argmax of a function allowing small violations - math

Question

I have a function that returns an object with the best rating among a set of objects.
Now I want to get the object with the best rating which is closest to a certain point. Basically I want to allow a small violation to the formula, such that if o_1 is closer to the given point than o_2 even if
if
for some small and fixed
Currently I achieve with a about 5 if statements in a program, but I was wondering if there's a way to represent this in a cleaner more elegant way with a mathematical formula perhaps? If you have ever had a problem like this, how did you code it?
Here is a snippet of the code I have, I know there are some refactorizations that could be done to the if statements, I wanted to make the cases clearer. Is there a more elegant way of solving this?
public VisualDevice MostSuitableDevice(GameObject u, GameObject t, List<VisualDevice> D)
{
float maxSuitability = 0;
VisualDevice mostSuitableDevice = null;
float epsMaxSuitability = 0;
VisualDevice epsMSD = null;
foreach (VisualDevice d in D)
{
float suitability = Rating(d, u, t);
bool closerDevice = mostSuitableDevice != null ? Vector3.Distance(d.GetLightPosition(), t.transform.position) < Vector3.Distance(mostSuitableDevice.GetLightPosition(), t.transform.position) : false;
if (suitability > maxSuitability && closerDevice)
{
maxSuitability = suitability;
mostSuitableDevice = d;
}
else if (suitability > maxSuitability + epsilon)
{
maxSuitability = suitability;
mostSuitableDevice = d;
}
else if (suitability + epsilon > maxSuitability && closerDevice)
{
closerDevice = epsMSD != null ? Vector3.Distance(d.GetLightPosition(), t.transform.position) < Vector3.Distance(epsMSD.GetLightPosition(), t.transform.position) : false;
if (suitability > epsMaxSuitability && closerDevice)
{
epsMaxSuitability = suitability;
epsMSD = d;
}
else if (suitability > epsMaxSuitability + epsilon)
{
epsMaxSuitability = suitability;
epsMSD = d;
}
else if (suitability + epsilon > epsMaxSuitability && closerDevice)
{
epsMaxSuitability = suitability;
epsMSD = d;
}
}
}
bool epsCloser = epsMSD != null && mostSuitableDevice!=null ? Vector3.Distance(epsMSD.GetLightPosition(), t.transform.position) < Vector3.Distance(mostSuitableDevice.GetLightPosition(), t.transform.position) : false;
if (epsMaxSuitability + epsilon > maxSuitability && epsCloser)
{
maxSuitability = epsMaxSuitability;
mostSuitableDevice= epsMSD;
}
return mostSuitableDevice;
}

Answer 1

First find maximum rating rmax. Then select the object closest to the given point from all objects o with Rating(o) >= rmax - epsilon.
Like that:
float maxSuitability = D.Max( d => Rating(d, u, t));
float minDistance = float.PositiveInfinity;
VisualDevice mostSuitableDevice = null;
foreach (VisualDevice d in D)
{
float suitability = Rating(d, u, t);
if (suitability < maxSuitability - epsilon)
continue;
float distance = Vector3.Distance(...
if (distance < minDistance)
{
minDistance = distance;
mostSuitableDevice = d;
}
}