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Row #2
For the second test case, the number of newly-covered target combinations
in
π
is 1 for value 1 and 2 for value 2, so we choose value 2 and append it to the
test case, which now becomes
(
1
,
2
,
2
,
−
)
. And then, we remove the newly-covered
target combinations from
π
, which now becomes:
π
={
(
2
,
−
,
1
,
−
), (
2
,
−
,
2
,
−
),
(
−
,
1
,
2
,
−
), (
−
,
2
,
1
,
−
)
}
.
Row #3
For the third test case, if we choose value 1 for
p
3
, the partial test case will be
ˆ
(
2
,
1
,
1
,
−
)
, which violates constraint “
p
3
=
ˆ
1
→ˆ
p
1
=
1”, so we can only choose
value 2. The resulting test case is
(
2
,
1
,
2
,
−
)
, and
π
becomes:
π
={
(
2
,
−
,
1
,
−
),
(
−
,
2
,
1
,
−
)
}
.
Row #4
For the fourth test case, if we choose value 1 for
p
3
, the partial test case will
ˆ
be
(
2
,
2
,
1
,
−
)
, which also violates constraint “
p
3
=
ˆ
1
→ˆ
p
1
=
1”, so we can only
choose value 2. The resulting test case is
(
2
,
2
,
2
,
−
)
, and
π
stays unchanged.
Now the test suite is
⎛
⎞
111
122
212
222
⎝
⎠
,
and
π
={
(
2
,
−
,
1
,
−
),
(
−
,
2
,
1
,
−
)
}
.
Then we enter the vertical extension stage:
(1) For target combination
(
2
,
−
,
1
,
−
)
, we check its validity and find it violates
constraint “
p
3
=
ˆ
1
→ˆ
p
1
=
1”, so it is marked as unsatisfiable and neglected.
(2) For target combination
(
−
,
2
,
1
,
−
)
, it satisfies the constraint, so we add a new
test case
(
∗
,
2
,
1
,
−
)
into the test suite.
ˆ
After extending the covering array for
p
3
, the resulting test suite is as follows:
⎛
⎝
⎞
⎠
111
122
212
222
∗
,
21
which is a covering array of strength 2 for
p
1
,
ˆ
ˆ
p
2
,
ˆ
p
3
.
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