Lecture 4 Hamming Code,

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CS147
Lecture 4 Hamming Code, Circuit Minimization and
Karnaugh Maps
Prof. Sin-Min Lee Department of Computer Science
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Since functions can be represented by a sum of
product form of minterms, any function can be
shown in the map by placing each a 1 in each
square which represents a minterm in the
function. EXAMPLE Draw the map for f X
Y 1. Determine which minterms are needed by
using truth table. X Y f X Y
minterm 0 0 0 m0 0 1 1 m1 1 0
1 m2 1 1 1 m3
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2. Draw the map and place a 1 in each square for
required minterms.
Y
Y
X
1
0
X
1
f X Y m1 m2 m3 X Y X Y X Y
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Use the 2-Variable Karnaugh Map for Minimization
Example Given f (X,Y) ? (0, 2) Find
Simplified sum of products
Y
Y
X
m1
m0
m2
m3
X
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2-Variable Karnaugh Map
1. Place the 1s corresponding to the minterms
on the map.
f (X,Y) ? (0, 2)
Y
Y
X
0
1
1
0
X
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2-Variable Karnaugh Map
2. Now group the 1s into columns or rows in
this case we can group them in the first column.
f (X,Y) ? (0, 2)
Y
Y
X
0
1
1
0
X
3. This column is Y so the simplified function
f (X,Y) Y
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2-Variable Karnaugh Map
Example 2 Given f (X,Y) ? (0, 2, 3) Find
Simplified sum of products
Y
Y
X
m1
m0
m2
m3
X
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2-Variable Karnaugh Map
Example 3 Given f (X,Y) ? (0, 1) Find
Simplified sum of products
Y
Y
X
m1
m0
m2
m3
X
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Three Variable Map
m0 m1 m3 m2 m4 m5 m7 m6
Y
YZ
X
0 0 0 1 1 1
10
X Y Z X Y Z X Y Z X YZ
0
X Y Z XYZ X Y Z X Y Z
1
X
Z
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Three Variable Map
There are 2N 23 8 squares. The minterms are
arranged so that only one variable changes from
0 to 1 or from 1 to 0 as you move from square to
square in the vertical or horizontal direction.
For any two adjacent squares, only one literal
changes from complemented to non-complemented
(normal). From this property the left and right
ends of the map are adjacent.
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Y
YZ
X
0 0 0 1 1 1
10
XYZ X Y Z X Y Z X Y Z
0
1
X
X YZ X Y Z X Y Z X YZ
Z
Using the distributive and complement properties,
any two adjacent minterms can be combined and
simplified to a single term with one less
literal.
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Combining terms on the Karnaugh Map simplifies
Boolean functions.
Example combine adjacent squares for X Y Z and
XYZ XY Z XYZ Y Z ( X X) Y Z 1
Y Z
Example Combine adjacent squares for X YZ and
X YZ XYZ X YZ X Z ( Y Y) X Z
1 X Z
Y
YZ
X
0 0 0 1 1 1
10
XYZ X Y Z X Y Z X Y Z
0
1
X
X YZ X Y Z X Y Z X YZ
Z
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Example Simplify f XY Z X Y Z XY Z
X Y Z
Y
YZ
0 0 0 1 1 1
10
X
1
1
1
0
1
1
X
Z
f XY Y Z
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Even if the function is not in its simplest form,
we can still use the map to simplify it
further. Example Simplify f X Y Y Z
X Z X Y Z
Y
YZ
0 0 0 1 1 1
10
X
1
1
1
0
1
1
1
X
Z
f Z X Y (by further grouping of minterms)
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3-Variable Karnaugh Map
Example Given f (X,Y,Z) ? (0, 2, 3, 4,
7) Find Simplified sum of products
Y
YZ
X
m2
m3
m1
m0
m6
m5
m7
m4
X
Z
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3-Variable Karnaugh Map
Example Given f (X,Y,Z) ? (0, 2, 3, 4,
7) Simplified sum of products f YZ YZ
XY
Y
YZ
X
1
1
0
1
0
0
1
1
X
Z
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Truth Table
f (X,Y,Z) ? (0, 2, 3, 4, 7) Simplified sum of
products f YZ YZ XY
The truth table for f
X Y Z f
0 0 0 1 0 0 1
0 0 1 0 1 0 1
1 1 1 0 0 1 1
0 1 0 1 1 0
0 1 1 1 1
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f (X,Y,Z) ? (0, 2, 3, 4, 7) YZ YZ XY
Y

Z
f
Y
Z
X
Y
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3-Variable Karnaugh Map
Example 2 Given f (X,Y,Z) ? (2, 3, 4,
5) Find Simplified sum of products
Y
YZ
X
m2
m3
m1
m0
m6
m5
m7
m4
X
Z
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3-Variable Karnaugh Map
Example 2 Given f (X,Y,Z) ? (1, 2, 5, 6,
7) Find Simplified sum of products
Y
YZ
X
m2
m3
m1
m0
m6
m5
m7
m4
X
Z
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3-Variable Karnaugh Map
Example 3 Given f (X,Y,Z) XZ XY XYZ
YZ Find Sum of minterms expression
Y
YZ
X
m2
m3
m1
m0
m6
m5
m7
m4
X
Z
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Exclusive OR XOR

• f XY XY X ? Y

X Y f 0 0 0 0 1
1 1 0 1 1 1 0
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Exclusive OR
f (X,Y) ? (1, 2) X ? Y
Y
Y
X
1
0
0
1
X
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XOR Truth Table
f (X,Y,Z) X ? Y ? Z ? (1, 2, 4, 7)
The truth table for f
X Y Z f
0 0 0 0 0 0 1
1 0 1 0 1 0 1
1 0 1 0 0 1 1
0 1 0 1 1 0
0 1 1 1 1
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Exclusive OR
X
Y
f (X,Y,Z) X ? Y ? Z
Z
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Four Variable Map
Y
YZ
0 0 0 1 1 1 10
WX
m0 m1 m3 m2 m4 m5 m7 m6
00
01
m12 m13 m15 m14 m8 m9
m11 m10
X
11
w x yz
W
10
Z
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Four Variable Map
N 4 variables 2N 24 16 square (minterms)
Row and column are numbered using a
reflected-code sequence. The minterm number can
be obtained by concatenation of the row and
column number . Example Row 4 10, Column 2
01 giving 1001 9 decimal for W X Y Z.
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Y
YZ
0 0 0 1 1 1 10
wxyz
wxy z
00
01
X
11
W
10
Z
Notice that top and bottom edges and right and
left edges are adjacent.
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1 square a term with 4 literals 2 square a
term with 3 literal 4 square a term with 2
literals 8 square a term with 1 literal 16
square a function equal to 1
Y
YZ
0 0 0 1 1 1 10
WX
00
01
X
11
W
10
Z
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Simplify f (W, X, Y, Z) W X Z W X Z W Y
Z f W Z Y Z f Z ( W Y)
Y
YZ
0 0 0 1 1 1 10
WX
00
01
X
11
W
10
Z
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Simplify f (W, X, Y, Z) ? ( 0, 1, 4, 5, 6, 8,
9, 12, 13, 14) f Y WZ XZ
Y
YZ
0 0 0 1 1 1 10
WX
00
01
X
11
W
10
Z
8 square 4 square
4 square
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Simplify f W X Y XY Z W X Y Z W
X Y f X Z X Y
W Y Z
Y
YZ
0 0 0 1 1 1 10
WX
00
01
X
11
10
Z
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Simplification of kmap

• Generate all PIs
• Find EPIs
• If EPIs can cover all minterms, then it is
  answer. Otherwise choose some non-essential PIs
  (which has less cost) such that all minterms are
  cover.

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Step 1

• Generate PIs
• Blue circle are PIs
• They are the largest circle you can drawn on kmap

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Step 2

• Find EPIs
• Red circle are EPIs
• Minterm 5, 14, 11 can only be cover by these 3
  red circle

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Step 3

• EPIs cannot cover minterm 7
• Choose between green/blue circle to cover minterm
  7
• Green is chosen as it is larger
• Less cost

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Final

• Final result is obtained
• x3x4 x2x3 x1x3 x2x3x4

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Complement and Product of Sums
On the K Map a 1 was placed in the squares which
represented minterms for a function. The squares
not used represent the complement of the
function. Mark these squares with 0, combine
them and read the complement as a sum of
products function, f. Using De Morgans Law on
f will give f, the original Function. It will be
in a product of sums form. Example Determine
the Product of sums form for f ( W, X, Y, Z) ?
(0, 1, 2, 5, 8, 9, 10) f W X YZ X Z
from the map f (W X) ( Y Z) ( X Z)
by De Morgans
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Complement and Product of Sums
Also, can determine the product of sums form
directly from The map by recognizing that the 0s
on the map represent maxterms. Remember that a 1
on the edge of the map Represents a complemented
literal for maxterms. Thus, for f ( W, X, Y, Z)
? (0, 1, 2, 5, 8, 9, 10).
(combine 0s on k-map) f ( Y Z) (W X)
( X Z)
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Y
YZ
0 0 0 1 1 1 10
WX
1 1 0 1
00
0 1 0 0
01
11
0 0 0 0
W
10
1 1 0 1
Z
f ( Y Z ) ( W X ) ( X Z)