CIS 115
Compression & Error Checking
What is Compression?
Why is it useful?
Compression
- Resources are expensive
- Storage space
- Transmission bandwidth
- Time to read/send
- Does require more computation; can be a tradeoff
Run Length Encoding
- Find "runs" (repeated sequences) in data
- Replace them with a shorter version
- Usually the sequence and a count
RLE Example
WWWWWWWWWWWWBWWWW WWWWWWWWBBBWWWWWW WWWWWWWWWWWWWWWWW WBWWWWWWWWWWWWWWW
Image Source: Wikipedia
RLE Example
WWWWWWWWWWWWBWWWW WWWWWWWWBBBWWWWWW WWWWWWWWWWWWWWWWW WBWWWWWWWWWWWWWWW 12W1B12W3B24W1B15W
Image Source: Wikipedia
Wait, isn't text just numbers (ASCII)?
How can we tell which numbers are text and which are not?
RLE Example
WWWWWWWWWWWWBWWWW WWWWWWWWBBBWWWWWW WWWWWWWWWWWWWWWWW WBWWWWWWWWWWWWWWW
Image Source: Wikipedia
Escape Coding
WWWWWWWWWWWWBWWWW WWWWWWWWBBBWWWWWW WWWWWWWWWWWWWWWWW WBWWWWWWWWWWWWWWW WW12BWW12BB3WW24BWW15
Image Source: Wikipedia
Huffman Coding
- Count the occurrences of each character
- Make a binary tree with the data
- The paths of the tree give the codes
Huffman Example
THIS IS AN EXAMPLE OF
A HUFFMAN TREE
- space: 7
- a: 4
- e: 4
- f: 3
- h: 2
- i: 2
- m: 2
- n: 2
- s: 2
- t: 2
- l: 1
- o: 1
- p: 1
- r: 1
- u: 1
- x: 1
Image Source: Wikipedia
Huffman Example
THIS IS AN EXAMPLE OF
A HUFFMAN TREE 0110 1010 1000 1011 111 1000 1011 111 010 0010 111 000 10010 ... (135 bits)
Image Source: Wikipedia
What is Error Checking?
Why is it useful?
What is Error Checking?
Why is it useful?
Intuition - Squeezing
Hello there
Hello there
Hello ghere
Hello ghere
Simple - Addition
Hello there
72 101 108 108 111 32
116 104 101 114 101 33
= 1101
Simple - Addition
- If we only have the sum, can we recover the original?
- How well does this detect errors?
- Are there errors it cannot detect?
- 1101 is larger than 8 bits. How should we handle that?
- Can we do better?
Better - Pinpoint
4837543622563997
| 4 | 8 | 3 | 7 | ? |
| 5 | 4 | 3 | 6 | ? |
| 2 | 2 | 5 | 6 | ? |
| 3 | 9 | 9 | 7 | ? |
| ? | ? | ? | ? |
Better - Pinpoint
| 4 | 8 | 3 | 7 | 2 |
| 5 | 4 | 3 | 6 | 8 |
| 2 | 2 | 5 | 6 | 5 |
| 3 | 9 | 9 | 7 | 8 |
| 4 | 3 | 0 | 6 |
483725436822565399784306
Better - Pinpoint
483725436827565399784306
| 4 | 8 | 3 | 7 | 2 | 2 |
| 5 | 4 | 3 | 6 | 8 | 8 |
| 2 | 7 | 5 | 6 | 5 | 0 |
| 3 | 9 | 9 | 7 | 8 | 8 |
| 4 | 3 | 0 | 6 | ||
| 4 | 8 | 0 | 6 |
Better - Fletcher's Checksum
INPUT: a data word (e.g., a sequence of ASCII-numbers)
OUTPUT: two checksums for the word, each sized to fit
in one byte
ALGORITHM:
1. divide the Word into a sequence of equally-sized
blocks, b1 b2 ... bn
2. define two checksums, starting at C1 = 0 and C2 = 0
3. for each block, bi,
add bi to C1
add the new value of C1 to C2
4. compute Checksum1 = C1 mod 255 and
Checksum2 = C2 mod 255
5. return Checksum1 and Checksum2
Better - Fletcher's Checksum
72 101 108 108 111 32 116 104 101 114 101 33
| Block | C1 | C2 |
| 72 | 72 | 72 |
| 101 | 173 | 245 |
| 108 | 281 | 526 |
| 108 | 389 | 915 |
| Total: | 1101 (81) | 7336 (96) |
Testing Fletcher's
- 72 101 108
- 72 108 101
- 74 99 108
- 72 101 0 108
- Which ones does it catch?
Can we do better?
Cyclic Redundancy Check (CRC)
Input: 010100001001
Check: 1011
010100001001
XOR 1011
------------
10001001
XOR 1011
---------
00111001
XOR 1011
-------
10101
XOR 1011
------
Checksum: 011
Hash Codes
Choose a "hash base", b (e.g., b= 2 or b= 10 or b= 37)
For a word of integers of length n+1:
w = x0 x1 x2 ... xn-1 xn,
Compute this hash number:
hash(w) = (x0 * bn) + (x1 * bn-1) + (x2 * bn-2) + ...
(xn-1 * b1) + xn
Hash Codes
For word = 4 5 6,
when b = 10,
hash(word) = (4 * 102) + (5 * 101) + 6 =
400 + 50 + 6 = 456
when b = 100,
hash(word) = (4 * 1002) + (5 * 1001) + 6 =
40000 + 500 + 6 = 40506
when b = 5,
hash(word) = (4 * 52) + (5 * 51) + 6 =
100 + 20 + 6 = 126
when b = 2,
hash(word) = 16 + 10 + 6 = 32
when b = 1,
hash(word) = 4 + 5 + 6 = 15
Hash Codes
For word = 12 33 08,
when b = 10,
hash(word) = (12 * 102) + (33 * 101) + 8 =
1200 + 330 + 8 = 1538
when b = 100,
hash(word) = (12 * 1002) + (33 * 1001) + 8 =
120000 + 3300 + 8 = 123308
when b = 5,
hash(word) = (12 * 52) + (33 * 51) + 8 =
300 + 165 + 8 = 473
when b = 2,
hash(word) = 48 + 66 + 8 = 122
Hamming Codes
BIT#: 1 2 3 4 5 6 7
-------------------------------------------
PURPOSE: P3,5,7 P3,6,7 D P5,6,7 D D D
where D is a data bit, and
Pa,b,c,... is the parity bit for data bits at a,b,c,...
Examples:
DATA 3 5 6 7 P3,5,7 P3,6,7 P5,6,7 HAMMING CODE
------------------------------------------------------
0 1 0 0 1 0 1 1 0 0 1 1 0 0
1 0 1 1 0 1 0 0 1 1 0 0 1 1
Assignments
- Read and be prepared to discuss:
- 9ALG 7: Data Compression
- Blog 7 - Due 12/5 10:00 PM
- Cryptography - Due 11/15 10:00 PM
- Mars Rover - Due 11/29 10:00 PM
- Video Game - Due 12/8 10:00 PM
Blog 7: 9 Algorithms - Making Meaning
Now that we’ve finished reading the last textbook, it is time to step back and think about what we read. Write about your reactions to it and what you learned from it. I’d recommend almost treating this like an in-depth book review for others who are interested in reading the book, but don’t mind some spoilers. Some questions I’d like you to answer:
- How did you feel reading this book? Engaged? Bored? Interested?
- What was the most interesting thing you learned?
- Were there any parts of the book you didn’t like?
- Were there any terms or concepts that you looked up (Googled) to find more information about? What were they? What did you find?
- Did this book help explain things you didn’t know about computers?
- Did this book leave you with any questions unanswered?
- Would you recommend this book to a friend that wanted to know more about computers?
- What is significant about this book? Why do you think it was chosen as a textbook?