procedural programming & fundamentals of...

06.06.2018

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Procedural Programming & Fundamentals of ProgrammingLecture 4 - Summer Semester 2018

Prof. Dr.-Ing. Axel Hunger & Joachim Zumbrägel

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Writing a computer program is only one step in the Program Development Process

Prof. Dr.-Ing. Axel Hunger

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What we know so far... (1/3)

Computers should help us to solve problems

First we need to understand the problem, which requires to get all relevant information

Then we create a step-wise solution to it, which at its best is following a problem-solving strategy


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By means of flowcharts (for algorithm) and pseudocode (for program) we document and structure our solution approach

A computer executes the program, which we write in a specific programming language

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Control structures serve to manage the order of instructions, which change data in the main memory of a computer.

To organize this data and its respective manipulation, we use data types and data structures.

The challenge is to develop a proper abstraction of the system in the real world into a problem model to be solved by the computer.

We are responsible for the programs that we write.


Struct(ures)

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Records (Structures)

A record permits a programmer to handle a group of variables as one variableThe fields / members / components of a record can be any built-in or programmer-defined data typeA record can have values assigned to and read from it just like the built-in variable types

A record can also be passed as an argument to a function and serve as the return value for a function


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Structures in C

• A struct is a mechanism for grouping together related data items of different types.

Recall: An array groups items of a single type

• Example: We want to represent an airborne aircraft:

• We can use a struct to group these data together for each plane.


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Defining a Struct

• We first need to define a new type for the compiler and tell it what our struct looks like.

• This tells the compiler how big our struct is and how the different data items (“members”) are laid out in memory.

• But it does not allocate any memory.


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Declaring and Using a Struct

time

altitud

e

flightNum: 409

flightNum: 1767

flightType planeA ;

planeA

planeB

flightType planeB ;

planeA . flightNum planeB . flightNum

planeA . airSpeed

planeA . altitude

planeA . time

planeB . airSpeed

planeB . altitude

planeB . time

Access: Access:

==


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Declaring and Using a Struct

• To allocate memory for a struct, we declare a variable“plane” using our new struct data type “struct flight type”.

• Memory is allocated, and we can access individual members of this variable:

• A struct’s members are laid outin the order specified by the definition.


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Airplane

Coordinates

Position time x y z

100s x0 y0 z0

200s x1 y0 z1

300s x2 y0 z2

400s x3 y0 z3

500s x4 y0 z4

600s x5 y0 z5

700s x6 y0 z6

800s x7 y0 z7

900s x8 y0 z8

1000s x9 y0 z9


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Airplane

Coordinates(in km)

Position time x y z

100s 2 5 9

200s 4 6 8

300s

400s

500s

600s

700s

800s

900s

1000s

………………………………………

………………………………………

………………………………………

………………………………………


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Airplane

From the previous lecture: User‐defined data type

Define an array of type flightType:

flightType t[10];

time IndexElement

ofArray t

Cells in Main

Memory

100s 0 t [0]

200s 1 t [1]

300s 2 t [2]

400s 3 t [3]

500s 4 t [4]

600s 5 t [5]

700s 6 t [6]

800s 7 t [7]

900s 8 t [8]

1000s 9 t [9]At

100sAt

200sAt

300sAt

400sAt

500sAt

600sAt

700sAt

800sAt

900sAt

1000s


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Declaration of User-defined Data Type

struct student

{

int matr_num;

char family_name [20];

char first_name [20];

char study_course [20];

} ;

components of the structure withdifferent data types

keyword to start the structure declaration

name of the structure


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Access to Structure Components

…

student_inStudyCourse . matr_num

student_inStudyCourse . family_name

student_inStudyCourse . first_name

student_inStudyCourse . study_course

…

Cells in Main

Memory

Access to each component

student_inStudyCourse is a structured variableconsisting of four components.

struct student

{

int matr_num;

char family_name [20];

char first_name [20];

char study_course [20];

};

struct student student_inStudyCourse ;


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Dynamic Data Structures- Linked List -

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Linked List

• is a data structure in which each element is dynamically allocated and in which elements point to each other to define a linear relationship

• Singly- or doubly-linked

• Stack, queue, circular list


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Definition of Linked List (1/2)

Prof. Dr.-Ing. Axel HungerProf. Dr.-Ing. Axel Hunger

struct listItem {struct student student_inStudyCourse;struct listItem *next;

};

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Definition of Linked List (2/2)

• Items of list are usually same type

• Each item points to next item

• Last item points to null

• Need “head” to point to first item!


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Head of a Linked List


struct listItem {struct student student_inStudyCourse;struct listItem *next;

};struct listItem *head;

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Usage of Linked Lists

• Not massive amounts of data

• Linear search is okay

• Sorting not necessary

• or sometimes not possible

• Can add and delete data “on the fly”

• Even from middle of list


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Adding an Item to a List (1/3)

• Add an item pointed to by q after item pointed to by p• Neither p nor q is NULL


q

p

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• Add an item pointed to by q after item pointed to by p


q

p

q.next = p.next

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• Add an item pointed to by q after item pointed to by p



q

p

p.next = q

Data structuresapplied in problem solving

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How to cross the river?


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Mapping information to data


START

Boat: 0 Fox: 0 Goose: 0 Corn: 0

a state indicating thelocation of items

• 0: item is at ‚near shore‘• 1: item is at ‚far shore‘

int SHORE[4]={0,0,0,0};

Condition on state transition: Boat can only carry one or none item at a time!‐> relevant to identify neighboring states

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State space model (24 = 16 states)


(1): START

Boat: 0

Fox: 0

Goose: 0

Corn: 0

(16): END

Boat: 1

Fox: 1

Goose: 1

Corn: 1

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0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

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State space model (10 valid/ 6 invalid states)


START

Boat: 0

Fox: 0

Goose: 0

Corn: 0

END

Boat: 1

Fox: 1

Goose: 1

Corn: 1

2

0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

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Breadth-first search

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Breadth-first search (BFS)

• belongs to uninformed (or blind) searching strategies in a graph (a structure made up of nodes connected with edges)

• does not make use of heuristics for minimizing its computational expensiveness in solving a problem, such asin informed searching approaches

• begins at a root node and inspects all the neighboring nodes before expanding next level

• does not explore graph as far as possible along one path before backtracking, such as in depth-first search

• makes use of queue to manage graph traversal


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Queue

• is a data structure following the FIFO (First In First Out) principle to its elements

• can be implemented by linked list


HEAD TAIL1 TAIL… TAILnFRONT BACK

TAILn+1enqueue

HEADdequeue

Queue

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An algorithm for breadth-first search

1. Determine START state and add it to QUEUE.

2. Pick head state from QUEUE, add it to QUEUE _visited.

a) IF ‚head==END‘ then stop and „success“.b) ELSE add all neighboring states of head

to (tail of) QUEUE.3. IF ‚QUEUE == empty‘ then „fail“.

4. Repeat 2.

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1. Determine START state and add it to LIST.

2. Pick head state from LIST and add it to LIST_visited.

a) IF ‚head==END‘ then stop and „success“.

b) ELSE add all (valid) next states to (tail of) LIST.

3. IF ‚LIST == empty‘ then „fail“.

4. Repeat 2.

Step 1 of 10


START


END

Boat: 1 Fox: 1 Goose:

2

0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

QUEUE START

QUEUE_visited

Head of QUEUE Tail of QUEUE

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4. Repeat 2.

Step 2 of 10


QUEUE START 5 6 7 8

QUEUE_visited START

START


END


2

0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

Strategy #1: Do not consider invalid states.

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4. Repeat 2.

Step 3 of 10


QUEUE 7 START 3

QUEUE_visited START 7

START


END


2

0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

Strategy #2: Do not consider already visited states again.

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4. Repeat 2.

Step 4 of 10


QUEUE 3 7 9 10

QUEUE_visited START 7 3

START


END


2

0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

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4. Repeat 2.

Step 5 of 10


QUEUE 9 10 12 3 4

QUEUE_visited START 7 3 9

START


END


2

0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

Note: QUEUE_visited does NOT equals THE solution path!

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4. Repeat 2.

Step 6 of 10


QUEUE 10 4 14 2 3

QUEUE_visited START 7 3 9 10

START


END


2

0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

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Step 7 of 10


QUEUE 4 2 6 9 15

QUEUE_visited START 7 3 9 10 4

START


END


2

0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

already close to END, but...

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4. Repeat 2.

Step 8 of 10


QUEUE 2 15 8 (15) 10

QUEUE_visited START 7 3 9 10 4 2

START


END


2

0 0 0 1

3

0 0 1 0

4

0 1 0 0

5

1 0 0 0

6

1 1 0 0

7

1 0 1 0

8

1 0 0 1

9

1 1 1 0

10

1 0 1 1

12

0 1 1 0

13

0 1 0 1

14

0 0 1 1

15

1 1 0 1

11

0 1 1 1

Strategy #3: Do not consider same head state twice.

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4. Repeat 2.

Step 9 of 10


START


END


2:0001

0 0 0 1

3:0011

0 0 1 0

4:0100

0 1 0 0

5:0101

1 0 0 0

6:0110

1 1 0 0

7:0111

1 0 1 0

8:1000

1 0 0 1

9:1001

1 1 1 0

10:1010

1 0 1 1

12:1100

0 1 1 0

13:1101

0 1 0 1

14:1110

0 0 1 1

15:1111

1 1 0 1

11:1011

0 1 1 1

QUEUE 15 13 2 4

QUEUE_visited START 7 3 9 10 4 2 15

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4. Repeat 2.

Step 10 of 10


START


2:0001

0 0 0 1

3:0011

0 0 1 0

4:0100

0 1 0 0

5:0101

1 0 0 0

6:0110

1 1 0 0

7:0111

1 0 1 0

8:1000

1 0 0 1

9:1001

1 1 1 0

10:1010

1 0 1 1

12:1100

0 1 1 0

13:1101

0 1 0 1

14:1110

0 0 1 1

15:1111

1 1 0 1

11:1011

0 1 1 1

END

Boat: 1

Fox: 1

Goose: 1

Corn: 1

QUEUE 13 15 END

QUEUE_visited START 7 3 9 10 4 2 15 13

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4. Repeat 2.

Final Step: „Success“


START


END


2:0001

0 0 0 1

3:0011

0 0 1 0

4:0100

0 1 0 0

5:0101

1 0 0 0

6:0110

1 1 0 0

7:0111

1 0 1 0

8:1000

1 0 0 1

9:1001

1 1 1 0

10:1010

1 0 1 1

12:1100

0 1 1 0

13:1101

0 1 0 1

14:1110

0 0 1 1

15:1111

1 1 0 1

11:1011

0 1 1 1

QUEUE END

QUEUE_visited START 7 3 9 10 4 2 15 13 END

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4. Repeat 2.

Final Step: „Success“


QUEUE END

QUEUE_visited START 7 3 9 10 4 2 15 13 END

START


END


2:0001

0 0 0 1

3:0011

0 0 1 0

4:0100

0 1 0 0

5:0101

1 0 0 0

6:0110

1 1 0 0

7:0111

1 0 1 0

8:1000

1 0 0 1

9:1001

1 1 1 0

10:1010

1 0 1 1

12:1100

0 1 1 0

13:1101

0 1 0 1

14:1110

0 0 1 1

15:1111

1 1 0 1

11:1011

0 1 1 1

the „tricky“ step

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Finally…

• Applying strategies can reduce time and space complexity of a solution approach

• Developing solutions is no mechanic procedure, it is a creative act

• For problem solvers and programmers to be effective:

There is not just one way of doing it.


END of 6th lecture

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