Yeah... I also agree... Although you might not agree because dictionary existed at the first place, we have to respect the definition of the one giving riddle... I know you have concern for him, but he had already given a nice definition about it so it might be also good to just take it as it is... (Cancel means possibilities of capture.)
Chess with Maths
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Because of that, it is good if you double it because White Queen captures Black Queen and on the other hand Black Queen captures White Queen...
I think I understood it this way... q If capturing a piece made them cancel each other hence both of them are removed from the board, How many possible ways that both of them are not removed from the board if you use a turn based movement assuming that you need to use one move only while the other one would not be able to but the other one can also do the same at the different scenario... q
Mar 16, 2024
0
#85
Mr_Mathematician wrote:
CoDCVN wrote:
Mr_Mathematician đã viết:
StockOfHey wrote:
In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place , assume white plays first ).
I found the words "that they never get cancel by each other". It is hard to get... If you did not combine the definition of the second sentence "assume white plays first", it is hard to know the real definition of the word cancel...
There are two definition:
1. decide or announce that (a planned event) will not take place.
2. (of a factor or circumstance) neutralize or negate the force or effect of (another).
The first one says that, capturing a piece is included in the problem... Because, if you cancel the effects of the other piece, you must capture it while the ones who capture remain hence "they did not cancel each other"...
The second one says that, capturing a piece is not included in the problem... Because, if you capture the other, it will not exist...
Because most of us agreed that the second definition is what we are talking about... I guess, the second definition seems more practical...
Well as per your question
I have used the word cancel because if I would write capture then definitely some will come up and will say capturing has has only one ways
Also capturing is the choice whether the queens would like to do capture or not
As per your concern I will edit question for less confusions
I think you can change the cancel to "in a line of sight" just to be clear and please tell me what queen color is being used in this puzzle
Its clear that two queens of same color cannot capture or cancel
It's clear they cannot capture. Whether they can cancel is anybody's guess, it depends what you intended to mean by the term. If you intended to mean just "capture" then two queens of the same colour satisfy your criterion wherever they are placed, so all such diagrams should be included in the count unless you explicitly say the queens are of opposite colour.
If you consider queens of opposite colour, then the answer would depend on what you intended by the phrase "such that they never get cancel by each other". The inclusion of the word "never" seems to imply that the game continues. In that case, assuming "cancel" is synonymous with "capture", there are 0 diagrams where that is the case in all possible continuations (though it could be the case from any diagram if e.g. a draw has been agreed).
Edit: I notice you have have clarified that "cancel" is to be taken as an approximate synonym of "capture" and removed the reference to White having the move. The question is probably about only the diagram and any game continuation is irrelevant, in which case
In how many ways can a queen of each colour be placed on an otherwise empty chessboard in such a way that neither queen occupies a square attacked by the other?
would have been clearer.
Edit2: Just realised that my first two paragraphs don't address the question as currently posed. Any position with only two queens on the board is necessarily dead, because neither side can checkmate. Therefore neither side has any possibility of capture, because a dead position immediately terminates the game. So all such positions satisfy your criterion.
The correct answer to the question as currently posed is therefore 64.63.(1+½+½)=8064.
If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled .
No. You may have to change your id.
As you know chess board have coordinates so placing on that coordinates from any of the sides are totally different positions
MARattigan wrote:
Mr_Mathematician wrote:
CoDCVN wrote:
Mr_Mathematician đã viết:
StockOfHey wrote:
In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place , assume white plays first ).
I found the words "that they never get cancel by each other". It is hard to get... If you did not combine the definition of the second sentence "assume white plays first", it is hard to know the real definition of the word cancel...
There are two definition:
1. decide or announce that (a planned event) will not take place.
2. (of a factor or circumstance) neutralize or negate the force or effect of (another).
The first one says that, capturing a piece is included in the problem... Because, if you cancel the effects of the other piece, you must capture it while the ones who capture remain hence "they did not cancel each other"...
The second one says that, capturing a piece is not included in the problem... Because, if you capture the other, it will not exist...
Because most of us agreed that the second definition is what we are talking about... I guess, the second definition seems more practical...
Well as per your question
I have used the word cancel because if I would write capture then definitely some will come up and will say capturing has has only one ways
Also capturing is the choice whether the queens would like to do capture or not
As per your concern I will edit question for less confusions
I think you can change the cancel to "in a line of sight" just to be clear and please tell me what queen color is being used in this puzzle
Its clear that two queens of same color cannot capture or cancel
It's clear they cannot capture. Whether they can cancel is anybody's guess, it depends what you intended to mean by the term. If you intended to mean just "capture" then two queens of the same colour satisfy your criterion wherever they are placed, so all such diagrams should be included in the count unless you explicitly say the queens are of opposite colour.
If you consider queens of opposite colour, then the answer would depend on what you intended by the phrase "such that they never get cancel by each other". The inclusion of the word "never" seems to imply that the game continues. In that case, assuming "cancel" is synonymous with "capture", there are 0 diagrams where that is the case in all possible continuations (though it could be the case from any diagram if e.g. a draw has been agreed).
Edit: I notice you have have clarified that "cancel" is to be taken as an approximate synonym of "capture" and removed the reference to White having the move. The question is probably about only the diagram and any game continuation is irrelevant, in which case
In how many ways can a queen of each colour be placed on an otherwise empty chessboard in such a way that neither queen occupies a square attacked by the other?
would have been clearer.
If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled .
No. You may have to change your id.
As you know chess board have coordinates so placing on that coordinates from any of the sides are totally different positions
I apologize if this question really created such confusion
But if I really mark your statement
"In how many ways can a queen of each colour be placed on an otherwise empty chessboard in such a way that neither queen occupies a square attacked by the other?"
Then we shall consider all the colors VIBGYOR
Then in this we will consider 7 different colors queen on board which was not in questions
By the way thanks For next time I will make sure to correct my sentence in more efficient way.
MARattigan wrote:
Mr_Mathematician wrote:
CoDCVN wrote:
Mr_Mathematician đã viết:
StockOfHey wrote:
In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place , assume white plays first ).
I found the words "that they never get cancel by each other". It is hard to get... If you did not combine the definition of the second sentence "assume white plays first", it is hard to know the real definition of the word cancel...
There are two definition:
1. decide or announce that (a planned event) will not take place.
2. (of a factor or circumstance) neutralize or negate the force or effect of (another).
The first one says that, capturing a piece is included in the problem... Because, if you cancel the effects of the other piece, you must capture it while the ones who capture remain hence "they did not cancel each other"...
The second one says that, capturing a piece is not included in the problem... Because, if you capture the other, it will not exist...
Because most of us agreed that the second definition is what we are talking about... I guess, the second definition seems more practical...
Well as per your question
I have used the word cancel because if I would write capture then definitely some will come up and will say capturing has has only one ways
Also capturing is the choice whether the queens would like to do capture or not
As per your concern I will edit question for less confusions
I think you can change the cancel to "in a line of sight" just to be clear and please tell me what queen color is being used in this puzzle
Its clear that two queens of same color cannot capture or cancel
It's clear they cannot capture. Whether they can cancel is anybody's guess, it depends what you intended to mean by the term. If you intended to mean just "capture" then two queens of the same colour satisfy your criterion wherever they are placed, so all such diagrams should be included in the count unless you explicitly say the queens are of opposite colour.
If you consider queens of opposite colour, then the answer would depend on what you intended by the phrase "such that they never get cancel by each other". The inclusion of the word "never" seems to imply that the game continues. In that case, assuming "cancel" is synonymous with "capture", there are 0 diagrams where that is the case in all possible continuations (though it could be the case from any diagram if e.g. a draw has been agreed).
Edit: I notice you have have clarified that "cancel" is to be taken as an approximate synonym of "capture" and removed the reference to White having the move. The question is probably about only the diagram and any game continuation is irrelevant, in which case
In how many ways can a queen of each colour be placed on an otherwise empty chessboard in such a way that neither queen occupies a square attacked by the other?
would have been clearer.
If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled .
No. You may have to change your id.
As you know chess board have coordinates so placing on that coordinates from any of the sides are totally different positions
Also the reason for doubling and answer without changing id is we are placing two queens one by one outside, it's not a chess tournament .
MARattigan wrote:
Mr_Mathematician wrote:
CoDCVN wrote:
Mr_Mathematician đã viết:
StockOfHey wrote:
In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place , assume white plays first ).
I found the words "that they never get cancel by each other". It is hard to get... If you did not combine the definition of the second sentence "assume white plays first", it is hard to know the real definition of the word cancel...
There are two definition:
1. decide or announce that (a planned event) will not take place.
2. (of a factor or circumstance) neutralize or negate the force or effect of (another).
The first one says that, capturing a piece is included in the problem... Because, if you cancel the effects of the other piece, you must capture it while the ones who capture remain hence "they did not cancel each other"...
The second one says that, capturing a piece is not included in the problem... Because, if you capture the other, it will not exist...
Because most of us agreed that the second definition is what we are talking about... I guess, the second definition seems more practical...
Well as per your question
I have used the word cancel because if I would write capture then definitely some will come up and will say capturing has has only one ways
Also capturing is the choice whether the queens would like to do capture or not
As per your concern I will edit question for less confusions
I think you can change the cancel to "in a line of sight" just to be clear and please tell me what queen color is being used in this puzzle
Its clear that two queens of same color cannot capture or cancel
It's clear they cannot capture. Whether they can cancel is anybody's guess, it depends what you intended to mean by the term. If you intended to mean just "capture" then two queens of the same colour satisfy your criterion wherever they are placed, so all such diagrams should be included in the count unless you explicitly say the queens are of opposite colour.
If you consider queens of opposite colour, then the answer would depend on what you intended by the phrase "such that they never get cancel by each other". The inclusion of the word "never" seems to imply that the game continues. In that case, assuming "cancel" is synonymous with "capture", there are 0 diagrams where that is the case in all possible continuations (though it could be the case from any diagram if e.g. a draw has been agreed).
Edit: I notice you have have clarified that "cancel" is to be taken as an approximate synonym of "capture" and removed the reference to White having the move. The question is probably about only the diagram and any game continuation is irrelevant, in which case
In how many ways can a queen of each colour be placed on an otherwise empty chessboard in such a way that neither queen occupies a square attacked by the other?
would have been clearer.
If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled .
No. You may have to change your id.
As you know chess board have coordinates so placing on that coordinates from any of the sides are totally different positions
Also pieces got canceled or captured then that case counted you have to count others too
TOTAL cases
Mar 16, 2024
0
#89
@Mr_Mathematician
English is an imprecise language and almost anything that is said depends on context. I think the great majority of readers would take the phrase "a queen of each colour" to mean a black queen and a white queen in this context.
I can't understand "Also the reason for doubling and answer without changing id is we are placing two queens one by one outside, it's not a chess tournament ." in any way, but doubling the figure 2576 in @Optimissed's calculation which has already accounted for the same positions that would be arrived at by starting instead with the black queen is incorrect.
Try it with a black and white knight on a chessboard reduced to two adjacent squares. If you come up with the answer 4 using the same reasoning, you need to keep working on it.
Mar 16, 2024
0
#90
Mr_Mathematician wrote:
MARattigan wrote:
Mr_Mathematician wrote:
CoDCVN wrote:
Mr_Mathematician đã viết:
StockOfHey wrote:
In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place , assume white plays first ).
I found the words "that they never get cancel by each other". It is hard to get... If you did not combine the definition of the second sentence "assume white plays first", it is hard to know the real definition of the word cancel...
There are two definition:
1. decide or announce that (a planned event) will not take place.
2. (of a factor or circumstance) neutralize or negate the force or effect of (another).
The first one says that, capturing a piece is included in the problem... Because, if you cancel the effects of the other piece, you must capture it while the ones who capture remain hence "they did not cancel each other"...
The second one says that, capturing a piece is not included in the problem... Because, if you capture the other, it will not exist...
Because most of us agreed that the second definition is what we are talking about... I guess, the second definition seems more practical...
Well as per your question
I have used the word cancel because if I would write capture then definitely some will come up and will say capturing has has only one ways
Also capturing is the choice whether the queens would like to do capture or not
As per your concern I will edit question for less confusions
I think you can change the cancel to "in a line of sight" just to be clear and please tell me what queen color is being used in this puzzle
Its clear that two queens of same color cannot capture or cancel
It's clear they cannot capture. Whether they can cancel is anybody's guess, it depends what you intended to mean by the term. If you intended to mean just "capture" then two queens of the same colour satisfy your criterion wherever they are placed, so all such diagrams should be included in the count unless you explicitly say the queens are of opposite colour.
If you consider queens of opposite colour, then the answer would depend on what you intended by the phrase "such that they never get cancel by each other". The inclusion of the word "never" seems to imply that the game continues. In that case, assuming "cancel" is synonymous with "capture", there are 0 diagrams where that is the case in all possible continuations (though it could be the case from any diagram if e.g. a draw has been agreed).
Edit: I notice you have have clarified that "cancel" is to be taken as an approximate synonym of "capture" and removed the reference to White having the move. The question is probably about only the diagram and any game continuation is irrelevant, in which case
In how many ways can a queen of each colour be placed on an otherwise empty chessboard in such a way that neither queen occupies a square attacked by the other?
would have been clearer.
If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled .
No. You may have to change your id.
As you know chess board have coordinates so placing on that coordinates from any of the sides are totally different positions
Also pieces got canceled or captured then that case counted you have to count others too
TOTAL cases
Sorry. Total incomprehension.
Perhaps if you post it again in your native language I can run it thro' Google translate.
(By the way I posted an answer for the current wording in an edit (Edit2) here.)
Since two queens have a mutual line of sight, then if one can take so the other can take.
Is this, in effect, the same puzzle as a Queen and a King, and asking "how many safe square positions are there"?
Mar 16, 2024
0
#92
GumboStu wrote:
Since two queens have a mutual line of sight, then if one can take so the other can take.
...
Not quite true. Only one player can have the move.
MARattigan wrote:
@Mr_Mathematician
English is an imprecise language and almost anything that is said depends on context. I think the great majority of readers would take the phrase "a queen of each colour" to mean a black queen and a white queen in this context.
I can't understand "Also the reason for doubling and answer without changing id is we are placing two queens one by one outside, it's not a chess tournament ." in any way, but doubling the figure 2576 in @Optimissed's calculation which has already accounted for the same positions that would be arrived at by starting instead with the black queen is incorrect.
Try it with a black and white knight on a chessboard reduced to two adjacent squares. If you come up with the answer 4 using the same reasoning, you need to keep working on it.
I may but you are seeing the final result as outcome but in questions there are no. Of ways
Let we fix black position first there are ways for white
And we again fix white position for black there are ways for black.
Ok let me clarify you more suppose there are three cities A B C we have to go to the another city D which have 3 ways going via A , B , C
Now a person would be 3 possibilities whether he would like to go with A , B or C cities finally he reaches on city D . So as we are observer, with context to your thoughts he would have must followed only one way since his final position is same No
The two opposite colours are taken to be equally likely even we would consider both of them.
W B
BW
are different, even in making most probable move by computers they also prefer them as separate
Probablity of Occurrence of each event is ½ here
Mar 16, 2024
0
#94
This is often referred to as the classic Eight Empresses problem. It first appeared in a German chess magazine in 1848. A few years later, people found its answer, which was 92.
Mar 16, 2024
0
#95
A mathematician provided an almost certain answer. Harvard mathematician Michael Simkin calculated that there are approximately (0.143n)ⁿ ways to place queens that prevent them from attacking each other on an n x n grid chessboard.
Mar 16, 2024
0
#96
Capriccioninth wrote:
A mathematician provided an almost certain answer. Harvard mathematician Michael Simkin calculated that there are approximately (0.143n)ⁿ ways to place queens that prevent them from attacking each other on an n x n grid chessboard.
Very easy to provide an exact answer in that case for just two queens (no almost). Just use @Optimissed's algorithm (up to the point where he failed to quit while he was ahead). It will work for any n.
Mar 16, 2024
0
#97
Mr_Mathematician wrote:
MARattigan wrote:
@Mr_Mathematician
English is an imprecise language and almost anything that is said depends on context. I think the great majority of readers would take the phrase "a queen of each colour" to mean a black queen and a white queen in this context.
I can't understand "Also the reason for doubling and answer without changing id is we are placing two queens one by one outside, it's not a chess tournament ." in any way, but doubling the figure 2576 in @Optimissed's calculation which has already accounted for the same positions that would be arrived at by starting instead with the black queen is incorrect.
Try it with a black and white knight on a chessboard reduced to two adjacent squares. If you come up with the answer 4 using the same reasoning, you need to keep working on it.
I may but you are seeing the final result as outcome but in questions there are no. Of ways
Let we fix black position first there are ways for white
And we again fix white position for black there are ways for black.
Ok let me clarify you more suppose there are three cities A B C we have to go to the another city D which have 3 ways going via A , B , C
Now a person would be 3 possibilities whether he would like to go with A , B or C cities finally he reaches on city D . So as we are observer, with context to your thoughts he would have must followed only one way since his final position is same No
The two opposite colours are taken to be equally likely even we would consider both of them.
W B
BW
are different, even in making most probable move by computers they also prefer them as separate
Probablity of Occurrence of each event is ½ here
This comes back to the point I made about context.
If the question, "how many ways are there of choosing 2 things from 6", or similar questions are asked then people expect the answer to be based on the final outcomes.
Of course you can say
I can take two with my left hand, that's 1 way,
I can take the first one with my left hand and the second with my right, that's 2 ways.
I can sniff the first one up against my left nostril then shake the rest up until one jumps out, that's 3 ways.
I can let the dog run through them and choose two that have moved furthest (with my right nostril), that's 4 ways.
I can glue on labels 1,2,...6 and stick pins in an old fashioned telephone directory until I've got two matches ...
And the answer may well come out differently.
But in the context people will generally assume that a count of final outcomes is what is required. If your question intends something else, it should make that explicit.
Mar 16, 2024
0
#98
Mr_Mathematician wrote:
🎯Brain teaser '‼️
- In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place).Cancel means possibilities of capture.
Can you answer this question with logic
If it was rooks, I'd answer it with 'logic' - a rook always takes away 15 squares, so the answer is 64 * (64-15) / 2. The 2 is because the rooks are not distinguished.
For queens it's more a matter of counting. Classify squares based on the number of diagonal moves available. Then add up:
N_squares * (64 - 15 - N_diagonal_moves) / 2
where N_squares is the number of squares with a specific number N_diagonal_moves of diagonal moves. The values of N_diagonal_moves are 7, 9, 11, and 13, depending only on the distance to the edge of the board.
[In truth both are a combination of logic and counting, it's just that the first is narrower, with only one type of square].
Mar 16, 2024
0
#99
Yes, I was adding that as you wrote! Doubtless it has appeared earlier in the discussion as well (I didn't read it).
Mar 16, 2024
0
#100
Elroch wrote:
Mr_Mathematician wrote:
🎯Brain teaser '‼️
- In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place).Cancel means possibilities of capture.
Can you answer this question with logic
If it was rooks, I'd answer it with 'logic' - a rook always takes away 15 squares, so the answer is 64 * (64-15) / 2. The 2 is because the rooks are not distinguished.
For queens it's more a matter of counting. Classify squares based on the number of diagonal moves available. Then add up:
N_squares * (64 - 15 - N_diagonal_moves) / 2
where N_squares is the number of squares with a specific number of diagonal moves.
[In truth both are a combination of logic and counting, it's just that the first is narrower, with only one type of square].
To quibble, a rook takes away 15 squares if you assume the rooks cannot occupy the same square. That would be illegal, of course, but then it's illegal anyway.
In how many ways two queens shall be placed into 8×8 chessboard such that they never get cancel by each other ( there are only two queens on board place , assume white plays first ).
I found the words "that they never get cancel by each other". It is hard to get... If you did not combine the definition of the second sentence "assume white plays first", it is hard to know the real definition of the word cancel...
There are two definition:
1. decide or announce that (a planned event) will not take place.
2. (of a factor or circumstance) neutralize or negate the force or effect of (another).
The first one says that, capturing a piece is included in the problem... Because, if you cancel the effects of the other piece, you must capture it while the ones who capture remain hence "they did not cancel each other"...
The second one says that, capturing a piece is not included in the problem... Because, if you capture the other, it will not exist...
Because most of us agreed that the second definition is what we are talking about... I guess, the second definition seems more practical...
Well as per your question
I have used the word cancel because if I would write capture then definitely some will come up and will say capturing has has only one ways
Also capturing is the choice whether the queens would like to do capture or not
As per your concern I will edit question for less confusions
I think you can change the cancel to "in a line of sight" just to be clear and please tell me what queen color is being used in this puzzle
Its clear that two queens of same color cannot capture or cancel
If the game has 2 queens you have to count for the both first and second that's the reason answer get doubled .
As you know chess board have coordinates so placing on that coordinates from any of the sides are totally different positions