# CS1104 Computer Organisation by wpr1947

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```									  CS1104: Computer Organisation

http://www.comp.nus.edu.sg/~cs1104

Lecture 5: Karnaugh Maps
Lecture 5: Karnaugh Maps
   Function Simplification
   Algebraic Simplification
   Introduction to K-maps
   Venn Diagrams
   2-variable K-maps
   3-variable K-maps
   4-variable K-maps
   5-variable and larger K-maps

CS1104-5                 Lecture 5: Karnaugh Maps   2
Lecture 5: Karnaugh Maps
   Simplification using K-maps
   Converting to Minterms Form
   Simplest SOP Expressions
   Getting POS Expressions
   Don‟t-care Conditions
   Review
   Examples

CS1104-5              Lecture 5: Karnaugh Maps   3
Function Simplification (1/2)
 Why simplify?
 Simpler expression uses less logic gates.
 Thus: cheaper, less power, faster (sometimes).

 Simplification techniques:
 Algebraic Simplification.
simplify symbolically using theorems/postulates.
requires skill but extremely open-ended.
 Karnaugh Maps.
 diagrammatic technique using „Venn-like diagram‟.
 easy for humans (pattern-matching skills).
 simplified standard forms.
 limited to not more than 6 variables.

CS1104-5                     Function Simplification          4
Function Simplification (2/2)
 Simplification techniques:
 Quine-McCluskey tabulation technique.
 tabulation technique based on certain „cancellation theorems‟.
 simplified standard forms.
 tedious, repetitive step-by-step technique.
 boring to humans BUT suitable for computers.
 larger variables possible, but computationally intensive.

CS1104-5                     Function Simplification                       5
Algebraic Simplification (1/5)
 Algebraic simplification aims to minimise
(i) number of literals, and
(ii) number of terms
 But sometimes conflicting.
 Let‟s aim at reducing the number of literals.

CS1104-5                  Algebraic Simplification   6
Algebraic Simplification (2/5)
 Example:
(x+y).(x+y').(x'+z)                     (6 literals)
= (x.x+x.y'+x.y+y.y').(x'+z)            (assoc.)
= (x+x.(y'+y)+0).(x'+z)                 (idemp.,assoc., compl.)
= (x+x.(1)+0).(x'+z)                    (complement)
= (x+x+0).(x'+z)                        (identity 1)
= (x).(x'+z)                            (idemp, identity 0)
= (x.x'+x.z)                            (assoc.)
= (0+x.z)                               (complement)
= x.z                                   (identity 0)
Number of literals reduced from 6 to 2.

CS1104-5                     Algebraic Simplification                        7
Algebraic Simplification (3/5)
 Find minimal SOP and POS expressions of
f(x,y,z) = x'.y.(z + y'.x) + y'.z
x'.y.(z+y'.x) + y'.z
= x'.y.z + x'.y.y'.x + y'.z   (distributivity)
= x'.y.z + 0 + y'.z           (complement, null element 0)
= x'.y.z + y'.z               (identity 0)
= x'.z + y‟.z                 (absorption)
= (x' + y').z                 (distributivity)
Minimal SOP of f = x'.z + y'.z (2 2-input AND gates and
1 2-input OR gate)
Minimal POS of f = (x' + y').z (1 2-input OR gate and
1 2-input AND gate)

CS1104-5                         Algebraic Simplification               8
Algebraic Simplification (4/5)
 Find minimal SOP expression of
f(a,b,c,d) = a.b.c + a.b.d + a'.b.c' + c.d + b.d'
a.b.c + a.b.d + a'.b.c' + c.d + b.d'
= a.b.c + a.b + a'.b.c' + c.d + b.d' (absorption)
= a.b.c + a.b + b.c' + c.d + b.d' (absorption)
= a.b + b.c' + c.d + b.d'            (absorption)
= a.b + c.d + b.(c' + d')            (distributivity)
= a.b + c.d + b.(c.d)'               (DeMorgan)
= a.b + c.d + b                      (absorption)
= b + c.d                            (absorption)
Number of literals reduced form 13 to 3.

CS1104-5                       Algebraic Simplification            9
Algebraic Simplification (5/5)
 Difficulty – needs good algebraic manipulation skills.

CS1104-5               Algebraic Simplification             10
X,Y) together, to produce a result of two bits (called C,
S).
 A black-box representation of this circuit is:
Truth table representation is:

X       Half       S                   X    Y    C   S
adder           (X+Y)           0    0    0   0
Y                  C
0    1    0   1
1    0    0   1
1    1    1   0

X    Y       C   S
 In sum-of-minterms forms:                  0    0       0   0
C = X.Y                            0    1       0   1
1    0       0   1
S = X'.Y + X.Y'                    1    1       1   0
 Algebraic simplification could simplify S to:
S = X'.Y + X.Y'
= XY
X
 Giving:                        Y
S

C

Introduction to K-maps
 Systematic method to obtain simplified sum-of-
products (SOPs) Boolean expressions.
 Objective: Fewest possible terms/literals.
 Diagrammatic technique based on a special form of
Venn diagram.
 Advantage: Easy with visual aid.
 Disadvantage: Limited to 5 or 6 variables.

CS1104-5               Introduction to K-maps          13
Venn Diagrams (1/2)
 Venn diagram to represent the space of minterms.
 Example of 2 variables (4 minterms):

a'.b'

a.b'      a.b      a'.b
a
b

CS1104-5                Venn Diagrams                 14
Venn Diagrams (2/2)
 Each set of minterms represents a Boolean
function. Examples:

{ a.b, a.b' }       a.b + a.b' = a.(b+b') = a
{ a'.b, a.b }       a'.b + a.b = (a'+a).b = b
{ a.b }             a.b
a'.b'
{ a.b, a.b', a'.b }  a.b + a.b' + a'.b = a + b
{}                 0                             a.b' a.b   a'.b
a
{ a'.b',a.b,a.b',a'.b }  1                                              b

CS1104-5                           Venn Diagrams                         15
2-variable K-maps (1/4)
 Karnaugh-map (K-map) is an abstract form of Venn
diagram, organised as a matrix of squares, where
 each square represents a minterm
 adjacent squares always differ by just one literal (so
that the unifying theorem may apply: a + a' = 1)
 For 2-variable case (e.g.: variables a,b), the map can
be drawn as:

CS1104-5                       2-variable K-maps                16
2-variable K-maps (2/4)
 Alternative layouts of a 2-variable (a, b) K-map
Alternative 1:                          Alternative 2:
a
b                    b                    a
OR                                        OR
a'b'   a'b             m0   m1             a'b'   ab'        m0   m2

a            ab          a        m3         b                 b   m1   m3
ab'                   m2                  a'b    ab

Alternative 3:
a                      a
OR
b      ab    a'b         b   m3   m1             and others…
ab'    a'b'            m2   m0

CS1104-5                                2-variable K-maps                               17
2-variable K-maps (3/4)
 Equivalent labeling:
b                               b
a           0   1
0
equivalent to:
a
1

a                                   a
b           1   0

equivalent to:         0
b
1

CS1104-5                 2-variable K-maps                       18
2-variable K-maps (4/4)
 The K-map for a function is specified by putting
 a „1‟ in the square corresponding to a minterm
 a „0‟ otherwise

 For example: Carry and Sum of a half adder.
b                           b

0   0                       0   1

a    0   1                  a    1   0

C = a.b                 S = a.b' + a'.b

CS1104-5                       2-variable K-maps                19
3-variable K-maps (1/2)
 There are 8 minterms for 3 variables (a, b, c).
Therefore, there are 8 cells in a 3-variable K-map.
b
b
bc
bc
a                                                         a        00   01       11       10
00      01          11     10
a'b'c'   a'b'c       a'bc       a'bc'                0    m0   m1       m3       m2
0
OR
a            m4   m5       m7       m6
a             ab'c'    ab'c        abc        abc'                 1
1
c
c

Above arrangement ensures that minterms                                   Note Gray code sequence
of adjacent cells differ by only ONE literal.
(Other arrangements which satisfy this
criterion may also be used.)

CS1104-5                                             3-variable K-maps                                     20
3-variable K-maps (2/2)
 There is wrap-around in the K-map:
 a'.b'.c' (m0) is adjacent to a'.b.c' (m2)
 a.b'.c' (m4) is adjacent to a.b.c' (m6)
bc
a
00   01   11    10
0    m0   m1   m3    m2

1    m4   m5   m7    m6

Each cell in a 3-variable K-map has 3 adjacent neighbours.
In general, each cell in an n-variable K-map has n adjacent
neighbours. For example, m0 has 3 adjacent neighbours:
m1, m2 and m4.

CS1104-5                             3-variable K-maps                   21
Quick Review Questions (1)
Textbook page 104.
5-1. The K-map of a 3-variable function F is shown below.
What is the sum-of-minterms expression of F?
b
bc
a
00     01       11   10
0     1      0        0        1

a             0      1        0        0
1

c

5-2. Draw the K-map for this function A:
A(x, y, z) = x.y + y.z' + x'.y'.z

CS1104-5                     Quick Review Questions (1)    22
4-variable K-maps (1/2)
 There are 16 cells in a 4-variable (w, x, y, z) K-map.

y
yz
wx        00    01        11        10
00   m0    m1        m3        m2

01   m4    m5        m7        m6
x
m12   m13       m15       m14
11
w
m8    m9        m11       m10
10
z

CS1104-5                         4-variable K-maps                 23
4-variable K-maps (2/2)
 There are 2 wrap-arounds: a horizontal wrap-around
and a vertical wrap-around.
 Every cell thus has 4 neighbours. For example, the
cell corresponding to minterm m0 has neighbours m1,
m2, m4 and m8.
yz                         y
wx
m0    m1        m3        m2

m4    m5        m7        m6
x
m12   m13       m15       m14
w
m8    m9        m11       m10

z

CS1104-5                 4-variable K-maps                                            24
5-variable K-maps (1/2)
 Maps of more than 4 variables are more difficult to use
because the geometry (hyper-cube configurations) for
combining adjacent squares becomes more involved.
 For 5 variables, e.g. vwxyz, need 25 = 32 squares.

CS1104-5                 5-variable K-maps                   25
5-variable K-maps (2/2)
 Organised as two 4-variable K-maps:
v'                                                  v
y                                                   y
yz                                                  yz
wx        00    01        11        10              wx        00    01        11        10
00   m0    m1        m3        m2                   00   m16   m17       m19       m18

01   m4    m5        m7        m6                   01   m20   m21       m23       m22
x                                                 x
m12   m13       m15       m14                       m28   m29       m31       m30
11                                                  11
w                                                   w
m8    m9        m11       m10                       m24   m25       m27       m26
10                                                  10
z                                                   z

Corresponding squares of each map are adjacent.
Can visualise this as being one 4-variable map on TOP of the
other 4-variable map.

CS1104-5                                           5-variable K-maps                                             26
Larger K-maps (1/2)
 6-variable K-map is pushing the limit of human
“pattern-recognition” capability.
 K-maps larger than 6 variables are practically
unheard of!
 Normally, a 6-variable K-map is organised as four
4-variable K-maps, which are mirrored along two
axes.

CS1104-5                   Larger K-maps                27
Larger K-maps (2/2)                            b

ef         a'.b'                     a'.b               ef
cd         00   01   11    10       10    11       01   00         cd
00    m0   m1   m3    m2       m18 m19 m17 m16               00

01    m4   m5   m7    m6       m22 m23 m21 m20               01

11    m12 m13 m15 m14          m30 m31 m29 m28               11

10    m8   m9   m11 m10        m26 m27 m25 m24               10

10    m40 m41 m43 m42          m58 m59 m57 m56               10
11    m44 m45 m47 m46          m62 m63 m61 m60               11
a
01    m36 m37 m39 m38          m54 m55 m53 m52               01
00    m32 m33 m35 m34          m50 m51 m49 m48               00
cd         00   01   11    10       10    11       01   00         cd
ef                                                     ef
a.b'                          a.b
Try stretch your recognition capability by finding simplest
sum-of-products expression for S m(6,8,14,18,23,25,27,29,41,45,57,61).
CS1104-5                                    Larger K-maps                                 28
Simplification Using K-maps (1/9)
 Based on the Unifying Theorem:
A + A' = 1
 In a K-map, each cell containing a „1‟ corresponds to a
minterm of a given function F.
 Each group of adjacent cells containing „1‟ (group must
have size in powers of twos: 1, 2, 4, 8, …) then
corresponds to a simpler product term of F.
 Grouping 2 adjacent squares eliminates 1 variable, grouping 4
squares eliminates 2 variables, grouping 8 squares eliminates 3
variables, and so on. In general, grouping 2n squares eliminates
n variables.

CS1104-5                     Simplification Using K-maps                 29
Simplification Using K-maps (2/9)
 Group as many squares as possible.
 The larger the group is, the fewer the number of literals in the
resulting product term.

 Select as few groups as possible to cover all the squares
(minterms) of the function.
 The fewer the groups, the fewer the number of product terms in
the minimized function.

CS1104-5                      Simplification Using K-maps                 30
Simplification Using K-maps (3/9)
 Example:
F (w,x,y,z) = w‟.x.y'.z' + w'.x.y'.z + w.x'.y.z'
+ w.x'.y.z + w.x.y.z' + w.x.y.z
= S m(4, 5, 10, 11, 14, 15)
y
yz
wx        00   01       11       10
00

01   1    1
x   (cells with „0‟ are not
1        1
w
11                                    shown for clarity)
1        1
10
z

CS1104-5                             Simplification Using K-maps                     31
Simplification Using K-maps (4/9)
 Each group of adjacent minterms (group size in
powers of twos) corresponds to a possible product
term of the given function.
y
yz
wx        00   01       11       10
00
A
01   1    1
x
11                 1        1
w
10                 1        1        B

z

CS1104-5                Simplification Using K-maps                  32
Simplification Using K-maps (5/9)
 There are 2 groups of minterms: A and B, where:
A     = w'.x.y'.z' + w„.x.y'.z
= w'.x.y'.(z' + z)
= w'.x.y'

B     =   w.x'.y.z' + w.x'.y.z + w.x.y.z' + w.x.y.z
y
=   w.x'.y.(z' + z) + w.x.y.(z' + z)          yz
wx          00   01       11       10
=   w.x'.y + w.x.y
00
=   w.(x'+x).y                        A
=   w.y                                      01        1    1
x
11                 1        1
w
10                 1        1         B
z
CS1104-5                       Simplification Using K-maps                                        33
Simplification Using K-maps (6/9)
 Each product term of a group, w'.x.y' and w.y,
represents the sum of minterms in that group.
 Boolean function is therefore the sum of product terms
(SOP) which represent all groups of the minterms of
the function.
F(w,x,y,z) = A + B = w'.x.y' + w.y

CS1104-5              Simplification Using K-maps           34
Simplification Using K-maps (7/9)
 Larger groups correspond to product terms of fewer
literals. In the case of a 4-variable K-map:
1 cell   = 4 literals, e.g.: w.x.y.z, w'.x.y'.z
2 cells = 3 literals, e.g.: w.x.y, w.y'.z'
4 cells = 2 literals, e.g.: w.x, x'.y
8 cells = 1 literal, e.g.: w, y', z
16 cells = no literal, e.g.: 1

CS1104-5               Simplification Using K-maps              35
Simplification Using K-maps (8/9)
 Other possible valid groupings of a 4-variable K-map
include:

1           1   1        1
1                                          1   1   1

1             1   1   1

1

1               1   1        1         1           1
1

P                      P                   P

CS1104-5            Simplification Using K-maps                   36
Simplification Using K-maps (9/9)
 Groups of minterms must be
(1) rectangular, and
(2) have size in powers of 2‟s.
Otherwise they are invalid groups. Some examples of
invalid groups:
1    1     1                          1

1    1     1                      1

1                                             1   1

1   1    1                                1   1

O                                    O
CS1104-5                Simplification Using K-maps                   37
Converting to Minterms Form (1/2)
 The K-map of a function is easily drawn when the
function is given in canonical sum-of-products, or sum-
of-minterms form.
 What if the function is not in sum-of-minterms?
 Convert it to sum-of-products (SOP) form.
 Expand the SOP expression into sum-of-minterms
expression, or fill in the K-map directly based on the SOP
expression.

CS1104-5                   Converting to Minterms Form              38
Converting to Minterms Form (2/2)
 Example:
f(A,B,C,D) = A.(C+D)'.(B'+D') + C.(B+C'+A'.D)
= A.(C'.D').(B'+D') + B.C + C.C' + A'.C.D
= A.B'.C'.D' + A.C'.D' + B.C + A'.C.D
A
A.B'.C'.D' + A.C'.D' + B.C + A'.C.D                      AB
CD         00   01       11       10
= A.B'.C'.D' + A.C'.D'.(B+B') + B.C + A'.C.D
00                  1        1
= A.B'.C'.D' + A.B.C'.D' + A.B'.C'.D' +
B.C.(A+A') + A'.C.D                                  01
D
= A.B'.C'.D' + A.B.C'.D' + A.B.C + A'.B.C +             11    1    1        1
A'.C.D                                          C
10         1        1
= A.B'.C'.D' + A.B.C'.D' + A.B.C.(D+D') +
A'.B.C.(D+D') + A'.C.D.(B+B')                                        B
= A.B'.C'.D' + A.B.C'.D' + A.B.C.D + A.B.C.D'
+ A'.B.C.D + A'.B.C.D' + A'.B„.C.D
CS1104-5                    Converting to Minterms Form                                          39
Simplest SOP Expressions (1/3)
 To find the simplest possible sum of products (SOP)
expression from a K-map, you need to obtain:
 minimum number of literals per product term; and
 minimum number of product terms

 This is achieved in K-map using
 bigger groupings of minterms (prime implicants) where
possible; and
 no redundant groupings (look for essential prime implicants)

Implicant: a product term that could be used
to cover minterms of the function.

CS1104-5                  Simplest SOP Expressions                    40
Simplest SOP Expressions (2/3)
 A prime implicant is a product term obtained by
combining the maximum possible number of minterms
from adjacent squares in the map.
 Use bigger groupings (prime implicants) where
possible.

1   1   1                         1    1   1

1   1   1
O
1    1   1
P

CS1104-5                    Simplest SOP Expressions               41
Simplest SOP Expressions (3/3)
 No redundant groups:
1    1                               1   1

1
1

1
1
O                      1
1

1
1
P
1   1                                1   1

Essential prime implicants
 An essential prime implicant is a prime implicant that
includes at least one minterm that is not covered by
any other prime implicant.

CS1104-5                   Simplest SOP Expressions                   42
Quick Review Questions (2)
Textbook page 104.
5-3. Identify the prime implicants and the essential
prime implicants of the two K-maps below.
A
b
AB
bc                                         CD         00   01       11       10
a
00   01       11     10
00    1    1                  1
0    1    1         0          1
01                  1         1
a                                                                                         D
1    0    1         0          0
11    1    1                  1
C
c                                10    1             1         1

B

CS1104-5                               Quick Review Questions (2)                                    43
Simplest SOP Expressions (1/5)
 Algorithm 1 (non optimal):
1. Count the number of adjacencies for each minterm on the K-
map.
2. Select an uncovered minterm with the fewest number of
adjacencies. Make an arbitrary choice if more than one choice
is possible.
3. Generate a prime implicant for this minterm and put it in the
cover. If this minterm is covered by more than one prime
implicant, select the one that covers the most uncovered
minterms.
4. Repeat steps 2 and 3 until all the minterms have been
covered.

CS1104-5                   Simplest SOP Expressions                     44
Simplest SOP Expressions (2/5)
 Algorithm 2 (non optimal):
1. Circle all prime implicants on the K-map.
2. Identify and select all essential prime implicants for the cover.
3. Select a minimum subset of the remaining prime implicants to
complete the cover, that is, to cover those minterms not
covered by the essential prime implicants.

CS1104-5                    Simplest SOP Expressions                        45
Simplest SOP Expressions (3/5)
 Example:
f(A,B,C,D) =  m(2,3,4,5,7,8,10,13,15)
A
AB
CD         00   01       11       10

00         1                  1
All prime implicants
01         1        1
D
11    1    1        1
C
10    1                       1

B

CS1104-5                        Simplest SOP Expressions                          46
Simplest SOP Expressions (4/5)
A
AB                                                                              A
CD        00    01       11       10                     AB
CD            00       01       11       10
00
1                 1
01
1        1                           00                1                  1        Essential prime
D

C
11       1     1        1                           01                1        1
implicants
10       1                       1                                                           D
11       1        1        1
B                          C
10       1                           1

B
A
AB
CD                 00       01       11       10

00                     1                  1                               Minimum cover
01                     1        1
D
11           1         1        1
C
10           1                            1

B

CS1104-5                                                   Simplest SOP Expressions                                            47
Simplest SOP Expressions (5/5)
A
AB
CD         00   01       11       10
A'BC'
00         1                  1       AB'D'
01         1        1
D
11    1    1        1
C
10    1                       1       B.D
B

A'B'C

f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C'

CS1104-5                        Simplest SOP Expressions                 48
Quick Review Questions (3)
Textbook page 104.
5-4. Find the simplified expression for G(A,B,C,D).
A
AB
CD         00   01       11       10

00         1

01         1        1         1
D
11    1    1        1
C
10                  1

B

5-5 to 5-7.

CS1104-5                Quick Review Questions (3)            49
Getting POS Expressions (1/2)
 Simplified POS expression can be obtained by grouping
the maxterms (i.e. 0s) of given function.
 Example:
Given F=m(0,1,2,3,5,7,8,9,10,11), we first draw
the K-map, then group the maxterms together:
A
AB
CD         00   01       11       10

00    1    0        0         1

01    1    1        0        1
D
11    1    1        0        1
C
10    1    0        0         1

B

CS1104-5                    Getting POS Expressions                50
Getting POS Expressions (2/2)
A                                          A
AB                                         AB
CD         00   01       11       10       CD         00   01       11       10

K-map         00    1    0        0        1             00    0    1        1        0        K-map
of F          01    1    1        0        1             01    0    0        1        0        of F'
D                                          D
11    1    1        0        1             11    0    0        1        0
C                                          C
10    1    0        0        1             10    0    1        1        0

B                                          B

 This gives the SOP of F' to be:
F' = B.D' + A.B
 To get POS of F, we have:
F = (B.D' + A.B)'
= (B.D')'.(A.B)'                      DeMorgan
= (B'+D).(A'+B')                      DeMorgan
CS1104-5                                 Getting POS Expressions                                             51
Don’t-care Conditions (1/3)
 In certain problems, some                     No.   A   B   C   D   P
0    0   0   0   0   1
outputs are not specified.                    1    0   0   0   1   0
 These outputs can be either „1‟ or             2
3
0
0
0
0
1
1
0
1
0
1
„0‟.                                          4    0   1   0   0   0
5    0   1   0   1   1
 They are called don‟t-care                     6    0   1   1   0   1
conditions, denoted by X (or                  7    0   1   1   1   0
8    1   0   0   0   0
sometimes, d).                                9    1   0   0   1   1

 Example: An odd parity generator              10
11
1
1
0
0
1
1
0
1
X
X
for BCD code which has 6                     12    1   1   0   0   X
unused combinations.                         13    1   1   0   1   X
14    1   1   1   0   X
15    1   1   1   1   X

CS1104-5                Don‟t-care Conditions                             52
Don’t-care Conditions (2/3)
 Don‟t-care conditions can be used to help simplify
Boolean expression further in K-maps.
 They could be chosen to be either „1‟ or „0‟,
depending on which gives the simpler expression.
 We usually use the notation Sd to denote the set of
don‟t-care minterms. For example, the function P in
the odd-parity generator for BCD can be written as:

P = Sm(0, 3, 5, 6, 9) + Sd(10, 11, 12, 13, 14, 15)

CS1104-5                   Don‟t-care Conditions                53
Don’t-care Conditions (3/3)
 For comparison:                                             CD
C

 WITHOUT don‟t-cares:                              AB         00   01      11       10
00
P = A'.B'.C'.D‟ + A'.B'.C.D + A'.B.C'.D                 1            1

+ A'.B.C.D' + A.B'.C'.D                     01
1                1
B
11
A
10          1

 WITH don‟t-cares:                                                     D

P = A'.B'.C'.D' + B'.C.D + B.C'.D                  CD
C

+ B.C.D' + A.D                         AB         00   01      11       10
00     1            1
01
1                1
B
11    X    X       X        X
A
10          1       X        X

D

CS1104-5                        Don‟t-care Conditions                                             54
Review – The Techniques
 Algebraic Simplification.
 requires skills but extremely open-ended.

 Karnaugh Maps.
 can obtain simplified standard forms.
 easy for humans (pattern-matching skills).
 limited to not more than 6 variables.

 Other computer-aided techniques such as Quine-
McCluskey method (not covered in this course).

CS1104-5                          Review              55
Review – K-maps (1/4)
 Characteristics of K-map layouts:
(i) each minterm in one square/cell
(ii) adjacent/neighbouring minterms differ by only 1 literal
(iii) n-literal minterm has n neighbours/adjacent cells

 Valid 2-, 3-, 4-variable K-maps
b                       b

a'b'   a'b                m0   m1
OR
a   ab'    ab             a   m2   m3

CS1104-5                            Review                           56
Review – K-maps (2/4)
b                                                             b
bc                                                                     bc
a                                                                     a
00      01          11      10                                        00   01       11       10
0    a'b'c'   a'b'c       a'bc        a'bc'                            0    m0   m1       m3       m2

a             ab'c'    ab'c        abc         abc'                    a             m4   m5       m7       m6
1                                                                      1

c                                                               c
y
yz
wx           00    01        11            10
00    m0     m1        m3            m2

01    m4     m5        m7            m6
x
11    m12    m13       m15       m14
w
10    m8     m9        m11       m10

z

CS1104-5                                                     Review                                                    57
Review – K-maps (3/4)
 Groupings to select product-terms must be:
 (i) rectangular in shape
 (ii) in powers of twos (1, 2, 4, 8, etc.)
 (iii) always select largest possible groupings of minterms
(i.e. prime implicants)
 (iv) eliminate redundant groupings

 Sum-of-products (SOP) form obtained by selecting
groupings of minterms (corresponding to product
terms).

CS1104-5                            Review                           58
Review – K-maps (4/4)
 Product-of-sums (POS) form obtained by selecting
groupings of maxterms (corresponding to sum terms)
and by applying DeMorgan‟s theorem.
 Don‟t cares, marked by X (or d), can denote either 1 or
0. They could therefore be selected as 1 or 0 to further
simplify expressions.

CS1104-5                     Review                         59
Examples (1/11)
 Example #1:
f(A,B,C,D) =  m(2,3,4,5,7,8,10,13,15)
A
AB
CD         00   01       11       10

00         1                  1
Fill in the 1‟s.
01         1        1
D
11    1    1        1
C
10    1                       1

B

CS1104-5                                           Examples                      60
Examples (2/11)
 Example #1:
f(A,B,C,D) =  m(2,3,4,5,7,8,10,13,15)
A
AB
CD         00   01       11       10
These are all the
00         1                  1         prime implicants; but
01         1        1                   do we need them
D
11    1    1        1
all?
C
10    1                       1

B

CS1104-5                                           Examples                           61
Examples (3/11)
 Example #1:
f(A,B,C,D) =  m(2,3,4,5,7,8,10,13,15)
A
AB
CD         00   01       11       10

00         1                  1
Essential prime implicants:
01         1        1
D
11    1    1        1
B.D
C
10    1                       1          A'.B.C'
B                        A.B'.D'

CS1104-5                                           Examples                           62
Examples (4/11)
 Example #1:
f(A,B,C,D) =  m(2,3,4,5,7,8,10,13,15)
A
AB
CD         00   01       11       10

00         1                  1
Minimum cover.
01         1        1
D     EPIs: B.D, A'.B.C', A.B'.D'
11    1    1        1
C
1                       1
+
10

B
A'.B'.C

f(A,B,C,D) = B.D + A'.B.C' + A.B'.D' + A'.B'.C
CS1104-5                                           Examples                            63
Examples (5/11)
A
AB                                                                              A
CD        00        01       11       10                     AB
CD            00   01       11       10
00
1                   1
01
1          1                           00            1                  1        Essential prime
D

C
11       1       1          1                           01            1        1
implicants
10       1                           1                                                       D
11       1    1        1
B                          C
10       1                       1
SUMMARY
A                                B
AB
CD                 00        01           11       10

00                  1                      1
Minimum cover
01                  1            1
D
11       1          1            1
C
10       1                                 1
f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C'
B

CS1104-5                                                                     Examples                                          64
Examples (6/11)
 Example #2:
f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D'

A
AB
CD         00   01       11       10

00    1                      1
Fill in the 1‟s.
01                  1        1
D
11                  1        1
C
10    1             1        1

B

CS1104-5                                          Examples                      65
Examples (7/11)
 Example #2:
f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D'
A
AB
CD         00   01       11       10

00    1                      1
Find all PIs:
01                  1        1
D      A.D
11                  1        1
C                                             A.C
10    1             1        1
B'.D'
B                        A.B'

A.D, A.C and B'.D' are EPIs, and they cover all the minterms.
So the answer is: f(A,B,C,D) = A.D + A.C + B'.D'
CS1104-5                                          Examples                   66
Examples (8/11)
 Example #3 (with don‟t cares):
f(A,B,C,D) =  m(2,8,10,15) +  d(0,1,3,7)
A
AB
CD         00   01       11       10

00    X                      1
Fill in the 1‟s and X‟s.
01    X
D
11    X    X        1
C
10    1                      1

B

CS1104-5                                          Examples                              67
Examples (9/11)
 Example #3 (with don‟t cares):
f(A,B,C,D) =  m(2,8,10,15) +  d(0,1,3,7)

A
AB
CD         00   01       11       10         Do we need to have an
00    X                      1          additional term A'.B' to
01    X                                 cover the 2 remaining x‟s?
D
11    X    X        1                   No, because all the 1‟s
C
1                      1          (minterms) have been
10
covered.
B

f(A,B,C,D) = B'.D' + B.C.D

CS1104-5                                          Examples                                68
Examples (10/11)
 To find simplest POS expression for example #2:
f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D'
   Draw the K-map of the complement of f, f '.

A                   From K-map,
AB
CD         00   01       11       10
f ' = A'.B + A'.D + B.C'.D'
00         1        1
Using DeMorgan‟s theorem,
01    1    1
D
11    1    1                                  f = (A'.B + A'.D + B.C'.D')'
C
10         1                                   = (A+B').(A+D').(B'+C+D)
B

CS1104-5                                          Examples                                   69
Examples (11/11)
 To find simplest POS expression for example #3:
f(A,B,C,D) =  m(2,8,10,15) +  d(0,1,3,7)
 Draw the K-map of the complement of f, f '.
f '(A,B,C,D) =  m(4,5,6,9,11,12,13,14) +  d(0,1,3,7)
A
AB
CD         00   01       11       10
From K-map,
00    X    1        1                        f ' = B.C' + B.D' + B'.D
01    X    1        1        1           Using DeMorgan‟s theorem,
D
11    X    X                 1
C                                                 f = (B.C' + B.D' + B'.D)'
10         1        1
= (B'+C).(B'+D).(B+D')
B

CS1104-5                                          Examples                               70
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