// Questions

1. Reduce the following expressions using K-maps. Show your maps.
   • F(A,B,C)=ABC+ABC+AB+ABC+A B
   • F(A,B,C,D)=A BCD+ A BCD+ ABCD+ ABCD+ ABCD+ ABC D
   • F(A,B)=A B+AB
   • F(A,B,C)=A B C+ABC+ABC+ABC
   
   
   

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2. Use a K-map to derive the simplest expression based on the following table:

   A	B	C	F
   0	0	0	0
   0	0	1	1
   0	1	0	1
   0	1	1	0
   1	0	0	1
   1	0	1	0
   1	1	0	0
   1	1	1	1

   
   
   
   

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3. Use a K-map to derive the simplest expression based on the following table:

   A	B	C	F
   0	0	0	1
   0	0	1	0
   0	1	0	1
   0	1	1	1
   1	0	0	1
   1	0	1	0
   1	1	0	0
   1	1	1	1

   
   
   
   

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4. Use a K-map to derive the simplest expression based on the following table:

   A	B	C	D	F
   0	0	0	0	1
   0	0	0	1	0
   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	0
   1	0	0	0	1
   1	0	0	1	0
   1	0	1	0	1
   1	0	1	1	0
   1	1	0	0	0
   1	1	0	1	0
   1	1	1	0	0
   1	1	1	1	0

   
   
   
   

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5. Use a K-map to derive the simplest expression based on the following table:

   A	B	C	D	F
   0	0	0	0	1
   0	0	0	1	0
   0	0	1	0	1
   0	0	1	1	0
   0	1	0	0	1
   0	1	0	1	0
   0	1	1	0	1
   0	1	1	1	1
   1	0	0	0	1
   1	0	0	1	0
   1	0	1	0	1
   1	0	1	1	0
   1	1	0	0	1
   1	1	0	1	0
   1	1	1	0	1
   1	1	1	1	1

   
   
   
   

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6. Design a combinational circuit that when given a 3-bit binary number as an input will produce a binary number equal to the square of the number as an output. The truth table for such a circuit follows. Find the simplest expressions for the output quantities in SOP form:

   x	y	z	A	B	C	D	E	F
   0	0	0	0	0	0	0	0	0
   0	0	1	0	0	0	0	0	1
   0	1	0	0	0	0	1	0	0
   0	1	1	0	0	1	0	0	1
   1	0	0	0	1	0	0	0	0
   1	0	1	0	1	1	0	0	1
   1	1	0	1	0	0	1	0	0
   1	1	1	1	1	0	0	0	1

   
   
   
   

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// Answers

1. 	AB	AB	AB	AB
   C	√		√	√
   C	√	√	√	√

   
   
   C+A+A B

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2. 	AB	AB	AB	AB
   CD		√	√
   CD
   CD	√
   CD	√

   
   
   A B C+BC D

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3. 	A	A
   B	√	√
   B

   
   
   B

   ---

4. 	AB	AB	AB	AB
   C	√		√
   C		√		√

   
   A B C+ABC+ABC+ABC
   

   ---

5. 	AB	AB	AB	AB
   C		√		√
   C	√		√	√

   
   ABC+AC+AB+C

   ---

6. 	AB	AB	AB	AB
   C	√	√		√
   C		√	√

   
   A C+ AB+BC+BC

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7. 	AB	AB	AB	AB
   CD	√	√		√
   CD
   CD
   CD	√			√

   
   
   ACD +ABD +BCD +ABD

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8. 	AB	AB	AB	AB
   CD	√	√	√	√
   CD
   CD		√	√
   CD	√	√	√	√

   
   
   BC+BD+BCD +CD +ABD+ ABD+ BD

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9. A+B+C

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Source: Passafine, John and Michael Douglas, Digital Logic Design