Question 8 of 8

Question 8 (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The unit under test is the Appendix A main function's nested-if leap-year logic: three sequential decisions — year % 4 == 0, then (nested inside it) year % 100 == 0, then (nested inside that) year % 400 == 0 — each guarding one of four printf outcomes.

Find. The cyclomatic complexity V(G) of the unit, and a minimal basis path testing suite of V(G) independent test cases (one input year each) that together execute every independent path through the control-flow graph.

Approach. Draw the flow graph of the triple-nested decision, compute V(G) two independent ways (they must agree), then derive V(G) independent basis paths — flipping one prior decision outcome at a time from a baseline — and pick the classic century-year test inputs that force each path.

1. Build the flow graph and count nodes/edges. Nine nodes (1: read; 2,3,4: the three nested decisions; 5,6,7,8: the four printf outcomes; 9: exit) and eleven edges: 1→2; 2→3(T); 2→8(F); 3→4(T); 3→7(F); 4→5(T); 4→6(F); 5→9; 6→9; 7→9; 8→9.
2. Compute V(G) two independent ways. $$V(G) = E - N + 2 = 11 - 9 + 2 = \boxed{4}$$ $$V(G) = 1 + \sum_{\text{predicates}}(\text{out-degree}-1) = 1 + (2-1) + (2-1) + (2-1) = \boxed{4}$$ Both formulas agree at V(G) = 4, matching the three nested binary decisions (each predicate node contributes one extra independent path). Exactly 4 basis-path test cases are required.
3. Path 1: not divisible by 4 (nodes 1-2-8-9). The outer decision alone determines the outcome, both inner decisions are never reached.
   Input: year = 2015 → 2015 % 4 = 3 ≠ 0 → Expected output: "2015 is not a leap year."
4. Path 2: divisible by 4, not by 100 (nodes 1-2-3-7-9). An ordinary (non-century) leap year — the inner %400 test is never reached.
   Input: year = 2016 → 2016 % 4 = 0, 2016 % 100 = 16 ≠ 0 → Expected output: "2016 is a leap year."
5. Path 3: divisible by 4, 100 and 400 (nodes 1-2-3-4-5-9). A century year that IS a leap year — all three decisions are true.
   Input: year = 2000 → 2000 % 4 = 0, 2000 % 100 = 0, 2000 % 400 = 0 → Expected output: "2000 is a leap year."
6. Path 4: divisible by 4 and 100, not 400 (nodes 1-2-3-4-6-9). A century year that is NOT a leap year — the case the algorithm exists specifically to catch (a naive "divisible by 4" check alone would wrongly call this a leap year).
   Input: year = 1900 → 1900 % 4 = 0, 1900 % 100 = 0, 1900 % 400 = 300 ≠ 0 → Expected output: "1900 is not a leap year."