Lecture 07

1
Lecture 07 October 11, 2001

• Programming our toy machine
• Review of hardware
• Storing images and sounds
• Compression
• Administrivia
• send questions to cs111_at_cs.princeton.edu
• not cs111_at_princeton.edu
• NO lab sessions on Fridays
• (do the lab before Friday if you want help from
  the lab TAs)
• Collaboration policy
• Everyone does their own writeup

2
CPU
Memory Registers
Control Unit(State Machine)
3
Our Example CPU

• Well look at a toy CPUs Machine Language
• Our CPU has
• 8 Registers each holds 16 bits (2 bytes)
• Our RAM
• Works in blocks of 16 bits (2 bytes)
• Has 28 256 addresses
• Total Memory 512 bytes

4
Machine Core

• Everything is in binary (hex)

Registers
R0 0000 R1 0CA8 R2 A9DB R3 0705 R4 1011 R5
90A0 R6 0807 R7 00A0
5
Representing Instructions

• Machine instructions will also be representedby
  16 bits.
• We divide those bits into 4 groups - 4 bits
  each
• The first group of 4 bits is called the Op-Code
• Tells what instruction it is.

6 C A 5
6
Instructions

• 16 possible Op-Codes
• Well only look at a fewin detail

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
7
Halt Instruction

• 0 Halt
• This just tells the machineto stop.
• Other 12 bits are ignored
• All have same effect.

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
0 0 0 0
0 F F F
0 9 A C
8
Data Instructions

• B Load Direct / Address
• Sets a Register to a fixedSpecified Value
• Encoding
• Effect Register A becomesthe specified value

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
B regA value (8 bits)
9
Arithmetic/Logic Instructions

• 1 Add
• Adds contents of two registerstogether and puts
  result in thirdregister
• Encoding
• Effect Register A becomesRegister B Register C

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
1 regA regB regC
10
Adding Numbers

• Simple Program to calculate1 2 3 4 5 6
  21 15 (hex)

10 Load R0 ? 0001 (always 1) 11 Load R2 ?
0000 (running total) 12 Load R1 ? 0001
(current number) 13 Add R2 ? R2 R1
(R21) 14 Add R1 ? R1 R0 (R12) 15 Add
R2 ? R2 R1 (R23) 16 Add R1 ? R1 R0
(R13) 17 Add R2 ? R2 R1 (R26) 18 Add
R1 ? R1 R0 (R14) 19 Add R2 ? R2 R1
(R2A) 1A Add R1 ? R1 R0 (R15) 1B Add
R2 ? R2 R1 (R2F) 1C Add R1 ? R1 R0
(R16) 1D Add R2 ? R2 R1 (R215) 1E halt
11
Adding Numbers

• Simple Program to calculate1 2 3 4 5 6
  21 15 (hex)
• Lets watch it execute

12
Arithmetic/Logic Instructions

• 2 Subtract
• Similar to Add
• Encoding
• Effect Register A becomesRegister B - Register C

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
2 regA regB regC
13
Control Instructions

• 6 Jump if Positive
• Jump to a different placein memory if Register
  is gt 0
• Encoding
• Effect If Register A gt 0,Go To Instruction at
  specifiedaddress

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
6 regA address (8 bits)
14
Adding Numbers

• Use a Loop to calculate1 2 3 4 5 N

10 Load R1 ? 0006 (N) 11 Load R2 ?
0000 (running total) 12 Load R0 ? 0001
(always 1) 13 Add R2 ? R2 R1
(add in N) 14 Sub R1 ? R1 - R0 (N N-1)
15 Jump to 13 if (R1gt0) (If N isnt 0
yet, go back) 16 halt (N is
now 0, and R2 12N)
15
Adding Numbers

• Use a Loop to calculate1 2 3 4 5 N
• Execution

16
Control Instructions

• 7 Jump and count (backwards)
• Jump to a different placein memory if Register
  is gt 0,AND decrement by 1
• Encoding
• Effect If Register A gt 0,Go To Instruction at
  specifiedaddress, and A A - 1

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
7 regA address (8 bits)
17
Adding Numbers

• Alternate Loop using Jump Count for1 2 3
  4 5 N

10 Load R1 ? 0006 (N) 11 Load R2 ?
0000 (running total) 12 Add R2 ? R2
R1 (add in N) 13 Jump to 12 and
(If N isnt 0 yet, decrease R1 if (R1gt0)
Let NN-1, and go back) 14 halt
(N is now 0, and R2 12N)
18
Adding Numbers

• Alternate Loop using Jump Count for1 2 3
  4 5 N
• Execution

19
Memory Instructions

• 9 Load (from memory)
• Load from Memory into Register
• Encoding
• Effect Load from RAM atspecified address
  intoRegister A

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
9 regA address (8 bits)
20
Memory Instructions

• A Store (into memory)
• Store Register into Memory
• Encoding
• Effect Store contents ofRegister A into
  memoryat specified address

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
A regA address (8 bits)
21
Memory Instructions Indirect

• 9 Load (from memory)
• Load from Memory into Register
• Encoding
• Effect Load from RAM ataddress BC into
  Register A

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
9 regA regB regC
22
Memory Instructions Indirect

• How can CPU tell these apart?
• Sneaky trick
• CPU has 8 Registers,so only need 3 bits to
  specifyRegister A
• Use extra bit to tell which type of LOAD

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
9 regA address (8 bits)
9 regA regB regC
23
Hacks

• This is called a HACK!
• Hacks are terrible! Generally should be avoided.
• Sadly, they are everywhere.
• Some CS people get perversepleasure out of
  hacks.
• This is a major source of bugs!

0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
24
A Virus!

• Copies itself on top of other programs!

10 Load R1 ? 0006 (Length of
Virus) 11 Load R2 ? 001F (Memory
Location to copy self To -1) 12 Load R3 ? 000F
(Memory Location to copy self From
-1) 13 Load R0 ? MemoryR3R1 (Load
instruction from source) 14 Store MemoryR2R1
? R0 (Store instruction in destination) 15
Jump to 13 and (If havent copied
whole virus, decrease R1 if (R1gt0)
decrease R1, and go back) 16 halt
(Virus has replicated) ...
__________________________________________________
____________ 20 . . . .
(Program about to be destroyed.) 21 . . . .
...
25
A Virus!

• Copies itself on top of other programs!
• Execution

26
Summary

• We looked at a typical Machine Languageand how
  instructions are represented.
• Halt Instruction
• Data Instructions
• Arithmetic/Logic Instructions
• Control Instructions
• Memory Instructions

27
Summary (cont.)

• We saw a few simple examples of programsyou can
  write in Machine Language.
• You now have a pretty solid understandingof how
  computers really work !

28
Hardware Wrap-up
29
Layers of Abstraction

• Build more and more powerful tools out of simpler
  ones.

Really Simple Stuff
Computers
30
Logic Gates
X
Z
Y
X
Z
Y
Y
X
X
Y
31
Universal Method
First, make circuitsfor each row in Truth Table
witha 1 in the output.
A B C F 0 0 0 0 0 0 1 1 0 1 0 1 0 1
1 0 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 0
A
B
C
A
B
C
A
B
C
32
Universal Method (cont.)
A
Finally, combine themwith an OR gate.
B
C
A
F
B
C
A
B
C
33
Universality

• For ANY Truth Table we can write down,we can
  make a circuit for it using only 3 Logic Gates
  AND, OR, NOT
• In real life, make AND/OR/NOT gatesout of
  Transistors.

34
Representing Information

• Information in the world comes in manyforms
  letters, numbers, pictures, sounds
• How can we represent these thingswith 0s and
  1s?
• Start with numbers.

35
The Flip-Flop
Set
M
Reset

• M becomes 1 if Set is turned on
• M becomes 0 if Reset is turned on
• Otherwise (if Set and Reset are both 0), M just
  remembers its value

36
Snapshot Memory
D
M
Write

• When Write changes from 1 to 0,the value of D
  gets stored in M. (snapshot)
• Otherwise, M just remembers its value.(It
  ignores D.)

37
Design for Any State Machine
Current State
Memory Register
Logic For Next State Output
Clock
Inputs
Outputs
38
Review
On our way to building a computer
Represent Manipulate Info. with 0s 1s (aka
bits)
Memory Circuits
State Machines

Computers
We are here
39
Putting it all together

• Computer has many parts, connected by a Bus
• Each part is a state machine.

CPU
Keyboard
Display
Bus
Hard Disk
CD-ROM
RAM
40
Putting it all together

• This cycle is orchestrated by the Control Unitin
  the CPU.

Memory Registers
Control Unit(State Machine)
41
Instructions
0 halt 1 add 2 subtract 3 multiply 4 bus
output 5 jump 6 jump if positive 7 jump
count 8 bus input 9 load A store B load
direct/addr. C NAND D AND E Shift Right F
Shift Left
42
Adding Numbers

• Simple Program to calculate1 2 3 4 5 6
  21 15 (hex)

10 Load R0 ? 0001 (always 1) 11 Load R2 ?
0000 (running total) 12 Load R1 ? 0001
(current number) 13 Add R2 ? R2 R1
(R21) 14 Add R1 ? R1 R0 (R12) 15 Add
R2 ? R2 R1 (R23) 16 Add R1 ? R1 R0
(R13) 17 Add R2 ? R2 R1 (R26) 18 Add
R1 ? R1 R0 (R14) 19 Add R2 ? R2 R1
(R2A) 1A Add R1 ? R1 R0 (R15) 1B Add
R2 ? R2 R1 (R2F) 1C Add R1 ? R1 R0
(R16) 1D Add R2 ? R2 R1 (R215) 1E halt
43
Adding Numbers

• Use a Loop to calculate1 2 3 4 5 N

10 Load R1 ? 0006 (N) 11 Load R2 ?
0000 (running total) 12 Load R0 ? 0001
(always 1) 13 Add R2 ? R2 R1
(add in N) 14 Sub R1 ? R1 - R0 (N N-1)
15 Jump to 13 if (R1gt0) (If N isnt 0
yet, go back) 16 halt (N is
now 0, and R2 12N)
44
A Virus!

• Copies itself on top of other programs!

10 Load R1 ? 0006 (Length of
Virus) 11 Load R2 ? 001F (Memory
Location to copy self To -1) 12 Load R3 ? 000F
(Memory Location to copy self From
-1) 13 Load R0 ? MemoryR3R1 (Load
instruction from source) 14 Store MemoryR2R1
? R0 (Store instruction in destination) 15
Jump to 13 and (If havent copied
whole virus, decrease R1 if (R1gt0)
decrease R1, and go back) 16 halt
(Virus has replicated) ...
__________________________________________________
____________ 20 . . . .
(Program about to be destroyed.) 21 . . . .
...
45
Further Directions

• Weve covered the basics of computers.
• What weve left out
• How to make computers faster.
• How to make circuits more efficient
• Large machine languages
• Pipelining make the computer execute more
  than one instruction at once.
• Parallelism having many CPUs work together
• To learn more, take CS courses (217,306,471)

46
Sounds and Images

• How do we represent sounds?
• How do we represent images?
• These representations are too big, can we
  compress?
• How do we manipulate them?
• Next few lectures
• Next 2 labs

47
Representing sounds

• Start from waveform of the sound
• Then digitize
• Eventually, consider frequencies

48
The waveform
49
Sampling
50
The approximation
51
Finer sampling
52
Even finer sampling
53
Analog vs. digital

• In analog technology, a wave is recorded or used
  in its original form. For example, in an analog
  tape recorder, a signal is taken straight from
  the microphone and laid onto tape. The wave is
  analog. It can be read, amplified and sent to a
  speaker to produce the sound.
• In digital technology, the analog wave is sampled
  at some interval, and then turned into numbers
  that can be stored digitally.

54
Digital sound

• On a CD, the sampling rate is 44,100 samples per
  second. Each sample is quantized into a number in
  the range from 0 to 65,535, So a CD has 44,100
  numbers (each 2 bytes) per second of music.
• Numbers are used to make a voltage wave that
  approximates the original wave.
• Advantages of digital technology
• Recording does not degrade over time.
• Groups of numbers can often be compressed by
  finding patterns in them.
• It is easy to use special computers called
  digital signal processors (DSPs) to process
  streams of numbers.

55
Numeracy

• CD has 44,100 2 byte samples per second.
• 44,100 x 2 x 60 x 2 (channels) 10.09 megabytes
  per CD minute
• A CD is 74 minutes (746 megabytes) long.
• MP3 players typically use memory 32 or 64
  megabytes in size. (or 3-6 minutes)
• Compression is needed

56
Frequency domain

• Sounds are made up of frequencies.
• We often want to filter out certain frequencies
  scratches, background noise, or because we are
  unable to hear them.
• How do we do such things?

57
Enter Jean Baptiste Joseph Fourier
He lived 1768-1830 in France Developed the idea
of building trigonometric series for turning
signals measured over time into signals measured
by frequency His ideas are the basis of
digital signal processing (DSP)
58
Demonstration

• Fourier transform

59
Representing Images

• Start from analog representation (waveform) of
  the image
• Then digitize
• Eventually, consider frequencies

60
Digital resolution

• A printer will display 300 or 600 or 1200 dots
  per inch (dpi).
• Visually
• Less than 150 dots per inch is tiring
• 200 dots per inch makes a very crisp screen
  display
• 600 dots per inch is now the norm
• 1200 dots per inch was what phototypesetters did
• Photograph negatives typically have 2000 dots per
  inch
• Each dot can have 256 shades of selected colors.
• Printers use Cyan, Yellow, Magenta and sometimes
  blacK
• Monitors use Red, Green and Blue .

61
Numeracy

• An 8x10 picture at 600 dpi has
• 80 square inches
• 360,000 dots per square inch
• 28,800,000 dots per picture
• Each dots can be 3 bytes
• 86.4 Megabytes of storage
• So, compression is needed

62
Representing Images

• Each dot is a picture element (or a pixel)
• So, a picture is just a stream of bytes.
• When we process the picture (as we will in the
  next lab) we manipulate the bytes.

63
Frequency domain for images

• Images are made up of frequencies.
• We often want to filter out certain frequencies
  scratches, background of image,
• By checking for frequencies, we can find places
  where the image changes.
• How do we do such things?

64
The need for compression

• A CD is 746 megabytes long
• A printed page requires 86.4 Megabytes of storage

65
Compression

• Start with text
• Eliminate duplication
• Be more clever
• Move to sounds and images
• Remove frequencies that wont be heard/seen
• Apply text ideas
• Be willing to lose part of the image/sound that
  doesnt matter

66
Simple compression

• A sentence to compress
• "Ask not what your country can do for you -- ask
  what you can do for your country."

67
Simple compression (cont.)

• "Ask not what your country can do for you -- ask
  what you can do for your country."
• The quote has 17 words, made up of 61 letters, 16
  spaces, one dash and one period. Storing this,
  takes 79 units of memory.

68
How to shrink

• Use redundancy
• Immediately, we notice that
• "ask" , "what" , "your" , "country" , "can" ,
  "do", "for" , and "you" appear two times.

69
Shrink via redundancy

• Immediately, we notice that
• "ask" , "what" , "your" , "country" , "can" ,
  "do", "for" , and "you" appear two times.
• Build a dictionary
• 1 gt ask 2 gt what 3 gt your
• 4 gt country 5 gt can 6 gt do
• 7 gt for 8 gt you
• Our sentence now reads
• "1 not 2 3 4 5 6 7 8 -- 1 2 8 5 6 7 3 4

70
Using redundancy

• Build a dictionary
• 1 gt ask 2 gt what 3 gt your
• 4 gt country 5 gt can 6 gt do
• 7 gt for 8 gt you
• Our sentence now reads
• "1 not 2 3 4 5 6 7 8 -- 1 2 8 5 6 7 3 4
• Ask what your country can do for you1 not 2 3 4
  5 6 7 8 -- 1 2 8 5 6 7 3 4
• Memory usage
• 37 units and 37 units for dictionary 74 units
• Need a symbol to tell when the dictionary ends
• Need to know that there are no numbers or in
  text

71
Use patterns to further shrink

• New dictionary
• Ask_
• what_
• you
• r_country
• _can_do_for_you
• New encoding of sentence
• 1not_2345_--_12354"
• Ask_what_your_country _can_do_for_you1not_234
  5_--_12354
• Memory usage
• 18 units and 41 units for dictionary 59 units

72
Typical savings

• Text files (unstructured)
• Usually can compress by half
• Text files (structured e.g. computer programs)
• Compression is greater
• Music/Image files
• Can usually reduce by factor of 10 with some loss
• Jpeg, mp3, files
• Usually no savings theyre already compresed

73
More complex compression

• Lossy compression
• Can save space by eliminating parts of the
  image/sound that dont matter very much
• Big savings compared to lossless
• Ideas
• Dropping frequencies you cant hear
• Dropping small differences you wont notice
• Generally large savings compared to lossless

74
Formats

• GIF original graphics compression format
• Can be lossy or lossless
• Limits numbers of colors to 256 (28 vs. 224)
• Can be progressive
• JPEG (Joint Photographic Experts Group)
• Can adjust level of lossy-ness
• Most popular format
• MPEG (Moving Picture Experts Group)
• Built on JPEG
• Works also in the time dimension

75
MPEG-3 Audio layer

• This is MP3
• Compresses sound by about a factor of 10
• Preserves CD quality sound