Do you know example where recent chess engines (Houdini, Rybka, Komodo, ...) failed to find a forced checkmate.
I guess they must find mate in 1 or 2 all the time but maybe are they failing to be as creative as human some times.
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Positing this answer in addition to the answers/comments on tablebases and endgames with ridiculously long forced mates.
White to move and win (I do not know the author of this study, unfortunately). The engines will tend to fail on this and similar problems. As far as I tried, all recommend to move hxg8=Q. Some show 0.00 and are soon thereafter putting black in a stalemate; some show slight advantage for white and are trying to play the game out by letting black to activate his pieces. Further explanation (and hint to the solution) given in the spoiler block below the diagram.
6r1/p1p1p1pP/P1PpP1P1/8/2P1B3/2K3pp/3P2rq/R5bk w - - 0 1
The forced mate is not found because the engines use pruning heuristic. It removes certain branches from search tree, after deeming them as irrelevant to the search result (see: http://chessprogramming.wikispaces.com/Pruning). In case of this puzzle, the solution consists of many subsequent sacrifices and its branch tends to be discarded when searching. Note: Possibly, with tweaked parameters and mate-search heuristic when exact number of moves required is provided, they could find the solution, but I have not tried that.
Depending on black's move choices, a few solutions exist (the idea is always the same, though and up until move 12 the line is always the same, too). Here's an example: 1. hxg8=N d5 2. Bf3 d4+ 3. Kb4 d3 4. Nh6 gxh6 5. g7 h5 6. g8=N h4 7. Nf6 exf6 8. e7 f5 9. e8=N f4 10. Nd6 cxd6 11. c7 d5 12. c8=N dxc4 13. Nb6 c3 14. dxc3 d2 15. Kb3 d1=Q 16. Rxd1 axb6 17. a7 b5 18. a8=Q b4 19. Be2 bxc3 20. Bf1 c2 21. Rc1 f3 22. Qxf3 Bf2 23. Bxg2+#
There are some tricky positions, where it looks like there is no solution, but then it turns out white must be able to take en passant. In these positions engines might overlook the mate, because they don't have the information about the last black move, whereas a human can deduce this information by retrograde analysis.
4k2r/8/5B1P/3R1KpP/8/8/8/8 w - - 0 1
In this position it is impossible to determine which variation leads to mate in two. It is only possible to show that there must be a mate in two-variation. Either black moved the pawn on his last move - then taking en passant leads to mate. Or he moved his king or rook - then king e6 leads to mate, as there is no more castling possible.
Edit: Another answer, that is just as irrelevant for all practical purposes: As we all know thanks to tablebases, there is a big number of forced checkmates out there, that are far beyond the calculation horizon of any engine. Of course we can use tablebases to detect those mates, but it doesn't take any stretch of the imagination, to acknowledge the existence of forced checkmates over thousands of moves, that aren't now, and likely never will be stored in a tablebase.