Steiner Tree
Algorithms and Networks 2014/2015 Hans L. Bodlaender Johan M. M. van Rooij
1 The Steiner Tree Problem
Let G = (V,E) be an undirected graph, and let N µ V be a subset of the terminals. A Steiner tree is a tree T = (V’,E’) in G connecting all terminals in N V’ µ V, E’ µ E, N µ V’ We use k=|N|.
Streiner tree problem: Given: an undirected graph G = (V,E), a terminal set N µ V, and an integer t. Question: is there a Steiner tree consisting of at most t edges in G.
2 My Last Lecture
Steiner Tree. Interesting problem that we have not seen yet.
Introduction Variants / applications NP-Completeness Polynomial time solvable special cases. Distance network. Solving Steiner tree with k-terminals in O*(2k)-time. Uses inclusion/exclusion. Algorithm invented by one of our former students.
3 Steiner Tree – Algorithms and Networks INTRODUCTION
4 Variants and Applications
Applications: Wire routing of VLSI. Customer’s bill for renting communication networks. Other network design and facility location problems.
Some variants: Vertices are points in the plane. Vertex weights / edge weights vs unit weights. Different variants for directed graphs.
5 Steiner Tree is NP-Complete
Steiner Tree is NP-Complete. Membership of NP: certificate is a subset of the edges. NP-Hard: reduction from Vertex Cover. Take an instance of Vertex Cover, G=(V,E), integer k. Build G’=(V’,E’) by subdividing each edge. Set N = set of newly introduced vertices. All edges length 1. Add one superterminal connected to all vertices. G’ has Steiner Tree with |E|+k edges, if and only if, G has vertex cover with k vertices. = terminal
6 Steiner Tree – Algorithms and Networks POLYNOMIAL-TIME SOLVABLE SPECIAL CASES
7 Special Cases of Steiner Tree
k = 1: trivial. k = 2: shortest path. k = n: minimum spanning tree.
k = c = O(1): constant number of terminals, polynomial- time solvable (next slides).
8 Distance Networks
Distance network D(X) of G=(V,E) (induced by the set X). Take complete graph with vertex set X. Cost of edge {v,w} in distance network is length shortest path from v to w in G.
Observations: Let W be the set of vertices of degree larger than two for an optimal Steiner tree T in G with terminal set N. The Steiner tree T consists of a series of shortest paths between vertices in N [ W. The cost of T equals the cost of the minimum spanning tree in D(N[W). The cost of the optimal Steiner tree in D(V) equals the cost of T.
9 Steiner Tree with O(1) Terminals
Suppose |N|= k is constant c. Compute distance network D(V). There is a minimum cost Steiner tree in D(V) that contains at most k – 2 non-terminals. Any Steiner tree that has one that is no longer without non- terminal vertices of degree 1 and 2. A tree with r leaves and internal vertices of degree at least 3 has at most r – 2 internal vertices.
Polynomial time algorithm for k = O(1) terminals: Enumerate all sets W of at most k – 2 non-terminals in G. For each W, find a minimum spanning tree in the distance network D(NW). Take the best over all these solutions Takes polynomial time for fixed k = O(1).
10 Steiner Tree – Algorithms and Networks O*(2K) ALGORITHM BY INCLUSION/EXCLUSION
11 Some background on the algorithm
Algorithm invented by Jesper Nederlof. Just after he finished his Master thesis supervised by Hans (and a little bit by me). Master thesis on Inclusion/Exclusion algorithms.
12 A Recap: Inclusion/Exclusion Formula
General form of the Inclusion/Exclusion formula:
Let N be a collection of objects (anything). Let 1,2, ...,n be a series of requirements on objects. Finally, let for a subset W µ {1,2,...,n}, N(W) be the number of objects in N that do not satisfy the requirements in W.
Then, the number of objects X that satisfy all requirements is: |W| X (1) N(W) W {1,2,...,n}
13 The Inclusion/Exclusion formula: Alternative proofs
|W| X (1) N(W) W {1,2,...,n}
Various ways to prove the formula. 1. See the formula as a branching algorithm branching on a requirement: required = optional – forbidden
2. If an object satisfies all requirements, it is counted in N(). If an object does not satisfy all requirements, say all but those in a set W’, then it is counted in all W µ W’ With a +1 if W is even, and a -1 if W is odd. W’ has equally many even as odd subsets: total contribution is 0.
14 Using the Inclusion/Exclusion Formula for Steiner Tree (problematic version)
|W| One possible approach: X (1) N(W) Objects: trees in the graph G. W {1,2,...,n} Requirements: contain every terminal. Then we need to compute 2k times the number of trees in a subgraph of G. For each W µ N, compute trees in G[V\W].
However, counting trees is difficult: Hard to keep track of which vertices are already in the tree. Compare to Hamiltonian Cycle: We want something that looks like a walk, so that we do not need to remember where we have been.
15 Branching Walks
Definition: Branching walk in G=(V,E) is a tuple (T,Á): Ordered tree T. Mapping Á from nodes of T to nodes of G, s.t. for any edge {u,v} in the tree T we have that {Á(u),Á(v)} 2 E.
The length of a branching walk is the number of edges in T. When r is the root of T, we say that the branching walk starts in Á(r) 2 V. For any n 2 T, we say that the branching walk visits all vertices Á(n) 2 V.
Some examples on the blackboard...
16 Branching Walks and Steiner Tree
Definition: Branching walk in G=(V,E) is a tuple (T,Á): Ordered tree T. Mapping Á from nodes of T to nodes of G, s.t. for any edge {u,v} in the tree T we have that {Á(u),Á(v)} 2 E.
Lemma: Let s 2 N a terminal. There exists a Steiner tree T in G with at most c edges, if and only if, there exists a branching walk of length at most c starting in s visiting all terminals N.
17 Using the Inclusion/Exclusion Formula for Steiner Tree
|W| Approach: X (1) N(W) Objects: branching walks from W {1,2,...,n} some s 2 N of length c in the graph G. Requirements: contain every terminal in N\{s}. We need to compute 2k-1 times the number of branching walks of length c in a subgraph of G. For each W µ N\{s}, compute branching walks from s in G[V\W].
Next: how do we count branching walks? Dynamic programming (similar to ordinary walks).
18 Counting Branching Walks
Let BW(v,j) be the number of branching walks of length j starting in v in G[W].
BW(v,0) = 1 for any vertex v. B (v,j) = B (u,j ) B (v,j ) W u2(N(v)ÅW) j1 + j2 = j-1 W 1 W 2 j2 = 0 covers the case where we do not branch / split up and walk to vertex u.
Otherwise, a subtree of size j1 is created from neighbour u, while a new tree of size j2 is added starting in v. This splits off one branch, and can be repeated to split of more branches.
We can compute BW(v,j) for j = 0,1,2,....,t. All in polynomial time.
19 Putting It All Together
|W| Algorithm: X (1) N(W) Choose any s 2 N. W {1,2,...,n} For t = 1, 2, … Use the inclusion/exclusion formula to count the number of branching walks from s of length t visiting all terminals N. This results in 2k-1 times counting branching walks from s of length c in G[V\W]. If this number is non-zero: stop the algorithm and output that the smallest Steiner tree has size t.
20 Steiner Tree – Algorithms and Networks THAT’S ALL FOLKS…
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