have been assigned. So, if a language offers the conditioning transformation in polytime but not the undominated query, there cannot be any polynomial algorithm for performing the strong optimistic projection within this language (unless $P= NP$).

A direct corollary is that GAI and GAI$_k$ ($k > 1$) fail to provide strong and weak optimistic projections in polynomial time (unless $\textsf {P} = \textsf {NP}$) - since (i) the extraction of an undominated alternative is $\textsf {NP}$-hard for these languages and (ii) they support conditioning in polytime.

Proposition 22

The strong (resp. weak) projection cannot be computed in polytime for GAI and GAI$_k (k > 1)$ (unless $\textsf {P} = \textsf {NP}$).

Proof

The extraction of an undominated alternative is $\textsf {NP}$-hard for GAI and GAI$_k, k > 1$ (Proposition 18) while conditioning can be done in polytime for these languages. From Proposition 21 we deduce the strong optimistic projection cannot be computed in polytime unless $P = NP$. Because GAI’s encode complete and transitive relations, the strong and weak optimistic projections are identical [6] - hence the weak optimistic projection cannot be computed in polytime unless $P = NP$.