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We could try something like Qf4+ followed by Qf7+, and on Black’s capture, promote to a knight for a fork. I think White then has a known win in the endgame, but I’m playing through some variations to see.
‘Can you find the winning continuation for White?’
The real question is: Can YOU find the winning continuation for White?
Found it eventually, but this is a tough one! I won’t spoil the enjoyment by revealing the solution yet. It deserves the accolade.
searching for a winning fork with an underpromotion… maybe:
1.Qf4+ Ke6
2.Qf7+ Kxf7
3.d8N
1. Qf4+ Ke6/e7 2. Qf7+ Kxf7 3. d8=Q+ wins.
If this is the solution, I have no idea why this is called the study of the year. What is so great in this ordinary puzzle?
An improvement to my last comment.
1. Qf4+, and now:
A) 1…Ke7 2. Qf7+ Kd8 (2… Kxf7 3. d8N+; 2… Kd6 3. d8Q+) 3. Qe8+ Kc7 4. d8Q#.
B) 1…Ke6 2. Qf7+ Ke5 (otherwise 3. d8Q+) 3. Qg7+ Kf4 4. d8Q wins.
What is ‘on’:
Qf4+ Ke6/Ke7
d8=N wins if there’s a Q/K fork
…
1.Qf4+ Ke6/Ke7
2.Qf7+!! and then
2….Kd6 3.d8=Q+ wins
2…Ke5 3.Qe7+ K? 4.d8=Q wins
2…Kd8 3.Qe8+ Kc7 4.d8=Q#
so 2…Kxf7
but this is a tablebase loss …
3.d8=N!! K? 4.Nxb7
Since there was no restriction I referred “Shredder” & got the answer but will refrain from posting it now.I
will consider later if Susan replies to this post.
1. Qf4+
if 1… Ke6
2. Qf7+ Kd6
3. d8(N) QxQ
4. NxQ
now, need to make sure that this still wins for white. I’m assuming it does.
if 2… KxQ
3. d8(N) + and 4. Nxb7
hopefully, here too, the N is able to take black’s a-pawn and protect white’s own a-pawn.
if 1… Ke7
2. Qf7+ Kd8
(2… KxQ transposes to the line above)
3. Qe8+ Kc7
4. d8(Q) wins.
The main issue is can the N win the game for white — i.e., can it at least defend the white a-pawn?
I guess
1. Qf4+ Ke7(e6)
2. Qf7+ Kxf7
3. d8=N+ Kf6
4. Nxb7
Than I thought Na5-c4-a3-c1 would be win but Black is to fast.
Therefore
4. … Ke5
5. Kg6 Kd4
6. Kf5 Kc3
7. Ke4 Kb2
8. Kd3 Kxa2 (… Ka1, Kc3)
9. Kc2 Ka1
10. Nc5 Ka2
11. Nd3 Ka1
12. Nc1 a2
13. Nb3#
Qf4 Ke6
Qf7 KxQ
d8=N K any
NxQ White wins
Arctic Knight in 5 secs. (which means it might be wrong)
May be???
1. Qf4+ Ke6 [or 1…Ke7]
2. Qf7+ Kxf7
3. d8=N+
I strongly suspect a pawn underpromotion for a knight while the black king is in f7, forking King and Queen.
The challenge after is either taking the a3 pawn without losing the white remaining pawn, or losing elegantly the pawn a2 without apparent compensation… to transpose in a classic checkmate demonstrated by Saint Nicholai 8 centuries ago !
Now, the hard part : calculation !
1. Qf4 Ke6
2. Qf7+ Ke5
3. d8=N and white should be on the way to winning?
1. Qf4+ Ke6
2. Qf7+!!Ke5
3. Kg8 and white promotes the pawn i think is a better solution …
At least Rainer showed the line that leads to the smothered mate! lol
o.k. i think i got this one… i’m new to this so…
i’m thinking first qf4 which brings ke6 THEN qf7 when the queen is taken….then d8 knight then whatever then take d7. now i’m thinking the black king goes after the pawn so the black king e5 then you kg6 black will go d4 kf5 so on until black takes the pawn, the the knight comes down and the check should be obvious from there!
correct me if i’m wrong in there
thanks rainer for showing us how hard it was, especially the kn moves to hit b3# just right
Best Study of 2008?
I have a deep doubt about it. The position is very natural, game-like, few pieces and all, and this is uncommon about modern studies IMHO. But…
An under-promotion in knight, following by a classic checkmate manoeuvre was already realized by the strong player and study composer Jan Timman in Vrij Nederland, 1992. Win
Here is the FEN (so you can paste it in your favorite chess GUI, engine, or database) and solution to Timman’s Study :
FEN : 2q3B1/R1P1k3/1P3bK1/2Rp4/8/p1P5/P6B/8 w – – 0 1
Solution :
1. Bd6+! Kxd6
2. Rxd5+ Ke7
3. Be6 Kxe6
4. Td8! (third sacrifice)Bxd8
5. Ta8 (fourth one) Qxa8
6. b7
(A known trapezoidal configuration of white and black pieces which occurred for the first time in a study of Nadareishvili, Vecherni Leningrad, 1965, first prize ex-aequo, draw
FEN : K2b3q/3kP3/5P2/8/2Pp4/6R1/4P3/7B w – – 0 1)
6… Qxb7
7. cxd8N+ (here is our under-promotion)
7… Kd5
8. Nxb7 Kc4
9. Kf5 Kxc3
and now you know the finish, because the position is the very same as in the Gurgenidze study composed 27 years later…
Originality? Perhaps the seemingly dead draw Queen ending (because the d pawn is pinned). But is this enough for such an award?
This is not the place for such debates. But I feel obliged to mention the much superior Timman’s Study.
Thank you Susan and bravo Rainer to bring us the precise and complete solution.
Umesh::ഉമേഷ് said…
1. Qf4+ Ke6/e7 2. Qf7+ Kxf7 3. d8=Q+ wins.
wins so much that 3. …;Qh1! is checkmate
Errata
——-
I’ve done a slight mistake due to my French mother-tongue and the hour of my intervention, early in the morning.
Please read, instead of
4. Td8! Bxd8
5. Ta8
naturally
4. Rxd8! Bxd8
5. Ra8
My apologize
I can’t see a straightforward white win here. Say:
1. Qf4+ K~
2. Qf7+ Kxf7
3. d8N+ Kf6
4. Nxb7
Why is this a certain win for white? The black king can reach b2 easily, and white’s only way of protecting its a2 pawn is by putting his knight on b4. Then all I see is a draw by Nb4 Kc3 Na6 Kb2 Nb4… What am I missing here?