Originally posted by David113
First, please see here this mate in 127 by O. T. Blathy and it's solution:
Now, J. Halumbirek composed another version of this problem, to make the solution longer. This is Halumbirek's mate in 130:
[fen]8/p6p/7p/p6p/b2Q3p/K6p/p1r5/rk3n1n w - - 0 1[/fen]
But I don't know the solution. Of course I assu ...[text shortened]... the original one's solution, but still I can't solve it. Can someone help, please? thanks.
This one should be easy, given the knowledge of Blathy's problem.
First, clean up the 'extraneous' pieces with checks:
1.Qd1+ Rc1 2.Qd3+ Rc2 3.Qxf1+ Rc1 4.Qd3+ Rc2 5.Qd1+ Rc1 6.Qd2 Rc2 7.Qe1+ Rc1 8.Qe4+ Rc2 9.Qxh1+ Rc1 10.Qe4+ Rc2 11.Qe1+ Rc1 12.Qd2 Rc2 13.Qd1+ Rc1 14.Qd3+ Rc2 15.Qe4 (zugzwang)
After every non-promoting pawn move, White has to make 5 moves to restore zugzwang (Qe1+/Qd2/Qd1+/Qd3+/Qe4).
After every promotion (except the last) White has to make 7 moves to restore zugzwang (Qxh1+/Qe4+ before the above 5-move sequence).
After the last promotion, White has the much stronger Qxh1+/Qh7!/Qe4 to restore zugzwang.
When Ba4 finally moves, White finishes up with Q(e4)-e1+ Rc1 Qd2 Rc2 Qd1+ Rc1 Q(x)b3#.
Add this all up, and you get:
15 (initial moves)
16*5=80 (restore zug after 16 non-promoting pawn moves)
4*7=28 (restore zug after 4 pawn promotions)
3 (restore zug after the last pawn promotes)
4 (finish up after Ba4 moves)