Transcript

COSC 3101N J. Elder

Lecture 1. Introduction

COSC 3101N J. Elder

So you want to be a computer scientist?

COSC 3101N J. Elder

Is your goal to be a mundane programmer?

COSC 3101N J. Elder

Or a great leader and thinker?

COSC 3101N J. Elder

Original Thinking

COSC 3101N J. Elder

Given today’s prices of pork, grain, sawdust, …

Given constraints on what constitutes a hotdog.

Make the cheapest hotdog.

COSC 3101N J. Elder

Um? Tell me what to code.

With more suffocated software engineering systems,the demand for mundane programmers will diminish.

COSC 3101N J. Elder

I learned this great algorithm that will work.

Soon all known algorithms will be available in libraries.

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I can develop a new algorithm for you.

Great thinkers will always be needed.

COSC 3101N J. Elder

Content: An up to date grasp of fundamental problems and solutions

Method: Principles and techniques to solve the vast array of unfamiliar problems that arise in a rapidly changing field

The future belongs to the computer scientist who has

Rudich www.discretemath.com

COSC 3101N J. Elder

Course Content

A list of algorithms.

Learn their code.

Trace them until you are convinced that they work.

Implement them. class InsertionSortAlgorithm extends SortAlgorithm {

void sort(int a[]) throws Exception {

for (int i = 1; i < a.length; i++) {

int j = i;

int B = a[i];

while ((j > 0) && (a[j-1] > B)) {

a[j] = a[j-1];

j--; }

a[j] = B;

}}

COSC 3101N J. Elder

Course Content

A survey of algorithmic design techniques.

Abstract thinking.

How to develop new algorithms for any problem that may arise.

COSC 3101N J. Elder

Study:

Many experienced programmers were asked to code up binary search.

COSC 3101N J. Elder

Study:

Many experienced programmers were asked to code up binary search.

80% got it wrong

Good thing is was not for a nuclear power plant.

COSC 3101N J. Elder

What did they lack?

COSC 3101N J. Elder

What did they lack?

Formal proof methods?

COSC 3101N J. Elder

What did they lack?

Formal proof methods?

Yes, likely

Industry is starting to realize that formal methods are important.

But even without formal methods …. ?

COSC 3101N J. Elder

What did they lack?

Fundamental understanding of algorithmic design techniques.

Abstract thinking.

COSC 3101N J. Elder

Course Content

Notations, analogies, and abstractions

for developing,

and describing algorithms

so correctness is transparent

COSC 3101N J. Elder

Slide Credits

Some materials also due to Prof. A. Mirzaian (York).

These in turn include some slides created by Prof. S. Rudich of Carnegie Mellon University.

The slides for these 12 lectures have been adapted from slides created by Prof. J. Edmonds (York).

COSC 3101N J. Elder

COSC 3101N J. Elder

Useful Learning Techniques

COSC 3101N J. Elder

You are expected to read the textbook and lecture notes before the lecture.

This will facilitate more productive discussion during class.

COSC 3101N J. Elder

Explaining

We are going to test you on your ability to explain the material.

Hence, the best way of studying is to explain the material over and over again out loud to yourself, to each other, and to your stuffed bear.

COSC 3101N J. Elder

Day Dream

Mathematics is not all linear thinking.

Allow the essence of the material to seep into your subconscious

Pursue ideas that percolate up and flashes of inspiration that appear.

While going along with your day

COSC 3101N J. Elder

Be Creative

Why is it done this way and not that way?

COSC 3101N J. Elder

Guesses and Counter Examples

Guess at potential algorithms for solving a problem.

Look for input instances for which your algorithm gives the wrong answer.

COSC 3101N J. Elder

Refinement:

Rudich www.discretemath.com

The best solution comes from a process of repeatedly refining and inventing alternative solutions

COSC 3101N J. Elder

Grade School Revisited:How To Multiply Two Numbers

2 X 2 = 5

A Few Example A Few Example AlgorithmsAlgorithms

Rudich

COSC 3101N J. Elder

Complex Numbers

Remember how to multiply 2 complex numbers?

(a+bi)(c+di) = [ac –bd] + [ad + bc] i

If a real multiplication costs \$1 and an addition cost a penny. What is the cheapest way to obtain the output from the input?

Can you do better than \$4.02?

COSC 3101N J. Elder

m1 = ac

m2 = bd

A1 = m1 – m2 = ac-bd

m3 = (a+b)(c+d) = ac + ad + bc + bd

A2 = m3 – m1 – m2 = ad+bc

Gauss’ \$3.05 Method:

COSC 3101N J. Elder

Question:

The Gauss method saves one multiplication out of four. It requires 25% less work.

Could there be a context where performing 3 multiplications for every 4 provides a more dramatic savings?

COSC 3101N J. Elder

Odette Bonzo

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How to add 2 n-bit numbers.

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How to add 2 n-bit numbers.

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How to add 2 n-bit numbers.

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COSC 3101N J. Elder

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COSC 3101N J. Elder

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COSC 3101N J. Elder

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= θ(n) = linear time.

On any reasonable computer adding 3 bits can be done in constant time.

COSC 3101N J. Elder

# of bits in numbers

tim

ef = θ(n) means that f can be

sandwiched between two lines

COSC 3101N J. Elder

Is there a faster way to add?

QUESTION: Is there an algorithm to add two n-bit numbers whose time grows sub-linearly in n?

COSC 3101N J. Elder

Suppose there is a mystery algorithm that does not examine each bit

Give the algorithm a pair of numbers. There must be some unexamined bit position i in one of the numbers

If the algorithm is not correct on the numbers, we found a bug

If the algorithm is correct, flip the bit at position i and give the algorithm the new pair of numbers. It give the same answer as before so it must be wrong since the sum has changed

COSC 3101N J. Elder

So any algorithm for addition must use time at least linear in

the size of the numbers.

COSC 3101N J. Elder

How to multiply 2 n-bit numbers.

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n2

COSC 3101N J. Elder

How to multiply 2 nHow to multiply 2 n--bit numbers.bit numbers.

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n2

I get it! The total time is bounded by

cn2.

COSC 3101N J. Elder

No matter how dramatic the difference in the constants the quadratic curve will eventually dominate the linear curve

# of bits in numbers

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COSC 3101N J. Elder

Neat! We have demonstrated that as things scale multiplication is a

Mathematical confirmation of our common sense.

COSC 3101N J. Elder

grade school addition. This is a comparison of the complexity of two

algorithms.

To argue that multiplication is an inherently harder problem than

addition we would have to show that no possible multiplication algorithm runs in

linear time.

COSC 3101N J. Elder

Is there a clever algorithm to multiply two numbers

in linear time?

COSC 3101N J. Elder

Despite years of research, no one knows! If you resolve this question,

York will give you a PhD!

COSC 3101N J. Elder

Is there a faster way to multiply two numbers

than the way you learned in grade school?

COSC 3101N J. Elder

Divide And Conquer (an approach to faster algorithms)

DIVIDE a problem into smaller subproblems

CONQUER them recursively

GLUE the answers together so as to obtain the answer to the larger problem

COSC 3101N J. Elder

Multiplication of 2 n-bit numbers

X =

Y =

X = a 2n/2 + b Y = c 2n/2 + d

XY = ac 2n + (ad+bc) 2n/2 + bd

a b

c d

COSC 3101N J. Elder

Multiplication of 2 n-bit numbers

X =

Y =

XY = ac 2n + (ad+bc) 2n/2 + bd

a b

c d

MULT(X,Y):

If |X| = |Y| = 1 then RETURN XY

Break X into a;b and Y into c;d

RETURN

MULT(a,c) 2n + (MULT(a,d) + MULT(b,c)) 2n/2 + MULT(b,d)

COSC 3101N J. Elder

Time required by MULT

T(n) = time taken by MULT on two n-bit numbers

What is T(n)? What is its growth rate? Is it θ(n2)?

COSC 3101N J. Elder

Recurrence Relation

T(1) = k for some constant k

T(n) = 4 T(n/2) + k’ n + k’’ for some constants k’ and k’’

MULT(X,Y):

If |X| = |Y| = 1 then RETURN XY

Break X into a;b and Y into c;d

RETURN

MULT(a,c) 2n + (MULT(a,d) + MULT(b,c)) 2n/2 + MULT(b,d)

COSC 3101N J. Elder

Let’s be concrete

T(1) = 1

T(n) = 4 T(n/2) + n

How do we unravel T(n) so that we can determine its growth rate?

COSC 3101N J. Elder

Technique 1: (Substitution)

Recurrence Relation:

T(1) = 1 & T(n) = 4T(n/2) + n

Guess: G(n) = 2n2 – n

Verify: Left Hand Side Right Hand Side

T(1) = 2(1)2 – 1

T(n) = 2n2 – n

1

4T(n/2) + n

= 4 [2(n/2)2 – (n/2)] + n

= 2n2 – n

COSC 3101N J. Elder

Technique 2: Recursion Tree

• T(n) = n + 4 T(n/2)

•T(1) = 1

n

T(n/2) T(n/2) T(n/2) T(n/2)

= n

T(n/2) T(n/2) T(n/2) T(n/2)

=T(n)T(n)

T(1) 1=

COSC 3101N J. Elder

n

T(n/2) T(n/2) T(n/2) T(n/2)

T(n) =

COSC 3101N J. Elder

n

T(n/2) T(n/2) T(n/2)

T(n) =

n/2

T(n/4)T(n/4)T(n/4)T(n/4)

COSC 3101N J. Elder

nT(n) =

n/2

T(n/4)T(n/4)T(n/4)T(n/4)

n/2

T(n/4)T(n/4)T(n/4)T(n/4)

n/2

T(n/4)T(n/4)T(n/4)T(n/4)

n/2

T(n/4)T(n/4)T(n/4)T(n/4)

COSC 3101N J. Elder

nT(n) =

n/2 n/2 n/2n/2

11111111111111111111111111111111 . . . . . . 111111111111111111111111111111111

n/4 n/4 n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4 n/4

COSC 3101N J. Elder

Level i is the sum of 4i copies of n/2i

1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

. . . . . . . . . . . . . . . . . . . . . . . . . .

n/2 + n/2 + n/2 + n/2

n

n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4

0

1

2

i

log n2

COSC 3101N J. Elder

Level i is the sum of 4 i copies of n/ 2 i

1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

. . . . . . . . . . . . . . . . . . . . . . . . . .

n/ 2 + n/ 2 + n/ 2 + n/ 2

n

n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4

0

1

2

i

lo g n2

=1n

= 4n/2

= 16n/4

= 4i n/2i

Total: θ(nlog4) = θ(n2)

= 4lognn/2logn

=nlog41

COSC 3101N J. Elder

Divide and Conquer MULT: θ(n2) time Grade School Multiplication: θ(n2) time

All that work for nothing!

COSC 3101N J. Elder

MULT revisited

MULT calls itself 4 times. Can you see a way to reduce the number of calls?

MULT(X,Y):

If |X| = |Y| = 1 then RETURN XY

Break X into a;b and Y into c;d

RETURN

MULT(a,c) 2n + (MULT(a,d) + MULT(b,c)) 2n/2 + MULT(b,d)

COSC 3101N J. Elder

Gauss’ Idea: Input: a,b,c,d Output: ac, ad+bc, bd

A1 = ac

A3 = bd

m3 = (a+b)(c+d) = ac + ad + bc + bd

A2 = m3 – A1- A3 = ad + bc

COSC 3101N J. Elder

Gaussified MULT (Karatsuba 1962)

T(n) = 3 T(n/2) + n

Actually: T(n) = 2 T(n/2) + T(n/2 + 1) + kn

MULT(X,Y):

If |X| = |Y| = 1 then RETURN XY

Break X into a;b and Y into c;d

e = MULT(a,c) and f =MULT(b,d)

RETURN e2n + (MULT(a+b, c+d) – e - f) 2n/2 + f

COSC 3101N J. Elder

nT(n) =

n/2

T(n/4)T(n/4)T(n/4)T(n/4)

n/2

T(n/4)T(n/4)T(n/4)T(n/4)

n/2

T(n/4)T(n/4)T(n/4)T(n/4)

n/2

T(n/4)T(n/4)T(n/4)T(n/4)

COSC 3101N J. Elder

n

T(n/2) T(n/2) T(n/2)

T(n) =

COSC 3101N J. Elder

n

T(n/2) T(n/2)

T(n) =

n/2

T(n/4)T(n/4)T(n/4)

COSC 3101N J. Elder

nT(n) =

n/2

T(n/4)T(n/4)T(n/4)

n/2

T(n/4)T(n/4)T(n/4)

n/2

T(n/4)T(n/4)T(n/4)

COSC 3101N J. Elder

Level i is the sum of 3i copies of n/2i

1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

. . . . . . . . . . . . . . . . . . . . . . . . . .

n/2 + n/2 + n/2

n

n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4

0

1

2

i

log n2

COSC 3101N J. Elder

=1n

= 3n/2

= 9n/4

= 3i n/2i

Total: θ(nlog3) = θ(n1.58..)

Level i is the sum of 3 i copies of n/ 2 i

1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1

. . . . . . . . . . . . . . . . . . . . . . . . . .

n/ 2 + n/ 2 + n/ 2

n

n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4 + n/ 4

0

1

2

i

lo g n2 = 3lognn/2logn

=nlog31

COSC 3101N J. Elder

Dramatic improvement for large n

Not just a 25% savings!

θ(n2) vs θ(n1.58..)

COSC 3101N J. Elder

Multiplication Algorithms

Kindergarten n2n