Answer the following questions for the method intersection() below: public Set i

Question

Answer the following questions for the method intersection() below:

public Set intersection (Set s1, Set s2)

// Effects: If s1 or s2 is null throw NullPointerException

// else return a (non null) Set equal to the intersection

// of Sets s1 and s2

Characteristic: Validity of s1

>s1 = null

>s1 = {}

>s1 has at least one element

Characteristic: Relation between s1 and s2

>s1 and s2 represent the same set

>s1 is a subset of s2

>s2 is a subset of s1

>s1 and s2 do not have any elements in common

(a) Does the partition “Validity of s1” satisfy the completeness property? If not, give a value for s1 that does not fit in any block.

(b) Does the partition “Validity of s1” satisfy the disjointness property? If not, give a value for s1 that fits in more than one block.

(c) Does the partition “Relation between s1 and s2” satisfy the completeness property? If not, give a pair of values for s1 and s2 that does not fit in any block.

(d) Does the partition “Relation between s1 and s2” satisfy the disjointness property? If not, give a pair of values for s1 and s2 that fits in more than one block.

(e) If the “Base Choice” criterion were applied to the two partitions (exactly as written), how many test requirements wouldresult?

Explanation / Answer

(a) Yes, the partition validity of Set s1 satisfies the completeness property as Set s1 can either be null or empty or has at least 1 element in it. Any value given to the Set s1 will fit in one of the three conditions and there cannot be any value for s1 which doesn't satisfy any of the 3 conditions.

(b) Yes, the partition validity of Set s1 satisfies the disjointness property i.e. no value of Set s1 can qualify for more than 1 condition. The values will either give a null set, an empty set or a set with at least 1 element in it. The value cannot fit in more than 1 partition.

(c) Yes, it displays the completeness, if s1,s2 are not null sets then there can't be any condition outside the given four.

(d) No it doesn't display the disjointness property as when s1= {} and s2={} i.e. both sets are empty will one consider the condition s1 & s2 represent the same set or the condition that s1 & s2 do not have any elements in common.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.