Question

3. (a) Let R be a binary relation on the set X = {1,2,3,4,5,6,7}, defined by R= {(1,3), (2,3), (3, 4), (4,4),(4,5), (5,6), (5,7)} (1) (6 pts) Find Rk for all k = 2, 3, 4, 5,... (2) (3 pts) Find the transitive closure t(R) of R by Washall's algorithm and draw the directed graph of t(R).

Answer 1

LetR be a binary relation on the set x = {1, 2, 3, 4, 5, 6 17 defined by a l . (4,4), (4,5), R = 5cl ago (2,3), (3, 4), (4,4), (4,5), (5,6), (6,7)} we can draw adjacency matriae of cabove Relation R a Definition of adjacency matrix- an adjacency manae is a square matrae used to represent a finite graph. The element of the matriae indicate wheather pairs of vestices are adjacent on not in a graph. . A = T0 0 To 0 0 0 lo Too 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 adjacency matrise of This is Relation the R. cebove

I then Ak represents the kth power of A Tie kth power of relation R. LR re presents the number of walks Then At represents the k power of A. In particular. entry ci) in Ak gives the number of Walks from i to i of length K. k times product of Relation R. Lie whenever we have to find Rh e simply find Ak. - formina for Rk is AR for every k

closure to R) by 2 Transitive Algorithm warshall's Р= CC1,3), 2 ) Сэ, ц), сч,4), Cais), CS, 6) , L617) o o I o o o o o o o o o o o o o 1 тоо o o o o o | | To o o o o o o o o o o o o o

let wo-MR & we successively compute wi, W2,---- Wz Wk for k=1,2, - - -. 7 is calculated using column is f now it of weg. This w, is completed using column 1 5 row 1 of Worme ... step 1 - Transfer all of is in writo the corresponding positions of wr Step 2 - list separately I have is in column t value is in now k. the nows & column that that step 3 - pair each of the now numbers I with each of the column numbers, & put a lin the corresponding position of we if there isn't one there already put o's in any remaining empty positions.

Fort, there is no lts in column 1 & there is column that have is in in column I as there in no t's w is same as wo I simlary as there is no ls in column 2 I Wa is same as wo. for We there is 'S in column thired at position 1&2. 4 there is column that have is in row Brd cut position 4 we put i's et position - To obtain Wa 7 (14) & (24) 1 100 000 0 0 1 0 0 0 Too oi o oT Too 0 0 0 1 1 To ooo ooo To ooo ooo

for k= 4 in column 4 of Wz I is at position I 1, 2, 3, 4 in now 4 of WB I is cet position I at the . To obtain wy we put position f(1,4), (2,4), (3, 4), (4,4)} looooo 1001 0 0 0 To oo 0 0 0 To oo oo oo To 0 0 0 0 0 0 for k=5 in columns of wy I is at position in now s of Wy 1 is at position ws TO position we put Oftal'n (4,6) l at the

о о т 1 o o o о от To o o o o o o o o o 0 : 1 т т о o | o | o o o o o o o o o o o o o .

for k=6 in column 6 of Ws 1 is at Position 4 fs in row 6 of Wy there is no position no exetre positions of I in WG Wo is similar aus Ws . & Wz is also similar as Wo since row z of Wg there is no position W6 - Wz = 0 or 1 0 0 0 To o I i o o o 0 0 0 0 0 0 0 0 0 1 1 0 To oo old To o o o o oo Loooooool a The Transitive closure of R is given as R=% (1,3), (lu,(2, 3), (214), (3, 4) cuiu, cus) (416) (5,6), (5,7)}