2 votes 2 votes The number of totally ordered sets compatible to the given POSET are ________ Shadan Karim asked Dec 27, 2018 Shadan Karim 1.1k views answer comment Share Follow See all 13 Comments See all 13 13 Comments reply Hemanth_13 commented Dec 27, 2018 reply Follow Share 4.. not sure about it 0 votes 0 votes Ashwani Kumar 2 commented Dec 28, 2018 reply Follow Share A total order set is a chain in every two elements are comparable. Here 4 chains are there 0 votes 0 votes Priyanka Agarwal commented Dec 28, 2018 reply Follow Share 40 0 votes 0 votes Shadan Karim commented Dec 28, 2018 reply Follow Share @Priyanka Agarwal please explain 0 votes 0 votes Priyanka Agarwal commented Dec 28, 2018 reply Follow Share Toset compatible with poset means find no of topological sort possible 1 votes 1 votes Deepanshu commented Dec 28, 2018 reply Follow Share Priyanka Agarwal little ellaborate plss 0 votes 0 votes mehul vaidya commented Jan 14, 2019 reply Follow Share I think 4, answer given by made easy is wrong 0 votes 0 votes kumar.dilip commented Jan 14, 2019 reply Follow Share mehul vaidya Question is noting but find the number of topological ordering. $2 * \frac{6!}{3! * 3!} = 40$ 0 votes 0 votes mehul vaidya commented Jan 14, 2019 reply Follow Share can you please explain how you found above equation ,. a c b d e g i k can be topological order but not TOS because for TOS every element must be comparable to every other element in TOS , but here c and b and not comparable Please reply 0 votes 0 votes snaily16 commented Jan 14, 2019 reply Follow Share @kumar.dilip so how are the dependencies preserved (e->g), (g->i), (f->h), (h->j), if we place this six letters in 6C3 ways??? 0 votes 0 votes Nirmal Gaur commented Jan 25, 2019 reply Follow Share Here we have 11 nodes hence 11 positions. To satisfy topological ordering we have to fix node a at first position, node k at 11th position, and node d at 4th position. Now we have 2,3,5,6,7,8,9,10 as empty positions. Among them we can fill positions 2 and 3 from nodes b or c only (hence 2 ways) and remaining 6 places can be filled by e,g,i,f,h,j such that relative ordering e-g-i and f-h-j must be maintained, this can be done in 6!/(3!3!) = 20 ways. Total topological orderings = 2 x 20 = 40 1 votes 1 votes snaily16 commented Jan 25, 2019 reply Follow Share thanks @Nirmal Gaur 0 votes 0 votes balchandar reddy san commented Jan 25, 2019 reply Follow Share refer to this: https://gateoverflow.in/285489/me-test-series 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes https://medium.com/@WindUpDurb/on-partial-ordering-total-ordering-and-the-topological-sort-9f9c0d0d812f You may refer this blog and read few topics A DAG Is A Poset Total Ordering A Total Ordering of a Poset actually, the incomparable elements ( or you can say simultaneous action) can be ordered in any order. So topological sort is the best way to perform one action after other. Made easy answer is correct. Answer must be 40. arti111 answered Jan 21, 2019 arti111 comment Share Follow See all 0 reply Please log in or register to add a comment.