1. Math for Liberal Studies

2.  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)

3.  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  Mostly, whenever information needs to be transmitted from one location to another, a binary code is used

4.  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

5.  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”?

6.  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 0000010010001100 0001010110011101 0010011010101110 0011011110111111

7.  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

8.  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

9.  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 be chosen so that each circle contains an even number of ones III IV II I VII VVI

10.  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 that the last circle has two ones  Our message is now 0100101 0 0 1 0 1 10

11.  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 is only one way to fix it: change it from 1 to 0 0 1 1 0 1 10

12.  Encode the message 1110 using this method  You have received the message 0011101. Find and correct the error in this message.

13.  Binary codes can be used to represent more conventional information, but 4 digits only gives us 16 possible messages  That’s not even enough to represent the alphabet!  If we have n digits, then we can make 2 n different messages  5 digits -> 32 messages  6 digits -> 64 messages, etc.

14.  The idea we’re using is a specific example of a parity check sum  The parity of a number is either odd or even  For example, digit V is 0 if I + II + III is even, and odd if I + II + III is odd

15.  Instead of using Roman numerals, we’ll use a 1 to represent the first digit of the message, a 2 to represent the second digit, and so on  We’ll use c 1 to represent the first check digit, c 2 to represent the second, etc.

16.  Using this notation, our rules for our check digits become  c 1 = 0 if a 1 + a 2 + a 3 is even  c 1 = 1 if a 1 + a 2 + a 3 is odd  c 2 = 0 if a 1 + a 3 + a 4 is even  c 2 = 1 if a 1 + a 3 + a 4 is odd  c 3 = 0 if a 2 + a 3 + a 4 is even  c 3 = 1 if a 2 + a 3 + a 4 is odd

17.  Under this new way of thinking about our system, how do we decode messages?  Simply compare the message with the list of possible correct messages and pick the “closest” one  What should “closest” mean?  If you have two messages of the same length, the distance between the two messages is the number of digits in which they differ

18.  What is the distance between 1100101 and 1010101?  The messages differ in the 2 nd and 3 rd digits, so the distance is 2  What is the distance between 1110010 and 0001100?  The messages differ in all but the 7 th digit, so the distance is 6

19.  The nearest neighbor decoding method decodes a received message as the code word that agrees with the message in the most positions

20.  In this example, our messages are three digits long: a 1 a 2 a 3  We have three check digits  c 1 = 0 if a 1 + a 2 + a 3 is even  c 1 = 1 if a 1 + a 2 + a 3 is odd  c 2 = 0 if a 1 + a 3 is even  c 2 = 1 if a 1 + a 3 is odd  c 3 = 0 if a 2 + a 3 is even  c 3 = 1 if a 2 + a 3 is odd

21.  Using these rules, we can find all of our code words  By analyzing this list, we see that the smallest distance between two code words is 3  That means we can use these code words to either detect two errors or correct one error MessageCode Word 000000000 001001111 010010101 011011010 100100110 101101001 110110011 111111100