Graphs with prescribed degree sequences. Given a list of n...

Graphs with prescribed degree sequences. Given a list of $n$ positive integers $d_{1}, d_{2}, \ldots, d_{n},$ we want to efficiently determine whether there exists an undirected graph $G=(V, E)$ whose nodes have degrees precisely $d_{1}, d_{2}, \ldots, d_{n} .$ That is, if $V=\left\{v_{1}, \ldots, v_{n}\right\},$ then the degree of $v_{i}$ should be exactly $d_{i} .$ We call $\left(d_{1}, \ldots, d_{n}\right)$ the degree sequence of $G .$ This graph $G$ should not contain self-loops (edges with both endpoints equal to the same node) or multiple edges between the same pair of nodes. (a) Give an example of $d_{1}, d_{2}, d_{3}, d_{4}$ where all the $d_{i} \leq 3$ and $d_{1}+d_{2}+d_{3}+d_{4}$ is even, but for which no graph with degree sequence $\left(d_{1}, d_{2}, d_{3}, d_{4}\right)$ exists. (b) Suppose that $d_{1} \geq d_{2} \geq \cdots \geq d_{n}$ and that there exists a graph $G=(V, E)$ with degree sequence $\left(d_{1}, \ldots, d_{n}\right) .$ We want to show that there must exist a graph that has this degree sequence and where in addition the neighbors of $v_{1}$ are $v_{2}, v_{3}, \ldots, v_{d_{1}+1} .$ The idea is to gradually transform $G$ into a graph with the desired additional property. i. Suppose the neighbors of $v_{1}$ in $G$ are not $v_{2}, v_{3}, \ldots, v_{d_{1}+1} .$ Show that there exists $i < $ $j \leq n$ and $u \in V$ such that $\left\{v_{1}, v_{i}\right\},\left\{u, v_{j}\right\} \notin E$ and $\left\{v_{1}, v_{j}\right\},\left\{u, v_{i}\right\} \in E$. ii. Specify the changes you would make to $G$ to obtain a new graph $G^{\prime}=\left(V, E^{\prime}\right)$ with the same degree sequence as $G$ and where $\left(v_{1}, v_{i}\right) \in E^{\prime}$. iii. Now show that there must be a graph with the given degree sequence but in which $v_{1}$ has neighbors $v_{2}, v_{3}, \ldots, v_{d_{1}+1}$. (c) Using the result from part (b), describe an algorithm that on input $d_{1}, \ldots, d_{n}$ (not necessarily sorted) decides whether there exists a graph with this degree sequence. Your algorithm should run in time polynomial in $n$ and in $m=\sum_{i=1}^{n} d_{i}$.

Graphs with prescribed degree sequences. Given a list of $n$ positive integers $d_{1}, d_{2}, \ldots, d_{n},$ we want
to efficiently determine whether there exists an undirected graph $G=(V, E)$ whose nodes have degrees precisely $d_{1}, d_{2}, \ldots, d_{n} .$ That is, if $V=\left\{v_{1}, \ldots, v_{n}\right\},$ then the degree of $v_{i}$ should be exactly $d_{i} .$ We call $\left(d_{1}, \ldots, d_{n}\right)$ the degree sequence of $G .$ This graph $G$ should not contain self-loops (edges with both endpoints equal to the same node) or multiple edges between the same pair of nodes.
(a) Give an example of $d_{1}, d_{2}, d_{3}, d_{4}$ where all the $d_{i} \leq 3$ and $d_{1}+d_{2}+d_{3}+d_{4}$ is even, but for which no graph with degree sequence $\left(d_{1}, d_{2}, d_{3}, d_{4}\right)$ exists.
(b) Suppose that $d_{1} \geq d_{2} \geq \cdots \geq d_{n}$ and that there exists a graph $G=(V, E)$ with degree sequence $\left(d_{1}, \ldots, d_{n}\right) .$ We want to show that there must exist a graph that has this degree sequence and where in addition the neighbors of $v_{1}$ are $v_{2}, v_{3}, \ldots, v_{d_{1}+1} .$ The idea is to gradually transform $G$ into a graph with the desired additional property.
i. Suppose the neighbors of $v_{1}$ in $G$ are not $v_{2}, v_{3}, \ldots, v_{d_{1}+1} .$ Show that there exists $i < $ $j \leq n$ and $u \in V$ such that $\left\{v_{1}, v_{i}\right\},\left\{u, v_{j}\right\} \notin E$ and $\left\{v_{1}, v_{j}\right\},\left\{u, v_{i}\right\} \in E$.
ii. Specify the changes you would make to $G$ to obtain a new graph $G^{\prime}=\left(V, E^{\prime}\right)$ with the same degree sequence as $G$ and where $\left(v_{1}, v_{i}\right) \in E^{\prime}$.
iii. Now show that there must be a graph with the given degree sequence but in which $v_{1}$ has neighbors $v_{2}, v_{3}, \ldots, v_{d_{1}+1}$.
(c) Using the result from part (b), describe an algorithm that on input $d_{1}, \ldots, d_{n}$ (not necessarily sorted) decides whether there exists a graph with this degree sequence. Your algorithm should run in time polynomial in $n$ and in $m=\sum_{i=1}^{n} d_{i}$.

Key Concepts

Degree Sequence

A degree sequence is a list of nonnegative integers that represent the degrees of the nodes in a graph. In graph theory, determining whether a degree sequence is graphical—that is, whether there exists a simple graph that realizes the sequence—is a fundamental problem. The degree sequence encapsulates basic properties of a graph’s structure, and many necessary conditions (such as the even sum condition) and sufficient theorems (like Erdos–Gallai) have been developed to address this problem.

Even Sum Condition

The even sum condition is a necessary criterion for a degree sequence to be graphical. Since each edge in an undirected graph contributes exactly two to the sum of the vertex degrees, the overall sum must be even. This fundamental observation is often the first check when determining the viability of constructing a graph from a given degree sequence.

Graph Realizability Problem

The graph realizability problem asks whether there exists a simple graph that has a given degree sequence. This challenge is central in both theoretical and computational graph theory. It comes with various necessary and sufficient conditions—most notably the Erdos–Gallai theorem and the Havel-Hakimi algorithm—which provide methods to test and construct such graphs in a systematic way.

Havel-Hakimi Algorithm

The Havel-Hakimi algorithm is a greedy procedure used to determine if a given degree sequence is graphical. It operates by repeatedly removing the vertex with the highest degree and reducing the degrees of the next largest vertices accordingly. This iterative process not only provides a decision on the realizability of the degree sequence but also effectively constructs a corresponding graph if the sequence is graphical.

Edge Switching / Graph Transformations

Edge switching or graph transformation techniques allow for the modification of an existing graph’s structure while preserving the original degree sequence. By carefully swapping edges and non-edges, one can adjust the graph to obtain desirable properties, such as ensuring a particular vertex is adjacent to a specific set of vertices. This method is key in proving that a graph can be rearranged to satisfy additional constraints while maintaining its prescribed degree sequence.

Algorithmic Graph Construction

Algorithmic graph construction refers to the systematic process of using polynomial-time procedures to decide the realizability of degree sequences and to construct corresponding graphs if they exist. Algorithms like Havel-Hakimi not only serve as decision procedures but also exhibit the practical aspect of constructing graphs with specific structures and properties, which is essential in many applied contexts.

Transcript

00:02 We are given families of graphs and we are asked to determine whether the graph family is sparse, dense, or neither.

00:21 So in part a, we're given the family of graphs kn.

00:33 Now we know that kn is a simple graph with n vertices and an edge between two vertices.

00:48 And so it follows by the handshaking theorem that two times the number of edges is going to be sum over the number of edges.

00:58 Of vertices of the degree of a vertices, which is going to be the sum from i equals 1 to n, and the degree of each vertex is going to be n minus 1.

01:17 So this is n times n minus 1.

01:20 So the number of edges is going to be n times n minus 1 over e, mean over 2.

01:28 And therefore it follows that for each n, we have the density of kn, is equal to two times the number of edges.

01:46 So two times n times n minus 1 over 2 over the number of vertices, which is n times number of vertices minus 1, n minus 1.

02:01 And this is simply equal to 1.

02:10 And so therefore, it follows that the limit, as n approaches positive infinity on the density of kn, is simply the limit as n approaches infinity of 1.

02:28 And since 1 is a constant, this is simply 1 again.

02:36 And so we've shown that the limit of the density approaches a positive real number.

02:47 And so it follows that the family of graphs is dense.

03:00 In part b, we are given the family of graphs cn.

03:07 So wheels with n vertices, or cycles with n vertices, i mean.

03:14 We have that the number of vertices is by definition n.

03:17 And we have the number of edges, which we could prove, but as easy to see also, is going to be n.

03:28 And so it follows that the density for this graph, since it's a simple graph, is well defined, and is given by two times the number of edges, which is n, over number of vertices n times the number of vertices minus 1, n, which is 2.

03:46 2 over n minus 1.

03:48 And we see that this expression goes to 0 as n approaches infinity.

04:01 And so it follows that the family cycles is sparse, since the limit is 0.

04:19 In part c, we are given the family of graphs, knn.

04:26 These are complete bipartite graphs with two vertices, two vertex sets each with n vertices.

04:38 So we have v1 has n vertices and v2 has n vertices.

04:44 And of course, v1 and v2 are disjoint.

04:46 So it follows that total number of vertices v is 2n.

04:56 And to find the number of edges, notice that two times the number of edges is equal to the sum overall vertices in v of the degree of v, which is the same as the sum of all vertices in v1, plus the sum of all vertices in v2, the degree of v.

05:30 And notice that if a vertex belongs to v1, since this is a complete bipartite graph, it's going to be linked to n vertices in v2.

05:49 Equal to the sum and there are n vertices in v1, from i equals 1 to n of n plus the sum from i equals 1 to n again of n.

06:04 This is because if you have a vertex in v2, well there's any of these vertices and each vertex is connected to the n vertices in v1.

06:30 And so it follows of the number of edges.

06:33 Well, this is going to be n -sortices.

06:35 Squared plus n squared, which is 2n squared.

06:39 So the number of edges is n squared.

06:44 And so it follows that the density, since this is a simple graph, is well defined for each n, and is given by two times number of edges.

06:56 So 2 times n squared over the number of vertices, 2n, times the number of vertices minus 1, 2n minus 1.

07:10 So we have in the numerator, while we have 2n squared, in the denominator we have 4 n squared minus 2n.

07:22 And if you're pretty familiar with calculus, this is an easy limit, but just to elaborate, if we divide the top and bottom by n squared, we get 2 over 4 minus 2 over n...

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