The 5-variables K-Map

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 07458 All rights reserved.

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 07458 All rights reserved.

Figure 4–44 Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

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

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

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.

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’

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 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 E = 0

How many k-map is needed? 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10

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’

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 = 00000  1 = 00001 

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

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 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0 E = 1

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

4th- look for similar grouping that can be done in both k-map 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 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

4th- look for similar grouping that can be done in both k-map 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 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

The combination will look like this… 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 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

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

6th- group the remaining 1’s 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 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.

6th- group the remaining 1’s 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 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,

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

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

END OF Chapter 1