Digitalteknik EIT020. Lecture 10: Karnaugh Maps
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1 Digitalteknik EIT020 Lecture 10: Karnaugh Maps September 30, 2014
2 Idea of minimisation Theorem There exists a prime cover that is a minimal cover. Minimal prime cover Find all prime implicants. Find the essential prime implicants. These must be part of a minimal prime cover a part of on-set. Cover the rest of on-set with a minimal number of prime implicants., Digitalteknik L10:2, Ch 5.1.2
3 Gray code In a Gray code only one variable is changed for each (cyclic) step. Example Example of NBCD and Gray code for N = 4 and N = 8. N = 4 N = 8 n NBCD Gray (NBCD = Natural Binary Coded Decimal.) n NBCD Gray , Digitalteknik L10:3, Ch 5.1.2
4 Karnaugh maps A Karnaugh map is a function table in matrix form with at most two variables per dimension. the input combinations are listed as Gray code. Then, between two consecutive positions (horizontal or vertical) there is only a change in one variable. The Karnaugh map can be seen as a graphical interpretation of B n. The maps useful on paper (2-dimensions) are for functions with 2, 3, or 4 variables., Digitalteknik L10:4, Ch 5.1.2
5 Karnaugh maps (cont'd) f (x 1 x 2 ): f (x 1 x 2 x 3 ): f (x 1 x 2 x 3 x 4 ): f x f x 2 x f x 3 x f (0) f (1) 0 f (0) f (1) f (3) f (2) 00 f (0) f (1) f (3) f (2) x 1 1 f (2) f (3) x 1 1 f (4) f (5) f (7) f (6) 01 f (4) f (5) f (7) f (6) f 00 x f (0) f (1) x 1 x f (12) f (13) f (15) f (14) f (8) f (9) f (11) f (10) 01 x 1 x f (2) f (3) f (6) f (7) f (4) f (5), Digitalteknik L10:5, Ch 5.1.2
6 Minimisation Cubes In a Karnaugh map a cube is a (cyclic) rectangular block covering 2 k positions. Minimisation by Karnaugh maps Find as big rectangular blocks as possible with 2 k 1s and. These represent the prime implicants. Find the essential implicants. These must be part of the function. Chose a minimal number of the rest of the implicants such that the function is covered., Digitalteknik L10:6, Ch 5.1.2
7 Karnaugh minimisation (Ex) Example Consider the function f with f 1 (1) = {0, 2, 3, 7}, f 1 ( ) = {5, 6} The Karnaugh map is f x 1 x x A B C Prime implicants: A = c (0B0) (x) = x 1x 3 B = c (B1B) (x) = x 2 C = c (1B1) (x) = x 1 x 3 A and B are essential, and covers the function, i.e., f min = A B = x 1x 3 x 2 is a minimal function (disjunctive form)., Digitalteknik L10:7, Ch 5.1.2
8 Karnaugh minimisation (Conjunctive form) Example Consider the function f with f 1 (1) = {0, 2, 3, 7}, f 1 ( ) = {5, 6} The Karnaugh map is f x 1 x 2 11 C 10 x A B 0 - (Anti) prime implicants: ( A = c (x)) (B01) ( ) = x 2 x 3 = x2 x 3 ( B = c (x)) (10B) ( = x1 x 2) = x 1 x 2 ( C = c (x)) (1B0) ( = x1 x 3) = x 1 x 3 A is essential. Minimal conjunctive form: f min = A B = (x 2 x 3)(x 1 x 2 ) f min = A C = (x 2 x 3)(x 1 x 3 ), Digitalteknik L10:8, Ch 5.1.2
9 Minimal disjunctive form Example 5.7 (disjunctive form) The function f (x 1 x 2 x 3 x 4 ) has f 1 (1) = {0, 4, 5, 8, 9, 10, 11, 13, 15} Prime implicants: f A 00 B 01 x 1 x 2 11 C 10 x 3 x D E F A = x 2x 3x 4 B = x 1x 3x 4 C = x 1x 2 x 3 D = x 2 x 3x 4 E = x 1 x 4 F = x 1 x 2, Digitalteknik L10:10, Ch 5.1.2
10 Prime table A prime table (P-table) shows which minterms the prime implicants cover. Rows correspond to prime implicants. Columns correspond to minterms. A cross shows when a minterm is covered by the implicant. Then an essential row contains a that is not represented in any other rows. Essential rows must be in the cover. row r dominates row s, r s, if all in s also exist in r. The dominated row, s, can be deleted. column c dominates column d, c d, if all in d also exist in c. The dominating column, c, can be deleted., Digitalteknik L10:11, Ch 5.1.2
11 Prime table (Ex 5.7) A x x B x x C x x D x x E x x x x F x x x x Essential: E from 15 F from 10 All columns that have in row E and F are deleted from the table, they are covered by E and F., Digitalteknik L10:12, Ch 5.1.2
12 Prime table (Ex 5.7) B A x B x x C x x C D x Here, B dominates A and C dominates D B x x C x x Now we see that 4 dominates both 0 and 5. Hence, the column 4 is deleted. Then, both B and C are (secondary) essential. f = E F B C Remark: From the Karnaugh map we can nd two other minimal covers: f = E F B D and f = E F A C, Digitalteknik L10:13, Ch 5.1.2
13 Minimal conjunctive form Example 5.7 (conjunctive form) The function f has f 1 (0) = {1, 2, 3, 6, 7, 12, 14} Karnaugh map: f x 1 x 2 11 x 3 x 4 A B C 10 D, Digitalteknik L10:15, Ch 5.1.2
14 Minimal conjunctive form Example (cont'd) (Dual of) implicants: A = (x 1x 2x 4 ) = x 1 x 2 x 4 B = (x 1x 3 ) = x 1 x 3 C = (x 2 x 3 x 4) = x 2 x 3 x 4 D = (x 1 x 2 x 4) = x 1 x 2 x 4 Essential: A, B, and D The essentials cover f. f = A B D = (x 1 x 2 x 4)(x 1 x 3)(x 1 x 2 x 4 ), Digitalteknik L10:16, Ch 5.1.2
15 Cyclic prime table Example The function f i specied by f 1 (1) = {5, 6, 15} f 1 (0) = {0, 2, 8, 10, 12} A f x 1 x x 3 x B C, Digitalteknik L10:18, Ch 5.1.2
16 Cyclic prime table Example (cont'd) A x x B x x C x x The table is cyclic (No essential rows, dominating rows or dominating columns). f = A B = x 1 x 2 x 4 f = A C = x 1 x 2 x 2 x 3 f = B C = x 4 x 2 x 3, Digitalteknik L10:19, Ch 5.1.2
17 5 variables Example 5.9 Minimise f (x 1 x 2 x 3 x 4 x 5 ) where f 1 (1) = {4, 5, 6, 7, 13, 15, 20, 21, 23, 26, 27, 29, 30, 31} To solve this we need three dimensions. Use two separate maps and view them as the third dimension (on top of each other)., Digitalteknik L10:20, Ch 5.1.2
18 Ex 5.9 (solution) Split the function in two Karnaugh maps, one for x 1 = 0 and one for x 1 = 1: x 4 x 5 x 4 x 5 f x f x A x 2 x B C D x 2 x x 1 = 0 x 1 = 1 Prime implicants: A = x 2 x 3x 4 C = x 3 x 5 B = x 1 x 2 x 3 D = x 1 x 2 x 4 All prime implicants are essential, f = A B C D, Digitalteknik L10:21, Ch 5.1.2
19 More than one function Example 5.10 Minimise a minimal circuit that realises the functions f 1 1 (1) = {1, 3, 5, 7, 8, 9, 12, 14} f 1 2 (1) = {1, 3, 5, 7, 8, 9, 10, 14, 15}, Digitalteknik L10:22, Ch 5.1.2
20 Ex 5.10 (solution) Karnaugh maps for f 1 and f 2 to get implicants for the separated functions: f x 2 x D x 4 x 5 A B C f x 2 x x 4 x 5 A B F H G E E I, Digitalteknik L10:23, Ch 5.1.2
21 Ex 5.10 (solution) Implicants that can be used in both f 1 and A f 2 : x 3 x 4 f 1 f B 01 x 1 x J E, Digitalteknik L10:24, Ch 5.1.2
22 P-table (Ex 5.10) f 1 f A x x x x B x x x x x x x x C x x D x x E x x x x F x x G x x H x x I x x J x x, Digitalteknik L10:25, Ch 5.1.2
23 1st reduction f 1 f A x x C x x D x x E x x x x F x G x x H x x I x x J x x, Digitalteknik L10:26, Ch 5.1.2
24 2nd reduction A E delete A. Then E is (sec.) essential in both f 1 and f 2. C D delete D I G delete I F H delete F f 1 f C x x G x x H x x J x x Then C is (sec.) essential in f 1 and G and H are (sec.) essential in f 2. f 1 = B E C f 2 = B E G H, Digitalteknik L10:27, Ch 5.1.2
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