4x4 chessboard, bishops and knights are removed, 4 moves to checkmate
$begingroup$
Four-by-four chessboard
Bishops and knights are removed
Four moves to checkmate
Minor spoiler:
Many ways to win
I counted nine solutions
But one is special
Source: Me, I think this website needs more haiku :)
chess poetry
asked Jan 10 at 0:09
deep thoughtdeep thought
3,1641738
$endgroup$
add a comment |
$begingroup$
Four-by-four chessboard
Bishops and knights are removed
Four moves to checkmate
Minor spoiler:
Many ways to win
I counted nine solutions
But one is special
Source: Me, I think this website needs more haiku :)
chess poetry
asked Jan 10 at 0:09
deep thoughtdeep thought
3,1641738
$endgroup$
$begingroup$
there's not enough room for 8 pawns aside. can i assume 4 pawns a side are missing too?
$endgroup$
– SteveV
Jan 10 at 0:22
$begingroup$
No further comment / Time for explaining is past / Death of the author :-)
$endgroup$
– deep thought
Jan 10 at 3:23
add a comment |
$begingroup$
Four-by-four chessboard
Bishops and knights are removed
Four moves to checkmate
Minor spoiler:
Many ways to win
I counted nine solutions
But one is special
Source: Me, I think this website needs more haiku :)
chess poetry
asked Jan 10 at 0:09
deep thoughtdeep thought
3,1641738
$endgroup$
Four-by-four chessboard
Bishops and knights are removed
Four moves to checkmate
Minor spoiler:
Many ways to win
I counted nine solutions
But one is special
Source: Me, I think this website needs more haiku :)
chess poetry
chess poetry
asked Jan 10 at 0:09
deep thoughtdeep thought
3,1641738
asked Jan 10 at 0:09
deep thoughtdeep thought
3,1641738
edited Jan 10 at 0:21
deep thought
asked Jan 10 at 0:09
deep thoughtdeep thought
3,1641738
asked Jan 10 at 0:09
deep thoughtdeep thought
3,1641738
asked Jan 10 at 0:09
deep thoughtdeep thought
3,1641738
3,1641738
$begingroup$
there's not enough room for 8 pawns aside. can i assume 4 pawns a side are missing too?
$endgroup$
– SteveV
Jan 10 at 0:22
$begingroup$
No further comment / Time for explaining is past / Death of the author :-)
$endgroup$
– deep thought
Jan 10 at 3:23
add a comment |
$begingroup$
there's not enough room for 8 pawns aside. can i assume 4 pawns a side are missing too?
$endgroup$
– SteveV
Jan 10 at 0:22
$begingroup$
No further comment / Time for explaining is past / Death of the author :-)
$endgroup$
– deep thought
Jan 10 at 3:23
$begingroup$
there's not enough room for 8 pawns aside. can i assume 4 pawns a side are missing too?
$endgroup$
– SteveV
Jan 10 at 0:22
$begingroup$
there's not enough room for 8 pawns aside. can i assume 4 pawns a side are missing too?
$endgroup$
– SteveV
Jan 10 at 0:22
$begingroup$
No further comment / Time for explaining is past / Death of the author :-)
$endgroup$
– deep thought
Jan 10 at 3:23
$begingroup$
No further comment / Time for explaining is past / Death of the author :-)
$endgroup$
– deep thought
Jan 10 at 3:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm going to assume this is the intended puzzle:
Mate in 4.
Here's one solution (or four, depending on how you count), I think:
Take twice on b3.
B-pawn takes; discovered check.
Queen to a2, mate.
Or in a more common notation:
1. axb3+ Qxb3
2. cxb3+ Kxb3
3. bxa3+ Kc4
4. Qa2#
If2. - Kb4?instead, then3. bxc3#
After ruling out pretty much everything else, here are four more solutions: (found the special one before these, but saving it for last in order to maintain dramatic tension..)
Start just like before.
Queen to A2 on move three,
then mate with a pawn
1. axb3+ Qxb3 (again, white can reorder the first two moves)
2. cxb3+ Kxb3
3. Qa2+ Kb4
4. bxc3# (or dxc3#)
And finally, the special one:
Sac queen on d3:
Exclamation marks galore.
Easy mate in two.
1. cxd3+! Rxd3
2. Qxd3+!! Kxd3
3. dxc3+ Kc4
4. Rd4#
answered Jan 10 at 1:21
BassBass
27.7k467170
$endgroup$
$begingroup$
Correct, that's one. Actually, that's two. The original notation can be translated into common notation two ways.
$endgroup$
– deep thought
Jan 10 at 1:44
$begingroup$
Changing the second line to "b-pawn takes: discovered check" brings that up to four, even :-)
$endgroup$
– Bass
Jan 10 at 1:50
$begingroup$
That's right, plus the nice one makes five, so you've got most of them!
$endgroup$
– deep thought
Jan 10 at 2:06
$begingroup$
... And now nine. Well done! And extra exclamation points for your notation!!
$endgroup$
– deep thought
Jan 10 at 2:44
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
I'm going to assume this is the intended puzzle:
Mate in 4.
Here's one solution (or four, depending on how you count), I think:
Take twice on b3.
B-pawn takes; discovered check.
Queen to a2, mate.
Or in a more common notation:
1. axb3+ Qxb3
2. cxb3+ Kxb3
3. bxa3+ Kc4
4. Qa2#
If2. - Kb4?instead, then3. bxc3#
After ruling out pretty much everything else, here are four more solutions: (found the special one before these, but saving it for last in order to maintain dramatic tension..)
Start just like before.
Queen to A2 on move three,
then mate with a pawn
1. axb3+ Qxb3 (again, white can reorder the first two moves)
2. cxb3+ Kxb3
3. Qa2+ Kb4
4. bxc3# (or dxc3#)
And finally, the special one:
Sac queen on d3:
Exclamation marks galore.
Easy mate in two.
1. cxd3+! Rxd3
2. Qxd3+!! Kxd3
3. dxc3+ Kc4
4. Rd4#
answered Jan 10 at 1:21
BassBass
27.7k467170
$endgroup$
$begingroup$
Correct, that's one. Actually, that's two. The original notation can be translated into common notation two ways.
$endgroup$
– deep thought
Jan 10 at 1:44
$begingroup$
Changing the second line to "b-pawn takes: discovered check" brings that up to four, even :-)
$endgroup$
– Bass
Jan 10 at 1:50
$begingroup$
That's right, plus the nice one makes five, so you've got most of them!
$endgroup$
– deep thought
Jan 10 at 2:06
$begingroup$
... And now nine. Well done! And extra exclamation points for your notation!!
$endgroup$
– deep thought
Jan 10 at 2:44
add a comment |
$begingroup$
I'm going to assume this is the intended puzzle:
Mate in 4.
Here's one solution (or four, depending on how you count), I think:
Take twice on b3.
B-pawn takes; discovered check.
Queen to a2, mate.
Or in a more common notation:
1. axb3+ Qxb3
2. cxb3+ Kxb3
3. bxa3+ Kc4
4. Qa2#
If2. - Kb4?instead, then3. bxc3#
After ruling out pretty much everything else, here are four more solutions: (found the special one before these, but saving it for last in order to maintain dramatic tension..)
Start just like before.
Queen to A2 on move three,
then mate with a pawn
1. axb3+ Qxb3 (again, white can reorder the first two moves)
2. cxb3+ Kxb3
3. Qa2+ Kb4
4. bxc3# (or dxc3#)
And finally, the special one:
Sac queen on d3:
Exclamation marks galore.
Easy mate in two.
1. cxd3+! Rxd3
2. Qxd3+!! Kxd3
3. dxc3+ Kc4
4. Rd4#
answered Jan 10 at 1:21
BassBass
27.7k467170
$endgroup$
$begingroup$
Correct, that's one. Actually, that's two. The original notation can be translated into common notation two ways.
$endgroup$
– deep thought
Jan 10 at 1:44
$begingroup$
Changing the second line to "b-pawn takes: discovered check" brings that up to four, even :-)
$endgroup$
– Bass
Jan 10 at 1:50
$begingroup$
That's right, plus the nice one makes five, so you've got most of them!
$endgroup$
– deep thought
Jan 10 at 2:06
$begingroup$
... And now nine. Well done! And extra exclamation points for your notation!!
$endgroup$
– deep thought
Jan 10 at 2:44
add a comment |
$begingroup$
I'm going to assume this is the intended puzzle:
Mate in 4.
Here's one solution (or four, depending on how you count), I think:
Take twice on b3.
B-pawn takes; discovered check.
Queen to a2, mate.
Or in a more common notation:
1. axb3+ Qxb3
2. cxb3+ Kxb3
3. bxa3+ Kc4
4. Qa2#
If2. - Kb4?instead, then3. bxc3#
After ruling out pretty much everything else, here are four more solutions: (found the special one before these, but saving it for last in order to maintain dramatic tension..)
Start just like before.
Queen to A2 on move three,
then mate with a pawn
1. axb3+ Qxb3 (again, white can reorder the first two moves)
2. cxb3+ Kxb3
3. Qa2+ Kb4
4. bxc3# (or dxc3#)
And finally, the special one:
Sac queen on d3:
Exclamation marks galore.
Easy mate in two.
1. cxd3+! Rxd3
2. Qxd3+!! Kxd3
3. dxc3+ Kc4
4. Rd4#
answered Jan 10 at 1:21
BassBass
27.7k467170
$endgroup$
I'm going to assume this is the intended puzzle:
Mate in 4.
Here's one solution (or four, depending on how you count), I think:
Take twice on b3.
B-pawn takes; discovered check.
Queen to a2, mate.
Or in a more common notation:
1. axb3+ Qxb3
2. cxb3+ Kxb3
3. bxa3+ Kc4
4. Qa2#
If2. - Kb4?instead, then3. bxc3#
After ruling out pretty much everything else, here are four more solutions: (found the special one before these, but saving it for last in order to maintain dramatic tension..)
Start just like before.
Queen to A2 on move three,
then mate with a pawn
1. axb3+ Qxb3 (again, white can reorder the first two moves)
2. cxb3+ Kxb3
3. Qa2+ Kb4
4. bxc3# (or dxc3#)
And finally, the special one:
Sac queen on d3:
Exclamation marks galore.
Easy mate in two.
1. cxd3+! Rxd3
2. Qxd3+!! Kxd3
3. dxc3+ Kc4
4. Rd4#
answered Jan 10 at 1:21
BassBass
27.7k467170
edited Jan 10 at 2:53
answered Jan 10 at 1:21
BassBass
27.7k467170
answered Jan 10 at 1:21
BassBass
27.7k467170
answered Jan 10 at 1:21
BassBass
27.7k467170
27.7k467170
$begingroup$
Correct, that's one. Actually, that's two. The original notation can be translated into common notation two ways.
$endgroup$
– deep thought
Jan 10 at 1:44
$begingroup$
Changing the second line to "b-pawn takes: discovered check" brings that up to four, even :-)
$endgroup$
– Bass
Jan 10 at 1:50
$begingroup$
That's right, plus the nice one makes five, so you've got most of them!
$endgroup$
– deep thought
Jan 10 at 2:06
$begingroup$
... And now nine. Well done! And extra exclamation points for your notation!!
$endgroup$
– deep thought
Jan 10 at 2:44
add a comment |
$begingroup$
Correct, that's one. Actually, that's two. The original notation can be translated into common notation two ways.
$endgroup$
– deep thought
Jan 10 at 1:44
$begingroup$
Changing the second line to "b-pawn takes: discovered check" brings that up to four, even :-)
$endgroup$
– Bass
Jan 10 at 1:50
$begingroup$
That's right, plus the nice one makes five, so you've got most of them!
$endgroup$
– deep thought
Jan 10 at 2:06
$begingroup$
... And now nine. Well done! And extra exclamation points for your notation!!
$endgroup$
– deep thought
Jan 10 at 2:44
$begingroup$
Correct, that's one. Actually, that's two. The original notation can be translated into common notation two ways.
$endgroup$
– deep thought
Jan 10 at 1:44
$begingroup$
Correct, that's one. Actually, that's two. The original notation can be translated into common notation two ways.
$endgroup$
– deep thought
Jan 10 at 1:44
$begingroup$
Changing the second line to "b-pawn takes: discovered check" brings that up to four, even :-)
$endgroup$
– Bass
Jan 10 at 1:50
$begingroup$
Changing the second line to "b-pawn takes: discovered check" brings that up to four, even :-)
$endgroup$
– Bass
Jan 10 at 1:50
$begingroup$
That's right, plus the nice one makes five, so you've got most of them!
$endgroup$
– deep thought
Jan 10 at 2:06
$begingroup$
That's right, plus the nice one makes five, so you've got most of them!
$endgroup$
– deep thought
Jan 10 at 2:06
$begingroup$
... And now nine. Well done! And extra exclamation points for your notation!!
$endgroup$
– deep thought
Jan 10 at 2:44
$begingroup$
... And now nine. Well done! And extra exclamation points for your notation!!
$endgroup$
– deep thought
Jan 10 at 2:44
add a comment |
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$begingroup$
there's not enough room for 8 pawns aside. can i assume 4 pawns a side are missing too?
$endgroup$
– SteveV
Jan 10 at 0:22
$begingroup$
No further comment / Time for explaining is past / Death of the author :-)
$endgroup$
– deep thought
Jan 10 at 3:23