An accelerating PSO algorithm based test data generator for data-flow dependencies using dominance concepts

Abstract

One of the most important and effort intensive activity of the entire software development process is software testing. The effort involved chiefly increases because of the need to obtain optimal test data out of the entire search space of the problem under testing. Software test data generation is one area that has seen tremendous research in terms of automation and optimization. Generating or identifying an optimal test set that satisfies a more robust adequacy criteria, like data flow testing, is still a challenging task. A number of heuristic and meta-heuristics like GA, PSO have been applied to optimize the test data generation problem. GA, although more popular, has its own difficulties such as complex to implement and slow convergence rate. In this paper an accelerating particle swarm optimization algorithm (APSO) is applied to generate test data for data-flow dependencies of a program guided by a new fitness function. APSO is used because of its capability of balancing in exploration and exploitation. A new fitness function is designed based on the concepts of dominance relations, weighted branch distance for APSO to guide the search direction. A set of benchmark programs and four modules of Krishna Institute of Engineering and Technology ERP system were taken for the experimental analysis. The experimental results show that the proposed APSO based approach performed significantly better than random search, genetic algorithm and PSO in enhancing the convergence speed.

Appendix 1: Case study on the calculation of fitness value

Consider dpu-path# 13 (8, 11, 12) for ‘Equal roots’ condition. The two objectives to be covered by an input test case are the definition node 8 and the p-use edge (11, 12). Fitness value of the input test case is computed using Eqs. (7) and (8). The various test case scenarios are discussed below. The dominance paths of the nodes are:

$${\text{dom}}\left( 8 \right) = \left\{ {1,2,7,8} \right\}$$

$${\text{dom}}\left( {11} \right) = \left\{ {1,2,7,8,9,11} \right\}$$

$${\text{dom}}\left( {12} \right) = \left\{ {1,2,7,8,9,11,12} \right\}$$

• 1. Input test case t₁ < 1, 4, 2>

Path traversed is {1, 2, 3, 5, 6, 7, 8, 9, 10, 13, 18, 19}

Dominance path of the def-node 8 is covered.

Dominance path of node 11 as well as node 12 of the p-use edgeis not covered. Branch distance for node 11 and node 12 at the critical node 9 is 0.125 (Eq. 10). Branch weight for node 11 and node 12 at the critical node 9 is 0.08958 (Table 5). Therefore wbd (x, t₁) = branch distance * branch weight(w_i) (Eq. 9).

$${\text{wbd }}\left( {11, \, t_{1} } \right) = {\text{wbd}}\left( {12,{\text{ t}}_{1} } \right) = 0.125*0.08958 = 0.0112.$$

$${\text{ft}}\left( {8, \, 11, \, 12,{\text{ t}}_{1} } \right) = 1/3*\left[ {\left( {1*1} \right) + \left( {\left( {5/6} \right)*0.0112} \right) + \left( {\left( {5/7} \right)*0.0112} \right)} \right] = 1/3*\left[ {1 + 0.00933 + 0.008} \right] = 1/3*1.01733 = 0.33911$$

The fitness value of test case t₁ is ft (8, 11, 12, t₁) = 0.33911 (Eq. 8)

• 2. Input test case t₂ < 1, 2, 2>

Path traversed is {1, 2, 3, 5, 6, 7, 8, 9, 11, 13, 18, 19}

Dominance path of the def-node 8 is covered.

Dominance path of node 11 of the p-use edge is covered.

Dominance path of node 12 of the p-use edge is not covered; the critical node is node 11. Branch distance for node 12 at the critical node 11 is 0.25 (Eq. 3). Branch weight for node 12 at critical node 11 is 0.06204 (Table 5). Therefore wbd (x, t₂) = branch distance * branch weight(w_i) (Eq. 9).

$${\text{wbd}}\left( {12,{\text{t}}_{2} } \right) = 0.25*0.06204 = 0.01551$$

$${\text{ft}}\left( {8,11,12,{\text{t}}_{2} } \right) = 1/3*[\left( {1*1} \right) + \left( {1*1} \right) + \left( {\left( {6/7} \right)*0.01551} \right) = 1/3*2.01329 = 0.6711$$

The fitness value of test case t₂ is ft (8, 11, 12, t₂) = 0.6711 (Eq. 8).

• 3. Input test case t₃ < 3, 6, 3>

Path traversed is {1, 2, 3, 5, 6, 7, 8, 9, 11, 12, 18, 19}

Dominance path of the def-node 8 is covered.

Dominance path of node 11 of the p-use edge is covered.

Dominance path of node 12 of the p-use edge is covered.

Using Eq. (8), the fitness value of test case t₃ is 1 (optimal test case).It can be seen that ft (8, 11, 12, t₁) < ft (8, 11, 12, t₂) < ft (8, 11, 12, t₃). Thus, the input test cases are also ranked according to their fitness values.