Given the following pseudocode, derive the recurrence relation that represents its time complexity. def fibonacci(n): if (n==0 or n==1): return n return fibonacci(n - 1) + fibonacci(n - 2)Let b and c represent constant values. What is the base case and recurrence step?

Question

Given the following pseudocode, derive the recurrence relation that represents its time complexity. def fibonacci(n): if (n==0 or n==1): return n  return fibonacci(n - 1) + fibonacci(n - 2)Let b and c represent constant values. What is the base case and recurrence step?

BlackTom AI · Accepted Answer

Each call to fibonacci(n) makes two recursive calls (fibonacci(n-1) and fibonacci(n-2)) plus a constant amount of work for the if-check and return, so the recurrence must include both subproblems and a constant term.

1) Relevant knowledge points and concepts
- Recurrence relations model time cost T(n) of recursive functions: T(n) = cost of recursive subcalls + local work.
- Base cases occur when recursion stops (here n == 0 or n == 1).
- b represents the constant cost of a base-case invocation; c represents the constant non-recursive overhead for a non-base invocation.

2) Step-by-step reasoning
- If n == 0 or n == 1 the function returns immediately, performing only constant work: T(0) = T(1) = b.
- For n > 1 the function performs:
  - one call costing T(n-1),
  - another call costing T(n-2),
  - plus constant overhead c for the conditional and returning the sum.
  Hence T(n) = T(n-1) + T(n-2) + c for n > 1.

3) Intermediate calculations (examples)
- T(2) = T(1) + T(0) + c = b + b + c = 2b + c.
- T(3) = T(2) + T(1) + c = (2b + c) + b + c = 3b + 2c.
These show the recurrence expands by summing previous terms plus constants.

4) Why the correct answer is correct
- It includes both recursive calls and the constant overhead for the non-base case, and it assigns a constant cost to each base case, matching the function’s behavior exactly.

5) Why other options are incorrect
- Options that set T(0)=T(1)=n misuse n instead of a constant b.
- Options that omit one of the recursive terms (e.g., only T(n-1) + c or only T(n-2) + c) fail to model the fact the code makes two recursive calls.

6) Conclusion
Therefore the correct recurrence is T(n) = T(n-1) + T(n-2) + c with base cases T(0) = T(1) = b, which matches the provided answer.

Given the following pseudocode, derive the recurrence relation that represents its time complexity. def fibonacci(n): if (n==0 or n==1): return n return fibonacci(n - 1) + fibonacci(n - 2)Let b and c represent constant values. What is the base case and recurrence step?Single choice

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