zyBooks_17_1
Suppose a recursive function's runtime is
T(n−1)+Θ(n){"version":"1.1","math":"T(n-1) + \Theta(n)"}
How many levels will the recursion tree have?

Question

zyBooks_17_1
Suppose a recursive function's runtime is 
T(n−1)+Θ(n){"version":"1.1","math":"T(n-1) + \Theta(n)"}
How many levels will the recursion tree have?

BlackTom AI · Accepted Answer

To determine how many levels the recursion tree has, we look at how the function calls itself: T(n) = T(n-1) + Θ(n). Each recursive step reduces the problem size by 1, so the depth of the recursion corresponds to how many times we can subtract 1 from n until we reach the base case. This means the recursion runs for n, n-1, n-2, ..., down to the base case, giving a total of n levels.

Option 1: n. This matches the intuitive depth of the tree since each level decreases the input by 1, resulting in a linear number of levels.

Option 2: nlogn. This would suggest a depth that grows faster than linear, which is not supported by a constant decrement of 1 per level. The log factor does not arise from the step n -> n-1.

Option 3: n^2. A quadratic depth would imply many more levels than the problem provides; the recurrence reduces n by 1 each time, so depth cannot be quadratic in n.

Option 4: logn. A logarithmic depth would occur if the size halved (or divided by a constant factor) at each step, not when it decreases by 1. Therefore, depth is not logarithmic here.

zyBooks_17_1 Suppose a recursive function's runtime is T(n−1)+Θ(n){"version":"1.1","math":"T(n-1) + \Theta(n)"} How many levels will the recursion tree have?

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