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Saturday, December 21, 2013

Autodesk 123D test

Merry Xmas, I used a lamb from a nativity scene...
original object, 
hi-res triangulated mesh from 123D, 
automatically created texture map from 123D, 
and test render from Houdini with 500 instanced lambs.

Friday, December 20, 2013

Thursday, December 19, 2013

links updated

some renderman / RSL /OSL links added to my list:

thanks to Jimmy Gunawan and Michael Anders

Saturday, August 24, 2013

Process Scheduling Algoithms:

Round Robin:
each process gets a strict amount of time on the CPU 
(its quantum is typically 10 to 100 ms)
seems fair
easy to implement
Round Robin is pre-emptive - its specifically designed for time-sharing systems
there is no risk of starvation*

but if the timeslice is too big round robin tends to a FCFS system
if the quantum is too small, RR tends to become 

Typically Round Robin has a better average response time than a SJF - Shortest Job First system.

A potential deficiency of Round Robin: processes that are IO-bound* often do not use their full quantum before being interrupted by an IO operation. Once blocked and evicted, they then get moved to the back of the queue, so their spot in the queue is essentially lost to another process. they go 'hungry' for CPU time

so Round Robin as a scheduling algorithm is not really as fair as it might initially seem.

* Starvation / indefinite blocking : can occur when a scheduling system leaves a lower priority job indefinitely

* IO bound Process - the process spends more time frequently interrupted by IO operations, and subject to context switches thus often doesn't use its full quantum of CPU time.

* CPU bound process - spends more time on average being executed in the CPU, 
than being interrupted by IO operations.

Youtube videos of scheduling algorithms in action:

(Priority Scheduling, a 'pre-emptive verison' of SJF)

Round Robin

Friday, August 23, 2013

Edsger W. Dijkstra.

...the manuscripts of Edsger W. Dijkstra, 1930–2002

example of pathfinding from macha

Vimeo Animation:

Monday, July 22, 2013

aberration post updated

Chromatic Aberration post from aeons ago updated with link to After Effects tutorial
by Cyril Kotecky (Jul 19, 2013.)

Tuesday, April 23, 2013

tips and tricks for digital number formats, logic gates and Boolean Algebra.

from one of my colleagues - thanks Anon.
  • You must be really comfortable in interchanging between types of number representation.
  • Write down the RANGE for every number representation available in your notes.
  • A XOR B = ~AB + ~BA
  • De Morgan’s rule, De Morgan’s rule everywhere. Not only on Boolean algebra, some ALU questions have them.
  • If there seems to be no answer, try using De Morgan’s rule if appropriate.
  • From my experience, (tutors please correct me if I am wrong), using De Morgan’s rule on 3 variable also works.
  • BOOLEAN IDENTITIES, the easiest and most important including:
AA = A, A + A = A, (A+ BC) = (A+ B)(A+C), A(A+B) = A, A + AB = A

Knowing this only is not enough, you NEED practice. For those of you who are going to say “challenge accepted”, go on, make my day.

  • How to convert from one’s complement to two’s complement? add the sign to the number. e.g. 1110 in one’s complement: add the sign to the number, 1110 + 1 = 1111 in two’s complement. second example, 0110 in one’s complement: add the sign to the number, 0110 + 0 = 0110 in two’s complement.
  • Conversely, from two’s complement to one’s complement, take the sign number instead of adding.
  • Negative numbers also have odd and even.
  • Two’s complement, signed number, unsigned, excess-128: don’t affect whether it is odd or even.
  • One’s complement however, when negative: the LSB is 1 for even, 0 for odd. When positive à 0 for even, 1 for odd.
  • Binary Coded Decimal: e.g. 789, become 0111 1000 1001. Always use 4 bits for each number (since max is 9, 1001).
  • I wasted a lot of time simplifying the sum of products on questions 53 – 57 on sample midsem paper. IMO it is quicker to just sub in A B C with their value on the table and see if the result matches.
  • Use ruler for questions like q110.
  • 2 inputs XOR gate and NAND gate can be used as an inverter. If one input is 1 and is fixed, no matter what the other input is, the output will be that other input inverted. e.g. 1 XOR A = ~A. 1 NAND A = ~A.

Friday, April 5, 2013

Designing circuits in logisim...

Logisim is great to prototype your circuits. This one worked, thankfully - as I also designed it on paper a couple of times which took a while :D

If any 2 of 3 bits of input in an incoming bitstream (B) are 1's then X will be 1,
(the X output LED will light up)
and if 3 1's re detected - the Y output will be 1
(the Y LED will light up)

You can 'poke' the clock to make it tick over after you've adjusted your inputs, or you can mke it tick automatically at a given Hz.  

Saturday, March 30, 2013

I have been invited to attend an award ceremony at Flinders University to accept a Chncellor's "Letter of Commendation" for my 1st yr of studies, Computer Science.

Sunday, March 3, 2013

IEEE-754 (anon. post)

Summary of IEEE-754

This is a summary of what I understand about IEEE-754 notation. 
For the purpose of explanation, single precision is chosen.

What is being written?
IEEE-754 is another way to represent the scientific notation.
Example of scientific notation : 1.00 x 10^5
Where 1.00 represents the fraction part (Mantissa), and 5 represent the exponent part. 10 represents the radix.
Note that the explanation below uses binary.

Notation :

First bit = Sign of the number, 0 as positive, 1 as negative - the 'sign bit'
The next 8 bits = the Exponent part. Represented by 8-bit (in excess-127 format)
FORBIDDEN à 00000000 and 11111111 are forbidden for normalised single precision, will be explained later.
00000000 as an exponent is ONLY FOR DENORMALISED numbers
11111111 as exponent is ONLY FOR INFINITY.
The last 23 bits = the Fraction part (aka the Mantissa).
The next 23 bits represent the bits to the right of the binary point. To the left of the binary point is always a 1 (strictly for normalised - and therefore it can be disregarded and assumed to be a one).
All 0 (twenty-three 0's) represents the value of 1.0 while all 1 (twenty-three 1's) represents the value of approximately 2.0.
Note : Binary point is the dot between the number of the fraction part,
like a decimal point - but in binary :)

Range :

For negative : From -2.0 x 2^(127) to -1.0 x 2^(-126)
For positive : From 1.0 x 2^(-126) to 2.0 x 2^(127)

Example :

1.125 (decimal) in IEEE-754 :
In scientific notation of binary: 1.125 = 1.001 x 2^(0).
Sign : Positive (0)
Fraction part : 1.001 (2^0 + 2^(-3))
Exponent part : 0
Solve the Exponent part first; exponent part must be in the form of 8-bit excess-127.
0 in 8-bit excess-127 à 10000000

Fraction part à In IEEE-754, ignore the 1 to the left of the binary point, and COPY the numbers to the right of the binary point (001 in this example) and then fulfill the 23 bits requirement (in this case 001 + twenty more 0s).

Therefore the 1.125 in IEEE-754 notation become:
0 10000000 00100000000000000000000

(Spaces between the sign, exponent and fraction were used to make things more visible)
To make it shorter, convert to Hexadecimal
1.125 in IEEE-754 notation in hexadecimal = 40100000

Basically the same as normalised. However, exponent part is ALWAYS 00000000. Although in excess-127 00000000 is considered to be -127, FOR denormalised form, 00000000 indicates that the exponent is -126, but the number to the left of the binary point of the fraction part is ALWAYS 0.

Note :

For denormalised, all 0 in the fraction part represents the value of 0.0 and all 1 in the fraction part represents the value of approximately 1.0.

What can be written with denormalised?
Example : 0.001 x 2^(-126)
Fraction part is ALWAYS BELOW 1.0, exponent part is ALWAYS -126.
Range :
Negative : From 0 to -1 x 2^(-126)
Positive : From 0 to 1 x 2^(-126)
Note : Two zeros exist: +0 and -0.
Smallest possible non zero value à -2^(-23) * 2^(-126) and 2^(-23) * 2^(-126)

Basically the same thing except the exponent part is 11111111 with all 0 as fraction part.
Note : There exist positive infinity and negative infinity.

32 bit and 64 bit floating point formats:

Saturday, March 2, 2013

handy summary of digital Number formats:

one of my fellow students posted (anon.) this handy summary of
digital Number formats:
8-bit unsigned
Range : 0 to 255
Note : Always positive . Start from 00000000 (0) to 11111111 (255).

Example :
69 = 01000101

8-bit signed magnitude
Range : -127 to 127
Note : Two zeros, +0 (00000000) and -0 (-00000000).
MSB is only the positive or negative sign, with 0 as positive and 1 as negative, contain no value.
Start from 10000000 (-127) to 01111111 (127).

Example :
+69 = 01000101
-69 = 11000101

8-bit One's Complement
Range : -127 to 127
Note : Two zeros, +0 (00000000) and -0 (11111111).
MSB is always negative and has value of -127. Start from 10000000 (-127) to 01111111 (127).

Example :
+69 = 01000101
-69 = 10111010

How to get this from signed number (!!) :
If MSB of signed number is 0 (positive), same as signed number
If MSB of signed number is 1 (negative), keep MSB as 1, invert the rest of the bits.

NB. on my current practice exam questions there are a number of questions such as the following:

Which of the following statements about the 8 bit binary number 10001111 is true ?
a) the number is odd when considered to be in signed magnitude, 2's complement, 1's complement, or excess 128 bit formats. 

In general a binary number ending in a zero is usually even unless it is in 1's complement format and its negative, in which case ending in a zero will cause it to be negative and a bin number ending in 1 will be positive, but for all other cases , as per normal - a bin ending in 0 will be even and ending in a 1 will be negative.

8-bit Two's Complement
Range : -128 to 127
Note : -128 written as 100000000, only one zero; +0 (00000000).
MSB is always negative and has value of -128. Start from 10000000 (-128) to 01111111 (127).

Example :
+69 = 01000101
-69 = 10111011

How to get this from signed number (?) :
If MSB of signed number is 0 (positive), same as signed number
If MSB of signed number is 1 (negative), keep MSB as 1, invert the rest of the bits, add 1 to the LSB.

Addition in 2's compliment format 

when adding bit sequences (on paper) and you have a 'carry bit' on the left at the MS end 
(most significant end) you can simply discard this carry bit.

8-bit excess-128
Range : -128 to 127
Note : From decimal number, add 128 to the value and write the 8-bit unsigned.
Start from 00000000 (-128) to 11111111 (127).

Example :
+69 = 69 + 128 = 197 = 11000101
-69 = -69 + 128 = 59 = 00111011

Binary to Octal/Hex and vice versa
Start from the right hand side (or the least significant bit) and group the bits in groups of three for octal, and four for hexadecimal. Then treat each group as an individual binary string and the result is the

Octal/Hexadecimal representation. 
For Binary strings that don't divide evenly into groups of three or four, just add as many zeros as needed to the left side.

Example :
1000101 can become 001 | 000 | 101 for the purpose of conversion (note the
added zeros to the left side)

Hence 69 (or 1000101) is 105 in Octal.

To convert Octal or Hexadecimal to Binary, all you need to do is expand each character into the appropriate number of binary digits (three for octal or four for Hex)

Also, if you want to convert from Hex to Octal or vise-versa, the easiest
way to do it is use conversion to binary in between.

Tuesday, February 26, 2013


I have applied for credit for 1st yr Comp Science at the University of QLD,
and I hope to continue with second yr subjects.

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