Search algorithm to find integer input that produces the first 'True' (bool: 1) occurence of a computationally expensive boolean function

Question

A boolean-valued function defined in the set of positive integers, $\mathcal Z$.

$$f(n) = \begin{cases} 0, &n_{min}\le n < n\ast\\1, &n\ast\le n\le n_{max} \end{cases} ; n \in \mathcal Z $$

$f(n)$ is computationally expensive and it's analytical form (a closed-form equation producing these binary-valued outputs is unknown, since it is the result of a time-domain simulation of complicated system of non-linear PDEs meeting certain exit conditions etc.). Furthermore, $\frac{\partial{f}}{\partial{n}}$ is also obviously not available.

The goal is to compute $n\ast$, i.e. the first occurrence of the integer-valued function input, $n$ that produces a Boolean $1$. Needless to say, this can be trivially achieved in computer code by using a simple $for$ loop and rejecting the values of $n$ one by one, until we hit $n\ast$.

The lower and upper bounds, $n_{min}$ and $n_{max}$ are known apriori.

However, given the expensive nature of computing $f(n)$, I am looking for an efficient algorithm beyond this naive implementation, especially one that requires small number of function evaluations. A pseudo-code implementation of your proposed solution would be much welcome too.

A conceptual schematic/visualisation of the problem is also shown here, purely to aid the understanding of the task at hand

Answer 1

Use binary search.

You know that the correct value is somewhere in the range $[n_\min,n_\max]$, so pick $t = \lfloor (n_\min+n_\max)/2 \rfloor$ (the point halfway in the middle), and evaluate $f(t)$. This will either tell you that $n^*$ is in $[n_\min,t]$ or that $n^*$ is in $[t+1,n_\max]$. Now you can repeat on the smaller interval. Each time, the width of the interval decreases by a factor of two, so after $\lg (n_\max-n_\min)$ iterations, you will find the correct value of $n^*$. This is much faster than trying all $n_\max-n_\min$ possible values of $n^*$.