DS.txt

Answer author note:
0b prefix means binary
0o prefix means octal
0x prefix means hexadecimal

Group Assignment  (3 per group) 

1. Hexadecimal conversions:
  a. Convert to hexadecimal: 2345.67, Round to two digits past the hexadecimal point.
  b Convert your answer to binary. and then to octal
  c. Devise a scheme for converting hexadecimal directly to base 4 and convert your answer to base 4
  d. Convert to decimal. 0xCAD.B
2. Add, subtract, and multiply in binary
  (b) 0b101110 and 0b11101
  (c) 0b100100 and 0b10110
3. Subtract in binary. Place a 1 over each column from which it was necessary to borrow.
  (a) 0b11001100 - 0b1001011
  (b) 0b11001100 - 0b101001
4. Add the following numbers in binary using 2's complement to represent negative numbers. Use a word length of 6 bits (including sign) and indicate if an ovediow occurs.
  (a) 20 + 12
  (b) (-15) + (-31)
  Repeat (a) and (b) using 1's complement to represent negative number.
5. Convert to octal. Convert to hexadecimal. Then convert both of your answers to decimal. and verify that they are the same
  (a) 0b101010100100.101
  (b) 0b101001101001.01
6. Convert the decimal number 78.9 into a number with exactly the same value represented in the following bases. The exact value requires an infinite repeating part in the fractional part of the number. Show the steps of your derivation.
  (a) binary
  (b) octal
  (c) hexadecimal
  (d) base 3
  (e) base 5
7. Divide in binary:
  (a) 0b11101011 / 0b101
  (b) 0b110000011 / 0b1110
  Check your answers by multiplying out in binary and adding the remainder. 


1.
  a.
    2345.67 to Hexadecimal

    2345

    2345 9
    146 2
    9

    929

    0.67

    0.67 * 16 = 10.72 10 A
    0.72 * 16 = 11.52 11 B
    0.52 * 16 = 8.32 8 8
    0.32 * 16 = 5.12 5 5
    0.12 * 16 = 1.92 1 1

    AB851

    0x929.AB851

  b. NOT COMPLETED
    0x929.AB851 to Decimal

    929
    9 1001
    2 0010
    9 1001

    100100101001

  c. NOT COMPLETED

  d.
    0xCAD.B to decimal

    D 13*16^0 = 13
    A 10*16^1 = 160
    C 12*16^2 = 3072

    13 + 160 + 3072 = 3245

    B 11*16^-1 = 0.6875

    0.6875

    3245 + 0.6875 = 3245.6875

2.
  Add, subtract, multiply
  a.
    101110 and 11101

    111    <- carry
    101110
    011101
    ======= +
    1001011

    02  02 <- borrow
    101110
    011101
    ====== -
    010001

              1 0 1 1 1 0
              0 1 1 1 0 1
              =========== *
    1 1 2 2 2 1 1     <- carry
              1 0 1 1 1 0
            0 0 0 0 0 0
          1 0 1 1 1 0
        1 0 1 1 1 0
      1 0 1 1 1 0
    0 0 0 0 0 0
    ===================== +
    1 0 1 0 0 1 1 0 1 1 0

  b.
    100100 and 10110

      1    <- carry
    100100
    010110
    ====== +
    111010

    0112
      02  <- borrow
    100100
    010110
    ====== -
    001110

              1 0 0 1 0 0
              0 1 0 1 1 0
              =========== *
        1 1             0 <- carry
              0 0 0 0 0 0
            1 0 0 1 0 0
          1 0 0 1 0 0
        0 0 0 0 0 0
      1 0 0 1 0 0
    0 0 0 0 0 0
    ===================== +
    0 1 1 0 0 0 1 1 0 0 0

3.
  a.
        012 <- borrow
    11001100
    01001011
    ======== -
    10000001

  b.
    02  012 <- borrow
    11001100
    00101001
    ======== -
    10100011

4.
  a.
    20 + 12

    1s complement
    20 0b10100
    12 0b1100

    With 6 bit + sign
    20 0b010100
    12 0b001100

    1 1 1       <- carry
    0 1 0 1 0 0
    0 0 1 1 0 0
    =========== +
    1 0 0 0 0 0

    0b100000 = 0b011111 = -31
      ^          ^
      | minus    | 31

    Overflow occur because the carry attempt to change the sign

    2s complement
    20 0b10100
    12 0b1100

    With 6 bit + sign
    20 0b010100
    12 0b001100

    1 1 1       <- carry       1 1 1 1 1   <- carry
    0 1 0 1 0 0                0 1 1 1 1 1
    0 0 1 1 0 0                0 0 0 0 0 1
    =========== +              =========== +
    1 0 0 0 0 0                1 0 0 0 0 0

    Overflow occur because the carry attempt to change the sign

    0b100000 = 0b011111 + 0b1 = 0b100000 = -32
      ^                           ^
      | minus                     | 32

  b.
    (-15) + (-31)

    1s complement

    31 0b11111
    15 0b1111

    With 6 bit + sign
    31 0b111111
    15 0b101111

      1 1 1 1 1   <- carry
      1 1 1 1 1 1
      1 0 1 1 1 1
      =========== +
    1 1 0 1 1 1 0

    Overflow occur because the carry attempt to change the sign

    0b1101110 = 0b0010001 = -17
      ^           ^
      | minus     | 17

    2s complement

    31 0b11111
    15 0b1111

    With 6 bit + sign
    31 0b111111
    15 0b101111

      1 1 1 1 1   <- carry
      1 1 1 1 1 1
      1 0 1 1 1 1
      =========== +
    1 1 0 1 1 1 0
      ^
      | clamp the bits

    Overflow occur because the carry attempt to change the sign

    0b1101110 = 0b0010001 + 0b1 = 0b0010010 = 18
      ^                             ^
      | minus                       | 18