Algorithm 2.

Stopping-Criterion

$\gamma =1+\frac{\log (n)}{N}$ // ratio of the growth must be γ > 1

k = log(N) // number of top oop

v = 0 // number of successive iterations without progress $\mathcal{A}=\text{a}$ list containing k zeros // keeping the top k largest values of oop up to each iteration

Stopping-Criterion (oop(t))

1) Detect the best iteration:

if ${\mathrm{oop}}^{(t)}>\max (\mathcal{A})$, set T = t

2) Check the stopping criteria:

if v > k or t > Tmax, halt Algorithm 1 // Tmax : maximum number of iterations

3) Check the progress of oop:

$v=\{\begin{array}{ll}v+1\hfill & {\text{ oop }}^{(t)}<\gamma \cdot \min (\mathcal{A})\hfill \\ 0\hfill & \text{ otherwise }\hfill \end{array}$

4) Update $\mathcal{A}$:

if ${\mathrm{oop}}^{(t)}>\min (\mathcal{A})$, then replace the minimum entry with oop(t)