 ### Challenge from 9. December

Superfluous Presents

Author: Max Klimm

Challenge:

In order to facilitate the transport of presents from the North Pole, a pipeline network from the North Pole (N) to the continents of Africa (F), Antarctica (T), Asia (A), Australia (U), Europe (E), North America (M) and South America (S) was started to be constructed already in 1547. Before their transport, the presents are transformed into the gaseous phase by a secret process at the North Pole. Then, they are transported via the illustrated pipeline network to the respective continents, where they are condensed to their original state of aggregation. On Christmas Eve, a gift volume of 1 unit is to be discharged from the network at each of the continents (except at the North Pole N). Accordingly, a gift volume of 7 units needs to be fed in at the North Pole. In pipeline networks, moving gases approximately satisfy the so-called Weymouth equations. These equations state that for each pipe between two nodes, the square of the gas flow f is equal to the difference of the squares of the pressures at the respective end nodes, i. e.

f2 = p+2 - p-2,
where the gas flows from the node with higher pressure p+ in the direction of the node with lower pressure p-.

For example, suppose that there is a pressure of 5 at node E, a pressure of 4 at node M, and a pressure of 3 at node S. Then, √(52 - 42) = 3 units move from E to M; as well as √(42 - 32) = √7 ≈ 2.65 units move from M to S. Consequently, 3 - √7 units are discharged at point M: Help Santa Claus to find correct pressures for all nodes A, E, F, M, N, S, T, and U such that at each node (except at N) exactly one unit is discharged, at N exactly seven units are fed in, and all edges in the network satisfy the Weymouth equations. Furthermore, at U, the pressure should be zero.

Which of the following answers is correct?

Artwork: Julia Nurit Schönnagel 1. The pressure at node E is √1.

2. The pressure at node E is √2.

3. The pressure at node E is √3.

4. The pressure at node E is √4.

5. The pressure at node E is √5.

6. The pressure at node E is √6.

7. The pressure at node E is √7.

8. There exists no solution, such that all requirements are fulfilled.

9. Another node in the network has the same pressure as is present at node E.

10. Deleting the edge between A and F and computing a new solution on this altered graph, will yield an increased pressure at the node E.