
Test Adequacy Assessment
cover all c-uses. This is because dcu(x, 1) will be covered when dcu(z,5)
is covered. Continuing in this manner, one can show that dcu(count,
1) can also be ignored. This leads us to a minimal set of c-uses to be
covered as shown in the table below. Note that the total number of c-
uses to consider has now been reduced from 17 to 12. The total number
of p-uses to consider has been reduced from 10 to 4. Similar analysis for
removing p-uses from consideration is left as an exercise (see Exercise
6.30).
Defined at
Variable (v) node (n) dcu (v, n) dpu (v, n)
y 6 {4, 6} {(3, 4), (3, 6)}
z 1 {4, 6, 7} { }
z 4 {4,6,7} {}
z 5 {4, 6, 7} { } ...