1. Median/Order Statistics Algorithms
Minimum and Maximum
Selection in expected linear time
Selection in worst-case linear time

2. Minimum and Maximum
How many comparisons are sufficient to find minimum/maximum?
How many comparisons are sufficient to find both minimum AND maximum?
Show n + log n - 2 comparisons are sufficient to find second minimum (and minimum)

3. Median Problem
How quickly can we find the median (or in general the kth largest element) of an unsorted list of numbers?
Two approaches
Quicksort partition algorithm expected Q (n) time but W(n2) time in the worst-case
Deterministic Q(n) time in the worst-case

4. Quicksort Approach int Select(int A[], k, low, high)
Choose a pivot item
Determine rank of pivot element in current partition
Compare all items to this pivot element
If pivot is kth item, return pivot
Else update low and high and recurse on partition that contains kth item

5. Example k=5 low high rank 17 12 6 23 19 8 5 10 1 8
found:

6. Probabilistic Analysis
Assume each of n! permutations is equally likely
Modify earlier indicator variable analysis of quicksort to handle this k-selection problem
What is probability ith smallest item is compared to jth smallest item?
If k is contained in (i..j)?
If k ≤ i?
If k ≥ j?
2/(j-i+1)
2/(j-k+1)
2/(k-i+1)

7. Cases where (i..j) do not contain k
Case k ≥ j:
S(i=1 to k-1)Sj = i+1 to k 2/(k-i+1) = Si=1 to k-1 (k-i) 2/(k-i+1)
= Si=1 to k-1 2i/(i+1) [replace k-i with i]
= 2 Si=1 to k-1 i/(i+1)
≤ 2(k-1)
Case k ≤ i:
S(j=k+1 to n)Si = k to j-1 2/(j-k+1) = Sj=k+1 to n (j-k) 2/(j-k+1)
= Sj = 1 to n-k 2j/(j+1)
[replace j-k with j and change bounds]
= 2 Sj=1 to n-k j/(j+1)
≥ 2(n-k)
Total for both cases is ≤ 2n-2

8. Case where (i..j) contains k
At most 1 interval of size 3 contains k
i=k-1, j=k+1
At most 2 intervals of size 4 contain k
i=k-1, j=k+2 and i=k-2, j= k+1
In general, at most q-2 intervals of size q contain k
Thus we get S(q=3 to n) (q-2)2/q ≤ S(q=3 to n) 2 = 2(n-2)
Summing together all cases we see the expected number of comparisons is less than 4n

9. Best case, Worst-case Best case running time?
What happens in the worst-case?
Pivot element chosen is always what?
This leads to comparing all possible pairs
This leads to Q(n2) comparisons

10. Deterministic O(n) approach
Need to guarantee a good pivot element while doing O(n) work to find the pivot element
int Select(int A[], k, low, high)
Choosing pivot element
Divide into groups of 5
For each group of 5, find that group’s median
Use median of the medians as pivot element
Determine rank of pivot element
Compare some remaining items directly to median
Update low and high and recurse on partition that contains kth item (or return kth item if it is pivot)

11. Guarantees on the pivot element
Median of medians is guaranteed to be smaller than all the red colored items
Why?
How many red items are there?
Likewise, median of medians is guaranteed to be larger than the blue colored items
Thus median of medians is in the range:
What elements do we need to compare to pivot to determine its rank?
How many of these are there?
3n/10 ignoring non-perfect division issues

12. Analysis of number of comparisons
int Select(int A[], k, low, high)
Choosing pivot element
For each group of 5, find that group’s median
Find the median of the medians
Compare remaining items directly to median
Recurse on correct partition
Analysis
Choosing pivot element
c1 n/5
c1 for median of 5
Recurse on problem of size n/5
c2 n comparisons
Recurse on problem of size 7n/10
T(n) =

13. Solving recurrence relation
T(n) = T(7n/10) + T(n/5) + O(n)
Key observation: 7/10 + 1/5 = 9/10 < 1
Prove T(n) ≤ cn for some constant n by induction on n
T(n) = 7cn/10 + cn/5 + dn
= 9cn/10 + dn
Need 9cn/10 + dn ≤ cn
Thus c/10 ≥ d  c ≥ 10d