Chess with Maths - Chess Forums - Page 6

MARattigan

Mar 16, 2024

0

#101

@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.

MARattigan

Mar 16, 2024

0

#102

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.)

GumboStu

Mar 16, 2024

0

#103

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"?

MARattigan

Mar 16, 2024

0

#104

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.

V-A-C

Mar 16, 2024

0

#105

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

Milch_Tee

Mar 16, 2024

0

#106

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.

Milch_Tee

Mar 16, 2024

0

#107

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.

MARattigan

Mar 16, 2024

0

#108

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.

MARattigan

Mar 16, 2024

0

#109

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.

Elroch

Mar 16, 2024

0

#110

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].

Optimissed

Mar 16, 2024

0

#111

<<where N_squares is the number of squares with a specific number of diagonal moves>>
Which depends on how close to the centre of the board a square is. Hence the concentric fields method identifies them.

Elroch

Mar 16, 2024

0

#112

Yes, I was adding that as you wrote! Doubtless it has appeared earlier in the discussion as well (I didn't read it).

MARattigan

Mar 16, 2024

0

#113

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.

V-A-C

Mar 16, 2024

0

#114

MARattigan wrote:

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.

That is another case choosing 2 balls together , choosing 2 balls one by one answer will come up different

I agree that in these cases final outcomes but in chess board placing black after white or placing white after black is equally probable but you have to count both case

Ok let's go to probality suppose my question was find the probability of the same problem if white plays first answer would be 1/2 here in numerator 1 is the outcomes of only white but in the denominator there is the case of white + black both

suppose now according you only we have to count one type of case either of black or white so your answer would be 1 ( probality) but this contradicts that fact that if I play with black then also we get same probality which is not possible ( an event cannot have probality greater than one here we are getting two by adding)

MARattigan

Mar 16, 2024

0

#115

Mr_Mathematician wrote:

MARattigan wrote:

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.

That is another case choosing 2 balls together , choosing 2 balls one by one answer will come up different

But not if you're counting final outcomes, which would be the normal interpretation.

I agree that in these cases final outcomes but in chess board placing black after white or placing white after black is equally probable but you have to count both case

Similarly. If you're counting final outcomes it doesn't matter which way round you place the pieces. Do you understand where the flaw was in @Optimissed's post? If not try the problem with two knights on a 2x1 board and you should see.

Ok let's go to probality suppose my question was find the probability of the same problem if white plays first answer would be 1/2 here in numerator 1 is the outcomes of only white but in the denominator there is the case of white + black both

How you define the probability of a problem? And you removed the White to move specification - is it meant to be a purely combinitorial problem or are you envisaging playing on? (As I mentioned previously, not actually legal because the positions are dead.)

suppose now according you only we have to count one type of case either of black or white so your answer would be 1 ( probality) but this contradicts that fact that if I play with black then also we get same probality which is not possible ( an event cannot have probality greater than one here we are getting two by adding)

You appear to be confusing the order in which the pieces are added to the board to make the position with side to move or some order of moves played from the position. You will have to clarify what you're saying if it is to make sense to me. If you want to include side to move as an attribute of what is to be counted then naturally the number of cases doubles, but if so I think you need to make that explicit in your question. Similarly if you wanted to include the ply count under the 75 move rule you need to multiply by 150, but again you would need to make that explicit.

V-A-C

Mar 16, 2024

0

#116