I have invented a new light switch. In contrast to a normal light switch (which is either ‘off’ or ‘on’), it has three positions: Off, Low and High. That means with each switch I can represent three different pieces of information (a ternary situation). I am going to build a computer using this as the micro-switch and applying the same principles as a binary computer.

 

My basic unit (one grouping) will be a ‘chomp’ (as opposed to a byte). I am going to include the entire extended ASCII character set in the data type we call ‘char’ using one chomp. I don’t have to worry about parity, since my machine will be perfect.

 

I will only have one type of integer.  It will consist of two (2) chomps. All my integers will be signed, and I plan to use one of my three position light switches to represent the sign.

 

My real numbers will require three (3) chomps. One (1) switch will be used for the sign (just like my integers). I am going to assign 5 of my light switches to the characteristic of the exponent.

 

(Note: with the exception of the first question (below) each of these can be answered using a formula (only), although the numeric answers are not too difficult)

 

1.      How many of my ternary light switches do I need to group together to include the entire extended ASCII character set (i.e., how many switches are in a ‘chomp’)?

 

2.      What will be the range of integers using 2 chomps?

 

3.      What will be my range of magnitude using my ‘chomp’ machine?

 

4.      What will be my level of precision using my ‘chomp’ machine?

 

Given 1 light switch, I can represent 3 messages. The formula I  = Bn (where I is the amount of messages I can produce given n switches, where each switch has B positions (in this case, 3)). Therefore:

 

n

I

n

I

1

31 = 3

6

36 = 729

2

32 = 9

7

37 = 2,187

3

33 = 27

8

38 = 6,561

4

34 = 81

9

39 = 19,683

5

35 = 243

10

310 = 59,049

 

1.      Therefore, in order to represent all of the characters in the extended ASCII character set I will need 6 light switches (28 = 256; 5 will not be enough).

 

2.      If I have 2 chomps (12 light switches). Since 1 light switch will be assigned for the sign, I have 11 (eleven light switches left). Therefore the range of my integers will be:

 

                  -311 through 311 – 1 = -177,147 to 177,147

 

3.      Given that five light switches are assigned to the characteristic of the exponent there are a total of 35 (= 243) possible combinations. However, since the characteristic of the exponent can take on negative and non-negative values (i.e., one light switch is assigned for the sign), the range is:

 

                  -34 through 34 – 1 = -81 to 80

 

4.      Given three chomps, I have 3 * 6 = 18 light switches.

 

Component

Light Switches

Total Number:

18

Sign Bit:

1

Characteristic of the Exponent:

5

Remainder (Mantissa)

12

 

         Therefore, I can have whatever precision is available with 312 combinations, or:

 

         312 = 531,441 (five (5) decimals of precision, some to six (6))