Domain Testing: Divide and Conquer - Testing Education

For example, consider a function that has five variables a, b, c, d and e. ‘a’ has 3
test cases corresponding to it and each of the remaining four have two test cases
each. Let us apply all-pairs combination to this function and see how we could
achieve all-pairs, that is, each test case value of each of the five variables is paired
(at least once) with each of the test case values of the remaining four variables.
Let’s assume that all the five variables are useful for all-pairs combination, since
we really don’t have any information about the variables other than the number of
test cases they have. Let us also assume that each of their test cases produces an
acceptable result by the function. Let us give symbolic representations to the test
cases of each of the variables:
Variable Test Cases
a A1, A2, A3
b B1, B2
c C1, C2
d D1, D2
e E1, E2
The following is the result of all-pairs combination:
Test
case#
a(3) b(2) c(2) d(2) e(2)
1. A1 B1 C1 D1 E1
2. A1 B2 C2 D2 E2
3. A2 B1 C2 D2 E1
4. A2 B2 C1 D1 E2
5. A3 B1 C1 D2 E2
6. A3 B2 C2 D1 E1
If you observe carefully, you will see that every test case of ‘a’ has been paired at
least once with every test case of ‘b’, ‘c’, ‘d’ and ‘e’. This holds true for ‘b’, ‘c’,
‘d’ and ‘e’ also. Hence, all-pairs of the five variables have been achieved in the
above arrangement. Also, note that the number of combinations needed to arrive
at this all-pairs configuration is six, which is the product of the number of test
cases of ‘a’ and ‘b’ (3x2).
In general, if we were to sort the variables in descending order in terms of the
number of representatives (values chosen to be tested) and if Max1 and Max2 are
the first two in this sorted list (the two highest) then the minimal number of
combinations in all-pairs combination would be (Max1) x (Max2). We might not
be able to fit in all pairs within Max1 x Max2 combinations, in which case,
additional combinations might have to be added.
© Sowmya Padmanabhan, 2003
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