[Let us assume there are D days in a year, in order to be more general.]

Actually, the expected number of unique birthdays is very easy to calculate. Let e(n) denote the expected number of unique birthdays when there are n people.

Choose one of the n people. Then either they have a different birthday than everybody else, or they don’t. They have a different birthday than the other n-1 with probability . Let us write for , and so the probability that they have a different birthday to everybody else is . In this case, the expected number of unique birthdays is 1+e(n-1). In the other case, which thus has probability  , the expected number is just e(n-1).

Hence,

As e(0) = 1, it follows that:

 (As one should expect, as n tends to infinity this tends to D, as it is almost certain that every birthday gets taken)

For instance, with D = 365, you should expect about 21 different birthdays when there are 22 people, or 46 different birthdays when there are 50 people. When there are 1000 people, there will be around 241 different birthdays. Perhaps it is quite surprising that you should expect there to be so many unclaimed birthdays with so many people!

Consider the following problem:

Suppose a factory is in a country with the following law: Every worker must work every day, unless it is one of the workers birthdays. In this case, everybody has the day off. How many people should you hire in order to maximize the expected number of man-days worked every year?

Naturally, if there is only a single person he will certainly work 364 days (D-1) a year. If there are two, they are likely to have different birthdays and so each will work 363 days a year (D-2), giving a total of 726 days.

The expected number of man days worked will be the number of people, n, multiplied by the expected number of non-birthdays. The number of non-birthdays is just D – e(n). Thus we are trying to maximise the following:

Considering n as a real number instead of an integer, one can find that the function W has a maximum at . Thus the maximum must be around this, and so for all but the smallest D (where the approximation is not valid), the maximum will be either D-1 or D.

In fact, it is easily seen that W(D) = W(D-1). It follows that one should employee either D or D-1 people. Since you will probably have to pay these workers, it is better to go for the D-1 people.

A simple change of this problem is to consider that for each man-day of work in the factory, you earn £a. However, you must pay each worker £y per year. Then instead you are maximising the yearly income of:

The derivative of this is . Hence if  we must have:

Define so that:

And so we have that:

Let us note that is the ratio between what you pay the worker per year, and what he earns for you per year if he were to work every day. Let us call this ratio P.

Note that when z is positive, the equation always has a unique solution. The function which takes z to w is called the ‘Lambert W Function‘. Hence we have that:

It follows that:

As P will be small, and in particular less than 1, an approximation of the Lambert W function may be useful. We will use the Taylor Series approximation . Along with the usual log approximation, this gives:

Or, in terms of P,

For instance, suppose that I pay my workers £18,000 per year. They generate about £350 in profit for me each day they work. They work every day, except for when one of them is a birthday. By evaluating the W function exactly, I find out that the maximum expected profit is when there are around 260 workers. The approximation yields 269 workers. If the profit they generate per day were to be decreased to £75then the approximation is completely inaccurate. Whilst I should only employee 72 people, the approximation gives over 7,000!

Using the former interpretation, it is possible to generalise the theorem a bit. This generalisation is in fact of practical importance. (This importance is not described here. Essentially it naturally arises because if the covariance matrix of a random vector X is R, then the covariance matrix of AX is ARA’, where a dash denotes conjugate-transpose).

This document explores these implications.

I read the ‘Reasons’ report, which I did think was interesting (and surprisingly readable). Comments on the judgement are easy to come by, I’m sure, so I won’t give any more here.  However, I did like the following little story. It is paragraph 329, and it was from evidence put forward from the defence to show claim that Evra may have been angry enough to make the whole thing up. The evidence is mostly provided by Marriner, the referee for the game:

Mr Evra was seen to dispute the outcome of the coin toss with the referee. Mr Marriner explained that he used a FIFA coin which is blue on one side and yellow on the other. He asked Mr Evra, as the visiting captain, to call the colour. Mr Marriner tossed the coin, it came down yellow, and he awarded it to Steven Gerrard who elected to stay in their current ends. Manchester United had kick off. Mr Evra remonstrated that he had called correctly but, Mr Marriner said, he had not. Mr Evra then spoke to Ryan Giggs about it, and Mr Marriner walked over to Mr Evra to assure him that he (Mr Marriner) had got it right. Mr Evra’s evidence was that when such a coin was used, he always called yellow given that the alternative, blue, is a Manchester City colour, which he would never call. The toss came down yellow and so Mr Evra knew that he had won it. He particularly wanted to change ends at the start, he explained to the referee that he had called yellow, and why he had done so. Mr Evra was angry but the referee did not change his mind.

I liked the reason at the end. It seems funny to me that such things would influence him. It’s good to hear it, even though I’m an Arsenal fan.

The known problems are:

Chart Object

• No support for HWA, which means it doesn’t work in the latest version of MMF
• No support for Unicode titles, etc.

Ini++

• Crash to do with encryption

String Replacer

• Unknown crash

• Lack of support for additional characters (such as accented characters)

None of these will be much fun at all to work on, or particularly rewarding, so don’t expect progress to be quick. However, hopefully it will happen someday.

The answer is: There’s not.

The situation is that there are versions of the Ini++ object. The original, and Ini++ v1.5. The latter was made when the Ini++ object behaviour changed. In particular, it no longer enforced items and groups to be in sorted order. Therefore I could not release this as an update to the original, as it could break peoples code. To make things worse, this was not something which could be an option either (without some rewriting of code in the original). It was called Ini++ v1.5 as a nod to Multimedia Fusion v1.5, which was Multimedia Fusion v1.0 plus many new features.

Naturally, there have been different releases of the ‘Ini++ Object’ and the ‘Ini++ v1.5′ objects, but both kept calling themselves that name. There is no ‘Ini++ v1.6′ object. There is no  ’Ini++ v1.1′. Just the two mentioned. They have the file names ‘Ini++.mfx’ and ‘Ini++15.mfx’ respectively.

So why the confusion?

The problem stems from the fact that although the behaviour of Ini++ v1.5 changed from that of the original, it was still compatible with the original Ini++ object. That is, if you took ‘Ini++15.mfx’ and renamed it to ‘Ini++.mfx’, then any MMF application using the old version would suddenly be ‘upgraded’ to the new version. If you knew your program wasn’t going to stop working because of the changes, this was a great way to make use of the new features without copying over all the events to the new object. Quite a lot of people did this.

It was entirely deliberate that you could do this. In fact, Ini++ v1.5 has a few actions which have no menu items exactly for this purpose. For if they were removed from the code, then this renaming method might not work.

You would expect that if you did this you would see two identical looking objects in the ‘Create new object’ dialog. The same icon, the same text: ‘Ini++ v1.5 Object’. In fact, this is not what you see. Multimedia Fusion will rename the second object to ensure it doesn’t have the same name, so you actually end up with ‘Ini++ v1.5 Object’ and ‘Ini++ v1.5 Object 2′.

The first one (selected in the picture) actually has the file name ‘Ini++.mfx’ and the second has the file name ‘Ini++15.mfx’.

An earlier version of the Ini++ v1.5 Object called itself simply the ‘Ini++ v1.5′ object. Therefore if both ‘Ini++.mfx’ and ‘Ini++15.mfx’ were that version, you would in fact get ‘Ini++ v1.5′ and ‘Ini++ v1.6′ in the dialog, as it intelligently renames. Therefore any object calling itself ‘Ini++ v1.6′ is just an illusion. (Indeed, the title of the object was renamed after it was noticed this happened to avoid this confusion)

But perhaps you are seeing this anyway, despite not being one of the people who renamed ‘Ini++15.mfx’ to ‘Ini++.mfx’? Unfortunately, this is common. The first installer for the first release of Ini++ v1.5 actually installed the file as ‘Ini++.mfx’, which was obviously complicated the situation. (I didn’t make this installer. Later installers install it correctly.) I believe that the one of the extension updaters for MMF will also distribute Ini++ v1.5 with the filename ‘Ini++.mfx’. Combined, this means that nearly everybody will have this problem.

Perhaps I will solve this in future by having the next installer install the original version of Ini++ as well. Or perhaps by making sure it is uninstalled (as you shouldn’t really be using it anymore anyway – it exists only for compatibility reasons now).

So which version is the latest version? Well, it is quite possible they are both the same version. Make sure you are using the version with the file name ‘Ini++15.mfx’ (you can tell this by selecting the object and clicking the ‘About’ tab in the properties viewer). Ensure this is up to date by installing the latest release. If you don’t use ‘Ini++.mfx’, perhaps you should delete it.

In fact, it isn’t true that the diaeresis is not used in English. Some people will use them on the second vowel when there are two vowels together, to indicate they are said as two separate sounds. For instance, one might write ‘coöperate’, to show that it is said ‘co-operate’, rather than ‘coop-erate’. ‘Naïve’ is another one.

So, I solved this problem by writing a program called Umalut. It sits in your system tray and when you either click the icon, or press a shortcut (hard coded to Windows Key + S at the moment), it will insert an umlauted character if a vowel or ‘y’ is pressed, or an eszett if ‘s’ is pressed.

It can generate the character in two ways – either by generated a WM_CHAR message directly to the window that it stole the keypress from, or by simulating input and typing out the Alt-code for the character. It is written in C++ and consists of two projects – a GUI and the hook DLL.

The problem is, I’ve broken it at the moment somehow, and I’ve just realised that when you are using Microsoft Word, it is massively easy to insert characters. The default shortcut to insert an umlaut is to press Control+Shift+; and then press the required vowel. For the eszett, it is Control+&, and then ‘s’ (on my keyboard, this is the same as Control+Shift+7 and then ‘s’).

By going into the Symbol shortcut dialog (Insert tab, Symbol, More symbols…, select the eszett and then click ‘Shortcut’) you can change this shortcut. I’ve set it to Control+Shift+; and then ‘s’, so it is the same to access any of the characters I am interested in.

What is more, you can set up auto-correct text. I’ve made it so that ‘..a’ (for instance) is translated into ‘ä’, which is very convenient. I reccomend these to everybody else, too!

For now, I’m not going to bother with releasing the Umlaut program. However I might in the future if there is any demand for it.

A group in the Champions League has 4 teams in it. Each team player each other twice. There are three points for a win, one for a draw, none for a loss. Since 1999-2000, the top 2 teams (ordered by points, then in the case of ties head-to-head goal difference, and then general goal difference) in each group quality to the next round.

Is it true that ten points has guaranteed qualification 99% of the time? Well, it’s not too far out at 2.5%. On 158 occasions has a team achieved 10 or more points, since the 1999-2000 season (but excluding this season). Only four of those teams failed to qualify, and each of those got exactly 10 points.

These were:

• Werder Bremen in 2003-4 (Barcelona got second play with 11 points)
•  Olympiacos and Dynamo Kyiv in 2005-6 (Liverpool and Real Madrid went through in second place with 10 and 11 points respectively)
• PSV Eindhoven in 2006-7 (Deportivo had 10 points but went through instead of them).

Whilst in practice no team has ever failed to go through with 11 points, it is in fact possible to still miss out with 12 points. Imagine three teams A, B and C who win all their home games against each other (and so lose all their away games against each other), but all beat D both home and away. Then you get the following table, where C misses out on qualification:

13 points does ensure you qualification. As suppose you fail to qualify with 13 points. Then another two teams must get at least 13 points, and so there are at least 39 points on the board. But it is only possible to get 36 points on the board, as there are 12 games and each game can add up to three points on the board.

In reality, no team has qualified with less than 7 points. However, teams have qualified with exactly 7 points. Werder Bremen in 2005-6 and Lokomotiv Moscow in 2002-3 both came runners up to a high-scoring Barcelona. In theory, it is possible to get four points and still qualify. Imagine team A wins every game and teams B, C and D draw against each other. Then you get the following table:

The question is to find the determinate of J_k. It’s a straight forward question, but I am writing a blog post on it because it is interesting in that it allows a few different approaches to easily be taken, and so it is interesting to see which you select. This post will document five solutions to the problem.

Solution 1: Calculation via minors

The usual way of actually calculating the determinate of matrices is to use ‘minors‘. Let us use the minors expansion in the first column. All the terms in the first column are zero except for the bottom one, and so you get that:

With the extra condition that J₁ = 1, it follows that:

which is equivalent to the previous result.

Solution 2: Using Eigenvalues and the trace

This method is particularly elegant.

Note that as  and so all Eigenvalues of J_k have , and thus all the Eigenvalues are either 1 or -1.

Let Tr(A) be the trace of a matrix A. We have that Tr(J_k) = 0 when k is even, and Tr(J_k) = 1 when k is odd. The trace is the sum of the Eigenvalues, and so in either of the cases k = 2r or k = 2r + 1, we must have r pairs of -1 and 1 Eigenvalues. (In the odd case we have an addition 1 Eigenvalue.) As the determinate is the product of the Eigenvalues, it must be .

I like this method the best, but if it had a flaw it would be that it doesn’t easily generalise to the case when the anti-diagonal elements are arbitrary, instead of being ones. However, this doesn’t really matter, as the properties of a determinate just mean you can ‘pull out’ the values anyway, and so it is just the product of the diagonal elements multiplied by |J_k|.

Solution 3: Via Swaps

The determinate is such that if you swap two adjacent columns (or rows), you reverse the sign. We also know that the determinate of the identity matrix is 1. Therefore it should be possible to turn J_k into I using a series of swaps.

Therefore the problem reduces to finding how many swaps can turn the sequence [k,k-1,...,2,1] into the sequence [1,2,...,k-1,k]. To solve this, note it takes k-1 swaps to ‘push the k to the right’, and another k-2 to push the 1 back to the left. Therefore after 2k-3 swaps you get  [k-1,k-2,...,1,k] . After that, you can apply the same result for k’ = k-2. Therefore we have that |J_k| = -|J_k-2|.

Given the base cases for k=1 and k=2, the result follows.

Solution 4: Via Permutations

This method is essentially the same as the above, but carried out in a different way.  We use the following definition of a determinate:

Using this definition to calculate |J_k|, we note that only one permutation contributes anything to the sum: The one which takes i to k-i. In this case, the determinate is just the sign of the aforementioned permutation.

The permutation swaps the first and last, second and second-last, etc. Therefore, if k = 2r, there are r permutations. If k = 2r + 1, there are still r permutations, as the central element (r+1) remains fixed. Hence the determinate of k is (-1) to the power r.

Solution 5: Directly via the Eigenvalues

Whilst this is a difference approach to Solution 4, it gives fairly similar calculations. The determinate of a matrix is the product of the Eigenvalues. Thus we can just determine the Eigenvalues.

J_k swaps coordinate 1 with coordinate k. Therefore [1,0,...,0,1] is an Eigenvector with Eigenvalue 1. Also, [1,0,...,0,-1] is an Eigenvector which goes to negative itself, and so -1 is an Eigenvalue.

As above, there are r coordinates which are swapped when k = 2r or k = 2r + 1. Therefore the determinate is (-1) to the power r.

The solutions do have similarities, but they also have differences, and I think it is interesting the number of ways there are to approach this simple problem.

The Man In Seat 61 briefly describes how to travel from Albania to Macedonia, however the point of this blog post is to eleberate on this.

The Albania train network is fairly basic.  There is a train that once a day (at 6am, at the time of writing) goes from Tiranë, the capital of Albania, to Pogradec, which is on the eastern border near Lake Ohrid. The train goes via Durrës, however, which is on the west cost, so it is not very direct. In fact, we split the journey up by going to Durrës for one day (there are many more trains that go just from Tirana to Durres – there is an 11:45 and a 14:20, for instance) and then catching the early train the next day (at 7am).

The approximate route of the train (100 miles). Google map link

It is worth noting that the trains are very, very slow. We got in around 14:30 (it is scheduled for 13:30), so it took about 7.5 hours to travel what is, as the crow flies, 55 miles – so for every hour you are on the train, you are only getting around 8 miles closer. Indeed, when the train ran along side roads, buses would easily overtake the train. I estimate it was travelling at about 20mph.

The smashed windows are not of any practical concern

The train isn’t actually uncomfortable, but they look bad. Children apparently throw rocks at the train (although we did not witness this ourselves), and so the windows have many fractures in them. However, for us at least, they were all solid and so did not let any of the cold in.

The train and carriages have a charm about them. Moving slowly through the beautiful scenery is enjoyable

When you get to Pogradec station, there isn’t much there. There is a station, a gate, and a car park area, as you can see in this photo. You will know you are approaching the station, as the line moves south on the west edge of Lake Ohrid, so there is water for a about twenty minutes before arriving.

The train station is a fair distance away from the main town, as can be seen on the Google Map. Some people from the train will be picked up cars waiting for them at the station, but most people get on a kind of minibus thing, called a furgon. A lot of people pack into these things, but it isn’t completely uncomfortable. This will get you into the centre of Pogradec. I cannot recall how much I paid for the furgon, but it wasn’t too much, and even then I must have overpaid, as when I offered the driver the money he seemed very pleased by it!

From Pogradec you can take a taxi to the monastery of St Naum (Sveti Naum), except it won’t be able to get you there. For about €5, however, a taxi will take you to the Macedonian border. From there, you must walk down to another border control point (it is not too far). I didn’t quite work out why there were two, but it is easy enough to navigate.

A taxi will take you as far as the border control point on the left, and then you must walk to the one on the right

From the second border point you could, if you were so inclined, look at Sveti Naum, which is rather near the crossing point. In either case, you will eventually which to get to Ohrid. There are a number of furgons which run the route between the border crossing point and Ohrid, so just jump on one of those. I cannot recall the price of this – it may have been around 500 lek each.

This rather bumpy ride will take you to Ohrid bus station (which is slightly outside of Ohrid, but only a 5 minute walk). It is also possible to stop at a museum between the two points, and get back on later. However, I did not try that.

I hope this has been of use to some people.