Lessons from a grandmaster

The next game is the only tournament game I’ve ever played against a grandmaster. My opponent was Georgian GM Mikheil Kekelidze. Not surprisingly, he outplayed me in every phase of the game. Obviously, I’ve still got a lot of work to do!

[Event "Marchand Open"]
[Site "Rochester, NY"]
[Date "2012.05.30"]
[Round "3"]
[White "Mikheil Kekelidze"]
[Black "Aaron Demby Jones"]
[Result "1-0"]
[WhiteELO "2563"]
[BlackELO "2106"]

1.d4 Nf6 2.c4 e6 3.Nf3 c5
{Not my usual repertoire, but I hadn't been getting good results with
the Queen's Indian Defense lately.}
4.d5 exd5 5.cxd5 d6 6.h3
{White plays to increase the value of his space advantage by avoiding
trades. (His move prevents a future Bc8-g4.)}
6...g6 7.e4 Bg7
( 7...Nxe4 $4 8.Qa4+ )
8.Bd3 O-O 9.O-O Re8
{Pressuring the e-pawn.}
{White defends. Now Black has to figure out how to develop his
remaining minor pieces. It's always a struggle in the Modern Benoni.}
{The start of the typical queenside expansion plan.}
{The usual reply. White has no hurry in this position and first plays
to restrict Black's possibilities before embarking on his own plans.}
{A bit clumsy, but there was no clear alternative.}
{White puts his finger on Black's awkward coordination. The d6-pawn
needs defense.}
{An undesirable move from a future tactical point of view. (See
White's 17th move.)}
13.Re1 Nh5
{Dangerous, since if White lands g2-g4 at the right moment, Black will
lose a lot of time.}
( 13...Rb8 )
( 13...b6 )
14.Bh2 Ne5
( 14...Bh6 {was an interesting alternative.} )
{Retaining the bishop pair and threatening Nf3xe5, discovering an
attack on the knight on h5.}
15...Nxf3+ 16.Bxf3 Nf6
{Forced. Now Black has traded off a pair of knights but at the cost of
17.e5 $1
{Very energetic play. White takes advantage of Black's unfortunate
queen placement.}
17...dxe5 18.d6
{The point.}
18...Qb6 19.Bxe5 Be6 $2
( 19...Bd7 {was necessary to keep the rook eyeing the bishop on e5.} )
20.a5 $1 Qb4
( 20...Qxb2 $2 21.Nd5 {and White wins. Had Black played 19. ... Bd7,
this resource would not be available to White.} )
21.Ra4 Bb3
{The point of 19. ... Be6. But this is based on faulty calculation.}
{Starting a favorable forcing sequence.}
22...Bxd1 23.Rxb7 Bxf3 24.gxf3 Nh5
{I had foreseen this sequence on move 19 and thought that I was
winning material here. However...}
{... I overlooked this defense in my calculations. White has won a
clean pawn.}
25...Red8 26.Ne4 Bxe5 27.Rxe5 f5
{Trying to eliminate the dangerous d-pawn.}
28.Nxc5 Rxd6 29.Re6 Rxe6 $2
{Overly cooperative.}
( 29...Rd2 30.Rxa6 Rxa6 31.Nxa6 Rxb2 {and Black has some chances to
draw.} )
30.Rxe6 Nf4
{Black no longer has any meaningful counterplay. White is clearly
winning now.}
31.Rxa6 Rd8 32.Ne6 Rd1+ 33.Kh2 Nd3
{Desperately hoping for some sort of mating net.}
34.Rd6 $1
{A nice tactical solution. White forces trades.}
34...Re1 35.Rd8+ Kf7 36.Ng5+ Kf6 37.Nxh7+ Kg7 38.Rxd3 Kxh7 39.a6
{And I resigned. White's pawn promotes without much resistance.}
( 39.a6 Ra1
( 39...Re7 40.Ra3 Ra7 41.b4 Kg7 42.b5 Kf7 43.b6 )
40.Ra3 )

Bishop stifling

I played the following game with the White pieces against WFM Anna Levina. The middlegame shows the thematic idea of playing with a good knight against a bad bishop. The finale shows off a beautiful twist on the same idea.

[Event "Grand Prix"]
[Site "Rochester, NY"]
[Date "2011.02.10"]
[Round "3"]
[White "Aaron Demby Jones"]
[Black "Anna Levina"]
[Result "1-0"]
[WhiteElo "2106"]
[BlackElo "2111"]
[ECO "D31"]
[PlyCount "85"]
[Variant "Standard"]
[Opening "Semi-Slav Defense: Accelerated Move Order"]

1.d4 d5 2.c4 e6 3.Nc3 c6 4.e3 Bd6 5.Nf3 Nd7 6.e4
{Since Black isn't playing her knight out to f6, why not?}
6...dxe4 7.Nxe4 Bb4+ 8.Bd2 Bxd2+ 9.Qxd2
{The trade of these bishops seems to help White, since Black's
remaining bishop is worse than White's.}
9...Ndf6 10.Nc3
{Trying to make Black suffer from redundant knights.}
10...Ne7 11.Be2
( 11.Bd3 {looks more aggressive.} )
11...O-O 12.O-O Qc7 13.Rad1 Ng6 14.Rfe1 b6
{Black plans to slowly unroll with Bc8-b6, c6-c5, etc.}
15.b3 $6
{White drifts here for a few moves.}
15...Bb7 16.Qb2 $6 Nf4 17.c5 $1
{White wakes up and plays with a plan. Black's bishop shall not free
itself so easily after all.}
17...Nxe2+ $6
{This trade doesn't seem to help Black much. White can easily end up
with a favorable knight vs. bishop scenario.}
18.Qxe2 bxc5 19.dxc5 Qa5 20.Qe3 Nd5 21.Nxd5 exd5 22.b4 $1
{A rather deep pawn sacrifice.}
22...Qxa2 23.Nd4 Rab8
{If Black passes, say, with}
( 23...h6 {, White wins material via} 24.Ra1 Qc4 25.Rec1 Qxb4 26.Rab1
{The knight, having feinted to the queenside, now goes for the throat
on the kingside.}
24...Qb2 25.Rd4
{Blocking the line and threatening Qe3-e5.}
25...Bc8 26.Nxg7 $6
{A tactical solution, but simpler was}
( 26.Ne7+ Kh8 27.Nxc6 {and Black's position collapses.} )
26...Kxg7 27.Qg5+ Kh8 28.Qf6+ Kg8 29.Rg4+ Bxg4 30.Qxb2
{After the dust clears, White has a queen against rook and bishop. As
usual, the queen manages to prove superior by exploiting the color
complex opposite the bishop.}
30...a5 31.Qf6 Rxb4 32.h3
{Getting some luft with tempo.}
32...Be6 33.Re3 Rfb8 34.Qh6
{Trying to keep the king cornered by cutting off f8.}
{Intending Bf5-g6, plugging up the g-line.}
{Trying to cross her plans with h4-h5.}
35...Rb3 36.Rxb3 Rxb3 37.Qg5+ Bg6 38.h5
{The bishop is caught. But all of a sudden, Black is whipping up some
counterplay with a passed a-pawn.}
38...a4 39.h6 $1
{The 'drive-by' in action! White doesn't need to capture the bishop.
The pawn is actually more valuable in this position since it
coordinates in deadly fashion with the queen on the dark squares.}
39...f6 40.Qxf6 Rb7 41.Qxc6 Rb1+ 42.Kh2 Ra1 $2
{Losing on the spot, but otherwise White wins easily by pushing the
{Black resigns. Qf6-g7 mate can't be stopped. A thematic finish!}

My chessvideos.tv analysis can be found here.

Probability and number theory, part II

In this post, we will explore the following question: If we choose two integers uniformly at random, what is the probability that they are relatively prime?

We start with a preliminary result. The Riemann zeta function, \( \zeta(s) \), is ubiquitous in number theory. Although the most interesting version of it is the complex-valued one, today we only need the real version. It is defined as follows: for any real number \( s \) greater than \( 1, \)

\( \zeta(s) = \displaystyle\sum_{n=1}^{\infty} n^{-s} \).

The Euler product  for the zeta function is the following alternative expression:

\( \zeta(s) = \displaystyle\prod_{p} \frac{1}{1 – p^{-s}}, \)

where the product is taken over every prime \( p \). To see why this formula is true, we start with the right-hand side and use a geometric series expansion to obtain

\(\displaystyle\prod_{p}\frac{1}{1-p^{-s}}=\displaystyle\prod_{p}(1+p^{-s}+p^{-2s}+\cdots). \)

By the Fundamental Theorem of Arithmetic, every positive integer can be expressed as a unique product of primes, up to order. (We consider \( 1 \) to be the product of no primes; i.e., the empty product.) Thus, there is a one-to-one correspondence between the terms in the sum \( \sum_{n=1}^{\infty} n^{-s} \) and the terms in the expansion of the product \( \prod_{p}(1+p^{-s}+p^{-2s}+\cdots).\) Hence,

\( \displaystyle\prod_{p}\frac{1}{1-p^{-s}}=\displaystyle\sum_{n=1}^{\infty}n^{-s}=\zeta(s).\)

Now back to our original problem! Suppose we choose \( a, b \in \mathbb{Z} \) uniformly at random. Fix an arbitrary prime \( p \). The probability that both \( a \) and \( b \) are divisible by \( p \) is \( \left(\frac{1}{p}\right)\left(\frac{1}{p}\right) = p^{-2} \). Hence, the probability that they are not both divisible by \( p \) is \( 1 – p^{-2} \).

The condition that \( a \) and \( b \) are relatively prime is equivalent to the condition that, for every prime \( p \), \( a \) and \( b \) are not both divisible by \( p \). Furthermore, for distinct primes \( p \) and \( q \), the events that \( a \) and \( b \) are not both divisible by \( p \) and \( a \) and \( b \) are not both divisible by \( q \) are mutually independent. (In fact, more precisely, the events that \( a \) and \( b \) are not both divisible by \( m \) and \( a \) and \( b \) are not both divisible by \( n \), for relatively prime integers \( m \) and \( n \), are mutually independent.)

Thus, the probability that \( a \) and \( b \) are relatively prime is equal to

\( \displaystyle\prod_{p}(1-p^{-2}), \)

where the product is taken over every prime \( p \). But

\( \displaystyle\prod_{p}(1-p^{-2}) = \displaystyle\prod_{p}(\frac{1}{1-p^{-2}})^{-1}=(\displaystyle\prod_{p}\frac{1}{1-p^{-2}})^{-1} = \frac{1}{\zeta(2)}.\)

Hence the answer is \( \frac{1}{\zeta(2)}\approx 0.608. \, \spadesuit \)

The closed-form expression for \( \zeta(2) \) is actually well-known (it equals \( \frac{\pi^{2}}{6}) \), but that’s another story for another day. The result also generalizes to more than two integers, a nice exercise for the reader.

In-between moves

Here is a miniature I played with Black against Class B player Jack Oleksyn. This game highlights the power of what FM Carsten Hansen calls “ESTs” (i.e., equal or stronger threats). White’s threats find themselves dominated by Black’s ESTs throughout.

[Event "Saturday tournament"]
[Site "Rochester, NY"]
[Date "2010.10.17"]
[Round "1"]
[White "Jack Oleksyn"]
[Black "Aaron Demby Jones"]
[Result "0-1"]
[WhiteELO "1675"]
[BlackELO "2106"]

1.d4 Nf6 2.Nf3 g6 3.Nc3
{Playing for e2-e4, which would transpose to a Pirc/Modern.}
{Holding up the center.}
( 4.Bf4 {is the Barry Attack, and probably more accurate.} )
4...Bg7 5.Qd2 $6
{This move doesn't mesh well with the bishop on g5.}
5...Ne4 $1
{Black goes for immediate equality by trying to snag the bishop pair.}
6.Qf4 $2
{A serious error. Now the White pieces trip over each other.}
( 6.Nxe4 $2 dxe4 {and the d4 pawn hangs.} )
( 6.Qe3 {was best.} )
6...h6 $6
{Winning material, but not the most accurate.}
( 6...f6 {was more precise--see the followup note.} )
7.Nxe4 $6
( 7.Bxe7 Kxe7 {and Black has been a bit discomboulated, although he
remains up a piece.}
( 7...Qxe7 $2 8.Nxd5 {is unpleasant.} )
7...dxe4 8.Ne5
{Threatening mate on f7.}
( 8...Bxe5 $2 9.Qxe5 {hitting the rook in the corner is an unpleasasnt
surprise.} )
9.Bh4 $6
{White expected g6-g5, but Black has other plans.}
9...Qxd4 $1
{Black doesn't bother picking off the useless White pieces on the
kingside, instead focusing his attack on the exposed White queenside.
Now e5 and b2 are hanging, and g6-g5 is still in the cards.}
{Probably expecting f7xg6.}
10...Qxb2 $1
{With White's queen away from the defense, his queenside is falling
apart. }
{Capitulation, but alternatives weren't much better.}
( 11.Rd1
( 11.Rc1 Bc3+ 12.Kd1 Nc6 {and a rook comes to d8 with devastating
effect.} )
( 11.Qc1 Qc3+ 12.Kd1 fxg6 {and Black has an overwhelming position
with too many threats.} )
11...Bc3+ 12.Rd2 Qc1# )
{Cashing in.}
{White tries to hide on the kingside.}
{Another in-between move! Black seals the door on any hope of White
development. Now the king and the bishop on f1 become permanently
( 13.Qxe3 Bd4 )
( 13.Kxe3 fxg6 {might have been better, but it's all rather depressing
for White.} )
{Finally there was nothing better to do than take the knight! Black is
now up a whole rook and a knight.}
{Of course White could resign, but who can resist threatening mate?}
{Staying alert. Black doesn't care about any of his pawns if he can
quickly mobilize his extra pieces.}
15.Qxb7 Nd7 16.Bxe7 Rab8
{Suddenly White's back rank looks vulnerable.}
{Threatening Qe4xe6+, winning back some material. But Black has a very
pretty counter.}
17...Qxf1+ $1
{Remeniscent of a puzzlebook checkmate!}
18.Kxf1 Rb1#
{The pawn on e3 pulls its weight after all. Black's pieces completely
overpowered their White counterparts.}

Here’s my chessvideos.tv video analysis.

Chaos on the board

This game is a crazy encounter between myself playing the black pieces and six-time Venezuelan chess champion László Tapasztó. It features several mutual blunders, but also many twists and turns, and explosive tactics. Not to be missed!

[Event "Wednesday night chess"]
[Site "Rochester, NY"]
[Date "2012.05.26"]
[Round "?"]
[White "László Tapasztó"]
[Black "Aaron Demby Jones"]
[Result "1/2-1/2"]
[WhiteELO "2301"]
[BlackELO "2106"]

1.d4 Nf6 2.Bg5
{The Trompowsky attack.}
{A combative continuation. Black immediately attempts to counterattack
the dark squares.}
{This move was already unknown to me. Out of book on move 3!}
3...cxd4 4.Qxd4 Nc6 5.Qh4
{The position has some features resembling the Open Sicilian. Black
has more center pawns, but lags behind in piece mobility. White is
prepared to castle long and potentially attack in the kingside or the
{I decided to set up my pieces similar to a Sicilian Dragon.}
6.e4 Qa5
{Threatening 7...Nxe4 8.Qxe4 Qxg5.}
{Breaking the pin.}
{Since White has committed his king position to the queenside, I
decided to immediately focus my attack there. Black can mobilize very
quickly with Ra8-c8.}
{Safeguarding the a2 pawn. A typical tidying move.}
( 8...Nxe4 9.Nxe4 Qxa2+ 10.Kc1 Qa1+ 11.Kd2 Qxb2 {was also plausible,
where Black has three pawns for the sacrificed knight. I couldn't
easily evaluate that position over the board, so I settled for
increasing the pressure without making any sacrifices for now.} )
{Probably forced, otherwise Nc6-b4 was very strong. (Black would be
ready to sacrifice the exchange on c3.)}
9...a6 $5
{It seems more natural to try to start mobilizing the kingside, but in
fact it is not so easy to do so. I decided to leave my king in the
middle and launch an immediate assault on White's king. White's pieces
are anticipating that Black will eventually castle short, so they are
now temporarily misplaced. The idea of 9...a6 is of course to play
b7-b5-b4 and open lines.}
10.f4 $1
{White wastes no time in changing plans, starting his attack in the
middle rather than the kingside.}
10...b5 11.f5 b4 $5
( 11...Bd7 {was more sedate, but possibly better.} )
( 12.axb4 $2 Nxb4 {and Black's attack is suddenly overwhelming with
the threat of Rc8xc3.} )
12...bxc3 13.exf7+ Kd8
( 13...Kxf7 {is also possible, but it appeared dangerous to step onto
the a2-g8 diagonal with White's bishop ready to pounce.} )
14.Rxd6+ $2
{This move wins a few pawns, but should lose by force! White's pieces
become hopelessly disorganized in their greed.}
14...exd6 15.Bxf6+ gxf6 16.Qxf6+
{The point. But White's queen will be buried away from the action on
16...Kd7 $6
( 16...Kc7 {was more accurate to avoid a later queen check on h3.} )
{With best play, White is now lost.}
17...Rb8 $5
{This move doesn't throw away the win, but }
( 17...cxb2 {was simplest.} )
18.b3 Ne5 $2
{Plausible, but now White can defend! }
( 18...Bh6 {was a shot!} 19.Qxh7
( 19.Qf6 Qxa3 20.Qxc3 Qc1+ 21.Ka2 Nb4+ {and Black mates soon.} )
19...Rxb3+ 20.cxb3 c2+ {and Black will eventually mate.} )
19.Qxh7 $1
{With the defensive idea of Qh7-h3+ and Qh3xc3, should Black abandon
the defense of the c3 pawn. If White can consolidate, he has a winning
material advantage.}
19...Qxa3 20.Qh3+ Ng4 $5
{An amusing idea. It's not every day you can "hang" a piece with
( 21.Qxg4+ $4 Kc7 {and White's queen is on the wrong defensive
circuit.} )
{Black at least tries to win some material back. Unfortunately, the
rook on h1 is not a very useful piece to win ...}
22.e5 $6
( 22.Nf3 {was simplest and more natural. } )
22...Nd1 $5
{An excellent twist! Black spurns the rook in favor of extra pressure
on the king. The point is that the White queen must stay on the a1-h8
diagonal to prevent Qb2#.}
23.Qd4 $2
{Allowing an immediate draw. }
( 23.Qa1 $4
( 23.e6+ $1 Ke7 24.Qh8 {and by covering the long diagonal as well
as h6, White wins!} )
23...Rxb3+ 24.cxb3 Qxb3+ 25.Kc1 Bh6# )
23...Rxb3+ $1
{Black wastes no time in forcing a perpetual check. Any other
continuation is likely losing.}
24.cxb3 Qxb3+ 25.Ka1 Qa3+ 26.Kb1 Qb3+ 27.Ka1
{The king can never escape to c1 because of Bf8-h6+. Of course, if
Black pushes his luck by playing for a win, White will be able to
mobilize his pieces with tempo and consolidate. For instance, }
( 27.Ka1 Nc3 $4 28.e6+ Ke7 29.Qh4+ Kxe6 30.Bc4+ {and White wins.} )

Here’s my chessvideos.tv video commentary as well.

Probabilistic approach to the Euler totient function

Euler’s totient function, \( \varphi(n) \) is a well-known arithmetic function from number theory. For any \(n \in \mathbb{N} \), \( \varphi(n) \) is defined to be the number of integers \( k \) in the range \( 1 \leq k \leq n \) for which \( \gcd(n, k) = 1 \). There is a formula for computing \( \varphi(n) \):

\( \varphi(n) = n \displaystyle\prod_{p|n} (1 – \frac{1}{p}), \)

where the product is taken over all the distinct primes \( p \) dividing \( n \). This formula can be proved in a few ways. One rests on the fact that \( \varphi(n) \) is multiplicative. Another uses an elaborate form of inclusion-exclusion. The one I am presenting now is based on probability.

Suppose we fix a positive integer \( n \) and choose uniformly at random an integer \( k \) in the range  \( 1 \leq k \leq n \). What is the probability that \( \gcd(n, k) = 1 \)? We answer this question in two ways. On one hand, we know by definition that there are \( \varphi(n) \) successful outcomes and \( n \) total outcomes, so the probability is equal to \( \frac{\varphi(n)}{n} \). On the other hand, we proceed more constructively. In order for the integer \( k \) to be relatively prime to \( n \), it is necessary and sufficient for each prime divisor \( p \) of \( n \) that \( k \) must not be divisible by \( p \).

The probability that \( k \) is divisible by \( p \) is \( \frac{1}{p} \). Hence, the probability that \( k \) is not divisible by \( p \) is \( 1 – \frac{1}{p} \). Furthermore, for distinct primes \( p \) and \( q \), the events that \( k \) is not divisible by \( p \) and \( k \) is not divisible by \( q \) are mutually independent. (More precisely, the events that \( a \) and \( b \) are not both divisible by \( u \) and \( a \) and \( b \) are not both divisible by \( v \), for relatively prime integers \( u \) and \( v \), are mutually independent.)

Thus, the probability that \( k \) is relatively prime to \( n \) is

\( \displaystyle \prod_{p|n} \left(1 – \frac{1}{p}\right). \)

Equating our two results, we have

\( \displaystyle\frac{\varphi(n)}{n} = \displaystyle\prod_{p|n}(1-\frac{1}{p}).\)

In other words,

\( \varphi(n) = n \displaystyle\prod_{p|n}(1-\frac{1}{p}). \, \spadesuit \)

We’ll use a similar approach in the future to show that the probability of choosing uniformly at random a pair of relatively prime integers is \( \frac{1}{\zeta(2)}. \)

Everything you ever wanted to know about temperaments (but were afraid to ask)

Here’s a historical paper I wrote on different musical temperaments from Antiquity until the late Renaissance.

A History of Temperaments

Recommended reading for anyone who is confused about the differences between just, mean, and equal temperament.

Here are some supplementary audio samples:

Just major third:

Pythagorean major third (ditone):

Syntonic comma:

Just major third cycle:

The resulting comma (“diesis”):

Transforming advantages

Here’s a nice positional game I played recently as White against USCF expert Robert Radford. What started out as a nice space advantage for me first changed into a structural advantage, then into a dynamic advantage, and then finally into a decisive structural and dynamic endgame advantage. (Had my opponent not resigned, I would have concluded with a material advantage!)

[Event "Ventura Quads"]
[Site "Ventura, CA"]
[Date "2011.10.04"]
[Round "1"]
[White "Aaron Demby Jones"]
[Black "Robert Radford"]
[Result "1-0"]
[WhiteELO "2106"]
[BlackELO "2030"]

1.d4 Nf6 2.c4 b6
{A rare move order. If Black wants to play the English defense, more
standard is 1... e6 and 2...b6 in order to have the later possibility
of f7-f5.}
3.Nc3 Bb7 4.Nf3
{White decides to steer the game back into standard Queen's Indian
Defense lines. The more testing options try to control the e4 square
( 4.Qc2 )
( 4.f3 )
4...e6 5.a3
{Transposing to the Petrosian variation of the Queen's Indian Defense.
White's move prevents Bf8-b4 and thus indirectly fights for the e4
5...Be7 $6
( 5...d5 {is considered more accurate. Now White can make a space
grab.} )
{Playing against the bishop on b7.}
6...O-O 7.e4 exd5 8.cxd5 d6
{The pawn structure favors White. Black's bishop on b7 doesn't have
much scope.}
9.Bd3 c6
{Trying to free the bishop.}
( 10.dxc6 $6 Nxc6 {develops Black's knight for him.} )
{A concession. Now when White captures on c6, Black simply loses time.
( 10...cxd5 11.exd5 Nxd5 12.Nxd5 Bxd5 13.Bxh7+ Kxh7 14.Qxd5 {is
pleasant for White.} )
11.dxc6 Bxc6 12.Nd4 Bb7
{The bishop on b7 is no longer stymied, but now the pawn on d6 is a
potential target and White's knight is beautifully placed on d4.}
( 13.Bf4 {looks more concrete, eyeing d6.} )
{Pressuring the e4 pawn.}
{Natural, but unnecessary. }
( 14.Nf5 {defended tactically:} 14...Ncxe4 $2 15.Nxe4 Nxe4 16.Qg4 {and
the double threat wins for White.} )
{Grabbing the bishop pair while trading a pair of minor pieces helps
Black ease his spatial disadvantage.}
15.Qxd3 Nd7
( 15...d5 {is premature due to} 16.e5 )
{Pressuring d6.}
16...Nc5 17.Qd4 $6
( 17.Qe2 {was most accurate. } )
( 17.Qd2 $2 Nb3 )
17...Bf6 $1
{Liquidating into a queenless middlegame where Black is much closer to
equality. I completely missed this move, believing Nc5-e6 was forced.}
18.Qxd6 Qxd6 19.Nxd6 Ba6 20.Rfd1 Nd3 21.Nd5 $1
{The only try for an advantage. Now, despite the soon-to-be equal
material and lack of weaknesses, Black lags in piece activity.}
( 21.Rab1 {trying to keep the extra pawn fails to the simple tactic}
21...Nxb2 )
21...Bxb2 22.Ra2
{Threatening Rd1xd3.}
( 22...Rad8 $2 23.Rxb2 Nxb2 24.Ne7+ Kh8 25.Nxf7+ Rxf7 26.Rxd8+ Rf8 27.
Rxf8# {was a fun fantasy variation.} )
( 23.Nc7 Bxd6 $6
( 23...Rad8 {I eventually noticed, which immediately equalizes.} )
24.Nxa6 Nc5 25.Bxc5 Bxc5+ 26.Nxc5 bxc5 27.Rc2 Rac8 28.Rdc1 {and White
picks up a pawn and has some winning chances in the rook endgame.} )
{The knights were getting out of hand, so Black trades one off.}
24.Rad2 Bxd5 25.Ne7+ $6
( 25.exd5 {is probably better. I couldn't decide over the board
whether the newly created d-pawn would be strong or weak.} )
25...Kh8 26.Nxd5 Nb2 27.Rc1
{f2-f4 is a threat, removing the guard.}
{Trying to activate the rooks.}
28.Rdc2 Rxc2 29.Rxc2 Nd3 30.Nb4 $1
{A nice positional idea. White forces a favorable change in the pawn
{More or less forced. }
( 30...Nf4 31.Nc6 Rc8 32.Rd2 $1 )
{Now b4-b5 cannot be prevented. By fixing Black's queenside majority,
White maintains an advantage.}
31...h6 $6
( 31...a5 32.bxa5 bxa5 33.Rc5 {wins a pawn, although White would still
have to prove that his 4-3 advantage on the kingside is sufficient to
win with no other pawns on the board. The game continuation creates
luft, but perhaps Kh8-g8 was more natural.} )
{As planned.}
32...Rd8 33.Ra2 $2
( 33.f4 {to shut down any tactical ideas Black has involving Rd8-d1+
and Be5xh2 was stronger. Don't rush in the endgame!} )
33...Rd3 $6
( 33...Rd1+ 34.Kf2 Bxh2 {was probably the best try, playing for
complications.} )
34.Kf2 Bb8 $2
{My opponent offered me a draw with this move. Ironically, I believe
it loses by force!}
35.Rd2 $1
{Now a rook trade is forced due to the potential threat of Rd2-d8+.
The resulting bishop endgame is winning for White.}
35...Rxd2+ 36.Bxd2
{White's advantage is threefold: more active king, more active bishop,
and superior pawn structure.}
36...h5 $2
( 36...Kg8 37.Ke3 Kf8 38.Kd4 Ke7 39.Kd5 Kd7 {and at least Black has
kept the White king out of c6. But White would win regardless by
slowly advancing his kingside pawns. He is effectively up a pawn
thanks to the bind on the queenside.} )
37.Ke3 $1
{Now the White king reaches c6 with decisive effect.}
{It makes no difference whether Black grabs this pawn since he cannot
mobilize his kingside.}
38.Kd4 Bd6 $6
{As usual, all moves are bad in a lost position, but allowing Kd4-d5
with tempo certainly cannot be best. Black has completely lost the
thread of this endgame.}
39.Kd5 Bg3
( 39...Bc5 {was probably his first intention, but after} 40.Bf4 {,
Bf4-b8 is unstoppable.} )
{Simply threatening Kc6-b7.}
40...Bf2 41.Bf4 $1
{And Black resigned in view of Bf4-b8. White will win both Black's
queenside pawns, after which Black will have to sacrifice his bishop
for White's b pawn. The resulting position is then easily winning for
White. White's strategic bind with b4-b5 paid off handsomely!}

Combinatorics, Contour Integrals, and Bessel Functions … oh my?!

Here’s an interesting exercise that can be solved using complex integration and basic combinatorics. Consider the following integral:

 \( \displaystyle\int_{0}^{2\pi} \! \exp{(2\cos \theta)} \, \mathrm{d} \theta. \)

No apparent tricks from ordinary calculus appear to be helpful, so we turn instead to complex variables. The limits of integration suggest a parametrization of the unit circle. Let \( C \) denote the unit circle, oriented counterclockwise. We’d like to convert our original integral into a contour integral of the form \( \oint_C \! f(z) \, \mathrm{d} z \) for an appropriate function \( f(z) \). We can parametrize the contour by letting \( z = e^{i\theta}, 0 \le \theta \le 2\pi. \) Then \( \mathrm{d} z = ie^{i\theta} \mathrm{d} \theta \), so that \( \mathrm{d} \theta = \frac{1}{i} e^{-i\theta} \mathrm{d} z \), or in other words, \( \mathrm{d} \theta = \frac{1}{i} z^{-1} \mathrm{d} z. \) We also observe that, on our contour, \( z + z^{-1} = e^{i\theta} + e^{-i\theta} = (\cos \theta + i \sin \theta) + (\cos \theta – i \sin \theta) = 2 \cos \theta \). Hence,

\( \displaystyle\int_{0}^{2\pi} \! \exp{(2\cos \theta)} \, \mathrm{d} \theta = \displaystyle\oint_C \! \exp{(z + z^{-1})}\frac{1}{i} z^{-1} \, \mathrm{d} z. \)

Within the region bounded by the contour, the integrand has exactly one singularity. More precisely, it has an essential singularity at \( z=0 \). Thus, by Cauchy’s residue theorem, the value of the integral is equal to \( 2\pi i \) times the integrand’s residue at \( z=0 \).

Let \( f(z) = z^{-1} \exp{(z + z^{-1})} \). To find the residue of \( f \) at \( 0 \), we consider the Laurent series for \( f \) centered at \( 0 \). (A simple approach here is to write \( \exp{(z + z^{-1})} = \exp{(z)} \exp{(z^{-1})} \), but we shall take a more roundabout path. The reader is invited to attempt the direct route.) Using the standard series for the exponential, we obtain

\( f(z) = z^{-1} \displaystyle\sum_{n=0}^{\infty} \frac{(z + z^{-1})^n} {n!}. \)

The residue of \( f \) at \( z = 0 \) is defined as the coefficient of the \( z^{-1} \) term in the Laurent series for \( f \) centered at \( 0 \). From the above expansion, we see that in order to determine this coefficient, it suffices to determine the coefficient of \( z^0 \) within the infinite sum.

Fix a nonnegative integer \( n \) and consider the binomial \( (z + z^{-1})^n \). We wish to determine the coefficient of \( z^0 \) in the expansion of this binomial. In order to obtain a \( z^0 \) term in the first place, we see that the terms \( z \) and \( z^{-1} \) must multiply together in pairs. Hence, if \( n \) is odd, the coefficient of \( z^0 \) is \( 0 \). Thus, we may focus our attention on the nonnegative even integers.

Consider the binomial \( (z + z^{-1})^{2n} \). What is the coefficient of \( z^0 \) in its expansion? If we imagine \( 2n \) copies of the expression \( (z + z^{-1}) \) being multiplied together, the only way we obtain a nonzero \( z^0 \) term is when we choose the \( z \) and \( z^{-1} \) terms in pairs. There will be a total of \( n \) pairs. Of the \( 2n \) copies of \( z \), we can choose any \( n \) of them, and then the remaining \( n \) choices will have to be \( z^{-1} \) terms. Hence, the coefficient of \( z^0 \) is \( {{2n}\choose{n}}\).

Since, in our original series, each term \( (z + z^{-1})^n \) is divided by \( n ! \), we must divide \( {{2n}\choose{n}} \) by \( (2n)! \). By definition,  \( {{2n}\choose{n}} = \frac{(2n)!}{n! n!} \), so  \( {{2n}\choose{n}} \) divided by \( (2n)! \) is simply \( \frac{1}{n! n!} \).

Thus, the desired residue is seen to be

\( \displaystyle\sum_{n=0}^{\infty} \frac{1}{n! n!}. \)

It follows by the residue theorem that the value of the integral we seek is equal to

\( \displaystyle 2\pi i \frac{1}{i} \sum_{n=0}^{\infty} \frac{1}{n! n!} = 2\pi \displaystyle\sum_{n=0}^{\infty} \frac{1}{n! n!}. \, \spadesuit\)

 It turns out that the original integral is not entirely unmotivated, as it can also be written as \( 2\pi I_{0}(2) \), where \( I_{n}(z) \) denotes the modified Bessel function of the first kind. This result can be seen from the integral formulas for these functions.


Percussion quartet

Here’s my senior composition project from Brown, written in 2008–2009. Titled “Percussion Quartet,” it’s scored for two pianists and two percussionists, much in the spirit of Bartók’s well-known piece for the same instruments. Lasting approximately 16 minutes, the piece is divided into four movements, the first two being played attacca.

Movement I Score | Movement II Score

Movement III Score

Movement IV Score

An analytic commentary is also available.