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Home/ Questions/Q 7659185
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T13:17:59+00:00 2026-05-31T13:17:59+00:00

This algorithm finds a given value n by iterating between a lower bound and

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This algorithm finds a given value n by iterating between a lower bound and upper bound, similar to binary search. How can I improve this algorithm? A few details. The value n is almost always less than 1.0, but this isn’t a boundary condition. It is, however, never less than 0.

def approx_value(n, lower_bound, upper_bound, precision = 0.001):
    approx = (lower_bound + upper_bound) / 2.0
    while (abs(approx - n) > precision):
        if n > approx:
            lower_bound = approx
        else:
            upper_bound = approx
        approx = (lower_bound + upper_bound) / 2.0
    return approx

def approx_value_no_bounds(n, precision = 0.0001):
    approx = 1.0
    while (abs(approx - n) > precision):
        if approx > n: #decrease volatility
            return approx_value(n, 0.0, approx)
        else:
            approx *= 2
    return approx

I know this seems odd because the value of n is already supplied, but the algorithm is not complete. Basically, n is the solution of a complex equation that cannot be solved for approx in closed form, so I’m doing it iteratively. Eventually, this algorithm will compare the value of the function when using approx with the value n and return approx is it approximates the input variable well enough.

I’m hoping to keep the running time in O(log(n)), which is why I modelled it somewhat after binary search, albeit only somewhat because the upper bound is not necessarily known immediately.

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  1. Editorial Team
    Editorial Team
    2026-05-31T13:18:00+00:00Added an answer on May 31, 2026 at 1:18 pm

    This sounds like a classic optimization problem. These are well studied, and there are several well-known algorithms. Some simple but reasonably efficient ones are Newton’s method or gradient descent.

    You could also try the nonlinear-simplex algorithm if the above don’t work very well for your function.

    The running time vs accuracy tradeoff here depends on the nature of the function you are trying to solve for.

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