Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Functional programming, design and analysis of algorithms / 8 / c

Table of contents
- Solution c

Solution

The earliest end is

1

, so strategy 2 first meets at $t_{1} = 1$ , removing

I_{1}

and

I_{2}

. Among the remaining intervals the earliest end is

4

, so it next meets at $t_{2} = 4$ , removing

I_{3}

and

I_{4}

. This is the two-meeting optimum proved in part (b). The closed endpoints are essential: attendance at exactly an interval's endpoint is permitted.

 Ancestors (11)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version