Algorithm Review: Medians, Order Stats, Radix & Bucket Sort, Selection, Dynamic Sets, Lecture notes of Algorithms and Programming

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CS 332: Algorithms

Medians and Order Statistics

Structures for Dynamic Sets

Homework 3

● On the web shortly…

■ (^) Due Wednesday at the beginning of class (test)

Review: Bucket Sort ● (^) Bucket sort ■ (^) Assumption: input is n reals from [0, 1) ■ (^) Basic idea: ○ (^) Create n linked lists ( buckets ) to divide interval [0,1) into subintervals of size 1/ n ○ (^) Add each input element to appropriate bucket and sort buckets with insertion sort ■ (^) Uniform input distribution  O(1) bucket size ○ (^) Therefore the expected total time is O(n) ■ (^) These ideas will return when we study hash tables

Review: Order Statistics ● (^) The i th order statistic in a set of n elements is the i th smallest element ● (^) The minimum is thus the 1st order statistic ● (^) The maximum is (duh) the n th order statistic ● (^) The median is the n /2 order statistic ■ (^) If n is even, there are 2 medians ● (^) Could calculate order statistics by sorting ■ (^) Time: O(n lg n) w/ comparison sort ■ (^) We can do better

Review: Randomized Selection

● Key idea: use partition() from quicksort

■ (^) But, only need to examine one subarray ■ (^) This savings shows up in running time: O(n)  A[q]  A[q] p q r

Review: Randomized Selection RandomizedSelect(A, p, r, i) if (p == r) then return A[p]; q = RandomizedPartition(A, p, r) k = q - p + 1; if (i == k) then return A[q]; // not in book if (i < k) then return RandomizedSelect(A, p, q-1, i); else return RandomizedSelect(A, q+1, r, i-k);  A[q]  A[q] k p q r

Worst-Case Linear-Time Selection

● Randomized algorithm works well in practice

● What follows is a worst-case linear time

algorithm, really of theoretical interest only

● Basic idea:

■ (^) Generate a good partitioning element ■ (^) Call this element x

Worst-Case Linear-Time Selection ● (^) The algorithm in words:

1. Divide n elements into groups of 5
2. Find median of each group ( How? How long? )
3. Use Select() recursively to find median x of the  n/5 medians
4. Partition the n elements around x. Let k = rank( x )
5. if (i == k) then return x if (i < k) then use Select() recursively to find i th smallest element in first partition else (i > k) use Select() recursively to find ( i-k )th smallest element in last partition

Worst-Case Linear-Time Selection

● Thus after partitioning around x , step 5 will

call Select() on at most 3 n /4 elements

● The recurrence is therefore:

 (^)                 if is big enough

cn c cn cn n cn n cn cn n T n T n n T n T n T n n 

??? ??? ??? ??? ???  n/5   n/ Substitute T(n) = cn Combine fractions Express in desired form What we set out to prove

Worst-Case Linear-Time Selection

● Intuitively:

■ (^) Work at each level is a constant fraction (19/20) smaller ○ (^) Geometric progression! ■ (^) Thus the O(n) work at the root dominates

Linear-Time Median Selection

● Worst-case O(n lg n) quicksort

■ (^) Find median x and partition around it ■ (^) Recursively quicksort two halves ■ (^) T(n) = 2T(n/2) + O(n) = O(n lg n)

Structures…

● Done with sorting and order statistics for now

● Ahead of schedule, so…

● Next part of class will focus on data structures

● We will get a couple in before the first exam

■ (^) Yes, these will be on this exam

Binary Search Trees

● Binary Search Trees (BSTs) are an important

data structure for dynamic sets

● In addition to satellite data, eleements have:

■ (^) key : an identifying field inducing a total ordering ■ (^) left : pointer to a left child (may be NULL) ■ (^) right : pointer to a right child (may be NULL) ■ (^) p : pointer to a parent node (NULL for root)

Binary Search Trees

● BST property:

key[left(x)]  key[x]  key[right(x)]

● Example:

F B H A D K