Computer Organization and Design: The Hardware/Software Interface – RISC-V Edition (2021) – Patterson & Hennessy – Solutions to Exercises (PDF)

ALL 6 CHAPTERS & APPENDIX A COVERED




Solutions to Exercises

,Solutions for Appendix A Exercises 1




Solutions for Appendix A Exercises
A.1

A B A B A+B A·B A·B A+B
0 0 1 1 1 1 1 1
0 1 1 0 0 0 1 1
1 0 0 1 0 0 1 1
1 1 0 0 0 0 0 0



A.2 Here is the first equation:
E = ((A ⋅ B) + (A ⋅ C) + (B ⋅ C)) ⋅ (A ⋅ B ⋅ C) .
Now use DeMorgan’s theorems to rewrite the last factor:
E = ((A ⋅ B) + (A ⋅ C) + (B ⋅ C)) ⋅ (A + B + C)
Now distribute the last factor:
E = ((A ⋅ B) ⋅ (A + B + C)) + ((A ⋅ C) ⋅ (A + B + C)) + ((B ⋅ C) ⋅ (A + B + C))
Now distribute within each term; we show one example:
((A ⋅ B) ⋅ (A + B + C)) = (A ⋅ B ⋅ A) + (A ⋅ B ⋅ B) + (A ⋅ B ⋅ C) = 0 + 0 + (A ⋅ B ⋅ C)
(This is simply A . B . C .) Thus, the equation above becomes
E = ( A ⋅ B ⋅ C ) + ( A ⋅ B ⋅ C ) + ( A ⋅ B ⋅ C ) , which is the desired result.

,2 Solutions for Appendix A Exercises



A.7 Four inputs A0–A3 & F (O/P) = 1 if an odd number of 1s exist in A.

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



A.8 F = A3′A2′A1′A0 + A3′A2′A1 A0′ + A3′A2 A1′A0' + A3′A2 A1 A0 +
A3 A2′A1′A0′ + A3 A2′A1 A0 + A3′A2′A1 A0′ + A3 A2 A1 A0′
Note: F = A0 XOR A1 XOR A2 XOR A3. Another question can ask the students to
prove that.

, Solutions for Appendix A Exercises 3




F
A0
A1
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A0
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A0
A1
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A3

A0
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A0
A1
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A0
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A3

A0
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A0
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