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There are many interesting questions here, and one can probably answer some asymptotically, e.g. the tail behavior of the size of the connected component assuming it is finite (i.e., probability that the diameter is larger than $L$ for $L$ large, conditioned on being finite), or the question of percolation i.e is there a positive probability that the cluster of the origin is infinite (the cluster contains by coupling the percolation cluster of $p=1/2$ bond percolation on the 2D lattice, which is finite a.s., so I do not see a simple domination argument).

However the specific question you asked is indeed I believe easier. Here is an attempt.

Edit: Note that the question "is a vertex the center of a component of size 1" has a simple answer - it is $p_1=(1/2)^4$ (since it is determined by the neighboring squares, i.e. by 4 independent diagonals). Now apply the ergodic theorem to conclude that the asymptotic fraction of such squares is $p_1$. The reason this does not answer your question is that you asked for the asymptotic number of components, but it does give a lower bound on the fraction you asked about.

To get the answer to your question, repeat the computation for all 2x1 components, 3x1, 3x2, etc that contain the origin. That is, for a shape $Q$, compute $p_Q=P(C(0)\sim Q)$ where $C(0)$ is the shape containing $0$ and $\sim$ is up to translation of the shape. Then the events in the definition of $p_Q$ are disjoint, and $\sum p_Q\leq 1$. Now the fraction you ask about is $p_1/\sum_Q(p_Q/|Q|)$, where $|Q|$ is the number of vertices contained in the interior of $Q$.

(Former version, which looked at a somewhat different notion of component, had: Note that the question "is the square A having (0,0) and (1,1) as vertices a component" has a simple answer - it is $p_1=(1/2)^{6}$ (since it is determined by the neighboring squares, and for a given orientation of the diagonal in $A$, only 6 of these neighbors are relevant).

There are many interesting questions here, and one can probably answer some asymptotically, e.g. the tail behavior of the size of the connected component assuming it is finite (i.e., probability that the diameter is larger than $L$ for $L$ large, conditioned on being finite), or the question of percolation i.e is there a positive probability that the cluster of the origin is infinite (the cluster contains by coupling the percolation cluster of $p=1/2$ bond percolation on the 2D lattice, which is finite a.s., so I do not see a simple domination argument).

However the specific question you asked is indeed I believe easier. Here is an attempt.

Edit: Note that the question "is a vertex the center of a component of size 1" has a simple answer - it is $p_1=(1/2)^4$ (since it is determined by the neighboring squares, i.e. by 4 independent diagonals). Now apply the ergodic theorem to conclude that the asymptotic fraction of such squares is $p_1$. The reason this does not answer your question is that you asked for the asymptotic number of components, but it does give a lower bound on the fraction you asked about.

To get the answer to your question, repeat the computation for all 2x1 components, 3x1, 3x2, etc that contain the origin. That is, for a shape $Q$, compute $p_Q=P(C(0)\sim Q)$ where $C(0)$ is the shape containing $0$ and $\sim$ is up to translation of the shape. Then the events in the definition of $p_Q$ are disjoint, and $\sum p_Q\leq 1$. Now the fraction you ask about is $p_1/\sum_Q(p_Q/|Q|)$, where $|Q|$ is the number of vertices contained in the interior of $Q$.

(Former version, which looked at a somewhat different notion of component and answered a different notion of component, had: Note that the question "is the square A having (0,0) and (1,1) as vertices a component" has a simple answer - it is $p_1=(1/2)^{6}$ (since it is determined by the neighboring squares, and for a given orientation of the diagonal in $A$, only 6 of these neighbors are relevant).

There are many interesting questions here, and one can probably answer some asymptotically, e.g. the tail behavior of the size of the connected component assuming it is finite (i.e., probability that the diameter is larger than $L$ for $L$ large, conditioned on being finite), or the question of percolation i.e is there a positive probability that the cluster of the origin is infinite (the cluster contains by coupling the percolation cluster of $p=1/2$ bond percolation on the 2D lattice, which is finite a.s., so I do not see a simple domination argument).

However the specific question you asked is indeed I believe easier. Here is an attempt.

Note that the question "is the square A having (0,0) and (1,1) as vertices a component" has a simple answer - it is $p_1=(1/2)^{6}$ (since it is determined by the neighboring squares, and for a given orientation of the diagonal in $A$, only 6 of these neighbors are relevant). Now apply the ergodic theorem to conclude that the asymptotic fraction of such squares is $p_1$. The reason this does not answer your question is that you asked for the asymptotic number of components, but it does give a lower bound on the fraction you asked about.

To get the answer to your question, repeat the computation for all 2x1 components, 3x1, 3x2, etc that contain the square A. That is, for a shape $Q$, compute $p_Q=P(C(A)\sim Q)$ where $C(A)$ is the shape containing $A$ and $\sim$ is up to translation of the shape. Then the events in the definition of $p_Q$ are disjoint, and $\sum p_Q\leq 1$. Now the fraction you ask about is $p_1/\sum_Q(p_Q/|Q|)$, where $|Q|$ is the area of $Q$.

There are many interesting questions here, and one can probably answer some asymptotically, e.g. the tail behavior of the size of the connected component assuming it is finite (i.e., probability that the diameter is larger than $L$ for $L$ large, conditioned on being finite), or the question of percolation i.e is there a positive probability that the cluster of the origin is infinite (the cluster contains by coupling the percolation cluster of $p=1/2$ bond percolation on the 2D lattice, which is finite a.s., so I do not see a simple domination argument).

However the specific question you asked is indeed I believe easier. Here is an attempt.

Edit: Note that the question "is a vertex the center of a component of size 1" has a simple answer - it is $p_1=(1/2)^4$ (since it is determined by the neighboring squares, i.e. by 4 independent diagonals). Now apply the ergodic theorem to conclude that the asymptotic fraction of such squares is $p_1$. The reason this does not answer your question is that you asked for the asymptotic number of components, but it does give a lower bound on the fraction you asked about.

To get the answer to your question, repeat the computation for all 2x1 components, 3x1, 3x2, etc that contain the origin. That is, for a shape $Q$, compute $p_Q=P(C(0)\sim Q)$ where $C(0)$ is the shape containing $0$ and $\sim$ is up to translation of the shape. Then the events in the definition of $p_Q$ are disjoint, and $\sum p_Q\leq 1$. Now the fraction you ask about is $p_1/\sum_Q(p_Q/|Q|)$, where $|Q|$ is the number of vertices contained in the interior of $Q$.

(Former version, which looked at a somewhat different notion of component, had: Note that the question "is the square A having (0,0) and (1,1) as vertices a component" has a simple answer - it is $p_1=(1/2)^{6}$ (since it is determined by the neighboring squares, and for a given orientation of the diagonal in $A$, only 6 of these neighbors are relevant).

There are many interesting questions here, and one can probably answer some asymptotically, e.g. the tail behavior of the size of the connected component assuming it is finite (i.e., probability that the diameter is larger than $L$ for $L$ large, conditioned on being finite), or the question of percolation i.e is there a positive probability that the cluster of the origin is infinite (the cluster contains by coupling the percolation cluster of $p=1/2$ bond percolation on the 2D lattice, which is finite a.s., so I do not see a simple domination argument).

However the specific question you asked is indeed I believe easier. Here is an attempt.

Note that the question "is the square A having (0,0) and (1,1) as vertices a component" has a simple answer - it is $p_1=(1/2)^{6}$ (since it is determined by the neighboring squares, and for a given orientation of the diagonal in $A$, only 6 of these neighbors are relevant). Now apply the ergodic theorem to conclude that the asymptotic fraction of such squares is $p_1. The reason this does not answer your question is that you asked for the asymptotic number of components, but it does give a lower bound on the fraction you asked about.

To get the answer to your question, repeat the computation for all 2x1 components, 3x1, 3x2, etc that contain the square A. That is, for a shape $Q$, compute $p_Q=P(C(A)\sim Q)$ where $C(A)$ is the shape containing $A$ and $\sim$ is up to translation of the shape. Then the events in the definition of $p_Q$ are disjoint, and $\sum p_Q\leq 1$. Now the fraction you ask about is $p_1/\sum_Q(p_Q/|Q|)$, where $|Q|$ is the area of $Q$.

There are many interesting questions here, and one can probably answer some asymptotically, e.g. the tail behavior of the size of the connected component assuming it is finite (i.e., probability that the diameter is larger than $L$ for $L$ large, conditioned on being finite), or the question of percolation i.e is there a positive probability that the cluster of the origin is infinite (the cluster contains by coupling the percolation cluster of $p=1/2$ bond percolation on the 2D lattice, which is finite a.s., so I do not see a simple domination argument).

However the specific question you asked is indeed I believe easier. Here is an attempt.

Note that the question "is the square A having (0,0) and (1,1) as vertices a component" has a simple answer - it is $p_1=(1/2)^{6}$ (since it is determined by the neighboring squares, and for a given orientation of the diagonal in $A$, only 6 of these neighbors are relevant). Now apply the ergodic theorem to conclude that the asymptotic fraction of such squares is $p_1$. The reason this does not answer your question is that you asked for the asymptotic number of components, but it does give a lower bound on the fraction you asked about.

To get the answer to your question, repeat the computation for all 2x1 components, 3x1, 3x2, etc that contain the square A. That is, for a shape $Q$, compute $p_Q=P(C(A)\sim Q)$ where $C(A)$ is the shape containing $A$ and $\sim$ is up to translation of the shape. Then the events in the definition of $p_Q$ are disjoint, and $\sum p_Q\leq 1$. Now the fraction you ask about is $p_1/\sum_Q(p_Q/|Q|)$, where $|Q|$ is the area of $Q$.