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 Show 10 previous comments 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.