Three pairs of edges cross in the drawing, which is enough to make a direct count unreliable. Add up the degrees instead: 3 + 3 + 2 + 2 + 2 + 2 = 14, so there are 7 edges, with no counting at all.
The degree sum and a direct count are two independent routes to the edge total, and only one of them can be miscounted.
| Word | What it means |
|---|---|
| Vertex | A point of the graph. Plural vertices. |
| Edge | A connection between two vertices. |
| Adjacent vertices | Two vertices joined by an edge. |
| Adjacent edges | Two edges that share a vertex. |
| Degree | How many edge-ends meet at a vertex. A loop counts twice. |
| Simple | No loops, and no two vertices joined more than once. |
| Complete | Every pair of vertices joined. Written Kn. |
| Weighted | Each edge carries a number: a distance, a cost, a time. |
| Directed | Edges have arrows. Then each vertex has an in degree and an out degree. |
| Connected | Every vertex can be reached from every other by some route. A graph in two separate pieces is not. |
| Strongly connected | For a DIRECTED graph: every vertex can be reached from every other following the arrows. A one-way system can be connected and not strongly connected. |
| Subgraph | Some of the vertices and some of the edges between them. |
| Tree | Connected, with no cycles. Always exactly n − 1 edges. |
| Cycle | A closed route that repeats no vertex except its start. |
None of this is drawn information. The same graph can be drawn with edges crossing or not, in a circle or in a line, and it is the same graph. What matters is which pairs are joined, which is why the next sub-topic writes it as a matrix.
Every edge has two ends. Adding up the degrees counts every edge exactly twice, once at each end. So:
sum of the degrees = 2 × (number of edges)
For the graph in the figure, with degrees 3, 3, 2, 2, 2, 2:
sum = 14, so edges = 14/2 = 7
and for the complete graph on 5 vertices, where every vertex is joined to the other four:
sum = 5 × 4 = 20, so edges = 10
In general Kn has n(n − 1)/2 edges: K4 has 6 and K6 has 15.
The degree sum is always even. It is twice something, so it has to be. And that gives you a test no drawing can:
Can a graph have degrees 3, 3, 3, 2, 2, 2?
The sum is 15, which is odd. So no: no such graph exists, and no amount of trying to draw one will help. It is the one kind of question here you can settle without looking at a graph at all.
Trees and cycles, counted. A tree on 6 vertices has exactly 5 edges. The figure's graph has 7, which is 2 more, and in a connected graph every edge beyond n − 1 forces a cycle. This one is connected, so it has 2 independent cycles, and you knew that before finding either of them. Here they are: A-B-C-A, and A-B-E-F-D-A. The connectedness matters: six vertices and seven edges split into a K₄ and a separate single edge would give 3 independent cycles instead, so count the pieces before you count the cycles.
Put an arrow on each edge and each vertex gets two numbers:
Every edge contributes one to somebody's in degree and one to somebody's out degree, so
sum of in degrees = sum of out degrees = number of edges
which for 7 directed edges is 7 each, not 14. The doubling is gone, because the two ends are now being counted in two different totals rather than in one.
The Graph menu on both machines plots functions. It knows nothing about vertices and edges, and there is no graph-theory mode on either. What the calculator is for in 3.14 is arithmetic and, from 3.15 onwards, matrix powers.
When you may use it. Applications. A calculator is allowed in every paper, and in this sub-topic it earns its place on exactly one job: n(n − 1)/2 for a large complete graph, where K20 having 190 edges is faster typed than reasoned.
The mark people lose. Counting edges off a drawing. On the figure's graph, three pairs of edges cross and the usual miscounts are 6 and 8. The degree sum takes the same information and halves it, and a miscounted degree shows up as an odd total, which tells you immediately that you have gone wrong. An odd degree sum is never the graph's fault; it is always yours.
1. A graph has degrees 3, 3, 2, 2, 2 and 2. What is the sum of the degrees?
2. How many edges does it have?
3. How many edges does the complete graph K5 have?
4. And K6?
5. Can a graph have degrees 3, 3, 3, 2, 2, 2?
1 markThe degrees, read off correctly.
1 markThe handshake lemma stated or used.
1 markThe edge count, or the impossibility with its reason.
On an impossibility question the reason is the mark. "No" on its own scores nothing; "the degree sum is odd and every degree sum is even" scores all of it.
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