Forced merging of large components ($N_{\ge K}(G_m)\ge an\Rightarrow L_1(G_{m+\lceil A(a)n/K\rceil })\ge an/3$)

For the two-choice Achlioptas process, fix $a>0$ and $K\ge 1$. Conditional on any current graph with $N_{\ge K}\ge an$, put $W$ equal to the vertices in its large graph components. There are at most $n/K$ such initial graph components. If after $s$ more steps no graph component contains $an/3$ vertices of $W$, greedily assign final graph components to two groups so that each meets at least $|W|/3\ge an/3$ vertices of $W$. This induces a set partition of the initial large graph components, of which there are at most $2^{n/K}$ possibilities.

For any fixed such split $W=A\cup B$, a uniform candidate edge joins $A$ to $B$ with probability at least $2a^2/9$. If both candidates do so, the selected edge cannot avoid the crossing. Thus the probability that the split survives all $s$ steps is at most $(1-a^4/81{)}^s$. The deliberately weaker constant also covers distinct candidate sampling for sufficiently large $n$. In the absent-edge version, as long as the split has survived, every $A$--$B$ edge is absent, so conditioning on absence can only increase this crossing probability. A union bound with $s=\lceil A(a)n/K\rceil$ and $A(a)=81(\log 2+2)/a^4$ bounds failure by $e^{-2n/K}$. For fixed $K$, a further union bound makes the conclusion simultaneous over all starting steps $m\le 3n$. This proof is independent of how the rule chooses between the two candidates.