seacucumber: a syntheszing compiler
TRANSCRIPT
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Karnaugh-Maps
September 14, 2006
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What are Karnaugh Maps?
A simpler way to handle most (but not all) jobs of manipulatinglogic functions.
– Typeset by FoilTEX – 1
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What are Karnaugh Maps?
A simpler way to handle most (but not all) jobs of manipulatinglogic functions.
Hooray!!
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Karnaugh Map Advantages
• Can be completed more systematically
• Much simpler to find minimum solutions
• Easier to see what is happening (graphical)
• Almost always use instead of boolean minimization...
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Truth Table Adjacencies
F = A’
A B F
0 0 1
0 1 1
1 0 0
1 1 0
�
��������9
These are adjacent in a gray code
sense – they differ by only one bit.
We can apply XY + XY’ = X
A’B’ + A’B = A’(B’+B) = A’(1)
= A’
F = B
A B F
0 0 0
0 1 1
1 0 0
1 1 1
XXXXXXXXy
��������9
Same idea:
A’B + AB = B
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Truth Table Adjacencies
Key Idea:
Gray code adjacency allows use of simplification theorem
(e.g., XY + XY’ = X).
Problem:
Physical adjacency in truth table is not equal to gray code adjacency.
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Two variable Karnaugh Map
0 1
0
1
BA
A=0,B=0
A=0,B=1
A=1,B=0
A=1,B=1
A different way to draw a truth table by folding it.
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K-Map
Physical adjacency does imply Gray code adjacency.
0 1
0
1
BA
1
1
0
0
F=A’B’+A’B=A’
1
0
0 1
0
1
BA
0
1
F = A’B + AB = B
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Two Variable Karnaugh Map
A B F
0 0 0
0 1 0
1 0 1
1 1 1
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Two Variable Karnaugh Map
A B F
0 0 1
0 1 1
1 0 0
1 1 0 1
0 1
0
1
BA
1
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Two Variable Karnaugh Map
A B F
0 0 1
0 1 1
1 0 0
1 1 0 1
0
0 1
0
1
BA
0
1
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Two Variable Karnaugh Map
A B F
0 0 1
0 1 1
1 0 0
1 1 0
1
0
0 1
0
1
BA
0
1
F = A’B’+A’B = A’
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Two Variable Karnaugh Map
A B F
0 0 1
0 1 1
1 0 0
1 1 0
0 1
0
1
BA
1
1
0
0A=0B=0 or 1
F = A’B’+A’B = A’
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Another Two Variable Karnaugh Map
A B F
0 0 0
0 1 1
1 0 1
1 1 1
1
0 1
0
1
BA
0
1
1
F = A’B + AB’+ AB = A+B
Each ’1’ must be covered at least once.
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Yet Another Two Variable Karnaugh Map
A B F
0 0 1
0 1 1
1 0 1
1 1 1
1
0 1
0
1
BA
1
1 1
F = A’B’ + A’B + AB’+ AB = 1
Groups of more than two 1’s can be combined as well.
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Three Variable Karnaugh MapShowing Minterm Locations
Note the order of the
B C Variables:
B C
0 0
0 1
1 1
1 0
0 1A
01
00
11
10
m0
m1
m3
m2
m4
m5
m7
m6
BC
ABC=010
ABC=101
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Adjacencies
Adjacent squares differ by exactly one variable.
0 1A
01
00
11
10
BC
A’B’C AB’C
ABC
AB’C’
There is wrap-around:
top and bottom rows are adjacent.
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Truth Table to Karnaugh Map
A B C F
0 0 0 0 0
0 0 1 0 1
0 1 0 1 2
0 1 1 1 3
1 0 0 0 4
1 0 1 1 5
1 1 0 0 6
1 1 1 1 7
-
0 1A
01
00
11
10
BC0
0
1
1 0
1
1
00
1
2
3
4
5
6
7
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Solution Example
A’BC+A’BC’ = A’B
→
0 1A
01
00
11
10
BC0
0
1
1 0
1
1
0
F = A’B + AC
AB’C+ABC = AC
←
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Another Example
A’B’C+A’BC=A’C AB’C’+ABC’=AC’
0 1A
01
00
11
10
BC0
1
1
0
0
0
1
1
F = A’C + AC’ = A ⊕ C
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Minterm Expansion to K-Map
F =∑
m(1, 3, 4, 6)
0 1A
01
00
11
10
m0
m1
m3
m2
m4
m5
m7
m6
BC
1
1
1
1
0
0
0
0
0 1A
01
00
11
10
BC
Minterms are the 1’s, everything else is 0.
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Maxterm Expansion to K-Map
F =∏
M(0, 2, 5, 7)
M0
M1
M3
M2
M4
M5
M7
0 1A
01
00
11
10
BC
M6
1
1
1
1
0
0
0
0
0 1A
01
00
11
10
BC
Maxterms are the 0’s, everything else is 1.
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Yet Another Example2n 1’s can be circled at a time. This includes 1,2,4,8 ... .
3,... not OK.
AB’C + ABC =AC
A’B’C’+AB’C’+A’B’C+AB’C =
B’
BC 0 1A
01
00
11
10
0 1
00
1 1
11
F=B’+AC
The larger the group of 1’s – the simpler the resulting product term.
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Boolean Algebra to Karnaugh Map
Plot:
0 1A
01
00
11
10
BC ab’c’ + bc + a’
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Boolean Algebra to Karnaugh Map
Plot:
0 1A
01
00
11
10
BC ab’c’ + bc + a’
1
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Boolean Algebra to Karnaugh Map
Plot:
0 1A
01
00
11
10
BC ab’c’ + bc + a’
1
1 1
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Boolean Algebra to Karnaugh Map
Plot:
0 1A
01
00
11
10
BC ab’c’ + bc + a’
1
1 1
1
1
1
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Boolean Algebra to Karnaugh Map
Plot:
Remaining spaces are ’0’.
0 1A
01
00
11
10
BC ab’c’ + bc + a’
1
1 1
1
1
1
0
0
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Boolean Algebra to Karnaugh Map
F = B’C’ + BC + A’
0 1A
01
00
11
10
BC ab’c’ + bc + a’
1
1 1
1
1
1
0
0
This is a simpler equation than we started with. How did we get
here?
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Mapping Sum of Product Terms
The three variable map has 12 possible groups of 2 spaces.
These become terms with 2 literals.
0 1A
01
00
11
10
BC 0 1A
01
00
11
10
BC 0 1A
01
00
11
10
BC
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Mapping Sum of Product Terms
The three variable map has 6 possible groups of 4 spaces.
These become terms with 1 literal.
0 1A
01
00
11
10
BC 0 1A
01
00
11
10
BC 0 1A
01
00
11
10
BC
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Four Variable Karnaugh Map
m12
m13
m15
m8
m9
m11
m10m14
01
00
11
10
m0
m1
m3
m4
m5
m7
ABCD 00 01 11 10
m6m2
D
A’BC
AB’C’
01
00
11
10
ABCD 00 01 11 10
0
1 1 1 1
1111
1
10 0
0 00
F = A’BC + AB’C + D
Note the row and column numbering. This is required for adjacency.
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Find a POS Solution
BC’
C’D
AB’CD’
01
00
11
10
ABCD 00 01 11 10
111
1
10
01
1
1
01
0 0 0 0
F’ = C’D + BC’ + AB’CD’
F = (C+D’)(B’+C)(A’+B+C’+D)
Find solutions to groups of 0’s to find F’. Invert to get F using DeMorgan’s.
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Dealing With Don’t Cares
F =∑
m(1, 3, 7) +∑
d(0, 5)
A’BC+AB’C+A’BC+ABC=
C
BC 0 1A
01
00
11
10
1
00
1
1
0
X
X
F = C
Circle the x’s that help get bigger groups of 1’s (or 0’s if POS).
Don’t circle the x’s that don’t help.
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Minimal K-Map Solutions
Some Terminology
and
An Algorithm to Find Them
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Prime Implicants
• A group of 1’s which are adjacent and can be combined on a
Karnaugh Map is called an implicant.
• The biggest group of 1’s which can be circled to cover a 1 is called
a prime implicant.
– They are the only implicants we care about.
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Prime Implicants
Prime Implicants Non−prime Implicants
01
00
11
10
ABCD 00 01 11 10
0
1 1
111
1
10 0
0
0 0
0
1 1
Are there any additional prime implicants in the map that are not
shown above?
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All the Prime Implicants
Prime Implicants
01
00
11
10
ABCD 00 01 11 10
0
1 1
111
1
10 0
0
0 0
0
1 1
When looking for a minimal solution – only circle prime implicants...
A minimal solution will never contain non-prime implicants
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Essential Prime Implicants
01
00
11
10
ABCD 00 01 11 10
0
1 1
111
1
10 0
0
0 0
0
1 1
Not all prime implicants are required ...
A prime implicant which is the only cover
of some 1’s is essential – a minimal
solution requires it.
Essential Prime Implicants Non−essential Prime Implicants
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A Minimal Solution Example
Minimum
F = AB’+BC+AD
Not required.01
00
11
10
ABCD 00 01 11 10
0
1 1
111
1
10 0
0
0 0
0
1 1
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Another Example
A’B’
F = A’D+BCD+B’D’Minimum
Not required.
01
00
11
10
ABCD 00 01 11 10
11
10 0
11 0
1
0
1
0
1
0
1
0
Every one one of F’s locations is covered by multiple implicants.
After choosing essentials, everything is covered...
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Finding the Minimum Sum of Products
1. Find each essential prime implicant and include it in the solution.
2. Determine if any minterms are not yet covered.
3. Find the minimal number of remaining prime implicants which
finish the cover.
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Yet Another Example
Using non-essential primes.
Essentials:
A’D’ and ADNon-essentials:
A’C and CDSolution:
A’D’+AD+A’Cor
A’D’+AD+CD
A’C
A’D’
AD
CD
01
00
11
10
ABCD 00 01 11 10
11
0
1 0
1
1
1
0 0
1
1
0
1
0
1
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K-Map Solution Summary
• Identify prime implicants.
• Add essentials to solution.
• Find minimum number non-essentials required to cover rest of map.
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Five and Six Variable K-Maps
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Five Variable Karnaugh Map
m12
m13
m15
m8
m9
m11
m10m14
m16
m17
m20
m21
m19 m23
m18 m22
m28
m29
m31
m30
m24
m25
m27
m26
This is the A=0 plane. This is the A=1 plane.
01
00
11
10
m0
m1
m3
m4
m5
m7
00 01 11 10
m6m2
01
00
11
10
00 01 11 10BC
DEBC
DE
The planes are adjacent to one another (one is above the other in
3D).
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Some Implicants in a Five Variable K-Map
ABCDEminimalnot
D’E’
B’C’E’AB’C’DA’BCD
A=1A=0
01
00
11
10
00 01 11 10
01
00
11
10
00 01 11 10BC
DEBC
DE
1
1 1 1 1
1
1
1 1 1 1
1
1
1
0 0
0 0
0
00
0
0
0 0
0
0 0 0
0
00
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Some Implicants in a Five Variable K-Map
Find the minimum sum-of-products for:
F =∑
m(0, 1, 4, 5, 11, 14, 15, 16, 17, 20, 21, 30, 31)
F = B’D’ + BCD + A’BDE
A=1A=0
01
00
11
10
00 01 11 10
01
00
11
10
00 01 11 10BC
DEBC
DE1 1
1
1
1 1
10 0
0
00
0
0
0 0
0
001 1
0 0
1
0 10
0
1 1
00
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Six Variable K-Map
m12
m13
m15
m8
m9
m11
m10m14
AB=00
AB=01
AB=10
AB=11
m16
m17
m20
m21
m19 m23
m18 m22
m28
m29
m31
m30
m24
m25
m27
m26
m32
m33
m35
m34
m36
m37
m39
m38
m44
m45
m47
m46
m40
m41
m42
m43
m48
m49
m51
m50
m52
m53
m55
m54
m60
m61
m63
m62
m56
m57
m59
m58
01
00
11
10
m0
m1
m3
m4
m5
m7
00 01 11 10
m6m2
EFCD
01
00
11
10
00 01 11 10CD
EF
01
00
11
10
00 01 11 10EFCD
01
00
11
10
00 01 11 10EFCD
– Typeset by FoilTEX – 48
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Six Variable K-Map
=CDEF
AC’D’=Solution= AC’D’+CDEF
AB=00
AB=01
AB=10
AB=11
1 1
11
1
1
1
1
1
1
1
1
0
0
0
0 0
0
0
0 0
0
0 0
0
0
0 0
0
0
0 0
0
0 0
0
0
0
000
0 0 0
0
000
0
0
0
0
00
0
00
0
0
00
0
0
0
01
00
11
10
00 01 11 10EFCD
01
00
11
10
00 01 11 10CD
EF
01
00
11
10
00 01 11 10EFCD
01
00
11
10
00 01 11 10EFCD
– Typeset by FoilTEX – 49
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K-Map Summary
• A K-Map is simply a folded truth table, where physical adjacency
implies logical adjacency.
• K-Maps are most commonly used hand method for logic
minimization.
• K-Maps have other uses for visualizing boolean equations.
– Typeset by FoilTEX – 50