1 votes 1 votes Graph Theory graph-theory + – Himanshu1 asked Jan 6, 2016 Himanshu1 573 views answer comment Share Follow See 1 comment See all 1 1 comment reply Pradip Nichite commented Jan 14, 2016 reply Follow Share is planarity in 2016 gate syllabus? 0 votes 0 votes Please log in or register to add a comment.
Best answer 4 votes 4 votes maximum no. Of edges 26 In a connected planner graph with v vertices e edges r regions or faces and min. Degree of region =k (no cycle of length less than 4 means no region can have degree less than 4) The following property should satisfy 1 v-e+r=2 2 k.r≤2e 3 e ≤ k(v-2)/(k-2) Here k=4 v=15 maximum no. Of edges e ≤ 2v-4 Max no. Of edges= 2*15-4= 26 khushtak answered Jan 6, 2016 • selected Jan 6, 2016 by Himanshu1 khushtak comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes e<=2v --4 so e<=2*15-4 e<=26 Pooja Palod answered Jan 6, 2016 Pooja Palod comment Share Follow See all 4 Comments See all 4 4 Comments reply Himanshu1 commented Jan 6, 2016 reply Follow Share when this formula is applicable e<= 2V - 4 ? 0 votes 0 votes resuscitate commented Jan 6, 2016 reply Follow Share when the planar graph doesnot conatain traigle 2 votes 2 votes Pooja Palod commented Jan 7, 2016 reply Follow Share when it contains no triangles u can derive it using 4r<=2e v-e+r=2 2 votes 2 votes Pradip Nichite commented Jan 14, 2016 reply Follow Share is planarity in 2016 gate syllabus? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Use following lemma for planer graph, If G = (V, E) is a planar graph with |E| ≥ g and no cycle of length < g, then: |E| ≤ (g /(g − 2)) (|V | − 2). Here g = 4; hence |E| ≤ (4 /(4 − 2)) (15 − 2) = 26. |E| ≤ 26 => Maximum no.of edges = 26. Shashank Kumar answered Jan 6, 2016 Shashank Kumar comment Share Follow See all 0 reply Please log in or register to add a comment.