# Problem 7.2

## Quicksort with equal element values

The analysis of the expected running time of randomized quicksort in section 7.4.2 assumes that all element values are distinct. In this problem. we examine what happens when they are not.

1. Suppose that all element values are equal. What would be randomized quick-sort's running time in this case?
2. The PARTITION procedure returns an index $q$ such that each element of $A[p \ldots q - 1]$ is less than or equal to $A[q]$ and each element of $A[q + 1 \ldots r]$ is greater than $A[q]$. Modify the PARTITION procedure to produce a procedure PARTITION'(A, p, r) which permutes the elements of $A[p \ldots r]$ and returns two indices $q$ and $t$ where $p \le q \le t \le r$, such that:
• all elements of $A[q \ldots t]$ are equal,
• each element of $A[p \ldots q - 1]$ is less than $A[q]$, and
• each element of $A[t + 1 \ldots r]$ is greater than $A[q]$.
Like PARTITION, your PARTITION' procedure should take $\Theta(r - p)$ time.
3. Modify the RANDOMIZED-QUICKSORT procedure to call PARTITION', and name the new procedure RANDOMIZED-QUICKSORT'. Then modify the QUICKSORT procedure to produce a procedure QUICKSORT'(p, r) that calls RANDOMIZED-PARTITION' and recurses only on partitions of elements not know to be equal to each other.
4. Using QUICKSORT', how would you adjust the analysis of section 7.4.2 to avoid the assumption that all elements are distinct?

### Running time

It will be $\Theta(n^2)$, because each split will be (n-1)-to-1 (see exercise 7.1-2).

### Implementation

The code is below.

PARTITION' is very similar to PARTITION, except that after it completes arranging the elements around a pivot $q$, it moves all elements $t > q: A[t] = x$ right after $q$. That way we get a chuck of equal elements after the pivot.

The procedure makes another pass at the array, which is at most $n$ more time and becuse $\Theta(n) + \Theta(n) = \Theta(n) = \Theta(r - p)$ we fulfill the condition.

### Analysis

The analysis does not change much. Section 7.4.2 uses the knowledge that the elements are distinct in order to determine when two elements cannot be compared. It will still be true that in any interval $Z_{ij}$, two elements will get compared only if $z_i$ or $z_j$ gets picked as a pivot first. This would not hold with PARTITION' if there are repeated elements.

Note that with this implementation, the number of comparisons increases, but only by a constant factor. The results from the analysis are the same.

### C code

#include <stdlib.h>

#define EXCHANGE(a, b) tmp = a; a = b; b = tmp;

typedef struct {
int q;
int t;
} pivot_t;

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

void quicksort(int A[], int p, int r) {
if (p < r - 1) {
pivot_t pivot = randomized_partition(A, p, r);
quicksort(A, p, pivot.q);
quicksort(A, pivot.t, r);
}
}

pivot_t randomized_partition(int A[], int p, int r) {
int i = rand() % (r - p) + p,
tmp;

EXCHANGE(A[i], A[r-1]);

return partition(A, p, r);
}

pivot_t partition(int A[], int p, int r) {
int x = A[r - 1],
q = p,
t,
tmp;

for (int i = p; i < r - 1; i++) {
if (A[i] < x) {
EXCHANGE(A[q], A[i]);
q++;
}
}

for (t = q; t < r && A[t] == x; t++);

for (int i = r - 1; i >= t; i--) {
if (A[i] == x) {
EXCHANGE(A[t], A[i]);
t++;
}
}

pivot_t result = {q, t};
return result;
}