Problem 7.4

Stack depth for quicksort

The QUICKSORT algorithm of Section 7.1 contains two recursive calls to itself. After QUICKSORT calls PARTITION, it recursively sorts the left subarray and then it recursively sorts the right subarray. The second recursive call in QUICKSORT is not really necessary; we can avoid it by using an iterative control structure. This technique, called tail recursion, is provided automatically by good compilers. Consider the following version of quicksort, which simulates tail recursion:

  while p < r
      // Partition and sort left subarray
      q = PARTITION(A, p, r)
      p = q + 1
  1. Argue that TAIL-RECURSIVE-QUICKSORT(A, 1, A.length) correctly sorts the array $A$.

Compilers usually execute recursive procedures by using a stack that contains pertinent information, including the parameter values, for each recursive call. The information for the most recent call is at the top of the stack, and the information for the initial call is at the bottom. Upon calling a procedure, its information is pushed onto the stack; when it terminates, its information is popped. Since we assume that array parameters are represented by pointers, the information for each procedure call on the stack requires $\O(1)$ stack space. The stack depth is the maximum amount of stack space used at any time during a computation.

  1. Describe a scenario in which TAIL-RECURSIVE-QUICKSORT's stack depth is $\Theta(n)$ on an $n$-element input array.
  2. Modify the code for TAIL-RECURSIVE-QUICKSORT so that the worst-case stack depth is $\Theta(\lg{n})$. Maintain the $\O(n\lg{n})$ expected running time of the algorithm.

Argue correctness

The original version partitions the array and then calls QUICKSORT one on each side. This version does the same, although in a different fashion - instead of calling TAIL-RECURSIVE-QUICKSORT, it just changes p and restarts the loop. It amounts to the same.

This is a straightforward tail-call unrolling.

Linear stack depth

This will happen whenever partition returns $r$. That is, whenever the array is sorted.

Modified algorithm

We are always doing a tail-recursive call on the second partition. We can modify the algorithm to do the tail recursion on the larger partition. That way, we'll consume less stack.

C code

#include <stdio.h>

int partition(int[], int, int);

static int stack_depth = 0;
static int max_stack_depth = 0;

void reset_stack_depth_counter();
void increment_stack_depth();
void decrement_stack_depth();

void tail_recursive_quicksort(int A[], int p, int r) {

    while (p < r - 1) {
        int q = partition(A, p, r);

        if (q < (p + r) / 2) {
            tail_recursive_quicksort(A, p, q);
            p = q;
        } else {
            tail_recursive_quicksort(A, q + 1, r);
            r = q;


int partition(int A[], int p, int r) {
    int x, i, j, tmp;

    x = A[r - 1];
    i = p;

    for (j = p; j < r - 1; j++) {
        if (A[j] <= x) {
            tmp = A[i]; A[i] = A[j]; A[j] = tmp;

    tmp = A[i]; A[i] = A[r - 1]; A[r - 1] = tmp;

    return i;

void increment_stack_depth() {
    if (max_stack_depth < stack_depth) {
        max_stack_depth = stack_depth;

void decrement_stack_depth() {

void reset_stack_depth_counter() {
    max_stack_depth = 0;
    stack_depth = 0;