# Sound and Fury

Signifying nothing

## Some thoughts, some musings and a ponder.

Here are a few oddities I have been thinking about in the past month.

When you are in a maze, if you stick to the left wall, always turning left when you are able, then you will eventually reach the end. Apparently. How would one go about proving something like this? What sort of mathematical resources could one make use of? Graph theory perhaps? I’ve no idea what that is, really, but it sounds like the sort of discipline which might help in these circumstances. Are there maze constructions where this strategy will not work?

Why do pound and euro coins come in 1,2,5,10,20,50,1,2 denominations while U.s. dollars and, I assume, other dollars come in 1,2(?),5,10,25,1 denominations? What sort of measures of efficiency or whatever can one put on a set of coin denominations? Obviously number of different denominations should be kept low. Fewer types of coins means less confusion. But on the other hand, being able to make any amount with the fewest number of coins would be a bonus. As would being able to have a larger number of values possible with some subset of a given collection of spare change, I suppose. Since these criteria are all essentially numerical, working out the measures of efficiency for various denomination sets would be childs’ play. For someone who knew a darn sight more about combinatorics than I do. Perhaps we can set bounds on the different weightings of the criteria the UK and US mints had in mind when deciding which coins to use. Wouldn’t that be fun?

If you have a circle with a continuous variable assigned over its circumference (a variable I shall call temperature) then there will be two points, directly opposite each other (ie endpoints of the same diameter) that have the same value. For any continuous variable. That’s pretty cool, no? It implies that on Earth there are always two points that are isothermic and antipodal. Here’s the proof for the circle. Now define a function of the variable which subtracts the value of one point from that of the point opposite it. This will be continuous over the circle. If you start at point A and have A’ opposite it, f(A) = A-A’. Now consider following the function as it travels round the circle. Eventually it gets to A’. f(A’) = A’-A = -(A-A’) = -f(A). So, by the intermediate value theorem, at some point between A and A’, the function f had value 0. For some X, f(X) = 0. Which means X=X’. (Perhaps I could be criticised for conflating the point and the value of the variable at the point, but whatever. The result still follows when you introduce all that crazy notation…)

This result, and its corollary (that there are always a pair of isothermic antipodes) got me thinking. Can you say more about the sphere case? Obviously, every great circle has a pair of points like this. So there are infinitely many of them. Cool. But can you prove anything about where they are placed? I can’t think how. Because the pair of points have to share a temperature, but the set of pairs don’t. So can you show any other cool stuff about continuous variables on a sphere? Like how there’s always somewhere where there is no wind on earth. Because of something to do with vector fields and stationary points, I think. (Because you can’t comb a sphere. But you can comb a doughnut…)

Well, that’s what’s been rattling aroung my brain recently.

Written by Seamus

April 17, 2008 at 5:07 pm

Posted in maths

### 4 Responses

1. Wotcha, Seamus! How’s life treating you?

So I thought I’d point out that your method for completing a maze doesn’t work. For a general maze it only holds when you begin at the entrance and want to find your way back – in fact, using the same method it is possible to return to any point you start at. But you can’t guarantee you’ll get to the exit if you start, for example, in the middle of the maze.

The method DOES work if all of the walls in the maze join up, in which case both the walls and the paths of the maze could be modelled as connected graphs, and thus graph theory probably could be used to solve the problem. There are probably easier ways. There are also algorithms you can use to solve more general mazes, but they usually involve marking passageways as you pass through them and are complicated, so I won’t go into them.

US dollars do not have 2 cent coins, I don’t think, but they technically do have \$2 notes in circulation that are very rare. I doubt I’ve met anyone that has ever seen one. You might like to know (if you’re still interested) that in the 1990s, some guys names Telser and Sumner tried to work out the optimal denomation and came out with powers of 3: 1, 3, 9, etc. I don’t know what method they used, but they supposedly took into account the factors you mentioned.

Your comment about the circle is very interesting. It took me completely by surprise. Your proof took me a while to follow, but I do see that the result holds. It’s pretty clever really. Not sure about the sphere. Could be interesting to look into. But not right now.

See you later. Be well!

Stephen Morffew

April 26, 2008 at 11:54 pm

2. Oh yeah. I was kind of assuming the walls of the maze were connected and that you start at the beginning (duh). So if the walls are connected then the path from entrance to exit will not only be connected, but also will be a tree. I think that’s the term. It won’t have any closed loops in it.

As for the coins thing, I’m now wondering whether Canadian, Australian and other Dollar currencies follow the American coin denomination system…

Thanks for the comments; good to hear from you. See you at the barbecue!

Seamus

April 28, 2008 at 10:14 pm

3. Yeah, tree. Well remembered.

The Canadian dollar uses the US system (1, 5, 10, 25, \$1, \$2) except that \$1 and \$2 are coins instead of notes. The Australian dollar mostly uses the sterling system, but without any 1 or 2 cent denominations (5, 10, 20, 50, \$1, \$2). I mostly think it’s weird that, however they deal with the coins and smaller denominations, all of these systems still use larger (over \$1 or £1) that are equivalent to 1, 2, 5, 10, 20, 50, 100. Same with the euro too, isn’t it? (By the way, I’m not actually this smart. I found all this out on wikipedia in about two minutes.)

Woo! Barbecue!

Stephen Morffew

April 29, 2008 at 9:36 am

4. Yes the Euro has 1 and 2 euro coins and then notes 5, 10, 20, 50, 100. In fact, the euro goes further with 200 and 500 denomination notes. And with exchange rates as they are currently that means almost four hundred pounds in one note. Or over 750 US dollars. Holy crap. I just did those calculations in google, and they surprised even me!