MC/DC requires practice. Let’s do one more example. Let’s suppose the expression we want to test is: (A && B) || C. 

That’s the truth table:

Let’s start with condition A. We have tests 2 and 6.

• If we flip A in row 1 (TTT), the counterpart is row 5 (FTT). As the outcome of both are the same TRUE, tests 1 and 5 are not able to evaluate the outcome of A independently of the other conditions.
• If we flip A in row 2 (TTF), the counterpart is row 6 (FTF). As the outcomes are different, tests 2 and 6 are able to evaluate the outcome of A independently of the other conditions. We keep them.
• We keep flipping. Row 3 and its counterpart 7 are not able to independently evaluate A, as their outcomes are the same.
• Row 4 and its counterpart 8 are not able to independently evaluate A, as their outcomes are the same.

Let’s look at condition B. We have tests 2 and 4.

• If we flip B in row 1 (TTT), the counterpart is row 3 (TFT). Outcomes are the same.
• If we flip B in row 2 (TTF), the counterpart is row 4 (TFF). Outcomes are different, so tests 2 and 4 are able to evaluate B independently of the other conditions.
• We keep flipping. Row 5 and 7 have the same outcomes.
• Rows 6 and 8 have the same outcomes.

Now, let’s look at C. If you repeat the same procedure, you will find three possible pairs: (3, 4), (5,6) and (7,8).

Now, if we try to minimize the number of tests (after all, we know that we need N+1 tests, where N is the number of conditions, thus, 4 tests), we have two options:

• 2, 3, 4, 6. So, A is tested by 2 and 6, B is tested by 2 and 4, and C is tested by 3 and 4.
• 2, 4, 5, 6. So, A is tested by 2 and 6, B is tested by 2 and 4, and C is tested by 5 and 6.

Good!

Now, an exercise for you: can you expand this idea to a problem that is not boolean? Let’s suppose you are calculating the price of your product, and you have the following decision table:

Can you apply MC/DC here? We’ll discuss this in future weeks.