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AQA COMP3 June 2012 Exam Thread

Reply 100

henrylee

Yeh thanks for the revision guide thing is really helpful .

Reply 101

Edwin Okli

3

Original post by Amirrryy

wow dude idk how to thank you this is made my job way easier thanks alot! now one question...is this all from our syllabus? (AQA COMP3) or is more than we need? many thanks :smile:

It's all for AQA. I went through the specification and effectively covered it all in order.

Reply 102

Amirrryy

1

is data streams and data channels the same thing? i wanna know if i can use the term data channels rather than data streams for broadband systems
thanks

Reply 103

SecondHand

Question 11 d) June 2010 - Are you meant to know what a class definition looks like? Is it in the textbook?

Reply 104

ngnav

2

Original post by Edwin Okli

I'm just checking out the WJEC specification right now, it seems to actually be quite a big larger than our COMP3 specification, with some things from our AS modules in there.

On another note. I kind of made a revision guide (from the textbook and other notes). I'll attach it here and people can tell me if there are any mistakes or content that needs to be added.

(Too large to be attached.)

OMG you explained hashing in such a way that it all makes sense to me now thank a billion. :biggrin:

Reply 105

exe

Original post by SecondHand

Question 11 d) June 2010 - Are you meant to know what a class definition looks like? Is it in the textbook?

i think it is, i had never seen one before that question though. it's quite easy to infer tbh.

Reply 106

sax100

1

can someone tell me how to represent -7.75 as floating point representation

Reply 107

username451352

11

Original post by sax100

can someone tell me how to represent -7.75 as floating point representation

I assume you mean two's compliment representation. I'm going to use an 8 bit mantissa and a 5 bit exponent.

So, you have

-7.75 = -111.11 = -00011111 \times 2^-2

now, in two's complement form, -00011111 = 11100001, and -2=-00010=11110

so you know that the floating point version of -7.75 = 1110000111110 (ie I concatenated the mantissa and the exponent.

Reply 108

Edwin Okli

3

Original post by sax100

can someone tell me how to represent -7.75 as floating point representation

-7.75 = -8 + 0.25 = 1000.01 (fixed point)

Let's have an 8 bit mantissa and 4 bit exponent.

10000100 - mantissa
Our decimal point is between the two most significant bits, so we need to move it 3 places to the right.

0011 - exponent
Therefore -7.75 = 10000100 0011

Or, as above.

(edited 11 years ago)

Reply 109

SecondHand

Original post by sax100

can someone tell me how to represent -7.75 as floating point representation

Write the number in two's complement, however many bits you like it doesn't matter, let's say 8.

Start by writing 7.75

000111.11

Convert to -7.75

111000.01

Move the decimal point so that the number starts 1.0

111.00001

We move the decimal point 3 places so the exponent is 3.
We can ignore leading 1s

1.000010 0011

10000100011

To check, convert back: 8 bits in the mantissa and 4 in the exponent.

1000010 0011

Convert the mantissa to a positive number -

0.111110

Raise by 3

111.11

=7.75
=-(7.75)

Reply 110

sax100

1

Original post by SecondHand

Write the number in two's complement, however many bits you like it doesn't matter, let's say 8.

Start by writing 7.75

000111.11

Convert to -7.75

111000.01

Move the decimal point so that the number starts 1.0

111.00001

We move the decimal point 3 places so the exponent is 3.
We can ignore leading 1s

1.000010 0011

10000100011

To check, convert back: 8 bits in the mantissa and 4 in the exponent.

1000010 0011

Convert the mantissa to a positive number -

0.111110

Raise by 3

111.11

=7.75
=-(7.75)

thanks

Reply 111

sax100

1

Original post by Edwin Okli

-7.75 = -8 + 0.25 = 1000.01 (fixed point)

Let's have an 8 bit mantissa and 4 bit exponent.

10000100 - mantissa
Our decimal point is between the two most significant bits, so we need to move it 3 places to the right.

0011 - exponent
Therefore -7.75 = 10000100 0011

Or, as above.

thanks for the help :smile:

Reply 112

sax100

1

Original post by Edwin Okli

-7.75 = -8 + 0.25 = 1000.01 (fixed point)

Let's have an 8 bit mantissa and 4 bit exponent.

10000100 - mantissa
Our decimal point is between the two most significant bits, so we need to move it 3 places to the right.

0011 - exponent
Therefore -7.75 = 10000100 0011

Or, as above.

sorry but can you tell me how to how to do 0.125 just like you showed me -7.75 using 8 bit mantissa and 4 bit exponent.

my calculations are: 00000010 - mantissa

but dont know where to put the decimal point? also how do you know where to start from with the decimal place like for -7.75 how did you know to move 3 places?

(edited 11 years ago)

Reply 113

sagederby

8

Original post by sax100

sorry but can you tell me how to how to do 0.125 just like you showed me -7.75 using 8 bit mantissa and 4 bit exponent.

my calculations are: 00000010 - mantissa

but dont know where to put the decimal point? also how do you know where to start from with the decimal place like for -7.75 how did you know to move 3 places?

The rule is that after you have moved the bicimal point if the number is positive the mantissa must start with
0.1

and if the number is negative the mantissa must start with
1.0

0.125 in fixed point is 0.001

The mantissa becomes 0.1000000 - fill up with zeros to the required number
The point has been moved two places right so the exponent is -2
Using 6 bit two's complement, the exponent will be 111110

(edited 11 years ago)

Reply 114

mfmdanny

In RegEx, does ^:
1) Negate the proceeding character, e.g. n[^t]+ n is followed by 1 or more character which isn't t
or
2) Match a character with nothing before it, i.e. ^t is followed by anything, but nothing is before it, e.g. tao, tion, tzzzzz, but NOT qt

Thanks

Reply 115

exe

Original post by mfmdanny

In RegEx, does ^:
1) Negate the proceeding character, e.g. n[^t]+ n is followed by 1 or more character which isn't t
or
2) Match a character with nothing before it, i.e. ^t is followed by anything, but nothing is before it, e.g. tao, tion, tzzzzz, but NOT qt

Thanks

first one

Reply 116

mfmdanny

Original post by exe

first one

Thanks

Reply 117

Edwin Okli

3

Original post by mfmdanny

In RegEx, does ^:
1) Negate the proceeding character, e.g. n[^t]+ n is followed by 1 or more character which isn't t
or
2) Match a character with nothing before it, i.e. ^t is followed by anything, but nothing is before it, e.g. tao, tion, tzzzzz, but NOT qt

Thanks

Wikipedia says both.
When not in square brackets it acts like 2), when in square brackets it acts like 1).

Reply 118

SecondHand

It's because the ^ is used for two different things in regex. It can mean negation in the square brackers but it can also mean the start of the string. Sometimes it's better to be explicit with these things when doing validation checks.

Like if you want to match only the words grey and gray then it would be best to use

^gr[ea]y$ rather than gr[ea]y because some programming languages when you ask them to match regex will interpret it as matching any substring of the string you are trying to match. ie the second would match aslkdjaskldjgrey which is clearly invalid.

(edited 11 years ago)

Reply 119

JSleek

QUICK QUESTION !:

In the AQA specification, under 3.3.6-Communication and Networking, there is a small paragraph which says:

"Compare and contrast thin-client computing (Saas, AJAX, Web 2.0,etc) vs rich client computing(client-server, peer-to-peer), web services as examples of 'systems architectures'."

Would anyone know how to go about doing this?(Briefly)

Thanks!

AQA COMP3 June 2012 Exam Thread

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