OCR MEI D2 (Decision Mathematics 2) 2010 June

1

1. Mickey ate the last of the cheese. Minnie was put out at this. Mickey's defence was "There wasn't enough left not to eat it all". Let "c" represent "there is enough cheese for two" and "e" represent "one person can eat all of the cheese".
   1. Which of the following best captures Mickey's argument? $$\mathrm { c } \Rightarrow \mathrm { e } \quad \mathrm { c } \Rightarrow \sim \mathrm { e } \quad \sim _ { \mathrm { c } } \Rightarrow \mathrm { e } \quad \sim _ { \mathrm { c } } \Rightarrow \sim \mathrm { e }$$ In the ensuing argument Minnie concedes that if there's a lot left then one should not eat it all, but argues that this is not an excuse for Mickey having eaten it all when there was not a lot left.
   2. Prove that Minnie is right by writing down a line of a truth table which shows that $$( c \Rightarrow \sim e ) \Leftrightarrow ( \sim c \Rightarrow e )$$ is false.
      (You may produce the whole table if you wish, but you need to indicate a specific line of the table.)
2. 1. Show that the following combinatorial circuit is modelling an implication statement. Say what that statement is, and prove that the circuit and the statement are equivalent. \includegraphics[max width=\textwidth, alt={}, center]{c3a528e4-b5fe-4bff-a77e-e3199bb225a1-2_188_533_1272_767}
   2. Express the following combinatorial circuit as an equivalent implication statement. \includegraphics[max width=\textwidth, alt={}, center]{c3a528e4-b5fe-4bff-a77e-e3199bb225a1-2_314_835_1599_616}
   3. Explain why \(( \sim \mathrm { p } \wedge \sim \mathrm { q } ) \Rightarrow \mathrm { r }\), together with \(\sim \mathrm { r }\) and \(\sim \mathrm { q }\), give p .

2 The network is a representation of a show garden. The weights on the arcs give the times in minutes to walk between the six features represented by the vertices, where paths exist. \includegraphics[max width=\textwidth, alt={}, center]{c3a528e4-b5fe-4bff-a77e-e3199bb225a1-3_483_985_342_539}

1. Why might it be that the time taken to walk from vertex \(\mathbf { 2 }\) to vertex \(\mathbf { 3 }\) via vertex \(\mathbf { 4 }\) is less than the time taken by the direct route, i.e. the route from \(\mathbf { 2 }\) to \(\mathbf { 3 }\) which does not pass through any other vertices? The matrices shown below are the results of the first iteration of Floyd's algorithm when applied to the network.

   \cline { 2 - 7 } \multicolumn{1}{c|}{}	\(\mathbf { 1 }\)	\(\mathbf { 2 }\)	\(\mathbf { 3 }\)	\(\mathbf { 4 }\)	\(\mathbf { 5 }\)	\(\mathbf { 6 }\)
   \(\mathbf { 1 }\)	\(\infty\)	15	\(\infty\)	\(\infty\)	7	8
   \(\mathbf { 2 }\)	15	30	6	2	6	23

\(\mathbf { 3 }\) & \(\infty\) & 6 & \(\infty\) & 3 & \(\infty\) & \(\infty\) \hline

\(\mathbf { 4 }\) & \(\infty\) & 2 & 3 & \(\infty\) & 10 & 17
\hline

\(\mathbf { 5 }\) & 7 & 6 & \(\infty\) & 10 & 14 & 8
\hline