1. Introduction to Algorithms
Counting Sort

2. Sorting So Far Insertion Sort Easy to code
Fast on small inputs (less than ~ 50 elements)
Fast on nearly-sorted inputs
O(n2) worst case
O(n2) average (equally-likely inputs) case
O(n2) reverse-sorted case

3. Sorting So Far Merge Sort Divide-and-Conquer O(n lg n) worst case
Split array in half
Recursively sort subarrays
Linear-time merge step
O(n lg n) worst case
It does not sort in place

4. Sorting So Far Heap Sort It uses the very useful heap data structure
Complete binary tree
Heap property: parent key > children’s keys
O(n lg n) worst case
It sorts in place

5. How Fast Can We Sort? We will provide a lower bound, then beat it
How do you suppose we’ll beat it?
First, an observation: all of the sorting algorithms so far are
comparison sorts
The only operation used to gain ordering information about a sequence is the pairwise comparison of two elements
Theorem: all comparison sorts are Ω(n lg n)

6. Decision Trees
5 7 3 35 35 7

7. Decision Trees
Decision trees provide an abstraction of comparison sorts
A decision tree represents the comparisons made by a comparison sort.
What do the leaves represent?
How many leaves must there be?

8. Decision Trees Decision trees can model comparison sorts
For a given algorithm:
One tree for each n
Tree paths are all possible execution traces
What’s the longest path in a decision tree for insertion sort?
For merge sort?
What is the asymptotic height of any decision tree for sorting n elements?
Answer: Ω(n lg n)
(Now let’s prove it…)

9. A Lower Bound for Comparison Sorts
Thus the time to comparison sort n elements is Ω(n lg n)
Corollary: Heapsort and Mergesort are asymptotically optimal comparison sorts
But the name of this lecture is “Sorting in linear time”!
How can we do better than Ω(n lg n)?

10. Sorting in Linear Time Counting Sort Basic Idea Algorithm Storage
No comparisons between elements !
But…depends on assumption about the numbers being sorted
Basic Idea
For each input element x,
determine the number of elements less than or equal to x
For each integer i (0 ≤ i ≤ k), count how many elements whose
values are i
Then we know how many elements are less than or equal to i
Algorithm Storage
A[1..n]: input elements
B[1..n]: sorted elements
C[0..k]: hold the number of elements less than or equal to i

11. Counting Sort Illustration
(Range from 0 to 5)
Spring 2006 Algorithm
Networking Laboratory 5-15/55

12. Counting Sort Illustration2 3 B
3 3 C 1 4 5 7 8
0 2
3 8 C C C B 2 C 3 4
7 7

13. Counting Sort
Θ(k)
Θ(n)
Θ(n+k)

14. Counting Sort Total time: O(n + k) But sorting is Ω(n lg n)!
Usually, k = O(n)
Thus counting sort runs in O(n) time
But sorting is Ω(n lg n)!
No contradiction
This is not a comparison sort
In fact, there are no comparisons at all !
Notice that this algorithm is stable

15. Bucket Sort Bucket sort Assumption: the keys are in the range [0, N)
Basic idea:
1. Create N linked lists (buckets) to divide interval [0,N) into subintervals of size 1
2. Add each input element to appropriate bucket
3. Concatenate the buckets
Expected total time is O(n + N), with n = size of original sequence
if N is O(n)  sorting algorithm in O(n) !

16. Bucket Sort
Each element of the array is put in one of the N “buckets”

17. Bucket Sort Now, pull the elements from the buckets into the array
At last, the sorted array
(sorted in a stable way):

18. Does it Work for Real Numbers?
What if keys are not integers?
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)
Distribution of keys in buckets similar with …. ?

19. Radix Sort How did IBM get rich originally?
Answer: punched card readers for census tabulation in early 1900’s.
In particular, a card sorter that could sort cards into different bins
Each column can be punched in 12 places
(Decimal digits use only 10 places!)
Problem: only one column can be sorted on at a time

20. Radix Sort
Intuitively, you might sort on the most significant digit, then the second most significant, etc.
Problem: lots of intermediate piles of cards to keep track of
Key idea: sort the least significant digit first
RadixSort(A, d)
for i=1 to d
StableSort(A) on digit i

21. Radix Sort Can we prove it will work? Inductive argument:
Assume lower-order digits {j: j<i}are sorted
Show that sorting next digit i leaves array correctly sorted
If two digits at position i are different, ordering numbers by that digit is correct (lower- order digits irrelevant)
If they are the same, numbers are already sorted on the lower-order digits. Since we use a stable sort, the numbers stay in the right order

22. Radix Sort What sort will we use to sort on digits?
Bucket sort is a good choice:
Sort n numbers on digits that range from 1..N
Time: O(n + N)
Each pass over n numbers with d digits takes time O(n+k), so total time O(dn+dk)
When d is constant and k=O(n), takes O(n) time

23. Radix Sort Example Problem: sort 1 million 64-bit numbers
Treat as four-digit radix 216 numbers
Can sort in just four passes with radix sort!
Running time: 4( 1 million )  4 million operations
Compare with typical O(n lg n) comparison sort
Requires approx lg n = 20 operations per number being sorted
Total running time  20 million operations

24. Radix Sort Sort N numbers, each with k bits
Cpt S 223
Radix sort achieves stable sorting
To sort each column, use counting sort (O(n))
=> To sort k columns, O(nk) time
Radix Sort
Sort N numbers, each with k bits
E.g, input {4, 1, 0, 10, 5, 6, 1, 8}
4 0100
1 0001
0 0000
5 0101
6 0110
8 1000
0100
0000
1010
0110
1000
0001
0101
0100
0000
1000
0001
0101
1010
0110
0000
1000
0001
1010
0100
0101
0110
0000 4
0001 1
0100 4
0101 5
0110 6
1000 8
lsb
msb
Washington State University

25. Radix Sort In general, radix sort based on bucket sort is
Asymptotically fast (i.e., O(n))
Simple to code
A good choice
Can radix sort be used on floating-point numbers?

26. Summary: Radix Sort Radix sort:
Assumption: input has d digits ranging from 0 to k
Basic idea:
Sort elements by digit starting with least significant
Use a stable sort (like bucket sort) for each stage
Each pass over n numbers with 1 digit takes time O(n+k), so total time O(dn+dk)
When d is constant and k=O(n), takes O(n) time
Fast, Stable, Simple
Doesn’t sort in place