Introduction to computing system (Patt & Patel):Chapter 2

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Content

Main content

  • Bits and Datatypes
  • Integer Datatype
    • Unsigned Interagers
    • Signed Interages with 1's and 2's complement code
  • Conversion between Binary and Decimal
  • Operation on bits
    • Arithmmetic operation
    • Logical operation
  • Other Representation
    • Float number
    • ASCII code
    • Hexadecimal Notation

Exercise

Made by myself. I'm not sure if it's correct.

2.1

there are \(2^n\) .

2.2

\(log_2 26 = 4\)
If we need both lowercase and uppercase we need 1 more bit, totally 5.

2.3

  1. \(log_2 400 = 9\)
  2. \(2^9 - 400 = 112\)

2.4

There are \(2^n\) unsigned intager numers,ranged from 0 to \(2^n - 1\).

2.5

number 2's complement signed magnitude 1' complement
-7 11001 10111 11000
7 00111 00111 00111

2.6

10000

2.7

number 2's complement
-8 1000
-7 1001
-6 1010
-5 1011
-4 1100
-3 1101
-2 1110
-1 1111
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111

2.8

  1. \(2^7 - 1 = 127\), binary: 01111111
  2. \(2^7 = 128\), binary: 10000000
  3. \(2^n - 1\)
  4. \(2^n\)

2.9

\(log_2 6.02 * 10^{23} = 591\)

2.10

  1. -6
  2. 90
  3. -2
  4. 14800

2.11

  1. 0110 0110
  2. 0100 0000
  3. 0010 0001
  4. 1000 0000
  5. 0111 1111

2.12

It can divided by 4.

2.13

  1. 1111 1010
  2. 1111 1000
  3. 0001 1001
  4. 0000 0001

2.14

  1. 1100
  2. 1010
  3. 1111
  4. 1011
  5. 0000 (overflowed)

2.15

elimate the last digit and divide by 2.

2.16

number 2's complement 1's complement magnitude
7 + -7 0111 + 1001 = (1)000 0111 + 1000 = 1111 1111 + 0111 = 0110

2.17

index binary dexcimal
a 1100 -4
b 0101 1000 84
c 0011 3
d 11 -1

2.18

index binary dexcimal
a 1100 12
b 0101 1000 88
c 1011 11
d 11 3

2.19

-27 to binary : 1110 0101 (8 bits)
1111 1111 1110 0101 (16bits)
1111 1111 1111 1111 1111 1111 1110 0101 (32 bits)
sign-extension helps bit-extension and don't change its value.

2.20

  1. 1111
  2. 0000
  3. 1000 (overflow)
  4. 0111
  5. 0000

2.21

When 2 number add with the same sign, but result in the other sign.

2.22

1000 0000 0000 0000 + 1000 0000 0000 0000

2.23

The sum is smaller than one of the origin.

2.24

1000 0000 0000 0000 + 1000 0000 0000 0000

2.25

Because the sum of them is in the range of the original one.

2.26

  1. seven
  2. 011 1111 63
  3. 111 1111 127

2.27

Yes. Sum of two positive numbers presents a negetive one. Overflow.

2.28

All the input is one.

2.29

X Y X AND Y
0 0 0
0 1 0
1 0 0
1 1 1

2.30

  1. 0101 0111
  2. 100
  3. 1010 0000
  4. 0001 0100
  5. 0000
  6. 0000

2.31

one of the input is 1.

2.32

X Y X OR Y
0 0 0
0 1 1
1 0 1
1 1 1

2.33

  1. 1101 0111
  2. 111
  3. 1111 0100
  4. 1011 1111
  5. 1101
  6. 1101

2.34

  1. 0111
  2. 0111
  3. 1101
  4. 0110

2.35

Use for change a specific bit.

2.36

  1. AND 1111 1011
  2. OR 0100 0100
  3. AND 0000 0000
  4. OR 1111 1111
  5. First we need a mask operation : And 0000 0100 Then times 32.
    If we add the final pattern, we always get a 0.

2.37

A1 = n AND 1000
A2 = (NOT n) AND 1000
B1 = m AND 1000
B2 = (NOT m) AND 1000
C1 = (NOT s) AND 1000
C2 = s AND 1000
RESLUT = (A1 AND B1 AND C1) OR (A2 AND B2 AND C2)

2.38

Using the symbol defined above.
RESULT = A1 AND B1 AND C1

2.39

a.3.75 = \(1.111 * 2^1\) , So the representation is 0 100 00000 1110 0000 0000 0000 0000 000
b. \(55\frac{23}{64} = 1.10111010111 * 2^5\) , So the representation is 1 100 00100 1011 1010 1110 0000 0000 0000
c. \(3.1415927 = 1.10010010000111111011010110100111111011010001100101110011 * 2^1\)
So the representation is 1 100 00000 1001 0010 0001 1111 1011 010 , not so precious
d. \(64000 = 1.1111010 * 2^{15}\) So the representation is 0 100 01110 1111 0100 0000 0000 0000 0000

2.40

  1. 2
  2. -17
  3. INF
  4. -2.25

2.41

  1. 127
  2. -149

2.42

He use the ASCII code of the number 5 and 8.
Actually ACSII code of 5 and 8 is 53 and 56. The sum of them is 109 which represents the character m in the ACSII code.

2.43

  1. Hello!
  2. hELLO!
  3. Computers!
  4. LC-2

2.44

Add, because the number in ASCII is continous. We just add some bias to get the ASCII number.

2.45

  1. 0xD1AF
  2. 0x1F
  3. 0x1
  4. 0xEDB2

2.46

  1. 1 0000
  2. 0100 0000 0001
  3. 1111 0111 0011 0001
  4. 0000 1111 0001 1110 0010 1101
  5. 1011 1100 1010 1101

2.47

  1. -16
  2. 2047
  3. 22
    d.-32768

2.48

  1. 0001 0000 0000
  2. 0110 1111
  3. 0111 0101 1011 1100 1101 0001 0101
  4. 1101 0100

2.49

  1. 0x2939
  2. 0x6E36
  3. 0x46F4
  4. 0xF1A8
  5. c and d is overflow

2.50

  1. 5468
  2. BBFD
  3. FFFF
  4. 32A3

2.51

  1. 0x644B
  2. 0x4428E800
  3. 0x48656c6c6f

2.52

rules 0x434F4D50 0x55544552
Unsigned binary 1,129,270,608 1,431,586,130
1's complement 1,129,270,608 1,431,586,130
2's complement 1,129,270,608 1,431,586,130
IEEE 754 floating point 207.302001953125 14587137097728
ASCII string COMP UTER

2.53

A B \(Q_1\) \(Q_2\)
0 0 1 0
0 1 1 1
1 0 1 1
1 1 0 1

\(Q_2\) = A OR B

2.54

X Y Z \(Q_1\) \(Q_2\)
0 0 0 0 1
0 0 1 0 1
0 1 0 0 1
0 1 1 0 1
1 0 0 1 1
1 0 1 1 1
1 1 0 1 1
1 1 1 0 0

2.55

  1. \(4^{3} - 1 = 63\)
  2. \(4^n -1\)
  3. 023 + 221 = 310
  4. \(222|_4\)
  5. 11011.11
  6. 0 100 00011 1011 1100 0000 0000 0000 000
  7. I don't know what's the question means. From my point of view, depends on the number of operations. Basically there are arithmetic operations like multiply and add and logical operations like AND, OR, NOT,XOR.

2.56

0xE5 = 1110 0101
sign : 1-> -1
exp: 8 + 4 - 7 -> 5
fraction: 101 -> 1.101 = 1.5625
totally: \(1.5625 * -1 * 2^5 = 52\)(shift first maybe faster)