1. The 5-variables K-Map

2. Figure 4–42 A 5-variable Karnaugh map.
Thomas L. Floyd Digital Fundamentals, 9e
Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.

3. Figure 4–43 Illustration of groupings of 1s in adjacent cells of a 5-variable map.
Thomas L. Floyd Digital Fundamentals, 9e
Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.

4. Figure 4–44 Thomas L. Floyd Digital Fundamentals, 9e
Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey All rights reserved.

5. 5 variable K-map
5 variables -> 32 minterms, hence 32 squares required

6. K-map Product of Sums simplification
Example: Simplify the Boolean function F(ABCD)=(0,1,2,5,8,9,10) in
(a) S-of-p (b) P-of-s
Using the maxterms (0’s) and complimenting F Grouping as if they were minterms, then using De Morgen’s theorem to get F.
F’(ABCD)= BD’+CD+AB
F(ABCD)= (B’+D)(C’+D’)(A’+B’)
Using the minterms (1’s)
F(ABCD)= B’D’+B’C’+A’C’D

7. 5 variable K-map
Adjacent squares. E.g. square 15 is adjacent to 7,14,13,31 and its mirror square 11.
The centre line must be considered as the centre of a book, each half of the K-map being a page
The centre line is like a mirror with each square being adjacent not only to its 4 immediate neighbouring squares, but also to its mirror image.

8. 5 variable K-map
Example: Simplify the Boolean function
F(ABCDE) = (0,2,4,6,11,13,15,17,21,25,27,29,31)
Soln: F(ABCDE) = BE+AD’E+A’B’E’

10. How many k-map is needed?
If you have 5 variables, you’ll need 2 k-map…
Let’s say the variables are A, B, C, D and E.
0 0
0 1
1 1
1 0
A B
C D
E = 0

11. How many k-map is needed?
0 0
0 1
1 1
1 0
A B
C D

12. Try this out…
Simplify the Boolean function F(A,B,C,D,E) = (0,1,4,5,16,17,21,25,29)
Soln: F(A,B,C,D,E) = A’B’D’+AD’E+B’C’D’

13. 1st step – convert the minterm into Boolean equation
F(A,B,C,D,E) = (0,1,4,5,16,17,21,25,29)
How to convert … ??
0 = 
1 = 

14. 1st step – convert the minterm into Boolean equation
F(A,B,C,D,E) = (0,1,4,5,16,17,21,25,29)

15. 2nd – prepare 2 k-map E = 0 E = 1 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B
A B
C D
0 0
0 1
1 1
1 0
A B
C D
E = 0
E = 1

16. 3rd- plug in the Boolean term/minterm into k-map
0 0
0 1
1 1
1 0
A B
C D
0 0
0 1
1 1
1 0
A B
C D 1 1 1 1 1 1 1 1 1
E = 0
E = 1

17. 4th- look for similar grouping that can be done in both k-map
0 0
0 1
1 1
1 0
A B
C D
0 0
0 1
1 1
1 0
A B
C D 1 1 1 1 1 1 1 1 1
E = 0
E = 1
That will give us

18. 4th- look for similar grouping that can be done in both k-map
0 0
0 1
1 1
1 0
A B
C D
0 0
0 1
1 1
1 0
A B
C D 1 1 1 1 1 1 1 1 1
E = 0
E = 1
That will give us

19. The combination will look like this…
0 0
0 1
1 1
1 0
A B
C D
0 0
0 1
1 1
1 0
A B
C D 1 1 1 1 1 1 1 1 1
E = 0
E = 1
That will give us

20. 5th- look for the remaining 1’s that has not been included yet
0 0
0 1
1 1
1 0
A B
C D
0 0
0 1
1 1
1 0
A B
C D 1 1 1 1 1 1 1 1 1
E = 0
E = 1

21. 6th- group the remaining 1’s
0 0
0 1
1 1
1 0
A B
C D
0 0
0 1
1 1
1 0
A B
C D 1 1 1 1 1 1 1 1 1
E = 0
E = 1
That new grouping will give us
Note that this time, E need to be included in the term, since the grouping is in the E = 1 k-map only.

22. 6th- group the remaining 1’s
0 0
0 1
1 1
1 0
A B
C D
0 0
0 1
1 1
1 0
A B
C D 1 1 1 1 1 1 1 1 1
E = 0
E = 1
Full expression will be,

23. 6 variable K-map
6 variables -> 64 minterms, hence 64 squares required

24. ICS217-Digital Electronics - Part 1.5 Combinational Logic
Tutorial 1.5
1. Simplify the Boolean function F(ABCDE) = (0,1,4,5,16,17,21,25,29)
Soln: F(ABCDE) = A’B’D’+AD’E+B’C’D’
2. Simplify the following Boolean expressions using K-maps.
(a) BDE+B’C’D+CDE+A’B’CE+A’B’C+B’C’D’E’
Soln: DE+A’B’C’+B’C’E’
(b) A’B’CE’+A’B’C’D’+B’D’E’+B’CD’+CDE’+BDE’
Soln: BDE’+B’CD’+B’D’E’+A’B’D’+CDE’
(c) F(ABCDEF) = (6,9,13,18,19,27,29,41,45,57,61)
Soln: F(ABCDEF) = A’B’C’DEF’+A’BC’DE+CE’F+A’BD’EF
ICS217-Digital Electronics - Part 1.5 Combinational Logic

25. END OF Chapter 1