Since working my through McDonald’s How to Play Against 1.e4 twice, I have revised Pandolfini’s Endgame Course, and worked my way through his 111 Winning Endgames (as reported previously). I have also continued to solve various problems of the day from the web, and keep a file of those that gave me more trouble than they should. (For the more complicated problems, I usually record the position where I failed to see something, rather than the start position itself.) I have subsequently revised most of the problems that I have recorded so far. This appears to be a worthwhile exercise. My memory of the troublesome problems is surprisingly good. I have also finished my latest pass through Weteschnik’s Understanding Chess Tactics. My memory retention of these problems is good. I am now using the same time slot to gradually work my way through Reninfeld’s 1,001 Winning Chess Sacrifices and Combinations again. I am only solving a few problems at a time on an irregular basis, but my memory retention here is also good.
More recently, I have worked my way through Coakley’s Winning Chess Exercises for Kids (the blue Coakley) again. I previously did as many as six passes for many of these problems, over a period of about two months, but that was more than a year ago. My memory of these problems had faded greatly, which reinforces the fact that closely spaced repetitions contribute little to long term memory. I started with the intention of writing down my own detailed solutions before looking them up in the book, as Coakley recommends. I previously found that this did not help much with Palliser’s Complete Chess Workout, but I thought that the technique would work better with Coakley. Nonetheless, I again found it very time consuming, and was not at all convinced that it helped. I soon decided to write down just sufficient analysis to convince myself that the combination worked. That seemed to work better. Most of my mistakes are arguably mistakes in analysis rather than in seeing things, but these mistakes are probably the result of failing to see things.
My next task was revising McDonald’s How to Play Against 1.e4. I began my third pass through the book about two months after I began the first. I decided that I had to limit the time that I spent on this pass, cutting it from three weeks to two. I had to compromise here. If I cut the time too much, I would end up memorising moves rather than studying games. As it is, I am building up long term memory, and my understanding is increasing with each pass. I have got Shredder on my ipad, which is very useful for playing through the illustrative games and analysing them either manually or with Shredder’s help. I can often find most of the games on the web. This avoids errors in entering the moves, and is useful when McDonald does not give the full game. I have recently also bought Chessbase Online for my ipad. This is very useful for exploring opening moves and playing over the games that began with those moves, which further increases my understanding. Chessbase Online has reputation for bugs, but I have not had any significant problems with it.
McDonald’s book is enough opening study for now. I cannot justify spending any more time on openings at the present time. The main sources that I have pencilled in for the future are McDonald’s Starting Out: Queen’s Gambit Declined, and Collins’ 1.e4 Repertoire. I have played the QGD previously (albeit a long time ago), so I should have a flying start there. I have watched Collins’ DVD a few times over the past year, and should not have too many problems there either.
The current task is revising Albert & Krogius’ Just the Facts! I also intend to revise Chessimo Endgames 01, and go over Coakley’s endgame problems in detail. My endgame skill has improved very significantly since I started this study plan. Revision is currently more than enough here too. When I do have time to take on more material, I am tempted to revisit Keres’ Practical Chess Endings. I tackled this book many years ago, but found it too detailed. It could be just what I need now.
Friday, 1 February 2013
Tuesday, 1 January 2013
Easy Server Rating
The purpose of this article is to discuss a very simple problem server rating method, so simple in fact that the user could check his rating with a pencil and paper. I am not suggesting that this method is better than the one that I recommend in More Rating, Time and Score, but it is certainly interesting. The method has four easy steps:
(1). For every problem tackled, find the time adjusted problem rating by adding the rating time adjustment given by the graph below to the problem rating.

(N.B. The x-axis of this graph is the number of seconds taken by the user tackling the problem concerned.)
(2). For every problem successfully solved, add 400 points to the time adjusted problem rating. Discard any problems for which the resulting rating is less than the user’s current rating.
(3). For every unsuccessful attempt to solve a problem, subtract 400 points from the time adjusted problem rating. Discard any problems for which the resulting rating is more than the user’s current rating.
(4). Find the average of the results that were not discarded from steps (3) and (4). This average is the new estimate for the user's rating.
For ease of pencil and paper calculation, we could use a table of values in place of the graph. I have assumed that target solution time at which the user and problem ratings are valid is 30 seconds. (N.B. The average solution time on Chess Tempo is about 30 seconds.) I have also assumed that the user’s effective rating increases by 125 points for every doubling of the time that he spends on a problem (see the earlier article).
We can better understand this calculation by looking at the extreme values:
* For a very rapid successful solution, we add a time bonus of up to 400 points to the problem rating, and also add a success bonus of 400 points.
* For a very slow successful solution, we subtract a time penalty of up to 400 points from the problem rating, and add a success bonus of 400 points. If the result is less than the user’s rating, we discard the result.
* For a very rapid failure, we add a time penalty of up to 400 points to the problem rating, and subtract a failure penalty of 400 points. The result should be less than the user’s rating if he is scoring over 50%. If the result is more than his rating, we discard this result.
* For a very slow failure, we subtract a time penalty of 400 points from the problem rating, and also subtract a failure penalty of 400 points. The result should again be less than the user’s rating if he is scoring more than 50%. If the result is less than the user’s rating, we discard the result.
The corresponding method for calculating the rating of a problem from user ratings is similar to that for calculating a user rating from problem ratings. When we rate a problem rather than a user, the time adjustment has the same magnitude but the opposite sign (see the earlier article). The success/failure adjustment for each user also has the same magnitude and the opposite sign.
The method above is essentially the same as that I recommended previously, except that a linear approximation is used for the logistic distribution. The graph below shows the logistic distribution (used by the USCF) in red, the normal distribution (used by FIDE) in blue, and a linear approximation (used by the ECF) in yellow:

The normal distribution used by FIDE is the cumulative normal distribution with standard deviation 282.84, see:
http://www.chessbase.com/newsdetail.asp?newsid=4326.
The logistic distribution used by the USCF has the equation:
1/(1+10^[-d/400]
Where d is the rating difference between the user and the problem that he tackles.
The linear approximation used by the ECF has the equation:
0.5 + d/800
(N.B. I have converted ECF points to Elo points. 50 ECF points is equivalent to 400 Elo points.)
There is clearly very little difference between the logistic and normal distributions over the range of the graph. The linear approximation is accurate up to an expected score of about 0.85, or a rating difference of 300 points. The histogram below shows the distribution of success probabilities for Chess Tactics Server (CTS):

Most of the CTS users have success rates (and therefore expected scores) between 50% and 90%. The linear approximation overestimates the expected score and underestimates the user’s rating for scores of more than 85%. However, users could probably be dissuaded from scoring over 85% by the promise of miserly ratings.
Let:
R = user’s rating.
T = the target time at which the user and problem ratings are valid.
K = number of rating points that the user gains for each doubling of his clock time.
N = number of problems in the problem set.
Ri = rating of problem i.
si = user’s score on problem i.
ti = time that the user takes on problem i.
As in the previous article, the time adjusted rating difference between the user and problem i is given by:
di = R - Ri + K*log2(ti/T)
The user’s expected score on the problem set is:
Σi {0.5 + di/800}
The user’s actual score on the problem set is:
Σi {si}
Equating the user’s expected and actual scores gives:
Σi {0.5 + di/800} = Σi {si}
Σi {di} = 800 * Σi {si - 0.5}
Substituting for di gives:
Σi {R - Ri + K*log2(ti/T)} = 800 * Σi {si - 0.5}
N*R = Σi {Ri - K*log2(ti/T) + 800*(si - 0.5)}
= Σi {Ri + 800*(si - 0.5)} + Σi {K*log2(ti/T)}
Let S be the set of indices for the problems that the user solved correctly, and F be the set of indices for the problems that he solved incorrectly. We can then rewrite the first term as:
Σi {Ri + 800*(si - 0.5)} =
Σi∈S {Ri + 400} + Σi∈F {Ri - 400}
We can rewrite the second term as:
Σi {K*log2(ti/T)} = N*K*log2(tGM/T)
Where tGM is the geometric mean of the ti values.
Alternatively, we have:
N*R = Σi {Ri + 800*(si - K*log2(ti/T)/800 - 0.5)}
We can interpret -K*log2(ti/T)/800 as a time adjustment to the user’s score on problem i.
The graph below shows the time adjustment to the score for K=125, and T=30:

On CTS, the user gets 1 point for a very rapid successful solution and 0 points for a very slow one. The adjusted score here is up to 1.5 points for a very rapid success and down to 0.5 points for a very slow success. On CTS, the user gets 0 points for a failure, irrespective of the time that he takes. The adjusted score here is 0.5 points for a very rapid failure, and down to -0.5 points for a very slow failure. If the user takes the target time, there is no time adjustment, and the user gets 1 for a success and 0 for a failure.
In the method above, I have added rules to ensure that the user cannot benefit from a fast failure, or lose from a slow success. (N.B. A slow solution time just increases the time adjusted problem rating.) The ECF goes further and ensures that a player always gains from a win, even against a beginner. In theory, it is possible to obtain an infinite ECF grade by beating a constant stream of beginners. Clearly, this adjustment is motivated more by politics than by statistics!
(1). For every problem tackled, find the time adjusted problem rating by adding the rating time adjustment given by the graph below to the problem rating.
(N.B. The x-axis of this graph is the number of seconds taken by the user tackling the problem concerned.)
(2). For every problem successfully solved, add 400 points to the time adjusted problem rating. Discard any problems for which the resulting rating is less than the user’s current rating.
(3). For every unsuccessful attempt to solve a problem, subtract 400 points from the time adjusted problem rating. Discard any problems for which the resulting rating is more than the user’s current rating.
(4). Find the average of the results that were not discarded from steps (3) and (4). This average is the new estimate for the user's rating.
For ease of pencil and paper calculation, we could use a table of values in place of the graph. I have assumed that target solution time at which the user and problem ratings are valid is 30 seconds. (N.B. The average solution time on Chess Tempo is about 30 seconds.) I have also assumed that the user’s effective rating increases by 125 points for every doubling of the time that he spends on a problem (see the earlier article).
We can better understand this calculation by looking at the extreme values:
* For a very rapid successful solution, we add a time bonus of up to 400 points to the problem rating, and also add a success bonus of 400 points.
* For a very slow successful solution, we subtract a time penalty of up to 400 points from the problem rating, and add a success bonus of 400 points. If the result is less than the user’s rating, we discard the result.
* For a very rapid failure, we add a time penalty of up to 400 points to the problem rating, and subtract a failure penalty of 400 points. The result should be less than the user’s rating if he is scoring over 50%. If the result is more than his rating, we discard this result.
* For a very slow failure, we subtract a time penalty of 400 points from the problem rating, and also subtract a failure penalty of 400 points. The result should again be less than the user’s rating if he is scoring more than 50%. If the result is less than the user’s rating, we discard the result.
The corresponding method for calculating the rating of a problem from user ratings is similar to that for calculating a user rating from problem ratings. When we rate a problem rather than a user, the time adjustment has the same magnitude but the opposite sign (see the earlier article). The success/failure adjustment for each user also has the same magnitude and the opposite sign.
The method above is essentially the same as that I recommended previously, except that a linear approximation is used for the logistic distribution. The graph below shows the logistic distribution (used by the USCF) in red, the normal distribution (used by FIDE) in blue, and a linear approximation (used by the ECF) in yellow:
The normal distribution used by FIDE is the cumulative normal distribution with standard deviation 282.84, see:
http://www.chessbase.com/newsdetail.asp?newsid=4326.
The logistic distribution used by the USCF has the equation:
1/(1+10^[-d/400]
Where d is the rating difference between the user and the problem that he tackles.
The linear approximation used by the ECF has the equation:
0.5 + d/800
(N.B. I have converted ECF points to Elo points. 50 ECF points is equivalent to 400 Elo points.)
There is clearly very little difference between the logistic and normal distributions over the range of the graph. The linear approximation is accurate up to an expected score of about 0.85, or a rating difference of 300 points. The histogram below shows the distribution of success probabilities for Chess Tactics Server (CTS):

Most of the CTS users have success rates (and therefore expected scores) between 50% and 90%. The linear approximation overestimates the expected score and underestimates the user’s rating for scores of more than 85%. However, users could probably be dissuaded from scoring over 85% by the promise of miserly ratings.
Let:
R = user’s rating.
T = the target time at which the user and problem ratings are valid.
K = number of rating points that the user gains for each doubling of his clock time.
N = number of problems in the problem set.
Ri = rating of problem i.
si = user’s score on problem i.
ti = time that the user takes on problem i.
As in the previous article, the time adjusted rating difference between the user and problem i is given by:
di = R - Ri + K*log2(ti/T)
The user’s expected score on the problem set is:
Σi {0.5 + di/800}
The user’s actual score on the problem set is:
Σi {si}
Equating the user’s expected and actual scores gives:
Σi {0.5 + di/800} = Σi {si}
Σi {di} = 800 * Σi {si - 0.5}
Substituting for di gives:
Σi {R - Ri + K*log2(ti/T)} = 800 * Σi {si - 0.5}
N*R = Σi {Ri - K*log2(ti/T) + 800*(si - 0.5)}
= Σi {Ri + 800*(si - 0.5)} + Σi {K*log2(ti/T)}
Let S be the set of indices for the problems that the user solved correctly, and F be the set of indices for the problems that he solved incorrectly. We can then rewrite the first term as:
Σi {Ri + 800*(si - 0.5)} =
Σi∈S {Ri + 400} + Σi∈F {Ri - 400}
We can rewrite the second term as:
Σi {K*log2(ti/T)} = N*K*log2(tGM/T)
Where tGM is the geometric mean of the ti values.
Alternatively, we have:
N*R = Σi {Ri + 800*(si - K*log2(ti/T)/800 - 0.5)}
We can interpret -K*log2(ti/T)/800 as a time adjustment to the user’s score on problem i.
The graph below shows the time adjustment to the score for K=125, and T=30:
On CTS, the user gets 1 point for a very rapid successful solution and 0 points for a very slow one. The adjusted score here is up to 1.5 points for a very rapid success and down to 0.5 points for a very slow success. On CTS, the user gets 0 points for a failure, irrespective of the time that he takes. The adjusted score here is 0.5 points for a very rapid failure, and down to -0.5 points for a very slow failure. If the user takes the target time, there is no time adjustment, and the user gets 1 for a success and 0 for a failure.
In the method above, I have added rules to ensure that the user cannot benefit from a fast failure, or lose from a slow success. (N.B. A slow solution time just increases the time adjusted problem rating.) The ECF goes further and ensures that a player always gains from a win, even against a beginner. In theory, it is possible to obtain an infinite ECF grade by beating a constant stream of beginners. Clearly, this adjustment is motivated more by politics than by statistics!
Saturday, 15 December 2012
Pandolfini’s 111 Winning Endgames
I have got a copy of Pandolfini’s Chess Challenges 111 Winning Endgames:

This book has been on my bookshelf for a while, but I finally found time to work through it, so it is time to give it a brief review. I previously reviewed another book by the same author: Pandolfini’s Endgame Course. I found that book very helpful. I have been using it as problem book, working through it multiple times, with increasing intervals between each repetition. Nonetheless, I was still short of endgame problems with complete solutions that I could find reasonably easily from the diagram. Pandolfini’s 111 Winning Endgames looked worth a try, and was recommended by Dan Heisman.
As the name suggests, the book provides only 111 problems, but it is not expensive. It presents one problem per page, with the solution overleaf on the next page. Pandolfini gives points for the first move, and for each variation in the solution. The problems become progressively harder throughout the book. (Nonetheless, I scored as well on the later problems as on the earlier ones.) I solved each problem from the diagram, and wrote down my solution, before looking at the solution in the book. I tried hard to write down replies to all the reasonable defences.
Pandolfini’s solutions appeared to be mostly accurate and reasonably complete. I did, however, find some shortcomings. Pandolfini did not always give the best solution. Here is problem 91:

Pandolfini gives 1...Kc3, which forces a K+B+N vs. K endgame. Nonetheless, I found a much better move (see below). Pandolfini also occasionally failed to document important defensive moves.
Early on in the book, I got one solution completely wrong and one mostly wrong, and later I was completely stumped by problem 101:

(See below for the solution.) I also failed to write down variations that would have earned me points for several problems. Nonetheless, Pandolfini also failed in this respect. Overall, I did not do much worse than Pandolfini and his team of checkers and proof readers!
Pandolfini gives the scoring table:
Points Rating Percent
400 1800 91%
350 1600 80%
300 1400 68%
250 1200 57%
200 1000 46%
(N.B. I added the final column - there were 438 points in all.) This table is rather odd. Going from 46% to 91% is an 800 point increase, which is about twice what it would be if we treated the problem set as an opponent, and used a standard rating calculation. I have serious doubts about this table, but Pandolfini’s assessment that an 1800 player should score about 90% looks reasonable to me. On that basis, I would expect a 1400 player to score about 50%.
Despite the minor shortcomings, I believe that this is a good book, and would recommend it. The problems are good, and the solutions appear to be reliable (with the exceptions noted). The book is well presented, easy to use and inexpensive. I expect that the book is most suitable for players in the 1400 to 1800 range. 111 problems is not going to make much difference to the stronger players, but it is a worthwhile contribution nonetheless, and good books like this are hard to find.
What about the solutions to the problems above?
For number 91, 1...Nc1 is mate in 8. I did not look as far as the mate, but I did see that I could force mate or promote my pawn to a queen
For number 101, the solution is 1...Re3+. If White plays 2.Rxe3, Black replies 2...Qa3+. 3.Qc2 or 3.Qd2 allows a skewer on the second rank. After 3.Qd4, Black plays 3.…Qc4+ which forces a skewer on White’s queen, wherever the king goes. Not easy to see!
This book has been on my bookshelf for a while, but I finally found time to work through it, so it is time to give it a brief review. I previously reviewed another book by the same author: Pandolfini’s Endgame Course. I found that book very helpful. I have been using it as problem book, working through it multiple times, with increasing intervals between each repetition. Nonetheless, I was still short of endgame problems with complete solutions that I could find reasonably easily from the diagram. Pandolfini’s 111 Winning Endgames looked worth a try, and was recommended by Dan Heisman.
As the name suggests, the book provides only 111 problems, but it is not expensive. It presents one problem per page, with the solution overleaf on the next page. Pandolfini gives points for the first move, and for each variation in the solution. The problems become progressively harder throughout the book. (Nonetheless, I scored as well on the later problems as on the earlier ones.) I solved each problem from the diagram, and wrote down my solution, before looking at the solution in the book. I tried hard to write down replies to all the reasonable defences.
Pandolfini’s solutions appeared to be mostly accurate and reasonably complete. I did, however, find some shortcomings. Pandolfini did not always give the best solution. Here is problem 91:
Pandolfini gives 1...Kc3, which forces a K+B+N vs. K endgame. Nonetheless, I found a much better move (see below). Pandolfini also occasionally failed to document important defensive moves.
Early on in the book, I got one solution completely wrong and one mostly wrong, and later I was completely stumped by problem 101:
(See below for the solution.) I also failed to write down variations that would have earned me points for several problems. Nonetheless, Pandolfini also failed in this respect. Overall, I did not do much worse than Pandolfini and his team of checkers and proof readers!
Pandolfini gives the scoring table:
Points Rating Percent
400 1800 91%
350 1600 80%
300 1400 68%
250 1200 57%
200 1000 46%
(N.B. I added the final column - there were 438 points in all.) This table is rather odd. Going from 46% to 91% is an 800 point increase, which is about twice what it would be if we treated the problem set as an opponent, and used a standard rating calculation. I have serious doubts about this table, but Pandolfini’s assessment that an 1800 player should score about 90% looks reasonable to me. On that basis, I would expect a 1400 player to score about 50%.
Despite the minor shortcomings, I believe that this is a good book, and would recommend it. The problems are good, and the solutions appear to be reliable (with the exceptions noted). The book is well presented, easy to use and inexpensive. I expect that the book is most suitable for players in the 1400 to 1800 range. 111 problems is not going to make much difference to the stronger players, but it is a worthwhile contribution nonetheless, and good books like this are hard to find.
What about the solutions to the problems above?
For number 91, 1...Nc1 is mate in 8. I did not look as far as the mate, but I did see that I could force mate or promote my pawn to a queen
For number 101, the solution is 1...Re3+. If White plays 2.Rxe3, Black replies 2...Qa3+. 3.Qc2 or 3.Qd2 allows a skewer on the second rank. After 3.Qd4, Black plays 3.…Qc4+ which forces a skewer on White’s queen, wherever the king goes. Not easy to see!
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