OCR D1 2005 January — Question 2 5 marks

Exam BoardOCR
ModuleD1 (Decision Mathematics 1)
Year2005
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGraph Theory Fundamentals
TypeDegree sum calculations
DifficultyModerate -0.8 This is a straightforward application of the handshaking lemma (sum of degrees = 2×edges) requiring basic arithmetic, followed by recall of Eulerian definitions and a simple logical argument about connectivity. All parts are standard D1 textbook exercises with no novel problem-solving required.
Spec7.02g Eulerian graphs: vertex degrees and traversability7.02h Hamiltonian paths: and cycles

2
  1. A graph has six vertices; two are of order 3 and the rest are of order 4. Calculate the number of arcs in the graph, showing your working.
  2. Is the graph Eulerian, semi-Eulerian or neither? Give a reason to support your answer. A simple graph is one in which any two vertices are directly connected by at most one arc and no vertex is directly connected to itself. A connected graph is one in which every vertex is connected, directly or indirectly, to every other vertex.
  3. Explain why a simple graph with six vertices, two of order 3 and the rest of order 4, must also be a connected graph.

2 (i) A graph has six vertices; two are of order 3 and the rest are of order 4. Calculate the number of arcs in the graph, showing your working.\\
(ii) Is the graph Eulerian, semi-Eulerian or neither? Give a reason to support your answer.

A simple graph is one in which any two vertices are directly connected by at most one arc and no vertex is directly connected to itself.

A connected graph is one in which every vertex is connected, directly or indirectly, to every other vertex.\\
(iii) Explain why a simple graph with six vertices, two of order 3 and the rest of order 4, must also be a connected graph.

\hfill \mbox{\textit{OCR D1 2005 Q2 [5]}}