Shortest-proof-game challenge

And you said you had done that for four such shortest solutions. I'm having trouble believing that. Could you post a game showing the four solutions?

They're almost exactly the same, but technically they each have a different collection of moves. These should be the only four solutions, and the collection of moves for each solution must be played in their specific orders.

Thank you. That's a very impressive accomplishment.

Find a game that can end with forced mate in 1, 2, or 3. Move count ends at the position before mate.

Solution in four moves. Did I interpret your challenge correctly?

Yes. EvinSung's three unique solutions are also nice.

Upon return, I find that the game has disappeared. It seems I will be unable to witness the three unique solutions.

c5

If you're interested in that conversation, you can ask Arisktotle for where to find it. He knows quite a lot about the situation.

Every proof game is always unique in the sense that it is the only proof game consisting of those exact moves in that exact order Some diagrams have 1 unique proof game some have 1,000,000 unique proof games.

When proof game composers honor different shortest proof games for the same position they do that on the basis of another attribute. Like, there is 0-0 in one proof game, and 0-0-0 in the second one. Or, in one proof game all captures are made by a knight, in the other one by a bishop. Or, one proof game features an e.p. move, one a promotion and one a castling move (known as the Valladao task). Obviously it is easy to cheat on these attributes as there is always something different about 2 proof games, like "the 6th and 8th move". It only works when the distincttions are clearly the result of "intelligent design" of the puzzle and require composing skill.

"where order isnt considered"

I missed that - and I am glad I did Let's assume this is serious for a second. So when I have 2 proof games with the same moves - and one exception: the bishop travels Bf1-g2-f3 in the one and Bf1-e2-f3 in the other proof game - then I would have two uniquely different proof games? And not if I switched the order between say Bc1-d2 and Nb1-c3? Well, I agree it's not the same but certainly both are equally sloppy up to the 3rd decimal on the sloppiness scale!

Note that I can do these things on any random diagram. I locate all proof games which are "shortest" - let's say 9,123,456 SPG's - and then keep 1 from all sets with the "same moves" but different moves orders. That leaves me e.g. with 1275 SPG's. And then I announce to the world I have designed a diagram with 1275 "unique" SPG's! And I close off with the challenge to anyone to produce a diagram with more than 1275 unique SPG solutions! What an achievement!

I would like to honor Leither's position, since it has exactly four SPGs, one without castling, one with castling by White, one with castling by Black, and one with castling by both sides.

By the way, I had challenged Leither's claim of four SPGs since, after giving the position to Stelvio 1.6, I received the result "Found 2 solutions. The problem is correct." After Leither showed his four SPGs, I sent a Stelvio problem report. In his reply, Reto Aschwanden replied that I had found a bug in Stelvio, which he has fixed, and a new Stelvio version will be available in January with this fix included.

Indeed; I had realized this some time after posting, which is why I had said: "In hindsight, I realize that it is remarkably easy to create positions of this nature, and that I sound very stupid".

Although, it seems our understanding is slightly different. In your explanation, it seems that you interpret what I said as that each game with a different collection of moves is a "unique solution", and there could be multiple different solutions with that same collection of moves; therefore, one could technically take one solution out of those "multiple solutions that share the same collection of moves", and state that that specific collection of moves is a "unique solution". The more moves one makes, the more different possible ways to reach that position, which ends up creating a ridiculous amount of solutions with a "unique collection of moves", decidedly not impressive at all. In fact, I could probably randomly make 100 moves to reach a position, and it would have hundreds of solutions with a "unique collection of moves".

But that would truly be too easy. What I meant was that each "unique collection of moves" has a specific order that it must be played in, so there aren't alternate solutions with the same collection of moves; then, you couldn't "keep 1 from all sets with the "same moves" but different moves orders", because there are no other solutions with the same collection of moves. The key here is that no two solutions can share the same collection of moves, and that each solution cannot be reached in a different order.

This is still remarkably easy to do, so I still sound stupid.

Edit: I sound stupid for a different reason. A high quality position of this nature is difficult to create. My position is not a high quality position, however, more of an experiment.

also, to leither.... how is this simple?

it looks like you have to ensure that all moves have to be in a specific order for every solution, which seems reasonably tough

Here's an example of what I think Arisktotle means.

This is what my original intention was, where all possible solutions do not have an alternate order (hence, no "set of solutions with the same moves"). A high quality position of this nature is difficult to create, but the one I created is not an example of that; it is quite basic in comparison. To be honest, I was not really thinking properly when making the position. Perhaps it is best to not take the position seriously.