Winter 2016COMP-250: Introduction

to Computer ScienceLecture 20, March 24, 2016

————— Public Announcement —————

————— ———————— —————

————— Public Announcement —————

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ES EXAM OFFICE: EXAM.ES@MCGILL.CA TEL. 514-398-4209

FINAL EXAMINATION CHECKLIST - April 2016 A signed original plus one copy of the exam is to be delivered to the Exams Office room MS-72 (3415 McTavish). In order to have your exam printed by the Exam Office, it must be submitted by: March 24th, 2016 Course : COMP 250 Sect: 1

Title: Intro to Computer Science

Exam Date: 4/28/2016

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ES EXAM OFFICE: EXAM.ES@MCGILL.CA TEL. 514-398-4209

FINAL EXAMINATION CHECKLIST - April 2016 A signed original plus one copy of the exam is to be delivered to the Exams Office room MS-72 (3415 McTavish). In order to have your exam printed by the Exam Office, it must be submitted by: March 24th, 2016 Course : COMP 250 Sect: 1

Title: Intro to Computer Science

Exam Date: 4/28/2016

Exam Time: 2:00:00 PM

Examiner: Name: Prof. C. Crépeau

Office #:

Email:

Examiner’s signature:

Cell #:

Associate Examiner:

Name: Tel #: Email:

Associate Examiner’s signature:

Cell #:

INFORMATION AND AUTHORIZATION OF MATERIALS: 1. Is this a CLOSED BOOK or OPEN BOOK exam? CLOSED OPEN

2. Is this a MULTIPLE-CHOICE Exam? Multiple versions must be submitted, regardless of the number of students, questions or whether the exam is only partially multiple-choice (4 versions for enrolments of 200+). YES NO

3. Is your MULTIPLE-CHOICE exam: ANSWER ON EXAM MACHINE SCORED ON ANSWER SHEETS

4. If MACHINE-SCORED, number of versions: _______

5. DICTIONARIES: Translation ONLY Regular ONLY ANY NONE

6. CALCULATORS: Non-programmable Faculty Standard ANY NONE

7. RETURN EXAM? If “YES”, include space for Student NAME and Student NUMBER on cover page YES NO

8. Answer in EXAM BOOKLET YES NO

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6.1Heaps I

HEAPS I• Heaps

• Properties

• Insertion and Deletion

6.2Heaps I

Heaps• A heap is a binary tree T that stores a collection of

keys (or key-element pairs) at its internal nodes andthat satisfies two additional properties:- Order Property: key(parent) ≤ key(child)- Structural Property: all levels are full, except the

last one, which is left-filled (complete binary tree)

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6.3Heaps I

Not Heaps• bottom level is not left-filled

• key(parent)> key(child)

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6.3Heaps I

Not Heaps• bottom level is not left-filled

• key(parent)> key(child)

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6.3Heaps I

Not Heaps• bottom level is not left-filled

• key(parent)> key(child)

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6.4Heaps I

Height of a HeapA heap T storing n keys has height h = log(n + 1),which is O(log n)

• n ≥ 1 + 2 + 4 + ... + 2h-2 + 1 = 2h-1 - 1 + 1 = 2h-1

• n ≤ 1 + 2 + 4 + ... +2h-1 = 2h - 1

• Therefore 2h-1 ≤ n ≤ 2h - 1

• Taking logs, we get log (n + 1) ≤ h ≤ log n + 1

• Which implies h = log(n+1)

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915

16h − 1

01

h − 2

h

h − 1

01

h − 2

h

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2516 2231

6.4Heaps I

Height of a HeapA heap T storing n keys has height h = log(n + 1),which is O(log n)

• n ≥ 1 + 2 + 4 + ... + 2h-2 + 1 = 2h-1 - 1 + 1 = 2h-1

• n ≤ 1 + 2 + 4 + ... +2h-1 = 2h - 1

• Therefore 2h-1 ≤ n ≤ 2h - 1

• Taking logs, we get log (n + 1) ≤ h ≤ log n + 1

• Which implies h = log(n+1)

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915

16h − 1

01

h − 2

h

h − 1

01

h − 2

h

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6.4Heaps I

Height of a HeapA heap T storing n keys has height h = log(n + 1),which is O(log n)

• n ≥ 1 + 2 + 4 + ... + 2h-2 + 1 = 2h-1 - 1 + 1 = 2h-1

• n ≤ 1 + 2 + 4 + ... +2h-1 = 2h - 1

• Therefore 2h-1 ≤ n ≤ 2h - 1

• Taking logs, we get log (n + 1) ≤ h ≤ log n + 1

• Which implies h = log(n+1)

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915

16h − 1

01

h − 2

h

h − 1

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h − 2

h

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6.5Heaps I

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21 10 20 8

22 28 13 25

Heap Insertion

So here we go ...

The key to insert is 6

19

6.6Heaps I

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21 10 20 8

22 28 13 19

Add the key in the next available position in theheap.

Now begin Upheap.

Heap Insertion

25 6

6.7Heaps I

Upheap• Swap parent-child keys out of order

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22 28 13 19 25 6

374

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6.7Heaps I

Upheap• Swap parent-child keys out of order

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22 28 13 19 25 6

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6.7Heaps I

Upheap• Swap parent-child keys out of order

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6.8Heaps I

Upheap Continues

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364

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6.8Heaps I

Upheap Continues

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6.8Heaps I

Upheap Continues

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6.9Heaps I

• Upheap terminates when new key is greaterthan the key of its parent or the top of the heapis reached

• (total #swaps) ≤ (h − 1), which is O(log n)

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6

End of Upheap

6.10Heaps I

Removal From a HeapRemoveMin()

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• The removal of the top key leaves a hole• We need to fix the heap• First, replace the hole with the last key in

the heap• Then, begin Downheap

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6.10Heaps I

Removal From a HeapRemoveMin()

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• The removal of the top key leaves a hole• We need to fix the heap• First, replace the hole with the last key in

the heap• Then, begin Downheap

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6.11Heaps I

Downheap20

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Downheap compares the parent with the smallestchild. If the child is smaller, it switches the two.

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6.11Heaps I

Downheap20

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Downheap compares the parent with the smallestchild. If the child is smaller, it switches the two.

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66.11Heaps I

Downheap20

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Downheap compares the parent with the smallestchild. If the child is smaller, it switches the two.

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6.12Heaps I

Downheap Continues

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6.12Heaps I

Downheap Continues

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6.12Heaps I

Downheap Continues

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6.13Heaps I

Downheap Continues

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6.13Heaps I

Downheap Continues

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6.13Heaps I

Downheap Continues

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6.14Heaps I

End of Downheap4

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• Downheap terminates when the key is greaterthan the keys of both its children or the bottomof the heap is reached.

• (total #swaps) ≤ (h − 1), which is O(log n)

6.1Heaps II

HEAPS II• Implementation

• HeapSort

• Bottom-Up Heap Construction

• Locators

6.2Heaps II

Implementation of a Heappublic class HeapPriorityQueue implements PriorityQueue{

BinaryTree T;Position last;Comparator comparator;...

}

lastheap

(4,C)

(6,Z)

(20,B)(7,Q)

(8,W)(11,S)

(5,A)

(9,F)

(12,H)(14,E)

(15,K)

(25,J)(16,X)

comp

<=>

6.3Heaps II

Implementation of a Heap(cont.)• Two ways to find the insertion position z in a heap:

(2,C)

(4,C)

(6,Z)(7,Q)

(8,W)(11,S)

(5,A)

(9,F)

(12,H)(14,E)

(15,K)

(25,J)(16,X) (20,B) (10,L)

z

(4,C)

(6,Z)

(20,B)(7,Q)

(5,A)

(9,F)

(12,H)(14,E)

(15,K)

(25,J)(16,X)w z

u

6.3Heaps II

Implementation of a Heap(cont.)• Two ways to find the insertion position z in a heap:

(2,C)

(4,C)

(6,Z)(7,Q)

(8,W)(11,S)

(5,A)

(9,F)

(12,H)(14,E)

(15,K)

(25,J)(16,X) (20,B) (10,L)

z

(4,C)

(6,Z)

(20,B)(7,Q)

(5,A)

(9,F)

(12,H)(14,E)

(15,K)

(25,J)(16,X)w z

u6.3Heaps II

Implementation of a Heap(cont.)• Two ways to find the insertion position z in a heap:

(2,C)

(4,C)

(6,Z)(7,Q)

(8,W)(11,S)

(5,A)

(9,F)

(12,H)(14,E)

(15,K)

(25,J)(16,X) (20,B) (10,L)

z

(4,C)

(6,Z)

(20,B)(7,Q)

(5,A)

(9,F)

(12,H)(14,E)

(15,K)

(25,J)(16,X)w z

u

6.4Heaps II

Vector Based Implementation• Updates in the underlying tree occur only at the “last

element”

• A heap can be represented by a vector, where thenode at rank i has- left child at rank 2i and- right child at rank 2i + 1

• The leaves do no need to be explicitly stored

• Insertion and removals into/from the heapcorrespond to insertLast and removeLast on thevector, respectively

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6.4Heaps II

Vector Based Implementation• Updates in the underlying tree occur only at the “last

element”

• A heap can be represented by a vector, where thenode at rank i has- left child at rank 2i and- right child at rank 2i + 1

• The leaves do no need to be explicitly stored

• Insertion and removals into/from the heapcorrespond to insertLast and removeLast on thevector, respectively

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2

5 6 7

3

4

8 9 10 11 12 13

6.4Heaps II

Vector Based Implementation• Updates in the underlying tree occur only at the “last

element”

• A heap can be represented by a vector, where thenode at rank i has- left child at rank 2i and- right child at rank 2i + 1

• The leaves do no need to be explicitly stored

• Insertion and removals into/from the heapcorrespond to insertLast and removeLast on thevector, respectively

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6.5Heaps II

Heap Sort• All heap methods run in logarithmic time or better

• If we implement PriorityQueueSort using a heap forour priority queue, insertItem and removeMin eachtake O(log k), k being the number of elements in theheap at a given time.

• We always have at most n elements in the heap, sothe worst case time complexity of these methods isO(log n).

• Thus each phase takes O(n log n) time, so thealgorithm runs in O(n log n) time also.

• This sort is known as heap-sort.

• The O(n log n) run time of heap-sort is much betterthan the O(n2) run time of selection and insertionsort.

In-Place Heap-Sort• Do not use an external heap

• Embed the heap into the sequence, using the vectorrepresentation

6.6Heaps II

Bottom-Up Heap Construction• build (n + 1)/2 trivial one-element heaps

• now build three-element heaps on top of them

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6.6Heaps II

Bottom-Up Heap Construction• build (n + 1)/2 trivial one-element heaps

• now build three-element heaps on top of them

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6.6Heaps II

Bottom-Up Heap Construction• build (n + 1)/2 trivial one-element heaps

• now build three-element heaps on top of them

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6.7Heaps II

Bottom-Up Heap Construction• downheap to preserve the order property

• now form seven-element heaps

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6.7Heaps II

Bottom-Up Heap Construction• downheap to preserve the order property

• now form seven-element heaps

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6.7Heaps II

Bottom-Up Heap Construction• downheap to preserve the order property

• now form seven-element heaps

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6.8Heaps II

Bottom-Up Heap Construction(cont.)

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6.8Heaps II

Bottom-Up Heap Construction(cont.)

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46.8Heaps II

Bottom-Up Heap Construction(cont.)

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6.9Heaps II

Bottom-Up Heap Construction(cont.)

The End

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6.10Heaps II

Analysis of Bottom-Up HeapConstruction

• Proposition: Bottom-up heap construction with nkeys takes O(n) time.- Insert (n + 1)/2 nodes- Insert (n + 1)/4 nodes and downheap them- Insert (n + 1)/8 nodes and downheap them- ...- visual analysis:

• n inserts, n/2 upheaps with total O(n) running time

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6.10Heaps II

Analysis of Bottom-Up HeapConstruction

• Proposition: Bottom-up heap construction with nkeys takes O(n) time.- Insert (n + 1)/2 nodes- Insert (n + 1)/4 nodes and downheap them- Insert (n + 1)/8 nodes and downheap them- ...- visual analysis:

• n inserts, n/2 upheaps with total O(n) running time

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Chapter 8

⟨a1,a2,a3⟩

⟨a3,a1,a2⟩⟨a1,a3,a2⟩⟨a1,a2,a3⟩

⟨a1,a3,a2⟩ ⟨a3,a1,a2⟩

⟨a3,a1,a2⟩⟨a1,a3,a2⟩

⟨a3,a1,a2⟩⟨a2,a1,a3⟩⟨a1,a2,a3⟩

⟨a3,a2,a1⟩⟨a2,a3,a1⟩⟨a1,a3,a2⟩

⟨a2,a1,a3⟩

⟨a3,a2,a1⟩⟨a2,a3,a1⟩

⟨a2,a1,a3⟩

⟨a3,a2,a1⟩⟨a2,a3,a1⟩

⟨a2,a3,a1⟩ ⟨a3,a2,a1⟩

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

N!

N!/2 N!/2

N!/4 N!/4 N!/4 N!/4

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

log N! ∈ 𝜽(N log N)

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩ ⟨a1,a2,a3⟩⟨a1,a2,a3⟩

⟨a1,a2,a3⟩ ⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

⟨a1,a2,a3⟩⟨a1,a2,a3⟩

⟨a1,a2,a3⟩

Winter 2016COMP-250: Introduction

to Computer ScienceLecture 20, March 24, 2016

Hardik Omar Faiz

Faizy Chris David B. David B.R.

Lekan

DoYeon