k-center

Chooses k centers so that the vertex farthest from its nearest center is as near as it can make it.

Time complexity O(k (V+E) log V), gograph function path.KCenter.

How it works

Where should a company open k warehouses so that no address is a long drive from all of them? Trying every choice is far too slow on a city, so path.KCenter chooses them farthest first: it opens the first warehouse, then keeps opening the next one at the address farthest from every warehouse so far. Each new warehouse needs a search only over the addresses it gets closer to, and k rounds cost O(k (V+E) log V).

When every road works both ways in the same time, the longest drive this leaves is never more than twice the best any k warehouses can do. With one-way streets the drives aren't symmetric, and that factor of two isn't guaranteed.

The second warehouse often lands at the edge of the map. The method shrinks the worst drive, so it serves the most remote address first. That is the difference from k-median, which minimizes the average drive and puts warehouses nearer the busy middle.

Warehouses that are already open can be passed as fixed centers. They count toward k and open first, in the order given. Without them, the first center is the first vertex added to the graph. Which one opens first changes the rest.

School run

Six places on the morning drive to school. Most streets go both ways, and each edge is the travel time in minutes.

A directed, weighted graph with 6 vertices and 13 edges.

With one warehouse at Home, where should a second open so that the longest drive to any of the six places is as short as it can be made?

Open warehouses at Home and School, in that order. The longest drive is 5. The best of all 15 ways to choose 2 is 4.

Step by step

No warehouse is open yet, so every drive time starts at ∞.
Open warehouse 1 at Home, which is given. A search from it gives every place it reaches a drive time. The search goes on to Bakery and Bridge.
Take Bridge from the queue. Warehouse 1 reaches it in 3, and nothing reaches it sooner. The search goes on to Park.
Take Bakery from the queue. Warehouse 1 reaches it in 4, and nothing reaches it sooner. Park already has a drive at least as quick.
Take Park from the queue. Warehouse 1 reaches it in 5, and nothing reaches it sooner. The search goes on to Library and School.
Take Library from the queue. Warehouse 1 reaches it in 9, and nothing reaches it sooner. The search goes on to School.

4 more steps lead to the last one:

Warehouse 2's search is done. 1 place moved to it. 2 warehouses are open. The farthest place from every warehouse is Park, 5 from warehouse 1, so the longest drive is 5. Trying all 15 ways to choose 2 places gives a best of 4. Some roads here are one way, so the factor of two isn't guaranteed.

Use it in Go

centers, longest, err := path.KCenter(g, 2, "Home")
if err != nil {
	return err
}
for _, v := range centers {
	fmt.Println(v.Label())
}
fmt.Println(longest)

Example graphs

New York streets, 295 vertices and 669 edges
London streets, 274 vertices and 598 edges
San Francisco streets, 248 vertices and 548 edges
Rasht streets, 298 vertices and 568 edges
School run, 6 vertices and 13 edges
Downtown grid, 20 vertices and 60 edges
One-way city, 45 vertices and 88 edges
Metro region, 100 vertices and 346 edges