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At each step offer a fixed number of independent uniform candidate edges on a fixed vertex set, and let a rule select one. The standard two-choice model offers two candidates. The rule may depend on the entire previously exposed history. Repeated already present edges make no change in the independent-candidate version; another usual version offers only absent edges. The forced merging of large components and persistence of unsampled graph components apply to both over a linear number of steps. Fixed choice is essential in the continuity of fixed-choice percolation argument.

A fixed-choice Achlioptas process cannot have its largest component of a graph jump from $o(n)$ to a fixed positive fraction of $n$ within $o(n)$ steps, with high probability. The proof combines forced merging of large components and persistence of unsampled graph components. An assumed jump forces positive vertex mass in each band $[k,Dk)$ slightly before the jump: insufficient smaller components can feed the final large component, while enough components above $Dk$ would merge too early. Persistence carries a fixed positive amount of each band to a common time. Taking sufficiently many disjoint bands with $k=1,D,D^2,\dots$ then counts more than $n$ vertices. The number of offered candidates is fixed throughout this argument. Riordan and Warnke establish this obstruction and broader continuity results in their original research paper.

Use the standard Achlioptas process with two independent uniform candidate edges per step and a rule choosing one. The component-band vertex count $N_{[k,Dk)}$ counts vertices, not the number of graph components. Integer time rounding changes the arguments below by $O(1)$ steps.

First, at $t_k$ the eventual graph component of order at least $\delta n$ is assembled from graph components already present at $t_k$. Contract those initial graph components. During $s=t_c-t_k$ steps, at most $s$ new edges are added, so the eventual connected contracted graph contains at most $s+1$ initial graph components. Initial graph components of order less than $k$ can therefore contribute at mostvertices, for sufficiently large $n$ and fixed $k$, since $s=\delta n/(2k)+O(1)$. For $k=1$ their contribution is zero. Consequently $N_{\ge k}(G_{t_k})\ge \delta n/2$.

We now use forced merging of large components. If at a time $m$ the graph components of order at least $K$ contain at least $an$ vertices, then after at most $A(a)n/K+O(1)$ steps there is a graph component of order at least $an/3$, except on an event of probability exponentially small in $n/K$. One explicit choice is $A(a)=81(\log 2+2)/a^4$. The proof is a union bound over set partitions of the at most $n/K$ initial large graph components: if all final graph components were small, a balanced component cut would split their initial vertices into sets of size at least $an/3$, and both candidate edges cross that fixed cut with probability at least $a^4/81$ in every step. The rule cannot avoid such a forced crossing. The failure bound is $e^{-2n/K}$, and a union bound makes the estimate simultaneous over all starting times at most $3n$, for each fixed $K$.

Set $a=\delta /4$ and choose a fixed integer $D\ge 2$ with $D\ge 4A(a)/\delta$. If $N_{\ge Dk}(G_{t_k})\ge \delta n/4$, this lemma would produce a graph component of order at least $\delta n/12$ by timeThat contradicts the assumed $L_1(G_{t_c^{\prime }})=o(n)$. Thus $N_{\ge Dk}(G_{t_k})<\delta n/4$, and subtraction givesThe same $D$ works for every fixed $k$; it depends only on $\delta$, with the number of offered edges fixed at two.