{"title":"Maximizing the number of spanning trees in a graph with n nodes and m edges","authors":"D. Shier","doi":"10.6028/JRES.078B.023","DOIUrl":null,"url":null,"abstract":"Consider the class of undirected graphs hav ing 11. nodes and m edges. The proble m to be addressed here is that of finding specific configurations of III edges on the given n nodes so that the resulting graph will contain the largest number of spanning trees. In particular, an explicit solution to this problem will be exhibited for graphs which have \"enough\" edges. To be specific, let the set E of k edges be deleted from the comp lete graph KII on 11. nodes: KII has an edge between every pair of distinct nodes and thus contains 11.(11.-1)/2 edges. For th e case when k ~ 11./2, we will demonstrate that the number of spanning trees T(n, E) in the res ulting graph is maximized by choosing the k deleted edges to be mutually nonadjacent. The (apparently more complicated) cases with k > 11./2 await resolution. Let Pic denote a set of k nonadjacent (\"parallel\") edges in K II , where k ~ n/2. We will show that lEI = k implies T(n, E) ~ T(n, Pic). First the case when the edge set E is disconnected will be di sposed of. This will be done in the context of the inductive hypothesis that if i < k and 151 = i, then T(n, S) ~ T(n, Pi) whether or not 5 is connected. Suppose lEI = k and that E can be decomposed into connected edge sets C I , C 2, •.• , C p with p 3 2. Certainly, when k = 2 the only disconnected set E possible consists of two nonadjacent edges, so that the inductive hypothesis holds. More generally, if lEI = k then IC,I < k and n 3 2k > 21C 11, whence 0 ~ T(n, C d ~ T(n, Pa ), (X= IC d, by the inductive hypothesis. Similarly, we have O~T(n, C,)~T(n, P(3), {3=IC ,I, where CI = C2 U ... U Co in the decomposition E=C 1 U C2 U ... U C/!. From [2, p. 106] it is known that ( 2)i T(n,P j )=nn2 1-;; ,","PeriodicalId":166823,"journal":{"name":"Journal of Research of the National Bureau of Standards, Section B: Mathematical Sciences","volume":"173 1","pages":"0"},"PeriodicalIF":0.0000,"publicationDate":"1974-10-01","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"33","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Journal of Research of the National Bureau of Standards, Section B: Mathematical Sciences","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.6028/JRES.078B.023","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 33
Abstract
Consider the class of undirected graphs hav ing 11. nodes and m edges. The proble m to be addressed here is that of finding specific configurations of III edges on the given n nodes so that the resulting graph will contain the largest number of spanning trees. In particular, an explicit solution to this problem will be exhibited for graphs which have "enough" edges. To be specific, let the set E of k edges be deleted from the comp lete graph KII on 11. nodes: KII has an edge between every pair of distinct nodes and thus contains 11.(11.-1)/2 edges. For th e case when k ~ 11./2, we will demonstrate that the number of spanning trees T(n, E) in the res ulting graph is maximized by choosing the k deleted edges to be mutually nonadjacent. The (apparently more complicated) cases with k > 11./2 await resolution. Let Pic denote a set of k nonadjacent ("parallel") edges in K II , where k ~ n/2. We will show that lEI = k implies T(n, E) ~ T(n, Pic). First the case when the edge set E is disconnected will be di sposed of. This will be done in the context of the inductive hypothesis that if i < k and 151 = i, then T(n, S) ~ T(n, Pi) whether or not 5 is connected. Suppose lEI = k and that E can be decomposed into connected edge sets C I , C 2, •.• , C p with p 3 2. Certainly, when k = 2 the only disconnected set E possible consists of two nonadjacent edges, so that the inductive hypothesis holds. More generally, if lEI = k then IC,I < k and n 3 2k > 21C 11, whence 0 ~ T(n, C d ~ T(n, Pa ), (X= IC d, by the inductive hypothesis. Similarly, we have O~T(n, C,)~T(n, P(3), {3=IC ,I, where CI = C2 U ... U Co in the decomposition E=C 1 U C2 U ... U C/!. From [2, p. 106] it is known that ( 2)i T(n,P j )=nn2 1-;; ,