CSC103 Homework 1 solutions 2012f

--D. Thiebaut 07:53, 11 September 2012 (EDT)

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This homework assignment was due on Tusday Sept. 25th, at 9:00 a.m (pushed back because of Mountain Day). Each problem was worth 1 point. The Smith grade scale is used to convert points into a letter grade. 4 points corresponds to A, 3.7 to A-, 3.3 to B+, 3.0 to B, etc.

Question #1

Count in binary and write down first 33 numbers of the series. In other words, complete the second column in the list below.

Decimal      Binary
 1               1
 2              10
 3              11
 4             100
 5             101
 6             110
 7             111
 8            1000
 9            1001
10            1010
11            1011
12            1100
13            1101
14            1110
15            1111
16           10000
17           10001
18           10010
19           10011
20           10100
21           10101
22           10110
23           10111
24           11000
25           11001
26           11010
27           11011
28           11100
29           11101
30           11110
31           11111
32          100000

Perform the following additions in binary:

    10011 + 10011 = 100110

    10111 + 00110 = 11101

Question #2

Assume that we live in a universe where everybody only has 4 fingers. Just as we did in class with a system of 2 digits (binary code), we invente a system for counting with only 4 digits: 0, 1, 2, and 3.

• Write the first 20 numbers of a system with 4 digits. To help you out, I will start with the first 7 numbers of the series:

0
1
2
3
10
11
12
13
20
21
22
23
30
31
32
33
100
101
102
103

Continue on until you have 20 consecutive numbers of a system in base 4.

Question #3

Perform the addition of the following numbers in base 4.

    1002 + 2301 = 3303

    2222 + 3301 = 12123

Question #4

Assume that we have several boolean expressions, labeled E1 to E6:

• E1: is a Smith student
• E2: is a senior
• E3: likes vanilla
• E4: has class on Monday
• E5: is a Hampshire student
• E6: is on the crew team

What is the boolean expression that is a combination of E1, E2, E3, E4, and/or E5, and the logic operators AND, OR, and NOT, that will be True whenever I find somebody on campus who is a Hamshire College student and who does not like vanilla?

We only need to worry about E1, E5, and E3. Some of you may have only used E5 and E3. I counted both correct.

Let's call our solution expression as Ehnv (Hampshire-Not-Vanilla)

   E1 E5 E3      Ehnv
   F   F   F           F
   F   F   T           F
   F   T   F           T  (Not Smith, Hampshire, doesn't like vanilla)
   F   T   T           F
   T   F   F           F
   T   F   T           F
   T   T   F           T  (Smith & Hampshire, doesn't like vanilla)
   T   T   T           F

   Ehnv = ( ( not E1 ) and E5 and ( not E3 ) ) or ( E1 and E5 and ( not E3 ) )

If you use only E3 and E5, you get

   E5 E3      Ehnv
    F   F           F
    F   T           F
    T   F           T  (Hampshire, doesn't like vanilla)
    T   T           F

   Ehnv = E5 and ( not E3 )