1. Bg5 Kf3 [Kg5?? 2. c8(Q) wins] 2. Be3 Ke3 [only move] 3. c8(Q) Kf2 [g1(Q)?? 4. Qc5 wins] 4. Qf5 Ke1 5. Qe4 Kf1 6. Qf3 Kg1 7. Kf6 Kh2 8. Qf2 Kh1 9. Qh4 Kg1 10.Kf5 Kf1 11.Qf4 Ke1 12.Qe3 Kf1 13.Qf3 Kg1 and the rest follows from move 7, with white bringing the king forward each time black king is forced to g1. Advancing the a pawn shouldn’t matter since any time the black king is at g1, he cannot queen the g-pawn and white will have one move to capture on either the third or fourth rank.
That’s cute. I think I got it. Bg5+. If the king captures, then when you queen, you have Qg8+ if he queens picking up his queen. If he doesnt take and plays Kf3, you play Be3. Now you are covering g1 and if he takes (Kxe3), you can queen and you are threatening Qc5+ if he queens, picking up his Queen in the xray. That is a really cool puzzle. I hope my answer is right.
Bg5+ looks clever, so that if Kxg5, then c8=Q, followed by a check from g8 if Black promotes. And if Black moves Kf3 instead, then Be3 by White either stops the pawn or allows another x-ray check from c5 after Kxe3, c8=Q, and g1=Q. Phil
Well, I think 1 Bg5! Kf3 (1…Kg5? 2 c8=Q g1=Q 3 Qg8+ +- is too easy for White) 2 Be3! Ke3 3 c8=Q Kf2 (3… g1=Q? 4 Qc5+ +-) 4 Qc5 and White’s queen can stop the knight-pawn by itself while the White king can enter the a-pawn’s queening square; but I am too lazy to analyze it fully to make sure this works.
Could one try
1. Bg5 Kf3 [Kg5?? 2. c8(Q) wins]
2. Be3 Ke3 [only move]
3. c8(Q) Kf2 [g1(Q)?? 4. Qc5 wins]
4. Qf5 Ke1
5. Qe4 Kf1
6. Qf3 Kg1
7. Kf6 Kh2
8. Qf2 Kh1
9. Qh4 Kg1
10.Kf5 Kf1
11.Qf4 Ke1
12.Qe3 Kf1
13.Qf3 Kg1 and the rest follows from move 7, with white bringing the king forward each time black king is forced to g1. Advancing the a pawn shouldn’t matter since any time the black king is at g1, he cannot queen the g-pawn and white will have one move to capture on either the third or fourth rank.
1.Bg5+ and Be3
1.Bg5+ Kf3 (1…Kxg5 c8Q) 2.Be3 Kxe3 3.c8Q g1Q 4.Qc5+
Nice problem indeed=)
I’ve seen a puzzle quite similar to this one lately, studying linear combinations.
1.Bg5+
1. …Kxg5 2.c8=Q g1=Q 3.Qg8+
1. …Kf3 (e4) 2.Be3 Kxe3 3.c8=Q g1=Q 4.Qc5+
1. Bg5+ Kf3 (1. … Kxg5 2. c8(Q) g1(Q) 3. Qg8+)
2. Be3 to stop the pawn. If 2. … Kxe3, then 3. c8(Q) g1(Q) 4. Qc5+ to pick up the queen.
That’s cute. I think I got it. Bg5+. If the king captures, then when you queen, you have Qg8+ if he queens picking up his queen. If he doesnt take and plays Kf3, you play Be3. Now you are covering g1 and if he takes (Kxe3), you can queen and you are threatening Qc5+ if he queens, picking up his Queen in the xray. That is a really cool puzzle. I hope my answer is right.
Bg5+ followed by Be3, c8=Q, Qc5
1.Bg5 Kg5 2.c8=Q g1=Q 3.Qg8 1-0
1… Kf3 2.Be3 Kxe3 3.c8=Q g1=Q 4.Qc5 1-0
1…Bg5+! 2.Kf3 – Ne3!
Bg5 Kf3 Be3 Kxe3 c8=Q g1=Q Qc5+
mg.
1.Bg5+, then Be3
Seems clear this probably begins with
Bg5+ Kf3
Be3 Kxe3
c8=Q Kf2
I’m too lazy to work out the remainder.
Bg5+ looks clever, so that if Kxg5, then c8=Q, followed by a check from g8 if Black promotes. And if Black moves Kf3 instead, then Be3 by White either stops the pawn or allows another x-ray check from c5 after Kxe3, c8=Q, and g1=Q. Phil
c8
Well, I think 1 Bg5! Kf3 (1…Kg5? 2 c8=Q g1=Q 3 Qg8+ +- is too easy for White) 2 Be3! Ke3 3 c8=Q Kf2
(3… g1=Q? 4 Qc5+ +-) 4 Qc5 and White’s queen can stop the knight-pawn by itself while the White king can enter the a-pawn’s queening square; but I am too lazy to analyze it fully to make sure this works.
-Justin Daniel
1. Bg5+
1. … Kxg5
2. c8/Q g1/Q
3. Qg8+ wins the Q
or-
1. … Ke4 or f3 (to prevent Be3)
2. Be3 KxB
3. c8/Q c1/Q
4. Qc5+ wins the Q
1.-Bg5 (if …Kxg5 2.- c8=Q g1 3.- Qg8+ and the black queen is lost)
…Kf3 2.- Be4 Kxe4 3.-c8=Q g1 4.-Qc5 and again the queen is lost
I think…
1. Bg5+! Kxg5?
2. c8=Q g1=Q
3. Qg8+ and black loses the queen
1. Bg5+! Kf3
2. Be3! Kxe3
3. c8=Q g1=Q
4. Qc5+ and black loses the queen
1. Bg5+ Kf3 (1…K:g5 2. c8Q g1Q 3. Qg8+) 2. Be3 wins (2…K:e3 3. c8Q g1Q 4. Qc5+)
1. Bg5+ Kf3
2. Be3 Kxe3
3. c8Q etc.
Bg5+; c8(Q); Qf8
I got some squares wrong before.
Looks like 1. Bg5+, then
if 1… Kxg5 2. c8=Q g1=Q 3. Qg8+ wins
if 1… Kf3 2. Be3 Kxe3 3. c8=Q g1=Q 4. Qc5+ wins.
Nobody has addressed the line:
1. Bg5 Kxg5
2. c8(Q) Kf4
It’s not clear to me how White proceeds from here.
Chris,
1. Bg5 Kg5
2. c8(Q)Kf4
3. Qc7 Kf3
4. Qa7 and g1 is covered, the second pawn is liquidated.