2 votes 2 votes Computer Networks computer-networks distance-vector-routing made-easy-test-series + – abhishek1995_cse asked Oct 16, 2018 • edited Mar 3, 2019 by Aditi Singh abhishek1995_cse 1.4k views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments anonymous commented Oct 21, 2018 reply Follow Share Yes, there is a shortcut. Take each Edge (Not node ) and see there is an Alternative better path available for that edge or not. If available then mark that otherwise mark chosen Edge.Finally, count the Unmarked Edge. 1 votes 1 votes csazad2702 commented Oct 16, 2019 reply Follow Share for each node check each of its neighbour if you get any other shorter path for that neighbour node rather than the adjacent edge .You can remove that edge . Here the answer will be 3 AG GC EF 0 votes 0 votes reboot commented Dec 23, 2020 reply Follow Share I think it will be BC GC EF 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes the answer is 2 paths- BC, CG only are unused rish1602 answered Apr 9, 2021 rish1602 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes BC , GC , EF = 3 sachin486 answered Dec 23, 2021 sachin486 comment Share Follow See all 0 reply Please log in or register to add a comment.
