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Introduction. Thinking about Algorithms Abstractly. Credits: Steve Rudich, Jeff Edmond, Ping Xuan. So you want to be a bioinformatician /mathematician /computer scientist? What is an Algorithm ? Grade school revisited: How to multiply two numbers. So you want to be a computer scientist?.

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Introduction

Introduction

Thinking about Algorithms Abstractly

Credits: Steve Rudich, Jeff Edmond, Ping Xuan


Introduction

  • So you want to be a bioinformatician /mathematician /computer scientist?

  • What is an Algorithm ?

  • Grade school revisited: How to multiply two numbers


So you want to be a computer scientist

So you want to be a computer scientist?


Is your goal to be a mundane programmer

Is your goal to be a mundane programmer?


Introduction

Or a great leader and thinker?


Original thinking

Original Thinking


Boss assigns task

Boss assigns task:

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

  • Given constraints on what constitutes a hotdog.

  • Make the cheapest hotdog.

Everyday industry asks these questions.


Your answer

Your answer:

  • Um? Tell me what to code.

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


Your answer1

Your answer:

  • I learned this great algorithm that will work.

Soon all known algorithms will be available in libraries.


Your answer2

Your answer:

  • I can develop a new algorithm for you.

Great thinkers will always be needed.


The future belongs to the computer scientist who has

The future belongs to the computer scientist who has

  • 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


Course content

Course Content

  • A list of algoirthms.

    • Learn their code.

    • Trace them until you are convenced that they work.

    • Impliment 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;

}}


Course content1

Course Content

  • A survey of algorithmic design techniques.

  • Abstract thinking.

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


Study

Study:

  • Many experienced programmers were asked to code up binary search.


Study1

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.


What did they lack

What did they lack?


What did they lack1

What did they lack?

  • Formal proof methods?


What did they lack2

What did they lack?

  • Formal proof methods?

Yes, likely

Industry is starting to realize that formal methods are important.

But even without formal methods …. ?


What did they lack3

What did they lack?

  • Fundamental understanding of the algorithmic design techniques.

  • Abstract thinking.


Course content2

Course Content

Notations, analogies, and abstractions

for developing,

thinking about,

and describing algorithms


You will see some math

Time Complexity

t(n) = Q(n2)

Recurrence Relations

T(n) = a T(n/b) + f(n)

You will see some Math …


What is an algorithm

What is an Algorithm?

  • A step-by-step description of procedures performing certain task.

    • Example: Sorting.

      Given a list of numbers, put them in increasing

      (non-decreasing) order.

  • Properties: Generality, Termination, Correctness, Feasibility.

  • Feasibility  Analysis of Algorithm

    Complexity theory


Analysis of algorithm

Analysis of Algorithm

  • Running time analysis

    • Input size N

    • Running time is a function of N: T(N)

      • Worst case, average case, … , etc

      • Time measured by number of computer operations

        • Each basic operation (add, load, write, etc) count as 1

      • Actual clock time differs by a constant factor

      • Usually very complex

    • The use of asymptotic bounds

      • To study the rate of growth of T(N) compared to simpler functions f(N), such as N, N2, log(N), etc

      • A constant factor can be ignored


Some definitions big o notation

Some Definitions: Big O Notation

  • T(N) = O( f(N) )

    • Exists constant c and n0 such that when N>n0, T(N) <= c * f(N)

    • Asymptotic Upper bound

c f(N)

T(N)

n0

N


Some definitions big omega

Some Definitions: Big Omega

  • T(N) = (g(N))

    • Exists constant c and n0 such that when N>n0, T(N) >= c * g(N)

    • Asymptotic Lower bound

T(N)

c g(N)

n0

N


Some definitions big theta

Some Definitions: Big Theta

  • T(N) = ( h(N) )

    • if and only if T(N)=O(h(N)) and T(N)= (h(N))

    • tight bound

c1 h(N)

T(N)

c2 h(N)

N


Some definitions little o

c p(N)

T(N)

N

Some Definitions: Little “o”

  • T(N) = o(p(N))

    if lim N  ∞ = o. E.g. log(N) = o(N).


Example insertion sort algorithm

Example: Insertion Sort Algorithm

  • 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;

  • }}


Iterative algorithms

0

i-1

i

T+1

i

i

9 km

5 km

Iterative Algorithms

<preCond>

codeA

loop

exit when <exit Cond>

codeB

codeC

<postCond>

One step at a time

Relay Race

Code


Introduction

52

88

14

14,23,25,30,31,52,62,79,88,98

31

98

25

30

23

62

79

Problem Specification

  • Precondition: The input is a list of n values with the same value possibly repeated.

  • Post condition: The output is a list consisting of the same n values in non-decreasing order.


Define step

30

14

25

23,31,52,88

98

79

Define Step

  • Select arbitrary element from side.

  • Insert it where it belongs.

30

14

62

25

98

79

23,31,52,62,88


Making progress while maintaining the loop invariant

Exit

79 km

75 km

6 elements

to school

5 elements

to school

30

14

25

23,31,52,88

98

79

Making progress while Maintaining the loop invariant

30

14

62

25

98

79

23,31,52,62,88

Sorted sub-list


Introduction

52

52

88

88

14

14

31

31

62

62

25

25

30

30

Exit

Exit

0 km

23

23

98

98

79

79

n elements

to school

0 elements

to school

14,23,25,30,31,52,62,79,88,98

14,23,25,30,31,52,62,79,88,98

Beginning &

Ending


Running time

30

14

25

23,31,52,88

98

79

Running Time

Inserting an element into a list of size i

takes O(i) time in the worst case.

Total = 1+2+3+…+n = n(n+1)/2 = (n2)

30

14

62

25

98

79

23,31,52,62,88


Explaining insertion sort

Explaining Insertion Sort

We maintain a subset of elements sorted within alist. The remaining elements are off to the sidesomewhere. Initially,think of the first element in the array as a sorted list of lengthone. One at a time, we take one of the elements that is off to theside and we insert it into the sorted list where itbelongs. This gives a sorted list that is one element longer than itwas before. When the last element has been inserted, the array iscompletely sorted.


Insertion sort

23,25,31,52,88

52,23,88,31,25,30,98,62,14,79

23,31,52,88

Insertion Sort

The input consists of an array of integers

We read in the i+1st object.

We will pretend that this larger prefix is the entire input.

We extend the solution we have by one for this larger prefix.


Insertion sort algorithm pseudo code

Insertion Sort Algorithm: Pseudo-Code

  • Input: an array (or list) a of numbers

    // data structure: a[0] …a[n-1]

  • Operations:

    • Leave the first element (location 0) untouched.

    • for each element from location 1 on

      insert it to the appropriate place in the

      (sorted) sub-array to its left.

a

0

1

2

n-1


Operations in an algorithm

Operations in an Algorithm

  • Decision Making

    • branching operation

    • if … then … else …

  • Repetition

    • loops: for each element do { … }

    • conditional loops: while ( …. ) do { … }


Insertion sort algorithm code

Insertion Sort Algorithm: Code

  • 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;

  • }}}


Grade school revisited how to multiply two numbers

2 X 2 = 5

Another Algorithm

Grade School Revisited:How To Multiply Two Numbers


Complex numbers

Complex Numbers

  • Remember how to multiply 2 complex numbers?

  • (a+bi)(c+di) = [ac –bd] + [ad + bc] i

  • Input: a,b,c,d Output: ac-bd, ad+bc

  • 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?


Gauss 3 05 method input a b c d output ac bd ad bc

Gauss’ $3.05 Method:Input: a,b,c,d Output: ac-bd, ad+bc

  • m1 = ac

  • m2 = bd

  • A1 = m1 – m2 = ac-bd

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

  • A2 = m3 – m1 – m2 = ad+bc


Question

Question:

  • The Gauss “hack” 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?


Introduction

Odette

Bonzo


How to add 2 n bit numbers

How to add 2 n-bit numbers.

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**

+


How to add 2 n bit numbers1

How to add 2 n-bit numbers.

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**

*

+


How to add 2 n bit numbers2

How to add 2 n-bit numbers.

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*

**

*

+


How to add 2 n bit numbers3

How to add 2 n-bit numbers.

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**

* **

* **

*

* **

*

**

*

+


How to add 2 n bit numbers4

How to add 2 n-bit numbers.

**

**

**

**

**

**

* **

* **

*

* **

*

* **

*

**

*

+


How to add 2 n bit numbers5

How to add 2 n-bit numbers.

*

*

* **

*

* **

*

* **

*

* **

*

* **

*

***

*

* **

*

* **

*

* **

*

* **

*

**

*

+


Time complexity of grade school addition

*

*

* **

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* **

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* **

*

* **

*

* **

*

* **

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+

Time complexity of grade school addition

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

T(n) = The amount of time grade school addition uses to add two n-bit numbers

= θ(n) = linear time.


Introduction

f = θ(n) means that f can be sandwiched between two lines

time

# of bits in numbers


Introduction

Please feel free

to ask questions!


Is there a faster way to add

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?


Any algorithm for addition must read all of the input bits

Any algorithm for addition must read all of the input bits

  • 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


Introduction

So any algorithm for addition must use time at least linear in the size of the numbers.

Grade school addition is essentially as good as it can be.


How to multiply 2 n bit numbers

n2

How to multiply 2 n-bit numbers.

* * * * * * * *

X

* * * * * * * *

* * * * * * * *

* * * * * * * *

* * * * * * * *

* * * * * * * *

* * * * * * * *

* * * * * * * *

* * * * * * * *

* * * * * * * *

* * * * * * * * * * * * * * * *


Introduction

I get it! The total time is bounded by cn2.


Grade school addition linear time grade school multiplication quadratic time

Grade School Addition: Linear timeGrade School Multiplication: Quadratic time

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

time

# of bits in numbers


Introduction

Neat! We have demonstrated that multiplication is a harder problem than addition.Mathematical confirmation of our common sense.


Introduction

Don’t jump to conclusions!We have argued that grade school multiplication uses more time than 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.


Grade school addition n time grade school multiplication n 2 time

Grade School Addition: θ(n) timeGrade School Multiplication: θ(n2) time

Is there a clever algorithm to multiply two numbers in linear time?


Introduction

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

Tunghai will give you a PhD!


Introduction

Is there a faster way to multiply two numbers than the way you learned in grade school?


Divide and conquer an approach to faster algorithms

Divide and Conquer(an approach to faster algorithms)

  • DIVIDE my instance to the problem into smaller instances to the same problem.

  • Have a friend (recursively) solve them.Do not worry about it yourself.

  • GLUEthe answers together so as to obtain the answer to your larger instance.


Multiplication of 2 n bit numbers

Multiplication of 2 n-bit numbers

a

b

  • X =

  • Y =

  • X = a 2n/2 + b Y = c 2n/2 + d

  • XY = ac 2n + (ad+bc) 2n/2 + bd

c

d


Multiplication of 2 n bit numbers1

Multiplication of 2 n-bit numbers

a

b

c

d

  • X =

  • Y =

  • XY = ac 2n + (ad+bc) 2n/2 + bd

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)


Time required by mult

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)?


Recurrence relation

Recurrence Relation

  • T(1) = k for some constant k

  • T(n) = 4 T(n/2) + k n + l for some constants k and l

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)


Let s be concrete

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?


Technique 1 guess and verify

Technique 1Guess and Verify

  • Recurrence Relation:

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

  • Guess: G(n) = 2n2 – n

  • Verify:


Technique 2 decorate the tree

  • T(1) = 1

Technique 2: Decorate The Tree

T(1)

=

1

  • T(n) = n + 4 T(n/2)

  • T(n) = n + 4 T(n/2)

T(n)

T(n)

n

n

=

=

T(n/2)

T(n/2)

T(n/2)

T(n/2)

T(n/2)

T(n/2)

T(n/2)

T(n/2)


Introduction

T(n)

n

=

T(n/2)

T(n/2)

T(n/2)

T(n/2)


Introduction

n/2

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n)

n

=

T(n/2)

T(n/2)

T(n/2)


Introduction

n/2

n/2

n/2

n/2

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n)

n

=


Introduction

T(n)

n

=

n/2

n/2

n/2

n/2

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

11111111111111111111111111111111 . . . . . . 111111111111111111111111111111111


Introduction

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


Introduction

=1×n

= 4×n/2

= 16×n/4

= 4i ×n/2i

= 4logn×n/2logn =nlog4×1

Total: θ(nlog4) = θ(n2)


Divide and conquer mult n 2 time grade school multiplication n 2 time

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

All that work for nothing!


Mult revisited

MULT revisited

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)

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


Gauss hack input a b c d output ac ad bc bd

Gauss’ Hack: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


Gaussified mult karatsuba 1962

Gaussified MULT(Karatsuba 1962)

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

  • T(n) = 3 T(n/2) + n

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


Introduction

n/2

n/2

n/2

n/2

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n)

n

=


Introduction

T(n)

n

=

T(n/2)

T(n/2)

T(n/2)


Introduction

n/2

T(n/4)

T(n/4)

T(n/4)

T(n)

n

=

T(n/2)

T(n/2)


Introduction

n/2

n/2

n/2

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n/4)

T(n)

n

=


Introduction

0

1

2

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

i


Introduction

=1×n

= 3×n/2

= 9×n/4

= 3i ×n/2i

= 3logn×n/2logn =nlog3×1

Total: θ(nlog3) = θ(n1.58..)


Dramatic improvement for large n

Dramatic improvement for large n

Not just a 25% savings!

θ(n2) vs θ(n1.58..)


Multiplication algorithms

Homework

Multiplication Algorithms

3*4=3+3+3+3


Introduction

You’re cool! Are you free sometime this weekend?

Not interested, Bonzo. I took the initiative and asked out a guy in my 4277 class.


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