Solution

Cyclomatic Complexity

Definition: $V(G) = E - N + 2P$ (McCabe) = number of linearly independent paths. Also $V(G) = \\text{decision nodes} + 1$.

Computation: Count if, for, while, case, &&, ||.

Example: Smallest of 3 numbers

int smallest(int a,int b,int c){
  int m = a;
  if(b < m) m = b;  // P1
  if(c < m) m = c;  // P2
  return m;
}

Control Flow Graph: 2 predicates → $V(G) = 2 + 1 = 3$

• Via formula: $E=7, N=6, P=1$ → $V=7-6+2=3$

Paths:

1. a is smallest (both false)
2. b is smallest (first true)
3. c is smallest (second true)

So 3 independent paths → needs 3 test cases.