Week 0 - Lectures

7/18/2019 Week 0 - Lectures

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BINARY NUMERATION SYSTEM.1Jean-Pierre Deschamps

University Rovira i Virgili, Tarragona, Spain



0 .11. Computer information representation

A computer receives, stores, processes, transmits data.

, , ,

(video), etc.

Data encoding: strings of zeroes and ones.

Computer technology is based on electronic circuits able to process vectors of 0’s and 1’s (the

so-called digital electronic circuits). For that reason all data are encoded by strings of 0’s and 1’s.

This type of information encoding is called binary encoding system.2



0 .12. Numeration systems

Most used systems:

• Decimal system

• Binary system

• Hexadecimal system

Conversion methods will be presented..

3



0 .12.1 Decimal system

• Uses ten digits:

0 1 2 3 4 5 6 7 8 9

• Positional system : a weight is associated to every digit position so that position is

relevant.

6 5 3Example: 653

653 = 6·102 + 5·101 + 3·100

6 hundreds, 5 tens, 3 units

4



0 .12.2 Binary system

• Uses two digits (binary digits, bits): 0, 1

• Positional system.

Example:

(weights) 23 22 21 20

• To compute the decimal representation, add up the weights corresponding to the

1’s of the binary representation:(1101)2 = 1·2 + 1·2 + 0·2 + 1·2 = 8 + 4 + 0 + 1 = 13

5



0 .1Exercise

Compute the decimal representation of the following binary number: (101001)2

6



0 .1Exercise solut ion

Compute the decimal representation of the following binary number: (101001)2

1 0 1 0 0 1

weights 25 24 23 22 21 20

(101001)2

= 1·25 + 0·24 + 1·23 + 0·22 + 0·21 + 1·20 = 32 + 8 + 1 = 41

7



0 .12.2 Binary system: representation range

• Pure binary system: non-negative number representation.

• With n bits: 2n distinct values.

• Representation range: 0 to 2n - 1.

EXAMPLE: Binary Decimal Binary Decimal

n = 4 bits16 different combinations

from 0 to 15 = 24-1

0000 0 1000 80001 1 1001 9

0010 2 1010 10

0100 4 1100 12

0101 5 1101 13

0111 7 1111 15

8



0 .12.2 Binary system: representation range

If n = 3 : 23 = 8 representable numbers, from 0 to 7.

If n = 4 : 24 = 16 representable , from 0 to 15.

If n = 5 : 25 = 32 representable , from 0 to 31.

6 , …

Example: how many bits do we need to represent 48?

31 ≤ 48 ≤ 63

n=5 n=6

=> to re resent decimal number 48 we need 6 bits: 110000

9



0 .12.3 Hexadecimal system

BINARY HEXA

0 0 0 0 0

0 0 0 1 1

• Uses sixteen dígits:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F0 0 1 0 2

0 0 1 1 3

0 1 0 0 4• Positional system

(weights) 163 162 161 160

0 1 1 0 6

0 1 1 1 7

1 0 0 0 8

• To compute the decimal representation, add up the digits

multiplied by the corresponding weights:

1 0 0 1 9

1 0 1 0 A

1 0 1 1 B

(3A9F)16 = 3·163 + 10·162 + 9·161 + 15·160 = 1500710 1 1 0 1 D

1 1 1 0 E

1 1 1 1 F



3. Base conversion (HEXADECIMAL to BINARY)

0 .1

5. Cambios de base (BINARY to HEXADECIMAL)

.

hexadecimal 3 A 9

y

• BINARY to HEXADECIMAL: 4 bits → 1 hexadecimal digit (starting from the four

binary 100 1011 1010 0101

r g mos s

hexadecimal 4 B A 511



0 .1Exercise

1101101010011002 = ????16

BINARY HEXA

0 0 0 0 0

0 0 0 1 1

Compute the hexadecimal representation:

0 0 1 0 2

0 0 1 1 3

0 1 0 0 4

0 1 0 1 5

5F2C16 = ?????2

0 1 1 0 60 1 1 1 7

1 0 0 0 8

Compute the binary representation:

1 0 0 1 9

1 0 1 0 A

1 0 1 1 B

1 1 0 0 C

1 1 0 1 D

1 1 1 0 E

1 1 1 1 F12



0 .1Exercise (so lu t ion )

BINARY HEXA

0 0 0 0 0

0 0 0 1 1

Compute the hexadecimal representation:

2 = 160 0 1 0 2

0 0 1 1 3

0 1 0 0 4

0 1 0 1 5

5F2C16 = ?01011111001011002

0 1 1 0 60 1 1 1 7

1 0 0 0 8

Compute the binary representation:

0101 1111 0010 11000101 1111 0010 11000101 1111 0010 11000101 1111 0010 1100

1 0 0 1 9

1 0 1 0 A

1 0 1 1 B

1 1 0 0 C

1 1 0 1 D

1 1 1 0 E

1 1 1 1 F13



0 .14. Base conversion (DECIMAL to BINARY)

5• Divide the decimal number by 2. Divide the obtained quotient by 2. Keep dividing

the obtained uotients b 2 until the obtained uotient is e ual to 1.

• The base 2 number consists of the last quotient 1 and the set of previously obtained

remainders.

Example: 18(10 = ? 10010(2

(10 = · 2 +

18 = 9·2 + 0

9 = 4·2 + 1

4 = 2·2 + 0 = · 18(10 = 10010(2 10010(2

14



Exercise

0 .1

43(10 = binary number?

15




43(10 = binary number?

43 = 21·2 + 1

21 = 10·2 + 1

= ·

5 = 2 · 2 + 12 = 1 · 2 + 0

43(10 = 101011

16



0 .16. Sum and difference of binar numbers

Sum of 2 bits:

0 + 0 = 0

Difference of 2 bits:

0 - 0 = 0 =

1 + 0 = 1

1 + 1 = 10 (current step bit: 0,

-

1 - 1 = 0

0 - 1 = 11 (current step bit: 1

borrow to the next ste : 1

==

Sum example: Difference example:

1 0 1 0 01 0 1 0 0 1 0 1 =1 0 1 = AA

++ 11 0 10 1 0 1 10 1 1 11 = B= B

-- 11 0 10 1 0 1 10 1 1 1 = B1 = B

1 1 1 1 1 1 0 01 1 1 1 1 1 0 0 0 1 0 0 1 1 1 00 1 0 0 1 1 1 0

1 0 1 1 1 11 0 1 1 1 1 →→ borrowborrow

17



0 .1Exercise

1 01 0 0 1 1 0 1 1 =0 1 1 0 1 1 = AA== 1 01 0 0 1 1 0 0 1 =0 1 1 0 0 1 = AA-- ==

18




1 01 0 0 1 1 0 1 1 =0 1 1 0 1 1 = AA1 01 0 0 1 1 0 0 1 =0 1 1 0 0 1 = AA

-- ==0 0 1 0 0 1 10 0 1 0 0 1 1

→→ carrycarry

1 1 1 0 1 1 1 01 1 1 0 1 1 1 00 1 0 0 0 1 1 00 1 0 0 0 1 1 0

++ 1 0 11 0 1 00 00 11 11 = B= B

11 0 0 1 11 1 →→ borrowborrow

19



SUMMARY0 .1

Computer information representation.

Numeration systems (decimal, binary, hexadecimal).

Pure binary system and representation range.

Base conversion.

um an erence o nary num ers.

20



PSEUDOCODE.2Jean-Pierre Deschamps

University Rovira i Virgili, Tarragona, Spain



1. Algorithm representation: pseudocode0 .2

Algorithm

Sequence of operations whose objective is the solution of some problem such as:

comp ex compu a on, con o o some p ocess, e c.

The result of the algorithm execution must be independent of the chosen type of

representation.

, ,

diagrams and programming languages.

Similar to programming language but more informal. It uses a mix of

natural language sentences,

,

key words that define basic structures.

22



2. Operations and control structures0 .2

• ASSIGNMENTS

• OPERATORS⁻ Compar son

⁻ Logic operations

⁻ Arithmetic operations

• SELECTION STRUCTURES (DECISIONS)⁻ Simple, Double, Multiple, Case

• ITERATION STRUCTURES (CYCLES)

⁻ While, For

•

23



2. 1 Assignments and types of operators0 .2

ASSIGNMENTS

Assignment instruction <= : allows to store a value within a variable. Examples:

x <= 15, x <= 2x + y + z, etc.

variable <= expression;

w ere x, y an z are var a es.

COMPARISON OPERATORS

Operator Meaning

< Smaller than

> Greater than

= Equal to

≤ Smaller than or e ual to

(1) Sometimes we can use <=

(2) Sometimes we can use >=

=

≥ (2) Greater than or equal to

≠(3) Different from

24



0 .22. 1 Assignments and types of operators

LOGIC OPERATORS

Operator Meaning

og c pro uc

or Logic sum

not Negation

ARITHMETIC OPERATORS

+ Sum

- Difference

* Product

/ Division25



3. Control structures0 .2

SELECTION STRUCTURES

Simple condition yes

If condition then action(s);

end if ;

noaction;

ou e

If condition then action1/s;else action2/s;

conditionno yes

action1;action2;

26



0 .23. Control structures

SELECTION STRUCTURES

Multiple condition1

yes no

If condition1 then action1/s;

elsif condition2 then action2/s;

end if ;

action2;

action1; condition2

Case

end if ;

x

when “value1" => action1;


“ " =

X

action2;action1; action3; action4;


end case; 27



0 .23. Control structures (example 1)

y = X/2 , rounded down

Different cases:

• When X is even, the result of the division is exact (whether X is positive or negative)

,

- if X is negative: (X+1)/2- if X is positive (X-1)/2

If (X is even) then y<= X/2;

elsif (X<0) then y <= (X+1)/2;

else y <= (X-1)/2;end if ;

end if ;28




X is a decimal digit: we calculate the binary representation

case x is

“ " = =

when “1" => y <= 0001;……..

“ " = =

end case;

29



0 .23. Control structures

ITERATION STRUCTURES

conditionyes no

While

While condition loop

action/s;

action;

for variable in min to max loo

For

action/s;

end loop;

30




Given two vectors of 8 positions:

, , ……

x(0), x(1), ……x(7)

We want to do the following calculation:

y = a(0) · x(0) + a(1) · x(1) + a(2) · x(2) + ………..+ a(7) · x(7)

for i in 0 to 7 loop

acc <= acc+ a(i) · x(i);

en oop;

31



4. Procedures0 .2

Procedure call name(parameters);

Procedure (subroutine): sequence of instructions (operations and control structures) that

execute some task (algorithm).

A name and a set of parameters are associated to every procedure.

A procedure can be called one or several times within a program. When called, values are given

to its parameters. Algorithm

---------

PROCEDURE

---------call procedure

---------

------------------

---------

call procedure ---------

---------32



4. Procedures (example 1)0 .2

Procedure square_root(x, y, n) is

z = a · √x + b · √y

----

----

end procedure

ALGORITHM

Begin

PROCEDURE

call square_root (x,u,16);

u <= a * u;

call square_root (y,r,16);

*

Algorithm

z <= u + r;

end

square_root

(x,y,n)

---------

---------

end procedure

call square_root (x, u, 16);

u <= a * u;

call square_root (y, r, 16);

r <= b * r;

z <= u + r;33



SUMMARY0 .2

Algorithm representation

Pseudocode

Operations and control structures

Procedures

34

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