Solution

Cyclomatic Complexity

$V(G)=E-N+2P$ or $V(G)=P+1$ where $P$ = predicate nodes. Measures independent paths.

C Program - GCD

#include <stdio.h>
int gcd(int a,int b){
  while(b!=0){          // P1
    int t=b;
    b=a%b;
    a=t;
  }
  return a;
}
int main(){
  int x,y; scanf("%d%d",&x,&y);
  if(x<=0 || y<=0){     // P2 (with || = 2 predicates)
    printf("Invalid");
    return 0;
  }
  printf("GCD=%d",gcd(x,y));
  return 0;
}

Complexity Calculation

For gcd function alone:

• Predicate: while(b!=0) → 1
• $V(G)=1+1=2$

For main with validation:

• Predicates: b!=0 (1), x<=0 (2), y<=0 (3) → $P=3$
• $V(G)=3+1=4$
• Via graph: $E=11, N=9, P=1$ → $V=11-9+2=4$

Paths: Need 4 test cases (e.g., (48,18), (0,5), (5,0), (18,48)).