CS442, Spring 2006, HW #1 Answers

1. (a)  191, 168, 137, 104, 75, ...
   (b)  1, 1, 1, 1, 1

2. (a) 1 image is 2856 x 2142 = 6,177,552 pixels = 6 Mpixels

   (b) each pixel is 3 bytes, so 1 image is 
       6,177,552 x 3 = 18,352,656 bytes 

   Our storage is 128 Mbytes = 128 x 1024 x 1024 = 134,217,728 bytes.

   So, we will be able to store 
   134,217,728 / 18,353,656 = 7 images (pics).
   
3.(a)
  MichaelEatsMammals = F
  chickensAreMammals = F
  MichaelEatsChicken = T

  so    chickensAreMammals and MichaelEatsChicken = F and T = F
	not (chickensAreMammals and MichaelEatsChicken) = T
        MichaelEatsMammals or
	   not(chickensAreMammals and MichaelEatsChicken) = F or T = T

  (b)
  MichaelEatsMammals = F
  batsAreMammals = F
  MichaelEatsBats = T

  so     batsAreMammals and MichaelEatsBats = F and T = F
         not(batsAreMammals and MichaelEatsBats) = T
         MichaelEatsMammals or 
	   not(batsAreMammals and MichaelEatsBats) = T

4. [ML: There was some confusion about how the relays work.  If there
   is power to the magnet, the switch is OFF.  If there is no power to
   the magnet, the switch is ON.]

   Each time either A or B is OFF (false), we can see that the power
   source is connected directly to the output, so it is ON.  When both
   A and B are ON (true), both switches that connect the power source
   to the output are "closed" (electricity doesn't flow through them),
   so the output will be OFF.  The relationship between A, B, and
   output C looks like not(A and B) or (not A) or (not B) or
   neither-nor, etc.  

  A  B  C
-----------
  F  F  T
  F  F  T
  T  F  T
  T  T  F

5. There are a lot of possibilities here, a lot of other fields of
   study can have their own way of implementing bits.  Anyway, here
   are some things to keep in mind when thinking about this problem: 
   - 0s and 1s have to be implemented by 2 different objects (when you
     see something, you have to know right away if it is a 1 or a 0).
     Example: you have 2 tubes, one is red, the other one is blue; if
     you have a ball rolling down in the red tube, that's a 1, of you
     have a ball rolling down the blue tube, it's a 0 .
  - The "NOT" gate has to reverse a bit, so if it gets a 1, it has to
    make that into a 0, and same with 0.  In "computer science
    language" that would be: if a NOT gate gets 0 as an input, it
    outputs a 1.  If it gets a 1 as an input it outputs a 0.  For the
    tube example, we'd need the tubes to cross over so that a ball in
    the red tube results in a ball in the blue tube and vice versa).
  - The "OR" gate has to function in a way that would be consistent
    with the truth table the "OR" operation, so it has 2 bits as an
    input and it outputs the result of the OR operation between the 2
    bits.  For the tube example, some sort of mechanism like the one
    shown in class is needed to ensure that we get a ball in the
    output red tube if and only if there was a ball in one of the
    input red tubes.

6. Take 2, take 2, take 2, take 2.

7. In order to have all the powers of 2, we would have to have an
   infinite sequence of 1-s as an input.   That's because the
   differences are the same as the sequence itself---it will never
   converge on differences of all zeroes.