Long-range pieces: queens, rooks, bishops.
Short-range pieces: pawns, knights, kings.
We can extend the definition of short-range pieces to include also fairy pieces like: Berolina pawns, wazirs, ferzes, dabbas, alfils, threeleapers, camels, zebras, trippers, etc. (see: http://en.wikipedia.org/wiki/Fairy_chess_piece).
Inspired by Richard Stanley’s question Joel David Hamkins with coauthors wrote the paper where they conjectured “that the general winning-position problem is undecidable and indeed, not even arithmetic”. The main argument preventing us from expanding the “mate-in-n” theorem to the winning position problem is a fact that “a player may have a winning strategy from a position, without there being any finite bound on the number of moves required”. This issue is directly related to the question here. The formulation of the question states that such a position “must involve a long-range piece for the losing side”.
Is it possible to apply the “mate-in-n” theorem to settle the winning-position problem in positions with only short-range pieces?
If so, how can we calculate the upper bound for $n$ in a given position? How can we be sure that we don’t have to keep solving the “mate-in-n” for larger and larger $n$?
If not, what are the promising ways which could possibly lead to a proof of the problem decidability?
With a finite board we know that repetitions of the positions have to occur. Then, we can use the non-repetition argument. If every path in the game tree leads to a repeated position then the initial position is a draw. With an infinite board the number of possible new positions is not a finite number. Heuristically, we can argue that if a white short-range piece is “too far away” from its nearest neighbor (any other piece), then such a position is not better for White than the one when the piece is still far away but not “too far away” from its nearest neighbor. The idea here is that if the piece is “too far away” it does not participate in the game. Thus, if we place it closer but still far away then it doesn’t change anything important – it still doesn’t participate in the checkmating process. Accepting the above argument, it is enough to consider only a finite number of positions to be sure that a position is a draw. I’m not sure if it is possible to make this heuristic argument rigorous.
UPDATE 28th March 2014: I’ve just came across a version of an infinite chess which was considered by Dénes Kőnig in 1927 (source: slide 26 here). It is played:
on an infinite chessboard,
with the rules of Chess, and
with the same moves as on a normal chessboard (i.e., the Queen, Rook, and Bishop move at most seven squares at a time).
Thus, the above version is the one with short-range pieces only. This version (let call it Kőnig version) has the rules for moves which are in perfect agreement with the rules of an ordinary chess. The same is true of the version with Queen, Rook, and Bishop able to move through any number of unoccupied squares (let call it Brumleve-Hamkins-Schlicht, or BHS for short).
It seems to be a plausible conjecture that while the winning position problem is undecidable for the BHS version it is decidable for the Kőnig version.