This anon reckons 1. Kb1, after wrestling with lots of corresponding-square theory. Fuller explanation later when at a decent machine, or if he hasn't been beaten to it by JMGB.
The long version. For more information, buy or borrow Jon Speelman's "Endgame Preparation" and turn to chapter four.
First, we need to note the zugzwang positions. This is best done on a piece of squared paper 4x8 labelled a1-d8, and writing the numbers on the squares.
1. Kc4 v Kb6 is the first one - Black must be able to meet White's advance to c4 with ... Kb6 - and not ... Ka6, as White would then make a dash for the kingside and win that race. Ink c4 = 1, b6 = 1. 2. Kd3 v Kc7 is the second zugzwang position: Black must meet 1. Kc4 with 1. ... Kb6, and still be in time on the kingside. 3. Kc3 v Kb7 follows as Black must be ready to meet 1. Kd3 and 1. Kc4 with 1. ... Kc7 and 1. ... Kb6. 4. Stepping back a rank, Kd2 v Kc8 is the next correspondence: Black needs to meet 1. Kd3, 1. Kc3 with 1. ... Kc7, 1. ... Kb7 and he must also be in time on the kingside, so b8 won't do. 5. Kc2 v Kb8 then follows straightaway. 6. Kb3 v Ka7 also follows: from b3, White can go to c4, c3 and c2 and Black needs to move to b6, b7, b8 respectively. 7. Kb2 v Ka8 then follows from that: from b2, White aims for b3, c3 and c2, and Black must go to a7, b7, b8 in reply.
On the back rank, d1 and c1 are the same as d3 and c3: Black must be ready to meet Kd2 and Kc2 and still be in time on the K-side, so his king must be on c7 and b7 when the White king is on these two squares - so label d1 = 2, c1 = 3.
Finally, from b1 White's king aims at b2, c2 and c1, meaning Black's king must be ready to go to a8, b8 and b7 respectively - so Kb1 v Ka7 is the last correspondence, label "6" as for b3.
Armed with all this ammo, we now look at the original position. Black's king is already on a7 so White must start with 1. Kb1!. 1. Kb2? fails to 1. ... Ka8!, and 1. Ka2? gains White nothing: his next move must go to the b-file, one of squares "6" or "7", and Black needs only to wait to see which it is by moving 1. ... Kb8! or 1. ... Kb7! and oscillating until he sees White move East, then stepping onto a8 or a7 as appropriate.
I think the previous "anonymous" gives enough to answer that (good analysis by the way)?
If I'm right, Kb7 is met by Kc1 (b7 and c1 are both labelled "3") then onto d1 if black goes to c7, b2 if black goes to a8, c2 if b8 etc. If black goes 2 ... Kb6 then I think Kd2 is the move.
2. Kc1 - the corresponding square to b7 (cf para "on the back rank"). It might go from there 2. ... Kc7 3. Kd1 Kc8 4. Kd2 Kc7 5. Kd3 Kb6 6. Kc4.
There are lots of variations depending on where Black goes and you can't list them all. White simply maintains the "opposition", as it were, until he reaches Kc4 v Kb6 with Black to move.
As per the short solution in comment 8 (7. Kd3) and the comment to 1. Kc4 v Kb6 in the longer one: "White would then make a dash for the kingside and win that race."
Good Lord, so he did. More than I did, mind, I just find the problems and reproduce the diagrams.
I also found it in Müller and Lamprecht, p.193-4, though my edition isn't Gambit, 2007, but rather Everyman, 2000, where the diagram is also covered in numbers indicating corresponding squares. I assume that was not the case when the problem was originally published in the Chicago Tribune, 1901.
You'll note that White never gets to c4 in their solution, which is why I was querying the answers given in solutions above. I didn't see the point in going Kd3-c4-d3!
17 comments:
Reichhelm, not Reichhlem!
Happy Christmas
Rather embarrassing, but I'm off to a bad start and struggling with this one.
This anon reckons 1. Kb1, after wrestling with lots of corresponding-square theory. Fuller explanation later when at a decent machine, or if he hasn't been beaten to it by JMGB.
It's unlikely to be me writting up a solution.
I thought 1 Kb1 Kb7, 2 Kc2 Kc8, 3 Kd3 Kc7 staying as far away from b6 as White is from c4.
Haven't had a chance to look more at this yet. Might get another go this evening. We shall see.
Shimmy to the left, shimmy to the right, shimmy shimmy shake...?
Ok, I think it involves Kb1-c1-d1-e1 from which white can play Ke2 (heading for b5) or Kf2 (heading for g5) and Black can't handle both threats...
Seani
The short version.
1. Kb1 Ka8 2. Kb2 Kb8 3. Kc2 Kb7 4. Kc3 Kc7 5. Kd3 Kb6 6. Kc4 Ka6 7. Kd3 12. Kh5 Kf6 13. Kh6 and wins.
The long version. For more information, buy or borrow Jon Speelman's "Endgame Preparation" and turn to chapter four.
First, we need to note the zugzwang positions. This is best done on a piece of squared paper 4x8 labelled a1-d8, and writing the numbers on the squares.
1. Kc4 v Kb6 is the first one - Black must be able to meet White's advance to c4 with ... Kb6 - and not ... Ka6, as White would then make a dash for the kingside and win that race. Ink c4 = 1, b6 = 1.
2. Kd3 v Kc7 is the second zugzwang position: Black must meet 1. Kc4 with 1. ... Kb6, and still be in time on the kingside.
3. Kc3 v Kb7 follows as Black must be ready to meet 1. Kd3 and 1. Kc4 with 1. ... Kc7 and 1. ... Kb6.
4. Stepping back a rank, Kd2 v Kc8 is the next correspondence: Black needs to meet 1. Kd3, 1. Kc3 with 1. ... Kc7, 1. ... Kb7 and he must also be in time on the kingside, so b8 won't do.
5. Kc2 v Kb8 then follows straightaway.
6. Kb3 v Ka7 also follows: from b3, White can go to c4, c3 and c2 and Black needs to move to b6, b7, b8 respectively.
7. Kb2 v Ka8 then follows from that: from b2, White aims for b3, c3 and c2, and Black must go to a7, b7, b8 in reply.
On the back rank, d1 and c1 are the same as d3 and c3: Black must be ready to meet Kd2 and Kc2 and still be in time on the K-side, so his king must be on c7 and b7 when the White king is on these two squares - so label d1 = 2, c1 = 3.
Finally, from b1 White's king aims at b2, c2 and c1, meaning Black's king must be ready to go to a8, b8 and b7 respectively - so Kb1 v Ka7 is the last correspondence, label "6" as for b3.
Armed with all this ammo, we now look at the original position. Black's king is already on a7 so White must start with 1. Kb1!. 1. Kb2? fails to 1. ... Ka8!, and 1. Ka2? gains White nothing: his next move must go to the b-file, one of squares "6" or "7", and Black needs only to wait to see which it is by moving 1. ... Kb8! or 1. ... Kb7! and oscillating until he sees White move East, then stepping onto a8 or a7 as appropriate.
What are you going to do on 1.Kb1 Kb7?
(Sorry for the delay in responding: have been unexpectedly busy, and then last night we had a power cut, and you know how it is, and...)
I think the previous "anonymous" gives enough to answer that (good analysis by the way)?
If I'm right, Kb7 is met by Kc1 (b7 and c1 are both labelled "3") then onto d1 if black goes to c7, b2 if black goes to a8, c2 if b8 etc. If black goes 2 ... Kb6 then I think Kd2 is the move.
2. Kc1 - the corresponding square to b7 (cf para "on the back rank"). It might go from there 2. ... Kc7 3. Kd1 Kc8 4. Kd2 Kc7 5. Kd3 Kb6 6. Kc4.
There are lots of variations depending on where Black goes and you can't list them all. White simply maintains the "opposition", as it were, until he reaches Kc4 v Kb6 with Black to move.
If that were to occur, what would White do after ...Ka6?
As per the short solution in comment 8 (7. Kd3) and the comment to 1. Kc4 v Kb6 in the longer one: "White would then make a dash for the kingside and win that race."
Ok. In which case, is it actually necessary to get to c4?
Sometimes, no - for example:
1. Kb1 Kb7 2. Kc1 Kc7 3. Kd1 Kb7 4. Ke2 reaches the kingside gap immediately.
Simply put: if White gets any opportunity at all to put his king three files east of his opponent's king, he takes it pronto.
[I appreciate you're crossing the i's and dotting the t's! It is rather a fiendish puzzle to start with. 1901 must have been a lean year?]
Well I never; Marcel Duchamp analysed this one - or so it says here
Good Lord, so he did. More than I did, mind, I just find the problems and reproduce the diagrams.
I also found it in Müller and Lamprecht, p.193-4, though my edition isn't Gambit, 2007, but rather Everyman, 2000, where the diagram is also covered in numbers indicating corresponding squares. I assume that was not the case when the problem was originally published in the Chicago Tribune, 1901.
You'll note that White never gets to c4 in their solution, which is why I was querying the answers given in solutions above. I didn't see the point in going Kd3-c4-d3!
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