1. Medians and Order Statistics
i-th order statistic: i-th smallest element
n elements: median is
n odd: (n+1)/2
n even: n/2 or n/2+1
Assume distinct numbers.
Input: A, n, 1<=i<=n
Output: element x of A larger than i-1 elements of A.

2. Solutions O(n log n) time based on … O(n) time average.
O(n) time worst case.

3. Minimum and Maximum How many comparisons?
At most n-1.
Examine each element and keep trach of smallest one:
Comparison based
Each element must be compared
Each must loose once (except winner).
What about simultaneous min and max?

4. Min & Max Can do with 2n-2 comparisons. Can do better
Form pairs of elements
Compare elements in each pair
Pair (ai, ai+1), assume ai < ai+1, then
Compare (min,ai), (ai+1,max)
3 comparisions for each pair.

5. Average Time Median Selection
Divide-and-Conquer (prune-and-search).
Randomized: behavior determined by output of random number generator.
Based on QuickSort:
Partition input array recursively, but
Work only on one side!

6. Randomized Selection QuickSort(A,p,r) RandSelect(A,p,r,i)
If p < r then
q=partition(A,p,r)
QuickSort(A,p,q)
QuickSort(A,q+1,r).
First call: QuickSort(A,1,n)
After partition(A,p,q):
A[i]<A[q}, i<q;
A[q]<A[j}, q<j.
RandSelect(A,p,r,i)
If p == r then return A[p]
q=RandPartition(A,p,r)
k=q-p+1 /* size of A[p..q]
If i ≤ k then return RandSelect(A,p,q,i)
Else return RandSelect(A,q+1,r,i-k).
First call: RandSelect(A,1,n,i).
Returns the i-th smallest element in A[p..r].

7. Selection (cont.)
RandPartition (see 8.3, 8.4 textbook) gives partition with low side:
1 element with probability 2/n
j elements with probability 1/n, for j=2,3,…,n.
Assume i-th element always on larger side:
T(n)≤(T(max(1,n-1)+Σk=1..n-1T(max(k,n-k)))/n+O(n)
≤(T(n-1)+2 Σk=n/2..n-1T(k))/n+O(n)
=2(Σk=n/2..n-1T(k))/n+O(n), since T(n-1)=O(n2).
Then T(n)=O(n) (proof by substitution).

8. Worst Case Linear Time Selection
O(n) worst case algorithm.
Works in similar way: recursively partition input array
Idea: guarantee good split
E.g., in QuickSort assume at each recursion level have T(n)=T(9n/10)+T(n/10)+O(n).
Then, T(n)=O(n log n).
Use deterministic partitioning:
Compute the element to partition around.

9. Steps to find i-th smallest element Algorithm Select
Divide elements in n/5 groups of 5 elements, plus at most one group with (n mod 5) elements.
Find median of each group:
Insertion sort: O(1) time (at most 5 elements).
Take middle element (largest if two medians).
Use Select recursively to find median x of medians.

10. Algorithm Select (cont.)
Partition input array around median-of-medians x. Let k be the number of elements on low side, n-k on high side.
a1,a2,…,ak | ak+1,ak+2,…,an
ai < aj, for 1 ≤ i ≤ k, k+1 ≤ j ≤ n.
Use Select recursively to:
Find i-th smallest element on low side, if i ≤ k
Find (i-k)-th smallest on high side, if i > k.

11. Analysis Find lower bound on number of elements greater than x.
At least half of medians in step 2 greater than x. Then,
At least half of the groups contribute 3 elements that are greater than x, except:
Last group (if less than 5 elements);
x own group.
Discard those two groups:
Number of elements greater than x is ≥ 3((n/5)/2-2)=3n/10-6.
Similarly, number of elements smaller than x is ≥3n/10-6.
Then, in worst case, Select is called recursively in Step 5 on at most 7n/10+6 elements (upper bound).

12. Analysis (cont.) Steps 1,2 and 4: O(n) time. Step 3: T(n/5)
Step 5: at most T(7n/10+6)
7n/10+6 < n for n > 20.
T(n) ≤ T(|¯n/5¯|)+T(7n/10+6)+O(n), n > n1.
Use substitution to solve:
Assume T(n) ≤ cn, for n > n1; find n1 and c.

13. Analysis (cont.) T(n) ≤ c|¯n/5¯| + c(7n/10+6) + O(n)
≤ cn/5 + c + 7cn/10 + 6c +O(n)
= 9cn/10 + 7c + O(n)
Want T(n) ≤ cn:
Pick c such that c(n/10-7) ≥ c1n, where c1 is constant from O(n) above (n1 = 80).

14. Questions Why not groups of 7 elements? Why not groups of 3 elements?
T(n)=O(?)