Year10ComputerScienceGCSE

ArkGlobeAcademyPhaseIII–WorkPack11thMay–22ndJune

DataRepresentationThisbookletalongwithdigitalversionsofalltheactivitiesandresourcescanbeaccessedanddownloadedfromwww.arkglobe.orginthestudentresourcessectionoftheVLE.Ifyouareunabletoworkdigitallythenpleasejustworkthroughthisbooklet.Whatyouwillneed

• PenapapertohelpwithworkingoutandactivitiesorComputertocompleteonscreen• CompletetheonlinequizbyFridayeachweek,linkedintheoutcomecolumn(canbe

doneonaphone).

Session Topics Worktobecompleted Outcome

11MayWeek1

Binaryarithmetic

Page26-33Activity–Addingbinarynumbers1Activity–Addingbinarynumbers2Activity–BinaryshiftleftActivity–BinaryShiftrightQuiz:tinyurl.com/datarep04

• Addtwo8bitbinarynumbers

• Recogniseoverflowerrors

• Performbinaryshiftstomultipleanddivide8bitbinarynumbers.

18MayWeek2

Charactersets

Page34–37Activity–ASCIItable1Activity–ASCIItable2Activity–UnicodeQuiz:tinyurl.com/datarep05

• Describetheneedforcharactersets

• DiscussthedifferencebetweenASCII,ExtendedASCIIandUnicode

• Convertcharactersusingacharactersheet

01JuneWeek3

FormativeAssessment

Contact-Foranyquestionsoradditionalsupportcontact:

MrHussainm.hussain@arkglobe.org

MrDimitriadis:j.dimitriadis@arkglobe.org

MrKoramoa:e.koramoa@arkglobe.org

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 26 of 61 by Paul Long

4) Binary arithmetic

Adding binary numbers

Some binary numbers are easy to add up:

0 + 0 = 0

1 + 0 = 1

0 + 1 = 1

Example – simple binary addition 0 0 1 0 0 0 0 1

+ 0 1 0 1 1 1 0 0

0 1 1 1 1 1 0 1

Activity – adding binary numbers 1

Add the following binary numbers. Show your working.

1) 0 0 1 1

+ 0 1 0 0

2) 0 0 0 1

+ 1 0 0 0

3) 0 0 1 1 0 1 0 1

+ 1 0 0 0 0 0 1 0

4) 1 1 0 1 0 0 1 0

+ 0 0 1 0 1 1 0 1

However, 1 + 1 = 2. Two is not a binary number. It is actually 1 0 (one zero).

Remember this:

1 0

x x

2 1

1 + 0 = 2

Therefore, 1 + 1 is not that simple. It

requires a digit to be carried like in

denary arithmetic.

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 27 of 61 by Paul Long

Maths Link – adding denary numbers

When you add denary numbers with more than one digit, you have to carry 10s,

100s, 1000s etc.

2 5 3 3 + 8 = 1 1.

+ 4 4 8

1 1 is written in the units column

1 and 1 is carried.

2 5 3 5 + 4 + 1 = 1 0.

+ 4 4 8

0 1 0 is written in the tens column and

1 1 1 is carried.

2 5 3 2 + 4 + 1 = 7.

+ 4 4 8

7 0 1 7 is written in the hundreds column.

1 1

When denary numbers are added, the maximum digit in a column is 9. The next

number is 10 (ten) and so 0 is put in the column with 1 (one) carried to the next

column.

Binary works the same way. The maximum digit in a column is 1. The next number

is 1 0 (one zero) and so if a column adds up to 1 0 (one zero) a 0 (zero) is put in the

column with 1 (one) carried to the next column.

Example – carrying binary numbers 1 1 0 1 1 + 1 = 1 0 (one zero, not ten, denary value 2).

+ 0 0 1

0 0 is written in the ones column

1 and 1 is carried.

1 0 1 0 + 0 + 1 = 1.

+ 0 0 1

1 0 1 is written in the twos column.

1

1 0 1 1 + 0 = 1

0 0 1

+ 1 1 0 1 is written in the fours column.

Check in denary: 5 + 1 = 6 (1 1 0 in binary)

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 28 of 61 by Paul Long

Video

Watch http://tiny.cc/binaryadd from Khan Academy.

Example – carrying binary numbers 2 0 1 1 1 + 1 = 1 0 (one zero, not ten, denary value 2).

+ 1 1 1

0 0 is written in the ones column

1 and 1 is carried.

0 1 1 1 + 1 + 1 = 1 1 (one one, not eleven).

+ 1 1 1

1 0 1 is written in the twos column.

1 1 and 1 is carried

0 1 1 0 + 1 + 1 = 1 0 (one zero).

+ 1 1 1

1 0 1 0 0 is written in the fours column

1 1 and 1 is carried.

Check in denary: 3 + 7 = 10 (ten, 1 0 1 0 in binary)

Activity – adding binary numbers 2

Add the following binary numbers. Show your working.

1) 0 0 1 1 0 0 0 1

+ 0 0 1 0 0 1 1 0

2) 0 0 1 1 1 0 0 1

+ 1 0 1 1 0 1 0 1

3) 0 0 1 1 0 1 0 1

+ 1 0 1 1 0 1 1 1

4) 0 1 1 1 1 0 1 1

+ 0 0 1 0 1 1 0 1

5) 0 0 1 1 1 1 1 1

+ 1 0 1 1 1 1 1 1

Extension: Use the random number generator to add more binary numbers.

You can check your answers by converting the binary numbers to denary.

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 29 of 61 by Paul Long

Overflow errors

Imagine you had a field in a database that stored the year in two digit format (eg

18 instead of 2018). Imagine what would happen when adding one to the year

99:

99 + 1 = ?

Is the answer 100, 00 or something else?

Example – millennium bug Before the year 2000, many computer systems used 2 digit dates (eg 94 for 1994).

Every new year, the year is increased by one. But 99 plus one equals 100 which

has 3 digits. This was commonly known as the ‘millennium bug’ which caused

some computers to crash.

A similar problem happens when adding numbers in binary. The largest number

that can be represented by a byte is 255. If only one byte is used for storing

numbers, then an overflow error can occur if binary numbers are added together

resulting in a total bigger than 255.

Example – overflow error 0 1 1 1 1 1 1 1 127 in denary

+ 1 1 0 0 0 0 0 0 192 in denary

1 1 1 1 1 1 So far everything is OK

0 1 1 1 1 1 1 1 1 + 1 = 1 0 (one zero)

+ 1 1 0 0 0 0 0 0

0 1 1 1 1 1 1 0 is written in the column and 1 carried

1

0 1 1 1 1 1 1 1 0 + 1 + 1 = 1 0 (one zero)

+ 1 1 0 0 0 0 0 0

0 0 1 1 1 1 1 1 0 is written in the column and 1 carried

1 1

Now there is a problem. The answer needs to include the 1 that was carried.

However, this would require a 9th bit. As only a byte has been allocated, only 8 bits

are available and so there is an overflow error.

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 30 of 61 by Paul Long

Binary shifts

Binary shifts are used for multiplication and division.

Binary multiplication

Maths Link – multiplying denarys by 10, 100, 1000

To multiply an integer by 10, you shift the number to the left and put a zero in its

place:

5 x 10 = 50

To multiply an integer by 100, you shift the number twice to the left and put 2 zeros

at the end:

5 x 100 = 500

To multiply an integer by 10000, you shift the number 3 places to the left and put 3

zeros at the end:

5 x 1000 = 5000

So in denary (base 10), you would shift to the left and add zeros when multiplying

integers by 10, 100, 1000.

In binary (base 2), you shift to the left and add zeros when multiplying by 2, 4, 8.

This only applies to binary integers.

Example – multiplying binary by 2, 4, 8 To multiply a binary number by 2, put a zero at the end of the number:

1 1 x 2 = 1 1 0 (check: 3 x 2 = 6)

0 0 1 1 0 1 0 1 x 2 = 0 0 1 1 0 1 0 1 0

To multiply a binary number by 4, put two zeros at the end of the number:

1 1 x 4 = 1 1 0 0 (check: 3 x 4 = 12)

To multiply a binary number by 8, put three zeros at the end of the number:

1 1 x 4 = 1 1 0 0 0 (check: 3 x 8 = 24)

This is known as the binary shift left because all bits are moved one place to the

left.

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 31 of 61 by Paul Long

Example – binary shift left To multiply a binary number by 2, shift 1 place to the left:

0 0 0 1 0 0 0 1

x2 = 0 0 1 0 0 0 1 0 (check: 17 x 2 = 34)

To multiply a binary number by 4, shift 2 places to the left:

0 0 0 1 0 0 0 1

x4 = 0 1 0 0 0 1 0 0 (check: 17 x 4 = 68)

To multiply a binary number by 8, shift 3 places to the left:

0 0 0 1 0 0 0 1

x8 = 1 0 0 0 1 0 0 0 (check: 17 x 8 = 136)

So the general rule for a left shift, is add a zero to the right most side of the binary

number and discard the bit at the left most side of the binary number. However,

this should not be done if the left most bit is a 1 (one).

Activity – binary shift left

Calculate the following. The first number in each question is shown in binary and

the second number is shown in decimal.

1) 0110 x 4 2) 1100 x 8 3) 0101 x 2 4) 1000 x 16

5) 0001 0010 x 8 6) 0101 1110 x 2

Extension:

Each answer should be shown in 8-bit binary format (one byte).

7) 0111 1101 x 4 8) 1011 1011 x 8

9) Explain the problem with your answers to questions 7 and 8.

Binary division

The opposite rule applies for a right shift which is used for division. Add a zero to

the left most side of the binary number and discard the bit at the right most side of

the binary number. This only works for the whole number (integer) part of the

answer as any denary parts are lost.

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 32 of 61 by Paul Long

Example – binary shift right To divide a binary number by 2, shift 1 place to the right:

0 0 1 0 0 0 1 0

÷2 = 0 0 0 1 0 0 0 1 (check: 34 ÷ 2 = 17)

To divide a binary number by 4, shift 2 places to the right:

0 1 0 0 0 1 0 0

÷4 = 0 0 0 1 0 0 0 1 (check: 68 ÷ 4 = 17)

To divide a binary number by 8, shift 3 places to the right:

1 0 0 0 1 0 0 0

÷8 = 0 0 0 1 0 0 0 1 (check: 136 ÷ 8 = 17)

Activity – binary shift right

Calculate the following. The first number in each question is shown in binary and

the second number is shown in decimal.

1) 0110 ÷ 2 2) 1100 ÷ 4 3) 0100 ÷ 2 4) 1000 ÷ 8

5) 0011 1000 ÷ 8 6) 0101 1110 ÷ 2

Extension:

7) 0001 0011 ÷ 2 8) 1011 1010 ÷ 8

9) Explain the problem with your answers to questions 7 and 8.

Note: a binary shift right is an integer division (DIV) rather than a regular arithmetic

division but you don’t need to know this for the exam.

Video

Watch http://tiny.cc/binaryshift from CatComputerTeacher on YouTube.

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 33 of 61 by Paul Long

Questions – follow me 1) What is one purpose of a binary shift right? [1]

2) What is one purpose of a binary shift left? [1]

3) Why might you perform a binary shift of 3 places to the left? [1]

4) Why might you perform a binary shift of 2 places to the right? [1]

5) a) Add together the following two 8 bit binary numbers. Express your

response in an 8 bit binary form. [1]

0 1 1 0 0 0 1 1

+ 1 0 1 1 1 1 0 0

b) Identify the problem this addition has created. [1]

c) Explain why this is a problem. [2]

5) a) Perform a two place right shift on the binary number 11001010. [1]

b) Explain the effect of performing this left shift. [2]

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 34 of 61 by Paul Long

5) Character sets

Computers only work with binary bits. Therefore, all characters need to be

represented by a binary number. A character set is the set of possible characters

that can be represented by a computer system.

ASCII

The American Standard Code for Information Interchange (ASCII) is a set of bit

patterns used for standard characters and control codes. In order that all

applications can understand the data being used, a standard set of character

codes must be used.

Example – ASCII character codes Character Denary Number Binary Number

[Null char] 0 0000 0000

[Horizontal Tab] 9 0000 1001

# 35 0010 0011

+ 43 0010 1011

5 53 0011 0101

A 65 0100 0001

Z 90 0101 1010

a 97 0110 0001

z 122 0111 1011

[Delete] 127 0111 1111

7-bit ASCII

7-bit ASCII is the original character set that was adopted internationally and only

uses 7 bits. Therefore, there are 128 codes from 0 to 127.

Activity – ASCII table 1

Search the World Wide Web for an ASCII table.

What are the denary codes for the following characters and controls?

1) 8 2) P (upper) 3) q (lower) 4) * 5) [Carriage Return]

What are the characters or controls represented by the following denary codes?

6) 27 7) 85 8) 68 9) 49 10) 4

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 35 of 61 by Paul Long

Sequential codes

Character codes appear in sequence. This means they are in order. Numbers run

from 0 to 9 and letters run from A to Z and a to z. Therefore, it is possible to work

out a character code if a similar character code is already known.

Example – sequential codes You are given these ASCII character codes:

Character ASCII Code

3 51

F 46

r 114

From this, you can work out that the number 4 will have the next ASCII code after

the number 3 which will be 52 (51 + 1). You can also work out that G will have the

next ASCII code after F which will be 47 (46 + 1).

To work out further codes, draw a small table. For example, to work out the code

for the character y:

Character ASCII Code

r 114 you have been told this above

s 115 add 1

t 116 add another 1

u 117 add another 1

v 118 add another 1

w 119 add another 1

x 120 add another 1

y 121 add another 1

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 36 of 61 by Paul Long

Activity – ASCII table 2

Here are some ASCII character codes:

Character ASCII Code

5 53

P (upper) 80

c (lower) 99

Without using a full ASCII table, work out the denary codes for the following

characters:

1) 2 2) 9 3) t 4) g 5) R 6) B

Without using a full ASCII table, work out the characters for the following character

codes:

7) 49 8) 57 9) 73 10) 88 11) 109 12) 97

Write out the sentence represented by these ASCII denary codes:

67 111 110 103 114 97 116 117 108 97 116 105 111 110 115 44 32 121 111 117 32 100

105 100 32 105 116 33

Extension: write out the sentence represented by these ASCII binary codes:

01010111 01101111 01110111 00100001 00100000 00100000 01010100 01101000

01100001 01110100 00100000 01110111 01100001 01110011 00100000 01101000

01100001 01110010 01100100 00101110

Extension: set your own challenge using the converter at

http://tiny.cc/BinaryConverter

8-bit Extended ASCII

ASCII only uses 128 characters which covers the English alphabet, numbers, basic

symbols and control codes. Extended ASCII was introduced to allow some foreign

characters to be included and additional symbols. One method of Extended

ASCII uses 8 bits and so there are 256 characters available.

Example – extended ASCII Here are some of the foreign characters and symbols available using one method

of 8-bit Extended ASCII:

ü (129) é (130) ç (135) ¬ (170) ½ (171) ¥ (190) ± (241) ¶ (244)

The Ultimate GCSE CS textbook – Chapter 11 Published by paullong.net

© paullong.net 2018 Page 37 of 61 by Paul Long

Unicode

Extended ASCII had many variants and there were still only 256 characters

available. A new character set emerged.

Unicode includes a lot more characters including a wider range of symbols and

characters for many different alphabets.

The first 128 characters (codes 0 to 127) of Unicode are the same as 7-bit ASCII.

After that, Unicode has its own set of codes. For example, character codes 880 to

1023 are Greek characters.

As there are many thousands of Unicode characters, they are usually shown as

hexadecimal codes rather than binary codes.

Activity - Unicode

Search the World Wide Web for a Unicode table. What are the characters

represented by the following codes?

1) 285 (011C) 2) 952 (03B8) 3) 2835 (0B13)

Bits and codes

The number of bits used in a character set determines the number of characters

that are available. For 7-bit ASCII, there are 128 possible characters (27). For 8-bit

ASCII there are 256 possible characters (28).

Unicode has many variations, but the UTF-32 version uses four bytes for each

character. This gives over 4 million possible characters (232)

Questions – follow me 1) Define the term character set. [1]

2) Explain why ASCII is needed. [2]

3) a) How many codes are available in 7-bit ASCII? [1]

b) How many codes are available in 8-bit Extended ASCII? [1]

4) The ASCII code for Q is 81. What is the ASCII code for T? [1]

5) The Unicode code for g is 103. What is the Unicode code for c? [1]

6) Give two reasons why Unicode might be used instead of ASCII. [2]