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.

One-way city

Forty-five intersections, I1 to I45, where I stands for intersection, numbered row by row from the top left with nine to a row. Most streets are one way, and the outer streets form a clockwise loop. Only the middle row (I19 to I27) and the middle column (I5, I14, I23, I32 and I41) go both ways, and each edge is the travel time in minutes.

A directed, weighted graph with 45 vertices and 88 edges.

With one warehouse at I23, where should three more open so that the longest drive to any intersection is as short as it can be made?

Open warehouses at I23, I2, I18 and I28, in that order. The longest drive is 16.

Step by step

No warehouse is open yet, so every drive time starts at ∞.
Open warehouse 1 at I23, which is given. A search from it gives every place it reaches a drive time. The search goes on to I14, I22, I24 and I32.
Take I14 from the queue. Warehouse 1 reaches it in 3, and nothing reaches it sooner. The search goes on to I5 and I13.
Take I32 from the queue. Warehouse 1 reaches it in 5, and nothing reaches it sooner. The search goes on to I33 and I41.
Take I22 from the queue. Warehouse 1 reaches it in 5, and nothing reaches it sooner. The search goes on to I21 and I31.
Take I24 from the queue. Warehouse 1 reaches it in 5, and nothing reaches it sooner. The search goes on to I15 and I25.

73 more steps lead to the last one:

Warehouse 4's search is done. 5 places moved to it. 4 warehouses are open. The farthest place from every warehouse is I7, 16 from warehouse 2, so the longest drive is 16. There are 148,995 ways to choose 4 of the 45 places, too many to try. Some roads here are one way, so the factor of two isn't guaranteed.

Use it in Go

centers, longest, err := path.KCenter(g, 4, "I23")
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