1 Algorithms and Analysis CS 2308 Foundations of CS II

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Algorithms

Step-by-step instructions that tell a computing agent how to solve some problem using only finite resources

Resources Memory CPU cycles

Time/Space

Types of instructions Sequential Conditional

If statements Iterative

Loops

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Pseudocode: The Interlingua for Algorithms an English-like description of the

sequential, conditional, and iterative operations of an algorithm

no rigid syntax. As with an essay, clarity and organization are key. So is completeness.

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Pseudocode Example

Find Largest Number Input: A list of positive numbers Output: The largest number in the list Procedure: 1. Set Largest to zero 2. Set Current-Number to the first in the list 3. While there are more numbers in the list 3.1 if (the Current-Number > Largest) then 3.1.1 Set Largest to the Current-Number 3.2 Set Current-Number to the next one in the list 4. Output Largest

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Pseudocode Example Find Largest Number Input: A list of positive numbers Output: The largest number in the list Procedure: 1. Set Largest to zero 2. Set Current-Number to the first in the list 3. While there are more numbers in the list 3.1 if (the Current-Number > Largest) then 3.1.1 Set Largest to the Current-Number 3.2 Set Current-Number to the next one in the list 4. Output Largest

conditionaloperation

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iterativeoperation

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Let’s “play computer” to review this algorithm…

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Algorithms vary in efficiency example: sum the numbers from 1 to n

• space requirement is constant (i.e. independent of n)• time requirement is linear (i.e. grows linearly with n). This is written “O(n)”

efficiency space= 3 memory cells time = t(step1) + t(step 2) + n t(step 4) + n t(step 5)

1. Set sum to 0

2. Set currNum to 1

3. Repeat until currNum > n

4. Set sum to sum + currNum

5. Set currNum to currNum + 1

Algorithm I

to see this graphically...

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Algorithm Is time requirements

0 1 2 … nsize of the problem

time y = mx + btime = m n + b

The exact equation for the line is unknown because we lack precise values for the constants m and b.

But, we can say:time is a linear function of the size of the problem

time = O(n)

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Algorithm II for summation

The “key insight”, due to Gauss:the numbers can be grouped into50 pairs of the form: 1 + 100 = 101 2 + 99 = 101 . . . 50 + 51 = 101

First, consider a specific case: n = 100.

} sum = 50 x 101This algorithm

requires a singlemultiplication!

Second, generalize the formula for any (even) n :sum = (n / 2) (n + 1)

Time requirement is constant. time = O(1)

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Sequential Search: A Commonly used Algorithm

Suppose you want to a find a student in the Texas State directory.

It contains EIDs, names, phone numbers, lots of other information.

You want a computer application for searching the directory: given an EID, return the student’s phone number.

You want more, too, but this is a good start…

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Sequential Search of a student database

algorithmto search database

by EID :

1. ask user to enter EID to search for2. set i to 13. set found to ‘no’4. while i <= n and found = ‘no’ do1. if EID = EIDi then set found to ‘yes’ else increment i by 1 7. if found = ‘no’ then print “no such student” else < student found at array index i >

.

.

.

.

.

.

name EID major credit hrs.John SmithPaula JonesLed Belly

Chuck Bin

JS456PJ123LEB900

CB1235

.

.

.

.

.

.

physicshistory

music

math

36

12572

89

12

n

3

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Time requirements for sequential search

• average case (expected amount of work): EID found in studentn/2

n/2 loop iterations

because the amount of work is a constant multiple of n, the time requirement is O(n)

in the worst case and the average case.

amountof work

• best case (minimum amount of work): EID found in student1

one loop iteration• worst case (maximum amount of work): EID found in studentn

n loop iterations

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O(n) searching is too slow

Consider searching Texas State’s student database using sequential search on a computer capable of 20,000 integer comparisons per second: n = 150,000 (students registered during past 10 years) average case

150,000 comparisons 1 seconds = 3.75 seconds 2 20,000 comparisons

x

worst case

150,000 comparisons x 1 seconds = 7.5 seconds 20,000 comparisons

Bad news for searching NYC phone book, IRS database, ...

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Searching an ordered list is faster: an example of binary search

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name student major credit hrs.John SmithPaula JonesLed Belly

Chuck Bin

245763679442356

93687

.

.

.

.

.

.

physicshistory

music

math

36

12572

89

12

n

3

student24576367943845341200437564598747865492775124358925598456001160367645968675693687

12345678910111213141516

note: thestudent arrayis sorted inincreasing

order

how would you search for 58925 ?

38453 ?46589 ?

Probe 1

Probe 2

Probe 3

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The binary search algorithmassuming that the entries in student are sorted in increasing order,

1. ask user to input studentNum to search for2. set found to ‘no’3. while not done searching and found = ‘no’ 4. set middle to the index counter at the middle of the student list5. if studentNum = studentmiddle then set found to ‘yes’6. if studentNum < studentmiddle then chop off the last half

of the student list7. If studentNum > studentmiddle then chop off the first half

of the student list8. if found = ‘no’ then print “no such student” else <studentNum found at array index middle>

How?

How?

What does this mean?

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The binary search algorithmassuming that the entries in student are sorted in increasing order,

1. ask user to input studentNum to search for2. set beginning to 13. set end to n4. set found to ‘no’5. while beginning <= end and found = ‘no’ 6. set middle to (beginning + end) / 2 {round down to nearest integer}

7. if studentNum = studentmiddle then set found to ‘yes’8. if studentNum < studentmiddle then set end to middle - 19. if studentNum > studentmiddle then set beginning to middle + 110.if found = ‘no’ then print “no such student” else <studentNum found at array index middle>

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Time requirements for binary search

At each iteration of the loop, the algorithm cuts the list(i.e. the list called student) in half.

In the worst case (i.e. when studentNum is not in the list called student)

how many times will this happen?

n = 161st iteration 16/2 = 82nd iteration 8/2 = 43rd iteration 4/2 = 24th iteration 2/2 = 1

the number of times a number ncan be cut in half and not go below 1

is log2 n.Said another way:

log2 n = m is equivalent to 2m = n

In the average case and the worst case, binary search is O(log2 n)

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This is a major improvementn sequential search binary search O(n) O(log2 n)

100 100 7 150,000 150,000 1820,000,000 20,000,000 25

27=128

218=262,144

225 is about 33,000,000

number of comparisons needed in the worst case

150,000 comparisons x 1 second = 7.5 seconds 20,000 comparisons

in terms of seconds...

18 comparisons x 1 second > .001 seconds 20,000 comparisons

sequential search:

binary search:

vs.

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Sorting a listFirst, an algorithm that’s easy to write, but is badly inefficient...

unsorted sorted3546767854467811352887341

12345

initially

unsorted sorted3546767854467811352887341

12345

13528

1st iteration

unsorted sorted3546767854467811352887341

12345

1352835467

2nd iteration

unsorted sorted3546767854467811352887341

12345

135283546746781

etc.

3rd iteration

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The “Simple Sort” Algorithm

1. set i to 12. repeat until i > n 1. set indexForSmallest to the index of the smallest positive value in unsorted4. set sortedi to unsortedindexForSmallest

5. set unsortedindexForSmallest to 06. increment i

given a list of positive numbers, unsorted1, …, unsortedn andanother list, sorted1, …, sortedn, with all values initially set to zero

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This algorithm is expensive!Time requirement

total time = n iterations x time per iteration

time per iteration = time to find smallest value in a list of length n = O(n)

total time = n x O(n) = O(n2)

Space requirement

total space = space for unsorted + space for sorted = O(2n)

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Creating Algorithms is the Challenge of Computer Science

A salesperson wants to visit 25 cities while minimizing the total number of miles driven, visiting each city exactly once, and returning home again. Which route is best?

It’s not easy; try this one:The Traveling Salesperson problem:

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Simplify the Problem to get an intuition about it

A B

C D

four cities connected by roads

Q: Starting at A, what’s the shortest route that meets the requirements?A: Obvious to anyone who looks at the entire map. Not so obvious to an algorithm that “sees”, at any one time, only one city and its roads

One algorithm to answer the question: 1. generate all possible routes of length 5 2. check each path to determine whether it meets the requirement

start at A, visit B, C, and D in some order,

then return to A

How much time does this algorithm require?

25

All Paths from A of length 5 A

A D A D A D A D A D A D A D A D

B C B C B C B C

A D A D

B CA B

C D

Number of paths to generate and check is 24 = 16.

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Can you Improve the Algorithm?

Prune bad routes as soon as possible. What’s a “bad route?”

Look for good solutions, even if they’re sub-optimal. What’s a “good solution?”

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This gets real bad, real fast!

In general, the algorithm’s time requirement is: (the number of roads in&out of a city) number of cities

Assuming the number of roads is only 2, the time requirement is 2number of cities , given by the powers of 2 table.

Assuming a computer could evaluate 10 million routes per second, finding the best route for 25 cities would require about 3.5 seconds. No problem!

However, finding the best route for 64 cities would require about seconds, or hours!

20 million5000

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Comparing the time requirements

0 5 10 15n

work

35302520151050

n

log2n

n22n

order 10 50 100 1,000log2n .0003 .0006 .0007 .001n .0001 .005 .01 .1n2 .01 .25 1 1.67 min2n .1024 3570 4x1016 forget it! years centuries

time requirements for algorithmsof various orders of magnitude.

Time is in seconds, unless stated otherwise

n

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Conclusions

Algorithms are the key to creating computing solutions.

The design of an algorithm can make the difference between useful and impractical.

Algorithms often have a trade-off between space (memory) and time.

Documents

1 Algorithms and Analysis CS 2308 Foundations of CS II