CS E6831 Solutions to HW #6IN ASYNC: 3.11. List dichotomies in each column: (1, 25) (1, 34) (25, 34) (12, 3) (12, 45) (3, 45) (14, 23) (14, 35) None of the above are covered by any others. 1, 3, and 4 are the most common, each appearing in all but 3 dichotomies. Choose 1 and discard all ordered dichotomies in which 1 appears in the right set. This leaves 10 ordered dichotomies. Note that every pair of rows is separated by at least one of the dichotomies. e.g., 12 is separated by (1, 25), 13 by (1, 34) and 23 by (25, 34). If some pair pq were not so separated, we would have to add a dichotomy (a, b) to the list. Now set up the pair chart as below: a[1,25] ...... b[1,34] .............. X c[12,45] ...................... X d[12,3] ............................. X X X e[14,23] ..................................... X X f[14,35] ............................................. X X X X g[25,34] ..................................................... X X X X X X g'[25,34] .............................................................. X X X X X X h[3,45] ..................................................................... X X X X X X X h'[45,3] ............................................................................. The MC's are aef, eh', df, bdg, dh', bcd, ch We now need to find a minimal set of the MC's that cover each UNORDERED dichotomy, e.g., we need cover h or h', but need not cover both. Finding a minimal covering set of the MC's can be done using the covering table techniques that are the second part of the Quine-McCluskey method. In this case it is not hard to discover that the optimal solution is {aef, bdg, ch}. These correspond to the dichotomies, (14, 235), (125, 34), and (123, 45) respectively. This leads to the state assignment below (where the intermediate y-states are added with appropriate next-state entries). A B C y1 y2 y3 1 (1) 2 (1) 0 0 0 2 (2) (2) 3 1 0 0 3 4 (3) (3) 1 1 0 4 (4) 5 1 0 1 1 5 2 (5) 3 1 0 1 4 - 1 0 1 0 4 5 3 1 1 1 - 5 - 0 0 1 3.13. Dichotomies to be covered are (15, 24), (15, 34), (12, 45), (23, 45), (13, 2), (13, 5), (23, 4), (34, 5). Delete (34, 5), since it is covered by (15, 34). Every pair is separated by at least one dichotomy. 1 appears as often as any other row, so discard all ordered dichotomies with 1 in the right set. Generate the pair chart below. a[15,24] ...... b[15,34] .............. X X c[12,45] ...................... X X d[13,2] ............................. X X e[13,5] ..................................... X X X f[23,45] ............................................. X X X X X X f'[45,23] ..................................................... X X X X X g[2,34] .............................................................. X X X X X X g'[34,2] ..................................................................... MC's are cef, deg', cg, bg, ab, ad There are two minimal covers. Start with cef, since it uniquely covers f. Then we can cover the remaining rows a, b, d, g with either ad and bg, or with ab and dg'. Below we use ab->(15, 234), cef->(123, 45), dg'->(134, 2) This leads to the flow matrix below (the y1 value in row-5 could have been a 1, but this would have led to fewer don't cares in the NS entries). x1x2 00 01 11 10 y1 y2 y3 1 (1) 2 3 (1) 0 0 0 2 4 (2) (2) (2) 1 1 0 3 4 2 (3) (3) 1 0 0 4 (4) 5 3 (4) 1 0 1 5 1 (5) (5) (5) 0 0 1 - 2 - - 0 1 0 4 - - - 1 1 1 Non-Text Problem. (a) FROM 1 1/2 8 5 3 6 4 7 9 S we can get from 1 to any other row. TO 1 1/3 6 8 2 But no way to get from 4, 5, 7, or 9 back to 1, 2, 3, 6, or 8. So this table is NOT strongly connected. (b) FROM 1/4 6 3 9 7 2 5 8 So from 1 any other row can be reached. TO 1 1/5 7 3 8 6 9 4 2 So 1 can be reached from any row, and therefore, this table IS strongly connected.