Vtyjkyuk

Suppose we randomly choose a graph with $n$ vertices in the following manner: each edge is included with probability $frac12$. Thus, each graph has the same probability of being chosen, and there are $2^fracn(n-1)2$ possible graphs. Denote $H_n$ the event that such a graph contains a Hamiltonian path. Prove that
$$lim_n to infty P(H_n) = 1,$$
that is, as we consider larger and larger graphs, the proportion of graphs that contain a Hamiltonian path approaches $1$.

As a bonus, find an efficient algorithm for finding a Hamiltonian path in a graph. The algorithm must also be precise, that is, it must find a Hamiltonian path most of the time. More concretely, we are looking for an algorithm with the following two properties:

• Its average time complexity is polynomial with respect to the number of vertices. (But you should aim for as good as possible.)


• Denote $S_n$ the event that the algorithm succeeds if its input is a random graph with $n$ vertices. Then,
  $$lim_n to infty P(S_n) = 1.$$

So, what have I tried so far? There is a relatively easy way to obtain the lower bound of $0.25$ that holds for all $n$. Denote $X_n[a]$ the event that there exists a Hamiltonian path starting at $a$, in a random graph with $n$ vertices. Clearly, the probability of such an event occuring is the same for all vertices, and we will denote this probability $x$. We can bound $x = P(X_n[a])$ from below as follows:

• If there is at least one edge to some vertex $b$, we can consider Hamiltonian paths that start with vertices $a, b$. The probability that such a Hamiltonian path exists is $P(X_n-1[b]) = x_n-1$.


• If there is no edge from $a$, there is no Hamiltonian path starting at $a$. The probability of this occuring is $frac12^n-1$.

Thus we obtain the following inequality:
$$x_n geq (1 - frac 12^n-1) cdot x_n-1.$$

From this it is easy to prove that $x_n geq frac14$ for all $n$. To obtain a better lower bound, we can consider additional cases based on how many edges there are from $a$:

• If there are $0$ adjacent vertices, there is no hope.


• If there is one adjacent vertex, we can reduce to $x_n-1$.


• If there are at least $2$ adjacent vertices, denote them $b$ and $c$. The path can go through either $b$ or $c$. Thus, by the inclusion-exclusion principle, the probability that at least one of the paths exists is
  $$P(X_n-1[b]) + P(X_n-1[c]) - P(X_n-1[b] land X_n-1[c]).$$

Thus, if we were to find a good enough upper bound for $P(X_n-1[b] land X_n-1[c])$, the problem would be solved. However, such an upper bound eludes me, all I can come up with are lower bounds (which are not helpful).

Alternative forms of the above expression that may help:
$$P(X_n-1[b]) + P(X_n-1[c] mid neg X_n-1[b])$$

In this form, we are trying to find a lower bound for the second term, which corresponds to the probability that there exists a Hamiltonian path starting at $c$, given that we know that there is no Hamiltonian path from $b$.

Also, it may be better to consider the case with more vertices adjacent to $a$, such as three. Though with more vertices comes bigger trouble when applying the inclusion-exclusion principle.

For context: a result of Pósa says that a random graph with $C n log n$ edges already contains a Hamiltonian path with probability tending to $1$, and here we have a uniformly random graph (with $fracn^24$ edges on average). So we can be fairly wasteful.

As far as an algorithm goes, we can take the algorithm used to prove Dirac's theorem. The hypotheses of that theorem don't hold here, but the algorithm will still work with high probability. The strategy is this:

1. Pick a path greedily: start at a vertex $v_1$, pick a vertex $v_2$ it is adjacent to, then pick a vertex $v_3$ adjacent to $v_2$, and so on. Repeat until you get stuck.


2. Turn the path into a cycle: if it has endpoints $v_1$ and $v_ell$, find adjacent vertices $v_i$ and $v_i+1$ on the path such that $v_i+1$ is adjacent to $v_1$ and $v_i$ is adjacent to $v_ell$. Then the cycle is $v_1, dots, v_i, v_ell, dots, v_i+1, v_1$.


3. Turn the cycle into a longer path: if $v_ell+1$ is a vertex not on the cycle, find a vertex $v$ on the cycle adjacent to $v_ell+1$, and take the path starting next to $v$, going around the cycle, and then going to $v_ell+1$.


4. Repeat steps 2 and 3 until the path contains all the vertices.

All of these steps are quite likely to work individually. The problem is that to analyze how likely they are to work as a whole, we'd have to check edges of the graph multiple times, which doesn't preserve independence.

So instead, we will write our uniformly random graph $G$ as the union of graphs $G_0, G_1, dots, G_10 log n$, where $G_0$ contains each edge independently with probability $frac14$, and $G_1, dots, G_10 log n$ contain each edge independently with probability $p$, chosen so that $frac34(1-p)^10 log n = frac12$. Then the union will contain each edge with probability $frac12$, so it will be the uniformly random graph. If we solve for $p$, we get $p = O(frac1log n)$, but all we'll really need is for $p$ to be asymptotically bigger than $frac1sqrt n$.

Then I claim that:

1. A greedy path chosen in $G_0$ will reach length $n - 5log n$ with very high probability.


2. Doing step 2 of the algorithm in any of the $G_1, dots, G_10 log n$ will work with very high probability.


3. Doing step 3 of the algorithm in any of the $G_1, dots, G_10 log n$ will work with very high probability.

Then we can just look at graphs $G_0, G_1, dots$ sequentially as we go through the algorithm. This preserves independence, and if the edges we need will be in the graph $G_i$ we're looking at, they will be in the union $G$.

For the first claim: the probability that a greedy path will get stuck while there's still $5 log n$ vertices to pick from is $(frac34)^5log n = n^-5log frac43$, so the probability is at most $n^1 - 5log frac43 approx n^-0.43$ that it ever gets stuck.

For the second claim: we have at least $frac n2$ options for vertices $v_i$ and $v_i+1$, and each one works with probability $p^2$. The probability that none of them work is $(1-p^2)^n/2 le e^-p^2n/2$, which approaches $0$ as $nto infty$. (This is where we want $p gg frac1sqrt n$, so that the exponent goes to $-infty$ as $ntoinfty$. We actually want a bit better than that, because we want all $O(log n)$ of these steps to work, so the probability should go to $0$ faster than $frac1log n$. But our $p$ is actually quite a bit larger than $frac1sqrt n$, so we have that wiggle room.)

For the third claim: we have at least $frac n2$ options for the vertex $v$, and each works with probability $p$. The probability that none of them work is $(1-p)^n/2 le e^-pn/2$, which approaches $0$ as $nto infty$.

These are pretty well-known.

Indeed the threshold for a graph to be Hamiltonian is $p=fraclog n + log log nn ll frac 12$. Indeed a with high probability a random graph became Hamiltonian as soon as its minimum degree raise to $2$. You may find the proof in every classic book for random graphs for example here.

The algorithm is well-known as well. You can find a great lecture note about that in Alistiar Sinclair's course. I propose you to download it cause he will remove lecture notes after the semester. Grab Lecture 15. I suggest you to follow the entire lecture notes as well.