Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

402 BOOLEAN ALGEBRA [CHAP. 15

(a) Check the two squares corresponding to each ofx′yzandy′zt′, and check the square corresponding to each of

x′yz′t,xyzt′, andxy′z′t′. This gives the Karnaugh map in Fig. 15-36(a). A minimal cover consists of the three

designated loops. Thus a minimal sum forE 1 follows:

E 1 =zt′+xy′t′+x′yt

(b) Check the four squares corresponding tozt′, check the two squares corresponding to each ofy′z′tandyzt′,

and check the square corresponding tox′y′zt. This gives the Karnaugh map in Fig. 15-36(b). A minimal cover

consists of the three designated maximal basic rectangles. Thus a minimal sum forE 2 follows:

E 2 =zt′+xy′t′+x′yt

Fig. 15-36

15.42. Find a minimal sum-of-products form for the Boolean expressionEwith the following truth tables:

(a) T( 00001111 , 00110011 , 01010101 )=10100110.

(b) T( 00001111 , 00110011 , 01010101 )=00101111.

(a) From the given truth tableT(and the truth tables in Example 15.13 for the minterms in variablesx,y,z)wecan

read off the complete sum-of-products form forE:

E=x′y′z′+x′yz′+xy′z+xyz′

Its Karnaugh map appears in Fig. l5-37(a). There are three prime implicants, as indicated by the three loops,

which form a minimal cover ofE. Thus a minimal form forEfollows:

E=yz′+x′z′+xy′z

(b) From the given truth table we can read off the complete sum-of-products form forE:

E=x′yz′+x′yz+xy′z+xyz′+xyz

Its Karnaugh map appears in Fig. 15-37(b). There are two prime implicants, as indicated by the two loops, which

form a minimal cover ofE. Thus a minimal form forEfollows:

E=xz+y

Fig. 15-37