# 16-3 Acyclic subgraphs

a.Thefor an undirected graph $G = (V, E)$ is a $|V| \times |E|$ matrix $M$ such that $M_{ve} = 1$ if edge $e$ is incident on vertex $v$, and $M_{ve} = 0$ otherwise. Argue that a set of columns of $M$ is linearly independent over the field of integers modulo $2$ if and only if the corresponding set of edges is acyclic. Then, use the result of Exercise 16.4-2 to provide an alternate proof that $(E, \mathcal I)$ of part (a) is a matroid.incidence matrix

b.Suppose that we associate a nonnegative weight $w(e)$ with each edge in an undirected graph $G = (V, E)$. Give an efficient algorithm to find an acyclic subset of $E$ of maximum total weight.

c.Let $G(V, E)$ be an arbitrary directed graph, and let $(E, \mathcal I)$ be defined so that $A \in \mathcal I$ if and only if $A$ does not contain any directed cycles. Give an example of a directed graph $G$ such that the associated system $(E, \mathcal I)$ is not a matroid. Specify which defining condition for a matroid fails to hold.

d.Thefor a directed graph $G = (V, E)$ with no self-loops is a $|V| \times |E|$ matrix $M$ such that $M_{ve} = -1$ if edge $e$ leaves vertex $v$, $M_{ve} = 1$ if edge $e$ enters vertex $v$, and $M_{ve} = 0$ otherwise. Argue that if a set of columns of $M$ is linearly independent, then the corresponding set of edges does not contain a directed cycle.incidence matrix

e.Exercise 16.4-2 tells us that the set of linearly independent sets of columns of any matrix $M$ forms a matroid. Explain carefully why the results of parts (d) and (e) are not contradictory. How can there fail to be a perfect correspondence between the notion of a set of edges being acyclic and the notion of the associated set of columns of the incidence matrix being linearly independent?

**a.** First, suppose that a set of columns is not linearly independent over $\mathbb F_2$ then, there is some subset of those columns, say $S$ so that a linear combination of $S$ is $0$. However, over $\mathbb F_2$, since the only two elements are $1$ and $0$, a linear combination is a sum over some subset.

Suppose that this subset is $S'$, note that it has to be nonempty because of linear dependence. Now, consider the set of edges that these columns correspond to. Since the columns had their total incidence with each vertex $0$ in $\mathbb F_2$, it is even. So, if we consider the subgraph on these edges, then every vertex has a even degree. Also, since our $S'$ was nonempty, some component has an edge. Restrict our attention to any such component. Since this component is connected and has all even vertex degrees, it contains an Euler Circuit, which is a cycle.

Now, suppose that our graph had some subset of edges which was a cycle. Then, the degree of any vertex with respect to this set of edges is even, so, when we add the corresponding columns, we will get a zero column in $\mathbb F_2$. Since sets of linear independent columns form a matroid, by problem 16.4-2, the acyclic sets of edges form a matroid as well.

**b.** One simple approach is to take the highest weight edge that doesn't complete a cycle. Another way to phrase this is by running Kruskal's algorithm (see Chapter 23) on the graph with negated edge weights.

**c.** Consider the digraph on [3] with the edges $(1, 2), (2, 1), (2, 3), (3, 2), (3, 1)$ where $(u, v)$ indicates there is an edge from $u$ to $v$. Then, consider the two acyclic subsets of edges $B = (3, 1), (3, 2), (2, 1)$ and $A = (1, 2), (2, 3)$. Then, adding any edge in $B - A$ to $A$ will create a cycle. So, the exchange property is violated.

**d.** Suppose that the graph contained a directed cycle consisting of edges corresponding to columns $S$. Then, since each vertex that is involved in this cycle has exactly as many edges going out of it as going into it, the rows corresponding to each vertex will add up to zero, since the outgoing edges count negative and the incoming vertices count positive. This means that the sum of the columns in $S$ is zero, so, the columns were not linearly independent.

**e.** There is not a perfect correspondence because we didn't show that not containing a directed cycle means that the columns are linearly independent, so there is not perfect correspondence between these sets of independent columns (which we know to be a matriod) and the acyclic sets of edges (which we know not to be a matroid).