Vertices, Edges and Degree
Degree is the first number a network offers you, and it turns out to be tightly constrained. Add up the degrees of every vertex and the total can never be odd, which is a genuine theorem obtained by counting one thing in two ways.
Draw four dots and join some pairs. At each dot, count how many lines meet it. Now add those four counts. Compare your total with the number of lines you drew. What is the relationship?
The degree of a vertex is the number of edge-ends meeting it. Every edge has two ends, so adding the degrees counts every edge exactly twice. That gives a result that holds for every network without exception.
$$\sum \text{degrees} = 2 \times \text{number of edges}$$
One immediate consequence matters more than the result itself. Since the total is always even, the odd degrees must pair up, so the number of odd-degree vertices is always even. Lesson 9 turns that observation into a decision procedure.
Know
- the definitions of vertex, edge and degree, and that a loop contributes $2$ to the degree of its vertex
- that the sum of all the degrees in a network equals twice the number of edges
- that the number of vertices of odd degree in any network is even
Understand
- why the degree sum counts each edge twice, and why this makes the total necessarily even
- why odd-degree vertices must occur in pairs
- why the list of degrees can prove two networks different but cannot prove them the same
Can Do
- find the degree of every vertex in a given network, including networks with loops
- use the degree-sum relationship to find a missing degree or the number of edges
- compare two drawings and decide whether they represent the same network
A vertex is one of the objects in a network, drawn as a dot. An edge is a connection between two vertices, drawn as a line.
The degree of a vertex is the number of edge-ends that meet it. For an ordinary vertex with no loops that is simply the number of lines you can see arriving there.
In the network in the diagram, $A$ has three lines meeting it, so its degree is $3$. Vertices $B$, $C$ and $E$ each have two, and $D$ has three.
Degree belongs to a vertex, not to the network. A network has many degrees, one per vertex, and questions about "the degree" always name which vertex they mean.
Degrees are written as $\deg(A) = 3$ when a compact notation is wanted.
A loop joins a vertex to itself, so both of its ends arrive at the same place.
That means a loop contributes $2$ to the degree of its vertex, not $1$. A vertex with one ordinary edge and one loop has degree $3$.
This is not an arbitrary convention chosen to make a formula work. Imagine walking along the network: arriving at the vertex along the loop and leaving along it again uses the loop twice, once in each direction, and the degree is counting the ways in and out.
The phrase "edge-ends" in the definition is doing this work. Defining degree as "the number of edges meeting the vertex" would give a loop a count of $1$ and would break everything that follows, so the definition is written in terms of ends deliberately.
Here is the central result of the lesson, and it is proved by counting one quantity in two different ways.
Count all the edge-ends in a network. Every edge has exactly two ends, so if there are $e$ edges then there are $2e$ ends altogether.
Now count the same ends a second way, by going round the vertices and adding up how many arrive at each. That total is the sum of the degrees.
Both counts count the same collection of things, so they are equal:
$\text{sum of degrees} = 2e$
Checking on the diagram: the degrees are $3$, $2$, $2$, $3$, $2$, adding to $12$, and the network has $6$ edges, with $2 \times 6 = 12$. The result holds for every network, loops and multiple edges included, because the argument never assumed otherwise.
The degree sum is $2e$, which is even. That single fact has a consequence worth more than the formula.
Split the vertices into those of even degree and those of odd degree. The even ones contribute an even total, since a sum of even numbers is even.
So the odd ones must also contribute an even total, because the whole is even and even minus even is even. But a sum of odd numbers is even only when there is an even number of them.
Therefore: every network has an even number of odd-degree vertices. Never one, never three, never five.
This is a strong claim about every network that will ever be drawn, obtained from nothing but counting ends. It is also the fact that decides the Königsberg bridge problem in Lesson 10, so it is worth being convinced by the argument rather than merely remembering the statement.
Lesson 1 noted that a network can be drawn many ways. Degrees give a quick test for telling drawings apart.
If two drawings represent the same network, then each vertex keeps its degree, so the two lists of degrees must match. Different degree lists therefore prove the networks are different, and that test is fast.
The converse fails. Two networks can have identical degree lists and still be different, because the degrees say how many connections each vertex has but not which vertices they go to.
So matching degree lists mean "not yet ruled out", and settling the question then requires comparing the edge lists properly, as in Lesson 1.
This is a general pattern in mathematics worth naming: a quick check that can only ever disprove is still useful, provided you do not read a pass as a proof.
Watch Me Solve It · 3 examples
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1Count the edge-ends at each vertex$\deg(P) = 3, \quad \deg(Q) = 2, \quad \deg(R) = 3, \quad \deg(S) = 2$$P$ appears in $PQ$, $PR$ and $PS$; $Q$ in $PQ$ and $QR$; $R$ in $PR$, $QR$ and $RS$; $S$ in $PS$ and $RS$.
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2Add the degrees$3 + 2 + 3 + 2 = 10$
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3Compare with twice the edges$2e = 2 \times 5 = 10$The two agree, so the degrees have been counted correctly.
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4Count the odd vertices$P$ and $R$ have odd degree, which is two of them. An even number, as it must be.
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1Find the required total$\text{sum of degrees} = 2e = 2 \times 9 = 18$The degree sum is fixed by the number of edges, whatever the network looks like.
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2Add the known degrees$4 + 3 + 3 + 2 + 2 + 1 = 15$
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3Subtract$18 - 15 = 3$The seventh vertex has degree $3$.
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4Check the parityThe odd degrees are now $3$, $3$, $1$ and $3$, which is four of them. Even, as required. Had the answer produced an odd count of odd vertices, it would have been wrong.
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1Count at X$\deg(X) = 2$Two separate edges join $X$ to $Y$, and each contributes one end at $X$.
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2Count at Y$\deg(Y) = 3$Two ends from the two $XY$ edges and one from $YZ$.
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3Count at Z, remembering the loop$\deg(Z) = 1 + 2 = 3$One end from $YZ$, and both ends of the loop, giving $2$ more.
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4Verify$2 + 3 + 3 = 8 = 2 \times 4$There are $4$ edges, counting the two $XY$ edges separately and the loop as one edge. The sum matches, which it would not have done had the loop been counted as $1$.
Brain Trainer · 5 problems
Five items on degree. Use the degree sum as a check wherever you can.
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1 A network has edges $AB$, $AC$, $AD$. What is $\deg(A)$?
$A$ appears in all three edges.$3$ -
2 A network has $8$ edges. What is the sum of all its degrees?
Every edge has two ends.$16$ -
3 A vertex has one ordinary edge and one loop. What is its degree?
The loop contributes both of its ends to the same vertex.$3$ -
4 Can a network have exactly three vertices of odd degree?
Odd-degree vertices must pair up, because the degree sum is even.No -
5 The degrees of a network are $3$, $3$, $2$, $2$. How many edges does it have?
The sum is $10$, which equals $2e$.$5$
Multiple Choice · 5 questions
In a network with edges $PQ$, $QR$, $QS$, $QT$ and $RS$, the degree of $Q$ is:
A network has $11$ edges. The sum of the degrees of all its vertices is:
Which of these lists of degrees is impossible for a network?
A vertex has two ordinary edges and one loop. Its degree is:
Two networks are found to have the same list of degrees. This shows that they:
Short Answer · 3 questions
Degree counts edge-ends
The degree of a vertex is how many edge-ends meet it. A loop contributes $2$, because both of its ends arrive at the same vertex.
The sum is always twice the edges
$\sum \deg = 2e$, proved by counting the edge-ends two ways. It holds for every network, loops and repeated edges included.
Odd vertices pair up
Since the total is even, an even number of vertices must have odd degree. Never one, never three. Lesson 9 depends on this.
Degrees disprove, never prove
Different degree lists show two networks differ. Matching lists leave the question open, because degrees do not record where the edges go.
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