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Domain Testing: Divide and Conquer - Testing Education

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Iteration 6:<br />

Test<br />

case#<br />

a (4) b (3) c (2) d (2)<br />

1. A1 B1 C1 D1<br />

2. A1 B2 C2 D2<br />

3. A1 B3 C3 D3<br />

4. A1 B4 C4 D4<br />

5. A2 B1 C2 D4<br />

6. A2 B2 C3 D1<br />

7. A2 B3 C4 D2<br />

8. A2 B4 C1 D3<br />

9. A3 B1 C3 D2<br />

10. A3 B2 C4 D3<br />

11. A3 B3 C1 D4<br />

12. A3 B4 C2 D1<br />

13. D4 B1 C4<br />

14. D4 B2 C1<br />

15. D4 B3 C2<br />

16. D4 B4 C3<br />

No ordering of test cases in set four for variable ‘d’ will yield all pairs of ‘d’ with all<br />

other variables. Even if we backtrack, we basically get stuck in the same combinations;<br />

this is because there are no more alternatives available. This means that we cannot fit<br />

all-pairs in sixteen combinations. Let us have the ordering in set four of variable ‘d’<br />

that makes all-pairs of ‘d’ with variable ‘c’, ‘b’ <strong>and</strong> ‘a’. After that let us add four<br />

additional combinations to take care of pairs of ‘d’ <strong>and</strong> ‘b’.

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