Exercice de contrôle
Si l'on suppose que la BC [KB] ne contient que
pm#relation \. (part pm#type: pm#transitive_relation_type)
pm#type pm#supertype
B1 part _[each->a]: (B2 part _[each->a]: (B3 part _[each->a]: B4)), /^ A1, \. C1;
//i.e.: B1 /^ A1, \. C1; each B1 part: a B2; any B2 part: a B3; any B3 part: a B4;
quelles réponses doit donner un moteur d'inférences aux requêtes FL suivantes
s'il connait la sémantique de pm#transitive_relation_type, pm#type et pm#supertype ?
//Queries + answers (here, with or without the "closed world assumption"):
/*1:*/ ? [a B1 part: a B4]
each B1 part: (a B2 part: (a B3 part: a B4)); //since "each" implies "a" part pm#type: pm#transitive_relation_type;
/*2.*/ ? [a B1 part .^1: a B4]
[] //no answer since "a B1" is not directly connected to "a B4"
/*3.*/ ? [a B1 part .^1: a ?x]
each B1 part: a B2; //not also [B1 /^ A1, \. C1] since the query is about "a B1", not "B1"
/*4.*/ ?x [a B1 part .^1: ?x]
?x = a B2; //with "a ?x" in the query, the answer would have been: ?x = B2
/*5.*/ ? [a B1 part: a ^(?x /^ A1)]
[] //since B2 is not subtype of A1 (it is B1 which is subtype of A1)
/*6.*/ ?x [a B1 part: ?x]
?x = a B2; ?x = a B3; ?x = a B4; //since part is transitive
/*7.*/ ? [B1 pm#relation .^1: a ?x] //i.e.: ? [B1 pm#relation _[.->.] .^1: a ?x]
[] //for most inference engines, since "a ?x" specifies that "a" must be explicit in the // searched graphs (+ note that the query is about the type B1, not about "a B1"). //However, since A1 and B1 are (implicitly) of type Class (= Concept_type; // so A1 and B1 both satisfy|match|specialize `a ?x´ with `?x = Class´) and // since the inference engine is here assumed to know the semantics of pm#type, // if the engine is good, it can infer that a solution for ?x is Class or, more precisely, // ^(? instance: B1) and ^(? instance: C1); hence, a good engine could answer: // B1 /^ A1, \. C1; pm#relation \. pm#type pm#supertype; // Note: the above 2nd part is needed since 1) pm#relation is in the query graph, // 2) pm#type and pm#supertype relations are exploited for finding the above solution, and // 3) it is the KB which tells that pm#relation is supertype of pm#type and pm#supertype.
/*8.*/ ?x [B1 pm#relation .^1: ?x]
?x = A1; ?x = C1;
/*9.*/ ? [B1 pm#relation .^1: ?x]
B1 /^ A1, \. C1; pm#relation \. pm#supertype;
/*10.*/ ? [every B1 pm#relation: a ?x]
[each B1 part: (a B2 part: (a B3 part: a B4))]; pm#relation \. (part pm#type: pm#transitive_relation_type); // pm#relation is in the query; transitivity needed. // Without [part pm#type: pm#transitive_relation_type] in the KB, // the 1st line of the answer would have been: [each B1 part: a B2];
Rappel (pour faciliter la compréhension des réponses ci-dessus) : la BC [KB] ne contient que
pm#relation \. (part pm#type: pm#transitive_relation_type)
pm#type pm#supertype
B1 part _[each->a]: (B2 part _[each->a]: (B3 part _[each->a]: B4)), /^ A1, \. C1;
Si une BC ne contient que le graphe
[B1 part _[each->a ?b2]: (B2 part _[?b2->a]: (B3 pm#type: A3), pm#type: A2)]
et que la sémantique de "part" est prédéfinie dans le moteur d'inférences utilisé,
quelles réponses ce moteur d'inférence doit retourner pour les 2 requêtes suivantes sur cette BC
(d'après la description donnée en cours pour le langage de requêtes utilisé ci-dessous) ?
Si le graphe entier est une réponse, vous pouvez utiliser [...] pour ne pas le ré-écrire entièrement.
Pour signaler qu'une requête n'a pas de réponse, utilisez [] (i.e. le graphe vide).
/*1:*/ ?s [a B1 part .^1: a ?x] <= ?s
?s = [B1 part _[each->a ?b2]: B2]
/*2:*/ ? [a B1 part: (a ?x pm#type: A2)]
[] /* Indeed, A2 is equivalent to the type Class (since B2 is a type and is
instance of A2); hence, [B2 pm#type: A2] does not imply [each B1 part: a B2].
The answer would have been [B1 part _[each->a ?b2]: (B2 pm#type: A2)]
if i) the query had been ? [a B1 part: a ^(?x pm#type: A2)]
or ii) there had been [a B2 pm#type: A2] instead of [B2 pm#type: A2].
*/