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there is a strong connection here to random graphs
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Aidan Rocke
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clarified my hypothesis
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Aidan Rocke
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It can be shown that the set of graphs with $N$ vertices $G_N$ has cardinality:

\begin{equation} \lvert G_N \rvert = 2^{N \choose 2} \tag{1} \end{equation}

Recently, I wondered how much bigger $\lvert G_N \rvert$ is compared to the number of graphs with $N$ vertices that are path-connected, $PC_N$.

If we denote the set of Hamiltonian paths by $H_N$ we can easily show that:

\begin{equation} H_N \subset PC_N \tag{2} \end{equation}

and by Stirling's approximation:

\begin{equation} \lvert H_N \rvert = N! \sim \sqrt{2\pi N} \big(\frac{N}{e}\big)^N \tag{3} \end{equation}

and therefore $\lvert PC_N \rvert$ must grow exponentially fast.

Now, I'm curious about asymptotic formulas for $\lvert PC_N \rvert$ and I'm fairly confident that for large $N$:

\begin{equation} \frac{\lvert PC_N \rvert}{\lvert G_N \rvert} \leq e^{-N} \tag{4} \end{equation}

but I suspect that a proof for this statement would be fairly subtle.

Note: I didn't make use of a computer before formulating this hypothesis but in retrospect I should have analysed the connectivity of random graphs, as suggested by Olivier Fouquet. This would have given me more insight into the problem.

It can be shown that the set of graphs with $N$ vertices $G_N$ has cardinality:

\begin{equation} \lvert G_N \rvert = 2^{N \choose 2} \tag{1} \end{equation}

Recently, I wondered how much bigger $\lvert G_N \rvert$ is compared to the number of graphs with $N$ vertices that are path-connected, $PC_N$.

If we denote the set of Hamiltonian paths by $H_N$ we can easily show that:

\begin{equation} H_N \subset PC_N \tag{2} \end{equation}

and by Stirling's approximation:

\begin{equation} \lvert H_N \rvert = N! \sim \sqrt{2\pi N} \big(\frac{N}{e}\big)^N \tag{3} \end{equation}

and therefore $\lvert PC_N \rvert$ must grow exponentially fast.

Now, I'm curious about asymptotic formulas for $\lvert PC_N \rvert$ and I'm fairly confident that:

\begin{equation} \frac{\lvert PC_N \rvert}{\lvert G_N \rvert} \leq e^{-N} \tag{4} \end{equation}

but I suspect that a proof for this statement would be fairly subtle.

It can be shown that the set of graphs with $N$ vertices $G_N$ has cardinality:

\begin{equation} \lvert G_N \rvert = 2^{N \choose 2} \tag{1} \end{equation}

Recently, I wondered how much bigger $\lvert G_N \rvert$ is compared to the number of graphs with $N$ vertices that are path-connected, $PC_N$.

If we denote the set of Hamiltonian paths by $H_N$ we can easily show that:

\begin{equation} H_N \subset PC_N \tag{2} \end{equation}

and by Stirling's approximation:

\begin{equation} \lvert H_N \rvert = N! \sim \sqrt{2\pi N} \big(\frac{N}{e}\big)^N \tag{3} \end{equation}

and therefore $\lvert PC_N \rvert$ must grow exponentially fast.

Now, I'm curious about asymptotic formulas for $\lvert PC_N \rvert$ and I'm fairly confident that for large $N$:

\begin{equation} \frac{\lvert PC_N \rvert}{\lvert G_N \rvert} \leq e^{-N} \tag{4} \end{equation}

but I suspect that a proof for this statement would be fairly subtle.

Note: I didn't make use of a computer before formulating this hypothesis but in retrospect I should have analysed the connectivity of random graphs, as suggested by Olivier Fouquet. This would have given me more insight into the problem.

improved question title
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Aidan Rocke
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Asymptotic formula for the number of path-connectedconnected graphs

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Aidan Rocke
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