CS1104: Computer Organisation cs1104 Lecture 5: Karnaugh Maps cs1104.
-
Upload
rosanna-price -
Category
Documents
-
view
219 -
download
0
Transcript of CS1104: Computer Organisation cs1104 Lecture 5: Karnaugh Maps cs1104.
CS1103-5
CS1104: Computer Organisation
http://www.comp.nus.edu.sg/~cs1104
Lecture 5: Karnaugh Maps
CS1104-5 Lecture 5: Karnaugh Maps 2
Lecture 5: Karnaugh Maps
Function Simplification
Algebraic Simplification
Half Adder
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 3
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 Function Simplification 4
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 5
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 Algebraic Simplification 6
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 7
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 8
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 9
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 10
Algebraic Simplification (5/5)
Difficulty – needs good algebraic manipulation skills.
Advantage – very open-ended (to your desired form!)
CS1104-5 Half Adder 11
Half Adder (1/2)
Half-Adder is a circuit which adds two single bits (called 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 Y C S0 0 0 00 1 0 11 0 0 11 1 1 0
Half adder
X
Y(X+Y)
S
C
CS1104-5 Half Adder 12
Half Adder (2/2)
In sum-of-minterms forms:
C = X.Y
S = X'.Y + X.Y'
Algebraic simplification could simplify S to:
S = X'.Y + X.Y'
= XY
Giving:
X Y C S0 0 0 00 1 0 11 0 0 11 1 1 0
XY
S
C
CS1104-5 Introduction to K-maps 13
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 Venn Diagrams 14
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.ba
b
CS1104-5 Venn Diagrams 15
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
{ } 0
{ a'.b',a.b,a.b',a'.b } 1
a.b' a'.b
a'.b'
a.ba b
CS1104-5 2-variable K-maps 16
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 17
2-variable K-maps (2/4) Alternative layouts of a 2-variable (a, b) K-map
a'b'
ab'
a'b abb
a
m0 m2
m1 m3b
a
OR
Alternative 2:
a'b'
a'b
ab' aba
b
m0 m1
m2 m3a
b
Alternative 1:
OR
ab a'b
ab' a'b'
b
a
m3 m1
m2 m0
b
aOR
Alternative 3:
and others…
CS1104-5 2-variable K-maps 18
2-variable K-maps (3/4)
Equivalent labeling:
a
b
equivalent to:
ab
0 1
0 1
b
a
equivalent to:
ba
1 0
0 1
CS1104-5 2-variable K-maps 19
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.
0 0
0 1a
b
0 1
1 0a
b
C = a.b S = a.b' + a'.b
CS1104-5 3-variable K-maps 20
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.
ab'c' ab'ca
b
abc abc'
a'b'c'
a'b'c a'bc a'bc'0 1
00 01 11 10
c
abc
ORm4 m5a
b
m7 m6
m0 m1 m3 m20 1
00 01 11 10
c
abc
Note Gray code sequenceAbove arrangement ensures that minterms of adjacent cells differ by only ONE literal. (Other arrangements which satisfy this criterion may also be used.)
CS1104-5 3-variable K-maps 21
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)
m4 m5 m7 m6
m0 m1 m3 m20 1
00 01 11 10abc
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 Quick Review Questions (1) 22
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?
5-2. Draw the K-map for this function A:A(x, y, z) = x.y + y.z' + x'.y'.z
0 1a
b
0 0
1 0 0 10 1
00 01 11 10
c
abc
CS1104-5 4-variable K-maps 23
4-variable K-maps (1/2)
There are 16 cells in a 4-variable (w, x, y, z) K-map.
m4 m5
w
y
m7 m6
m0 m1 m3 m200
01
11
10
00 01 11 10
z
wxyz
m12
m13
m15
m14
m8 m9 m11
m10
x
CS1104-5 4-variable K-maps 24
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.
m4 m5
w
y
m7 m6
m0 m1 m3 m2
z
wxyz
m12
m13
m15
m14
m8 m9 m11
m10
x
CS1104-5 5-variable K-maps 25
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 26
5-variable K-maps (2/2)
Organised as two 4-variable K-maps:
Corresponding squares of each map are adjacent.Can visualise this as being one 4-variable map on TOP of the other 4-variable map.
m20
m21
w
y
m23
m22
m16
m17
m19
m18
00
01
11
10
00 01 11 10
z
wxyz
m28
m29
m31
m30
m24
m25
m27
m26
x
m4 m5
w
y
m7 m6
m0 m1 m3 m200
01
11
10
00 01 11 10
z
wxyz
m12
m13
m15
m14
m8 m9 m11
m10
x
v ' v
CS1104-5 Larger K-maps 27
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 28
Larger K-maps (2/2)
Try stretch your recognition capability by finding simplest sum-of-products expression for m(6,8,14,18,23,25,27,29,41,45,57,61).
a'.b'
m000
01
11
10
00 01 11 10cdef
m1 m3 m2
m4 m5 m7 m6
m12
m13
m15
m14
m8 m9 m11
m10
m40
10
11
01
0000 01 11 10cd
ef
m41
m43
m42
m44
m45
m47
m46
m36
m37
m39
m38
m32
m33
m35
m34
m18
00
01
11
10
10 11 01 00 cd
ef
m19
m17
m16
m22
m23
m21
m20
m30
m31
m29
m28
m26
m27
m25
m24
m58
10
11
01
0010 11 01 00 cd
ef
m59
m57
m56
m62
m63
m61
m60
m54
m55
m53
m52
m50
m51
m49
m48
a'.b
a.b' a.b
a
b
CS1104-5 Simplification Using K-maps 29
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 30
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 31
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
= m(4, 5, 10, 11, 14, 15)
z
1 1
w
y
00
01
11
10
00 01 11 10wxyz
1 1
1 1
x (cells with ‘0’ are not shown for clarity)
CS1104-5 Simplification Using K-maps 32
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.
1 1
w
00
01
11
10
00 01 11 10
z
wxyz
1 1
1 1
x
A
B
y
CS1104-5 Simplification Using K-maps 33
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
= w.x'.y.(z' + z) + w.x.y.(z' + z)
= w.x'.y + w.x.y
= w.(x'+x).y
= w.y 1 1
w
00
01
11
10
00 01 11 10
z
wx
yz
1 1
1 1
x
A
B
y
CS1104-5 Simplification Using K-maps 34
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 35
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 36
Simplification Using K-maps (8/9)
Other possible valid groupings of a 4-variable K-map include:
1
11
1
1
1
1
1
1
11
1 1
111
1
11
1
CS1104-5 Simplification Using K-maps 37
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
11
1 1
111
1
1
1
1
1
1
1
1
CS1104-5 Converting to Minterms Form 38
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 39
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
11
C
A
00
01
11
10
00 01 11 10
B
CDAB
D1 1 1
1 1
A.B'.C'.D' + A.C'.D' + B.C + A'.C.D
= A.B'.C'.D' + A.C'.D'.(B+B') + B.C + A'.C.D
= A.B'.C'.D' + A.B.C'.D' + A.B'.C'.D' + B.C.(A+A') + A'.C.D
= A.B'.C'.D' + A.B.C'.D' + A.B.C + A'.B.C + A'.C.D
= A.B'.C'.D' + A.B.C'.D' + A.B.C.(D+D') + A'.B.C.(D+D') + A'.C.D.(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 Simplest SOP Expressions 40
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 41
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.
11 1
111
11 1
111
CS1104-5 Simplest SOP Expressions 42
Simplest SOP Expressions (3/3)
No redundant groups:
An essential prime implicant is a prime implicant that includes at least one minterm that is not covered by any other prime implicant.
1
1
1
11
1
1
1
1
1
1
11
1
1
1
Essential prime implicants
CS1104-5 Quick Review Questions (2) 43
Quick Review Questions (2)
Textbook page 104.
5-3. Identify the prime implicants and the essential prime implicants of the two K-maps below.
0 1a
b
0 0
1 1 0 10 1
00 01 11 10
c
abc
11
C
A
00
01
11
10
00 01 11 10
B
CDAB
D1 1 1
1 1
1
1 1
1
CS1104-5 Simplest SOP Expressions 44
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 45
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 46
Simplest SOP Expressions (3/5)
Example:
f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
All prime implicants
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
CS1104-5 Simplest SOP Expressions 47
Simplest SOP Expressions (4/5)
B
1
1
C
A
00
01
11
10
00 01 11 10CDAB
1
1
1
1
D
1
1
1
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
Essential prime implicants
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
Minimum cover
CS1104-5 Simplest SOP Expressions 48
Simplest SOP Expressions (5/5)
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
B.D
AB'D'A'BC'
A'B'C
f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C'
CS1104-5 Quick Review Questions (3) 49
Quick Review Questions (3)
Textbook page 104.
5-4. Find the simplified expression for G(A,B,C,D).
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
D1 1 1
1
1
1 1
5-5 to 5-7.
CS1104-5 Getting POS Expressions 50
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:
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
0
1
1
D
1
1
1 1
10
00
0 0
CS1104-5 Getting POS Expressions 51
Getting POS Expressions (2/2)
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
0
0
C
A
00
01
11
10
00 01 11 10
B
CDAB
0
1
0
0
D
0
0
0 0
01
11
1 1
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
0
1
1
D
1
1
1 1
10
00
0 0K-map of F
K-map of F'
CS1104-5 Don’t-care Conditions 52
Don’t-care Conditions (1/3) In certain problems, some
outputs are not specified.
These outputs can be either ‘1’ or ‘0’.
They are called don’t-care conditions, denoted by X (or sometimes, d).
Example: An odd parity generator for BCD code which has 6 unused combinations.
No. A B C D P0 0 0 0 0 11 0 0 0 1 02 0 0 1 0 03 0 0 1 1 14 0 1 0 0 05 0 1 0 1 16 0 1 1 0 17 0 1 1 1 08 1 0 0 0 09 1 0 0 1 1
10 1 0 1 0 X11 1 0 1 1 X12 1 1 0 0 X13 1 1 0 1 X14 1 1 1 0 X15 1 1 1 1 X
CS1104-5 Don’t-care Conditions 53
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 d 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 = m(0, 3, 5, 6, 9) + d(10, 11, 12, 13, 14, 15)
CS1104-5 Don’t-care Conditions 54
Don’t-care Conditions (3/3) For comparison:
WITHOUT don’t-cares:P = A'.B'.C'.D’ + A'.B'.C.D +
A'.B.C'.D + A'.B.C.D' + A.B'.C'.D
WITH don’t-cares:P = A'.B'.C'.D' + B'.C.D + B.C'.D + B.C.D' + A.D
1
A
C
00
01
11
10
00 01 11 10
D
ABCD
1
B
1
1
1
1
A
C
00
01
11
10
00 01 11 10
D
ABCD
1
B
1
1
1
X X
XXXX
CS1104-5 Review 55
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 56
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
a'b'
a'b
ab' aba
b
m0 m1
m2 m3a
b
OR
CS1104-5 Review 57
Review – K-maps (2/4)
ab'c' ab'ca
b
abc abc'
a'b'c'
a'b'c a'bc a'bc'0 1
00 01 11 10
c
abc
m4 m5a
b
m7 m6
m0 m1 m3 m20 1
00 01 11 10
c
abc
m4 m5
w
y
m7 m6
m0 m1 m3 m200
01
11
10
00 01 11 10
z
wxyz
m12
m13
m15
m14
m8 m9 m11
m10
x
CS1104-5 Review 58
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 59
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 Examples 60
Examples (1/11)
Example #1:
f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
Fill in the 1’s.
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
CS1104-5 Examples 61
Examples (2/11)
Example #1:
f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
These are all the prime implicants; but do we need them all?1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
CS1104-5 Examples 62
Examples (3/11)
Example #1:
f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
Essential prime implicants:
B.D
A'.B.C'
A.B'.D'
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
CS1104-5 Examples 63
Examples (4/11)
Example #1:
f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
Minimum cover.
EPIs: B.D, A'.B.C', A.B'.D'
+
A'.B'.C
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
f(A,B,C,D) = B.D + A'.B.C' + A.B'.D' + A'.B'.C
CS1104-5 Examples 64
Examples (5/11)
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
B
1
1
C
A
00
01
11
10
00 01 11 10CDAB
1
1
1
1
D
1
1
1
Essential prime implicants
Minimum cover
SUMMARY
f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C'
CS1104-5 Examples 65
Examples (6/11)
Example #2:
f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D'
Fill in the 1’s.1
1C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1D
1
1
11
CS1104-5 Examples 66
Examples (7/11)
Example #2:
f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D'
Find all PIs:
A.D
A.C
B'.D'
A.B'
1
1C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1D
1
1
11
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 67
Examples (8/11)
Example #3 (with don’t cares):
f(A,B,C,D) = m(2,8,10,15) + d(0,1,3,7)
Fill in the 1’s and X’s.1X
C
A
00
01
11
10
00 01 11 10
B
CDAB
X 1
XD
11
X
CS1104-5 Examples 68
Examples (9/11)
Example #3 (with don’t cares):f(A,B,C,D) = m(2,8,10,15) + d(0,1,3,7)
1X
C
A
00
01
11
10
00 01 11 10
B
CDAB
X 1
XD
11
X
f(A,B,C,D) = B'.D' + B.C.D
Do we need to have an additional term A'.B' to cover the 2 remaining x’s?
No, because all the 1’s (minterms) have been covered.
CS1104-5 Examples 69
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 '.
1
1C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
D
11
1
From K-map,
f ' = A'.B + A'.D + B.C'.D'
Using DeMorgan’s theorem,
f = (A'.B + A'.D + B.C'.D')'
= (A+B').(A+D').(B'+C+D)
CS1104-5 Examples 70
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)
From K-map,
f ' = B.C' + B.D' + B'.D
Using DeMorgan’s theorem,
f = (B.C' + B.D' + B'.D)'
= (B'+C).(B'+D).(B+D')
1
1C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
D
11
1
X
X
X
X
1