1 The infamous fictional detective Agatha Parrot is working on a case and decides to invite each of the four suspects to visit her for tea. She will invite one suspect to tea on Sunday (S), another on Monday (M), another on Tuesday (T) and the final suspect on Wednesday (W).
- Beryl Batty (B) could come to tea on Sunday or Monday but she is away for a few days after that.
- Colonel Chapman (C) could come to tea on Sunday or Tuesday, but is busy on the other days.
- Dimitri Delacruz (D) could only come to tea on Monday or Wednesday.
- Erina El-Sayed (E) could come to tea on Sunday, Monday or Tuesday, but will be away on Wednesday.
- Draw a bipartite graph to show which suspects could come to tea on which day.
Agatha initially intended to invite Beryl to tea on Sunday, Dimitri on Monday and Colonel Chapman on Tuesday. However this would mean that Erina would not be able to come to tea, as she will be away on Wednesday.
Draw a second bipartite graph to show this incomplete matching.Working from this incomplete matching, find the shortest possible alternating path, starting at Erina, to achieve a breakthrough. Write down which suspect is invited on which day with this matching.
Before issuing invitations, Agatha overhears Erina making arrangements to go to tea with the vicar one day. She knows that Erina will not want to accept two invitations to tea on the same day. Unfortunately, Agatha does not know which day Erina has arranged to go to tea with the vicar, other than that it was not Sunday. Agatha invites Erina to tea on Sunday and the others when she can.
After having tea with all four suspects she considers what they have said and invites them all back on Friday. She then accuses the person who came to tea on Monday of being guilty.Who does Agatha accuse of being guilty?