Retrograde Analysis Puzzles

In most chess puzzles, the goal is to mate in a x moves, or to win material. These puzzles are asking you to look into the FUTURE in order to find the solution.

The following puzzles require that you look into the PAST in order to find the solution.

Confused yet? Here is an easy example:

Black move last. What was his last move - and White's last move?

Answer is white text: 1.Nb6-a8 discovered check Ka7xa8

That wasn't too hard, was it? Here is where the hard problems begin.

In the endgame, often it is hard to tell which side is white and which side is black. In the position below, which side is white, and why?

In the position below, it is given that White gave Black odds of a queen, (unwisely it seems) that an original piece was captured on h6, and both Whites knights are original. White CAN castle. However, can he castle on both sides?

I have trouble adding comments after diagrams.

Have fun solving!

-Nate

Raymond Smullyan's puzzles, eh?

White can castle. Certainly he can't castle on both sides (since he can't castle again after castling, and also more seriously because at least one of the rooks has moved before.)

I'll leave the logic and/or proof game to someone else to deliver, since that's as much spoiler as I want to give.

On the first puzzle, if you assume the board is being looked at from blacks perperspective a solution could be:

...1 b8=B+

2. Kh1

For the third puzzle, you say an original piece (not a pawn) was captured there. Which means it had to be a black piece. White started without a queen, obviously the light squared bishop could not be captured on h6, and clearly the dark bishop was captured on its original squaure and he still has all his other pieces... ah wait, it could have been a knight and one of the remaining knights on the board is not orginal.

Its a crazy game, but for puzzle number three I believe I showed that its possible for white to achieve this position legally while retaining the right to castle either way.

@jdcannon, you have shown that the position can be accomplished with white having the right to castle in both directions, but you did not meet the requirement that both of whites knights have to be original (meaning not promoted.)

I believe that one satisfies it. I missed the part about both knights being original.

There's a second solution to the first one (white text): 1.g1=B Ka8

Edit: oops no there isn't, I had the board the wrong way round in my head!

Bravo jdcannon!

This problem can also be solved by reasoning: The piece on h6 is not the white queen (which was given as odds) nor the bishop from c1 (which never escaped), nor the bishop from f1 (which is on the wrong color) nor the missing pawn from e2 (which couldn't have made three captures to get to the h-file).

Therefore the pawn from e2 has promoted.Then it must be one of the rooks (since the White knights are both original). Since the King has not moved, the promoted rook could never get to a1, and hence must be on h1. So, White can castle only on the queen's side.

Now, suppose that the same problem was given, but it is said that a promoted piece was captured on h6, to which side can White castle?

-Nate

And to this second half of the question I'll again half-spoiler (white text):

Obviously the e2-pawn again promoted, but not on f8! (Since the bBe5 can't be from a1 because of wPb2.) Therefore the wPe2 made 2 captures to promote, therefore bPb7... [fill in the long blank here]

I believe this game provides the solution to the second verison of your problem.

Thanks for this interesting discussion.  You might perhaps also be interested in the following two new excellent retrograde chess problems that my nephew created: http://probability.ca/retrochess/

About the third problem, there was actually no need to precise that an original piece has been captured on h6. If it is an original one, white can only castle on one side, and if it is not an original one, white can only castle on the other side. So we can answer the question whithout knowing on what side he can castle.

Well analyzed! Actually it is almost unbelievable that Smullyan would have added the "original" condition for the h6 capture, as this throws away 75% of the content. What else are the careful placing of Pa3 and the missing black b-pawn for?

What might be playing here is an illusion of content. Possibly Smullyan explained in his book that white would have lost 0-0-0 right if the h6-unit would have been unoriginal with some extra analysis. However, that which is outside the conditions is not part of the problem and no solver would ever look at that unless he read the book (or his name was Bluzima).