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like the puzle with an underpromotion to each peice and then finaly you promote to queen wining the game.
Unfortunately, I cannot get the diagrams to work. Below is the FEN and PGN of the diagram, and if someoen can post the diagram, that would be great. If I can get it to work again, I'll post it.
FEN: 3rr3/2Kpp3/3pp3/3pp3/3PP3/3PP3/3PPk2/3RR3 w - - 0 1
PGN: [Date "????.??.??"] [result "*"] [fen "rnbqkbnr/pppppppp/8/8/8/8/PPPPPPPP/RNBQKBNR w KQkq - 0 1"][Result "*"][FEN "3rr3/2Kpp3/3pp3/3pp3/3PP3/3PP3/3PPk2/3RR3 w - - 0 1"]
It's a fairly simple puzzle.
That is not a reachable position.
Oops, my mistake, I believe the first one was reachable. Given how many of these I've done you'd think I wouldn't forget about pawn promotions.
Afteer you said that it was impossible, I tried to solve it. Although beforehand I had made sure that it was mathematically possible, I hadn't actually solved it. When I couldn't solve it after one try, I had assumed that you were correct.
I now have solved the puzzle (in 42 moves - I have no clue if that is good or not). I also deleted the second one.
The (first) puzzle goes until noon next Thursday (June 28).
I think 42 is pretty decent. My first attempt was 44, now I have 40. I think it is close to optimal. I'll post it when I decide I can't easily improve it anymore.
39 moves, I'd be surpised if someone improved this one.
They are both the same!
Hey guys, just looked at this thread and remembered seeing it a long time ago when it was just created. A lot of really good stuff here, I like these kind of retro problems in the forum.
browni, nice solution, but the problem here is that you go for 2 promotions for each side. The extra pawn moves cost you, each side can do with just 1 promotion.
Here is a 35.5 move solution:
I believe this is the fastest. Here is my analyse:
First of all, the symmetry (as always) makes it a bit more convenient to solve.
The knights have to be captured first to free up the bishops and the game. This is their most efficient route (6 total).
Next, for the bishops, the route they take is also the most efficient to be taken by one of the pawns (4 moves total).
Now we need to plan ahead. One of the a pawns have to be captured, and one of the h pawns (of opposite color) must be captured, to let each side promote a new piece to be taken by one of the pawns.
At first, my strategy was to capture those pawns with the queens, and I was able to reach a 36 move solution. Later I realized that the kings themselves, on the way towards their final position, can capture these pawns, which saved me half a move.
So the queens get into their positions to be taken by the pawns. This takes a minimum of 2 moves total. If the white queen could get to e5 in just two moves (or alternatively, if a newly promoted queen on a8 could reach e5 in just two moves) a 35 move solution would have been possible. But it is impossible as far as I can see.
One final trick is that white can castle long, which saves him a move. The final sequence is pretty much straight forward.
I didn't think to have the king capture the pawns, nice job!
I'm sorry shoopi, but I've come up with a better solution. Even though you haven't technically beaten me here, you have the way I see it.
There is no need to be sorry, this is great. I didn't consider waiting with the original queen before promoting to a new one. Nice final touch to solve that problem and reducing the solution by half a move.
Just goes to show that these things are really nice and very innovative, there are many things to consider and surprises lurk in every corner.
I'm fairly certain that this is the best possible. So I suppose you're next to post a diagram!
lets check it mathmatically
I think it is optimal though
It is optimal.
The knights have to be taken first - on c3/f3/c6/f6. While it is possible to sac just one knight for each side and use the bishops for the other pawn captures (instead of a second knight), it is extremely wasteful and inefficient.
Therefore, knights take a minimum of 6 moves total (3 for each knight).
Next for the bishops, a minimum of 4 moves total to be captured (2 for each bishop).
Queens - minimum of 2 moves total to be captured.
Kings - it takes a minimum of 6 moves for black to reach f2 and 7 moves for white to reach c7. However white, by castling long, saves a move getting his rook to d1.
So it takes black 2 moves to get his rooks to final position, and just 1 move for white to get his rooks to final position - 8 moves total for kings and rooks.
The pawns needed to be captured are captured within the kings journey to their final position, but it takes at least one move in order to bring them close enough (no way to capture those pawns withought using at least one move).
It takes 5 moves for the a and h pawns to promote.
The promoted pieces on a8 / h8 would take at least 2 moves to be captured by a pawn.
Finally, the pawn captures themselves take 6 moves total.
Total: 6 (knights) + 4 (bishops)+ 2 (queens) + 8 (kings and rooks) + 1 (pawns needed to be captured) + 5 (promoting) +2 (promoted pieces) + 6 (pawn captures) = 34 full moves, as per browni's sequence.
So browni, are you going to post the next diagram?
There is one more rule that we need to make: We need a time limit for how long the winner has to post his new position.
Question 1: Do people agree?
Question 2: How long should it be? - Suggestions are welcome.
Question 3: What is the new rule on who gets to post the new position
I would say that it probably should be between 3 days and a week (and a half??), and that whoever is the next on after the time is up should get to post.
I also agree that this is a good idea to keep the thread somewhat flowing.
I think that 3 days (72 hours) is more than enough time for the solver to post a new diagram.
And in the case he hasn't and the time limit has passed, anyone who is interested can post a valid diagram (or challenge) for others to solve.
I will post a diagram. I'd like to come up with something good, but I've got a lot to do so that might not happen. Give me a day and I'll come up with something.
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