Pawn Promotion Double ChecksThe separation of chess piecesHow Many Pieces from Both Sides can you Place on a Chess Board Without Attacking a Side?Equal Chess ControlEqual Chess Control With One ColorChess Game with Highest ScoringCan you stop the rambling rook?Checkmate with Pawn and KingThe Minimized DropmateWhat’s the most discovered checkmates in one in a position?Pawninos-44 Checks In A Row
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Pawn Promotion Double Checks
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Pawn Promotion Double Checks
The separation of chess piecesHow Many Pieces from Both Sides can you Place on a Chess Board Without Attacking a Side?Equal Chess ControlEqual Chess Control With One ColorChess Game with Highest ScoringCan you stop the rambling rook?Checkmate with Pawn and KingThe Minimized DropmateWhat’s the most discovered checkmates in one in a position?Pawninos-44 Checks In A Row
$begingroup$
In chess, it is possible to double check whilst giving promotion. An example of this would be a pawn promoting to a queen while a discovered attack from a rook behind the pawn checks the opponent’s king. Both the new queen and rook check at the same time, thus making it a double check.
Construct the longest possible sequence that you can of promotions with double check. Assume that white is to move first and black is to help. You must also prove your answer to be optimal via any reasonable method.
Good luck!
chess optimization
asked 3 hours ago
Rewan DemontayRewan Demontay
1,108120
$endgroup$
add a comment |
$begingroup$
In chess, it is possible to double check whilst giving promotion. An example of this would be a pawn promoting to a queen while a discovered attack from a rook behind the pawn checks the opponent’s king. Both the new queen and rook check at the same time, thus making it a double check.
Construct the longest possible sequence that you can of promotions with double check. Assume that white is to move first and black is to help. You must also prove your answer to be optimal via any reasonable method.
Good luck!
chess optimization
asked 3 hours ago
Rewan DemontayRewan Demontay
1,108120
$endgroup$
$begingroup$
Yes. That is it.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Sorry you answered exactly when I was deleting the question - got it.
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Oof. No problem.
$endgroup$
– Rewan Demontay
2 hours ago
add a comment |
$begingroup$
In chess, it is possible to double check whilst giving promotion. An example of this would be a pawn promoting to a queen while a discovered attack from a rook behind the pawn checks the opponent’s king. Both the new queen and rook check at the same time, thus making it a double check.
Construct the longest possible sequence that you can of promotions with double check. Assume that white is to move first and black is to help. You must also prove your answer to be optimal via any reasonable method.
Good luck!
chess optimization
asked 3 hours ago
Rewan DemontayRewan Demontay
1,108120
$endgroup$
In chess, it is possible to double check whilst giving promotion. An example of this would be a pawn promoting to a queen while a discovered attack from a rook behind the pawn checks the opponent’s king. Both the new queen and rook check at the same time, thus making it a double check.
Construct the longest possible sequence that you can of promotions with double check. Assume that white is to move first and black is to help. You must also prove your answer to be optimal via any reasonable method.
Good luck!
chess optimization
chess optimization
asked 3 hours ago
Rewan DemontayRewan Demontay
1,108120
asked 3 hours ago
Rewan DemontayRewan Demontay
1,108120
asked 3 hours ago
Rewan DemontayRewan Demontay
1,108120
asked 3 hours ago
Rewan DemontayRewan Demontay
1,108120
asked 3 hours ago
Rewan DemontayRewan Demontay
1,108120
1,108120
$begingroup$
Yes. That is it.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Sorry you answered exactly when I was deleting the question - got it.
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Oof. No problem.
$endgroup$
– Rewan Demontay
2 hours ago
add a comment |
$begingroup$
Yes. That is it.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Sorry you answered exactly when I was deleting the question - got it.
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Oof. No problem.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Yes. That is it.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Yes. That is it.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Sorry you answered exactly when I was deleting the question - got it.
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Sorry you answered exactly when I was deleting the question - got it.
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Oof. No problem.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Oof. No problem.
$endgroup$
– Rewan Demontay
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I now have
Four double checks by promotion in a row.
FEN: B1nQ2B1/1P1PPP2/2k5/8/8/8/8/7K w - -
Game: 1. b8=N+ Kd6 2. dxc8=N+ Ke6 3. f8=N+ Kf6 4. e8=N+
Apronus link
Proof that it is optimal: UNDER EDIT
A double check can occur either by a classic discovery, or by an en-passant capture. The latter is impossible here so it has to be a discovery. There are three cases:
Type A: Black King in the 8th rank right in front of the pawn to be promoted, promotion = Q or R, with another Q or R in the same column as the King.
Type B: Black King in the 7th rank, checked by Rook or Queen also on the 7th rank; all four types of promotions can occur.
Type C: Black King in the 6th rank, attacking-close to the pawn to be promoted. Promotion = Q or R or N, while the other checking piece can be anything.
Now here is why there can't be too many moves:
Fact 1: there can be only one type A in the sequence. Because (long story short) the King has to leave the 8th rank and it can never come back.
Fact 2: there can be only one type B in the sequence. Because (long story short) the King has to leave the 7th rank and it can never come back.
Therefore the only possible sequences are (possibly truncated) ABCCC... and ...CCCBA.
answered 2 hours ago
Arnaud MortierArnaud Mortier
2,747828
$endgroup$
$begingroup$
Nice try, but nope!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
You’re getting there! ;D
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I guess someone else will figure it out then.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
@RewanDemontay I meant "I can't clearly prove it yet". :)
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Ahhh okkkkkkay! I gotcha!
$endgroup$
– Rewan Demontay
2 hours ago
add a comment |
$begingroup$
I’ve managed to do
3 discovered checks, assuming that capture promotions are allowed.
The setup:
FEN: 3nnBB1/RP1kPP2/8/8/8/8/8/7K w - - 0 1
The game:
1. b8=N+ Kd6 2. exd8=R+ Ke6 3. fxe8=R+
Lichess link
I have a feeling we can do a bit better than that, but as of now this is the best I can get.
answered 2 hours ago
PiIsNot3PiIsNot3
3,160643
$endgroup$
$begingroup$
That’s allowed alright!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I now have one more, but a proof of optimality doesn't seem easy.
$endgroup$
– Arnaud Mortier
47 mins ago
$begingroup$
Yep. Whether or not 5 is possible is nagging at me. 6 is definitely impossible.
$endgroup$
– Rewan Demontay
42 mins ago
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
I now have
Four double checks by promotion in a row.
FEN: B1nQ2B1/1P1PPP2/2k5/8/8/8/8/7K w - -
Game: 1. b8=N+ Kd6 2. dxc8=N+ Ke6 3. f8=N+ Kf6 4. e8=N+
Apronus link
Proof that it is optimal: UNDER EDIT
A double check can occur either by a classic discovery, or by an en-passant capture. The latter is impossible here so it has to be a discovery. There are three cases:
Type A: Black King in the 8th rank right in front of the pawn to be promoted, promotion = Q or R, with another Q or R in the same column as the King.
Type B: Black King in the 7th rank, checked by Rook or Queen also on the 7th rank; all four types of promotions can occur.
Type C: Black King in the 6th rank, attacking-close to the pawn to be promoted. Promotion = Q or R or N, while the other checking piece can be anything.
Now here is why there can't be too many moves:
Fact 1: there can be only one type A in the sequence. Because (long story short) the King has to leave the 8th rank and it can never come back.
Fact 2: there can be only one type B in the sequence. Because (long story short) the King has to leave the 7th rank and it can never come back.
Therefore the only possible sequences are (possibly truncated) ABCCC... and ...CCCBA.
answered 2 hours ago
Arnaud MortierArnaud Mortier
2,747828
$endgroup$
$begingroup$
Nice try, but nope!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
You’re getting there! ;D
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I guess someone else will figure it out then.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
@RewanDemontay I meant "I can't clearly prove it yet". :)
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Ahhh okkkkkkay! I gotcha!
$endgroup$
– Rewan Demontay
2 hours ago
add a comment |
$begingroup$
I now have
Four double checks by promotion in a row.
FEN: B1nQ2B1/1P1PPP2/2k5/8/8/8/8/7K w - -
Game: 1. b8=N+ Kd6 2. dxc8=N+ Ke6 3. f8=N+ Kf6 4. e8=N+
Apronus link
Proof that it is optimal: UNDER EDIT
A double check can occur either by a classic discovery, or by an en-passant capture. The latter is impossible here so it has to be a discovery. There are three cases:
Type A: Black King in the 8th rank right in front of the pawn to be promoted, promotion = Q or R, with another Q or R in the same column as the King.
Type B: Black King in the 7th rank, checked by Rook or Queen also on the 7th rank; all four types of promotions can occur.
Type C: Black King in the 6th rank, attacking-close to the pawn to be promoted. Promotion = Q or R or N, while the other checking piece can be anything.
Now here is why there can't be too many moves:
Fact 1: there can be only one type A in the sequence. Because (long story short) the King has to leave the 8th rank and it can never come back.
Fact 2: there can be only one type B in the sequence. Because (long story short) the King has to leave the 7th rank and it can never come back.
Therefore the only possible sequences are (possibly truncated) ABCCC... and ...CCCBA.
answered 2 hours ago
Arnaud MortierArnaud Mortier
2,747828
$endgroup$
$begingroup$
Nice try, but nope!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
You’re getting there! ;D
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I guess someone else will figure it out then.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
@RewanDemontay I meant "I can't clearly prove it yet". :)
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Ahhh okkkkkkay! I gotcha!
$endgroup$
– Rewan Demontay
2 hours ago
add a comment |
$begingroup$
I now have
Four double checks by promotion in a row.
FEN: B1nQ2B1/1P1PPP2/2k5/8/8/8/8/7K w - -
Game: 1. b8=N+ Kd6 2. dxc8=N+ Ke6 3. f8=N+ Kf6 4. e8=N+
Apronus link
Proof that it is optimal: UNDER EDIT
A double check can occur either by a classic discovery, or by an en-passant capture. The latter is impossible here so it has to be a discovery. There are three cases:
Type A: Black King in the 8th rank right in front of the pawn to be promoted, promotion = Q or R, with another Q or R in the same column as the King.
Type B: Black King in the 7th rank, checked by Rook or Queen also on the 7th rank; all four types of promotions can occur.
Type C: Black King in the 6th rank, attacking-close to the pawn to be promoted. Promotion = Q or R or N, while the other checking piece can be anything.
Now here is why there can't be too many moves:
Fact 1: there can be only one type A in the sequence. Because (long story short) the King has to leave the 8th rank and it can never come back.
Fact 2: there can be only one type B in the sequence. Because (long story short) the King has to leave the 7th rank and it can never come back.
Therefore the only possible sequences are (possibly truncated) ABCCC... and ...CCCBA.
answered 2 hours ago
Arnaud MortierArnaud Mortier
2,747828
$endgroup$
I now have
Four double checks by promotion in a row.
FEN: B1nQ2B1/1P1PPP2/2k5/8/8/8/8/7K w - -
Game: 1. b8=N+ Kd6 2. dxc8=N+ Ke6 3. f8=N+ Kf6 4. e8=N+
Apronus link
Proof that it is optimal: UNDER EDIT
A double check can occur either by a classic discovery, or by an en-passant capture. The latter is impossible here so it has to be a discovery. There are three cases:
Type A: Black King in the 8th rank right in front of the pawn to be promoted, promotion = Q or R, with another Q or R in the same column as the King.
Type B: Black King in the 7th rank, checked by Rook or Queen also on the 7th rank; all four types of promotions can occur.
Type C: Black King in the 6th rank, attacking-close to the pawn to be promoted. Promotion = Q or R or N, while the other checking piece can be anything.
Now here is why there can't be too many moves:
Fact 1: there can be only one type A in the sequence. Because (long story short) the King has to leave the 8th rank and it can never come back.
Fact 2: there can be only one type B in the sequence. Because (long story short) the King has to leave the 7th rank and it can never come back.
Therefore the only possible sequences are (possibly truncated) ABCCC... and ...CCCBA.
answered 2 hours ago
Arnaud MortierArnaud Mortier
2,747828
edited 42 mins ago
answered 2 hours ago
Arnaud MortierArnaud Mortier
2,747828
answered 2 hours ago
Arnaud MortierArnaud Mortier
2,747828
answered 2 hours ago
Arnaud MortierArnaud Mortier
2,747828
2,747828
$begingroup$
Nice try, but nope!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
You’re getting there! ;D
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I guess someone else will figure it out then.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
@RewanDemontay I meant "I can't clearly prove it yet". :)
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Ahhh okkkkkkay! I gotcha!
$endgroup$
– Rewan Demontay
2 hours ago
add a comment |
$begingroup$
Nice try, but nope!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
You’re getting there! ;D
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I guess someone else will figure it out then.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
@RewanDemontay I meant "I can't clearly prove it yet". :)
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Ahhh okkkkkkay! I gotcha!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Nice try, but nope!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Nice try, but nope!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
You’re getting there! ;D
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
You’re getting there! ;D
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I guess someone else will figure it out then.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I guess someone else will figure it out then.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
@RewanDemontay I meant "I can't clearly prove it yet". :)
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
@RewanDemontay I meant "I can't clearly prove it yet". :)
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Ahhh okkkkkkay! I gotcha!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Ahhh okkkkkkay! I gotcha!
$endgroup$
– Rewan Demontay
2 hours ago
add a comment |
$begingroup$
I’ve managed to do
3 discovered checks, assuming that capture promotions are allowed.
The setup:
FEN: 3nnBB1/RP1kPP2/8/8/8/8/8/7K w - - 0 1
The game:
1. b8=N+ Kd6 2. exd8=R+ Ke6 3. fxe8=R+
Lichess link
I have a feeling we can do a bit better than that, but as of now this is the best I can get.
answered 2 hours ago
PiIsNot3PiIsNot3
3,160643
$endgroup$
$begingroup$
That’s allowed alright!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I now have one more, but a proof of optimality doesn't seem easy.
$endgroup$
– Arnaud Mortier
47 mins ago
$begingroup$
Yep. Whether or not 5 is possible is nagging at me. 6 is definitely impossible.
$endgroup$
– Rewan Demontay
42 mins ago
add a comment |
$begingroup$
I’ve managed to do
3 discovered checks, assuming that capture promotions are allowed.
The setup:
FEN: 3nnBB1/RP1kPP2/8/8/8/8/8/7K w - - 0 1
The game:
1. b8=N+ Kd6 2. exd8=R+ Ke6 3. fxe8=R+
Lichess link
I have a feeling we can do a bit better than that, but as of now this is the best I can get.
answered 2 hours ago
PiIsNot3PiIsNot3
3,160643
$endgroup$
$begingroup$
That’s allowed alright!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I now have one more, but a proof of optimality doesn't seem easy.
$endgroup$
– Arnaud Mortier
47 mins ago
$begingroup$
Yep. Whether or not 5 is possible is nagging at me. 6 is definitely impossible.
$endgroup$
– Rewan Demontay
42 mins ago
add a comment |
$begingroup$
I’ve managed to do
3 discovered checks, assuming that capture promotions are allowed.
The setup:
FEN: 3nnBB1/RP1kPP2/8/8/8/8/8/7K w - - 0 1
The game:
1. b8=N+ Kd6 2. exd8=R+ Ke6 3. fxe8=R+
Lichess link
I have a feeling we can do a bit better than that, but as of now this is the best I can get.
answered 2 hours ago
PiIsNot3PiIsNot3
3,160643
$endgroup$
I’ve managed to do
3 discovered checks, assuming that capture promotions are allowed.
The setup:
FEN: 3nnBB1/RP1kPP2/8/8/8/8/8/7K w - - 0 1
The game:
1. b8=N+ Kd6 2. exd8=R+ Ke6 3. fxe8=R+
Lichess link
I have a feeling we can do a bit better than that, but as of now this is the best I can get.
answered 2 hours ago
PiIsNot3PiIsNot3
3,160643
answered 2 hours ago
PiIsNot3PiIsNot3
3,160643
answered 2 hours ago
PiIsNot3PiIsNot3
3,160643
answered 2 hours ago
PiIsNot3PiIsNot3
3,160643
3,160643
$begingroup$
That’s allowed alright!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I now have one more, but a proof of optimality doesn't seem easy.
$endgroup$
– Arnaud Mortier
47 mins ago
$begingroup$
Yep. Whether or not 5 is possible is nagging at me. 6 is definitely impossible.
$endgroup$
– Rewan Demontay
42 mins ago
add a comment |
$begingroup$
That’s allowed alright!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I now have one more, but a proof of optimality doesn't seem easy.
$endgroup$
– Arnaud Mortier
47 mins ago
$begingroup$
Yep. Whether or not 5 is possible is nagging at me. 6 is definitely impossible.
$endgroup$
– Rewan Demontay
42 mins ago
$begingroup$
That’s allowed alright!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
That’s allowed alright!
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
I now have one more, but a proof of optimality doesn't seem easy.
$endgroup$
– Arnaud Mortier
47 mins ago
$begingroup$
I now have one more, but a proof of optimality doesn't seem easy.
$endgroup$
– Arnaud Mortier
47 mins ago
$begingroup$
Yep. Whether or not 5 is possible is nagging at me. 6 is definitely impossible.
$endgroup$
– Rewan Demontay
42 mins ago
$begingroup$
Yep. Whether or not 5 is possible is nagging at me. 6 is definitely impossible.
$endgroup$
– Rewan Demontay
42 mins ago
add a comment |
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$begingroup$
Yes. That is it.
$endgroup$
– Rewan Demontay
2 hours ago
$begingroup$
Sorry you answered exactly when I was deleting the question - got it.
$endgroup$
– Arnaud Mortier
2 hours ago
$begingroup$
Oof. No problem.
$endgroup$
– Rewan Demontay
2 hours ago