1. CS1104: Computer Organisation http://www. comp. nus. edu
CS1104: Computer Organisation 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
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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
Lecture 5: Karnaugh Maps

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.
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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.
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Function 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.
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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.
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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 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)
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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.
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Algebraic Simplification

10. Algebraic Simplification (5/5)
Difficulty – needs good algebraic manipulation skills.
Advantage – very open-ended (to your desired form!)
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Algebraic Simplification

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:
Half adder X Y
(X+Y) S C
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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:
= XY
Giving: X Y S C
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Half Adder

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.
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Introduction to K-maps

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.b a b
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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.b a b
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Venn Diagrams

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:
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2-variable K-maps

17. 2-variable K-maps (2/4)
Alternative layouts of a 2-variable (a, b) K-map
a'b'
a'b
ab' ab a b m0 m1 m2 m3
Alternative 1: OR
a'b'
ab'
a'b ab b a m0 m2 m1 m3 OR
Alternative 2: ab
a'b
ab'
a'b' b a m3 m1 m2 m0 OR
Alternative 3:
and others…
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2-variable K-maps

18. 2-variable K-maps (3/4) Equivalent labeling: equivalent to: 0 1 1 1 0 1
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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. 1 a b
C = a.b
S = a.b' + a'.b
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2-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'c a b
abc
abc'
a'b'c'
a'b'c
a'bc
a'bc' 1 c bc OR m4 m5 m7 m6 m0 m1 m3 m2
Note Gray code sequence
Above arrangement ensures that minterms of adjacent cells differ by only ONE literal. (Other arrangements which satisfy this criterion may also be used.)
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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 m2 1 a bc
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.
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3-variable K-maps

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 1 a b c bc
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Quick Review Questions (1)

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 m2 00 01 11 10 z wx yz
m12
m13
m15
m14 m8 m9
m11
m10 x
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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 wx yz
m12
m13
m15
m14 m8 m9
m11
m10 x
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4-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.
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5-variable K-maps

26. 5-variable K-maps (2/2) Organised as two 4-variable K-maps: v ' vy
m23
m22
m16
m17
m19
m18 00 01 11 10 z wx yz
m28
m29
m31
m30
m24
m25
m27
m26 x m4 m5 m7 m6 m0 m1 m3 m2
m12
m13
m15
m14 m8 m9
m11
m10
v ' v
Corresponding squares of each map are adjacent.
Can visualise this as being one 4-variable map on TOP of the other 4-variable map.
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5-variable 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.
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Larger K-maps

28. Larger K-maps (2/2) a'.b' a'.b a.b' a.b
00 01 11 10 cd ef m1 m3 m2 m4 m5 m7 m6
m12
m13
m15
m14 m8 m9
m11
m10
m40
10 11 01 00
m41
m43
m42
m44
m45
m47
m46
m36
m37
m39
m38
m32
m33
m35
m34
m18
m19
m17
m16
m22
m23
m21
m20
m30
m31
m29
m28
m26
m27
m25
m24
m58
m59
m57
m56
m62
m63
m61
m60
m54
m55
m53
m52
m50
m51
m49
m48
a'.b
a.b'
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).
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Larger 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.
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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.
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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) 1 w y 00 01 11 10 wx yz x
(cells with ‘0’ are not
shown for clarity) z
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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 w 00 01 11 10 z wx yz x A B y
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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 w 00 01 11 10 z wx yz x A B y
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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
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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
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Simplification Using K-maps

36. Simplification Using K-maps (8/9)
Other possible valid groupings of a 4-variable K-map include: 1 P
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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 O
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Simplification Using K-maps

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.
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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 C A 00 01 11 10 B CD AB D
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 1 1 1 1
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Converting to Minterms Form

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.
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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. 1 O P
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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 P O
Essential prime implicants
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Simplest SOP Expressions

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. 1 C A 00 01 11 10 B CD AB D 1 a b c bc
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Quick Review Questions (2)

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.
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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.
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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) 1 C A 00 01 11 10 B CD AB D
All prime implicants
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Simplest SOP Expressions

47. Simplest SOP Expressions (4/5)B 1 C A 00 01 11 10 CD AB D 1 C A 00 01 11 10 B CD AB D
Essential prime implicants 1 C A 00 01 11 10 B CD AB D
Minimum cover
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Simplest SOP Expressions

48. Simplest SOP Expressions (5/5)1 C A 00 01 11 10 B CD AB D
A'BC'
AB'D'
B.D
A'B'C
f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C'
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Simplest SOP Expressions

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 B CD AB D
5-5 to 5-7.
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Quick Review Questions (3)

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 C A 00 01 11 10 B CD AB D
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Getting POS Expressions

51. Getting POS Expressions (2/2)1 C A 00 01 11 10 B CD AB D C A 00 01 11 10 B CD AB 1 D
K-map of F
K-map of F'
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
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Getting POS Expressions

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.
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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 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)
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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 D AB CD B 1 A C 00 01 11 10 D AB CD B X
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Don’t-care Conditions

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).
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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' ab a b m0 m1 m2 m3 OR
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Review

57. Review – K-maps (2/4)
ab'c'
ab'c a b
abc
abc'
a'b'c'
a'b'c
a'bc
a'bc' 1 c bc m4 m5 a b m7 m6 m0 m1 m3 m2 1 c bc m4 m5 w y m7 m6 m0 m1 m3 m2 11 10 z wx yz
m12
m13
m15
m14 m8 m9
m11
m10 x
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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).
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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.
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Review

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 C A 00 01 11 10 B CD AB D
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Examples

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

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

63. Examples (4/11) Example #1: f(A,B,C,D) =  m(2,3,4,5,7,8,10,13,15)00 01 11 10 B CD AB D
Minimum cover.
EPIs: B.D, A'.B.C', A.B'.D' +
A'.B'.C
f(A,B,C,D) = B.D + A'.B.C' + A.B'.D' + A'.B'.C
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Examples

64. Examples (5/11) SUMMARY Essential prime implicants Minimum coverB 1 C A 00 01 11 10 CD AB D 1 C A 00 01 11 10 B CD AB D
Essential prime implicants
SUMMARY 1 C A 00 01 11 10 B CD AB D
Minimum cover
f(A,B,C,D) = B.D + A'.B'.C + A.B'.D' + A'.B.C'
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Examples

65. Examples (6/11) Example #2:
f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D' 1 C A 00 01 11 10 B CD AB D
Fill in the 1’s.
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Examples

66. Examples (7/11) Example #2:
f(A,B,C,D) = A.B.C + B'.C.D' + A.D + B'.C'.D' 1 C A 00 01 11 10 B CD AB D
Find all PIs:
A.D
A.C
B'.D'
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'
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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) 1 X C A 00 01 11 10 B CD AB D
Fill in the 1’s and X’s.
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) 1 X C A 00 01 11 10 B CD AB 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.
f(A,B,C,D) = B'.D' + B.C.D
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 '.
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) 1 C A 00 01 11 10 B CD AB 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) 1 C A 00 01 11 10 B CD AB D X
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')
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Examples

71. End of file