Difficulty: Easy
Correct Answer: Incorrect
Explanation:
Introduction / Context: In EER modeling, specialization/generalization connects a supertype to its subtypes. This connection is not a typical “relationship” with variable cardinality; rather, each subtype’s instances are by definition also instances of the supertype (an “is-a” relationship). The question asks whether we speak of cardinality between the supertype and subtype.
Given Data / Assumptions:
Concept / Approach: Because a subtype “is a” supertype, the mapping is inherently one-to-one: each subtype instance matches exactly one supertype instance. Modeling tools depict participation (double line for total) and disjointness/overlap, not variable cardinality. Therefore, saying “there is cardinality between supertype and subtype” is misleading—what exists are specialization constraints, not general 1:M or M:N cardinalities.
Step-by-Step Solution:
Recognize that a subtype inherits the supertype’s key; hence each subtype row maps to one supertype row.Identify applicable constraints: disjoint/overlapping, total/partial.Note that tools may still render a connector, but it does not carry ordinary cardinality symbols.Conclude the statement is incorrect; participation is modeled, not cardinality.Verification / Alternative check: When transforming to relational schemas, subtype tables share the supertype’s primary key (1:1). There is no 1:M option here, validating the point.
Why Other Options Are Wrong:
Common Pitfalls: Using crow’s foot symbols on specialization connectors; misreading total participation as “many.”
Final Answer: Incorrect
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