Building an MC/DC table is half the skill. The other half, and a favourite exam format, is being handed a set of rows and asked: is this a test set? Is it minimum? This lesson returns to A AND (B OR C) from 3.5, adds two speed skills the professor showed in class, and turns the checking into a fixed two-question routine.
Here is the completed table again (minimum sets: {2, 5, 6, 7} and {3, 5, 6, 7}):
| Row | A | B | C | Result | A varies | B varies | C varies |
|---|---|---|---|---|---|---|---|
| 1 | F | F | F | F | |||
| 2 | F | F | T | F | 6 | ||
| 3 | F | T | F | F | 7 | ||
| 4 | F | T | T | F | 8 | ||
| 5 | T | F | F | F | 7 | 6 | |
| 6 | T | F | T | T | 2 | 5 | |
| 7 | T | T | F | T | 3 | 5 | |
| 8 | T | T | T | T | 4 |
The professor used masking (3.4) to save time. In A AND (B OR C), A is joined by AND to the rest:
It is like a two-lock door: if the first lock is shut, you do not need to inspect the second one. Splitting the table at the variable that changes slowest (A, every 4 rows) turns an 8-row calculation into two very short ones.
Columns B and C have only one pair each. A column with only one pair gives you no choice: that pair must be in every answer. So rows 5, 6 and 7 are in every minimum set, and after that you only need to add one row from an A pair.
The general habit: start with the column that has the fewest pairs. It has no choice, so it is the safest place to begin.
The professor asked this in class. The easiest way: take one pair from each column and do not care about overlap.
It has 5 rows instead of 4. It is a real test set, but it is not minimum.
Then the professor gave sets and asked: is this a test set? Check each extended column for a whole pair inside the set:
| Set of rows | A pair inside? | B pair inside? | C pair inside? | Rows | Verdict |
|---|---|---|---|---|---|
| {4, 5, 6, 7, 8} | yes, {4, 8} | yes, {5, 7} | yes, {5, 6} | 5 | Test set, not minimum |
| {2, 4, 5, 6, 7} | yes, {2, 6} | yes, {5, 7} | yes, {5, 6} | 5 | Test set, not minimum |
| {3, 4, 5, 7, 8} | yes, {3, 7} | yes, {5, 7} | no, row 6 missing | 5 | Not a test set |
Look at the last set. Column C has only one pair, {5, 6}. Row 6 is missing, so column C has no pair. That means the set is not an MC/DC test set at all, even though it has plenty of rows.
The exam often gives you a set and asks you to check it. Always ask these two questions, in this order:
Diagram loads as you scroll.
Exam radar: the size trap. A set can have the correct size ( rows) and still not be a test set. The size tells you nothing until you have checked the pairs. The past exam question in 4.3 is exactly this trap.
How to write the answer: answer the question first, then justify it per column. For example: “No. {3, 4, 5, 7, 8} is not an MC/DC test set, because condition C has no independence pair: its only pair is {5, 6}, and row 6 is not in the set.” Name the failing column and the missing row.
For A AND (B OR C), is {2, 3, 5, 6, 7} a test set? Is it minimum?
Question 1 (pairs). A: {2, 6} is inside (and so is {3, 7}) ✓. B: {5, 7} is inside ✓. C: {5, 6} is inside ✓. Every column has a whole pair, so yes, it is a test set.
Question 2 (size). It has 5 rows, but . So it is not minimum. (Removing either row 2 or row 3 gives a minimum set.)
For A AND (B OR C), is {3, 4, 5, 7, 8} a test set?
No. Column C has only one pair, {5, 6}, and row 6 is missing. So condition C has no independence pair, and the set does not satisfy MC/DC. Columns A ({3, 7}, {4, 8}) and B ({5, 7}) are fine, but one failing column is enough. Since Question 1 already fails, there is no need to look at the size.
A classmate proposes {2, 3, 5, 6} for A AND (B OR C), saying “it has rows, so it is a minimum test set.” Check the claim.
Check the pairs before the size.
Not a test set, so it cannot be a minimum test set either. This is the size trap: 4 rows is the right size, but the size means nothing until every column has a pair. Rows 2 and 3 both come from A pairs, so one row is wasted; the classmate should have started from the no-choice columns B and C (forcing 5, 6, 7) and added only one A row.
Use masking to fill the result column of A OR (B AND C) quickly, without evaluating each row in full.
A is joined to the rest by OR.
So the result column, rows 1 to 8, is F, F, F, T, T, T, T, T. Check row 4: (F, T, T) → F OR (T AND T) = T ✓.
For A OR (B AND C) the pairs are A {1, 5}, {2, 6}, {3, 7}; B {2, 4}; C {3, 4}. Check each set with the two questions:
(a) {1, 2, 3, 4, 5} (b) {2, 3, 4, 5} (c) {2, 3, 4, 7}
Here , so a minimum set has 4 rows.
For (A OR B) AND C (pairs: A {2, 6}; B {2, 4}; C {3, 4}, {5, 6}, {7, 8}), give one non-minimum test set, and explain which column you should start from when looking for a minimum one.
Non-minimum set. Take one pair from each column without caring about overlap, e.g. A {2, 6} + B {2, 4} + C {7, 8} = {2, 4, 6, 7, 8}: 5 rows, every column has a whole pair, so it is a test set but not minimum.
Where to start. Start with the columns that give no choice: A and B each have only one pair, so rows 2, 4 and 6 are forced. Then add a C pair that reuses a forced row: {3, 4} (add 3) or {5, 6} (add 5). The minimum sets are {2, 3, 4, 6} and {2, 4, 5, 6}.