My Favourite Chess Game

I've covered a range of my Chess games in the past - some wins, some losses - but in this post I'd like to review my favourite Chess game, the strongest opponent where I scored a win.  This was within the Kidsgrove Chess Club's own internal league, and all six players played each other twice (once as White, once as Black).

This was my game against Jules H, the strongest player I've scored a win against.  I was White, and played my standard Queen's Gambit.

1. d4 d5
2. c4 e6
3. cxd5 exd5

Now I'm sure 3. cxd5 is regarded as a poor move, reducing the tension in the centre, but I thought it made sense to trade my c-pawn for my opponent's central e-pawn.

4. Nc3 Bb4
5. Bd2 Bxc3
6. Bxc3 Nf6
7. e3 0-0
8. Bd3 Re8
9. Ne2 Qd6
10. Ng3 Bg4

I can't play 11. f3 as I would immediately lose to Rxe3+ and have ongoing trouble on the now-open e-file.  However, I identified that Bg4 left b7 unprotected, so I decided to play Qb3 and look to castle as soon as possible.

11. Qb3 Qb6
12. Qxb6 axb6

No, I wasn't initially intending to trade queens, but since I could leave a scar on my opponent's pawn structure, with b7 as a fixed weakness, I decided to go for it... and then finally castled!
13. 0-0 g6
14. Bd2
Supporting the e-pawn, so that I can finally push play f3, which will kick the bishop and in future enable me to play e4.




















14. ... c5  My opponent looks to straighten out his kingside.
15. a3 c4
16. Bc2 Bd7
17. f3 Nc6
18. Rae1
Completing my development. Putting the rooks behind the e- and f-pawns should enable me, with the support of the minor pieces, me to push them forwards and make significant gains in space.

18.  ... b5
My opponent is grabbing space on the queenside, but I'm slowly and steadily preparing to advance my central pawns.

19.Bc3 Re6
20.e4 h5
21.Ne2? dxe4?
22.fxe4

After getting lucky with Ne2, I've now developed a triple threat:  

I can advance the d-pawn, forking knight and rook, and if the rook retreats I can then capture the knight on f6.  If the rook moves to d6, I can then also advance the e-pawn, forking the rook and the other knight (the pawn on e5 would be supported by the bishop on c3.  The bishop pair on c2 and c3 are beginning to see their diagonals open up, and they're pointing towards black's king.

22. ... Re7?
A blunder, since I can immediately play Rxf6.  I guess the complications of the position got to my opponent, and he figured he could save his pieces by retreating the rook.  The software I've consulted suggests ...Nxe4 as a better continuation for Black.

23.Rxf6 Nd8
24.Nf4 Ra6
25.d5 b6?

I'm not sure what my opponent was doing with these moves.  After threatening my rook on f6, he's now locked his rook out of the game, and is continuing to play on the wings, while I move through the centre.

26.e5 Nb7

Earlier that week, I'd been reading about 'clearing the barriers' towards your opponent's king, and I considered this very carefully.  I'm a knight ahead (after the capture on move 23),  so I have extra material to play with.  Also, I have both of my bishops pointing towards white's king, so I anticipated that after Nxg6 I would force white's king into the corner, and potentially get a discovered check to capture more material.  After 28. Rxg6+ Kh8 I can play e6 winning a bishop, or after 28. Rxg6+ Kh7 I can play Rxb7+ winning a rook.

So I went for it, trading my knight for two pawns and a direct attack.
27.Nxg6 fxg6
28.Rxg6+ 

After 28 Rxg6+ there are possibilities for me to win material through discovered checks

28.  ... Kf8

Avoiding both of the discovered checks, but enabling me to bring my other rook into play with a tempo.

29.Rf1+

I was expecting 29...Rf7 30.Rxf7+ Ke8 31.e6 Bxe6 32.dxe6 which lengthens the game but enables me to win more material.  Instead...

29. ...  Ke8
30.Rg8# 1-0

A surprisingly quick finish to a very pleasing game.  I appreciate that my opponent made a few suspect moves, but I'm pleased with the way I handled the game, the tactics and strategies I used (placing my rooks and bishops on squares that would maximise their range and usefulness) and as I said, this is probably one of my favourite games.

Here are a few of my other Chess games:

My earliest online Chess game
My very earliest Chess game (it was even earlier than I thought)
 
The strangest game of Chess I ever played - 1. d4 d5 2. c4 b5
I was not sure what I was supposed to do with that; apparently I was supposed to play 3. c4xb5, but played 3. c4xd5 and immediately and unintentionally took my opponent out of his prep.

The Chess game I'm least proud of

I got greedy, tried to hold onto a pawn that I should have given back, and expended a lot of time and effort on it, instead of protecting my King (on the other side of the board)

Revisiting Fibonacci Constants

In a previous post (four years ago), I explained the Fibonacci Series - where it comes from, how it originates, and how it can appear in nature.  I also talked about the golden ratio, which is the ratio between subsequent terms in the Fibonacci Series.

I've been doing some further reading and research on the Fibonacci Series, and have been doing some of my own calculations and investigations.

Firstly:  what happens if we extend the series, so that instead of just summing the two previous terms, we sum the three previous terms, or the four or five previous?

This has been done before (I wasn't too surprised), and these are known as the following:
Fibonacci - 2 terms
Tribonacci - 3 terms
Tetranacci (or quadranacci) - 4 terms
Quintanacci (or pentanacci) - 5 terms
Hexanacci - 6 terms


I struggled to find names for the higher-number terms, so I'm going to submit my own.

Heptanacci - 7 terms
Octanacci - 8 terms
Nonancci - 9 terms
Decanacci - 10 terms

I stopped at 10, as I found that the data I'd accumulated was enough to draw some interesting conclusions from.  Here are the first few terms of each of the series:

Fib:  0,1,1,2,3,5,8,13,21,34,55,89,144,233,377
Trib:  0,0,1,1,2,4,7,13,24,44,81,149,274,504,927

Tetra: 0,0,0,1,1,2,4,8,15,29,56,108,208,401,773

Quint: 0,0,0,0,1,1,2,4,8,16,31,61,120,236,464

Hex: ...0,0,1,1,2,4,8,16,32,63,125,248

Hept:  ...0,0,1,1,2,4,8,16,32,64,127

Oct: ...0,1,1,2,4,8,16,32,64,128,255,509,1016,2028,4048,8080,16128

Non:  ...0,1,1,2,4,8,16,32,64,128,256,511,1021,2040,4076,8144

Dec: ..,0,1,1,2,4,8,16,32,64,128,256,512,1023,2045,4088

Taking this raw data, I then started to look at the ratios between subsequent terms.  We've seen previously that the Fibonacci series has the golden ratio 1.61803... or (1+ sqrt(5))/2 but what about the other series?

Fib 1.61803
Trib 1.83929
Tetra 1.92756
Quint 1.965948
Hex 1.983583
Hept 1.991964
Oct 1.996031
Non 1.998029
Dec 1.999019

I haven't identified the expressions for each of these but have found from research that the tetranacci constant satisfies x + x^-4 = 2.  

Plotting the number of terms being summed (or the n-number for the series) against the ratio gives this graph.

The question is:  will the ratio ever reach 2?  It looks like the line will head towards 2 as an asymptote, but will it reach 2 if N increases?

The answer is no, and my proof is as follows:

Take the N=10 series as an example:
0,1,1,2,4,8,16,32,64,128,256,512,1023,2045,4088

We can see after the initial 0 and 1, that the first few terms are exactly double the previous one.  This is because each term is the sum of all the previous non-zero terms, including both of the 1s.  1+1 = 2, 1+1+2 = 4, 1+1+2+4=8 etc.  In this case, the ratio of each term to its previous term is 2, each term is exactly double the previous one.

However, this doubling of terms eventually ends:  when the sum no longer includes all the previous terms, that is to say, when the sum no longer includes both of the 1s and all subsequent terms, then the ratio falls below 2.  In the N=10 example above, the ratio falls below 2 when we reach 1023.

1+1+2+4+8+16+32+64+128+256 = 512
1+2+4+8+16+32+64+128+256+512 = 1023

At this point, the ratio falls below 2 (1023/512 = 1.998.  This fall will occur for any and all series which follow the "sum of previous terms" pattern; as N increases, it just takes more terms, and the final ratio will get closer to 2, but will remain below it.  As an aside, Wikipedia states that the ratio for an n-nacci series tends to the solution, x, of the equation  (no proof given, although my data confirms it).

Next:  I will look at what happens when we sum the previous term and half of the term before that...e.g.  N = 1.5, N= 2.5, N=3.5 etc.

Other maths articles on a similar theme:

The Fibonacci Series
Fibonacci Constants Revisited
The Collatz Conjecture

Ulam Sequences

How many photographs (permutations including zero)

My wife and I have recently had our third child - hence I've not been blogging much lately.  However, I've been thinking about blog ideas, or more specifically, I've been thinking of mathematical investigations that I could explore and then, if they were interesting, share on my blog.

Our house is full of family photographs - almost every room has family photographs in it - and in particular on one wall, we have three photographs:  one of our daughter; one of our older son, and one of the two of them together.

In mathematical notation, let's call my daughter A and my older son B; so we have A, B and AB.

Now we have three children, and I have started considering how many photos we would need to show the same range of variations... and it's more than I thought.  Let's introduce the baby as C.

We would have:
A, B, C - each child individually
AB, AC, BC - each child with one other sibling
ABC - all three children together.

So the total number of pictures has gone from 3 to 7.

Let's suppose we have four children, A, B, C and D, and we want the same range of photos, with all variations.  The list grows dramatically:

A, B, C and D - individual photos - 4
AB, AC, AD, BC, BD, CD - pairs - 6
ABC, ABD, ACD, BCD - trios - 4
ABCD - group - 1
Total = 15

Children  Photos
1   1
2   3
3   7
4   15

In order to work out the nature of the series, I looked at the differences between terms, and then the second differences (i.e. the differences between the differences).
3-1 = 2
7-3 = 4
15-7 = 8

4-2=2
8-4 = 4

What became clear to me at this point is that the sequence is expanding exponentially or logarithmically, and not quadratically.  And then it very quickly followed that each nth term is 2ⁿ -1 (the series 2ⁿ is immediately recognisable - 1, 2, 4, 8, 16, 32 etc).  The need to introduce the -1 suggests to me that we're excluding the photo which has no children in it.


I had not expected a logarithmic series from this starting point; in fact I had not expected expected the number of photographs to increase so quickly - the volume more than doubles each time, as we have to account for every combination incorporating the new child. I was expecting something similar to a Fibonacci series - but that's more about multiplying rabbits, not children!

BODMAS puzzles - what's the fuss?

There's a phase going round Facebook where unsolvable, convoluted maths problems are doing the rounds: The question starts: "93% give the wrong answer, only a genius can solve this."  And there's usually a picture of Albert Einstein. And then there's a maths question  - they vary in length but have the same general shape: single-digit numbers separated by various maths functions.

3×4+1-3+4 = ?

You don't have to be a genius, you just need to understand the rules of maths grammar.

And, since this has been shared on Facebook, you'll also gave to explain clearly why your answer is correct in the face increasing ire.

All the best with that.

Back to the Future: DeLorean Acceleration Rates

Towards the end of the film Back to the Future, Marty McFly has to accelerate the DeLorean time machine up to 88 mph in order to travel from 1955 to 1985.  

Doc Brown explains: "I've painted a white line waaayyy over there on the street. That's where you start out. I've calculated the distance and wind resistance to active the moment the lightning strikes. This alarm will go off and you hit the gas..."

Simple question:  how far away was the start line from the electric overhead wire and the clock tower?

We can use simple dynamics (or kinematics) to calculate the distance the the DeLorean would have needed to reach 88 mph, if we know the car's rate of acceleration, and take it as a given that the car is starting from rest (it's stationary).

The formula to use is v² = u² + 2as where v is the final velocity (88 mph), u is the start velocity (0 mph), a is the rate of acceleration, and s is the distance (which is what we want to know).

Inserting u=0 and rearranging, we have s = v² / 2a.

According to the Wikipedia entry for the DeLorean, acceleration from 0-60 mph took 8.8 seconds.  And this is where it gets tricky, because s = v²/2a requires consistent units for time in the velocity and acceleration.  I am going to make the simplifying assumption that the acceleration from 0-60 continues from 60-88 mph. In reality, it doesn't, but we'll assume it does.

60 mph = 60 / (60 × 60) miles per second = 0.01666 miles per second.

To reach this speed in 8.8 seconds, the acceleration rate is 0.01666 / 8.8 = 1.8939 ×10-4 miles per second per second. 

88 mph = 88/ (60×60) miles per second = 0.02444 miles per second. 

Time to plug in the numbers:

s= (0.02444) 2 / (2× 1.8939 ×10-4)

s = 1.572 miles

And, just for interest, how long (seconds) would it take? That's easier, using v=u+at and solving for t: t = v/a = 88/8.8 = 10 seconds.

So, the Start Line was just over a mile and a half from the clock tower, (perhaps you could describe that as "waaay over there" with some dramatic licence)  and the journey would have taken 10 seconds (all assumptions taken into consideration).  Somehow, though, it seems a little bit longer on screen.

Some other 'everyday maths' articles I've written:

A spreadsheet solution - the nearest point to the Red Arrows' flightpath from my house
The Twelve Days of Christmas - summing triangle and square numbers
Why are manhole lids usually circular?

Should Chelsea Sack Jose Mourinho?

In previous posts, during previous football seasons, I've monitored the performance of certain football managers - in particular David Moyes and Louis Van Gaal.  I'm not really targeting Manchester United specifically, it's just that over recent seasons, they've had a tough time and there's been some speculation about their managers' futures.  In fact, Moyes was sacked before his first season was finished.

This season, Chelsea's Jose Mourinho is coming under scrutiny.  At the time of writing (16 December 2015) his team have played 16 games and are in 16th place (out of 20) and sliding towards the relegation zone.  But is the situation really that bad?  It's time to compare his performance against some of the others I've mentioned.  The first comparison is cumulative points achieved through the season, and I'm comparing Mourinho with Moyes (his line is the 'this performance will get you fired' line).

Looking at this, it would appear that Mourinho is not going to last until the end of the season.  There's clearly more going on here - for example, Moyes was in his first season after Ferguson's era of success, while Mourinho is continuing after winning the league title last season.  And perhaps Chelsea (the club, staff and fans) are more loyal to their manager.

So, assuming that Mourinho is going to stay in place, at least for the short term, then let's look at what's going wrong for him (I guess he'll be more aware of this than me, but let's look impartially at the stats).

First of all: the performance during the first ten games of the season:

There's nothing obviously wrong with the number of goals his team is scoring: the problem is with the number being conceded.  Chelsea used to be famous for 'parking the bus' (i.e. scoring a goal and then defending with complete success) but it now appears that they've got a very leaky defence - and too leaky to be a serious challenger for the top position.  If we compare their current position (after 16 games) with some of the other teams in the league, we find some interesting points:

1. Only two teams - the bottom two, Aston Villa and Sunderland - have lost more games than Chelsea (Chelsea = 9, Sunderland = 10, Aston Villa = 12).  Previous analysis of Moyes and LVG in particular indicated that they were drawing too many matches that they needed to conver to wins.  Mourinho's task is different - it's not stop drawing, it's to recover more draws from losing situations and to stop losing. 

Comparison of Jose Mourinho to Alex Ferguson's
first season; final season; David Moyes
and Louis Van Gaal.

As at 16 December, the win/lose/draw rate for the Premier League, sorted by league position
from left to right.  Note the high lose-rate for Chelsea.

2. After 16 games, Chelsea have four clean sheets, ranking them joint 11th, mid-table.  Their issue is not the number of clean sheets they're keeping, it's conceding more goals than they're scoring (I know that seems obvious, but they don't have to keep clean sheets to help them improve their position).

3.  Chelsea's goal difference is not a significant factor. Or, to put it another way: on average, they're not losing by huge margins in their games.  The recommendation based on this (and their significant lose-rate) is to play more aggressively and play less cautiously when they concede a goal.  They can afford to lose by 3 or 4 goals without significantly denting their goal difference compared to the teams around them.

So, should Chelsea sack Mourinho?  Maybe, although perhaps he can be relied upon to change his team's style and go for a more attacking style - he has goals to play with, if not games.

More articles on data analysis in football:

Reviewing Manchester United's Performance
Should Chelsea Sack Jose Mourinho? (it was relevant at the time I wrote it)
How exciting is the English Premier League?  (quantifying a qualitative metric)
The Rollarama World Football Dice Game (a study in probability)

The Mathematics of Post Office Queues

I'm sure there's probably plenty of research (anecdotal, serious, genuine and some of it even humorous) around standing and waiting in queues.  Queuing is the stereotypical British national pastime, and we seem to be quite good at it.  However, if you think that being good at it means that we enjoy it, then you're mistaken.  In my own personal opinion, the worst place to queue is at a local Post Office, and as we approach Christmas and I send out parcels and so on, I'm getting to experience more of it.  Enough of it, in fact, to have done some empirical and informal research of my own.

Firstly, let's look at some assumptions about good customer service in a queue situation:

1.  If the queue gets longer, then introduce more staff to the counters.  Or, in mathematical terms, the number of people on the counters should rise in direct relation to the number of people queuing.  Not a one-to-one relationship (one assistant per customer would be expensive to staff and lead to very wide checkouts), but that there should be a consistent rise in assistants as the queue gets longer.  This continues until, of course, the desks are fully staffed and there's no more capacity.

1b.  If the queue gets longer, then the number of staff visible doing non-customer-facing administration tasks behind the desk should decrease.  There really is nothing more infuriating than a member of staff, who could be serving customers, is clearly visible doing something mundane like counting stamps.  Seriously.

2.  Consequently, if the queue gets shorter, it's reasonable to reduce the number of assistants on the counters.

3.  If the queue gets longer, then more effort should be spent to reduce the amount of time it takes to serve each customer.  Keep things formal, keep things concise and keep the queue moving.  The amount of time spent per customer should be inversely proportional to the length of the queue.

Let's have a look at graphical representations of the ideal situations outlined above:

1.The number of assistants increases as queue length increases.  There are two lines plotted on these graphs:  one (the blue line) for a larger office where more staff are employed, and another (the pink line) where there are fewer staff employed.  The principle is the same.

2.  As queue length increases, the number of staff doing non-customer facing work decreases.  Again, there are the two lines, with the pink link showing a smaller office.

3. Amount of time spent on each customer decreases with length of customer.  For smaller branches, the amount of time spent needs to fall more quickly as there are fewer assistants to handle the queue.  One interpretation from the graph shown below (which is a purely illustrative graph, with arbitrary numbers) is that as the queue length increases from 1 to 10 people, the amount of time that should be spent on each customer should fall from around five minutes to one minute.

Now let's look at what can happen when this breaks down.

1.  Increasing the number of staff on the desks when the queue length increases
Really?  More staff?  They're on their lunch break.  They're on training.  Or...
1b.  ...worse still, they're sitting at their desk counting out money, just behind a 'position closed' sign.  They're moving around behind the other assistants, putting paper into drawers and rearranging posters; fetching and distributing their business cards.  Infuriatingly visible to the ever-lengthening queue, and incomprehensibly not serving any of the people in it.

3.  There are some questions, such as, "Do you have a landline at home?" and "Do you know when your buildings and contents insurance is due for renewal?" and "Did you know that we offer travel insurance?" that are understandable when the queue is relatively short.  After all, the desks are well-staffed, people haven't been waiting for very long and there is potential for these up-sell opportunities.  However, when the queue is long, and has been long for many minutes, and when there are only one or two assistants at the desks, these questions really aren't helpful.  

As an additional consequence:  the likelihood of a potential up-sell falls sharply as the queue length increases.  People who've been waiting in the queue for ten minutes to do a three-minute job don't have the time to book an appointment to change their broadband provider.

Here's how I picture the graph of actual time spent per customer, and how it varies with queue length.  This is for the larger office I imagined above, but again, the principles are the same for smaller and larger outlets.

What happens here is that customers actually take it upon themselves to reduce the amount of time it takes to complete their task.  Having spent so long in a queue, they now wish to move through the execution of the task more quickly.  They dispense with formalities about the weather; the time of year; their plans for the weekend, and become much more focused on their task.  They have their money ready.  They have a pen in their hand, if needed.  Out of a sense of camaraderie with their fellow queuers, they aim to get served as quickly as possible, to alleviate any further unnecessary waiting for their fellow man. Moreover, they instinctively decline any offer of discounted insurance or other special deals.

And the consequence?  Here's the related 'probability of successful up-sell' versus queue length graph.

You'll notice that the graph actually falls below zero as the queue length extends beyond around 23 people.  This seems very strange - after all, there can't be less than a 0% chance of something happening.  What happens here is that the negative effect of standing in a queue with that many people in it is that people start to tell their friends about their queuing experience, and this negatively skews any future up-sell opportunities too.  Here, then, is perhaps a more sensible approach to how to make up-sell overtures.

But seriously...

Okay, so perhaps I'm being a little obtuse.  But you get the idea.  The Royal Mail is under threat from smaller couriers and delivery providers, but it has the advantage of a large branch network and brand familiarity.  We go to the Post Office because our parents did, and we know how the delivery system works.  However, the Post Office finds itself having to diversify into various financial products to keep itself profitable.

My worry is that this diversification is detracting from their core customer experience:  I know I've potentially exaggerated how frustrating queuing can be, but it seems to me that Post Office employees are not aware of this and continue to blanket-bomb everybody with the same sales lines, which are increasingly frustrating.  As more couriers become available online, and they continue to become more efficient, building relationships with other high-street stores (Argos is an example), I strongly suspect that more people will turn to them for their delivery needs.  At our house, I would estimate that 30-40% of the parcels we receive come via Hermes, but there's also DPD, parcel2go or Interlink.

So, Royal Mail have the history and the branch network, and the face-to-face engagement with their customers.  I just wonder how long it will last.

In fact, it seems that this is an ongoing problem.  And it's been going on for years!  During my search for a cartoon to add to my article (I borrowed the one just above from the Daily Mail, I found this article which lists everything I've said:  Our too-pushy post offices: Customers forced to queue for too long before being offered services they don’t want. There are some alarming stats in there (and this is three years ago):

- One post office user in three has to queue for at least five minutes, and for some the wait is more than twice that long said Consumer Focus (then Consumer Futures, now Citizen's Advice), which tested 448 post offices in urban high streets.
- For many customers, who are used to supermarkets opening extra tills if queues develop, the small number of counters which are open is exasperating.
- For example, Consumer Futures said a typical Crown office has 7.3 counters, but only 3.8 are actually open.

(And, while I think about it:  everything I've mentioned here applies just as equally to the high street banks.  The queues; the number of assistants at desks; the number of staff walking around not-quite-so-behind-the scenes, and of course the blanket approach to upselling financial products.  Why do you think I bank online?)

Some other 'everyday maths' articles I've written:

A spreadsheet solution - the nearest point to the Red Arrows' flightpath from my house
The Twelve Days of Christmas - summing triangle and square numbers
Why are manhole lids usually circular?