Analysis of AlgorithmsCS 477/677

Lecture 8Instructor: Monica Nicolescu

CS 477/677 - Lecture 8 2

Quick Announcement

• Hand imaging collection (hand shape)

• Reza Amayeh: amayeh@cse.unr.edu

• Lab address: LME building room 314

• Time:

– Mon, Wed, Fri: 10:00 am until 5:00pm

– Thu, Thr: 3:00 pm until 5:00pm

CS 477/677 - Lecture 8 3

How Fast Can We Sort?

• Insertion sort, Bubble Sort, Selection Sort

• Merge sort

• Quicksort

• What is common to all these algorithms?

– These algorithms sort by making comparisons between the

input elements

• To sort n elements, comparison sorts must make

(nlgn) comparisons in the worst case

(n2)(nlgn)(nlgn)

CS 477/677 - Lecture 8 4

Decision Tree Model• Represents the comparisons made by a sorting algorithm on an

input of a given size: models all possible execution traces

• Control, data movement, other operations are ignored

• Count only the comparisons

• Decision tree for insertion sort on three elements:

node

leaf:

one execution trace

CS 477/677 - Lecture 8 5

Counting Sort

• Assumption: – The elements to be sorted are integers in the range 0

to k• Idea:

– Determine for each input element x, the number of elements smaller than x

– Place element x into its correct position in the output array

303203521 2 3 4 5 6 7 8

A 774221 2 3 4 5

C 80

533322001 2 3 4 5 6 7 8

B

CS 477/677 - Lecture 8 6

Analysis of Counting SortAlg.: COUNTING-SORT(A, B, n, k)1. for i ← 0 to k2. do C[ i ] ← 03. for j ← 1 to n4. do C[A[ j ]] ← C[A[ j ]] + 15. C[i] contains the number of elements equal to i6. for i ← 1 to k7. do C[ i ] ← C[ i ] + C[i -1]8. C[i] contains the number of elements ≤ i9. for j ← n downto 110. do B[C[A[ j ]]] ← A[ j ]11. C[A[ j ]] ← C[A[ j ]] - 1

(k)

(n)

(k)

(n)

Overall time: (n + k)

CS 477/677 - Lecture 8 7

Analysis of Counting Sort

• Overall time: (n + k)• In practice we use COUNTING sort when k = O(n)

running time is (n)• Counting sort is stable

– Numbers with the same value appear in the same order in

the output array

– Important when satellite data is carried around with the

sorted keys

CS 477/677 - Lecture 8 8

Radix Sort

• Considers keys as numbers in a base-R number– A d-digit number will occupy a field of d

columns

• Sorting looks at one column at a time– For a d digit number, sort the least significant

digit first– Continue sorting on the next least significant

digit, until all digits have been sorted– Requires only d passes through the list

CS 477/677 - Lecture 8 9

RADIX-SORTAlg.: RADIX-SORT(A, d)

for i ← 1 to ddo use a stable sort to sort array A on digit i

• 1 is the lowest order digit, d is the highest-order digit

CS 477/677 - Lecture 8 10

Analysis of Radix Sort

• Given n numbers of d digits each, where each

digit may take up to k possible values, RADIX-

SORT correctly sorts the numbers in (d(n+k))

– One pass of sorting per digit takes (n+k) assuming

that we use counting sort

– There are d passes (for each digit)

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Correctness of Radix sort• We use induction on number of passes through each

digit• Basis: If d = 1, there’s only one digit, trivial• Inductive step: assume digits 1, 2, . . . , d-1 are sorted

– Now sort on the d-th digit– If ad < bd, sort will put a before b: correct

a < b regardless of the low-order digits– If ad > bd, sort will put a after b: correct

a > b regardless of the low-order digits– If ad = bd, sort will leave a and b in the

same order and a and b are already sorted on the low-order d-1 digits

CS 477/677 - Lecture 8 12

Bucket Sort• Assumption:

– the input is generated by a random process that distributes elements uniformly over [0, 1)

• Idea:– Divide [0, 1) into n equal-sized buckets– Distribute the n input values into the buckets– Sort each bucket– Go through the buckets in order, listing elements in each one

• Input: A[1 . . n], where 0 ≤ A[i] < 1 for all i • Output: elements ai sorted• Auxiliary array: B[0 . . n - 1] of linked lists, each list

initially empty

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BUCKET-SORT

Alg.: BUCKET-SORT(A, n)

for i ← 1 to n do insert A[i] into list B[nA[i]]for i ← 0 to n - 1

do sort list B[i] with insertion sort

concatenate lists B[0], B[1], . . . , B[n -1] together in order

return the concatenated lists

CS 477/677 - Lecture 8 14

Example - Bucket Sort

.78

.17

.39

.26

.72

.94

.21

.12

.23

.68

0

1

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8

9

1

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.21

.12 /

.72 /

.23 /

.78

.94 /

.68 /

.39 /

.26

.17

/

/

/

/

CS 477/677 - Lecture 8 15

Example - Bucket Sort

0

1

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3

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9

.23

.17 /

.78 /

.26 /

.72

.94 /

.68 /

.39 /

.21

.12

/

/

/

/

.17.12 .23 .26.21 .39 .68 .78.72 .94 /

Concatenate the lists from 0 to n – 1 together, in order

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Correctness of Bucket Sort

• Consider two elements A[i], A[ j]• Assume without loss of generality that A[i] ≤ A[j]• Then nA[i] ≤ nA[j]

– A[i] belongs to the same group as A[j] or to a group with a lower index than that of A[j]

• If A[i], A[j] belong to the same bucket:– insertion sort puts them in the proper order

• If A[i], A[j] are put in different buckets:– concatenation of the lists puts them in the proper order

CS 477/677 - Lecture 8 17

Analysis of Bucket Sort

Alg.: BUCKET-SORT(A, n)

for i ← 1 to n

do insert A[i] into list B[nA[i]]for i ← 0 to n - 1

do sort list B[i] with insertion sort

concatenate lists B[0], B[1], . . . , B[n -1] together in order

return the concatenated lists

O(n)

(n)

O(n)

(n)

CS 477/677 - Lecture 8 18

Conclusion

• Any comparison sort will take at least nlgn to sort an

array of n numbers

• We can achieve a better running time for sorting if we

can make certain assumptions on the input data:

– Counting sort: each of the n input elements is an integer in the

range 0 to k– Radix sort: the elements in the input are integers represented

with d digits

– Bucket sort: the numbers in the input are uniformly distributed

over the interval [0, 1)

CS 477/677 - Lecture 8 19

A Job Scheduling Application

• Job scheduling– The key is the priority of the jobs in the queue

– The job with the highest priority needs to be executed next

• Operations– Insert, remove maximum

• Data structures– Priority queues– Ordered array/list, unordered array/list

CS 477/677 - Lecture 8 20

Example

CS 477/677 - Lecture 8 21

PQ Implementations & Cost

Worst-case asymptotic costs for a PQ with N items

Insert Remove max

ordered arrayordered listunordered arrayunordered list

N

N

N

N

1

1

1

1

Can we implement both operations efficiently?

CS 477/677 - Lecture 8 22

Background on Trees• Def: Binary tree = structure composed of a finite

set of nodes that either:– Contains no nodes, or– Is composed of three disjoint sets of nodes: a root

node, a left subtree and a right subtree

2

14 8

1

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root

Right subtreeLeft subtree

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Special Types of Trees

• Def: Full binary tree = a binary tree in which each node is either a leaf or has degree exactly 2.

• Def: Complete binary tree = a binary tree in which all leaves have the same depth and all internal nodes have degree 2.

Full binary tree

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Complete binary tree

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The Heap Data Structure

• Def: A heap is a nearly complete binary tree with the following two properties:– Structural property: all levels are full, except

possibly the last one, which is filled from left to right– Order (heap) property: for any node x

Parent(x) ≥ x

Heap

5

7

8

4It doesn’t matter that 4 in level 1 is smaller than 5 in level 22

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Definitions• Height of a node = the number of edges on a longest

simple path from the node down to a leaf• Depth of a node = the length of a path from the root to

the node• Height of tree = height of root node

= lgn, for a heap of n elements

2

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1

16

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Height of root = 3

Height of (2)= 1 Depth of (10)= 2

CS 477/677 - Lecture 8 26

Array Representation of Heaps• A heap can be stored as an

array A.– Root of tree is A[1]– Left child of A[i] = A[2i]– Right child of A[i] = A[2i + 1]– Parent of A[i] = A[ i/2 ]– Heapsize[A] ≤ length[A]

• The elements in the subarray A[(n/2+1) .. n] are leaves

• The root is the maximum element of the heap

A heap is a binary tree that is filled in order

CS 477/677 - Lecture 8 27

Heap Types

• Max-heaps (largest element at root), have the max-heap property: – for all nodes i, excluding the root:

A[PARENT(i)] ≥ A[i]

• Min-heaps (smallest element at root), have the min-heap property:– for all nodes i, excluding the root:

A[PARENT(i)] ≤ A[i]

CS 477/677 - Lecture 8 28

Operations on Heaps

• Maintain the max-heap property– MAX-HEAPIFY

• Create a max-heap from an unordered array– BUILD-MAX-HEAP

• Sort an array in place– HEAPSORT

• Priority queue operations

CS 477/677 - Lecture 8 29

Operations on Priority Queues

• Max-priority queues support the following

operations:

– INSERT(S, x): inserts element x into set S– EXTRACT-MAX(S): removes and returns element of

S with largest key

– MAXIMUM(S): returns element of S with largest key

– INCREASE-KEY(S, x, k): increases value of element

x’s key to k (Assume k ≥ x’s current key value)

CS 477/677 - Lecture 8 30

Maintaining the Heap Property

• Suppose a node is smaller than a child– Left and Right subtrees of i are max-heaps

• Invariant: – the heap condition is violated only at that

node

• To eliminate the violation:– Exchange with larger child– Move down the tree– Continue until node is not smaller than

children

CS 477/677 - Lecture 8 31

Maintaining the Heap Property

• Assumptions:– Left and Right

subtrees of i are max-heaps

– A[i] may be smaller than its children

Alg: MAX-HEAPIFY(A, i, n)1. l ← LEFT(i)2. r ← RIGHT(i)3. if l ≤ n and A[l] > A[i]4. then largest ←l5. else largest ←i6. if r ≤ n and A[r] > A[largest]7. then largest ←r8. if largest i9. then exchange A[i] ↔ A[largest]10. MAX-HEAPIFY(A, largest, n)

CS 477/677 - Lecture 8 32

ExampleMAX-HEAPIFY(A, 2, 10)

A[2] violates the heap property

A[2] A[4]

A[4] violates the heap property

A[4] A[9]

Heap property restored

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MAX-HEAPIFY Running Time

• Intuitively:

– A heap is an almost complete binary tree must

process O(lgn) levels, with constant work at each

level

• Running time of MAX-HEAPIFY is O(lgn) • Can be written in terms of the height of the heap,

as being O(h)– Since the height of the heap is lgn

CS 477/677 - Lecture 8 34

Building a Heap

Alg: BUILD-MAX-HEAP(A)1. n = length[A]

2. for i ← n/2 downto 13. do MAX-HEAPIFY(A, i, n)

• Convert an array A[1 … n] into a max-heap (n = length[A])

• The elements in the subarray A[(n/2+1) .. n] are leaves

• Apply MAX-HEAPIFY on elements between 1 and n/2

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4 1 3 2 16 9 10 14 8 7A:

CS 477/677 - Lecture 8 35

Readings

• Chapter 8, 6