Automatic Generation of Combinatorial Test Data by Jian Zhang

By Jian Zhang

This publication reports the state of the art in combinatorial trying out, with specific emphasis at the computerized new release of try out information. It describes the main favourite techniques during this quarter - together with algebraic building, grasping equipment, evolutionary computation, constraint fixing and optimization - and explains significant algorithms with examples. furthermore, the e-book lists a few try out new release instruments, in addition to benchmarks and functions. Addressing a multidisciplinary subject, will probably be of specific curiosity to researchers and pros within the components of software program trying out, combinatorics, constraint fixing and evolutionary computation.

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IEEE Trans. Softw. Eng. 34(5), 633–650 (2008) 10. : Pairwise testing in the real world: practical extensions to test-case scenarios. In: Proceedings of the 24th Pacific Northwest Software Quality Conference (PNSQC’06), pp. 419–430 (2006) 11. : Automating test case generation for the new generation mission software system. In: Proceedings of the IEEE Aerospace Conference, vol. 1, pp. 431–443 (2000) 12. : Cascade: a test generation tool for combinatorial testing. In: Proceedings of the IEEE Sixth International Conference on Software Testing, Verification and Validation Workshops (ICSTW’13), pp.

5 The product of A ⊗ B (Reprinted with permission from [9]. ) Fig. 6 CA(N + (v − 1) · M, v2k , 3) AA 1 B Bπ 2 π BB ... 6 (Products of Strength-2 CAs) If a CA(N , vk , 2) and a CA(M, v , 2) both exist, then a CA(N + M, vk , 2) also exists. Let A = (ai j ) be CA(N , vk , 2) and B = (bi j ) be CA(M, v , 2). Then the (N + M) × k array C = (ci j ) = A ⊗ B in Fig. 5 is a CA(N + M, vk , 2). Similarly, there are product methods for pairwise mixed covering arrays. For details, see [4]. The methods can be used recursively.

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, −)}. 3 IPOG-C Example 45 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, −)}.

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