1. CPSC 311 Section 502 Analysis of Algorithm
Fall 2002
Department of Computer Science
Texas A&M University

2. CPSC 311-502 Instructor: O. Burchan Bayazit (pronounced as bourch han)
CPSC 311 O. Burchan Bayazit – Fall 02
CPSC
Instructor: O. Burchan Bayazit (pronounced as bourch han)
Office: 407 E Harvey R. Bright Bldg.
Office Hours: 4:00pm-5:30pm Tuesdays
Teaching Assistant: Kasthuri Srinivasan Kannan
Office Hours: 4:00pm-5:30pm Thursdays

3. CPSC 311 O. Burchan Bayazit – Fall 02
Online Material
Course homepage:
Mailing List: (announcements only)
Discussion Group: (announcements+discussions)
news://news.tamu.edu/tamu.classes.cpsc311

4. Class Meeting: 9:35-10:50am, Tuesdays and Thursdays ZACH 105B
CPSC 311 O. Burchan Bayazit – Fall 02
Course Information
Class Meeting: 9:35-10:50am, Tuesdays and Thursdays ZACH 105B
Textbook: Introduction to Algorithms, Cormen, Leiserson, Rivest, Stein, McGraw-Hill (and MIT Press), 2nd Edition (latest)

5. CPSC 311 O. Burchan Bayazit – Fall 02
Prerequisites
CPSC 211 (Data Structures) and Math 302 (Discrete Mathematics).
In particular, you should be familiar with:
mathematical solutions to recurrence relations
mathematical induction
data structures including linked lists, arrays, trees, and graphs
knowledge of a high level block structured language such as Pascal, C or Java

6. Mechanics Assignments (25%) Quizzes (15%) Exams (60%, 20% each)
CPSC 311 O. Burchan Bayazit – Fall 02
Mechanics
Assignments (25%)
Homework (6-10)
In-Class
Programming Assignments
CS Culture Activities
Quizzes (15%)
Exams (60%, 20% each)
Two midterms and a final

7. OK, what about algorithms?
CPSC 311 O. Burchan Bayazit – Fall 02
OK, what about algorithms?
Algorithm: a sequence of computational steps that transform given input to the output

8. Course Goals At the end of the semester you should:
CPSC 311 O. Burchan Bayazit – Fall 02
Course Goals
At the end of the semester you should:
be familiar with fundamental algorithms and algorithmic techniques,
given a particular application, be able to decide which algorithm among a set of possible choices is best,
be able to prove correctness and analyze the running time of a given algorithm,
be able to design efficient algorithms for new situations, using as building blocks the techniques learned,
be able to prove a problem is NP-complete using reduction.

9. CPSC 311 O. Burchan Bayazit – Fall 02
Course Outline
Introduction and Mathematical Fundamentals - (Ch. 1, 2, 3, 4 )
Sorting - (Ch. 6, 7, 8)
Selection - (Ch. 9)
EXAM 1
Hashing - (Ch. 11)
Basic Graph Algorithms - (Ch. 22 )
More Graph Algorithms - (Ch. 23, 24, 25)
Union-Find/Disjoint Set Data Structure - (Ch. 21)
Dynamic Programming - (Ch. 15)
EXAM 2
NP-Completeness - (Ch. 34, 35)
Special Topics TBD
EXAM 3 (FINAL) (Friday December 13, 12:30-2:30pm)

10. Quiz Time !!!!! Fill out the personal information in the front page
CPSC 311 O. Burchan Bayazit – Fall 02
Quiz Time !!!!!
Fill out the personal information in the front page
Answer the questions of the quiz in the back page

11. Answer to Question 1. Shown binary search for 13 1 2 3 4 5 6 7 8 9 10
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 1.
Shown binary search for 13 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

12. Answer to Question 1. Find the middle 1 2 3 4 5 6 7 8 9 10 11 12 13 14
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 1.
Find the middle 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

13. Answer to Question 1. Select the right side (8<13) 1 2 3 4 5 6 7 8
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 1.
Select the right side (8<13) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

14. Answer to Question 1. Select the right side (8<13) 1 2 3 4 5 6 7 8
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 1.
Select the right side (8<13) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

15. Answer to Question 1. Find the middle 1 2 3 4 5 6 7 8 9 10 11 12 13 14
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 1.
Find the middle 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Assume we have only integers (11 is also correct, if you take floor of the mid-point)

16. Answer to Question 1. Select right side (12<13) 1 2 3 4 5 6 7 8 9
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 1.
Select right side (12<13) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

17. Answer to Question 1. Find the middle 1 2 3 4 5 6 7 8 9 10 11 12 13 14
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 1.
Find the middle 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

18. Answer to Question 1. Select the left side (13<14) 1 2 3 4 5 6 7 8
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 1.
Select the left side (13<14) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

19. Answer to Question 1. Find the middle = 13 1 2 3 4 5 6 7 8 9 10 11 12
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 1.
Find the middle = 13 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

20. Answer to Question 2. What is returned on input 20?
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
What is returned on input 20?
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }

21. Answer to Question 2. x=20 Stack Subroutine whatami(integer x) {
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=20
Stack
whatami(20)

22. Answer to Question 2. x=20 Stack Subroutine whatami(integer x) {
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=20
Stack
whatami(20)

23. Answer to Question 2. x=20 Stack Subroutine whatami(integer x) {
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=20
Stack
whatami(20)

24. Answer to Question 2. x=10 Stack Subroutine whatami(integer x) {
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=10
Stack
whatami(10)
whatami(20)

25. Answer to Question 2. x=10 Stack Subroutine whatami(integer x) {
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=10
Stack
whatami(10)
whatami(20)

26. Answer to Question 2. x=10 Stack Subroutine whatami(integer x) {
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=10
Stack
whatami(10)
whatami(20)

27. Answer to Question 2. x=5 Stack Subroutine whatami(integer x) {
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=5
Stack
whatami(5)
whatami(10)
whatami(20)

28. Answer to Question 2. x=5 Stack Subroutine whatami(integer x) {
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=5
Stack
whatami(5)
whatami(10)
whatami(20)

29. Answer to Question 2. x=5 Stack Subroutine whatami(integer x) {
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=5
Stack
whatami(5)
whatami(10)
whatami(20)

30. Answer to Question 2. x=5 Stack Return 2
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=5
Stack
whatami(5)
whatami(10)
whatami(20)
Return 2

31. Answer to Question 2. x=10 Stack Return 2
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=10
Stack
whatami(10)
whatami(20)
Return 2

32. Answer to Question 2. x=20 Stack Return 2
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 2.
Subroutine whatami(integer x) {
if (x is even)
then return (whatami(x/2))
else return ((x-1)/2) }
x=20
Stack
whatami(20)
Return 2

33. Answer to Question 3. Show the link list after execution d e a b c tmp
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3.
Show the link list after execution d e a b c
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2
tmp

34. Answer to Question 3. d e a b c tmp tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

35. Answer to Question 3. d e a b c tmp tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

36. Answer to Question 3. d e a b c tmp tmp1 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

37. Answer to Question 3. d e a b c tmp tmp1 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

38. Answer to Question 3. d e a b c tmp tmp1 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

39. Answer to Question 3. d e a b c tmp tmp1 tmp2 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp2
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

40. Answer to Question 3. d e a b c tmp tmp1 tmp2 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp2
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

41. Answer to Question 3. d e a b c tmp tmp1 tmp2 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp2
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

42. Answer to Question 3. d e a b c tmp tmp1 tmp2 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp2
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

43. Answer to Question 3. d e a b c tmp tmp1 tmp2 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp2
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

44. Answer to Question 3. d e a b c tmp tmp1 tmp2 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp2
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

45. Answer to Question 3. d e a b c tmp tmp1 tmp2 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp2
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

46. Answer to Question 3. d e a b c tmp tmp1 tmp2 tmp1=next(next(tmp))
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 3. d e a b c
tmp
tmp1
tmp2
tmp1=next(next(tmp))
tmp2=next(tmp1)
next(tmp1)=next(tmp2)
next(tmp2)=next(tmp)
next(tmp)=tmp2

47. Answer to Question 4. Use induction to prove 1+2+…+n=n(n+1)/2
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 4.
Use induction to prove 1+2+…+n=n(n+1)/2

48. Answer to Question 4. Step 1: Verify that eq. holds for n=1
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 4.
Step 1: Verify that eq. holds for n=1
Left hand side is 1 and right hand site 1*(1+1)/2, so the equation holds for n=1

49. Answer to Question 4. Step 2: assume that eq. holds for n=k
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 4.
Step 2: assume that eq. holds for n=k
1+2+…+k= k(k+1)/2

50. CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 4.
Step 3: using assumption at Step2, show that eq. holds for n=k+1
From Step 2: 1+2+…+k= k(k+1)/2
If we add k+1 to each side
1+2+….+k+ (k+1) = k(k+1)/2 + (k+1) or
1+2+….+k+ (k+1) = (k(k+1) + 2(k+1))/2

51. Answer to Question 4. Step 3: cont.
CPSC 311 O. Burchan Bayazit – Fall 02
Answer to Question 4.
Step 3: cont.
1+2+….+k+ (k+1) = (k(k+1) + 2(k+1))/2
Becomes
1+2+….+k+ (k+1) = (k+1)( k+2))/2
Hence Eq. Holds for n=k+1

52. CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort 1 2 3 4 5 6 5 2 4 6 1 3

53. Insertion Sort 5 2 4 6 1 3 Insert 2 into sorted part sorted
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
sorted 5 2 4 6 1 3
Insert 2 into sorted part

54. Insertion Sort 5 4 6 1 3 2 After Insertion sorted
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
sorted 5 4 6 1 3 2
After Insertion

55. Insertion Sort 5 4 6 1 3 2 Insert 4 into the sorted part sorted
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
sorted 5 4 6 1 3 2
Insert 4 into the sorted part

56. Insertion Sort 4 5 6 1 3 2 Insert 6 into the sorted part sorted
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
sorted 4 5 6 1 3 2
Insert 6 into the sorted part

57. Insertion Sort 4 5 6 1 3 2 Insert 1 into the sorted part sorted
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
sorted 4 5 6 1 3 2
Insert 1 into the sorted part

58. Insertion Sort 2 4 5 6 3 1 Insert 3 into the sorted part sorted
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
sorted 2 4 5 6 3 1
Insert 3 into the sorted part

59. CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
sorted 1 2 3 4 5 6

60. CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort 6 1 2 3 4 5

61. CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
How to insert? (assume we inserted previous elements and we are at the last element)
j=6 (element’s position) 2 4 5 6 3 1
Start from i=j-1

62. Insertion Sort How to insert? 2 4 5 6 3 1 i=5 If(array[i]>Element)
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
How to insert? 2 4 5 6 3 1
i=5
If(array[i]>Element)
switch Array[i] and Array [i+1]

63. Insertion Sort How to insert? 2 4 5 3 6 1 i=4 If(array[i]>Element)
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
How to insert? 2 4 5 3 6 1
i=4
If(array[i]>Element)
switch Array[i] and Array [i+1]

64. Insertion Sort How to insert? 2 4 3 5 6 1 i=3 If(array[i]>Element)
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
How to insert? 2 4 3 5 6 1
i=3
If(array[i]>Element)
switch Array[i] and Array [i+1]

65. Insertion Sort How to insert? 2 3 4 5 6 1 i=2 If(array[i]<=Element)
CPSC 311 O. Burchan Bayazit – Fall 02
Insertion Sort
How to insert? 2 3 4 5 6 1
i=2
If(array[i]<=Element)
stop Insertion

66. Psuedo Algorithm for Insertion Sort
CPSC 311 O. Burchan Bayazit – Fall 02
Psuedo Algorithm for Insertion Sort
for j=2 to Length(Array) {
key=Array[j]
i=j-1
while (i>0 and A[i]>key) {
A[I+1]=A[I]
i=i-1 }
A[i+1]=key

67. Psuedo Algorithm for Insertion Sort
CPSC 311 O. Burchan Bayazit – Fall 02
Psuedo Algorithm for Insertion Sort
cost
for j=2 to Length(Array) {
key=Array[j]
i=j-1
while (i>0 and A[i]>key) {
A[I+1]=A[I]
i=i-1 }
A[i+1]=key c1 c2 c3 c4 c5 c6 c7

68. Psuedo Algorithm for Insertion Sort
CPSC 311 O. Burchan Bayazit – Fall 02
Psuedo Algorithm for Insertion Sort
cost times
for j=2 to Length(Array) {
key=Array[j]
i=j-1
while (i>0 and A[i]>key) {
A[I+1]=A[I]
i=i-1 }
A[i+1]=key c1 n
n-1 c2 c3
n-1 c4
sum(2,n)(tj) c5
sum(2,n)(tj-1)
sum(2,n)(tj-1) c6 c7
n-1

69. Cost of Insertion Sort cost times T(n)=c1.n+c2.(n-1)+c2.(n-1)+
CPSC 311 O. Burchan Bayazit – Fall 02
Cost of Insertion Sort
cost times c1 n
n-1 c2
T(n)=c1.n+c2.(n-1)+c2.(n-1)+
c4. sum(2,n)(tj)+c5. sum(2,n)(tj-1)+
c6. sum(2,n)(tj-1)+c7.(n-1) c3
n-1 c4
sum(2,n)(tj) c5
sum(2,n)(tj-1)
sum(2,n)(tj-1) c6 c7
n-1

70. Best-Case cost In the best case, the array is already sorted, hence
CPSC 311 O. Burchan Bayazit – Fall 02
Best-Case cost
In the best case, the array is already sorted, hence
tj =1
cost times c1 n
n-1 c2 c3
n-1 c4
sum(2,n)(tj)
T(n)=c1.n+c2.(n-1)+c2.(n-1)+
c3.(n-1)+c4.(n-1)+c7.(n-1)
= (c1+c2+c3+c4+c7)n-(c2+c3+c4+c7) c5
sum(2,n)(tj-1)
sum(2,n)(tj-1) c6 c7
n-1
T(n) is a linear function of n

71. CPSC 311 O. Burchan Bayazit – Fall 02
Worst-Case cost
In the worst case, the array is already sorted in the reverse order hence
tj =j
cost times c1 n
n-1 c2 c3
n-1 c4
sum(2,n)(tj)
sum(2,n)(tj) becomes sum(2,n)(j) =n*(n+1)/2-1
sum(2,n)(tj-1) becomes sum(2,n)(j-1) =n*(n-1)/2 c5
sum(2,n)(tj-1)
sum(2,n)(tj-1) c6 c7
n-1
T(n) is a linear function of n

72. Worst-Case cost cost times T(n) is a quadratic function of n
CPSC 311 O. Burchan Bayazit – Fall 02
Worst-Case cost
cost times
T(n)=c1.n+c2.(n-1)+c3.(n-1)+c4.(n.(n+1)/2-1)
+c5.n.(n-1)/2+c6.n.(n-1)/2+c7.(n-1)
=(c4+c5+c6)/2.n2+
(c1+c2+c3+(c4-c5-c6)/2+c7).n –(c2+c3+c4+c7) c1 n
n-1 c2 c3
n-1 c4
sum(2,n)(tj) c5
sum(2,n)(tj-1)
sum(2,n)(tj-1) c6 c7
n-1
T(n) is a quadratic function of n

73. CPSC 311 O. Burchan Bayazit – Fall 02
Average-Case cost
In the average case, we need the search half of the sorted part of the array hence
tj =j/2
cost times c1 n
n-1 c2 c3
n-1 c4
sum(2,n)(tj) c5
sum(2,n)(tj-1)
sum(2,n)(tj-1) c6 c7
n-1
T(n) is a quadratic function of n