Number of possible bishop moves on an $n times m$ chessboard
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
edited Dec 28 '18 at 8:52
Andreas Caranti
56.1k34295
asked Dec 27 '18 at 20:46
user514787
695210
add a comment |
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
edited Dec 28 '18 at 8:52
Andreas Caranti
56.1k34295
asked Dec 27 '18 at 20:46
user514787
695210
1
Expand and calculate?
– greedoid
Dec 27 '18 at 21:00
@greedoid, I mean something special, for example using floor function or nice looking sum.
– user514787
Dec 27 '18 at 22:23
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.
– bof
Dec 28 '18 at 7:03
add a comment |
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
edited Dec 28 '18 at 8:52
Andreas Caranti
56.1k34295
asked Dec 27 '18 at 20:46
user514787
695210
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
edited Dec 28 '18 at 8:52
Andreas Caranti
56.1k34295
asked Dec 27 '18 at 20:46
user514787
695210
edited Dec 28 '18 at 8:52
Andreas Caranti
56.1k34295
asked Dec 27 '18 at 20:46
user514787
695210
edited Dec 28 '18 at 8:52
Andreas Caranti
56.1k34295
edited Dec 28 '18 at 8:52
Andreas Caranti
56.1k34295
edited Dec 28 '18 at 8:52
Andreas Caranti
56.1k34295
56.1k34295
asked Dec 27 '18 at 20:46
user514787
695210
asked Dec 27 '18 at 20:46
user514787
695210
asked Dec 27 '18 at 20:46
user514787
695210
695210
1
Expand and calculate?
– greedoid
Dec 27 '18 at 21:00
@greedoid, I mean something special, for example using floor function or nice looking sum.
– user514787
Dec 27 '18 at 22:23
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.
– bof
Dec 28 '18 at 7:03
add a comment |
1
Expand and calculate?
– greedoid
Dec 27 '18 at 21:00
@greedoid, I mean something special, for example using floor function or nice looking sum.
– user514787
Dec 27 '18 at 22:23
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.
– bof
Dec 28 '18 at 7:03
1
1
Expand and calculate?
– greedoid
Dec 27 '18 at 21:00
Expand and calculate?
– greedoid
Dec 27 '18 at 21:00
@greedoid, I mean something special, for example using floor function or nice looking sum.
– user514787
Dec 27 '18 at 22:23
@greedoid, I mean something special, for example using floor function or nice looking sum.
– user514787
Dec 27 '18 at 22:23
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.
– bof
Dec 28 '18 at 7:03
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.
– bof
Dec 28 '18 at 7:03
add a comment |
2 Answers
2
active
oldest
votes
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?.
answered Dec 28 '18 at 7:23
bof
50.5k457119
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).
– user514787
Jan 1 at 0:40
1
You're welcome!
– bof
Jan 1 at 0:55
add a comment |
As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$
answered Dec 28 '18 at 3:53
Claude Leibovici
119k1157132
Thank you for answer! But this holds only for $n+2>m$.
– user514787
Dec 28 '18 at 11:11
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
votes
active
oldest
votes
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?.
answered Dec 28 '18 at 7:23
bof
50.5k457119
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).
– user514787
Jan 1 at 0:40
1
You're welcome!
– bof
Jan 1 at 0:55
add a comment |
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?.
answered Dec 28 '18 at 7:23
bof
50.5k457119
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).
– user514787
Jan 1 at 0:40
1
You're welcome!
– bof
Jan 1 at 0:55
add a comment |
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?.
answered Dec 28 '18 at 7:23
bof
50.5k457119
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?.
answered Dec 28 '18 at 7:23
bof
50.5k457119
edited Dec 31 '18 at 14:06
answered Dec 28 '18 at 7:23
bof
50.5k457119
answered Dec 28 '18 at 7:23
bof
50.5k457119
answered Dec 28 '18 at 7:23
bof
50.5k457119
50.5k457119
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).
– user514787
Jan 1 at 0:40
1
You're welcome!
– bof
Jan 1 at 0:55
add a comment |
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).
– user514787
Jan 1 at 0:40
1
You're welcome!
– bof
Jan 1 at 0:55
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).
– user514787
Jan 1 at 0:40
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).
– user514787
Jan 1 at 0:40
1
1
You're welcome!
– bof
Jan 1 at 0:55
You're welcome!
– bof
Jan 1 at 0:55
add a comment |
As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$
answered Dec 28 '18 at 3:53
Claude Leibovici
119k1157132
Thank you for answer! But this holds only for $n+2>m$.
– user514787
Dec 28 '18 at 11:11
add a comment |
As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$
answered Dec 28 '18 at 3:53
Claude Leibovici
119k1157132
Thank you for answer! But this holds only for $n+2>m$.
– user514787
Dec 28 '18 at 11:11
add a comment |
As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$
answered Dec 28 '18 at 3:53
Claude Leibovici
119k1157132
As greedoid commented, it is so simple to expand and simplify.
$$B(n,m)=4left(mbinom{n}{2}-binom{n+1}{3}+binom{n-m+1}{3}right)=frac{2}{3} m(m-1) (3 n-m-1)$$
answered Dec 28 '18 at 3:53
Claude Leibovici
119k1157132
answered Dec 28 '18 at 3:53
Claude Leibovici
119k1157132
answered Dec 28 '18 at 3:53
Claude Leibovici
119k1157132
answered Dec 28 '18 at 3:53
Claude Leibovici
119k1157132
119k1157132
Thank you for answer! But this holds only for $n+2>m$.
– user514787
Dec 28 '18 at 11:11
add a comment |
Thank you for answer! But this holds only for $n+2>m$.
– user514787
Dec 28 '18 at 11:11
Thank you for answer! But this holds only for $n+2>m$.
– user514787
Dec 28 '18 at 11:11
Thank you for answer! But this holds only for $n+2>m$.
– user514787
Dec 28 '18 at 11:11
add a comment |
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Expand and calculate?
– greedoid
Dec 27 '18 at 21:00
@greedoid, I mean something special, for example using floor function or nice looking sum.
– user514787
Dec 27 '18 at 22:23
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.
– bof
Dec 28 '18 at 7:03