Mate in 5:
i wanted something that isn't just check-check-check, but i guess it is possible to promote to other pieces but just imagine promotions are on auto queen:
Mate in 5:
i wanted something that isn't just check-check-check, but i guess it is possible to promote to other pieces but just imagine promotions are on auto queen:
I agree with your logic, but I think it's 3 squares and 4 squares. A minor detail I know, but it would make a difference if the movements were smaller.
Immediately after posting the same idea came to my mind ( that 3 and 4 squares would have been better ) but I did not edit my original post since still 4*4 = 16 < 3*3 + 3*3 = 18. You are right that it is more plausible to measure the length of a move by the distance that any point of a piece 'travels' during the move rather than the distance between the most distant points of the original and the target square along the diagonal or rank/file.
I agree with your logic, but I think it's 3 squares and 4 squares. A minor detail I know, but it would make a difference if the movements were smaller.
Immediately after posting the same idea came to my mind ( that 3 and 4 squares would have been better ) but I did not edit my original post since still 4*4 = 16 < 3*3 + 3*3 = 18. You are right that it is more plausible to measure the length of a move by the distance that any point of a piece 'travels' during the move rather than the distance between the most distant points of the original and the target square along the diagonal or rank/file.
I wouldn't have said anything but when moving 2 squares with the rook and 1 diagonally with the bishop, the longest move would be the rook. If the numbers were 1 higher, then the bishop would be mistakenly considered to move further.
mate in 3
That is not self solving; 3.c8N would not mate.
Auto queen is assumed at this solution. Otherwise it is not self solving not because some promotions are not mate but because there can only be one legal move in each position and with any promotion you have four legal moves. If you don't assume auto queen then you can't have promotions in self solving puzzles.
I wouldn't have said anything but when moving 2 squares with the rook and 1 diagonally with the bishop, the longest move would be the rook. If the numbers were 1 higher, then the bishop would be mistakenly considered to move further.
3*3 > 2*2 + 2*2
2*2 > 1*1 + 1*1
Using both conventions the rook move is longer.
mate in 3
That is not self solving; 3.c8N would not mate.
Auto queen is assumed at this solution. Otherwise it is not self solving not because some promotions are not mate but because there can only be one legal move in each position and with any promotion you have four legal moves. If you don't assume auto queen then you can't have promotions in self solving puzzles.
Auto-queen is too artificial for my taste.
ajttja, yours doesnt work:
oops, i meant there to be a white pawn in a6
So, thatt'll be 10 white pawns all together, then ?
Auto-queen is too artificial for my taste.
I have auto queen it in my blitz games. If blitz games are chess then auto queen can not be deemed too artificial.
On the other hand: if auto queen is too artificial then self solving puzzles should avoid promotion alltogether which in and out of itself can be viewed by some as artificial.
Auto-queen is too artificial for my taste.
I have auto queen it in my blitz games. If blitz games are chess then auto queen can not be deemed too artificial.
On the other hand: if auto queen is too artificial then self solving puzzles should avoid promotion alltogether which in and out of itself can be viewed by some as artificial.
No, that is fallacious; blitz does not inherently have auto-queen. That is just a convenience setting in the interface.
On the last point, those mysterious people who view lack of promotion in auto-solution problems as artificial are using the word 'artificial' with almost the opposite meaning of the usual one.
No, that is fallacious; blitz does not inherently have auto-queen. That is just a convenience setting in the interface.
It is also a convenience setting in self-solving puzzles :).
No, that is fallacious; blitz does not inherently have auto-queen. That is just a convenience setting in the interface.
It is also a convenience setting in self-solving puzzles :).
Yep, tacked on for convenience, and thus artificial.
Auto-promotion in queen looks assumed in the first puzzle of the OP (5...a1 could be any of the 4 promotions).
Anyways, here is my first and short try :
@ajtta - you realise your problem has 9 white pawns?!
And you guys are using bad maths to get the right result. It´s not
"4*4 + 4*4 = 32
The rook could make a move 5 squares long.
Since 5*5 = 25 < 32, the bishop move is longer."
It should be: the SQUARE ROOT of 4*4 + 4*4 is larger than 5 (it´s ca. 5,66).
Little excercise for you - which piece moves the greater actual distance, a knight which moves in 3 moves from a1 to g4, or a rook moving a1-h1? (assuming the N moves in straight lines between squares)
It should be: the SQUARE ROOT of 4*4 + 4*4 is larger than 5 (it´s ca. 5,66). Your method wouldn´t work for the diagonal of a 2x2 square, as someone pointed out. Using Pythagoras the diagonal is sqrt 8 (ca. 2,83).
x > y*y <=> sqrt(x) > y
The above is always true for natural numbers. Since you can calculate squares more easily in your head than square roots it is easier to use the former method ( but the two are equivalent ).
As to the definition of 'move length' you have to ask the creator of the puzzle or check move by move which definition fits the puzzle ( it may turn out that both but I have not checked it move by move ).
I have calculated the distance squares with the two methods, method (1) based on the square difference and method (2) based on total affected squares.
If the rook moves one more square along a file/rank than the bishop along a diagonal then the two definitions only yield a different result if the rook moves three squares and the bishop two squares ( if you look at all possible combinations the two definitions will differ in only two cases ):
bishop | rook | Bishop Dst Sq (1) | Rook Dst Sq (1) | Bishop Dst Sq (2) | Rook Dst Sq (2) |
1 | 2 | 2 | 4 | 8 | 9 |
2 | 3 | 8 | 9 | 18 | 16 |
3 | 4 | 18 | 16 | 32 | 25 |
4 | 5 | 32 | 25 | 50 | 36 |
5 | 6 | 50 | 36 | 72 | 49 |
6 | 7 | 72 | 49 | 98 | 64 |
Regarding auto queening assumption, since I wanted only one legal move to be possble at all times, you cannot actually promote under these restrictions. But it is true that the first puzzle I posted as an introduction uses promotion for black. Of course he's still going to get mated anyway, but he has four different legal moves.
So at the moment we have 6 moves with no queens. That is quite impressive, however the record we have is 7 moves. With 7 It's a little more tricky since a single line of checks is not enough (see the mate in 8 that I posted, the white king moves to create a new line of attack).
If you're interested in another challenge, try to do a self-solving puzzle where one side makes three consecutive knight moves. That is not as simple as the last one.
If no answers will be given I will give the solutions.
@kromhawk_31:
Because it is on a diagonal.
According to the Pythagorean theorem in any right triangle, the square of the length of the hypotenuse equals the sum of the squares of the lengths of the two other sides. If the hypotenuse has length c, and the legs have lengths a and b, then the theorem states that
Here the bishop makes a four square move along the diagonal:
4*4 + 4*4 = 32
The rook could make a move 5 squares long.
Since 5*5 = 25 < 32, the bishop move is longer.
I agree with your logic, but I think it's 3 squares and 4 squares. A minor detail I know, but it would make a difference if the movements were smaller.
These puzzles are wonderful by the way.