Solutions to HW #4Text: 6/9. The equivalence classes (special case of MC's) are: {1, 2356, 4, 7, 8} These lead to the following minimum equivalent table. A B C D (1) 1 5,0 4,0 3,0 2,0 (2356) 2 2,1 2,0 2,0 1,0 (4) 3 5,0 4,0 4,0 2,0 (7) 4 5,0 5,0 3,0 4,0 (8) 5 2,0 4,0 3,0 2,0 Async: 2.9 Construct the pair chart, immediately Xing all pairs which do not have identical patterns of don't cares (next-state as well as output), or which have clashing output specifications. This puts X's everywhere but in the positions corresponding to 17 (which implies 38), 38 (which implies 46), and 46 (which implies 17). We then find that the equivalence classes are {17, 2, 38, 46, 5). The resulting minimal EQUIVALENT table is: A B C D (17) 1 2,0 -,1 3,- 2,0 (2) 2 3,0 5,1 2,0 -,- (38) 3 3,0 4,1 -,- 5,0 (46) 4 -,- 1,1 2,- -,- (5) 5 -,- 1,1 1,1 -,- NT-1. CD 00 01 11 10 1 (1),0 3,- - 2,0 2 1,0 3,- 4,0 (2),0 3 1,- (3),1 5,1 - Primitive flow table 4 1,0 3,- (4),0 2,0 5 1,- 3,1 (5),1 6,1 6 1,- 3,1 5,1 (6),1 CD 00 01 11 10 1 (1),0 2,- (1),0 (1),0 Reduced flow table 2 1,- (2),1 (2),1 (2),1 NT-2. CD 00 01 11 10 y1 y2 1 (1),0 (1),0 4,- 2,0 0 0 FLOW MATRIX (Note error 2 1,0 1,0 (2),0 (2),0 0 1 in the flow matrix of 3 (3),1 (3),1 4,1 2,- 1 1 the problem statement. 4 3,1 3,1 (4),1 (4),1 1 0 Q for state 3-11 should be 1, not a don't care.) CD 00 01 11 10 y1 y2 ------------------------- 1 00 00 10 01 0 0 2 00 00 01 01 0 1 Y-MATRIX 3 11 11 10 01 1 1 4 11 11 10 10 1 0 Y1Y2 0010 0001 00-0 0000 0011 0000 K-MAPS 1110 1101 111- 1111 1100 1111 Y1 Y2 Q Y1 = Cy1 + Dy1 + y1y2' + CDy2' Y2 = Cy1 + Cy1'y2 + CD'y1' + {CD'y2 or D'y1y2} Q = y1