{"title":"增量瓶颈和瓶颈末端斯坦纳树问题的复杂性","authors":"Yen Hung Chen","doi":"10.1109/ICS.2016.0010","DOIUrl":null,"url":null,"abstract":"Given a graph G = (V, E) with non-negative edge lengths, a subset R ⊂ V, a Steiner tree for R in G is an acyclic subgraph of G interconnecting all vertices in R and a terminal Steiner tree is defined to be a Steiner tree in G with all the vertices of R as its leaves. A bottleneck edge of a Steiner tree is an edge with the largest length in the Steiner tree. The bottleneck Steiner tree problem (BSTP) (respectively, the bottleneck terminal Steiner tree problem (BTSTP)) is to find a Steiner tree (respectively, a terminal Steiner tree) for R in G with minimum length of a bottleneck edge. For any arbitrary tree T, lenb(T) denotes the length of a bottleneck edge in T. Let Topt(G, BSTP) and Topt(G, BTSTP) denote the optimal solutions for the BSTP and the BTSTP in G, respectively. Given a graph G = (V, E) with non-negative edge lengths, a subset E0 ⊂ E, a number h = |E \\E0|, and a subset R ⊂ V, the incremental bottleneck Steiner tree problem (respectively, the incremental bottleneck terminal Steiner tree problem) is to find a sequence of edge sets {E0 ⊂ E1 ⊂ E2 ⊂ … ⊂ Eh = E} with |Ei\\Ei-1| = 1 such that Σh i=1 lenb(Topt(Gi, BSTP)) (respectively, Σh i=1 lenb(Topt(Gi, BTSTP))) is minimized, where Gi = (V, Ei). In this paper, we prove that the incremental bottleneck Steiner tree problem is NP-hard. Then we show that there is no polynomial time approximation algorithm achieving a performance ratio of (1-ε) × ln |R|, 0","PeriodicalId":281088,"journal":{"name":"2016 International Computer Symposium (ICS)","volume":"2 1","pages":"0"},"PeriodicalIF":0.0000,"publicationDate":"2016-12-01","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"On the Complexities of the Incremental Bottleneck and Bottleneck Terminal Steiner Tree Problems\",\"authors\":\"Yen Hung Chen\",\"doi\":\"10.1109/ICS.2016.0010\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"Given a graph G = (V, E) with non-negative edge lengths, a subset R ⊂ V, a Steiner tree for R in G is an acyclic subgraph of G interconnecting all vertices in R and a terminal Steiner tree is defined to be a Steiner tree in G with all the vertices of R as its leaves. A bottleneck edge of a Steiner tree is an edge with the largest length in the Steiner tree. The bottleneck Steiner tree problem (BSTP) (respectively, the bottleneck terminal Steiner tree problem (BTSTP)) is to find a Steiner tree (respectively, a terminal Steiner tree) for R in G with minimum length of a bottleneck edge. For any arbitrary tree T, lenb(T) denotes the length of a bottleneck edge in T. Let Topt(G, BSTP) and Topt(G, BTSTP) denote the optimal solutions for the BSTP and the BTSTP in G, respectively. Given a graph G = (V, E) with non-negative edge lengths, a subset E0 ⊂ E, a number h = |E \\\\E0|, and a subset R ⊂ V, the incremental bottleneck Steiner tree problem (respectively, the incremental bottleneck terminal Steiner tree problem) is to find a sequence of edge sets {E0 ⊂ E1 ⊂ E2 ⊂ … ⊂ Eh = E} with |Ei\\\\Ei-1| = 1 such that Σh i=1 lenb(Topt(Gi, BSTP)) (respectively, Σh i=1 lenb(Topt(Gi, BTSTP))) is minimized, where Gi = (V, Ei). In this paper, we prove that the incremental bottleneck Steiner tree problem is NP-hard. Then we show that there is no polynomial time approximation algorithm achieving a performance ratio of (1-ε) × ln |R|, 0\",\"PeriodicalId\":281088,\"journal\":{\"name\":\"2016 International Computer Symposium (ICS)\",\"volume\":\"2 1\",\"pages\":\"0\"},\"PeriodicalIF\":0.0000,\"publicationDate\":\"2016-12-01\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"2016 International Computer Symposium (ICS)\",\"FirstCategoryId\":\"1085\",\"ListUrlMain\":\"https://doi.org/10.1109/ICS.2016.0010\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"\",\"JCRName\":\"\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"2016 International Computer Symposium (ICS)","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.1109/ICS.2016.0010","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
On the Complexities of the Incremental Bottleneck and Bottleneck Terminal Steiner Tree Problems
Given a graph G = (V, E) with non-negative edge lengths, a subset R ⊂ V, a Steiner tree for R in G is an acyclic subgraph of G interconnecting all vertices in R and a terminal Steiner tree is defined to be a Steiner tree in G with all the vertices of R as its leaves. A bottleneck edge of a Steiner tree is an edge with the largest length in the Steiner tree. The bottleneck Steiner tree problem (BSTP) (respectively, the bottleneck terminal Steiner tree problem (BTSTP)) is to find a Steiner tree (respectively, a terminal Steiner tree) for R in G with minimum length of a bottleneck edge. For any arbitrary tree T, lenb(T) denotes the length of a bottleneck edge in T. Let Topt(G, BSTP) and Topt(G, BTSTP) denote the optimal solutions for the BSTP and the BTSTP in G, respectively. Given a graph G = (V, E) with non-negative edge lengths, a subset E0 ⊂ E, a number h = |E \E0|, and a subset R ⊂ V, the incremental bottleneck Steiner tree problem (respectively, the incremental bottleneck terminal Steiner tree problem) is to find a sequence of edge sets {E0 ⊂ E1 ⊂ E2 ⊂ … ⊂ Eh = E} with |Ei\Ei-1| = 1 such that Σh i=1 lenb(Topt(Gi, BSTP)) (respectively, Σh i=1 lenb(Topt(Gi, BTSTP))) is minimized, where Gi = (V, Ei). In this paper, we prove that the incremental bottleneck Steiner tree problem is NP-hard. Then we show that there is no polynomial time approximation algorithm achieving a performance ratio of (1-ε) × ln |R|, 0