Today, we’re stepping into the world of algorithm analysis to determine the time complexity of a specific code segment. We’ll be using Big-O notation to provide an upper bound on the number of operations performed as the input size, n, grows.

Understanding the Problem

We are given a code segment, which is structured as a while loop. Inside the loop, we perform some arithmetic operations. Our task is to estimate the number of additions and multiplications (collectively referred to as “operations”) using Big-O notation. We’re ignoring the comparison step in the while loop, as is conventional. The code segment is:

i := 1
t := 0
while i <= n
  t := t + i
  i := 2 * i

The variable i starts at 1 and is doubled in each iteration, meaning it takes on the values 1, 2, 4, 8, and so on. The variable t accumulates a sum. Our goal is to determine how the number of operations grows as n increases.

Analyzing the Loop Iterations

Let’s trace the execution of the loop for a bit to get a handle on the pattern:

• Iteration 1: i = 1, t = 0 + 1, i becomes 2 (1 addition, 1 multiplication, where multiplication is i:=2*i).
• Iteration 2: i = 2, t = 1 + 2, i becomes 4 (1 addition, 1 multiplication).
• Iteration 3: i = 4, t = 3 + 4, i becomes 8 (1 addition, 1 multiplication).
• …

In general, in the k-th iteration, we have i = 2^(k-1). The loop continues as long as i ≤ n. So, the loop will continue as long as 2^(k-1) ≤ n.

Determining the Number of Iterations

To find the number of iterations, we want to find the largest value of k such that 2^(k-1) ≤ n. Taking the base-2 logarithm of both sides, we have:

k - 1 ≤ log₂(n)

k ≤ log₂(n) + 1

Since k represents the number of iterations, and k must be an integer, the maximum number of iterations is approximately log₂(n) + 1, which is a value that is proportional to log₂(n).

Counting the Operations

Within each iteration of the loop, we perform:

• One addition: t := t + i
• One multiplication: i := 2 * i

Therefore, the number of operations in a single iteration is 2. Since the number of iterations is proportional to log₂(n), the total number of operations is roughly 2 * log₂(n) + 2.

Big-O Notation

In Big-O notation, we ignore constant factors and lower-order terms. Thus:

O(2 log₂(n) + 2) becomes O(log₂(n)).

Furthermore, the base of a logarithm doesn’t matter in Big-O notation since different bases differ by only a constant factor. Therefore, log₂(n) and log(n) (base 10) or even ln(n) all represent the same time complexity. Therefore we can simply say:

O(log₂(n)) becomes O(log n)

Conclusion

The number of operations performed in this algorithm segment is O(log n). This means that the number of operations grows logarithmically with respect to the input n. This growth rate is relatively slow, indicating that this algorithm is efficient, especially compared to algorithms with linear or quadratic complexity.