= Solution
Four <closed intervals> suffice:
$$
I_1=[0,1],\qquad I_2=[1,3],\qquad
I_3=[2,4],\qquad I_4=[4,5].
$$
No time belongs to three intervals, so the maximum number of attendees at one meeting is two. Strategy 1 is allowed to choose $t=5/2$, which meets $I_2$ and $I_3$. The remaining intervals $I_1$ and $I_4$ are disjoint, requiring two further meetings. This run therefore uses three meetings.
However, times $1$ and $4$ meet everyone: $1\in I_1\cap I_2$ and $4\in I_3\cap I_4$. One meeting is impossible because $I_1\cap I_4=\varnothing$. Thus \b[the optimum is two, while strategy 1 can use three]. The counterexample deliberately uses an allowed tie between maximum-attendance choices: the instruction to pick “any” such time would require every tie choice to be safe.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/functional-main-interval-meetings.png]
{title=Four availability intervals: a permitted maximum-attendance choice at 2.5 leaves two extra meetings, while the endpoints 1 and 4 suffice}
Back to article page
