Abstract
Starting with the empty graph on $[n]$, at each round, a set of $K=K(n)$ edges is presented chosen uniformly at random from the ones that have not been presented yet. We are then asked to choose at most one of the presented edges and add it to the current graph. Our goal is to construct a Hamiltonian graph with $(1+o(1))n$ edges within as few rounds as possible. We show that in this process, one can build a Hamiltonian graph of size $(1+o(1))n$ in $(1+o(1))(1+(\log n)/2K) n$ rounds w.h.p. The case $K=1$ implies that w.h.p. one can build a Hamiltonian graph by choosing $(1+o(1))n$ edges in an online fashion as they appear along the first $(0.5+o(1))n\log n$ rounds of the random graph process. This answers a question of Frieze, Krivelevich and Michaeli. Observe that the number of rounds is asymptotically optimal as the first $0.5n\log n$ edges do not span a Hamilton cycle w.h.p. The case $K=\Theta(\log n)$ implies that the Hamiltonicity threshold of the corresponding Achlioptas process is at most $(1+o(1))(1+(\log n)/2K) n$. This matches the $(1-o(1))(1+(\log n)/2K) n$ lower bound due to Krivelevich, Lubetzky and Sudakov and resolves the problem of determining the Hamiltonicity threshold of the Achlioptas process with $K=\Theta(\log n)$. We also show that in the above process one can construct a graph $G$ that spans a matching of size $\lfloor V(G)/2) \rfloor$ and $(0.5+o(1))n$ edges within $(1+o(1))(0.5+(\log n)/2K) n$ rounds w.h.p. Our proof relies on a robust Hamiltonicity property of the strong $4$-core of the binomial random graph which we use as a black-box. This property allows it to absorb paths covering vertices outside the strong $4$-core into a cycle.