**Problem 1 - 6 points - 1 each for b-g**

a. F = ((A + B’)C + A’BC’)’

b. The truth table is shown below

A | B | C | (A+B') | (A+B')C | A'BC' | ((A + B’)C + A’BC’)’ |

0 | 0 | 0 | 1 | 0 | 0 | 1 |

0 | 0 | 1 | 1 | 1 | 0 | 0 |

0 | 1 | 0 | 0 | 0 | 1 | 0 |

0 | 1 | 1 | 0 | 0 | 0 | 1 |

1 | 0 | 0 | 1 | 0 | 0 | 1 |

1 | 0 | 1 | 1 | 1 | 0 | 0 |

1 | 1 | 0 | 1 | 0 | 0 | 1 |

1 | 1 | 1 | 1 | 1 | 0 | 0 |

d. It is much easier to reduce this when starting with the SOP expression you derived with the truth table. Because of the regular order of terms associated with SOP form it is often the case that it's easier to reduce

F(A,B,C) = A'B'C' + A'BC + AB'C' + ABC'

= A'B'C' + A'BC + AB'C' + ABC' + AB'C' Idempotent

= A'B'C' + AB'C' + A'BC + ABC' + AB'C' Commutative

= (A'+ A)B'C' + A'BC + AC'(B + B')
Distrubutive

= (1)B'C' + A'BC + AC'(1)
Complementarity

= B'C' + A'BC + AC'
Identity

e. f = AC' + A'BC + B'C'

f. After switching the 1's and 0's in the K-map :
f = AC + A'BC' + B'C

g.

**Problem 2 - 0 points - Katz 2.19a**

f(w,x,y,z) = w'x' + x'y'

**Problem 3 - 4 points - Katz 2.20c,d**

It's important to notice here that the question ask for the Product-of-Sums expression. Many of you didn't do this. To get obtain the product of sums expression from a k-map circle the rectangles of either the minterm 0's or the maxterm 1's (same set), then read their inverted values from the k-map, This is just an automatic way of applying Demorgans law. This problem asked you to use the K-map method, so if you just converted the SOP form you lost points. I didn't mark off for taking the complement of the SOP form and then apply demorgans by hand, this is the same thing you do from the K-map, it just involves an extra unnecessary step.

f(a,b,c,d) = d(a'+ c') (notice I circled the minterm
0's)

f(a,b,c,d) = (a + b' + c')(a + b + c) (notice I circled
the maxterm 1's)

**Problem 4 - 2 points - Katz 2.24**

It was not okay to use K-maps with don't cares in
them. The problem asked for two different expressions for the same k-map,
but when don't cares are used, the actual k-map changes depending on how
the don't cares are used. In the k-map below the vertical rectangle can
be located so it covers either the upper horizontal rectangle, resulting
in diffent expressions for the SAME k-map. This was the correct approach
to the problem.

**Problem 5 - 2 points - Katz 2.27**

a) Truth Table

A | B | C | D | 08 | 04 | 02 | 01 |

0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 |

0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |

0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 |

0 | 0 | 1 | 1 | 0 | 1 | 0 | 0 |

0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 |

0 | 1 | 0 | 1 | 0 | 1 | 1 | 0 |

0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 |

0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 |

1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 |

1 | 0 | 0 | 1 | 1 | 0 | 1 | 0 |

1 | 0 | 1 | 0 | 1 | 0 | 1 | 1 |

1 | 0 | 1 | 1 | 1 | 1 | 0 | 0 |

1 | 1 | 0 | 0 | 1 | 1 | 0 | 1 |

1 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |

1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 |

1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |

O8 SOP- f(a,b,c,d) = ab' + ac' + ad' + a'bcd'

O8 POS- f(a,b,c,d) = (a+b)(a+c)(a+d)(a'+b'+c'+d')

O4 SOP- f(a,b,c,d) = bc' + bd' + b'cd

O4 POS- f(a,b,c,d) = (b+c)(b+d)(b'+c'+d')

O2 SOP- f(a,b,c,d) = c'd + cd'

O2 POS- f(a,b,c,d) = (c+d)(c'+d')

O1 SOP- f(a,b,c,d) = d'

O1 POS- f(a,b,c,d) = d'

c. They both have the same number of literals (22).

**Problem 6 - 4 points - Katz 2.29 Except for an even number of 1'**

a. Truth Table:

A | B | C | D | F |

0 | 0 | 0 | 0 | 1 |

0 | 0 | 0 | 1 | 0 |

0 | 0 | 1 | 0 | 0 |

0 | 0 | 1 | 1 | 1 |

0 | 1 | 0 | 0 | 0 |

0 | 1 | 0 | 1 | 1 |

0 | 1 | 1 | 0 | 1 |

0 | 1 | 1 | 1 | 0 |

1 | 0 | 0 | 0 | 0 |

1 | 0 | 0 | 1 | 1 |

1 | 0 | 1 | 0 | 1 |

1 | 0 | 1 | 1 | 0 |

1 | 1 | 0 | 0 | 1 |

1 | 1 | 0 | 1 | 0 |

1 | 1 | 1 | 0 | 0 |

1 | 1 | 1 | 1 | 1 |

f(a,b,c,d) = a'b'c'd' + a'b'cd + a'bc'd +a'bcd' + abc'd' + ab'cd' + ab'c'd + abcd

Because each of these minterm differs by at least 2 bits from its nearest neighbor it cannot be simplified using a k-map. Spacing like this results from the XOR function and is a good clue that XOR's can be used to make a cheap implemetation.

c. f = (a XOR b XOR c XOR d)' == a XOR b XOR c XOR d XOR 1

**Problem 7 - 2 points**

a. Truth Table:

A | B | C | D | X |

0 | 0 | 0 | 0 | 1 |

0 | 0 | 0 | 1 | 1 |

0 | 0 | 1 | 0 | 1 |

0 | 0 | 1 | 1 | 0 |

0 | 1 | 0 | 0 | 1 |

0 | 1 | 0 | 1 | 1 |

0 | 1 | 1 | 0 | 1 |

0 | 1 | 1 | 1 | 1 |

1 | 0 | 0 | 0 | 1 |

1 | 0 | 0 | 1 | 1 |

1 | 0 | 1 | 0 | 1 |

1 | 0 | 1 | 1 | 1 |

1 | 1 | 0 | 0 | 1 |

1 | 1 | 0 | 1 | 1 |

1 | 1 | 1 | 0 | 1 |

1 | 1 | 1 | 1 | 1 |

F = a'b+a+c'+d' = ((a'b)'a'bc)'

See picture below.

**Problem 8 - 0 points**

This was the most popular solution, but the most optimal solution
is 0 (must be a misprint in the book - the ciruit goes to 0)