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=
ˆ
Column #4 For i
4, we extend the covering array for
p 4 . The set of combinations
π
is as follows:
π ={ (
1
, , ,
1
), (
1
, , ,
2
), (
2
, , ,
1
), (
2
, , ,
2
),
( ,
1
, ,
1
), ( ,
1
, ,
2
), ( ,
2
, ,
1
), ( ,
2
, ,
2
),
( , ,
1
,
1
), ( , ,
1
,
2
), ( , ,
2
,
1
), ( , ,
2
,
2
) } .
Then, we enter the horizontal extension stage:
Row #1 For the 1st test case, the number for newly-covered target combinations
in
is 2 for both value 1 and 2, so we choose value 1 and append it to the test
case, which now becomes
π
(
1
,
1
,
1
,
1
)
. And then we remove the newly-covered target
combinations from
π
, so it becomes:
π ={ (
, , ,
), (
, , ,
), (
, , ,
),
1
2
2
1
2
2
( ,
, ,
), ( ,
, ,
), ( ,
, ,
),
1
2
2
1
2
2
( , ,
1
,
2
), ( , ,
2
,
1
), ( , ,
2
,
2
) } .
Row #2 For the 2nd test case, the number for newly-covered target combinations in
π
is 2 for value 1 and 3 for value 2, so we choose value 2 and append it to the test
case, which now becomes
(
1
,
2
,
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
),
( , ,
1
,
2
), ( , ,
2
,
1
) } .
Row #3 For the 3rd test case, the number for newly-covered target combinations
in
is 2 for both value 1 and value 2, so we choose value 1 and append it to the
test case, which now becomes
π
(
2
,
1
,
2
,
1
)
. And then, we remove the newly-covered
target combinations from
π
, which now becomes:
π ={ (
2
, , ,
2
),
( ,
1
, ,
2
), ( ,
2
, ,
1
),
( , ,
1
,
2
) } .
Row #4 For the 4 th test case, the number for newly-covered target combinations in
π
is 1 for both value 1 and value 2, so we choose value 1 and append it to the test
case, which now becomes
(
2
,
2
,
2
,
1
)
. And then, we remove the newly-covered target
combinations from
π
, which now becomes:
π ={ (
, , ,
),
2
2
( ,
1
, ,
2
),
( , ,
1
,
2
) } .
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