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Gate Level Minimization
Lecture 06,07,08
By : Ali Mustafa
Outline
•Introduction
•The Karnaugh Map Method
–Two-Variable Map
–Three-Variable Map
–Four-Variable Map
–Five-Variable Map
•Product of Sums Simplification
•Don’t-Care Conditions
Introduction – The Map Method • K map provides a pictorial method of grouping together
expressions with common factors and therefore eliminating unwanted variables.
• The K map can also be described as a special arrangement of a truth table.
Choice of Blocks
• We can simplify function by using larger blocks –Do we really need all blocks? –Can we leave some out to further simplify
expression? • Function needs to contain special type of blocks –They are called Essential Prime Implicants • Need to define new terms –Implicant –Prime implicant –Essential prime implicant
Terminology
Procedure for Simplifying Boolean Functions
1.Generate all PIs of the function.
2.Include all essential PIs.
3.For remaining minterms not included in the essential PIs, select a set of other PIs to cover them, with minimal overlap in the set.
4.The resulting simplified function is the logical OR of the product terms selected above.
Examples to illustrate terms
2 variable map example
F = m3 F = m1 + m2 + m3
Two Variable Map
• Consider the following map. The function plotted is: Z = f(A,B) = A + AB
3-variable K Map
Example : 3-variable K Map (Cont..)
Example : 3-variable K Map
Another Example : 3-variable K Map
F = yz + xz’
3-variable K Map
Self Tasks
• Solve the expression using K MAP
Solution of Self Tasks
4-variable Karnaugh Map
4-variable Karnaugh Map • K map can be extended to 4 variables:
• Top cells are adjacent to bottom cells. Left-edge cells are adjacent to right-edge cells.
• Note variable ordering (WXYZ).
Four-variable Map Simplification
• One square represents a minterm of 4 literals.
• A rectangle of 2 adjacent squares represents a product term of 3 literals.
• A rectangle of 4 squares represents a product term of 2 literals.
• A rectangle of 8 squares represents a product term of 1 literal.
• A rectangle of 16 squares produces a function that is equal to logic 1.
Example : 4-variable K Map
Example
• Simplify the following Boolean function g(A,B,C,D) = Σm(0,1,2,4,5,7,8,9,10,12,13).
• First put the function g( ) into the map, and then group as many 1s as possible.
Example: 4-variable Karnaugh Map
• Example: F(w,x,y,z)=Σ(0,1,2,4,5,6,8,9,12,13,14)
Simplifying logic Function using 4-variable K-Map
5-variable Karnaugh Map
Example: 5-variable Karnaugh Map
• F(A,B,C,D,E) = Σ(0,2,4,6,9,13,21,23,25,29,31)
Product of Sums Minimization
• How to generate a product of sums from a Karnaugh map?
–Use duality of Boolean algebra (DeMorgan law)
• Look at 0s in map instead of 1s –Generate blocks around 0’s –Gives inverse of function –Use duality to generate product
of sums • Example: –F = Σ(0,1,2,5,8,9,10) –F’ = AB+ CD + BD’ –F = (A’+B’)(C’+D’)(B’+D)
You can easily do product‐of‐sum too
You can easily do product‐of‐sum too
Example: POS minimization
Don't Care Conditions
• There may be a combination of input values which will never occur if they do occur, the output is of no concern.
• The function value for such combinations is called a don't care.
• They are usually denoted with x. Each x may be arbitrarily assigned the value 0 or 1 in an implementation.
• Don’t cares can be used to further simplify a function
Minimization using Don’t Cares
• Treat don't cares as if they are 1s to generate PIs.
• Delete PI's that cover only don't care minterms.
• Treat the covering of remaining don't care minterms as optional in the selection process (i.e.they may be, but need not be, covered).
Minimization example
• F(w,x,y,z) = Σ(1,3,7,11,15) and d(w,x,y,z) = Σ(0,2,5)
• What are possible solutions?
Another Example
Self Tasks
• Self reading chapter 3 (Revision)
• 3.1 to 3.15 Chapter 3 exercise
• Self reading chapter 2 (Revision)
• 2.1 to 2.28 Chapter 2 exercise
Except 2.10,2.15,2.16,2.24-2.26
Recap
2 Variable K-Map
1
1
1
0
A B B’ B
A’
A
Group 1
Group 2
A’
B’
F= A’ + B’
3 Variable K-Map
m0
m1
1
m3
1
m2
m4
1 m5
1 m7
m6
A BC
A’ 0
A 1
B’C’ 00 B’C 01 BC 11 BC’ 10 Group 1
A’B Group 2
AC
IGNORE
F= A’B + AC
3 Variable K-Map
1
m0
1
m1
1
m3
1
m2
m4
1 m5
1 m7
m6
A BC
A’ 0
A 1
B’C’ 00 B’C 01 BC 11 BC’ 10 Group 1
A’ Group 2
C
F= A’ + C
3 Variable K-Map
1
m0
1
m1
1
m3
1
m2
1
M4
1
m5
1
m7
1
m6
A BC
A’ 0
A 1
B’C’ 00 B’C 01 BC 11 BC’ 10 Group 1
1
F= 1
3 Variable K-Map
1
m0
0
m1
1
m3
0
m2
0 m4
1 m5
0 m7
1 m6
A BC
A’ 0
A 1
B’C’ 00 B’C 01 BC 11 BC’ 10
A’BC
AB’C
ABC’ A’B’C’
F= A’B’C’ + A’BC + AB’C + ABC’
4 Variable K-Maps
0
m0
0
m1
0
m3
0
m2
0
m4
0
m5
0
m7
0
m6
0 m12
0 m13
0 m15
0 m14
1
m8
1
m9
1 m11
1 m10
AB CD
C’D’ 00 C’D 01 CD 11 CD’ 10
A’B’ 00
A’B 01
AB 11
AB’ 10
Group 1
AB’
F= AB’
4 Variable K-Maps
0
m0
0
m1
0
m3
0
m2
1
m4
1
m5
0
m7
0
m6
1 m12
1 m13
0 m15
0 m14
0
m8
0
m9
0 m11
0 m10
AB CD
C’D’ 00 C’D 01 CD 11 CD’ 10
A’B’ 00
A’B 01
AB 11
AB’ 10
Group 1
BC’
F= BC’
4 Variable K-Maps
0
m0
1
m1
1
m3
0
m2
0
m4
0
m5
0
m7
0
m6
0 m12
0 m13
0 m15
0 m14
0
m8
1
m9
1 m11
0 m10
AB CD
C’D’ 00 C’D 01 CD 11 CD’ 10
A’B’ 00
A’B 01
AB 11
AB’ 10
Group 1
B’D
F= B’D
4 Variable K-Maps
0
m0
0
m1
0
m3
0
m2
1
m4
0
m5
0
m7
1
m6
1 m12
0 m13
0 m15
1 m14
0
m8
0
m9
0 m11
0 m10
AB CD
C’D’ 00 C’D 01 CD 11 CD’ 10
A’B’ 00
A’B 01
AB 11
AB’ 10
Group 1
BD’
F= BD’
4 Variable K-Maps
0
m0
1
m1
1
m3
0
m2
1
m4
0
m5
0
m7
1
m6
1 m12
0 m13
0 m15
1 m14
0
m8
1
m9
1 m11
0 m10
AB CD
C’D’ 00 C’D 01 CD 11 CD’ 10
A’B’ 00
A’B 01
AB 11
AB’ 10
Group 1
BD’
Group 2
B’D
F= BD’ + B’D
5 Variable K-Maps
1
m0
1
m1
1
m3
1
m2
1
m4
0
m5
0
m7
0
m6
1
m12
0
m13
0
m15
0
m14
1
m8
0
m9
0
m11
0
m10
1
m16
0
m17
0
m19
1 m18
1 m20
0 m21
0 m23
0 m22
1 m28
0 m29
0 m31
1 m30
1 m24
1 m25
1 m27
1 m26
A’ 0 A 1
Adjacent Rows
BC DE
BC DE
5 Variable K-Maps
1
m0
1
m1
1
m3
1
m2
1
m4
0
m5
0
m7
0
m6
1
m12
0
m13
0
m15
0
m14
1
m8
0
m9
0
m11
0
m10
1
m16
0
m17
0
m19
1 m18
1 m20
0 m21
0 m23
0 m22
1 m28
0 m29
0 m31
1 m30
1 m24
1 m25
1 m27
1 m26
A’ 0 A 1
Adjacent Columns
BC DE
BC DE
5 Variable K-Maps
1
m0
0
m1
0
m3
1
m2
1
m4
0
m5
0
m7
1
m6
0
m12
1
m13
1
m15
0
m14
0
m8
1
m9
1
m11
0
m10
0
m16
1
m17
0
m19
0 m18
0 m20
1 m21
0 m23
0 m22
0 m28
1 m29
1 m31
0 m30
0 m24
1 m25
1 m27
0 m26
A’ 0 A 1
BC
DE
BC
DE
Simplify F = ∑m(0,2,4,6,9,11,13,15,17,21,25,27,29,31)
B’C’ 00
BC’ 10
BC 11
B’C 01
B’C’ 00
B’C 01
BC 11
BC’ 10
D’E’ 00 D’E 01 DE 11 DE’ 10 D’E’ 00 D’E 01 DE 11 DE’ 10
A’B’E’ + AD’E + BE F =
5 Variable K-Maps
1
m0
0
m1
0
m3
1
m2
0
m4
1
m5
1
m7
0
m6
0
m12
1
m13
1
m15
0
m14
0
m8
0
m9
0
M11
0
m10
0
m16
0
m17
0
m19
1 m18
1 m20
1 m21
1 m23
0 m22
1 m28
1 m29
1 m31
0 m30
0 m24
0 m25
0 m27
0 m26
A’ 0 A 1
BC
DE
BC
DE
Simplify F = ∑m(0,2,5,7,13,15,18,20,21,23,28,29,31)
B’C’ 00
BC’ 10
BC 11
B’C 01
B’C’ 00
B’C 01
BC 11
BC’ 10
D’E’ 00 D’E 01 DE 11 DE’ 10 D’E’ 00 D’E 01 DE 11 DE’ 10
A’B’C’E’ + ACD’ + CE + F = B’C’DE’ F = A’B’C’E’ + B’C’DE’ + ACD’ + CE F = B’C’E’(A’ + D) + ACD’ + CE
6 Variable K-Maps
m0
m15
A’B’00 EF CD
m31
A’B 01 EF CD
m47
AB 11 EF CD
m63
AB’10 EF CD
Simplify 4 Variable K-MAP
• Find the groups and write the desired equation
1
m0
1
m1
1
m3
1
M2
1
m4
0
m5
0
m7
0
M6
1 m12
0 m13
0 m15
0 m14
1
m8
0
m9
0 m11
0 m10
1
m0
0
m1
0
m3
1
m2
1
m4
0
m5
0
m7
0
m6
1 m12
0 m13
0 m15
1 m14
1
m8
1
m9
1 m11
1 m10
1
M0
1
m1
1
m3
1
m2
1
m4
1
m5
1
m7
1
m6
0 m12
0 m13
0 m15
0 m14
0
m8
1
m9
1 m11
0 m10
Self Task
• Y = ∑m(1,5,7,9,11,13,15)
• Y = ∑m(1,3,5,9,11,13)
• Y = ∑m(4,5,8,9,11,12,13,15)
• Minimize the following expression using K-Map and realize it using number of gates
Product of Sum K-MAP
Two ways to express
–Y = ∑ (0,1,2,5,8,9,10)
–Y = ∏ (0,2,3,5,7)
POS K-MAP Y = ∑ (0,1,2,5,8,9,10) Make POS expression
1
m0
1
m1
0
m3
1
m2
0
m4
1
m5
0
m7
0
m6
0 m12
0 m13
0 m15
0 m14
1
m8
1
m9
0 m11
1 m10
F = (C’+D’)(A’+B’)(B’+D)
AB CD
C’D’ 00 C’D 01 CD 11 CD’ 10
A’B’ 00
A’B 01
AB 11
AB’ 10
CD + AB + BD’ F’ =
Applying De Morgan Law
POS K-MAP
0
m0
m1
0
m3
0
m2
m4
0 m5
0 m7
m6
A BC
A’ 0
A 1
B’C’ 00 B’C 01 BC 11 BC’ 10
B’+C’
A’+C’
A+C
F= (B’+C’)(A+C)(A’+C’)
Simplify Y = ∏ (0,2,3,5,7)
But in POS we consider 0 as a positive logic SO
Self Tasks
1. Simplify the following expression with POS
A. Y = ∏ (0,2,3,7) B. Y = ∏ (0,2,8,10) C. Y = ∑ (0,1,2,5,8,10,13)
2. Simplify the following expression with SOP & POS
A. Y = X’Z’ + Y’Z’ + YZ’ + XY
3. Exercise questions 3.1 to 3.14
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