Solutions to 'Black holes' exercises

1. Black normal move threatening king relativistic capture of white king is 1...Hbb5+ threatening 2...Sb8-c6-b5xd1. White on move in diagram position can prevent this check either by checkmating immediately black (the answer below) or by Sd3! - pinning Hb6 on 'line' Sd3-h5-a4-b6-c8. White king moves are not enough as 1.Ke2 is selfcheck and 1.Kc2 allows another check by normal threatening knight relativistic capture of king, e.g. 1...H7a6+ for 2...Sb8-c6-a6xd2.
2. White mates with 1.Hd2# (check by Se1-c2-h7xc8), black mates 1...Hb7-b6-d6xd2# (check by Rc7-c6-d1xd1, e.g. 2.Kc2?? is not enough as black can then e.g. 2.Hc6-b6-d2xc2).
3. All original 10+10 pieces are still on board and some holes too, of course. How can white have two pawns on a-file? This can be explained by two ways: white pawn captured 'something', getting from e.g. b-file to a-file. 'Something' must be hole and as hole is normally uncapturable, it was captured by relativistic move by pawn. Second way of explaining is that white pawn entered hole anywhere and left hole on e.g. a1. In any case at least one relativistic move is connected with position of white pawns a2, a3. Second sure relativistic move was made by Ba8. It stays on light square, but it started the game on dark square. It means he made at least one relativistic move switching the colour. Last relativistic move that can be proved was surely made by pd7. It seems like it stays at home square silently from the beginning, but it's impossible due to presence of Be8 in the corner! Bishop first passed d7 on his way to e8 and only then pawn went to d7. And it was surely from 'white hole' on eight rank. Together we can prove at least 3 relativistic moves appeared in the game.