Question textConsider the following algorithm. Assume the input is a positive integer. [table]
1
2 3
4 5678 | function FUNC(n) if n

Question

Question textConsider the following algorithm. Assume the input is a positive integer. [table]
1
2 3
4 5678 | function FUNC(n) if n < 4 then return 3 * n  else count = 0 for i = 1 to n do count = count + i return count + FUNC(n/4) * FUNC(n/4)
[/table] The recurrence relation for the running time of this function can be expressed as ,
where and are expressions and and are their conditions under which they apply. Fill in the correct general case and base case to replace X/X' and Y/Y' below (no need to solve the relation, or give its time complexity). You may assume that a, b, c, d, e and f are constants of your choosing should you need them for the recurrence. Do not forget to include the condition on where each case would be applicable in the box on the right. General case (X if X'): Answer 1 Question 1[input] if Answer 2 Question 1[input] Base case (Y if Y'): Answer 3 Question 1[input] if Answer 4 Question 1[input]

BlackTom AI · Accepted Answer

The non-recursive work per call is Θ(n): the for-loop executes n iterations each doing O(1) work, so that cost can be written as a n + b.

1) Relevant knowledge points and concepts
- Cost of a simple for-loop of n iterations with O(1) body is linear in n (Θ(n)).
- Recursive cost adds the costs of recursive calls; here there are two calls on size n/4 each.
- Base cases for recurrences are constant-time when n is bounded by a small constant.

2) Step-by-step reasoning and intermediate steps
- Examine the code: when n < 4 the function does a single arithmetic operation return 3 * n (this is constant-time for each allowed n ∈ {1,2,3}). So the running time for any base-case input is a constant; we may choose that constant to be 3 (or any fixed c). Thus T(n) = 3 for n < 4.
- For n ≥ 4 the function:
  a) initializes count and runs for i = 1 to n: each loop iteration is O(1), so loop cost = c2 * n for some constant c2.
  b) performs constant overhead (initialization, return), contributing another constant c1.
  c) makes two recursive calls FUNC(n/4) and combines their results; the time for the recursive calls is T(n/4) + T(n/4) = 2T(n/4).
- Combine the non-recursive costs: c1 + c2*n = a n + b (rename constants a = c2, b = c1). So the full recurrence for n ≥ 4 is
  T(n) = a n + b + 2 T(n/4),
  where the non-recursive part is an + b.

3) Why the correct answers are correct
- The expression an + b correctly captures the per-call non-recursive work (linear loop plus constant overhead) used in the general-case recurrence.
- The base case is constant-time because n is limited to values < 4; choosing the constant 3 is a valid representation of that fixed-time base case.

4) Why other plausible expressions are incorrect
- An n^2 term would overstate the loop cost; there is no nested loop, only a single loop so cost is not quadratic.
- Using 3*n as the base-case running time is unnecessary: although the function returns the value 3*n, running time for n ∈ {1,2,3} is constant and can be represented by a constant (here 3) rather than a variable multiple of n.

Conclusion: The non-recursive work per general case is an + b, and the base-case running time can be represented as the constant 3. This matches the provided answers: "an + b" and "3".

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