Problem A:
What is the shortest sequence of legal moves which leads from the starting position back to the starting position, except that it is now black’s move at the end?
Please do not post solutions (spoilers) until Thursday, 4-15-2010, at about 11 pm. Just post snide comments, and remarks which expose your ignorance. Let others have a crack at the solution!
Bill Smythe
Problem B:
What is the shortest sequence of legal moves which leads from the starting position to the position shown here, with black to move?
As with the previous problem, please do not post spoilers until Thursday, 4-15-2010, at about 11 pm.
Bill Smythe
Problem C:
What is the shortest sequence of legal moves which leads from the starting position to the position shown here, with black to move?
As with the previous two problems, please do not post spoilers until Thursday, 4-15-2010, at about 11 pm.
Bill Smythe
Well, it’s the 16th, so here goes…
Either there is no solution to this problem, or there is something wrong with the argument below.
(If there is a flaw, let me know!)
Therefore, if the position is even-even, it is White’s move.– Randy “logic is a pretty flower that smells bad” Shane
I will humbly attempt to answer all three.
A. I believe that this can be shown to be an impossible situation. For both Knights of one side (say White) to end up on their starting squares, OR to switch places, requires an even number of moves. Likewise for both Rooks to end up on their starting squares - zero, or an even number of moves. No other pieces or pawns can have been moved. Thus White MUST have made an even number of moves (again, allowing for the trivial case of zero). Likewise for Black. Next, if both White and Black have made an even number of moves, there is no possibility that it can be Black’s move. I believe that is the answer.
B. This one is more interesting. You can “spend” a tempo by moving one Rook, and allowing the Knights to capture on different squares - one on the b-file, the other on the a-file. I think this is the minimum:
C. I started to respond that this is a trivial mirror-image of (B), but then I noticed the difference - you have to deal with (i.e., avoid) a possible check, which throws everything off! But still, it is not too different:
Success??
Yeah, that’s a simpler way of saying it. 
– Randy
LOL, thanks - I think we were typing our earlier responses at the same time. (I took a few extra minutes to make sure my notation on B and C was correct!)
Either Randy’s “smelly flower” approach (using a sledgehammer to crack a peanut), or Tom’s more colloquial version, does the trick for problem A.
As for B and C, Tom, in one or both cases (I won’t say which, just yet), you may have overlooked a shorter solution.
Continue feasting!
Bill Smythe
Hey, one thing about cracking peanuts with a sledgehammer – the peanut STAYS CRACKED.
And smashed. And probably spread out over a quarter acre.
– Randy
You’re assuming the knights return to their original squares, as opposed to trading places. I haven’t tried it, but I suspect that if either the white or black knights trade positions, but not BOTH of them, that will cause it to be black’s move.
Not so. Because the queenside knight and the kingside knight are on opposite color squares, each knight must make an odd number of moves to reach the other knight’s home square. So, for the two knights to trade squares, there must have been an even number of knight moves in total. There being at most one vacant square for either rook, each rook must have made an even number of moves to return to its home square. So, to return to the original position, each side must have made an even number of moves, and it must therefore be White’s move.
B.
Excellent - it looks like Eric has it. And it only works for (B), as you cannot avoid a check if you try that on (C).
Nicely done - another excellent puzzle by Bill. (I got a ton of mileage in my club with the 3-way underpromotion puzzle a few months back, btw.) I should have known better than to submit the mirror-image (non)solution! Thanks.
And, it could also be noted that there must be at least eight knight moves per player, because it always takes four moves (minimum) to get from the first rank to the eighth, and four to get back. Then, since we’re supposed to end with black to move, there must be at least a ninth move for white.
This leads to a way to find a shortest solution for problem B. To make it black’s move at the end, there must be an odd number of rook moves, and since each player must make at least eight knight moves, if there is an eight-and-a-half-move solution the number of rook moves must be 1, and this rook move must be white’s.
Thus, the question is, how does white get to a8 in four moves, and how does black get to b1 in four moves. Both players must use the knight that starts on the same color square as its desired destination, otherwise it would take 5 moves instead of 4. So, white must use the Nb1, while black must use the Ng8.
Further, we can note that black’s path must include the c3 square, since it can’t get there in time via a3 (it would take 6 moves instead of 4). This rules out a mirror-image solution for problem C, as the check at f3 would bollix things up. So a solution to problem C must take at least one full move longer. To avoid the check, while keeping the solution to only nine-and-a-half moves, we must now use 5 moves each way for one player, and 4 for the other. The player requiring 5 moves each way (10 moves round-trip) must be white, since black doesn’t have 10 moves. It follows that, now, the rook move must be black’s instead of white’s.
So, how does white get to g8 in 5 moves, and how does black get to h1 in 4 moves? The answer: white must use the Ng1, while black must use the Ng8.
Bill Smythe