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Nf3+…
Rd5+
looks good to me…
picks up room on e1 with mate
or bishop on d4 with mate
(i think)
1-Nxb3+ fails due to 2 cxb3 Rxe1 3 Rxe1 Rxe1 4 Bxc5 (4 Kxe1? Bxg1) 4-Re6 5 Bxd6 and black loses the pawn on b4 and the game.
But simply 1-Re2+ 2 Rxe2 (forced) Rxe2+ 3 Kd3 Rxg2 gives black excellent winning chances.
“But simply 1-Re2+ 2 Rxe2 (forced) Rxe2+ 3 Kd3 Rxg2 gives black excellent winning chances.”
Oh no, it doesn’t! What about the knight on d4? 4. Bxd4+ and 1:0.
Dodgy vicar did it. Though there is no mate on d4 (that is how I understand your last sentence) but 2. -, Bd4 3. Rxd4+ leads to the same mate on e1 as without Bd4. Bd4 means only a slowing of the inescapable.
Greetings
Jochen
That was a nice one. I immediately saw that either Nxb3+ or Nf3+ exposed an attack on white’s bishop, and I tried Nxb3+ first, as it snags a pawn. But I couldn’t find anything decisive. So I looked at the other move and after the pawn takes, the rook move to d5 jumped out at me; I only then saw the trick with the rook check and that white’s rooks get disconnected.
Another way to look at the problem (which would have led me to the solution much more quickly) would have been to see that if black simply didn’t have the knight, there would be a simple mate in two. Then it would be only a matter of getting rid of the knight with tempo: one of the two checks gives the king an escape square and one doesn’t — and so the problem is solved. This ability to see alternate counter-factual positions and ask “what if” or “if only… then…” is a crucial skill in solving so many of these problems.
I got this virtually immediately.
You want to know why ? It is virtually identical to a position (from Westerinen-Larsen 1967), in yesterday’s Chess Today from Westerinen-Larsen 1967, except Polgar’s position includes the bishops which are a non-factor.
Coincidence, or is a little attribution in order here ?