<s>[INST] Minimum steps adding edges to form a complete graph Consider a graph $G$ with $t$ vertices and $0$ edges. Turn it into the complete graph $K_t$ by repeatedly applying the following move $M$:
$M$: Choose $n$ vertices in $G$ and add edges between each of them to make a complete subgraph $K_n$ within $G$. This gives the new $G$.
Question: Given $t$ and $n$, what is the least number $m$ of times $M$ has to be applied before $G$ is $K_t$?

Notes:
If $n=2$ then $m$ is ${t \choose 2}$.
If $n=t-1$ then $m$ is 3.
I'd prefer a comprehensive derivation of some estimation over precise candidates for computation.
 [/INST] As mentioned in comments there is a trivial lower bound on $m$: $m \ge \left\lceil\frac{t(t - 1)}{n(n - 1)}\right\rceil$. This bounds is not tight even for $t = 4$ and $n = 3$ when $m = 3$ while the bound is $2$.
Let's think deeper: $\deg_{K_t} v = t - 1$ for any $v \in K_t$, and initially $\deg_G v = 0$. Each move $M$ increrases degree of $v$ by at most $n - 1$. Then each of $t$ vertices requires at least $\left\lceil\frac{t - 1}{n - 1}\right\rceil$ moves. And each move affects exactly $n$ vertices. Thus we get $m \ge \left\lceil\left\lceil\frac{t - 1}{n - 1}\right\rceil\cdot\frac{t}{n}\right\rceil$. It is exact up to $(t, n) = (7, 4)$ when $m = 5$ and the bound is $4$. (I mean my bound is exact for $t \le 6$ and for $t = 7$ and $n \le 3$.)
The most trivial upper bound for $n \ge 2$ is $\frac{t(t - 1)}2$ that is the number of edges. Let's try to get something better. To cover all edges incident to a single vertex it would be enought to make $\left\lceil\frac{t - 1}{n - 1}\right\rceil$ moves $M$. After that we can remove this vertex and get $K_{t - 1}$ on all other vertices. And so on until we left only $n$ vertices that need only one move:
$$m \le \left\lceil\frac{t - 1}{n - 1}\right\rceil + \left\lceil\frac{t - 2}{n - 1}\right\rceil + \cdots + \left\lceil\frac{n - 1}{n - 1}\right\rceil.$$
To simplify this bound let's use $k = \left\lfloor\frac{t - 2}{n - 1}\right\rfloor$ and $r = (t - 2) - k(n - 1)$. Noting that $\left\lceil\frac xy\right\rceil = \left\lfloor\frac{x + y - 1}{y}\right\rfloor$ for integers $x$ and $y$, we get:
$$m \le \left\lfloor\frac{t + n - 3}{n - 1}\right\rfloor + \left\lfloor\frac{t + n - 4}{n - 1}\right\rfloor + \cdots + \left\lfloor\frac{n + n - 3}{n - 1}\right\rfloor\\
= (k + 1) \cdot (r + 1) + k \cdot (n - 1) + (k - 1)\cdot(n - 1) + \cdots + 2 \cdot (n - 1) + 1\\
= (k + 1) \cdot (r + 1) + \left(\frac{k(k + 1)}{2} - 1\right)(n - 1) + 1\\
= \frac{k(k + 1)}{2}(n - 1) + kr + k + r - n + 3.$$
Here I combine these two bounds into one table:
$$\begin{array}{c|c|c|c|c|c|c|c}
n\backslash t & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
2 & 1 / 1 & 3 / 3 & 6 / 6 & 10 / 10 & 15 / 15 & 21 / 21 & 28 / 28\\\hline
3 & & 1 / 1 & 3 / 3 & \color{red}{5} / 4 & \color{red}{8} / 6 & \color{red}{11} / 7 & \color{red}{15} / 11\\\hline
4 & & & 1 / 1 & 3 / 3 & \color{red}{5} / 3 & \color{red}{7} / \color{red}{4}  & \color{red}{10} / 6\\\hline
5 & & & & 1 / 1 & 3 / 3 & \color{red}{5} / 3 & \color{red}{7} / \color{red}{3}\\\hline
6 & & & & & 1 / 1 & 3 / 3 & \color{red}{5} / 3\\\hline
7 & & & & & & 1 / 1 & 3 /3\\\hline
8 & & & & & & & 1 / 1
\end{array}$$
Non-exact bounds are marked with red color. As far as you see both bounds agree on $n = 2$, $n = t$ and $n = t - 1$ and the lower bound is much more close to the answer.
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