a 2006 chess puzzle anyone?

Noam Elkies is one of those persons I seem to bump into (figuratively speaking) wherever my interests take me. At the moment I’m reading (long overdue, I know, I know) the excellent book Notes on Fermat’s Last Theorem by Alf Van der Poorten. On page 48, Elkies figures as an innocent bystander in the 1994 April fools joke e-perpetrated by Henri Darmon in the midst of all confusion about ‘the gap’ in Wiles’ proof.

There has been a really amazing development today on Fermat’s Last Theorem. Noam Elkies has announced a counterexample, so that FLT is not true after all! He spoke about this at the institute today. The solution to Fermat that he constructs involves an incredibly large prime exponent (larger than $10^{20}$), but it is constructive. The main idea seems to be a kind of Heegner-point construction, combined with a really ingenious descent for passing from the modular curves to the Fermat curve. The really difficult part of the argument seems to be to show that the field of definition of the solution (which, a priori, is some ring class field of an imaginary quadratic field) actually descends to $\mathbb{Q}$. I wasn’t able to get all the details, which were quite intricate…

Elkies is also an excellent composer of chess problems. The next two problems he composed as New Year’s greetings. The problem is : “How many shortest sequences exists (with only white playing) to reach the given position?”

Here’s Elkies’ solution :

There are 2004 sequences of the minimal length 12. Each consists of the sin- gle move g3, the 3-move sequence c4,Nc3,Rb1, and one of the three 8-move sequences Nf3,Ne5,f3,Kf2,Ke3,Kd3(d4),Kc4(c5),Kb5. The move g3 may be played at any point, and so contributes a factor of 12. If the King goes through c5 then the 3- and 8-move sequences are independent, and can be played in $\binom{11}{3}$ orders. If the King goes through c4 then the entire 8-move sequence must be played before the 3-move sequence begins, so there are only two possibilities, depending on the choice of Kd3 or Kd4. Hence the total count is $12(\binom{11}{3}+2)=2004$ as claimed.

A year later he composed the problem

of which Elkies’ solution is :

There are 2005 sequences of the minimal length 14. This uses the happy coincidence $\binom{14}{4}=1001$. Here White plays the 4-move sequence f4,Kf2,Kg3,Kh3 and one of the five sequences Nc3,Na4,c3,Qc2,Qe4,d3,Bd2(e3,f4,g5,h6),Rc1,Rc2,Bc1 of length 10. If the Bishop goes to d2 or e3, the sequences are independent, and can be played in $\binom{14}{4}$ orders. Otherwise the Bishop must return to c1 before White plays f4, so the entire 10-move sequence must be played before the 4-move sequence begins. Hence the total count is $2 \binom{14}{4}+3 = 2005$.

With just a few weeks remaining, anyone in for a 2006 puzzle?