neutron : Ronald the newt

electron : vote for Ron!

boson : ship's officer

quark : duck noise

photon : imitation ton

particle : average tickle

To reiterate this question, imagine we are rowing on a round pond 200 m across when suddenly fog descends and visibility is effectively zero. You are lost and unfortunately the fog fell too quickly for you to have any idea of your location relative to the pond's edge. What is the shortest distance you can row to guarantee reaching the edge of the pond and what path should you follow?

To begin with, any straight path of length 200 m must be a diameter of the pond. Therefore, if we choose an direction at random (we do not know where the shore lies relative to our location so one direction is as good as another) and travel in a straight line, we are certain to reach the bank after at most 200 m. So this is am escape path.

It's natural to wonder if some ingenious non-linear path might guarantee hitting the pond's edge in a shorter distance, but we can show this is impossible.

Here's the proof.

Suppose we can guarantee reach the shore in a distance of less than 200 m by following some path starting at A and ending at M and let M be the point half-way along the path, so the path length for M to A and to B are both less than 100 m.

Consider an arbitrary point X somewhere along the path. The distance along the path from M to X must be less than 100 m and the straight line distance from M to X must therefore also be less than 100 m. Since X was an arbitrary point it follows that every point on the path is less than 100 m from M and we can draw a circle with centre M and radius less than 100 m.

But this is smaller than the pond, which has a radius of 100 m! So it's possible for the whole of AB to lie within the interior of the pond and it therefore cannot form a guaranteed escape path.

In a comment, Steven McDonald already gave a nice solution to the puzzle here. My own solution is equivalent but framed a bit differently; I went for a more visual explanation.

Each of the rectangles we want to count is determined by some two rows and two columns of the 10x10 grid, as shown in this example:

For the rows we have 10 choices for the first, which leaves 9 choices for the second, but that will have double-counted each combination, because for example row 5 and row 9 represents the same pair as row 9 and row 5. So we must divide by 2 and there are

(10 x 9) / 2  = 45

distinct combinations.

Similar reasoning gives 45 of ways to choose two distinct columns and so we find there are 45 x 45 = 2025 different rectangles.

On this 10 x 10 square grid I have drawn some rectangles with their corners on points of the grid and their sides parallel to the sides of the square. The colours have no particular significance I just used them to make the picture more interesting.

There are obviously many other rectangles that can be drawn on this grid. Can you work out how many there are altogether? Answer on Tuesday.

"Your decision to lower the pitch of the doorbell by a semitone was pure genius", said Tom flatteringly.

Merriam-Webster defines it as: "an unrealistic enterprise or prospect of prosperity". What is the expression? The picture above is a clue.

—You've probably seen this puzzle, which goes back to at least 1913 and was not new then.

Three utility companies, Gas, Water and Electricity, want to each connect a supply to three houses, Nos. 1 and 2 and 3 Topology View. For technical reasons they do not want anyone's connect to cross another's. The sketch below shows a partial solution, in which 8 of the 9 connections required have been made without any crossings. 

But making the last connection from E to No. 1 seems impossible without a crossing, and indeed it is. A particularly elegant proof is given in this Wikipedia article.

But all of this assumes we are working on a flat plane. What if we change the setting so that suppliers and houses are on a giant donut aka as a torus, as shown below?

Well now we are on a torus we can take a path for E to No.1 that goes round the back of the torus, up through the hole in the middle, and connects to No.1, as indicated by the dotted line in the next sketch.

This configuration of two sets of three points (aka vertices), with lines (aka edges) between point in the first set and every point in the second is called the complete K_3,3 which is reasonably self-explanatory. It is one of two "special" configurations, the other being K₅ . K₅ also cannot be drawn in the plane without a crossing, as we see in the next sketch.

Although I have labelled it K₅ that is not quite accurate, because it isn't complete; it ought to have 10 edges. However similar to K_3,3 it is impossible to draw it in the plane without two edges crossing.

These two configurations are special because they let us answer the question "What, not necessarily complete, point and line configurations is it possible to draw in the plane without any crossing edges?" The answer is: exactly those configurations that contain no sub-configurations equivalent (in a fairly simple way but I'll skip the detail) to K_3,3 or K₅.

This is amazing, if you think about it: there are just these two forbidden sub-configurations and if they are avoided any point and line configuration whatsoever of the kinds we have been looking at can be drawn,"embedded", in the plane with no crossings.

Back to the torus. We have seen that K_3,3 can be embedded in the surface of a torus, and so can K₅ it turns out. In fact we can embed K_4,4 (so we could fit in broadband too!) and K₇. So we might ask: is there some set of forbidden configurations that let us say what can and cannot be embedded in the surface of a torus, similarly to the way we answered the question for the plane?

The answer is yes there is, but non-one can say yet how many forbidden sub-configurations there are. All that is known at present is that they number at least 17,523. That is the surprise alluded to in the title of this post. When I read that going from the plane to the torus involved a jump from 2 to 17,523 I was staggered!

And of course a torus has only one hole, but we could have 2, 3, 4,... holes and so on, so whatever happens there? It's mind-boggling!