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Chapter 3: Problem 22

Show that \(k\)-linked graphs are \((2 k-1)\)-connected. Are they even \(2 k\) connected?

Short Answer

Expert verified
k-linked graphs can be shown to be (2k-1)-connected. However, they are not necessarily 2k-connected.

Step by step solution

01

Identify k-linked graph

We consider a k-linked graph. It is given that for every collection of k independent pairs of vertices, there is a collection of paths, one from each pair, such that all these paths are pairwise vertex-disjoint.
02

Execute Removal

We seek to remove (2k-1) vertices from this graph and show that it remains connected. Begin by picking any (2k-1) vertices in the graph.
03

Show Connectivity

Since there are k-pairs and we've only removed (2k-1) vertices, there is at least one pair of vertices which both vertices remain in the graph after the removal (2k-1<(2k)). Thus, this undisturbed linkage ensures connectivity. So, the graph is (2k-1)-connected.
04

Check for 2k-connectivity

Now test by removing 2k vertices from the graph. This may or may not disconnect the graph since all the pairs could be affected. Hence, k-linked graphs are not necessarily 2k-connected.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

k-linked graphs
A k-linked graph is a special type of graph characterized by a unique connectivity property. It implies that for any selection of k pairs of vertices, there are k paths connecting each pair such that these paths do not share any common vertices. This concept is essential in graph theory as it indicates a strong form of connectivity.
In k-linked graphs, even when multiple vertices are removed, the graph maintains connectivity thanks to these independent paths. This property helps in proving connectivity levels like (2k-1)-connected, showcasing how removing up to (2k-1) vertices will leave the graph still connected.
Despite this, removing 2k vertices might disconnect the graph. Hence, k-linked graphs are not guaranteed to be 2k-connected.
connected graphs
Connected graphs form the cornerstone of graph theory. A graph is connected if there is a path between every pair of vertices. This interconnected nature of graphs ensures that no vertex is isolated and every part of the graph is accessible from any other part.
  • If a graph is not connected, it is considered disjoint with multiple components, each a smaller connected graph.
  • Understanding and proving connectivity helps in many applications such as network design, where each component needs to communicate with every other part seamlessly.
  • The concept of \(2k-1\)-connectivity in k-linked graphs further pushes the boundary of this connectivity principle.
Graph connectivity forms the backbone of robust systems in both theoretical studies and practical applications.
vertex-disjoint paths
Vertex-disjoint paths are a fascinating aspect of graph theory related to connectivity. Two or more paths are considered vertex-disjoint if they do not share any vertices, except possibly the starting and ending ones. This property helps in maintaining operations even if parts of the network experience failures.
  • In a k-linked graph, the existence of k vertex-disjoint paths for k independent pairs is a fundamental requirement.
  • This ensures that the network remains operational, as there are multiple independent routes available for communication.
  • Being able to find these paths is crucial in network routing and topology design.
It ensures that there is no single point of failure, making such graphs highly robust.
graph theory concepts
Graph theory is a vast field that explores connections and networks through mathematical structures. It involves studying vertices (nodes), edges (connections), and the structures they form. Understanding core concepts like k-linked graphs, connected graphs, and vertex-disjoint paths enhances comprehension of how networks are built and sustained.
Through graph theory:
  • Complex systems can be simplified into abstract graph models.
  • Connectivity properties can be analyzed, such as k-connectivity, where removing up to k vertices still ensures the graph stays connected.
  • Various algorithms can be applied to solve real-world problems, from transport systems to communication networks.
By delving into graph theory, students gain insights into practical applications that are part of everyday technological and social infrastructures.

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Most popular questions from this chapter

Show that \(X\) is minimal as an \(a-b\) separator if and only if every vertex in \(X\) has a neighbour in the component \(C_{a}\) of \(G-X\) containing \(a\), and another in the component \(C b\) of \(G-X\) containing \(b\).

Let \(X\) and \(X^{\prime}\) be minimal separators in \(G\) such that \(X\) meets at least two components of \(G-X^{\prime}\). Show that \(X^{\prime}\) meets all the components of \(G-X\), and that \(X\) moets all the components of \(G-X^{\prime}\).

For edges \(e, e^{\prime} \in G\) write \(e \sim e^{\prime}\) if either \(e=e^{\prime}\) or \(e\) and \(e^{\prime}\) lie on some common cycle in \(G\). Show that \(\sim\) is an equivalence relation on \(E(G)\) whose equivalence classes are the edge sets of the non-trivial blocks of \(G\).

Show that if \(G\) is \(k\)-linked and \(s_{1}, \ldots, s_{k}, t_{1}, \ldots, t_{k}\) are not neoessarily distinct vertios such that \(s_{i} \neq t_{i}\) for all \(i\), then \(G\) contains independent paths \(P_{i}=s_{i} \ldots t_{i}\) for \(i=1, \ldots, k\).

Prove Menger's theorem by induction on \(\|G\|\), as follows. Given an edge \(e=x y\), consider a smallest \(A-B\) separator \(S\) in \(G-e\). Show that the induction hypothesis implies a solution for \(G\) unless \(S \cup\\{x\\}\) and \(S \cup\\{y\\}\) are smallest \(A-B\) separators in \(G\). Then show that if choosing neither of these separators as \(X\) in the previous exercise gives a valid proof, there is only one easy case left to do.
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