Number of possible bishop moves on an $n times m$ chessboard












1












$begingroup$


For rook we have obviously
$$R(n,m)=nm(n+m-2)$$
and for bishop
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)$$
if we assume $binom{n}{k}=0$ for $n<0$.



Is there a way to write it in more simple and nice form?






binomial-coefficients closed-form chessboard







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Expand and calculate?
    $endgroup$
    – Maria Mazur
    Dec 27 '18 at 21:00












  • $begingroup$
    @greedoid, I mean something special, for example using floor function or nice looking sum.
    $endgroup$
    – user514787
    Dec 27 '18 at 22:23










  • $begingroup$
    Maybe not an exact duplicate, but this question asked (among other things) for the number of ways you can put two bishops on a chessboard so that they don't attack each other, which is trivially equivalent to asking the number of ways you can put two bishops on a chessboard so that they do attack each other, which is half the number of possible bishop moves.
    $endgroup$
    – bof
    Dec 28 '18 at 7:03


















1












$begingroup$


For rook we have obviously
$$R(n,m)=nm(n+m-2)$$
and for bishop
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)$$
if we assume $binom{n}{k}=0$ for $n<0$.



Is there a way to write it in more simple and nice form?






binomial-coefficients closed-form chessboard







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Expand and calculate?
    $endgroup$
    – Maria Mazur
    Dec 27 '18 at 21:00












  • $begingroup$
    @greedoid, I mean something special, for example using floor function or nice looking sum.
    $endgroup$
    – user514787
    Dec 27 '18 at 22:23










  • $begingroup$
    Maybe not an exact duplicate, but this question asked (among other things) for the number of ways you can put two bishops on a chessboard so that they don't attack each other, which is trivially equivalent to asking the number of ways you can put two bishops on a chessboard so that they do attack each other, which is half the number of possible bishop moves.
    $endgroup$
    – bof
    Dec 28 '18 at 7:03
















1












1








1





$begingroup$


For rook we have obviously
$$R(n,m)=nm(n+m-2)$$
and for bishop
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)$$
if we assume $binom{n}{k}=0$ for $n<0$.



Is there a way to write it in more simple and nice form?






binomial-coefficients closed-form chessboard







share|cite|improve this question











$endgroup$




For rook we have obviously
$$R(n,m)=nm(n+m-2)$$
and for bishop
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)$$
if we assume $binom{n}{k}=0$ for $n<0$.



Is there a way to write it in more simple and nice form?






binomial-coefficients closed-form chessboard




binomial-coefficients closed-form chessboard






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 8:52









Andreas Caranti

57.4k34498




57.4k34498










asked Dec 27 '18 at 20:46









user514787user514787

749410




749410








  • 1




    $begingroup$
    Expand and calculate?
    $endgroup$
    – Maria Mazur
    Dec 27 '18 at 21:00












  • $begingroup$
    @greedoid, I mean something special, for example using floor function or nice looking sum.
    $endgroup$
    – user514787
    Dec 27 '18 at 22:23










  • $begingroup$
    Maybe not an exact duplicate, but this question asked (among other things) for the number of ways you can put two bishops on a chessboard so that they don't attack each other, which is trivially equivalent to asking the number of ways you can put two bishops on a chessboard so that they do attack each other, which is half the number of possible bishop moves.
    $endgroup$
    – bof
    Dec 28 '18 at 7:03
















  • 1




    $begingroup$
    Expand and calculate?
    $endgroup$
    – Maria Mazur
    Dec 27 '18 at 21:00












  • $begingroup$
    @greedoid, I mean something special, for example using floor function or nice looking sum.
    $endgroup$
    – user514787
    Dec 27 '18 at 22:23










  • $begingroup$
    Maybe not an exact duplicate, but this question asked (among other things) for the number of ways you can put two bishops on a chessboard so that they don't attack each other, which is trivially equivalent to asking the number of ways you can put two bishops on a chessboard so that they do attack each other, which is half the number of possible bishop moves.
    $endgroup$
    – bof
    Dec 28 '18 at 7:03










1




1




$begingroup$
Expand and calculate?
$endgroup$
– Maria Mazur
Dec 27 '18 at 21:00






$begingroup$
Expand and calculate?
$endgroup$
– Maria Mazur
Dec 27 '18 at 21:00














$begingroup$
@greedoid, I mean something special, for example using floor function or nice looking sum.
$endgroup$
– user514787
Dec 27 '18 at 22:23




$begingroup$
@greedoid, I mean something special, for example using floor function or nice looking sum.
$endgroup$
– user514787
Dec 27 '18 at 22:23












$begingroup$
Maybe not an exact duplicate, but this question asked (among other things) for the number of ways you can put two bishops on a chessboard so that they don't attack each other, which is trivially equivalent to asking the number of ways you can put two bishops on a chessboard so that they do attack each other, which is half the number of possible bishop moves.
$endgroup$
– bof
Dec 28 '18 at 7:03






$begingroup$
Maybe not an exact duplicate, but this question asked (among other things) for the number of ways you can put two bishops on a chessboard so that they don't attack each other, which is trivially equivalent to asking the number of ways you can put two bishops on a chessboard so that they do attack each other, which is half the number of possible bishop moves.
$endgroup$
– bof
Dec 28 '18 at 7:03












2 Answers
2






active

oldest

votes


















1












$begingroup$

The number of possible bishop moves on an $mtimes n$ chessboard is equal to $4$ times the number of $ktimes k$ squares, $2le klemin(m,n)$, on that chessboard, since each bishop move is a move from one corner to the opposite corner of one of those squares.



From this we get the formula
$$B(m,n)=4sum_{k=2}^{min(m,n)}(m+1-k)(n+1-k)$$$$=8binom{min(m,n)}3+4(|m-n|+1)binom{min(m,n)}2$$
which you may or may not consider "simple and nice". For square chessboards it becomes simpler and nicer:
$$B(n,n)=4sum_{k=2}^n(n+1-k)^2=4left[1^2+2^2+3^2+cdots+(n-1)^2right]$$$$=8binom n3+4binom n2=frac23(n-1)n(2n-1).$$



See also the answer to the question How many ways can you put: a) two bishops b) two knights c) two queens on a chessboard in such a way that one piece does not attack the other?.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for answer and for edition too! English not my native, sorry. Formulas in your answer which you linked need to be more popular. It also can be generalized to any figure for which some of 4 directions (|a|,|b|) are allowed. Another criteria - max length of move (in selected direction). Thinking about it show how primitive basic chess rules (but it still hard to play).
    $endgroup$
    – user514787
    Jan 1 at 0:40






  • 1




    $begingroup$
    You're welcome!
    $endgroup$
    – bof
    Jan 1 at 0:55





















3












$begingroup$

As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer! But this holds only for $n+2>m$.
    $endgroup$
    – user514787
    Dec 28 '18 at 11:11












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The number of possible bishop moves on an $mtimes n$ chessboard is equal to $4$ times the number of $ktimes k$ squares, $2le klemin(m,n)$, on that chessboard, since each bishop move is a move from one corner to the opposite corner of one of those squares.



From this we get the formula
$$B(m,n)=4sum_{k=2}^{min(m,n)}(m+1-k)(n+1-k)$$$$=8binom{min(m,n)}3+4(|m-n|+1)binom{min(m,n)}2$$
which you may or may not consider "simple and nice". For square chessboards it becomes simpler and nicer:
$$B(n,n)=4sum_{k=2}^n(n+1-k)^2=4left[1^2+2^2+3^2+cdots+(n-1)^2right]$$$$=8binom n3+4binom n2=frac23(n-1)n(2n-1).$$



See also the answer to the question How many ways can you put: a) two bishops b) two knights c) two queens on a chessboard in such a way that one piece does not attack the other?.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for answer and for edition too! English not my native, sorry. Formulas in your answer which you linked need to be more popular. It also can be generalized to any figure for which some of 4 directions (|a|,|b|) are allowed. Another criteria - max length of move (in selected direction). Thinking about it show how primitive basic chess rules (but it still hard to play).
    $endgroup$
    – user514787
    Jan 1 at 0:40






  • 1




    $begingroup$
    You're welcome!
    $endgroup$
    – bof
    Jan 1 at 0:55


















1












$begingroup$

The number of possible bishop moves on an $mtimes n$ chessboard is equal to $4$ times the number of $ktimes k$ squares, $2le klemin(m,n)$, on that chessboard, since each bishop move is a move from one corner to the opposite corner of one of those squares.



From this we get the formula
$$B(m,n)=4sum_{k=2}^{min(m,n)}(m+1-k)(n+1-k)$$$$=8binom{min(m,n)}3+4(|m-n|+1)binom{min(m,n)}2$$
which you may or may not consider "simple and nice". For square chessboards it becomes simpler and nicer:
$$B(n,n)=4sum_{k=2}^n(n+1-k)^2=4left[1^2+2^2+3^2+cdots+(n-1)^2right]$$$$=8binom n3+4binom n2=frac23(n-1)n(2n-1).$$



See also the answer to the question How many ways can you put: a) two bishops b) two knights c) two queens on a chessboard in such a way that one piece does not attack the other?.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for answer and for edition too! English not my native, sorry. Formulas in your answer which you linked need to be more popular. It also can be generalized to any figure for which some of 4 directions (|a|,|b|) are allowed. Another criteria - max length of move (in selected direction). Thinking about it show how primitive basic chess rules (but it still hard to play).
    $endgroup$
    – user514787
    Jan 1 at 0:40






  • 1




    $begingroup$
    You're welcome!
    $endgroup$
    – bof
    Jan 1 at 0:55
















1












1








1





$begingroup$

The number of possible bishop moves on an $mtimes n$ chessboard is equal to $4$ times the number of $ktimes k$ squares, $2le klemin(m,n)$, on that chessboard, since each bishop move is a move from one corner to the opposite corner of one of those squares.



From this we get the formula
$$B(m,n)=4sum_{k=2}^{min(m,n)}(m+1-k)(n+1-k)$$$$=8binom{min(m,n)}3+4(|m-n|+1)binom{min(m,n)}2$$
which you may or may not consider "simple and nice". For square chessboards it becomes simpler and nicer:
$$B(n,n)=4sum_{k=2}^n(n+1-k)^2=4left[1^2+2^2+3^2+cdots+(n-1)^2right]$$$$=8binom n3+4binom n2=frac23(n-1)n(2n-1).$$



See also the answer to the question How many ways can you put: a) two bishops b) two knights c) two queens on a chessboard in such a way that one piece does not attack the other?.






share|cite|improve this answer











$endgroup$



The number of possible bishop moves on an $mtimes n$ chessboard is equal to $4$ times the number of $ktimes k$ squares, $2le klemin(m,n)$, on that chessboard, since each bishop move is a move from one corner to the opposite corner of one of those squares.



From this we get the formula
$$B(m,n)=4sum_{k=2}^{min(m,n)}(m+1-k)(n+1-k)$$$$=8binom{min(m,n)}3+4(|m-n|+1)binom{min(m,n)}2$$
which you may or may not consider "simple and nice". For square chessboards it becomes simpler and nicer:
$$B(n,n)=4sum_{k=2}^n(n+1-k)^2=4left[1^2+2^2+3^2+cdots+(n-1)^2right]$$$$=8binom n3+4binom n2=frac23(n-1)n(2n-1).$$



See also the answer to the question How many ways can you put: a) two bishops b) two knights c) two queens on a chessboard in such a way that one piece does not attack the other?.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 31 '18 at 14:06

























answered Dec 28 '18 at 7:23









bofbof

52.6k559121




52.6k559121












  • $begingroup$
    Thank you for answer and for edition too! English not my native, sorry. Formulas in your answer which you linked need to be more popular. It also can be generalized to any figure for which some of 4 directions (|a|,|b|) are allowed. Another criteria - max length of move (in selected direction). Thinking about it show how primitive basic chess rules (but it still hard to play).
    $endgroup$
    – user514787
    Jan 1 at 0:40






  • 1




    $begingroup$
    You're welcome!
    $endgroup$
    – bof
    Jan 1 at 0:55




















  • $begingroup$
    Thank you for answer and for edition too! English not my native, sorry. Formulas in your answer which you linked need to be more popular. It also can be generalized to any figure for which some of 4 directions (|a|,|b|) are allowed. Another criteria - max length of move (in selected direction). Thinking about it show how primitive basic chess rules (but it still hard to play).
    $endgroup$
    – user514787
    Jan 1 at 0:40






  • 1




    $begingroup$
    You're welcome!
    $endgroup$
    – bof
    Jan 1 at 0:55


















$begingroup$
Thank you for answer and for edition too! English not my native, sorry. Formulas in your answer which you linked need to be more popular. It also can be generalized to any figure for which some of 4 directions (|a|,|b|) are allowed. Another criteria - max length of move (in selected direction). Thinking about it show how primitive basic chess rules (but it still hard to play).
$endgroup$
– user514787
Jan 1 at 0:40




$begingroup$
Thank you for answer and for edition too! English not my native, sorry. Formulas in your answer which you linked need to be more popular. It also can be generalized to any figure for which some of 4 directions (|a|,|b|) are allowed. Another criteria - max length of move (in selected direction). Thinking about it show how primitive basic chess rules (but it still hard to play).
$endgroup$
– user514787
Jan 1 at 0:40




1




1




$begingroup$
You're welcome!
$endgroup$
– bof
Jan 1 at 0:55






$begingroup$
You're welcome!
$endgroup$
– bof
Jan 1 at 0:55













3












$begingroup$

As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer! But this holds only for $n+2>m$.
    $endgroup$
    – user514787
    Dec 28 '18 at 11:11
















3












$begingroup$

As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for answer! But this holds only for $n+2>m$.
    $endgroup$
    – user514787
    Dec 28 '18 at 11:11














3












3








3





$begingroup$

As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$






share|cite|improve this answer









$endgroup$



As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 28 '18 at 3:53









Claude LeiboviciClaude Leibovici

126k1158135




126k1158135












  • $begingroup$
    Thank you for answer! But this holds only for $n+2>m$.
    $endgroup$
    – user514787
    Dec 28 '18 at 11:11


















  • $begingroup$
    Thank you for answer! But this holds only for $n+2>m$.
    $endgroup$
    – user514787
    Dec 28 '18 at 11:11
















$begingroup$
Thank you for answer! But this holds only for $n+2>m$.
$endgroup$
– user514787
Dec 28 '18 at 11:11




$begingroup$
Thank you for answer! But this holds only for $n+2>m$.
$endgroup$
– user514787
Dec 28 '18 at 11:11


















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