Test Selection, Minimization, and Prioritization
Step 2: PrT =< t
5
>.X
={t
1
, t
2
, t
3
, t
4
}. entitiesCov ={3, 14}.
Step 3: We continue the process as X and entitiesCov is not empty.
Step 3.1: Compute the residual coverage for each test in X
.
resCov(t
1
) =|{3, 14}\({1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 16}∩{3, 14})| = |∅| = 0
resCov(t
2
) =|{3, 14}\({1, 2, 4, 5, 12, 13, 15, 16}∩{3, 14})|=|{3, 14}| = 2
resCov(t
3
) =|{3, 14}\({1, 2, 3, 4, 5, 12, 13, 14, 16}∩{3, 14})| = |∅| = 0
resCov(t
4
) =|{3, 14}\({1, 2, 4, 5, 12, 13, 14, 16}∩{3, 14})|=|{14}| = 1.
Step 3.2: t
1
and t
3
have the least cost. We arbitrarily select t
3
. One
may instead use some other criteria to break the tie (see Exercise
5.35).
Step 3.3: PrT =< t
5
, t
3
>.X
={t
1
, t
2
}, entitiesCov =∅.
Step 3: There is no