Data Structures

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Data Structures

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The logical or mathematical model of a particular organization of data is called a data structure

DATA STRUCTURES

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A primitive data type holds a single piece of data–e.g. in Java: int, long, char, boolean etc.–Legal operations on integers: + - * / ...

A data structure structures data!–Usually more than one piece of data–Should provide legal operations on the data–The data might be joined together (e.g. in an array): a collection

DATA STRUCTURES

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Static vs. Dynamic StructuresStatic vs. Dynamic Structures

A static data structure has a fixed size

This meaning is different than those associated with the static modifier

Arrays are static; once you define the number of elements it can hold, it doesn’t change

A dynamic data structure grows and shrinks as required by the information it contains

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An Abstract Data Type (ADT) is a data type together with the operations, whose properties are specified independently of any particular implementation.

Abstract Data Type

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Abstract Data Type

In computing, we view data from three perspectives:

Application level

View of the data within a particular problem

Logical level

An abstract view of the data values (the domain) and the set of operations to manipulate them

Implementation level

A specific representation of the structure to hold the data items and the coding of the operations in a programming language

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Problem Solving: Main Steps

1. Problem definition

2. Algorithm design / Algorithm specification

3. Algorithm analysis

4. Implementation

5. Testing

6. [Maintenance]

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Problem Definition

What is the task to be accomplished?

Calculate the average of the grades for a given student

What are the time / space / speed / performance requirements?

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. Algorithm Design / Specifications

Algorithm: Finite set of instructions that, if followed, accomplishes a particular task.Describe: in natural language / pseudo-code / diagrams / etc. Criteria to follow:

Input: Zero or more quantities (externally produced)Output: One or more quantities Definiteness: Clarity, precision of each instructionFiniteness: The algorithm has to stop after a finite (may be very large) number of stepsEffectiveness: Each instruction has to be basic enough and feasible

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Implementation, Testing, Maintenances

Implementation

Decide on the programming language to use

C, C++, Lisp, Java, Perl, Prolog, assembly, etc. , etc.

Write clean, well documented code

Test, test, test

Integrate feedback from users, fix bugs, ensure compatibility across different versions Maintenance

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Algorithm Analysis

Space complexity

How much space is required

Time complexity

How much time does it take to run the algorithm

Often, we deal with estimates!

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Space Complexity

Space complexity = The amount of memory required by an algorithm to run to completion

[Core dumps = the most often encountered cause is “memory leaks” – the amount of memory required larger than the memory available on a given system]

Some algorithms may be more efficient if data completely loaded into memory

Need to look also at system limitations

E.g. Classify 2GB of text in various categories [politics, tourism, sport, natural disasters, etc.] – can I afford to load the entire collection?

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Space Complexity (cont’d)

1. Fixed part: The size required to store certain data/variables, that is independent of the size of the problem:- e.g. name of the data collection- same size for classifying 2GB or 1MB of texts

2. Variable part: Space needed by variables, whose size is dependent on the size of the problem:- e.g. actual text - load 2GB of text VS. load 1MB of text

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Space Complexity (cont’d)

S(P) = c + S(instance characteristics)c = constant

Example:float sum (float* a, int n) {

float s = 0; for(int i = 0; i<n; i++) { s+ = a[i]; } return s;}Space? one word for n, one for a [passed by reference!], one for i constant space!

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Time Complexity

Often more important than space complexityspace available (for computer programs!) tends to be larger and largertime is still a problem for all of us

3-4GHz processors on the market researchers estimate that the computation of various transformations for 1 single DNA chain for one single protein on 1 TerraHZ computer would take about 1 year to run to completion

Algorithms running time is an important issue

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Running Time

Problem: prefix averagesGiven an array XCompute the array A such that A[i] is the average of elements X[0] … X[i], for i=0..n-1

Sol 1At each step i, compute the element X[i] by traversing the array A and determining the sum of its elements, respectively the average

Sol 2 At each step i update a sum of the elements in the array ACompute the element X[i] as sum/I

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Running time

Input

1 ms

2 ms

3 ms

4 ms

5 ms

A B C D E F G

worst-case

best-case}average-case?

Suppose the program includes an if-then statement that may execute or not: variable running timeTypically algorithms are measured by their worst case

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Experimental Approach

Write a program that implements the algorithmRun the program with data sets of varying size.Determine the actual running time using a system call to measure time (e.g. system (date) );

Problems?

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Experimental Approach

It is necessary to implement and test the algorithm in order to determine its running time.

Experiments can be done only on a limited set of inputs, and may not be indicative of the running time for other inputs.

The same hardware and software should be used in order to compare two algorithms. – condition very hard to achieve!

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Use a Theoretical Approach

Based on high-level description of the algorithms, rather than language dependent implementations

Makes possible an evaluation of the algorithms that is independent of the hardware and software environments

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Algorithm Description

How to describe algorithms independent of a programming language Pseudo-Code = a description of an algorithm that is

more structured than usual prose but less formal than a programming language

(Or diagrams)Example: find the maximum element of an array.

Algorithm arrayMax(A, n):Input: An array A storing n integers.Output: The maximum element in A.currentMax A[0]for i 1 to n -1 do

if currentMax < A[i] then currentMax A[i]return currentMax

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Properties of Big-Oh

Expressions: use standard mathematical symbols use for assignment ( ? in C/C++)use = for the equality relationship (? in C/C++)

Method Declarations: -Algorithm name(param1, param2) Programming Constructs:

decision structures: if ... then ... [else ..]while-loops while ... do repeat-loops: repeat ... until ... for-loop: for ... do array indexing: A[i]

Methodscalls: object method(args)returns: return value

Use commentsInstructions have to be basic enough and feasible!

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Asymptotic analysis - terminology

Special classes of algorithms:

logarithmic: O(log n)

linear: O(n)

quadratic: O(n2)

polynomial: O(nk), k ≥ 1

exponential: O(an), n > 1

Polynomial vs. exponential ?

Logarithmic vs. polynomial ?

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Some Numbers

log n n n log n n2 n3 2n

0 1 0 1 1 21 2 2 4 8 42 4 8 16 64 163 8 24 64 512 2564 16 64 256 4096 655365 32 160 1024 32768 4294967296

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Relatives of Big-Oh

“Relatives” of the Big-Oh

(f(n)): Big Omega – asymptotic lower bound

(f(n)): Big Theta – asymptotic tight bound

Big-Omega – think of it as the inverse of O(n)

g(n) is (f(n)) if f(n) is O(g(n))

Big-Theta – combine both Big-Oh and Big-Omega

f(n) is (g(n)) if f(n) is O(g(n)) and g(n) is (f(n))

Make the difference:

3n+3 is O(n) and is (n)

3n+3 is O(n2) but is not (n2)

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More “relatives”

Little-oh – f(n) is o(g(n)) if for any c>0 there is n0 such that f(n) < c(g(n)) for n > n0.Little-omegaLittle-theta

2n+3 is o(n2) 2n + 3 is o(n) ?

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Example

Remember the algorithm for computing prefix averages compute an array A starting with an array X every element A[i] is the average of all elements X[j] with j < iRemember some pseudo-code … Solution 1Algorithm prefixAverages1(X):Input: An n-element array X of numbers.Output: An n -element array A of numbers such that A[i] is the average of elements X[0], ... , X[i].

Let A be an array of n numbers.for i 0 to n - 1 do

a 0for j 0 to i do

a a + X[j] A[i] a/(i+ 1)

return array A

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Example (cont’d)

Algorithm prefixAverages2(X):

Input: An n-element array X of numbers.

Output: An n -element array A of numbers such that A[i] is the average of elements X[0], ... , X[i].

Let A be an array of n numbers.

s 0

for i 0 to n do

s s + X[i]

A[i] s/(i+ 1)

return array A

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Back to the original question

Which solution would you choose?O(n2) vs. O(n)

Some math …properties of logarithms:

logb(xy) = logbx + logby

logb (x/y) = logbx - logby

logbxa = alogbx

logba= logxa/logxb–properties of exponentials:a(b+c) = aba c

abc = (ab)c

ab /ac = a(b-c)

b = a loga

b

bc = a c*loga

b

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Important Series

Sum of squares:

Sum of exponents:

Geometric series:

Special case when A = 2

20 + 21 + 22 + … + 2N = 2N+1 - 1

N largefor 36

)12)(1( 3

1

2 NNNNi

N

i

-1k and N largefor |1|

1

1

k

Ni

kN

i

k

1

11

0

A

AA

NN

i

i

N

i

NNiNNS1

2/)1(21)(

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Analyzing recursive algorithms

function foo (param A, param B) {

statement 1;

statement 2;

if (termination condition) {

return;

foo(A’, B’);

}

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Solving recursive equations by repeated substitution

T(n) = T(n/2) + c substitute for T(n/2)= T(n/4) + c + c substitute for T(n/4)= T(n/8) + c + c + c= T(n/23) + 3c in more compact form= …= T(n/2k) + kc “inductive leap”

T(n) = T(n/2logn) + clogn “choose k = logn”= T(n/n) + clogn= T(1) + clogn = b + clogn = θ(logn)

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Solving recursive equations by telescoping

T(n) = T(n/2) + c initial equation

T(n/2) = T(n/4) + c so this holds

T(n/4) = T(n/8) + c and this …

T(n/8) = T(n/16) + c and this …

…

T(4) = T(2) + c eventually …

T(2) = T(1) + c and this …

T(n) = T(1) + clogn sum equations, canceling the terms appearing on both sides

T(n) = θ(logn)

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RECURSION

Suppose P is a procedure containing either a CALL statement to itself or a CALL statement back to original procedure P .Then P is called a recursive procedure

Properties:

1. There must be certain criteria called basic criteria, for which the procedure does not call itself.

2. Each time the procedure does call itself (directly or indirectly), it must be closer to the base criteria.

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FACTORIAL WITHOUT RECURSION

FACTORIAL(FACT,N)

This procedure calculates N! and return the vale in the variable FACT .

1. If N ==0,then :Set FACT:=1, and Return.

2. Set FACT:=1[Initialize FACT for loop]

3. Repeat for K:=1 to N

Set FACT:=K*FACT

[END of loop]

4. Return.

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FACTORIAL WITH RECURSION

FACTORIAL(FACT,N)

This procedure calculates N! and return the vale in the variable FACT .

1. If N ==0,then :Set FACT:=1, and Return.

2. Call FACTORIAL(FACT,N-1).

3. Set FACT:=N*FACT.

4. Return.

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FACTORIAL EXAMPLE USING RECURSION

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42


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44


45


46


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Stack

A stack is a list that has addition and deletion of items only from one end.

It is like a stack of plates:

Plates can be added to the top of the stack.

Plates can be removed from the top of the stack.

This is an example of “Last in, First out”, (LIFO).

Adding an item is called “pushing” onto the stack.

Deleting an item is called “popping” off from the stack.

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STACK OPERATION (PUSH)

PUSH(STACK,TOP,MAXSTK,ITEM)

This procedure pushes an ITEM onto a stack.

1.[Stack already filled]

If TOP== MAXSTK, then: Print:OVERFLOW, and Return.

2. Set TOP:=TOP+1.[ Increases TOP by 1]

3. Set STACK[TOP]:=ITEM. [Inserting ITEM in new TOP position]

4. Return.

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STACK OPERATION (POP)

POP(STACK,TOP,ITEM)

This procedure deletes the top element of STACK and assigns it to the variable ITEM .

1.[Stack has an item to be to removed]

If TOP== 0, then: Print:UNDERFLOW, and Return.

2. Set ITEM:=STACK[top].[ Assigns TOP element to ITEM ]

3. Set TOP:=TOP-1. [Decreases TOP by 1]

4. Return.

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STACK EXAMPLES

Implementing a stack :

MAXSTK=7.

Data

push 43

push 23

push 53

pop

pop

push 100

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STACK EXAMPLES

7

6

5

4

2

1

TOP=0

MAXSTK=7

53

STACK EXAMPLES

7

6

5

4

2

1

TOP=1

MAXSTK=7

PUSH(43)

43

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STACK EXAMPLES

7

6

5

4

2

1

TOP=2

MAXSTK=7

PUSH(23)

43

23

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STACK EXAMPLES

7

6

5

4

2

1

TOP=3

MAXSTK=7

PUSH(53)

43

23

53

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STACK EXAMPLES

7

6

5

4

2

1

TOP=2

MAXSTK=7

POP( )

43

23

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STACK EXAMPLES

7

6

5

4

2

1

TOP=0

MAXSTK=7

POP( )

43

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STACK EXAMPLES

7

6

5

4

2

1

TOP=0

MAXSTK=7

PUSH(100)

43

100

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STACK EXAMPLES

1. INFIX to POSTFIX

2. POSTFIX expression solving

3. N-QUEEN’S problem

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INFIX to POSTFIX

Properties while transforming infix to postfix expression

1. besides operands and operators, arithmetic expression contains left and right parentheses

2. We assume that the operators in q consist only of

1. Exponent

2. Multiplication

3. Division

4. Addition

5. Subtraction

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INFIX to POSTFIX

3. We have three levels of precedence

precedence operators

high right parentheses

exponent

multiplication, division

low addition, subtraction

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INFIX to POSTFIX

POLISH(Q, P) Suppose Q is an arithmetic expression written in infix notation. This algorithm finds the equivalent postfix expression P.

1. Push “(“ into STACK and add “)” to the end of Q.

2. Scan Q from left to right and repeat Step 3 to 6 for each element of Q until the STACK is empty:

3. If an operand is encountered add it to P.

4. If a left parenthesis is encountered, push it onto STACK.

5. If an operator is encountered then:

(a) repeatedly pop from STACK and to P each operator (on the top of STACK) which has the same precedence as or higher precedence this operator

(b) Add operator to STACK

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INFIX to POSTFIX

6. If a right parenthesis is encountered then:

(a) Repeatedly pop from STACK and add to P each operator( on the top of STACK) until a left parenthesis is encountered

(b) Remove the left parenthesis .[Do not add the left parenthesis to P]

7. Exit

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INFIX to POSTFIX

INFIX Expression: 3 + 2 * 4

POSTFIX Expression:7

6

5

4

2

1

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INFIX to POSTFIX

INFIX Expression: + 2 * 4

POSTFIX Expression: 37

6

5

4

2

1

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INFIX to POSTFIX

INFIX Expression: 2 * 4

POSTFIX Expression: 37

6

5

4

2

1

+

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INFIX to POSTFIX

INFIX Expression: * 4

POSTFIX Expression: 3 27

6

5

4

2

1

+

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INFIX to POSTFIX

INFIX Expression: 4

POSTFIX Expression: 3 27

6

5

4

2

1

+

*

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INFIX to POSTFIX

INFIX Expression:

POSTFIX Expression: 3 2 47

6

5

4

2

1

+

*

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INFIX to POSTFIX

INFIX Expression:

POSTFIX Expression: 3 2 4 *7

6

5

4

2

1

+

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INFIX to POSTFIX

INFIX Expression:

POSTFIX Expression: 3 2 4 * +7

6

5

4

2

1

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POSTFIX EXPRESSION SOLVING

This algorithm finds the VALUE of an arithmetic expression P written in postfix notation

1. Add a right parenthesis “)” at the end of P

2. Scan P from left to right and repeat step 3 and 4 for each element of P until the sentinel “)” is encountered.

3. If an operand is encountered, put it on STACK.

4. If an operator is uncounted then:

a. Remove the two elements of Stack, where A is the top element and B is the next top element

b. Evaluate B operator A

c. Pace the result of (b) back on STACK

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5. Set VALUE equal to the top element on STACK

6. Exit

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POSTFIX Expression: 3 2 4 * +Push the numbers until operator comes 7

6

5

4

2

1

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POSTFIX Expression: 2 4 * +

7

6

5

4

2

1

3

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POSTFIX Expression: 4 * +

7

6

5

4

2

1

3

2

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POSTFIX Expression: * + Here we pop two time and perform multiplication on elements (4*2) and push the Result in to the stack

7

6

5

4

2

1

3

4

2

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POSTFIX Expression: +

7

6

5

4

2

1

3

8

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POSTFIX Expression: 3 2 4 * +

7

6

5

4

2

1

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The N-Queens Problem

Can the queens be placed on the board so that no two queens are attacking each other in chess board

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Two queens are not allowed in the same rowTwo queens are not allowed in the same row

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Two queens are not allowed in the same row, or in the same column...Two queens are not allowed in the same row, or in the same column...

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Two queens are not allowed in Two queens are not allowed in the same row, or in the same the same row, or in the same column, or along the same column, or along the same diagonal.diagonal.r along the same diagonal.

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The number of queens, and the size of the board can vary.

The number of queens, and the size of the board can vary.

N Queens

N columns

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The 3-Queens Problem

The program uses a The program uses a stack to keep track of stack to keep track of where each queen is where each queen is placed.placed.

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Each time the program decides to place a queen on the board, the position of the new queen is stored in a record which is placed in the stack.

ROW 1, COL 1

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We also have an integer variable to keep track of how many rows have been filled so far.

ROW 1, COL 1

1 filled

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Each time we try to place a new queen in the next row, we start by placing the queen in the first column

ROW 1, COL 1

1 filled

ROW 2, COL 1

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if there is a conflict with another queen, then we shift the new queen to the next column.

ROW 1, COL 1

1 filled

ROW 2, COL 2

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ROW 1, COL 1

1 filled

ROW 2, COL 3

When there are no When there are no conflicts, we stop and conflicts, we stop and add one to the value of add one to the value of filled.filled.

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When there are no conflicts, we stop and add one to the value of filled.

ROW 1, COL 1

2 filled

ROW 2, COL 3

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Let's look at the third row. The first position we try has a conflict

ROW 1, COL 1

2 filled

ROW 2, COL 3

ROW 3, COL 1

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so we shift to column 2. But another conflict arises

ROW 1, COL 1

2 filled

ROW 2, COL 3

ROW 3, COL 2

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and we shift to the third column.

Yet another conflict arises

ROW 1, COL 1

2 filled

ROW 2, COL 3

ROW 3, COL 3

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and we shift to column 4. There's still a conflict in column 4, so we try to shift rightward again

ROW 1, COL 1

2 filled

ROW 2, COL 3

ROW 3, COL 4

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but there's nowhere else to go.

ROW 1, COL 1

2 filled

ROW 2, COL 3

ROW 3, COL 4

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When we run out of

room in a row: pop the stack, reduce filled by 1 and continue

working on the previous row.

ROW 1, COL 1

1 filled

ROW 2, COL 3

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Now we continue working on row 2, shifting the queen to the right.

ROW 1, COL 1

1 filled

ROW 2, COL 4

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This position has no conflicts, so we can increase filled by 1, and move to row 3.

ROW 1, COL 1

2 filled

ROW 2, COL 4

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In row 3, we start again at the first column.

ROW 1, COL 1

2 filled

ROW 2, COL 4

ROW 3, COL 1

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In row 3, we start again at the first column.

ROW 1, COL 1

3 filled

ROW 2, COL 4

ROW 3, COL 2

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QUEUES

A queue is a data structure that maintains a “first-in first-out” (FIFO) ordering.

In contrast, a stack maintains a “last-in first-out” (LIFO) ordering.

A queue adds new elements at the end. An element can only be removed at the front.

This is an abstraction of the “first-come first-served” practice.

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QUEUE OPERATIONS

A queue has two operations:

QINSERT

QDELETE

An enqueue operation adds new elements at the end of the queue or its tail. This is similar to the stack operation push; only that push now is done at the end of the array instead of at the front (or top).

A dequeue operation removes an element from the front of the array or its head.

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QUEUE INSERTION

QINSERT(rear, item)

1. IF rear == MAX_QUEUE_SIZE_1 then Print queue_full Return;

2. INFO [++rear] = item;

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QUEUE DELETION

QDELETE(front, rear)

1. IF front == rear then Return queue_empty( ); Return queue [++ front];

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QUEUE

OFFSET

DATA

0 1 2 3

0

rear

0

OFFSET

DATA

0 1 2 3

A

front

1

rear

1

OFFSET

DATA

0 1 2 3

A B

front

1

rear

2

OFFSET

DATA

0 1 2 3

B

front

2

rear

2

front

Insert A

Insert B

delete

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CIRCULAR QUEUE

When a new item is inserted at the rear, the to rear moves upwards.

Similarly, when an item is deleted from the queue the front arrow moves downwards.

After a few insert and delete operations the rear might reach the end of the queue and no more items can be inserted although the items from the front of the queue have been deleted and there is space in the queue.

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CIRCULAR QUEUE

To solve this problem, queues implement wrapping around. Such queues are called Circular Queues.

Both the front and the rear wrap around to the beginning of the array when they reached the MAX size of the queue.

It is also called as “Ring buffer”.

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CIRCULAR QUEUE INSERTION

QINSERT (QUEUE, N, FRONT, ITEM) This procedure insert an element ITEM into a queue.

1. [Queue already filled?]

IF ( FRONT==1 and REAR==N ) or FRONT ==REAR + 1,then: write: overflow, and Return

2.[Find new value of REAR] IF FRONT==NULL then [Queue initially empty.]

Set FRONT=1 and REAR=1

ELSE IF REAR ==N then Set REAR=1

ELSE set REAR=REAR+1

[End of if structure]

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CIRCULAR QUEUE INSERTION

3. Set QUEUE[REAR]=ITEM.[This inserts new element]

4. Return.

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CIRCULAR QUEUE DELETION

QDELETE(QUEE, N, FRONT, REAR, ITEM)

This procedure deletes an element from a queue and assigns it to the variable ITEM

1.[Queue already empty]

if FRONT=NULL then write UNDERFLOW, and Return

2. Set ITEM=QUEUE[FRONT]

3. [Find new value of FRONT]

If FRONT =REAR then [Queue has only one element to start]

Set FRONT=NULL and REAR=NULL

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Else if FRONT ==N then

Set FRONT=1

Else

Set FRONT=FRONT +1

[End of if statement]

4.Return

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EMPTY QUEUE

[2] [3] [2] [3]

[1] [4] [1] [4]

[0] [5] [0] [5]

front = 0 front = 0 rear = 0 rear = 3

J2

J1

J3

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[2] [3] [2] [3]

[1] [4][1] [4]

[0] [5] [0] [5]

front =0rear = 5

front =4rear =3

J2 J3

J1 J4

J5 J6 J5

J7

J8 J9

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PRIORITY QUEUE

A priority queue is a collection of elements such that each element has been assigned a priority and such that the order in which elements are deleted and processed comes from the following rules

1. An element of a higher priority is processed before any elements of lower priority

2. Two elements with the same priority are processed according to the order in which they were added to the queue

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ARRAY REPRESENTATION PRIORITY QUEUE

Use a separate queue for each level of priority

Each such queue will appear in its own circular array and must have its own pair of pointers FRONT and REAR

In fact, if each queue is allocated the same amount of space in two-dimensional array QUEUE can be used instead of linear array

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DELETION ON PRIORITY QUEUE

Deletion

This algorithm deletes and processes the first element in a priority queue maintained by a two-dimensional array QUEUE

1.[Find the first nonempty queue] Find the smallest k that FRONT!=NULL

2.Delete and process the front element in row K of QUEUE

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INSERTION ON PRIORITY QUEUE

Insertion

This algorithm adds an ITEM with priority number M to a priority queue maintained by a two-dimensional array QUEUE

1. Insert ITEM as the rear element in row M of QUEUE

2. Exit

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LINKED LIST

Linked list

Linear collection of self-referential class objects, called nodes

Connected by pointer links

Accessed via a pointer to the first node of the list

Subsequent nodes are accessed via the link-pointer member of the current node

Link pointer in the last node is set to null to mark the list’s end

Use a linked list instead of an array when

You have an unpredictable number of data elements

Your list needs to be sorted quickly

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TYPES OF LINKED LIST

Types of linked lists:

Singly linked list

Begins with a pointer to the first node

Terminates with a null pointer

Only traversed in one direction

Circular, singly linked

Pointer in the last node points back to the first node

Doubly linked list

Two “start pointers” – first element and last element

Each node has a forward pointer and a backward pointer

Allows traversals both forwards and backwards

Circular, doubly linked list

Forward pointer of the last node points to the first node and backward pointer of the first node points to the last node

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linked list

Start Node

A placeholder node at the beginning of the list, used to simplify list processing. It doesn’t hold any data

Tail Node

A placeholder node at the end of the list, used to simplify list processing

122

Singly linked list

Single Linked ListConsists of data elements and reference to the

next Node in the linked list

First node is accessed by reference

Last node is set to NULL

A B CStart Tail

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SINGLE LINKED LIST INSERTION

INSFIRST(INFO, LINK, START, AVAIL, ITEM)

This algorithm inserts ITEM as the first node in the list

1. [OVERFLOW?] If AVAIL=NULL then :Write OVERFLOW and Exit

2. [Remove first node from AVIL list] Set NEW =AVAIL and AVAIL=LINK[AVAIL]

3. Set INFO[NEW]=ITEM [Copies new data into new node]

4. Set LINK[NEW]=START [new node now points to original first node]

5. Set START=NEW [Changes START so it points to the node]

6. Exit

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INSLOC(INFO, LINK,START, AVAIL, LOC, ITEM) This algorithm inserts ITEM so that ITEM follows the node with location LOC or inserts ITEM as the first node when LOC=NULL

1. [OVERFLOW ?] If AVAIL==NULL then write OVERFLOW and EXIT

2. [Remove first node from AVAIL list]

3. Set INFO[NEW]=ITEM[copies new data into new node]

4. If LOC==NULL then :[insert as first node] Set LINK[NEW]=Start and START=NEW

else [insert after node with location LOC] Set LINK[NEW]=LINK[LOC] and LINK[LOC]=NEW

end of if structure

5. exit

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SINGLE LINKED LIST DELETION

DELETE(INFO, LINK,START, AVAIL, ITEM) This algorithm deletes from a linked list the first node N which contains the given ITEM of information

1. [Use procedure FIND given after this algorithm] FIND(INFO, LINK, START, ITEM, LOC, LOCP)

2. If LOC==NULL then: Write ITEM not in list and exit

3. If LOCP==NUL then Set START=LINK[START]

else : Set LINK[LOCP]=LINK[LOC]

3. {Return deleted node to the AVAIL list] Set LINK[LOC]=AVAIL and AVAIL=LOC

4. Exit

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FINDB(INFO, START, ITEM, LOC, LOCP) This procedure finds the location LOC of first node N which contains ITEM and the location LOVP of the node preceding N. if ITEM does not appear in the list then the procedure sets LOC=NULL and if ITEM appears in the first node then it sets LOCP=NULL

1. [list empty?] if START==NULL then Set LOC =NULL and LOCP==NULL and return

2. [ITEM in first node ] if INFO[START]==ITEM then Set LOC=START and LOCP=NULL and return

3. Set SAVE=START and PRT=LINK[START]

4. Repeat steps 5 and 6 while PTR!=NULL

5. If INFO[PTR]==ITEM then Set LOC=PTR and LOCP=NULL

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6. Set SAVE=PTR and PTR=LINK[PTR]

7. Set LOC=NULL

8. Return

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Double Linked Lists

Consists of nodes with two linked references one points to the previous and other to the next node

Maximises the needs of list traversals

Compared to single list inserting and deleting nodes is a bit slower as both the links had to be updated

It requires the extra storage space for the second list

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Insertion Double Linked Lists

INSERTWL(INFO,FORW, BACK, STACK,AVAIL,LOCA,LOCB, ITEM)

1. [OVERFLOW?] If AVAIL==NULL then write OVER FLOW and exit

2. [Remove node from AVAIL list and copy new data into node] Set NEW=AVIAL, AVIAL=FORW[AVAIL] INFO[NEW]=ITEM

3. [Insert node into list] Set FORW[LOCA]=NEW, FORW[NEW]=LOCB BACK[LOCB]=NEW BACK[NEW]=LOCA

4. Exit

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LOC A LOC B

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Deletion Double Linked Lists

DELTWL(INFO, FORW, BACK, START, AVAIL, LOC)

1. [Delete node]

Set FORE[BACK[LOC]]=FORW[LOC]

Set BACK[FORW[LOC]]=BACK[LOC]

2. [Return node to AVAIL list]

Set FORW[LOC]=AVAIL and AVAIL=LOC

3. Exit

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Data Structures

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