Percolation related counting problem
$begingroup$
I was trying to look into the following problem, which I intend to use for a lemma for a bigger problem.
The question is: For the 2-dimensional integer lattice, what are some good lower and upper bounds for the number of connected components of size $k$ that contain the origin? (We say that two lattice points are connected if they are the two endpoints of an edge)
I tried to tackle the 1-dimensional case, but for that, the answer is much easier since you can actually compute the total number of such connected components.
I also tried to write a program that computes it and it seems like the boundaries should be some exponentials, but I didn't manage to find a proof for this.
Thanks
combinatorics percolation polyomino
edited Dec 20 '18 at 4:04
Alex Ravsky
42.7k32383
asked Dec 18 '18 at 1:08
user548645user548645
483
$endgroup$
add a comment |
$begingroup$
I was trying to look into the following problem, which I intend to use for a lemma for a bigger problem.
The question is: For the 2-dimensional integer lattice, what are some good lower and upper bounds for the number of connected components of size $k$ that contain the origin? (We say that two lattice points are connected if they are the two endpoints of an edge)
I tried to tackle the 1-dimensional case, but for that, the answer is much easier since you can actually compute the total number of such connected components.
I also tried to write a program that computes it and it seems like the boundaries should be some exponentials, but I didn't manage to find a proof for this.
Thanks
combinatorics percolation polyomino
edited Dec 20 '18 at 4:04
Alex Ravsky
42.7k32383
asked Dec 18 '18 at 1:08
user548645user548645
483
$endgroup$
add a comment |
$begingroup$
I was trying to look into the following problem, which I intend to use for a lemma for a bigger problem.
The question is: For the 2-dimensional integer lattice, what are some good lower and upper bounds for the number of connected components of size $k$ that contain the origin? (We say that two lattice points are connected if they are the two endpoints of an edge)
I tried to tackle the 1-dimensional case, but for that, the answer is much easier since you can actually compute the total number of such connected components.
I also tried to write a program that computes it and it seems like the boundaries should be some exponentials, but I didn't manage to find a proof for this.
Thanks
combinatorics percolation polyomino
edited Dec 20 '18 at 4:04
Alex Ravsky
42.7k32383
asked Dec 18 '18 at 1:08
user548645user548645
483
$endgroup$
I was trying to look into the following problem, which I intend to use for a lemma for a bigger problem.
The question is: For the 2-dimensional integer lattice, what are some good lower and upper bounds for the number of connected components of size $k$ that contain the origin? (We say that two lattice points are connected if they are the two endpoints of an edge)
I tried to tackle the 1-dimensional case, but for that, the answer is much easier since you can actually compute the total number of such connected components.
I also tried to write a program that computes it and it seems like the boundaries should be some exponentials, but I didn't manage to find a proof for this.
Thanks
combinatorics percolation polyomino
combinatorics percolation polyomino
edited Dec 20 '18 at 4:04
Alex Ravsky
42.7k32383
asked Dec 18 '18 at 1:08
user548645user548645
483
edited Dec 20 '18 at 4:04
Alex Ravsky
42.7k32383
asked Dec 18 '18 at 1:08
user548645user548645
483
edited Dec 20 '18 at 4:04
Alex Ravsky
42.7k32383
edited Dec 20 '18 at 4:04
Alex Ravsky
42.7k32383
edited Dec 20 '18 at 4:04
Alex Ravsky
42.7k32383
42.7k32383
asked Dec 18 '18 at 1:08
user548645user548645
483
asked Dec 18 '18 at 1:08
user548645user548645
483
asked Dec 18 '18 at 1:08
user548645user548645
483
483
add a comment |
add a comment |
1 Answer
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$begingroup$
The number of connected components of size $k$ that contain the origin is exactly the number $P_f(k)$ of fixed $k$-minoes multiplied by $k$. Clearly, $P(k)le P_f(k)le 8P(k)$, where $P(k)$ is number of $k$-polyominoes. The best currently known bounds on $P(k)$ are $3.72^k< P(k) <4.65^k$.
answered Dec 20 '18 at 4:02
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
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active
oldest
votes
$begingroup$
The number of connected components of size $k$ that contain the origin is exactly the number $P_f(k)$ of fixed $k$-minoes multiplied by $k$. Clearly, $P(k)le P_f(k)le 8P(k)$, where $P(k)$ is number of $k$-polyominoes. The best currently known bounds on $P(k)$ are $3.72^k< P(k) <4.65^k$.
answered Dec 20 '18 at 4:02
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
The number of connected components of size $k$ that contain the origin is exactly the number $P_f(k)$ of fixed $k$-minoes multiplied by $k$. Clearly, $P(k)le P_f(k)le 8P(k)$, where $P(k)$ is number of $k$-polyominoes. The best currently known bounds on $P(k)$ are $3.72^k< P(k) <4.65^k$.
answered Dec 20 '18 at 4:02
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
add a comment |
$begingroup$
The number of connected components of size $k$ that contain the origin is exactly the number $P_f(k)$ of fixed $k$-minoes multiplied by $k$. Clearly, $P(k)le P_f(k)le 8P(k)$, where $P(k)$ is number of $k$-polyominoes. The best currently known bounds on $P(k)$ are $3.72^k< P(k) <4.65^k$.
answered Dec 20 '18 at 4:02
Alex RavskyAlex Ravsky
42.7k32383
$endgroup$
The number of connected components of size $k$ that contain the origin is exactly the number $P_f(k)$ of fixed $k$-minoes multiplied by $k$. Clearly, $P(k)le P_f(k)le 8P(k)$, where $P(k)$ is number of $k$-polyominoes. The best currently known bounds on $P(k)$ are $3.72^k< P(k) <4.65^k$.
answered Dec 20 '18 at 4:02
Alex RavskyAlex Ravsky
42.7k32383
answered Dec 20 '18 at 4:02
Alex RavskyAlex Ravsky
42.7k32383
answered Dec 20 '18 at 4:02
Alex RavskyAlex Ravsky
42.7k32383
answered Dec 20 '18 at 4:02
Alex RavskyAlex Ravsky
42.7k32383
42.7k32383
add a comment |
add a comment |
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