MAT 105 Spring 2008
A binary code is a system for encoding data made up of 0’s and 1’s
Examples Postnet (tall = 1, short = 0) UPC (dark = 1, light = 0) Morse code (dash = 1, dot = 0) Braille (raised bump = 1, flat surface = 0) Movie ratings (thumbs up = 1, thumbs down = 0)
CD, MP3, and DVD players, digital TV, cell phones, the Internet, space probes, etc. all represent data as strings of 0’s and 1’s rather than digits 0-9 and letters A-Z
Whenever information needs to be digitally transmitted from one location to another, a binary code is used
What are some problems that can occur when data is transmitted from one place to another?
The two main problems are transmission errors: the message sent is not the
same as the message received security: someone other than the intended
recipient receives the message
Suppose you were looking at a newspaper ad for a job, and you see the sentence “must have bive years experience”
We detect the error since we know that “bive” is not a word
Can we correct the error?
Why is “five” a more likely correction than “three”?
Why is “five” a more likely correction than “nine”?
Suppose NASA is directing one of the Mars rovers by telling it which crater to investigate
There are 16 possible signals that NASA could send, and each signal represents a different command
NASA uses a 4-digit binary code to represent this information
0000 0100 1000 11000001 0101 1001 11010010 0110 1010 11100011 0111 1011 1111
The problem with this method is that if there is a single digit error, there is no way that the rover could detect or correct the error
If the message sent was “0100” but the rover receives “1100”, the rover will never know a mistake has occurred
This kind of error – called “noise” – occurs all the time
One way to try to avoid these errors is to send the same message twice
This would allow the rover to detect the error, but not correct it (since it has no way of knowing if the error occurs in the first copy of the message or the second)
There is a better way to allow the rover to detect and correct these errors, and only requires 3 additional digits
The original message is four digits long
We will call these digits I, II, III, and IV
We will add three new digits, V, VI, and VII
Draw three intersecting circles as shown here
Digits V, VI, and VII should bechosen so that each circlecontains an even number ofones
III IVII
I
VII
V VI
The message we want to send is “0100”
Digit V should be 1 so that the first circle has two ones
Digit VI should be 0 so that the second circle has zero ones (zero is even!)
Digit VII should be 1 so thatthe last circle has two ones
Our message is now 0100101
0 01
0
1
1 0
Now watch what happens when there is a single digit error
We transmit the message 0100101 and the rover receives 0101101
The rover can tell that the second and third circles have odd numbers of ones, but the first circle is correct
So the error must be in the digit that is in the second and third circles, but not the first: that’s digit IV
Since we know digit IV is wrong, there isonly one way to fix it: change it from 1 to 0
0 11
0
1
1 0
Encode the message 1110 using this method
You have received the message 0011101. Find and correct the error in this message.
This method only allows us to encode 16 possible messages, which isn’t even enough to represent the alphabet!
However, if we use more digits, we won’t be able to use the circle method to detect and correct errors
We’ll have to come up with a different method that allows for more digits
The circle method is a specific example of a “parity check sum”
The “parity” of a number is 1 is the number is odd and 0 if the number is even
For example, digit V is 0 if I + II + III is even, and 1 if I + II + III is odd
Instead of using Roman numerals, we’ll use a1
to represent the first digit of the message, a2 to represent the second digit, and so on
We’ll use c1 to represent the first check digit, c2 to represent the second, etc.
Using this notation, our rules for our check digits become c1 = 0 if a1 + a2 + a3 is even c1 = 1 if a1 + a2 + a3 is odd c2 = 0 if a1 + a3 + a4 is even c2 = 1 if a1 + a3 + a4 is odd c3 = 0 if a2 + a3 + a4 is even c3 = 1 if a2 + a3 + a4 is odd
a3 a4a2
a1
c3
c1 c2
If we want to have a system that has enough code words for the entire alphabet, we need to have 5 message digits: a1, a2, a3, a4, a5
We will also need more check digits to help us decode our message: c1, c2, c3, c4
We can’t use the circles to determine the check digits for our new system, so we use the parity notation from before
c1 is the parity of a1 + a2 + a3 + a4
c2 is the parity of a2 + a3 + a4 + a5
c3 is the parity of a1 + a2 + a4 + a5
c4 is the parity of a1 + a2 + a3 + a5
Using 5 digits in our message gives us 32 possible messages, we’ll use the first 26 to represent letters of the alphabet
On the next slide you’ll see the code itself, each letter together with the 9 digit code representing it
Letter Code Letter Code
A 000000000 N 011010101
B 000010111 O 011101100
C 000101110 P 011111011
D 000111001 Q 100001011
E 001001101 R 100011100
F 001011010 S 100100101
G 001100011 T 100110010
H 001110100 U 101000110
I 010001111 V 101010001
J 010011000 W 101101000
K 010100001 X 101111111
L 010110110 Y 110000100
M 011000010 Z 110010011
Now that we have our code, using it is simple
When we receive a message, we simply look it up on the table
But what happens when the message we receive isn’t on the list?
Then we know an error has occurred, but how do we fix it? We can’t use the circle method anymore
Using this new system, how do we decode messages?
Simply compare the (incorrect) message with the list of possible correct messages and pick the “closest” one
What should “closest” mean?
The distance between the two messages is the number of digits in which they differ
What is the distance between 1100101 and 1010101? The messages differ in the 2nd and 3rd digits, so the
distance is 2
What is the distance between 1110010 and 0001100? The messages differ in all but the 7th digit, so the
distance is 6
The nearest neighbor decoding method decodes a received message as the code word that agrees with the message in the most positions
Suppose that, using our alphabet code, we receive the message 010100011
We can check and see that this message is not on our list
How far away is it from the messages on our list?
Code Distance Code Distance
000000000 4 011010101 5
000010111 4 011101100 5
000101110 4 011111011 3
000111001 4 100001011 4
001001101 6 100011100 8
001011010 6 100100101 4
001100011 2 100110010 4
001110100 6 101000110 6
010001111 3 101010001 6
010011000 5 101101000 6
010100001 1 101111111 6
010110110 3 110000100 5
011000010 3 110010011 3
Since 010100001 was closest to the message that we received, we know that this is the most likely actual transmission
We can look this corrected message up in our table and see that the transmitted message was (probably) “K”
This might still be incorrect, but other errors can be corrected using context clues or check digits