COMPUTER ORGANIZATION+DESIGN >I<
5th Edition
ISBN: 9781541868397
Author: Patterson
Publisher: ZYBOOKS (CC)
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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 |
Expert Solution & Answer
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Chapter B Solutions
COMPUTER ORGANIZATION+DESIGN >I<
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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