Software CSSA Trial HSC (2 Viewers)

[harrisony]

We got a sheet from teacher. And its going to be in our exams.

Learn Boolean algebra! Saves you so much time guessing and checking. I spent 1 hour guessing/checking a 12 row truth table until I got the gates haha.

Something tells me you will like Discrete Mathematics

 

[SpiralFlex]

where the heck do you get your laws???? LOL. You engage it like a professional...and there's me: "Ok lets try this...nope?..okay how about this?..no? ok now this should work" LOL

but yeah do you know how to answer the ps2 question? cuz i have no clue...

I haven't seen the paper. Sorry.

 

[onedaysnotice]

We got a sheet from teacher. And its going to be in our exams.

Learn Boolean algebra! Saves you so much time guessing and checking. I spent 1 hour guessing/checking a 12 row truth table until I got the gates haha.

Yeah I plan to, after the trials

and omg I would just dropped dead if I saw table like that. Dx So much dedication...

Guessing you're gonna do CompSci at uni?

 

[onedaysnotice]

No. Advanced Mathematics. <333333333333333333 My love for SDD is marginal compared to maffs.

MATHS IS EVILLLLL!!!! Dx

It sucks, cuz I suck at maths and I'm gonna have to do maths at compsci....fahking hell. Math just isnt my thing yanno? xD

It seems like youre set for your uni course tho

oh yeah, would they accept an XNOR in the HSC? Cuz my friend used an xnor for that truth table question lol. It works out tho...i think.

 

[BlakeOlivo]

I ended up doing that first short response at the end with only 6 minutes left LOL. I didn't get to finish it xD Frkn system flowcharts....

and for your gates i interpreted it as this:

A------
`````OR-------
B------```````|-------
``|``````````````NAND------
``l-----------------------

which almost works, but doesn't.

Here was the truth table:

A B Output
0 0 1
0 1 0
1 0 1
1 1 1

Your gate design results in this:

A B Output
0 0 1
0 1 0
1 0 1
1 1 0

Because the B is 1 and the OR gate outputs 1, so when it enters the NAND gate it is a 1, but then is reversed to a 0.

I think my understanding of NAND is correct....LOL. A NAND gate gets the output of if it was an AND gate, and reverses it right?

You're right, except from memory, I thought the last values A=1 and B=1 produced a 0 as the output in the question? Maybe I'm just hoping I'm not wrong LOL.

 

[BlakeOlivo]

where the heck do you get your laws???? LOL. You engage it like a professional...and there's me: "Ok lets try this...nope?..okay how about this?..no? ok now this should work" LOL

but yeah do you know how to answer the ps2 question? cuz i have no clue...

hahahaha, floating on the same boat here!! I was guessing and checking SO many times, it just got to me and I burned the paper...in my dreams

 

[onedaysnotice]

You're right, except from memory, I thought the last values A=1 and B=1 produced a 0 as the output in the question? Maybe I'm just hoping I'm not wrong LOL.

Me and my friend remembered it to only have one output of 0, which was a=0 and b=1 aha. We shall see 8)

 

[ThievingSix]

You could only use 2 gates, that was specified in the question. Who ever wrote the previous proof is using the incorrect truth table. This is the correct diagram and table as below and produces the correct output.

In boolean algebraic terms the expression is as below. You don't need to know how to apply DeMorgan's theorems to the Sum Of Products. You simply had to know the truth table for an OR gate, and chuck in a NOT gate for the 2 inputs that needed to be different.

B' + A

For the PS2 Mouse question it required you to know that each chunk on 7 bits made up a byte, why? because the first bit was the header information separating each byte(and the nature of each byte) they didn't show it because it would have confused people way too much and it wasn't really relevant to the question, you just had to apply your knowledge to the 7 bits presented in the 3 bytes.(pg473 2012 davis textbook). They were even so kind to choose a question that had an error checking bit which was always zero, hence if you lost your place and the bit wasn't zero, it would be obvious.

The first byte gave you the fact the left click was pressed, the direction the mouse moved on the x plane was negative and the direction in the y plane was positive. The second byte gave you the amount the mouse moved in the x plane which was 72 in the negative. The third byte gave you the amount in the y plane which was positive 224.

So in summary the answer to the question was simply an interpretation of the binary based on the table given. i.e. the answer was;

Byte 1 - Left click pressed, x direction negative, y direction positive
Byte 2 - 72 in the x direction
Byte 3 - 224 in the y direction

Now apply the signs to each of the directions and graph it. Giving you the point (-72, 224). You had to either graph it and indicate the left click or write what each of the bytes gave for full marks, you did not have to do both.

 

[BlakeOlivo]

You could only use 2 gates, that was specified in the question. Who ever wrote the previous proof is using the incorrect truth table. This is the correct diagram and table as below and produces the correct output.

In boolean algebraic terms the expression is as below. You don't need to know how to apply DeMorgan's theorems to the Sum Of Products. You simply had to know the truth table for an OR gate, and chuck in a NOT gate for the 2 inputs that needed to be different.

B' + A

For the PS2 Mouse question it required you to know that each chunk on 7 bits made up a byte, why? because the first bit was the header information separating each byte(and the nature of each byte) they didn't show it because it would have confused people way too much and it wasn't really relevant to the question, you just had to apply your knowledge to the 7 bits presented in the 3 bytes.(pg473 2012 davis textbook). They were even so kind to choose a question that had an error checking bit which was always zero, hence if you lost your place and the bit wasn't zero, it would be obvious.

The first byte gave gave you the fact the the left click was pressed, the direction the mouse moved on the x plane was negative and the direction in the y plane was positive. The second byte gave you the amount the mouse moved in the x plane which was 72 in the negative. The third byte gave you the amount in the y plane which was positive 224.

So in summary the answer to the question was simply an interpretation of the binary based on the table given. i.e. the answer was;

Byte 1 - Left click pressed, x direction negative, y direction positive
Byte 2 - 72 in the x direction
Byte 3 - 224 in the y direction

Now apply the signs to each of the directions and graph it. Giving you the point (-72, 224). You had to either graph it and indicate the left click or write what each of the bytes gave for full marks, you did not have to do both.

Thanks for going to an effort to answer our questions . It makes sense now, however, doesn't it seem quite unfair on the people being tested to configure that out in test conditions? Especially since it is the last question of the paper.

 

[onedaysnotice]

You could only use 2 gates, that was specified in the question. Who ever wrote the previous proof is using the incorrect truth table. This is the correct diagram and table as below and produces the correct output.

In boolean algebraic terms the expression is as below. You don't need to know how to apply DeMorgan's theorems to the Sum Of Products. You simply had to know the truth table for an OR gate, and chuck in a NOT gate for the 2 inputs that needed to be different.

B' + A

For the PS2 Mouse question it required you to know that each chunk on 7 bits made up a byte, why? because the first bit was the header information separating each byte(and the nature of each byte) they didn't show it because it would have confused people way too much and it wasn't really relevant to the question, you just had to apply your knowledge to the 7 bits presented in the 3 bytes.(pg473 2012 davis textbook). They were even so kind to choose a question that had an error checking bit which was always zero, hence if you lost your place and the bit wasn't zero, it would be obvious.

The first byte gave gave you the fact the the left click was pressed, the direction the mouse moved on the x plane was negative and the direction in the y plane was positive. The second byte gave you the amount the mouse moved in the x plane which was 72 in the negative. The third byte gave you the amount in the y plane which was positive 224.

So in summary the answer to the question was simply an interpretation of the binary based on the table given. i.e. the answer was;

Byte 1 - Left click pressed, x direction negative, y direction positive
Byte 2 - 72 in the x direction
Byte 3 - 224 in the y direction

Now apply the signs to each of the directions and graph it. Giving you the point (-72, 224). You had to either graph it and indicate the left click or write what each of the bytes gave for full marks, you did not have to do both.

Good your gate design is exactly the same as mine

And thanks for clarifying the ps2 question, although I'm still kinda lost XD. Well it seems its time I actually read my textbook...after maths and physics trials Dx.

 

[onedaysnotice]

Thanks for going to an effort to answer our questions . It makes sense now, however, doesn't it seem quite unfair on the people being tested to configure that out in test conditions? Especially since it is the last question of the paper.

They like to screw with our minds I tell ya xD

 

[mickk794]

I stuffed up a few parts for this exam. Totally forgot my theory behind the logic gates question.
Hopefully i keep my 1st placing at my school though.

 

[harrisony]

We got a sheet from teacher. And its going to be in our exams.

Learn Boolean algebra! Saves you so much time guessing and checking. I spent 1 hour guessing/checking a 12 row truth table until I got the gates haha.

Actually, have you checked with your teacher that your able to use boolean algebra?

 

[SpiralFlex]

Actually, have you checked with your teacher that your able to use boolean algebra?

Yes he is attaching the sheet as a kind of standard integrals thing. It is introduced in the syllabus anyway

 

[harrisony]

Should have checked the latest syllabus

 

[DaHowdog]

wow...kinda glad we did programming paradigms.
SOFTY 2012 woof woof

 

[SpiralFlex]

^What...the..

Anyone else have their SDD exam tomorrow? Hopefully we'll get something that is not dodgy.

 

[Pelicanxo]

30/97 (3 questions were taken out of the binary section). consdering i didn't answer 40% of it, I think i did quite well

 

[onedaysnotice]

I got 74/100

My teacher said I made so many silly mistakes Dx fml.

EDIT: She gave another mark 75/100

 

[huynguyencab]

I got 74/100

My teacher said I made so many silly mistakes Dx fml.

EDIT: She gave another mark 75/100

WTH, you got the result just after a week?

Mine starts on Wednesday , functional language - paradigm :|. A question: How do you guys desk check? I usually desk check by drawing a table, put the lines correspondingly to the number of lines of the algorithm.