Analysis of Algorithms CS 477/677 Instructor: Monica Nicolescu
Dec 19, 2015
Analysis of AlgorithmsCS 477/677
Instructor: Monica Nicolescu
CS 477/677 2
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
30320352
1 2 3 4 5 6 7 8
A 77422
1 2 3 4 5
C 8
0
53332200
1 2 3 4 5 6 7 8
B
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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
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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
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Example - Bucket Sort
.78
.17
.39
.26
.72
.94
.21
.12
.23
.68
0
1
2
3
4
5
6
7
8
9
1
2
3
4
5
6
7
8
9
10
.21
.12 /
.72 /
.23 /
.78
.94 /
.68 /
.39 /
.26
.17
/
/
/
/
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Example - Bucket Sort
0
1
2
3
4
5
6
7
8
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
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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)
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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 11
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
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Example
CS 477/677 13
PQ Implementations & Cost
Worst-case asymptotic costs for a PQ with N items
Insert Remove max
ordered array
ordered list
unordered array
unordered list
N
N
N
N
1
1
1
1
Can we implement both operations efficiently?
CS 477/677 14
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
16
4
3
9 10
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
2
14 8
1
16
7
4
3
9 10
12
Complete binary tree
2
1
16
4
3
9 10
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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
14 8
1
16
4
3
9 10
Height of root = 3
Height of (2)= 1 Depth of (10)= 2
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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 heapA heap is a binary tree that is filled in order
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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]
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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
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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)
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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
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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)
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Example
MAX-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
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Building a Heap
Alg: BUILD-MAX-HEAP(A)
1. n = length[A]
2. for i ← n/2 downto 1
3. 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
2
14 8
1
16
7
4
3
9 10
1
2 3
4 5 6 7
8 9 10
4 1 3 2 16 9 10 14 8 7A:
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Example: A 4 1 3 2 16 9 10 14 8 7
2
14 8
1
16
7
4
3
9 10
1
2 3
4 5 6 7
8 9 10
14
2 8
1
16
7
4
10
9 3
1
2 3
4 5 6 7
8 9 10
2
14 8
1
16
7
4
3
9 10
1
2 3
4 5 6 7
8 9 1014
2 8
1
16
7
4
3
9 10
1
2 3
4 5 6 7
8 9 10
14
2 8
16
7
1
4
10
9 3
1
2 3
4 5 6 7
8 9 108
2 4
14
7
1
16
10
9 3
1
2 3
4 5 6 7
8 9 10
i = 5 i = 4 i = 3
i = 2 i = 1
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Correctness of BUILD-MAX-HEAP• Loop invariant:
– At the start of each iteration of the for loop, each node i + 1, i + 2,…, n is the root of a max-heap
• Initialization:– i = n/2: Nodes n/2+1, n/2+2, …, n are leaves
they are the root of trivial max-heaps
2
14 8
1
16
7
4
3
9 10
1
2 3
4 5 6 7
8 9 10
CS 477/677 29
Correctness of BUILD-MAX-HEAP• Maintenance:
– MAX-HEAPIFY makes node i a max-heap root and preserves the property that nodes i + 1, i + 2, …, n are roots of max-heaps
– Decrementing i in the for loop reestablishes the loop invariant
• Termination:– i = 0 each node 1, 2, …, n is the root of
a max-heap (by the loop invariant)
2
14 8
1
16
7
4
3
9 10
1
2 3
4 5 6 7
8 9 10
CS 477/677 30
Running Time of BUILD MAX HEAP
Running time: O(nlgn)
• This is not an asymptotically tight upper bound
Alg: BUILD-MAX-HEAP(A)
1. n = length[A]
2. for i ← n/2 downto 1
3. do MAX-HEAPIFY(A, i, n) O(lgn)O(n)
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Running Time of BUILD MAX HEAP
• HEAPIFY takes O(h) the cost of HEAPIFY on a node i is proportional to the height of the node i in the tree
Height Level
h0 = 3 (lgn)
h1 = 2
h2 = 1
h3 = 0
i = 0
i = 1
i = 2
i = 3 (lgn)
No. of nodes
20
21
22
23
hi = h – i height of the heap rooted at level ini = 2i number of nodes at level i
i
h
iihnnT
0
)( ihh
i
i 0
2 )(nO
CS 477/677 33
Readings
• Chapter 6• Chapter 8