Test Generation from Combinatorial Designs
HG. Step 4.1: t
1
= extend(t
1
, l
1
) = (a
1
, b
1
, c
1
). T
=
{(a
1
, b
1
, c
1
)}.
HG. Step 4.2: The run we just created by extending t
1
has cov-
ered the pairs (a
1
, c
1
) and (b
1
, c
1
). We now update AP as fol-
lows.
AP = AP −{(a
1
, c
1
), (b
1
, c
1
)}
={(a
1
, c
2
), (a
1
, c
3
), (a
2
, c
1
), (a
2
, c
2
), (a
2
, c
3
), (a
3
, c
1
), (a
3
, c
2
), (a
3
, c
3
)
(b
1
, c
2
), (b
1
, c
3
), (b
2
, c
1
), (b
2
, c
2
), (b
2
, c
3
)}
Repeating the above substeps for j = 2and j = 3, we obtain
the following.
HG. Step 4.1: t
2
= extend(t
2
, l
2
) = (a
1
, b
2
, c
2
). T
=
{(a
1
, b
1
, c
1
), (a
1
, b
2
, c
2
)}.
HG. Step 4.2:
AP = AP −{(a
1
, c
2
), (b
2
, c
2
)}
={(a
1
, c
3
), (a
2
, c
1
), (a
2
, c
2
), (a
2
, c
3
), (a
3
, c
1
), (a
3
, c
2
), (a
3
, c
3
)
(b
1
, c
2
), (b
1
, c
3
), (b
2
, c
1
), (b
2
, c
3
)}
HG. Step 4.1: t
3
= extend(t
3
, l
3
) = (a
2
, b
1
, c