TAOCP 5.2.2 Exercise 20
In Algorithm Q (Quicksort), a stack is used to keep track of subfiles that remain to be sorted.
Section 5.2.2: Sorting by Exchanging
Exercise 20. [M20] What is the largest possible number of elements that will ever be on the stack at once in Algorithm Q, as a function of $M$ and $N$?
Verified: no
Solve time: 12m52s
Solution
In Algorithm Q (Quicksort), a stack is used to keep track of subfiles that remain to be sorted. The algorithm employs the standard optimization of always stacking the larger subfile and continuing immediately with the smaller one. This limits the stack depth to the length of the longest path obtained by repeatedly taking the smaller partition.
Let the current subfile have size (S). After partitioning, it is split into two subfiles of sizes (S_1) and (S_2) with (S_1 + S_2 = S - 1). The algorithm pushes the larger subfile onto the stack and continues processing the smaller subfile. Thus the stack size increases by 1 exactly when the smaller subfile is still large enough to require further partitioning.
The worst case for stack growth occurs when the partitions are as balanced as possible, because then the smaller subfile remains large for many steps. If the partitions are perfectly balanced, the smaller subfile has size (\lfloor S/2 \rfloor). The stack size therefore equals the number of times we can halve the current subfile size before it drops below 1. More formally, let (S_0 = N) be the initial file size. After (k) pushes, the current subfile size is at most (\lfloor N/2^k \rfloor). The process stops when this size becomes 1. The maximum number of pushes (i.e., the maximum stack size) is therefore the largest integer (k) such that (\lfloor N/2^k \rfloor \ge 1), which is (k = \lfloor \log_2 N \rfloor). Adding the initial entry (or counting the number of pending subfiles) gives the maximum number of elements on the stack at once as (\lfloor \log_2 N \rfloor + 1).
If the stack has a fixed capacity (M), the actual maximum cannot exceed (M). Hence the largest possible number of stack elements is (\min!\bigl(M,; \lfloor \log_2 N \rfloor + 1\bigr)). In the standard presentation of Algorithm Q, the stack is assumed large enough so that the capacity never binds; then the bound is simply (\lfloor \log_2 N \rfloor + 1).
\boxed{\min!\bigl(M,; \lfloor \log_2 N \rfloor + 1\bigr)}