# Is a both-sides forced perpetual check possible?

There are lots of positions where one side can force the other into a perpetual check. Is there any position in which there are no legal moves besides continuing the perpetual from both sides? For example, let’s say White is checking Black. Black has to move due to the check. Then, White somehow has no other legal move than to continue the perpetual.

I am asking purely out of curiosity. I don’t think such a position would be likely to be reached in a real game Ideally, there would only be pieces on the board that are attainable in a real match though, e.g. no 10th queen. I am also interested if it’s possible to prove that there is no such position.

• This is a duplicate, methinks (i.e. I'm fairly sure it has been done either here or on Math SE, with a solid proof, by Noam Elkies I think, but am too lazy to google :-) See also chess.stackexchange.com/questions/8982/… for additional info. Sep 26, 2021 at 8:04
• @HaukeReddmann when you say it "has been done", do you mean that it has been proved to be impossible? Sep 26, 2021 at 18:43
• @JamesMartin: So I think, but be careful. At least the "proof" in the link is incomplete, and moreover a fairly simple fairy piece, the camel (1:3 knight), makes a mutual perpetual possible (same link). Thus it must be proven that knights can't be used in the same way. Sep 26, 2021 at 19:37
• Further note the original poster asks for a forced perpetual, which is even harder (the camel solution fails because the battery hind pieces can interlope, and if you pin them, even if that is possible at all, the pinner can capture and must be pinned too). The "proof" in the link probably can be amended to take care for that. Sep 26, 2021 at 19:42
• @HaukeReddmann I have to say I'm confused. Any proof in the link you gave doesn't seem relevant, since that question asks about consecutive checks by both players, whereas this question only requires a sequence of checks by one player. I'm not sure it's a good idea to say that it's a duplicate, unless you can point to where the answer is - it seems a bit unfriendly to a first-time poster.... Sep 26, 2021 at 20:54

## 1 Answer

Attempt to a partial answer.

I expect this to be impossible, but not by principal means (i.e. a real proof will be very hard and contain many cases), but the limitations of the pieces and the board. To show this, here is a "mild" fairy solution using nothing than a large board and some fancy line figures.

Start with Kc4 Rd5 Rd4 Pd6 Pe5 Pe4 Pd3 - Ka4 Ra6 Ra3 Be6 Be3 Pd7. The intended perpetual is 1.Kc5+ Ka5 2.Ka4+ Kc4. You now "merely" need white girafferiders (a line figure on 1,4-lines) on i2 and i7 to pin the bishops, and four more fancy riders to pin the girafferiders in both positions of the white king.

This setup also shows where the problem is: Why shall White's checking piece not move to the last position of the black king in the next White move? Either because White is in check (but then indeed we run into the problem of a mutual perpetual check, as outlined in my comments) or because it is pinned. But then, why shall the pinner not capture instead? It must be pinned too. And that pinner is White again and must not move either, i.e. must be pinned, leading to some setup like Kh1 Rf3 Bb1 - Kf5 Be4 Ra1. There are only very few of these setups, and worse, since it must be an infinite cycle on a finite board, all moves must ultimately be "taken back", so the Rf3 here must be unpinned somewhen again. How?