1 1 vote Graph Theory graph-theory + – Pawan Kumar 2 1.8k views answer comment Share Follow Print See all 7 Comments 7 7 Comments reply Show 4 previous comments joshi_nitish commented Dec 31, 2017 reply Follow flag @Pawan, 12 is correct answer, see how. it is well known for any graph that matching no. + edge cover = no. of vertices edge cover-> min no. of edges to cover all the vertices. for K25, with 12 edges we can cover max. of 24 vertices, and one more edge needed to cover 25th vertex, so edge cover = 12+1 = 13, now matching no. = no. of vertices - edge cover = 25 - 13 = 12. i think you are confusing yourself with no. perfect matching which is '0' for K2n+1(i.e complete grah with odd no. of vertices), but here in qsn it is asked just a matching no. not a perfect matching 2 2 replyShare Ashwin Kulkarni commented Dec 31, 2017 reply Follow flag Great !!! @nitish _/\_ 0 0 replyShare Pawan Kumar 2 commented Dec 31, 2017 reply Follow flag joshi_nitish Sir....a heartily thanks .... :) 1 1 replyShare Please log in or register to add a comment.
Best answer 4 4 votes matching number = floor (no of vertices / 2). In this case, floor (25/2) = 12 Avdhesh Singh Rana answered Dec 31, 2017 • selected Dec 31, 2017 by Pawan Kumar 2 Avdhesh Singh Rana comment Share Follow See all 3 Comments 3 3 Comments reply Pawan Kumar 2 commented Dec 31, 2017 reply Follow flag can u explain how ? 0 0 replyShare Avdhesh Singh Rana commented Dec 31, 2017 reply Follow flag Matching Number is the no of edges in maximal matching. So, There will be 24 vertices can be matched with 12 edges but one remains unmatched. 0 0 replyShare Avdhesh Singh Rana commented Dec 31, 2017 reply Follow flag I think the matching number is asked not the perfect matching number. You are talking about perfect matching. https://www.geeksforgeeks.org/mathematics-matching-graph-theory/ 0 0 replyShare Please log in or register to add a comment.
0 0 votes Yes it should be 12 only. Floor(25/2)=12 Aaditya Pundir answered Dec 31, 2017 Aaditya Pundir comment Share Follow 0 reply Please log in or register to add a comment.