OCR MEI D2 2013 June — Question 1

Exam BoardOCR MEI
ModuleD2 (Decision Mathematics 2)
Year2013
SessionJune
TopicGraph Theory Fundamentals

1
  1. A graph is simple if it contains neither loops nor multiple arcs, ie none of the following:
    \includegraphics[max width=\textwidth, alt={}, center]{a09472cd-8f65-4cca-9683-c386053e66aa-2_81_134_219_1683}
    or
    \includegraphics[max width=\textwidth, alt={}, center]{a09472cd-8f65-4cca-9683-c386053e66aa-2_79_589_301_328} In an examination question, students were asked to describe in words when a graph is simple. Mark the following responses as right or wrong, giving reasons for your decisions if you mark them wrong.
    1. A graph is simple if there are no loops and if two nodes are connected by a single arc.
    2. A graph is simple if there are no loops and no two nodes are connected by more than one arc.
    3. A graph is simple if there are no loops and two arcs do not have the same ends.
    4. A graph is simple if there are no loops and there is at most one route from one node to another.
  2. The following picture represents a two-way switch
    \includegraphics[max width=\textwidth, alt={}, center]{a09472cd-8f65-4cca-9683-c386053e66aa-2_149_138_932_1119} It can either be in the up state
    \includegraphics[max width=\textwidth, alt={}, center]{a09472cd-8f65-4cca-9683-c386053e66aa-2_104_104_1128_790}
    or in the down state • or in the down state . Two switches can be used to construct a circuit in which changing the state of either switch changes the state of a lamp.
    \includegraphics[max width=\textwidth, alt={}, center]{a09472cd-8f65-4cca-9683-c386053e66aa-2_309_543_1448_762} Georgios tries to connect together three two-way switches so that changing the state of any switch changes the state of the lamp. His circuit is shown below. The switches have been labelled 1,2 and 3.
    \includegraphics[max width=\textwidth, alt={}, center]{a09472cd-8f65-4cca-9683-c386053e66aa-2_496_547_1946_760}
    1. List the possible combination of switch states and determine whether the lamp is on or off for each of them.
    2. Say whether or not Georgios has achieved his objective, justifying your answer.
  3. Use a truth table to show that \(( \mathrm { A } \wedge ( \mathrm { B } \vee \mathrm { C } ) ) \vee \sim ( \sim \mathrm { A } \vee ( \mathrm { B } \wedge \mathrm { C } ) ) \Leftrightarrow \mathrm { A }\).