Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design)
5th Edition
ISBN: 9780124077263
Author: David A. Patterson, John L. Hennessy
Publisher: Elsevier Science
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Expert Solution & Answer
Chapter B, Problem 9E
Explanation of Solution
A3 A2 A1 A0 F 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 0 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 1 1 1 1 1 0 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 0 1 1 1 1 1 0
- In the above table, A0, A1, A2 and A3 represent four different inputs and F represents odd parity function.
- In odd parity, if the function includes odd number of ones then the function has value one and if the function includes even number of ones then the function has zero value.
- If A3=0, A2=0, A1=0 and A0=0 then F=0 because the number of 1 inputs is 0 which is an even number.
- If A3=0, A2=0, A1=0 and A0=1 then F=1 because the number of 1 inputs is 1 which is an odd number.
- If A3=0, A2=0, A1=1 and A0=0 then F=1 because the number of 1 inputs is 1 which is an odd number.
- If A3=0, A2=0, A1=1 and A0=1 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=0, A2=1, A1=0 and A0=0 then F=1 because the number of 1 inputs is 1 which is an odd number.
- If A3=0, A2=1, A1=0 and A0=1 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=0, A2=1, A1=1 and A0=0 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=0, A2=1, A1=1 and A0=1 then F=1 because the number of 1 inputs is 3 which is an odd number.
- If A3=1, A2=0, A1=0 and A0=0 then F=1 because the number of 1 inputs is 1 which is an odd number.
- If A3=1, A2=0, A1=0 and A0=1 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=1, A2=0, A1=1 and A0=0 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=1, A2=0, A1=1 and A0=1 then F=1 because the number of 1 inputs is 3 which is an odd number.
- If A3=1, A2=1, A1=0 and A0=0 then F=0 because the number of 1 inputs is 2 which is an even number.
- If A3=1, A2=1, A1=0 and A0=1 then F=1 because the number of 1 inputs is 3 which is an odd number.
- If A3=1, A2=1, A1=1 and A0=0 then F=1 because the number of 1 inputs is 3 which is an odd number.
- If A3=1, A2=1, A1=1 and A0=1 then F=0 because the number of 1 inputs is 4 which is an even number.
| A3 | A2 | A1 | A0 | F |
| 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 1 |
| 0 | 0 | 1 | 0 | 1 |
| 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 | 0 |
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Chapter B Solutions
Computer Organization and Design MIPS Edition, Fifth Edition: The Hardware/Software Interface (The Morgan Kaufmann Series in Computer Architecture and Design)
Ch. B - Prob. 1ECh. B - Prob. 2ECh. B - Prob. 3ECh. B - Prob. 4ECh. B - Prob. 5ECh. B - Prob. 6ECh. B - Prob. 7ECh. B - Prob. 8ECh. B - Prob. 9ECh. B - Prob. 10E
Ch. B - Prob. 11ECh. B - Prob. 14ECh. B - Prob. 15ECh. B - Prob. 16ECh. B - Prob. 17ECh. B - Prob. 19ECh. B - Prob. 20ECh. B - Prob. 25ECh. B - Prob. 26ECh. B - Prob. 27ECh. B - Prob. 28ECh. B - Prob. 29ECh. B - Prob. 30ECh. B - Prob. 31ECh. B - Prob. 32ECh. B - Prob. 33ECh. B - Prob. 34ECh. B - Prob. 35ECh. B - Prob. 36ECh. B - Prob. 37ECh. B - Prob. 38ECh. B - Prob. 39ECh. B - Prob. 40ECh. B - Prob. 41ECh. B - Prob. 42ECh. B - Prob. 43E
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