Originally posted by David113I may be wrong, but I believe this is only the first move of a (selfmate or helpmate) problem where the knights do play an essential role. It may even have been posted before here on RHP (biggdogproblem?)
Please see Fabel's famous "don't mate in 1" problem here:
In some sources the problem is printed without the black knights. In other sources - with the knights. Which is the original version of the problem? And what are the black knights for?
Originally posted by David113Well, if Sam Loyd can do it...
OK, someone changed there the stipulation a little bit; but the original stipulation is "don't mate in 1", and with this stipulation I don't know who added the black knights and why (or, if Fabel himself added them and someone else omitted them, why did Fabel think they are neccesary).
Originally posted by David113Nope. Just turn the board around!
About Loyd's and Fabel's problems:
I quote from the British Chess Problems Society site - "In the 19th Century it was common for composers to dress the board with useless pieces in an attempt to both confuse the solver and to ensure soundness. That practice has long since been seriously deprecated."
About Dawson's problem: White lost - the bet.
Originally posted by abejnoodIt's mate in 2, not in "one". And it has been posted before, no tricks needed.
Nope. Just turn the board around!
But Loyd's mate in one that Bigdog posted is tough. What's the solution?
Edit: Unless it IS the same idea, maybe turn the board around and then Qd1#.
Originally posted by crazybluehttp://www.chessbase.com/puzzle/puzzle11/puzz11-9b.htm
I have a question also now. On that first site, I found the bottom diagramm by T. R. Dawson, where the White player says he can lose the game. I only see forced Whi te could possibly lose it. Any ideas?