Connection Concepts

Section 1.3 Connection Concepts

Definition 1.3.1.

A graph $H$ is a subgraph of a graph $G$ if and only if $H$ consists of a non-empty subset of the vertices of $G$ and a subset of the edges of $G .$

🔗

Definition 1.3.2.

A graph is a path, denoted $P_{n},$ if and only if the vertices can be labeled $v_{1}, v_{2}, v_{3}, \dots, v_{n}$ such that the only edges are $v_{1} v_{2}, v_{2} v_{3}, v_{3} v_{4}, \dots, v_{n - 1} v_{n}$ and no vertex is reused.

🔗

Definition 1.3.3.

A graph is a cycle, denoted $C_{n}$ if and only if the vertices can be labeled $v_{1}, v_{2}, v_{3}, \dots, v_{n}$ such that the only edges are $v_{1} v_{2}, v_{2} v_{3}, v_{3} v_{4}, \dots, v_{n - 1} v_{n}, v_{n} v_{1}$ and only $v_{1}$ is reused.

🔗

Definition 1.3.4.

A graph is a complete graph on $n$ vertices, denoted $K_{n}$ if and only if every vertex is adjacent to every other vertex.

🔗

Problem 1.3.5.

Draw $K_{4} .$ Identify three distinct paths (subgraphs) contained in $K_{4} .$

Solution.

🔗

Problem 1.3.6.

For each of your paths in Problem 1.3.5 draw the path, write the matrix for the path, and write the set notation for the path.

Solution.

Subgraph 1

Matrix notation $[\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{matrix}]$

Set notation

V = {A, B, C}

E = {AB, AC}

Subgraph 2

Matrix notation $[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}]$

Set notation

P = {A, C}

Q = {q}

$q ⟷ A, B$

Subgraph 3

Matrix notation $[\begin{matrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{matrix}]$

Set notation

V = {A, B, C, D}

E = {AB, BC, CD}

🔗

Problem 1.3.7.

Identify a path contained in $K_{4}$ using the maximum number of vertices.

Solution.

So by the definitions we can form the following graph.

Therefore the a path on $K_{4}$ that has the maximum number of vertices is:

P_{n} = v_{1} v_{2}, v_{2} v_{3}, v_{3} v_{4}

Keep in mind this is not the only possible path on $K_{4} .$

🔗

Problem 1.3.8.

Identify a path contained in $K_{4}$ using the maximum number of edges.

Solution.

$v_{1} v_{2}, v_{2} v_{3}, v_{3} v_{4}$ Note, this is the same as the answer to Problem 1.3.7.

🔗

Problem 1.3.9.

🔗

(a)

Find a maximum length path contained in Figure 1.3.10.

Solution.

Labeled diagram

$P_{5} = (V_{1} V_{3}, V_{3} V_{5}, V_{5} V_{6}, V_{6} V_{7})$

🔗

(b)

Next find a path that cannot be extended, but is shorter than your maximum length path. Note this is called a maximal path.

Solution.

$P_{3} = (V_{2} V_{3}, V 3_{V} 4)$

🔗
Figure 1.3.10. Maximum and Maximal Paths

🔗

Problem 1.3.11.

Draw $C_{6} .$ Identify two, distinct paths contained in $C_{6} .$

🔗

Problem 1.3.12.

Identify a path contained in $C_{6}$ containing the maximum number of vertices.

Solution.

$v_{1} v_{2}, v_{2} v_{3}, v_{3} v_{4}, v_{4} v_{5}, v_{5} v_{6} .$

🔗

Problem 1.3.13.

Identify a path contained in $C_{6}$ containing the maximum number of edges.

Solution.

$v_{1} v_{2}, v_{2} v_{3}, v_{3} v_{4}, v_{4} v_{5}, v_{5} v_{6}$

🔗

Lemma 1.3.14.

The number of vertices in a cycle equals the number of edges in that cycle.

🔗

Proof.

Let $C_{n}$ be a cycle with $n$ vertices. Since $C_{n}$ is a cycle, it's vertices can be labeled $v_{0}, v_{1}, \dots, v_{n - 1}$ so that the only edges in $C_{n}$ are of the form $v_{i} v_{i + 1},$ with addition performed modulo $n .$ Furthermore, every such edge exists in $C_{n} .$ Since there are $n$ such edges, the number of edges is equal to the number of vertices.

🔗

Problem 1.3.15.

In the graph in Figure 1.3.16 find three, distinct cycles.

🔗

Definition 1.3.17.

A graph is a complete bipartite graph, denoted $K_{n, m}$ if and only if the vertices can be partitioned into two, non-empty sets of size $n$ and $m$ such that no pair of vertices in a set is adjacent and any pair with one vertex from each set is adjacent.

🔗

Theorem 1.3.18.

$K_{n, m}$ contains no odd cycles.

🔗

Proof.

Direct Proof Definition 1.3.17 Means No edge is incident with two nodes in $K_{n} .$ and also No edge is incident with two nodes in $K_{m} .$ Definition 1.3.3 implies a node is incident to zero or two edges in a cycle. Because no edge is incident with two nodes in $K_{n}$ / $K_{m},$ Each edge must be incident with some node in $K_{n}$ and some node in $K_{m} .$ For any path In $K_{n, m} :$ if the ith edge starts in kn and ends in km, the i+1th edge starts in km and ends in kn if the ith edge starts in km and ends in kn, the i+1th edge starts in kn and ends in km It follows: For a path starting in n: the nth edge where n is even, ends in n. and the nth edge where n is odd, ends in m. A cycle contains a path starting and ending at the same node. which means it must start and stop in the same n/m. From above for a path to start and stop in the same n/m it must be have an even number of edges. Therefor: A cycle contained in $K_{n, m}$ must be even. Which Proves: $K_{n, m}$ contains no odd cycles.

🔗

Definition 1.3.19.

A graph is connected if and only if there exists a path from any vertex to any other vertex.

🔗

Problem 1.3.20.

Determine if the following graphs are connected. Explain.

🔗

(a)

Solution.

This graph IS connected. The graph is one “blob,” so from one vertex you can simply “hop” from one vertex to another to reach any other vertex in the graph.

🔗

(b)

Solution.

This graph is NOT connected. There is no edge connecting the vertices on the left to the vertices on the right. This means we cannot form a path from a vertex on the left to a vertex on the right.

🔗

(c)

$V = {v_{1}},$ $E = \emptyset$

Solution.

This graph IS connected. This graph is a single point, and since there are no other points on the graph that this point needs to be connected to, this graph is not disconnected.

🔗

(d)

$K_{4} .$

Solution.

This graph IS connected. Each vertex shares an edge with every other vertex as part of the definition of being a complete graph, so we can definitely form a path from any vertex to any other vertex.

🔗

(e)

$C_{5} .$

Solution.

This graph IS connected. With the ring shape, we can see that each vertex has a path from itself to any other point in as many as 2 edges.

🔗

(f)

Let $G$ be the graph whose vertices are English words. Two words are adjacent if and only if they share at least one letter.

Solution.

This graph IS connected. Each word shares a letter with at least one other word, which will be adjacent to many other words with the same letters, which will have new letters. This makes them adjacent to more words with more letters, and so on. There does not exist a word that shares no letters with any other words, so you must be able to make a path from any word to every other word.

🔗

(g)

Let $G$ be the graph whose vertices are months of the year. Two months are adjacent if and only if they have the same number of days. Ignore leap years.

Solution.

This graph is NOT connected. There are three distinct cliques, and members of one clique do not have a path to members of another clique.

🔗

(h)

Let $G$ be the graph whose vertices are all people who have ever lived. Two people are adjacent if and only if one is a biological child of the other. State your assumptions (belief systems may effect the answer).

Solution.

This graph is NOT connected. You can make a path from sibling to sibling, and cousin to cousin, etc. Everyone that connects is either directly related since they share an ancestor, or they are related by marriage. There will be very large groups where there is an ancestor thousands of years ago and many branches down, but these groups will not always connect to each other.

🔗

Lemma 1.3.21.

$K_{n}$ is connected for all $n .$

🔗

Proof.

The definition of a $K_{n}$ graph is that all vertices are adjacent to every other vertice. The degree of all vertices in the graph is n, if an edge were removed, a vertex would still be adjacent to n-1 vertices in the graph. To make the graph disconnected, n-n edges would have to be removed from a single vertex.

🔗

Proof.

We will show that any two vertices of $K_{n}$ have a path between them. Let $u$ and $v$ be vertices of $K_{n} .$ Then, by the definition of $K_{n},$ $u$ and $v$ are adjacent, so $u v$ is a path between them. Since $u$ and $v$ are arbitrary, this suffices to show that $K_{n}$ is connected.

🔗

Lemma 1.3.22.

$K_{n, m}$ is connected for all $n$ and $m .$

🔗

Proof.

Let $K_{n, m}$ be a complete bipartite. And let X,Y be vertices in $K_{n, m} .$

I will prove that $K_{n, m}$ is connected for all n,m. This means there exists a path between X,Y.

There are two cases:

Case 1: X,Y are in different sets.

Given the definition of a $K_{n, m}$ graph, there exisits a path XY.

Case 2: X,Y are in the same set.

Let Z be a vertex that is in a different set than X,Y. From case 1, Z is both adjacent to X and Y. Thus the path XZY exists and X,Y are connected.

🔗

Problem 1.3.23.

Consider $K_{3, 4} .$ Let ${a, b, c}$ and ${d, e, f, g}$ be the two sets of vertices. Find all paths from $a$ to $f .$

Solution.

1 edge: we can draw a path from a to f in 1 edge -- af.

2 edges: we cannot draw a path from a to f in 2 edges since we must start at a, go to d, e, f, or g, then back to b or c, and finally to f.

3: start: a; next vertex: d, e, or g; next: b or c; end: f. This gives us 6 options since a and f are fixed and there are 6 ways to rearrange the middle.

ad, db, bf
ad, dc, cf
ae, eb, bf
ae, ec, cf
ag, gb, bf
ag, gc, cf

4: we cannot draw a path with an even number of edges, as discussed earlier.

5: start: a; next: d, e, or g; next: b or c; next: d, e, or g; next: b or c; last: f. This gives us 3 scenarios moving forward.

Scenario 1: start: a next: d next: b or c next: e or g next: b or c last: f

ad, db, be, ec, cf
ad, db, bg, gc, cf
ad, dc, ce, eb, bf
ad, dc, cg, gb, bf

Scenario 2: start: a next: e next: b or c next: d or g next: b or c last: f

ae, eb, bd, dc, cf
ae, eb, bg, gc, cf
ae, ec, cd, db, bf
ae, ec, cg, gb, bf

Scenario 3: start: a next: g next: b or c next: d or e next: b or c last: f

ag, gb, bd, dc, cf
ag, gb, be, ec, cf
ag, gc, cd, db, bf
ag, gc, ce, eb, bf

Answer: 19 paths.

🔗

Problem 1.3.24.

What is the smallest number of edges that can be removed that will make $K_{3, 4}$ disconnected?

Solution.

Least edges to remove to make disconected is $min (n, m) .$ $min (3, 4) = 3 .$

🔗

Definition 1.3.25.

A graph is $r$ -connected if and only if the removal of $r$ edges is necessary to disconnect the graph.

🔗

Theorem 1.3.26.

Conjecture and prove $K_{n, m}$ is $r$ -connected. (You conjecture the value of r).

🔗

Proof.

Suppose G is a disconnected graph that has been formed by removing some number of edges in $K_{n, m}$ where $n \leq m .$ To construct G, select $x$ elements from $n$ and $y$ elements from $m .$

Beginning with $K_{n, m}$ (in this case $n = 3$ and $m = 4$ we select two vertices to be in $x$ (green) and two vertices to be in $y$ (red).

Delete every connection from the elements in $x$ to the elements of $m$ that are not part of $y$

Delete the connections from the vertices of $x$ to the vertices in $m$ that are not in $y .$

Now delete every connection from the elements in $y$ to the elements of $n$ that are not part of $x$

Delete the connections from the vertices of $y$ to the vertices in $n$ that are not in $x .$

This results in a graph that is disconnected.

The graph is not disconnected.

Since $m - y$ edges were deleted from each element of $x$ and $n - x$ edges were deleted from each element of $y,$ an equation can be constructed for the total number of edges deleted from the graph.

\begin{matrix} (1.3.1) & r = x \cdot (m - y) + y \cdot (n - x) \end{matrix}

Our desire is to find the smallest number of edge deletions needed to disconnect $K_{n, m} .$ To find this, we begin by searching the equation for an absolute minimum by taking $\frac{\partial r}{\partial x}$ and $\frac{\partial r}{\partial y}$ and attempting to find some values of $x$ and $y$ that can simultaneously set both of these to zero.

\begin{matrix} (1.3.2) & \frac{\partial r}{\partial x} = (m - y) \end{matrix}

\begin{matrix} (1.3.3) & \frac{\partial r}{\partial y} = (n - x) \end{matrix}

It is immediately apparent that any combination of $x = 0, x = n, y = 0, y = m$ will set both the expressions to zero. However, these combinations are not valid inputs, as it would not result in a disconnected graph. Since this is the case, the minimum value of $r$ must be achieved by some pair of boundary values for x and y.

Suppose $x = 1$ and $y = 0 :$ then $r = 1 \cdot (m - 0) + 0 \cdot (n - 1) = m$

Suppose $x = 0$ and $y = 1 :$ then $r = 0 \cdot (m - 1) + 1 \cdot (n - 0) = n$

Suppose $x = n$ and $y = 0 :$ then $r = n \cdot (m - 0) + 0 \cdot (n - n) = n m$

Suppose $x = 0$ and $y = m :$ then $r = 0 \cdot (m - m) + m \cdot (n - 0) = m n$

As we traverse along any boundary between these corner values, we can see that the output increases monotonically as we hold $x$ or $y$ constant at a value of one and increase the other. Therefore, since $n \leq m,$ the minimum number of edge deletions is $n$ and $K_{n, m}$ is r-connected with a value of $n$