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Wednesday, October 11, 2017

A tale of two cities

In an island with 257 cities, there are 2018 railroad segments such that each segment connects exactly 2 cities, and each pair of cities are connected by at most 1 segment. A group of 10 cities is called "clustered" if each pair of those 10 is connected by a segment.

Prove that there exist 2 unconnected cities that we can connect without increasing the number of clustered groups.

Solution

Proof by contradiction. Suppose we cannot connect two cities without increasing the number of clustered groups.

Let $n=257$ be the number of cities. Take all possible groups of 10 cities, ${n \choose 10 }$ of them, and count how many segments there are among them. Of course, if the group is clustered, it should have ${10 \choose 2} = 45$ segments. Let $S$ be the sum of unconnected pairs, over all possible groups of 10 cities.

Now let us count $S$ in a different way. Take 2 connected cities $A$ and $B$. The segment $AB$ contributes to $S$ only when the chosen group of 10 cities include $AB$, so this segment is counted ${n-2 \choose 8}$ times. Because there are 2018 segments, then $S$ must be:

$$S = {n-2 \choose 8}.(2018)$$

Note that $2018 < 2020 = 8n-36$, so $S < {n-2 \choose 8}.(8n-26)$. That means, on average, there are $S / {n \choose 10 }$ segments.

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