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Testing Techniques

Testing Techniques. Equivalence class analysis Cause effect graphs All pair testing. Equivalence classes. Equivalence classes. Equivalence classes. Triangle Problem. Read 3 integer values in range [1..200].

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Testing Techniques

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  1. Testing Techniques Equivalence class analysis Cause effect graphs All pair testing

  2. Equivalence classes

  3. Equivalence classes

  4. Equivalence classes

  5. Triangle Problem • Read 3 integer values in range [1..200]. • These 3 values represent the length of the sides of a triangle. The programme displays a message which establishes that the triangle is isosceles, equilateral or scalene.

  6. Triangle Problem • we are interested in 4 questions: • Is it a triangle? • Is it an isosceles? • Is it a scalene? • Is it an equilateral? • We may define the input test data by defining the equivalence class through “viewing” the 4 output groups: • input sides <a, b, c> do not form a triangle • input sides <a, b ,c> form an isosceles triangle • input sides <a, b, c> form a scalene triangle • input sides <a, b, c> form an equilateral triangle

  7. Valid Classes

  8. Non Valid Classes

  9. Non Valid Classes Could you find other non valid classes ?

  10. Cause effect graph

  11. Methodology

  12. Triangle example

  13. Triangle Problem

  14. Cause Effect Graph

  15. Decision Table

  16. Test cases

  17. All pairs testing • Based on orthogonal Array • Drastically reduces number of test cases • A combination of worst cases and equivalence class testing • Well supported in the agile methods community • Supported by commercial and non-commercial products

  18. What Does All Pairs Do? • Input is a set of equivalence classes for each variable. • Sometimes the equivalence classes are those used in Boundary Value testing (min, min+, nominal, max-, max) • Output is a set of (partial) test cases that approximate an orthogonal array, plus some pairing information. • Why partial? No expected outputs are provided. natural enough, but a tester still needs to do this step.

  19. All Pairs Assumptions • Variables have clear equivalence classes. • Variables are independent. • Failures are the result of the interaction of a pair of variable values.

  20. Example • Suppose that we have three parameters. • For each parameter, there are two possible values. • Values are : • A, B for parameter 1. • J, K for parameter 2. • Y, Z for parameter 3. • Degree of interaction coverage is 2. • We want to cover all potential 2-way interactions among parameter values.

  21. A J Y A J Z A K Y A K Z B J Y B J Z B K Y B K Z Set of all possible test configurations Three parameters, two values for each. There are 23 = 8 possible test configurations.

  22. A J A Y J Y A K A Z J Z B J B Y K Y B K B Z K Z Set of all possible degree 2interaction elements There are ( )  2 2 = 12 possible interaction elements. 3 • Coverage measure: • Percentage of interaction elements covered. 2

  23. A J A Y J Y Test configurations assets of interactions A J Y One test configuration... … covers 3 possible interaction elements.

  24. A J Y A J Z A J A Y J Y A K Y A K A Z J Z A K Z B J B Y K Y B J Y B K B Z K Z B J Z B K Y B K Z Interaction test coverage goal Goal: cover all interaction elements… …using a subset of all test configurations.

  25. A J A Y A A J J Y Y A K A Z A J A Y J Y A J Z A K Y B J B Y J Y A K Z B J Y B K B Z J Z B J Z B K Y K Y B K Z K Z Selection of test configurations forcoverage of interaction elements Interaction elements Test configurations Degree 2 coverage: 3 / 12 = 25% Degree 3 coverage: 1 / 8 = 12.5%

  26. A J Y A K A Z A K Z K Z Selection of test configurations forcoverage of interaction elements Interaction elements Test configurations A J A Y J Y A J Z A K A Z J Z A K Y B J B Y K Y A K Z B K B Z K Z B J Y B J Z B K Y Degree 2 coverage: 6 / 12 = 50% B K Z Degree 3 coverage: 2 / 8 = 25%

  27. A A J J Y Y A J Z A K Y J Z A K Z B J A K Z B J Y B Z B J Z B J Z B K Y B K Z Selection of test configurations forcoverage of interaction elements Interaction elements Test configurations A J A Y J Y A K A Z J Z B J B Y K Y B K B Z K Z Degree 2 coverage: 9 / 12 = 75% Degree 3 coverage: 3 / 8 = 37.5%

  28. A A J J Y Y A J Z A K Y A K Z B Y K Y A K Z B J Y B K B J Z B J Z B K Y B K Y B K Z Selection of test configurations forcoverage of interaction elements Interaction elements Test configurations A J A Y J Y A K A Z J Z B J B Y K Y B K B Z K Z Degree 2 coverage: 12 / 12 = 100% Degree 3 coverage: 4 / 8 = 50%

  29. Choosing the degree of coverage • In one experiment, covering 2 way interactions resulted in the following average code coverage: • 93% block coverage. • 83% decision coverage. • 76% c-use coverage. • 73% p-use coverage. • Source: Cohen, et al, “The combinatorial design approach to automatic test generation”, IEEE Software, Sept. 1996. • Another experience report investigating interactions among 2-4 components: • Dunietz, et al, “Applying design of experiments to software testing”, Proc. Of ICSE ‘97.

  30. Online Mortgage Application(Thanks, Bernie Berger, STAREast) Closing Cost Bank Emp Intro Rate Residence Refinance Property tier Credit LTV NIV NAV Region

  31. Total number of combinations Twelve variables, with varying numbers of values, have 7 x 6 x 6 x 5 x 3 x 3 x 2 x 2 x 2 x 2 x 2 x 2 = 725,760 combinations of values. “All Pairs does it in 50.” (Bernie Berger, STAREast2003)

  32. 1 NY L 1 fam A+ Pri 80% Yes Yes Yes Cust Yes Yes 66 2 NY M 2 fam A Vac 90% No No No Bank No No 66 3 NJ L 2 fam A- Inv 100% Yes No Yes Bank Yes No 54 4 NJ M 1 fam B Pri 90% No Yes No Cust No Yes 46 47 NY H+1 ~Coop B ~Inv ~80% ~No ~No ~Yes ~Cust ~Yes ~Yes 1 48 Other H 2 fam ~B ~Vac ~100% ~No ~No ~Yes ~Cust ~No ~No 1 49 Other H+2 3 fam ~A- ~Pri ~80% ~Yes ~No ~No ~Cust ~No ~No 1 50 Other H+1 Condo <B ~Vac ~90% ~No ~No ~Yes ~Cust ~No ~No 2 Selected All Pairs Test Cases (note fault resolution!)

  33. Mortgage Application Observations • The example uses values, not equivalence classes. • Values are independent, • System level testing (apparently). • The efficiency comes at the expense of fault isolation. There are 2C12 = 12!/(2! x (12 - 2)!) = 66 pairs in the first row of the table. If that test case fails, which of the 66 pairs caused the failure? • NB: excellent candidate for regression testing

  34. What if the fault involves more than just a pair of variables? Number of combinations of n things taken p at a time is n ! pCn = p ! ( n p )! - Worst case (for first test case): 12 å pCn 2 ^ 12 1 4095 = - = 1

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