Problem 1: Euclid’s algorithm (or the Euclidean algorithm) is an algorithm that computes the greatest common divisor, denoted by gcd, of two integers. Below are the original versions of Euclid’s algorithm that uses repeated subtraction and another one that uses the remainder. int gcd_sub(int a, int b) {  if (!a)  return b;  while (b)  if (a > b)  a = a – b;  else  b = b – a;  return a; } int gcd_rem(int a, int b) {  int t;  while (b)  {  t = b;  b = a % b;  a = t;  }  return a; } 1. Trace each of the above algorithm using specific values for a and b. 2. Compare both algorithms. Problem 2: Given a fixed integer B (B ≥ 2), we demonstrate that any integer N (N ≥ 0) can be written in a unique way in the form of the sum of p+1 terms as follows: N = a0 + a1×B + a 2×B 2  + … + ap×B p where all ai, for 0 ≤ i ≤ p, are integer such that 0 ≤ ai ≤ B-1. The notation apap-1…a0 is called the representation of N in base B. Notice that a0 is the remainder of the Euclidean division of N by B. If Q is the quotient, a1 is the remainder of the Euclidean division of Q by B, etc. 1. Write an algorithm that generates the representation of N in base B.  2. Compute the time complexity of your algorithm. Problem 3: Consider the following recursive function: int Puzzle(int base, int limit) { if (base > limit) return -1; else if (base == limit) return 1; else return base*Puzzle(base+1, limit); } 1. Identify the base case(s) of the function Puzzle. 2. Identify the general case of the function Puzzle. 3. Show what would be written by the following calls to the recursive function Puzzle: a. cout << Puzzle(14,10); b. cout << Puzzle(4,7); c. cout << Puzzle(0,0);

Computer Networking: A Top-Down Approach (7th Edition)
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Problem 1: Euclid’s algorithm (or the Euclidean algorithm) is an algorithm that computes the
greatest common divisor, denoted by gcd, of two integers. Below are the original versions of
Euclid’s algorithm that uses repeated subtraction and another one that uses the remainder.
int gcd_sub(int a, int b)
{
 if (!a)
 return b;
 while (b)
 if (a > b)
 a = a – b;
 else
 b = b – a;
 return a;
}
int gcd_rem(int a, int b)
{
 int t;
 while (b)
 {
 t = b;
 b = a % b;
 a = t;
 }
 return a;
}
1. Trace each of the above algorithm using specific values for a and b.
2. Compare both algorithms.
Problem 2: Given a fixed integer B (B ≥ 2), we demonstrate that any integer N (N ≥ 0) can be
written in a unique way in the form of the sum of p+1 terms as follows:
N = a0 + a1×B + a
2×B
2
 + … + ap×B
p
where all ai, for 0 ≤ i ≤ p, are integer such that 0 ≤ ai ≤ B-1.
The notation apap-1…a0 is called the representation of N in base B. Notice that a0 is the
remainder of the Euclidean division of N by B. If Q is the quotient, a1 is the remainder of the
Euclidean division of Q by B, etc.
1. Write an algorithm that generates the representation of N in base B. 

2. Compute the time complexity of your algorithm.
Problem 3: Consider the following recursive function:
int Puzzle(int base, int limit)
{
if (base > limit)
return -1;
else
if (base == limit)
return 1;
else
return base*Puzzle(base+1, limit);
}
1. Identify the base case(s) of the function Puzzle.
2. Identify the general case of the function Puzzle.
3. Show what would be written by the following calls to the recursive function Puzzle:
a. cout << Puzzle(14,10);
b. cout << Puzzle(4,7);
c. cout << Puzzle(0,0); 

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