Lecture 13 Announcements HW5 up on course webpage Due on Tuesday 1021 in class Upcoming Exam on October 28 Will cover material from Chapter 4 Details to follow soon Agenda Last time ID: 414932
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Slide1
Digital Logic Design
Lecture 13Slide2
Announcements
HW5 up on course webpage. Due on
Tuesday
, 10/21 in class.
Upcoming: Exam on October 28. Will cover material from Chapter 4. Details to follow soon.Slide3
Agenda
Last time
Using 3,4 variable K-Maps to find minimal expressions (4.5)
This time
Minimal expressions for incomplete Boolean functions (4.6)
5 and 6 variable K-Maps (4.7)
Petrick’s
method of determining irredundant expressions (4.9)Slide4
Minimal Expressions of Incomplete Boolean Functions
Recall an incomplete Boolean function has a truth table which contains dashed functional entries indicating don’t-care conditions.
Idea: Can replace don’t-care entries with either 0s or 1s in order to form the largest possible
subcubes
.Slide5
Example
0
1
3
2
4
5
7
6
12
131514891110
00
01
11
10
00
01
11
1
0
Slide6
Example
1
1
0
1
0
1
--
--
0
010100--
00
01
11
10
00
01
11
1
0
Slide7
Example
Step 1: Find prime
implicants
(pretend don’t care cells set to 1)
1
1
0
1
0
1
----0
010100--00011110
00
01
11
10
Slide8
Example
Step 2: Find essential prime
implicants
(discount don’t care cells)
1
1
0
1
0
1
----
0010100--00011110
00
01
11
1
0
Essential prime
implicants
:
Slide9
Example
Step 3: Add prime
implicants
to cover all 1-cells (discount don’t care cells)
1
1
0
1
0
1
----
0010100--00011110
00
01
11
1
0
Essential prime
implicants
:
Add:
Slide10
Example
Step 3: Add prime
implicants
to cover all 1-cells (discount don’t care cells)
1
1
0
1
0
1
--
--0010100--00011110
00
01
11
1
0
Final minimal DNF:
Slide11
Five and Six Variable K-MapsSlide12
Five Variable K-Maps
We can visualize five-variable map in two different ways:Slide13
Five Variable K-Maps
0
1
3
2
8
9
11
10
24
252726
16171918000001011010
00
01
11
1
0
6
7
54
141513
123031
29
28
22
23
21
20
Subcubes: Two
subcubes
are possible about the mirror-image line. If there are two rectangular groupings of the same
dimensions on both halves and the two groupings are the mirror image of each other.
1
10
111
1
0
1
10
0Slide14
Five Variable K-Maps
0
1
3
2
4
5
7
6
12
131514
891110161719182021232228293130
2425
2726
Subcubes: If each layer contains a
subcube such that they can be viewed as being directly above and below each other, then the two subcubes collectively form a single
subcube consisting of
cells.
v=0
v=1
Slide15
Example
0
1
0
0
0
1
0
0
0
1000110
0
0
00
110
0111
100
11
v=0
v=1
Slide16
Example
Step 1: Find all Prime
Implicants
.
0
1
0
0
0
1
000
1000110000011001
11
10
011
v=0
v=1
Slide17
Example
Step 2: Find all Essential Prime
Implicants
.
0
1
0
0
0
1
000
100011000001100
11
11
0
011
Essential Prime
Implicants
:
v=0
v=1
00
01
11
1
0Slide18
Example
Step 2: Find all Essential Prime
Implicants
.
0
1
0
0
0
1
00
010001100000110
01
1
11
001
1
Essential Prime
Implicants
:
v=0
v=1
Slide19
Example
Step 2: Find all Essential Prime
Implicants
.
0
1
0
0
0
1
00
010001100000110
01
1
11
001
1
Final minimal DNF:
v=0
v=1
Slide20
Six Variable K-Maps
We can visualize a six-variable map in two different ways:Slide21
Six Variable K-Maps
0
1
3
2
8
9
11
10
24
252726
1617191867541415131230312928
2223
2120
48
49
51505657
595840
4143
4232
3335
34
5455
53
52
62
63
61
60
46
47
45
44
38
39
37
36
000
001
011
010
000
001
011
010
1
10
111
1
0
1
10
0
1
1
0
1
1
1
1
0
1
1
0
0
Subcubes: If each quadrant has a rectangular grouping of dimensions
and each grouping is a mirror image of the other about both the horizontal and vertical mirror-image lines.
Slide22
Six Variable K-Maps
0
1
3
2
4
5
7
6
12
131514
891110161719182021232228293130
2425
2726
48
49
51505253
555460
6163
6256
5759
58
3233
35
34
36
37
39
38
44
45
47
46
40
41
43
42
uv
=00
uv
=01
uv
=11
uv
=10
Subcubes:
Subcubes
occurring in corresponding positions on all four layers collectively form a single
subcube
.Slide23
An Algorithm for the Final Step in Expression MinimizationSlide24
Petrick’s
Method of Determining Irredundant Expressions
A
X
B
X
X
X
C
X
X
X
D
X
X
X
E
X
X
F
X
G
X
X
H
X
X
I
X
X
A
X
B
X
X
X
C
X
X
X
D
X
X
X
E
X
X
F
X
G
X
X
H
X
X
I
X
X
The covering problem: Determine a subset of prime
implicants
that covers the table.
A minimal cover is an irredundant cover that corresponds to a minimal sum of the function.Slide25
Petrick’s
Method of Determining Irredundant Expressions
A
X
B
X
X
X
C
X
X
X
D
X
X
X
E
X
X
F
X
G
X
X
H
X
X
I
X
X
A
X
B
X
X
X
C
X
X
X
D
X
X
X
E
X
X
F
X
G
X
X
H
X
X
I
X
X
p-expression: (G+H)(F+G)(A+B)(B+C)(H+I)(D+I)(C+D)(B+C+E)(D+E)
The p-expression equals 1
iff
a sufficient subset of prime
implicants
is selected.Slide26
P-expressions
If a p-expression is manipulated into its sum-of-products form using the distributive law, duplicate literals deleted in each resulting product term and subsuming product
temrms
deleted, then each remaining product term represents an irredundant cover of the prime
implicant
table.
Since all subsuming product terms have been deleted, the resulting product terms must each describe an irredundant cover.
The irredundant DNF is obtained by summing the prime
implicants
indicated by the variables in a product term.Slide27
Simplifying p-expressions
Slide28
Finding Minimal Sums
There are 10 irredundant expressions
Evaluate each one by the cost criteria to find the minimal sum.
Minimal DNFs correspond to the first, third and eighth terms.
Slide29Slide30Slide31