Best Party

Thanks to @Calvin Lin for this simple proof.

This is to prove that:

The middle house is always the best house to party --- in terms of total distance walked by the party goers.

Let there be an odd number of houses, $H_1, H_2, H_3, \cdots, H_n$ . First consider the two end houses $H_1$ and $H_n$ . We note that the total distance the two neighbors need to walk is the distance between them. Similarly for $H_2$ and $H_{n-1}$ , the total distance the two neighbors walk is the distance between $H_2$ and $H_{n-1}$ if the party is held in a neighbor's house in between their houses. If the party is held in $H_1$ or $H_2$ , the two neighbors of $H_2$ and $H_{n-1}$ need to walk an extra distance of twice the distance between $H_1$ and $H_2$ or $H_{n-1}$ and $H_n$ respectively. For the same reason, the shortest distances walked by the pair of neighbors $H_3$ and $H_{n-2}$ , $H_4$ and $H_{n-3}$ and so on are minimum if the party is held in house in between their houses. It is obvious that the overall distance walked by all the pairs of neighbors is minimum if the party is held in the house in the middle. In this problem the middle house is the $\color{#20A900}\boxed{\text{Green}}$ House. The minimum total distance walked is the sum of the distances between the paired houses.

For even number of houses either one of the two houses in the middle can hold the party to give the minimum total distance walked.

It has to be always the middle house. If you move the party from not the middle house to the next one being more in the middle you save the distance between those 2 houses for one more man.

If you change from the yellow to the green brown and yellow must go a longer distance but pink green and blue a shorter. So it is 2 to 3.

Since distances are all given. Simply calculate total distances and you'll find that Green house has lowest total distance.