White to move. Mate in 128.
I'm quite new here, so not sure if this one has been posted before, but I like it alot. I admit this position is not gonna appear too often in real games, but still it's funny to have a forced mate in no less than 128 moves. Computers will show only +-0.00 here for a long time :-)
edit: sorry, just saw that its only 127 moves ;-)
Well the beginning is pretty obvious:
1.Qe1+ Rc1 2.Qd2 which leaves me with just 126 1/2 moves to consider 🙂
EDITED: it appears the only possible move for Black now is 2...Rc2 (for 2...Bc2 is followed by 3.Qc3 and 4.Qb2🙄, so a few more moves:
2...Rc2 3.Qd1+ Rc1 4.Qxd3+ Rc2, still 122 moves left but it seems clear that the only two pieces that move are the White queen and the black rook.
Originally posted by crazyblueLOL, is this serious a forced mate in 128, and not earlier???
[fen]8/7p/7p/p4n1p/b3Q2p/K2p3p/p1r5/rk5n w - - 0 1[/fen]
White to move. Mate in 128.
I'm quite new here, so not sure if this one has been posted before, but I like it alot. I admit this position is not gonna appear too often in real games, but still it's funny to have a forced mate in no less than 128 moves. Computers will show only +-0.00 here for a long time :-)
Originally posted by Mephisto2Yes I am so stupid 🙂 Of course, it goes 8.Qd3+ Rc2 9.Qd1+ Rc1 10.Qd2 Rc2 11.Qe1+ Rc1 12. Qe4+ Rc2 13.Qxh1+ Rc1 14.Qe4+ Rc2 15.Qe1+ Rc1 16.Qd2 Rc2 17.Qd1+ Rc1 18.Qd3+ Rc2 19.Qe4!! h2 (repeat 15-19) 24...h3
8.Qe2? Rc3+ seems to spoil it
(repeat 15-19) 29...h4, and so on until, 39...h6 (repeat) 44...h1Q 45.Qxh1+ Rc1 46.Qe4+ Rc2 (repeat 15-19) 51...h2 (repeat 15-19) 56...h3 and so on, and so forth.
yup you got it. its the same mechanism until Q is on e4 once again and black moves his last pawn to h1. but what then? there has to be some checkmate then and if there is, why couldnt white do it before? thats the nice aspect about this problem, even though its not so difficult to figure out anymore now.
Originally posted by crazyblueAfter the last Qxh1+ Rc1 white goes to Qh7+ Rc2 Qe4! forcing a bishop move and then Qe1+ Rc1 Qd2 Rc2 Qd1+ Rc1 Qb3#, since the bishop is not on a4 anymore.
yup you got it. its the same mechanism until Q is on e4 once again and black moves his last pawn to h1. but what then? there has to be some checkmate then and if there is, why couldnt white do it before? thats the nice aspect about this problem, even though its not so difficult to figure out anymore now.
Originally posted by Mephisto2Sure.
Qb4+ wins immediately without a pawn on a5.
edit: as the posting above also indicates.
1.Qb4+ Bb3 and I thought 2.Qxb3+ Kc1 and 2.Kxb3 d2 would be good for black. But you are right:
2.Qe1+ Rc1 3.Qd2 Rc2 4.Qd1+ Rc1 5.Qxb3#
How could I miss that?
Today is not my day🙁