Essentials of Computer Organization and Architecture
5th Edition
ISBN: 9781284123036
Author: Linda Null
Publisher: Jones & Bartlett Learning
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Chapter 5, Problem 5E
a.
Explanation of Solution
Explanation:
The following table describes about the 2’s complement representation of an integer for the given value
| Value of an integer | 2’s complement |
| 2 | 0010 |
| A | 1010 |
| C | 1100 |
| 2 | 0010 | ...
b.
Explanation of Solution
Explanation:
- 2A of big endian represents the first bit as 0 which is a positive number and the most significant bit for the 2’s complement representation is SIGN...
c.
Explanation of Solution
Explanation:
The following table describes about the representation of Institute of Electrical and Electronic Engineers (IEEE) single precision floating point
| Sign | Exponent | Mantissa |
| -1 | -8 | -23 |
The following table describes about the decimal equivalent for the above values
| Sign | Exponent | Mantissa |
| -1 | -8 | -23 |
| 0 | 010101011 | 10000100000100000011011 |
The decimal equivalent of the address stored at address 100 is
d.
Explanation of Solution
Explanation:
The size of the data stored by the address 100 is 8 Bytes
The little endian is stored at last and the most significant bit is 0 and the absolute value of the integer is n-1 bits.
Therefore, the little endian of 32-bit integer number stored at address 100 is
e.
Explanation of Solution
Explanation:
- 1B of little endian represents the first bit as 0 which is a positive number and the most significant bit for the 2’s complement representation is SIGN...
f.
Explanation of Solution
Explanation:
The following table describes about the representation of Institute of Electrical and Electronic Engineers (IEEE) single precision floating point
| Sign | Exponent | Mantissa |
| -1 | -8 | -23 |
The following table describes about the decimal equivalent for the above values
| Sign | Exponent | Mantissa |
| -1 | -8 | -23 |
| 0 | 010101011 | 10000100000100000011011 |
Little endian is reverse of big endian and hence the decimal equivalent is calculated as follows
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Students have asked these similar questions
1.) Consider the problem of determining whether a DFA and a regular expression are
equivalent. Express this problem as a language and show that it is decidable.
ii) Let ALLDFA = {(A)| A is a DFA and L(A) = "}. Show that ALLDFA is decidable.
iii) Let AECFG = {(G)| G is a CFG that generates &}. Show that AECFG is decidable.
iv) Let ETM {(M)| M is a TM and L(M) = 0}. Show that ETM, the complement of
Erm, is Turing-recognizable.
Let X be the set {1, 2, 3, 4, 5} and Y be the set {6, 7, 8, 9, 10). We describe the
functions f: XY and g: XY in the following tables. Answer each part
and give a reason for each negative answer.
n
f(n)
n
g(n)
1
6
1
10
2
7
2
9
3
6
3
8
4
7
4
7
5
6
5
6
Aa. Is f one-to-one?
b. Is fonto?
c. Is fa correspondence?
Ad. Is g one-to-one?
e. Is g onto?
f.
Is g a correspondence?
vi) Let B be the set of all infinite sequences over {0,1}. Show that B is uncountable
using a proof by diagonalization.
Can you find the least amount of different numbers to pick from positive numbers (integers) that are at most 100 to confirm two numbers that add up to 101 when each number can be picked at most two times?
Can you find the formula for an that satisfies the provided recursive definition? Please show all steps and justification
Chapter 5 Solutions
Essentials of Computer Organization and Architecture
Ch. 5 - Prob. 1RETCCh. 5 - Prob. 2RETCCh. 5 - Prob. 3RETCCh. 5 - Prob. 4RETCCh. 5 - Prob. 5RETCCh. 5 - Prob. 6RETCCh. 5 - Prob. 7RETCCh. 5 - Prob. 8RETCCh. 5 - Prob. 9RETCCh. 5 - Prob. 10RETC
Ch. 5 - Prob. 11RETCCh. 5 - Prob. 12RETCCh. 5 - Prob. 13RETCCh. 5 - Prob. 14RETCCh. 5 - Prob. 15RETCCh. 5 - Prob. 16RETCCh. 5 - Prob. 17RETCCh. 5 - Prob. 18RETCCh. 5 - Prob. 19RETCCh. 5 - Prob. 20RETCCh. 5 - Prob. 21RETCCh. 5 - Prob. 22RETCCh. 5 - Prob. 23RETCCh. 5 - Prob. 24RETCCh. 5 - Prob. 25RETCCh. 5 - Prob. 1ECh. 5 - Prob. 2ECh. 5 - Prob. 3ECh. 5 - Prob. 4ECh. 5 - Prob. 5ECh. 5 - Prob. 6ECh. 5 - Prob. 7ECh. 5 - Prob. 8ECh. 5 - Prob. 9ECh. 5 - Prob. 10ECh. 5 - Prob. 11ECh. 5 - Prob. 12ECh. 5 - Prob. 13ECh. 5 - Prob. 14ECh. 5 - Prob. 15ECh. 5 - Prob. 16ECh. 5 - Prob. 17ECh. 5 - Prob. 18ECh. 5 - Prob. 19ECh. 5 - Prob. 20ECh. 5 - Prob. 21ECh. 5 - Prob. 22ECh. 5 - Prob. 23ECh. 5 - Prob. 24ECh. 5 - Prob. 25ECh. 5 - Prob. 26ECh. 5 - Prob. 27ECh. 5 - Prob. 28ECh. 5 - Prob. 29ECh. 5 - Prob. 1TFCh. 5 - Prob. 2TFCh. 5 - Prob. 3TFCh. 5 - Prob. 4TFCh. 5 - Prob. 5TFCh. 5 - Prob. 6TFCh. 5 - Prob. 7TFCh. 5 - Prob. 8TFCh. 5 - Prob. 9TFCh. 5 - Prob. 10TFCh. 5 - Prob. 11TFCh. 5 - Prob. 12TF
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