Anand vs. Gelfand world chess championship 2012 oldest pair of contenders since 1886

In 2012, World Chess Champion Viswanathan Anand will attempt to defend his title aqainst challenger Boris Gelfand. This is a very unusual match in that both players are fairly old by World Chess Champion contender standards. I decided to see just how unusual it was, and do so with some degree of rigor.

One tricky bit is that chess championships (usually) have (at least) two players, so we have to define an age metric for pairs of people. Creating a well-ordering on tuples is sometimes controversial. I chose to have the comparison routine be:
    age(contenders) = min(age(contender) : contender ∈ contenders)
which is to say, the age of the youngest contender.

Another tricky bit was deciding which matches were definitively “world chess championship matches.” I pulled the list of world chess championship matches from chessgames.com. For time periods where the organizational ownership of the title is in question, this includes matches sponsored by all contending organizations.

As a naïve first pass, I looked up the birth years for all the contenders and subtracted that from the year of the championship to get an estimated age. This could be off by a year if the youngest contender’s birthday comes after (or during?) the match. Nevertheless, this was accurate enough to give me a short list of matches to investigate further:.

Year Player 1 Estimated Age Player 2 Estimated Age Minimum
2012 Viswanathan Anand 43 Boris Gelfand 44 43
2010 Viswanathan Anand 41 Veselin Topalov 35 35
2008 Viswanathan Anand 39 Vladimir Kramnik 33 33
2006 Vladimir Kramnik 31 Veselin Topalov 31 31
2005 Veselin Topalov 30 Many N/A1 30
2004 Vladimir Kramnik 29 Peter Leko 25 25
2004 Rustam Kasimdzhanov 25 Michael Adams 33 25
2001 Ruslan Ponomariov 18 Vassily Ivanchuk 32 18
2000 Vladimir Kramnik 25 Garry Kasparov 37 25
2000 Viswanathan Anand 31 Alexey Shirov 28 28
1999 Alexander Khalifman 33 Vladimir Akopian 28 28
1998 Anatoly Karpov 47 Viswanathan Anand 29 29
1996 Anatoly Karpov 45 Gata Kamsky 22 22
1995 Garry Kasparov 32 Viswanathan Anand 26 26
1993 Garry Kasparov 30 Nigel Short 28 28
1993 Anatoly Karpov 42 Jan Timman 42 42
1990 Garry Kasparov 27 Anatoly Karpov 39 27
1987 Garry Kasparov 24 Anatoly Karpov 36 24
1986 Garry Kasparov 23 Anatoly Karpov 35 23
1985 Garry Kasparov 22 Anatoly Karpov 34 22
1984 Anatoly Karpov 33 Garry Kasparov 21 21
1981 Anatoly Karpov 30 Viktor Korchnoi 50 30
1978 Anatoly Karpov 27 Viktor Korchnoi 47 27
1972 Bobby Fischer 29 Boris Spassky 35 29
1969 Boris Spassky 32 Tigran Petrosian 40 32
1966 Tigran Petrosian 37 Boris Spassky 29 29
1963 Tigran Petrosian 34 Mikhail Botvinnik 52 34
1961 Mikhail Botvinnik 50 Mikhail Tal 25 25
1960 Mikhail Tal 24 Mikhail Botvinnik 49 24
1958 Mikhail Botvinnik 47 Vasily Smyslov 37 37
1957 Vasily Smyslov 36 Mikhail Botvinnik 46 36
1954 Mikhail Botvinnik 43 Vasily Smyslov 33 33
1951 Mikhail Botvinnik 40 David Bronstein 27 27
1948 Mikhail Botvinnik 37 Vasily Smyslov 27 27
1937 Alexander Alekhine 45 Max Euwe 36 36
1935 Max Euwe 34 Alexander Alekhine 43 34
1934 Alexander Alekhine 42 Efim Bogolyubov 45 42
1929 Alexander Alekhine 37 Efim Bogolyubov 40 37
1927 Alexander Alekhine 35 José Raúl Capablanca 39 35
1921 José Raúl Capablanca 33 Emanuel Lasker 53 33
1910 Emanuel Lasker 42 Dawid Janowski 42 42
1910 Emanuel Lasker 42 Carl Schlecter 36 36
1908 Emanuel Lasker 40 Siegbert Tarrasch 46 40
1907 Emanuel Lasker 39 Frank Marshall 30 30
1896 Emanuel Lasker 28 Wilhelm Steinitz 60 28
1894 Emanuel Lasker 26 Wilhelm Steinitz 58 26
1892 Wilhelm Steinitz 56 Mikhail Chigorin 42 42
1890 Wilhelm Steinitz 54 Isidor Gunsberg 36 36
1889 Wilhelm Steinitz 53 Mikhail Chigorin 39 39
1886 Wilhelm Steinitz 50 Johannes Zukertort 44 44

Closer investigation of each of the highlighted matches revealed that, astonishingly, in every case the youngest contender’s birthday came after the match:

  • 2012 Viswanathan Anand (43?) vs. Boris Gelfand: Anand’s birthday (December 11) comes after the match (starts in May) so he will still be 42.
  • 1993 Anatoly Karpov (42?) vs. Jan Timman (42?): Timman’s birthday (December 14) came after the match (finished November 1) so he was still 41.
  • 1934 Alexander Alekhine (42?) vs. Efim Bogolyubov: Alekhine’s birthday (October 31) came after the match (April to June) so he was still 41.
  • 1910 Emanuel Lasker (42?) vs. Dawid Janowski (42?): Janowski was born May 25; Lasker December 24. Lasker’s birthday came after the match (finished December 8) so he was still 41.
  • 1892 Wilhelm Steinitz vs. Mikhail Chigorin (42?): Chigorin’s birthday November 12 (October 31 old style) came after the match (finished February 28) so he was still 41.
  • 1886 Wilhelm Steinitz vs. Johannes Zukertort (44?): Zukertort’s birthday (September 7) came after the match (finished March 29) so he was still 43.

We conclude that Anand vs. Gelfand (2012) features the oldest contenders since the very first World Chess Championship Steinitz vs. Zukertort (1886) – and is within a year of even that! If the 2014 championship is a rematch, it will set the record.

1 Topalov was the clear winner of the 2005 FIDE World Championship Tournament so there was no need for a runoff.

Nitpicking Sam Loyd – a wheel within a wheel

In August 1878 Sam Loyd published this mate in two and dedicated it to a friend of his named Wheeler:

a-wheel-within-a-wheel
Mate in two; Black to move and mate in two; Selfmate in two; Black to move and selfmate in two

While the mates appear to stand up, the problem position is not legal. White has three a-pawns; this implies at least three Black pieces were captured by a White pawn. But Black has fifteen pieces on the board; only one is missing!

Looking at Black pawn captures – the b2-, c-, and d- pawns together account for three pawn captures. This seems OK at first glance since White has three pieces missing. But all the missing White pieces are pawns, and they are from the right half of the board… so they must have promoted. This implies more pawn captures to either get the Black pawns out of the way or to get the White pawns around them. (The promoted pieces could have been captured by the Black pawns, or the original pieces could have been captured in which case the promoted pieces are on the board now.)

Finally, the h-pawns on h5 and h6 could not have got into their present position without at least one pawn capture by White, or at least two pawn captures by Black.

Spock’s chess games from Star Trek Ishmael

In his post “A chess problem begging for a solution“, Michael Kaplan quotes Barbara Hambly’s Star Trek novel Ishmael.  In the quoted scene, Spock (AKA Ishmael) plays a couple of chess games against a stranger – rather unusual chess games.  The problem alluded to is to determine the moves of the games given certain information.

Let’s take the second game first.  There are two effective possibilities that meet the “Reverse Fool’s mate” and “three moves” criteria:

Reverse Fool’s mate, even material: $200 in 2½ moves

# Ishmael Stranger
1. e3 f6
2. a3 g5
3. Qh5#

spock-chess-1

… and…

Reverse Fool’s mate plus a pawn: $220 in 2½ moves

# Ishmael Stranger
1. e4 f5
2. exf5 g5
3. Qh5#

spock-chess-2

The task for the first game, then, is to win either $400 or $380 in seven moves.  Assuming the game ends in mate (this is reasonable) gives us a difference of either $200 or $180.  The first move can not be a capture, so we really only have six moves to capture $200 worth of material – going after the queen is obvious, but the rooks are quite well tucked away, and it is hard to go after them and simultaneously set up the ending mate.

I believe that the solution that Barbara Hambly had in mind is the following variation of the “other” mate that every chess student learns (Scholar’s Mate).  This particular setup is gated by White’s need to move his Bishop out of the way, and this nicely satisfies the common convention that players alternate colors in successive games.  Naturally Spock, being a gentleman, would let the stranger take White first.

Mate, a queen, and four pawns: $380 in 7 moves

# Stranger Ishmael
1. c4 e6
2. c5 Bxc5
3. d4 Bxd4
4. e3 Bxe3
5. Qf3 Qf6
6. Bc4 Qxf3
7. Kf1 Qxf2#

spock-chess-3

However, from a “chess problem” point of view, there’s a cook.  It is, in fact, possible to get mate plus a queen plus four pawns in a mere five and a half moves, rather than the seven full moves above:

Mate, a queen, and four pawns: $380 in 5½ moves

# Ishmael Stranger
1. e3 h5
2. Qxh5 d5
3. Qxd5 Bf5
4. Qxb7 Bh7
5. Qxc7 Qc8
6. Qxc8#

spock-chess-4

Even worse, there is a way to get $400 in a mere four and a half moves.

Mate, a queen, a bishop, and a knight: $400 in 4½ moves

# Ishmael Stranger
1. d4 Nh6
2. Bxh6 b7
3. Qd3 Ba6
4. Qxa6 Qc8
5. Qxc8#

spock-chess-5

The problem, as it stands, is therefore underdetermined… Spock and the stranger had a full two and a half moves to play around with in the first game, either to exchange material or perhaps to allow the stranger to pick up a free pawn (and return it in the second game.)

Nitpicking Sam Loyd’s “Organ Pipes” mate in two

Sam Loyd was a wacky guy.  His legacy of puzzles is well worth digging through.  (I’d stay away from the 15/14 puzzle though.)

A sprinkling of his puzzles were chess-related, and plenty of them stand up well in the greater lexicon of chess puzzles.  For example, his “Organ Pipes” mate-in-two problem is quite well known:

organ-pipes
White to move and mate in two (Sam Loyd, 1859)

(If you haven’t seen it before, go ahead and try it out.  It’s a cute little interference problem.)

OK, now we get to it.

From a technical point of view, there’s a slight flaw here.  Ideally, in a chess problem Black has very many options, and White has only a single path to victory in all variations.

This is not the case here.  If the Bishop on f8 wanders away (1. … Bg7, 1. … Bh6) or is blocked by the Rook on e8 (1. … Re7) then White has two mates: 2. Qb6# or 2. Qxb4#.

1. Qa5 1. … Bb7 2. Nf5#
1. … Bd7 2. Qd5#
1. … Be6 2. Qe5#
1. … Bf5 2. Nxf5#
1. … Rd7 2. Nf5#
1. … Rd6 2. Qxb4#
1. … Rd5 2. Qxd5#
1. … Re7 2. Qb6# or 2. Qxb4#
1. … Rd6 2. Nf5#
1. … Re5 2. Qxe5#
1. … Bc5 2. Qa1#
1. … Bd6 2. Qd5#
1. … Be7 2. Qe5#
1. … Bg7 2. Qb6# or 2. Qxb4#
1. … Bh6 2. Qb6# or 2. Qxb4#

Luckily, the problem can be easily “fixed” – add a black pawn on a7:

organ-pipes-fixed
White to move and mate in two
(Sam Loyd, 1859; version by Matthew van Eerde, 2008)

This covers the b6 square.  Now the only mate is to take on b4.

Beauty is a tricky beast though.  This is now a more “technically correct” problem, but the position is no longer quite as “natural”-looking.  (I quote “natural” because I’ve seen plenty of ugly over-the-board positions.)  The pawn on a6 must have come from b7, and the pawn on b4 must have come from c7 (or perhaps from further away.)

I’m not sure which version I prefer, but it’s nice that the flaw is not serious.