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BCS - Compute & Network Technology
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Computer & Network Technology
Chamila FernandoBSc(Eng) Hons,MBA,MIEEE
04/12/23 Information Representation 1
Lecture 6: 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
04/12/23 Quick Review Questions (3) 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
04/12/23 Quick Review Questions (3) 3
Function Simplification
Why simplify? Simpler expression uses less logic gates. Thus: cheaper, less power, faster (sometimes).
Simplification techniques: Algebraic SimplificationAlgebraic Simplification.
simplify symbolically using theorems/postulates. requires skill but extremely open-ended.
Karnaugh MapsKarnaugh Maps. diagrammatic technique using ‘Venn-like diagram’. easy for humans (pattern-matching skills). simplified standard forms. limited to not more than 6 variables.
04/12/23 Quick Review Questions (3) 4
Function Simplification
Simplification techniques: Quine-McCluskey tabulation techniqueQuine-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.
04/12/23 Quick Review Questions (3) 5
Algebraic Simplification
Algebraic simplificationAlgebraic 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.
04/12/23 Quick Review Questions (3) 6
Algebraic Simplification Example:
(x+y).(x+y').(x+y).(x+y').(x'+z) (6 literals)= (x.x+x.y'+x.yx.y'+x.y+y.y'y.y').(x'+z) (assoc.)= (x+x.(y'+y)(y'+y)+0).(x'+z) (idemp.,assoc., compl.)= (x+x.(1)x.(1)+0).(x'+z) (complement)= (x+x+0x+x+0).(x'+z) (identity 1)= (x).(x'+z)(x).(x'+z) (idemp, identity 0)= (x.x'x.x'+x.z) (assoc.)= (00+x.z) (complement)= x.z (identity 0)
Number of literals reduced from 6 to 2.
04/12/23 Quick Review Questions (3) 7
Algebraic Simplification Find minimal SOP and POS expressions of
f(x,y,z) = x'.y.(z + y'.x) + y'.z x'.y.(z+y'.x)x'.y.(z+y'.x) + y'.z= x'.y.z + x'.y.y'.xx'.y.y'.x + y'.z (distributivity)= x'.y.z + 0x'.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'.zx'.z + y'.z (2 2-input AND gates and 1 2-input OR gate)
Minimal POS of f = (x' + y').z(x' + y').z (1 2-input OR gate and 1 2-input AND gate)
04/12/23 Quick Review Questions (3) 8
Algebraic Simplification 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.da.b.d + a'.b.c' + c.d + b.d'b.d'= a.b.c + a.b + a'.b.c'a.b + a'.b.c' + c.d + b.d' (absorption)= a.b.c + a.ba.b.c + a.b + b.c' + c.d + b.d' (absorption)= a.b + b.c' + c.d + b.d'c.d + b.d' (absorption)= a.b + c.d + b.(c' + d')(c' + d') (distributivity)= a.b + c.d + b.(c.d)'c.d + b.(c.d)' (DeMorgan)= a.ba.b + c.d + bb (absorption)= b + c.d (absorption)
Number of literals reduced form 13 to 3.
04/12/23 Quick Review Questions (3) 9
Algebraic Simplification
Difficulty – needs good algebraic manipulation skills.
Advantage – very open-ended (to your desired form!)
04/12/23 Quick Review Questions (3) 10
Half Adder
Half-AdderHalf-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:
04/12/23 Quick Review Questions (3) 11
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
Half Adder
In sum-of-minterms forms:C = X.YS = X'.Y + X.Y'
Algebraic simplification could simplify S to:S = X'.Y + X.Y' = XY
Giving:
04/12/23 Quick Review Questions (3) 12
X Y C S0 0 0 00 1 0 11 0 0 11 1 1 0
XY
S
C
Introduction to K-maps Systematic method to obtain simplified simplified sum-of-productssum-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.
04/12/23 Quick Review Questions (3) 13
Venn Diagrams
Venn diagram to represent the space of minterms. Example of 2 variables (4 minterms):
04/12/23 Quick Review Questions (3) 14
ab' a'b
a'b'
aba
b
Venn Diagrams
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
04/12/23 Quick Review Questions (3) 15
ab' a'b
a'b'
aba b
2-variable K-maps
Karnaugh-mapKarnaugh-map (K-map) is an abstract form of Venn diagram, organised as a matrix of squares, where each square represents a mintermminterm adjacent squares always differ by just one literaldiffer by just one literal (so that the
unifying theorem may apply: a + a' = 1a + a' = 1)
For 2-variable case (e.g.: variables a,b), the map can be drawn as:
04/12/23 Quick Review Questions (3) 16
2-variable K-maps Alternative layouts of a 2-variable (a, b) K-map
04/12/23 Quick Review Questions (3) 17
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…
2-variable K-maps
Equivalent labeling:
04/12/23 Quick Review Questions (3) 18
a
b
equivalent to:
ab
0 1
0 1
b
a
equivalent to:
ba
1 0
0 1
2-variable K-maps
The K-map for a function is specified by putting a ‘1’ in the square corresponding to a minterma ‘1’ in the square corresponding to a minterm a ‘0’ otherwisea ‘0’ otherwise
For example: Carry and Sum of a half adder.
04/12/23 Quick Review Questions (3) 19
0 0
0 1a
b
0 1
1 0a
b
C = ab S = ab' + a'b
3-variable K-maps
There are 8 minterms for 3 variables (a, b, c). Therefore, there are 8 cells in a 3-variable K-map.
04/12/23 Quick Review Questions (3) 20
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.)
3-variable K-maps
There is wrap-aroundwrap-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)
04/12/23 Quick Review Questions (3) 21
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.
Quick Review Questions (1)
Textbook page 104.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
04/12/23 Quick Review Questions (3) 22
0 1a
b
0 0
1 0 0 10 1
00 01 11 10
c
abc
4-variable K-maps
There are 16 cells in a 4-variable (w, x, y, z) K-map.
04/12/23 Quick Review Questions (3) 23
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
4-variable K-maps
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.
04/12/23 Quick Review Questions (3) 24
m4 m5
w
y
m7 m6
m0 m1 m3 m2
z
wxyz
m12
m13
m15
m14
m8 m9 m11
m10
x
5-variable K-maps
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.
04/12/23 Quick Review Questions (3) 25
5-variable K-maps Organised as two 4-variable K-maps:
04/12/23 Quick Review Questions (3) 26
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
Larger K-maps
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.
04/12/23 Quick Review Questions (3) 27
Larger K-maps
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).
04/12/23 Quick Review Questions (3) 28
w
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
Simplification Using K-maps Based on the Unifying TheoremUnifying Theorem:
A + A' = 1A + 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 powersin powers of twosof twos: 1, 2, 4, 8, …) then corresponds to a simpler product termsimpler 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.
04/12/23 Quick Review Questions (3) 29
Simplification Using K-maps
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.
04/12/23 Quick Review Questions (3) 30
Simplification Using K-maps 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)
04/12/23 Quick Review Questions (3) 31
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)
Simplification Using K-maps
Each group of adjacent minterms (group size in powers of twos) corresponds to a possible product termproduct term of the given function.
04/12/23 Quick Review Questions (3) 32
1 1
w
00
01
11
10
00 01 11 10
z
wxyz
1 1
1 1
x
A
B
y
Simplification Using K-maps
There are 2 groups of minterms: A and B, where: AA = w'.x.y'.z' + w‘.x.y'.z
= w'.x.y'.(z' + z)= w'.x.y'w'.x.y'
BB = 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.yw.y
04/12/23 Quick Review Questions (3) 33
1 1
w
00
01
11
10
00 01 11 10
z
wx
yz
1 1
1 1
x
A
B
y
Simplification Using K-maps
Each product term of a group, w'.x.y' w'.x.y' and w.yw.y, represents the sum of mintermssum 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
04/12/23 Quick Review Questions (3) 34
Simplification Using K-maps
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
04/12/23 Quick Review Questions (3) 35
Simplification Using K-maps
Other possible valid groupings of a 4-variable K-map include:
04/12/23 Quick Review Questions (3) 36
1
11
1
1
1
1
1
1
11
1 1
111
1
11
1
Simplification Using K-maps 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:
04/12/23 Quick Review Questions (3) 37
1
11
1 1
111
1
1
1
1
1
1
1
1
Converting to Minterms Form
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.
04/12/23 Quick Review Questions (3) 38
Converting to Minterms Form Example:
f(A,B,C,D) = A(C+D)'(B'+D') + C(B+C'+A'D) = A(C'D')(B'+D') + BC + CC' + A'CD = AB'C'D' + AC'D' + BC + A'CD
04/12/23 Quick Review Questions (3) 39
11
C
A
00
01
11
10
00 01 11 10
B
CDAB
D1 1 1
1 1
AB'C'D' + AB'C'D' + AC'D'AC'D' + BC + A'CD + BC + A'CD
= AB'C'D' + AC'D'(B+B') + = AB'C'D' + AC'D'(B+B') + BCBC + A'CD + A'CD
= AB'C'D' + ABC'D' + AB'C'D' + = AB'C'D' + ABC'D' + AB'C'D' + BC(A+A') + A'CDBC(A+A') + A'CD
= AB'C'D' + ABC'D' + ABC + A'BC + = AB'C'D' + ABC'D' + ABC + A'BC + A'CDA'CD
= AB'C'D' + ABC'D' + ABC(D+D') + = AB'C'D' + ABC'D' + ABC(D+D') + A'BC(D+D') + A'CD(B+B')A'BC(D+D') + A'CD(B+B')
= AB'C'D' + ABC'D' + ABCD + ABCD' + = AB'C'D' + ABC'D' + ABCD + ABCD' + A'BCD + A'BCD' + A'B'CDA'BCD + A'BCD' + A'B'CD
Simplest SOP Expressions 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 implicantsprime implicants) where possible;
and no redundant groupings (look for essential prime implicantsessential prime implicants)
ImplicantImplicant: a product term that could be used to cover minterms of the function.
04/12/23 Quick Review Questions (3) 40
Simplest SOP Expressions
A prime implicantprime implicant is a product term obtained by combining the maximum possible number of minterms from adjacentadjacent squares in the map.
Use bigger groupings (prime implicants) where possible.
04/12/23 Quick Review Questions (3) 41
11 1
111
11 1
111
Simplest SOP Expressions
No redundant groups:
An essential prime implicantessential prime implicant is a prime implicant that includes at least one minterm that is not covered by any other prime implicant.
04/12/23 Quick Review Questions (3) 42
1
1
1
11
1
1
1
1
1
1
11
1
1
1
Essential prime implicants
Quick Review Questions (2)
Textbook page 104.Textbook page 104.5-3. Identify the prime implicants and the essential prime
implicants of the two K-maps below.
04/12/23 Quick Review Questions (3) 43
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
Simplest SOP Expressions 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.
04/12/23 Quick Review Questions (3) 44
Simplest SOP Expressions 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.
04/12/23 Quick Review Questions (3) 45
Simplest SOP Expressions
Example:f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
04/12/23 Quick Review Questions (3) 46
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
Simplest SOP Expressions
04/12/23 Quick Review Questions (3) 47
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
Simplest SOP Expressions
04/12/23 Quick Review Questions (3) 48
1
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1
1
D
1
1
1
BD
AB'D'A'BC'
A'B'C
f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C'
Quick Review Questions (3)
Textbook page 104.Textbook page 104.5-4. Find the simplified expression for G(A,B,C,D).
04/12/23 Quick Review Questions (3) 49
1
C
A
00
01
11
10
00 01 11 10
B
CDAB
D1 1 1
1
1
1 1
5-5 to 5-7.5-5 to 5-7.
Getting POS Expressions
Simplified POS expressionSimplified 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:
04/12/23 Quick Review Questions (3) 50
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
Getting POS Expressions
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')(B'+D).(A'+B') DeMorgan
04/12/23 Quick Review Questions (3) 51
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'
Don’t-care Conditions In certain problems, some outputs
are not specified.
These outputs can be either ‘1’ or ‘0’.
They are called don’t-care don’t-care conditionsconditions, denoted by X (or sometimes, d).
Example: An odd parity generator for BCD code which has 6 unused combinations.
04/12/23 Quick Review Questions (3) 52
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
Don’t-care Conditions
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.
04/12/23 Quick Review Questions (3) 53
Don’t-care Conditions 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
04/12/23 Quick Review Questions (3) 54
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
Review – The Techniques
Algebraic Simplification. requires skill 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).
04/12/23 Quick Review Questions (3) 55
Review – K-maps
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
04/12/23 Quick Review Questions (3) 56
a'b'
a'b
ab' aba
b
m0 m1
m2 m3a
b
OR
Review – K-maps
04/12/23 Quick Review Questions (3) 57
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
Review – K-maps
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).
04/12/23 Quick Review Questions (3) 58
Review – K-maps
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.
04/12/23 Quick Review Questions (3) 59
Examples
Example #1:f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
04/12/23 Quick Review Questions (3) 60
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
Examples
Example #1:f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
04/12/23 Quick Review Questions (3) 61
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
Examples
Example #1:f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
04/12/23 Quick Review Questions (3) 62
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
Examples
Example #1:f(A,B,C,D) = m(2,3,4,5,7,8,10,13,15)
04/12/23 Quick Review Questions (3) 63
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
Examples
04/12/23 Quick Review Questions (3) 64
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'
Examples
Example #2:f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D'
04/12/23 Quick Review Questions (3) 65
Fill in the 1’s.1
1C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1D
1
1
11
Examples
Example #2:f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D'
04/12/23 Quick Review Questions (3) 66
Find all PIs:
A.D
A.C
B'.D'
1
1C
A
00
01
11
10
00 01 11 10
B
CDAB
1
1
1D
1
1
11
Are all ‘1’s covered by the PIs? Yes, so the answer is: f(A,B,C,D) = A.D + A.C + B'.D'
Examples
Example #3 (with don’t cares):f(A,B,C,D) = m(2,8,10,15) + d(0,1,3,7)
04/12/23 Quick Review Questions (3) 67
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
Examples
Example #3 (with don’t cares):f(A,B,C,D) = m(2,8,10,15) + d(0,1,3,7)
04/12/23 Quick Review Questions (3) 68
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.
Examples
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 '.
04/12/23 Quick Review Questions (3) 69
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)
Examples
• 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)
04/12/23 Quick Review Questions (3) 70
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
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