In Counterplay, someone mused over the fact that there seems to be little chance of needing to under promote to a bishop.
I’ve seen at least one contrived position in which under promoting to a bishop was needed for a mate in 3, if I recall. That being said, just about anybody could easily see that promoting to a queen assured a mate in 5. I suppose the position was created so that it couldn’t be solved by a chess engine.
I’m sure there are plenty of ways to contrive positions which under promoting to a bishop prevents stalemate, or allows a faster line of winning play. But I suspect in actual chess games, excluding errors due to time pressure, assuming an average chess player or higher skilled, it would be rare to have to under promote to a bishop in order to assure a victory.
Hold a (non-rated) tournament with throwback rules that disallowed a player from having multiple queens (bigamy) and you’d see more underpromotions of all three types.
I’ve seen an IL state K-8 final round board one game where underpromotion to a rook let the player force mate in two (winning the state championship) while promotion to a queen was stalemate (with a player on a different board getting the championship on tie-breaks). There are probably similar cases with the bishop (and admittedly as rare as you already indicated they would be).
What if a player already has a queen, two rooks, two knights, and two bishops? What does he promote to in that case, to avoid bigamy, trigamy, trigamy, and trigamy, respectively?
I had the craziest game I’ve ever played at the 2012 US Open in Vancouver, WA where promoting to a bishop isn’t necessary but would have been just as good as promoting to a queen. Here is the game:
Donald Poston-Micah Smith (2012 US Open round 5, 4-day schedule)
1.Nf3 d5 2.e4 (the so called Tenneson Gambit, a king of reversed Budapest Gambit) 2…dxe4 3.Ng5 Bf5 4.d3 exd3 5. Qf3 dxc2 6.Nc3 Qc8 7.Bc4 Nh6 8.Nb5 c6 9.Nd4 Bg6 10.Bf4 Nd7 11.Nge6 fxe6 12.Nxe6 Bf5 13.Nc7+ Kd8 14. Be6 Ne5 15.Rd1+ and now I played 15…cxd1=Q+ but 15…cxd1=B would have been just as good!
Promotion to a bishop gives white a mate in 12. Promotion to a knight gives a long draw. Promotion to a queen or rook gives a quick stalemate. Anything other than promotion gives a draw or loses.
Although I’m not sure I’m remembering the position correctly, I am sure what the intended solution is. Your proposed change (with pawns on f2 and f4 instead of e2 and e4) would cook the intended solution.
I’m also sure that the c1-h6 diagonal is irrelevant to the intended solution.
It’s a chestnut. I’ve never been able to track down the original source. I first saw it at the old No Exit Café, where John Tyler (and a few others, including masters) were trying mightily to solve it. Finally, I made one flippant suggestion, which turned out (inadvertently) to lead directly to the part of the answer they hadn’t found yet. If I say any more, I’ll give it away.
Sam Loyd, maybe?
Hint: The title of this thread is relevant.
It’s a completely legitimate problem, in all the classical senses. For example, there is only one solution. There may be different branches depending on black’s response to each move, but to each such, there is only one white move which still leads to mate in 5.
Also, every piece and pawn (except the white king, I suppose) is essential to the problem, either by being part of the solution or by preventing multiple solutions (cooks).
The exact position of the white king may not be important. Also, I think the h-pawns could be on h3 and h5 instead of h2 and h4, without changing the solution. And it may be that the white bishop at h8 was actually on g7 in the original.
Are you sure? It seems to me the same white moves (especially in that part of the solution relating to the original thread title) work either way.
Of course, you won’t be able to respond to that question without giving away (much of) the solution.
Good grief! How on earth did you find it? You must have used the “exact material” option. (That might not have worked, had I misremembered the position even slightly.)
