Time and Clocks

Earth completes one rotation relative to the fixed stars (a sidereal day) in 23 hours, 56 minutes, and 4.09 seconds. This is slightly less than 24 hours because during Earth's rotation, it also moves along its orbit. The solar day (noon to noon) averages 24 hours, but the actual rotation period is shorter.

Using Master Theorem: a=3, b=4, f(n)=n^2. Since n^(log_4(3)) ≈ n^0.79 < n^2, we're in Case 3 where f(n) dominates. Therefore T(n) = O(n^2). The recursive calls at each level contribute polynomial work, but the n^2 term at the root determines overall complexity.

Using Master Theorem: a=3, b=4, f(n)=n log n. Since n^(log_4(3)) ≈ n^0.79 < n log n, we're in Case 3 where f(n) dominates by a polynomial factor. Therefore T(n) = O(n log n), which matches the claimed answer B.

Using Master Theorem: a=9, b=3, f(n)=n. Critical exponent: log_3(9)=2, so n^(log_b(a))=n^2. Since f(n)=n < n^2, we're in Case 1 where the recursive part dominates. Therefore T(n) = O(n^2). This matches the claimed answer A - this recurrence represents 9 recursive calls each on 1/3 of the input.

Using the Master Theorem: a=4, b=2, so n^(log_b a) = n^2. The function f(n) = n²√n = n^2.5 is polynomially larger than n^2 (by factor n^0.5). Since the regularity condition a·f(n/b) ≤ c·f(n) holds, case 3 applies: T(n) = Θ(f(n)) = Θ(n²√n). The n log n option would only apply if f(n) = O(n^2), which is false here.

The recursion tree has log n levels, each contributing n/log(n/2^i). Level 0: n/log n, Level 1: n/log(n/2), ..., Level log n: O(log n). Using the integral approximation, the sum of n/log(n/2^i) for i=0 to log n is O(n log log n). This is because we're essentially summing 1/log(x) which gives the log log factor.

At each level, the total work is n + n/2 + n/4 + n/8 + ... (sum of T(n/2), T(n/4), T(n/8) contributions) plus the constant n at the root. The sum n(1 + 1/2 + 1/4 + 1/8 + ...) converges to 2n = O(n). The O(n²) option would require quadratic growth at each level, and O(log n) ignores the linear work at each node.

This is the classic exponential recurrence. Expanding: T(n) = 2T(n-1) + 1 = 2(2T(n-2) + 1) + 1 = 4T(n-2) + 3 = 8T(n-3) + 7 = ... = 2^n T(0) + (2^n - 1). The dominant term is 2^n, so T(n) = O(2^n). The n² option would apply to T(n)=T(n-1)+n, and n^log 3 is for divide-and-conquer recurrences.

Using the Master Theorem: T(n) = 2T(n/2) + n^2, we have a=2, b=2, f(n)=n^2. Since n^log_b(a) = n^1 and f(n) = n^2 = Ω(n^(1+ε)) for ε<1, this is Case 3. The regularity condition 2f(n/2) = n^2/2 ≤ cn^2 holds, so T(n) = Θ(n^2). Options B, C, and D do not match this result.

The recurrence T(n) = 2T(√n) + n log n + n requires a different technique. Let m = log n, then T(2^m) = 2T(2^(m/2)) + 2^m * m + 2^m. This recurrence doesn't yield any of the simple forms in options A, B, or C. The complexity involves polynomial-logarithmic terms that don't match the given options, making 'None of these' the correct choice.