IntroductionPizzanu Kanongchaiyos, Ph.D.pizzanu@cp.eng.chula.ac.th

What is Algorithm?A detailed sequence of actions to perform to accomplish some task. Named after an Iranian mathematician, Al-Khawarizmi.algorithm[N] algorithm[N] algorithm(' ) n. ., S. algorism, -algorithmic adj.

So you want to be a computer scientist?

Is your goal to be a mundane programmer?

Or a great leader and thinker?

Original Thinking

Boss assigns task:Given todays prices of pork, grain, sawdust, Given constraints on what constitutes a hotdog.Make the cheapest hotdog.

Everyday industry asks these questions.

Your answer:Um? Tell me what to code.With more suffocated software engineering systems, the demand for mundane programmers will diminish.

Your answer:I learned this great algorithm that will work.Soon all known algorithms will be available in libraries.

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

Content: An up to date grasp of fundamental problems and solutionsMethod: Principles and techniques to solve the vast array of unfamiliar problems that arise in a rapidly changing field

Rudich www.discretemath.com

Course ContentA 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 ContentA survey of algorithmic design techniques.Abstract thinking.How to develop new algorithms for any problem that may arise.

Study:Many experienced programmers were asked to code up binary search.

Study:Many experienced programmers were asked to code up binary search.80% got it wrongGood thing is was not for a nuclear power plant.

What did they lack?

What did they lack?Formal proof methods?

What did they lack?Formal proof methods?Yes, likelyIndustry is starting to realize that formal methods are important.But even without formal methods . ?

What did they lack?Fundamental understanding of the algorithmic design techniques.Abstract thinking.

Course ContentNotations, analogies, and abstractions for developing, thinking about, and describing algorithms so correctness is transparent

A survey of fundamental ideas and algorithmic design techniquesFor example . . .

Start With Some Math

Iterative Algorithms Loop Invariants codeA loop exit when codeBcodeC

CodeRelay RaceOne step at a time

Recursive Algorithms

Graph Search Algorithms

Network Flows

Greedy Algorithms

Recursive Back Tracking

Dynamic Programing

Reduction=Rudich www.discretemath.com

Useful Learning Techniques

Read AheadYou are expected to read the lecture notes before the lecture.This will facilitate more productive discussion during class.

ExplainingWe 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.

Day DreamMathematics 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

Be CreativeAsk questions. Why is it done this way and not that way?

Guesses and Counter ExamplesGuess at potential algorithms for solving a problem. Look for input instances for which your algorithm gives the wrong answer.

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

Grade School Revisited:How To Multiply Two NumbersA Few Example Algorithms

Complex NumbersRemember 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 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: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?

OdetteBonzo

How to add 2 n-bit numbers. * * * * * * * * * * * * * * * * * * * * * * +

How to add 2 n-bit numbers. * * * * * * * * * * * * * * * * * * * * * * * * +

How to add 2 n-bit numbers. * * * * * * * * * * * * * * * * * * * * * * * * * * +

How to add 2 n-bit numbers. * * * * * * * * * * * * * * * * * * * * * * * * * * * * +

How to add 2 n-bit numbers. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * +

How to add 2 n-bit numbers. * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * +* *

Time complexity of grade school additionT(n) = The amount of time grade school addition uses to add two n-bit numbers= (n) = linear time.On any reasonable computer adding 3 bits can be done in constant time.

# of bits in numberst imef = (n) means that f can be sandwiched between two lines

Please feel free to ask questions!Rudich www.discretemath.com

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 bitsSuppose 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 numbersIf the algorithm is not correct on the numbers, we found a bugIf 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

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.X* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

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

How to multiply 2 n-bit numbers.

X

* * * * * * * *

* * * * * * * *

Grade School Addition: Linear timeGrade School Multiplication: Quadratic timeNo matter how dramatic the difference in the constants the quadratic curve will eventually dominate the linear curve# of bits in numberst ime

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

Dont 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) timeGrade School Multiplication: (n2) timeIs there a clever algorithm to multiply two numbers in linear time?

Despite years of research, no one knows! If you resolve this question, Chula will give you a PhD!

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 a problem into smaller subproblems

CONQUER them recursively

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

Multiplication of 2 n-bit numbersX = Y =

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

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

Multiplication of 2 n-bit numbersX = Y = XY = ac 2n + (ad+bc) 2n/2 + bd abcdMULT(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 MULTT(n) = time taken by MULT on two n-bit numbers What is T(n)? What is its growth rate? Is it (n2)?

Recurrence RelationT(1) = k for some constant k

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

Lets be concreteT(1) = 1T(n) = 4 T(n/2) + n

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

Technique 1Guess and VerifyRecurrence Relation: T(1) = 1 & T(n) = 4T(n/2) + nGuess: G(n) = 2n2 nVerify:

Left Hand SideRight 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

Technique 2: Decorate The TreeT(n) = n + 4 T(n/2)nT(n/2)T(n/2)T(n/2)T(n/2)T(n)=nT(n/2)T(n/2)T(n/2)T(n/2)T(n)=T(1)T(1) = 11=

nT(n/2)T(n/2)T(n/2)T(n/2)T(n)=

nT(n/2)T(n/2)T(n/2)T(n)=

nT(n)=

nT(n)=n/2n/2n/2n/211111111111111111111111111111111 . . . . . . 111111111111111111111111111111111n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4n/4

012i

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

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

=1n= 4n/2= 16n/4= 4i n/2iTotal: (nlog4) = (n2) = 4lognn/2logn =nlog41

0

1

2

i

Divide and Conquer MULT: (n2) time Grade School Multiplication: (n2) timeAll that work for nothing!

MULT revisitedMULT 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)

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)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

nT(n)=

nT(n/2)T(n/2)T(n/2)T(n)=

nT(n/2)T(n/2)T(n)=

nT(n)=

n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4 + n/4012i

nn/2 + n/2 + n/2

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+1

=1n= 3n/2= 9n/4= 3i n/2iTotal: (nlog3) = (n1.58..) = 3lognn/2logn =nlog31

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

0

1

2

i

Dramatic improvement for large nNot just a 25% savings! (n2) vs (n1.58..)

Multiplication Algorithms

Brute ForceSolving a problem by methodically plowing through all the possibilities. For example, solving the EightQueensProblem by generating all possible board positions. Does not scale well.

EightQueensProblemPlace eight queens on a chess board so that no queen attacks any other.

To solve the eight queens problem, you must place 8 queens (black squares) onto the chess board below. No queen can be attacking any other queen. (A queen can attack all cells on its vertical, horizontal, and diagonal lines of sight) Using recursion, the board can be solved from the top down, trying all possibilities and backtracking (known as a depth first search). http://spaz.ca/aaron/SCS/queens/index.html

Check StringCheck Pattern gcagagag in string gcatcgcagagagtatacagtacg

Brute Force algorithmMain featuresno preprocessing phase; constant extra space needed; always shifts the window by exactly 1 position to the right; comparisons can be done in any order; searching phase in O(mn) time complexity; 2n expected text characters comparisons

Cvoid BF(char *x, int m, char *y, int n) { int i, j; /* Searching */ for (j = 0; j = m) OUTPUT(j); } }

Brute Force algorithmThe brute force algorithm consists in checking, at all positions in the text between 0 and n-m, whether an occurrence of the pattern starts there or not. Then, after each attempt, it shifts the pattern by exactly one position to the right.The brute force algorithm requires no preprocessing phase, and a constant extra space in addition to the pattern and the text. During the searching phase the text character comparisons can be done in any order. The time complexity of this searching phase is O(mn) (when searching for am-1b in an for instance). The expected number of text character comparisons is 2n.To sum up ! A straightforward approach to solving a problem, usually directly based on the problems statement and definitions of the concepts involved. Examples: selection sort, sequential search Important special case: exhaustive search

Transform and ConquerSolve the problem by transforming it to: a simpler instance of the same problem presorting Gaussian elimination another representation of the same problem. Horners rule FFT another problem altogether path counting by computing powers of adjacency Matrix math modeling such as linear programming

The Game of FifThe board consists of a row of nine squares numbered 1 through 9. Players take turns selecting squares. The goal of the game is to select squares such that among those selected by a single player there will be a triplet that sums up to 15 1 2 3 4 5 6 7 8 9

Theory of the GameThe game is equivalent to playing a regular TicTacToe game on the magic square board. (The square is magic in that the sums of three numbers on any straight line (vertical, horizontal or diagonal) is always 15.) There are eight such lines:

Theory of the GameThese are all possible representations of 15 with three smaller positive integers.As is well known, a TicTacToe game ends in a stalemate unless one of the players makes a mistake. In the Fif game, computer has the advantage of knowing the origin of the games and having in front of its Mind's eye the magic square board. To make the game more appealing computer errs with the probability of 1/20.

ReductionThe process of transforming an expression according to certain reduction rules.

Find medianThe statistical median is middle number of a group of numbers that have been arranged in order by size. If there is an even number of terms, the median is the mean of the two middle numbers

Median of mediansLine up elements in groups of five (this number 5 is not important, it could be e.g. 7 without changing the algorithm much). Call each group S[i], with i ranging from 1 to n/5. Find the median of each group. (Call this x[i]). This takes 6 comparisons per group, so 6n/5 total (it is linear time because we are taking medians of very small subsets). Find the median of the x[i], using a recursive call to the algorithm. If we write a recurrence in which T(n) is the time to run the algorithm on a list of n items, this step takes time T(n/5). Let M be this median of medians. Use M to partition the input and call the algorithm recursively on one of the partitions, just like in quickselect.

Divide-and-Conquer Given an instance of the problem to be solved, split this into several, smaller, sub-instances (of the same problem) independently solve each of the sub-instances and then combine the sub-instance solutions so as to yield a solution for the original instance. This description raises the question:

By what methods are the sub-instances to be independently solved?

Divide-and-ConquerConsider the following: We have an algorithm, alpha say, which is known to solve all problem instances of size n in at most c n^2 steps (where c is some constant). We then discover an algorithm, beta say, which solves the same problem by: Dividing an instance into 3 sub-instances of size n/2. Solves these 3 sub-instances. Combines the three sub-solutions taking d n steps to do this.

Divide-and-Conquer Suppose our original algorithm, alpha, is used to carry out the `solves these sub-instances' step 2. Let T(alpha)( n ) = Running time of alpha T(beta)( n ) = Running time of beta Then, T(alpha)( n ) = c n^2 (by definition of alpha) But

Divide-and-Conquer T(beta)( n ) = 3 T(alpha)( n/2 ) + d n = (3/4)(cn^2) + dn

So if dn < (cn^2)/4 (i.e. d < cn/4) then beta is faster than alpha

Divide-and-Conquer In particular for all large enough n, (n > 4d/c = Constant), beta is faster than alpha. This realization of beta improves upon alpha by just a constant factor. But if the problem size, n, is large enough then n > 4d/c n/2 > 4d/c ... n/2^i > 4d/c which suggests that using beta instead of alpha for the `solves these' stage repeatedly until the sub-sub-sub..sub-instances are of size n0 < = (4d/c) will yield a still faster algorithm.

Divide-and-Conquer So consider the following new algorithm for instances of size n procedure gamma (n : problem size ) is begin if n

Divide-and-Conquer Let T(gamma)(n) denote the running time of this algorithm. cn^2 if n < = n0 T(gamma)(n) = 3T(gamma)( n/2 )+dn otherwise We shall show how relations of this form can be estimated later in the course. With these methods it can be shown that T(gamma)( n ) = O( n^{log3} ) (=O(n^{1.59..}) This is an asymptotic improvement upon algorithms alpha and beta.

Divide-and-Conquer The improvement that results from applying algorithm gamma is due to the fact that it maximizes the savings achieved beta. The (relatively) inefficient method alpha is applied only to "small" problem sizes. The precise form of a divide-and-conquer algorithm is characterized by: The threshold input size, n0, below which the problem size is not sub-divided. The size of sub-instances into which an instance is split. The number of such sub-instances. The algorithm used to combine sub-solutions.

Divide-and-Conquer it is more usual to consider the ratio of initial problem size to sub-instance size. In our example this was 2. The threshold is sometimes called the (recursive) base value. In summary, the generic form of a divide-and-conquer algorithm is: procedure D-and-C (n : input size) is begin if n < = n0 then Solve problem without further sub-division; else Split into r sub-instances each of size n/k; for each of the r sub-instances do D-and-C (n/k); Combine the r resulting sub-solutions to produce the solution to the original problem; end if; end D-and-C; Such algorithms are naturally and easily realized as: Recursive Procedures

EndRestart IntroductionRelevant Mathematics