Simplifying Boolean Expressions

1
Simplifying Boolean Expressions

• In this session we examine various ways of
  simplifying Boolean expressions, in particular,
  the use of Karnaugh maps.

2
A Boolean (0,1) modelfor ?

• p T or F q T or F (T on F
  off)pT, q T, p ? q T ? T T p T,
  q F , p ? q T ? F T p F, q T, p
  ? q F ? T Tp F, q F, p ? q F ?
  F F
• p 0 or 1 q 0 or 1 1 1 1
  ? 1 1 1 0 1 ? 0
  1 0 1 0 ? 1 1 0 0
  0 ? 0 0

The "arithmetic" looks likea special addition pq

3
A Boolean (0,1) modelfor ?

• p T or F q T or F (T on F
  off)pT, q T, p ? q T ? T T p T,
  q F , p ? q T ? F F p F, q T, p
  ? q F ? T Fp F, q F, p ? q F ?
  F F
• p 0 or 1 q 0 or 1 1 1 1
  ? 1 1 1 0 1 ? 0 0 0
  1 0 ? 1 0 0 0
  0 ? 0 0

The "arithmetic" looks like
multiplication pq

4
A Boolean (0,1) modelfor ?

• p T or F (T on F off)p T , ? p
  F p F , ? p T
• p 0 or 1 1 0 0 1

It is ( 1 - p) p'

5
Analysing a simple circuit

• In terms of logic clearly we have ? p ? q
  pT, q T, ? p ? q F ? T F p T, q
  F , ? p ? q F ? F F p F, q T, ? p ?
  q T ? T T p F, q F, ? p ? q T ? F
  F
• 1 1 0 ? 1 0 1
  0 0 ? 0 0 0 1
  1 ? 1 1 0 0 1 ? 0
  0

p'q
6
Finding the Boolean expression for a circuit

• In terms of logical operators (p ? q) ? ? (p
  ? q) ? (pq)(pq)' ? (p ? q) ? (?p ??q)
  ? (pq)(p'q') ? (p?(?p ??q)) ? (q?(?p??q)
  ? p(p'q')q(p'q') ? ((p??p)?(p ??q))?
  ((q??p)?(q??q)) ? pp'pq'qp'qq'? (c?(p
  ??q))?((q??p)?c) ? 0 pq'qp' 0 ?
  (p??q)?(q??p) ? pq'qp' (the symmetric
  difference)

7
Convention

• We will leave out the multiplication sign, ?, and
  assume that multiplication is executed before
  addition . Thus we shall write p q r for
  (p ? q) rbut we will continue to write (p ?
  q) ? rsince it is not usual to give ? precedence
  over ?

8
Some laws of Boolean Algebra

• x x' 0
• x'' x
• (x y)' x'y' (xy)' x'y'
• x 1 1 x 0 x
• x 1 x x 0 0
• x(xy) x x (xy) x
• 0' 1 1' 0
• x x x x(1-x) x- xx 0

9
The algebra
10
"Simplifying" Boolean Expressions

• Using the axioms and derived laws
• xz'x'y(yz)' xz'x'yy'z'
  xz'1z'x'yy' (x1)z' x'yy' 1z'
  x'y'' y' z'(xy')'y' z' ((xy')y)'
  z' (xyy'y)' z' (xy0)'
  z'(xy)' x'y'z' (xyz)'
• (Identical to logic reductions but in different,
  more compact, notation.)

11
"Simplifying"

• x'yy' x'y'' y' (xy')'y'
  ((xy')y)' (xy0)' (xy)' x'y'
• andxy' y x yyx' xx'y x(cf
  Grossmanp.144)

12
Picking out when f is "true"

• Consider the table x y f (x,y) 1 1
  1 xy 1 1 0 1 xy' 1 0
  1 1 x'y 1 0 0 0If any one
  of the three cases is true then f is true, or,f
  (x,y,z) xy xy' x'y(xx')yxy'yxy' x y
• Recall that 11 1

Consider the table x y f (x,y) 1 1
1 xy 1 1 0 0 0 1
0 0 0 0In only one case is f is
true f (x,y,z) xy The sum of minterms is the
disjunctive normal form. (Not necessarily the
simplest form.)
13
Minterms

• x y z w f(x,y,z,w)1 1 1 1 1
  ?xyzw1 1 1 0 01 1 0 1
  01 1 0 0 1 ?xyz'w' 1 0 1 1
  1 ?xy'zw1 0 1 0 01 0
  0 1 01 0 0 0 1
  ?xy'z'w'0 1 1 1 00 1 1 0
  0 0 1 0 1 00 1 0 0 00 0
  1 1 1 ?x'y'zw0 0 1 0
  00 0 0 1 1 ?x'y'z'w0 0 0 0
  0

• Minterms are the products of factors, one for
  each variable, which return 1 for each particular
  case where f 1
• e.g. the minterms to right arexyzw, xyz'w',
  xy'zw, xy'z'w', x'y'zw and x'y'z'w

14
The Disjunctive Normal Form

• The disjunctive normal form for a Boolean
  function is the sum of its minterms.
• In the previous example, the disjunctive normal
  form isf(x,y,z,w) xyzwxyz'w'xy'zw xy'z'w'
  x'y'zw x'y'z'w
• It is not necessarily the simplest expression for
  the function.

15
Karnaugh map for f (x,y)

• Suppose the disjunctive normal form for f (x,y)
  is xyx'yxy'. In the table y y' x
  1 1 x' 1 0the 1 entries
  identify the presence of minterms. Boxing in
  rows and columns of 1's identifies minterms with
  common factors. Since xxx xyx'yxy'
  xyx'yxyxy'(xx')yx(yy')xy which we can
  deduce from the diagram.

16
Karnaugh map for f (x,y,z)

• The table is set up as below (order of columns is
  important) yz
  y'z y'z' yz' x 1 0 0
  1
• x' 0 1 0 1
• We search for
• four 1's in a row
• four 1's in a square -- assuming right and left
  edges are adjacent
• two adjacent 1's in a row or column -- "
• single 1's
• in that order. Loops may overlap and all 1's
  must be covered, hopefully with as few loops as
  possible with loops as large as possible.

17
f (x,y,z) xyz xyz' x'y'z x'yz'

• The Karnaugh map is shown above and is boxed
  accordingly. From the two-loops we get y and
  yz' and from the single 1, x'y'z. Thus our
  simplification is xy yz' x'y'z

18
Karnaugh map for f (x,y,z,w)

• The table is set up as below (order of rows and
  columns is important)
  zw z'w z'w' zw' xy 1 0
  0 0 x'y 0 1 0
  0 x'y' 1 1 0 1 xy'
  1 0 0 1
• We search for
• eight 1's in four by two or two by four
  rectangles
• four 1's in a row or a column (top and bottom,
  left and right are adjacent
• four 1's in a square
• two adjacent 1's in a row or column
• single 1's

19
f (x,y,z,w) xyzw x'yz'w x'y'zw x'y'z'w
x'y'zw' xy'zw xy'zw'

• The table is set up as below (order of rows and
  columns is important)
  zw z'w z'w' zw' xy 1 0
  0 0 x'y 0 1 0
  0 x'y' 1 1 0 1
  xy' 1 0 0 1
• The reduction is y'z xzw x'z'w

20
The "don't care" option

• Occasionally the function table does not list
  all possible inputs, possibly because some will
  never occur e.g. x (agtb) y (altb) will
  never have the inputs 1 1 or 0 0.In these
  cases we can place a d in the Karnaugh map
  and include it in a loop if desirable or ignore
  it if preferred.

21
Writing any given expression in disjunctive
normal form

• Note that x y is not in disjunctive normal
  form. To obtain that form we re-write it as
  x(yy') y(xx') xy xy' xy
  x'y xy xy' x'y(as obtained earlier
  from the function table).
• Apply the same procedure, together with
  distributive and deMorgan laws (etc) to any
  expression to bring it to disjunctive normal form.

22
f (x,y,z) (xy)'z (x'z z')y

• To write f in disjunctive normal form
  (xy)'z (x'z z')y (x'y')z x'yz yz'
  x'z y'z x'yz yz' x'(yy')z (xx')y'z
  x'yz (xx')yz' x'yzx'y'z xy'zx'y'z x'yz
  xyz'x'yz' xyz' xy'z x'yzx'y'z x'yz'

23
Adding two bits again

• The input/output tables for adding two bits
  are x y sum x y carry 0 0 0 0 0 0
  1 0 1 1 0 0 0 1 1 0 1 0 1 1 0 1 1 1
• Exercise produce a pair of circuits to do it.

24
Adding three bits again

• The input/output tables for adding three bits
  are x y z sum x y z carry
  0 0 0 0 0 0 0 0 0 0
  1 1 0 0 1 0 0 1 0 1
  0 1 0 0 1 0 0 1 1 0 0
  0 0 1 1 0 0 1 1 1
  Exercise 1 0 1 0 1 0 1 1
  produce a pair1 1 0 0 1 1
  0 1 of circuits to do it 1 1 1
  1 1 1 1 1

25
Whats the big idea ?

• Logical reductions can be achieved by analysis
  of the Boolean expressions that describe them
  using the laws of logic and Karnaugh maps or
  truth tables or the Venn diagrams that represent
  those tables geometrically.