OCR D1 2006 June — Question 2 7 marks

Exam BoardOCR
ModuleD1 (Decision Mathematics 1)
Year2006
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGraph Theory Fundamentals
TypeGraph isomorphism
DifficultyModerate -0.8 This is a straightforward D1 question testing basic graph theory definitions. Part (i) requires constructing simple graphs with given degree sequence (routine but requires care), and part (ii) only needs recall of the Eulerian graph condition (all vertices must have even degree). No problem-solving or proof required, just application of standard definitions.
Spec7.02a Graphs: vertices (nodes) and arcs (edges)7.02b Graph terminology: tree, simple, connected, simply connected7.02g Eulerian graphs: vertex degrees and traversability

2
  1. Draw three mathematically different graphs, labelled graph \(A\), graph \(B\) and graph \(C\), each with four vertices, of orders 1, 3, 3 and 3, and five arcs.
  2. Explain how you know that none of the graphs from part (i) is Eulerian.

Question 2:
Part (i)
AnswerMarks Guidance
AnswerMarks Guidance
Graph \(A\), Graph \(B\), Graph \(C\) (as shown)M1, A1 [2] For a reasonable attempt; For a graph topologically equivalent to one of these graphs
Second graphM1, A1 [2] For a different reasonable attempt; For a graph topologically equivalent to one of these graphs
Third graphM1, A1 [2] For another different reasonable attempt; For a graph topologically equivalent to one of these graphs
Part (ii)
AnswerMarks Guidance
AnswerMarks Guidance
The graphs each have four odd nodes, but Eulerian graphs have no odd nodes.B1 [1] For any recognition that the nodes are not all even
# Question 2:

## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Graph $A$, Graph $B$, Graph $C$ (as shown) | M1, A1 [2] | For a reasonable attempt; For a graph topologically equivalent to one of these graphs |
| Second graph | M1, A1 [2] | For a different reasonable attempt; For a graph topologically equivalent to one of these graphs |
| Third graph | M1, A1 [2] | For another different reasonable attempt; For a graph topologically equivalent to one of these graphs |

## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| The graphs each have four **odd** nodes, but Eulerian graphs have no odd nodes. | B1 [1] | For any recognition that the nodes are not all even |

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2 (i) Draw three mathematically different graphs, labelled graph $A$, graph $B$ and graph $C$, each with four vertices, of orders 1, 3, 3 and 3, and five arcs.\\
(ii) Explain how you know that none of the graphs from part (i) is Eulerian.

\hfill \mbox{\textit{OCR D1 2006 Q2 [7]}}