OCR D1 2009 June — Question 2 9 marks

Exam BoardOCR
ModuleD1 (Decision Mathematics 1)
Year2009
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicGraph Theory Fundamentals
TypeImpossibility proofs for degree sequences
DifficultyEasy -1.2 This is a straightforward D1 question testing basic graph theory concepts. Part (i) requires simple recall of the handshaking lemma (sum of degrees must be even). Parts (ii) require drawing graphs with specified properties and applying definitions. Part (iii) is a guided proof of Ramsey's theorem R(3,3)=6, but the question provides substantial scaffolding through its structure. All parts involve standard textbook material with no novel problem-solving required.
Spec7.02b Graph terminology: tree, simple, connected, simply connected7.02d Complete graphs: K_n and number of arcs7.02g Eulerian graphs: vertex degrees and traversability

  1. Explain why it is impossible to draw a graph with four vertices in which the vertex orders are 1, 2, 3 and 3. [1]
A simple graph is one in which any two vertices are directly joined by at most one arc and no vertex is directly joined to itself. A connected graph is one in which every vertex is joined, directly or indirectly, to every other vertex. A simply connected graph is one that is both simple and connected.
    1. Draw a graph with five vertices of orders 1, 1, 2, 2 and 4 that is neither simple nor connected. [2]
    2. Explain why your graph from part (a) is not semi-Eulerian. [1]
    3. Draw a semi-Eulerian graph with five vertices of orders 1, 1, 2, 2 and 4. [1]
Six people (Ann, Bob, Caz, Del, Eric and Fran) are represented by the vertices of a graph. Each pair of vertices is joined by an arc, forming a complete graph. If an arc joins two vertices representing people who have met it is coloured blue, but if it joins two vertices representing people who have not met it is coloured red.
    1. Explain why the vertex corresponding to Ann must be joined to at least three of the others by arcs that are the same colour. [2]
    2. Now assume that Ann has met Bob, Caz and Del. Bob, Caz and Del may or may not have met one another. Explain why the graph must contain at least one triangle of arcs that are all the same colour. [2]

\begin{enumerate}[label=(\roman*)]
\item Explain why it is impossible to draw a graph with four vertices in which the vertex orders are 1, 2, 3 and 3. [1]
\end{enumerate}

A \textit{simple graph} is one in which any two vertices are directly joined by at most one arc and no vertex is directly joined to itself.

A \textit{connected graph} is one in which every vertex is joined, directly or indirectly, to every other vertex.

A \textit{simply connected graph} is one that is both simple and connected.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item 
\begin{enumerate}[label=(\alph*)]
\item Draw a graph with five vertices of orders 1, 1, 2, 2 and 4 that is neither simple nor connected. [2]

\item Explain why your graph from part (a) is not semi-Eulerian. [1]

\item Draw a semi-Eulerian graph with five vertices of orders 1, 1, 2, 2 and 4. [1]
\end{enumerate}
\end{enumerate}

Six people (Ann, Bob, Caz, Del, Eric and Fran) are represented by the vertices of a graph. Each pair of vertices is joined by an arc, forming a complete graph. If an arc joins two vertices representing people who have met it is coloured blue, but if it joins two vertices representing people who have not met it is coloured red.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item 
\begin{enumerate}[label=(\alph*)]
\item Explain why the vertex corresponding to Ann must be joined to at least three of the others by arcs that are the same colour. [2]

\item Now assume that Ann has met Bob, Caz and Del. Bob, Caz and Del may or may not have met one another. Explain why the graph must contain at least one triangle of arcs that are all the same colour. [2]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR D1 2009 Q2 [9]}}