Client suggestion: make #closest() return the actual closest entity

In short, this is a suggestion to change the calculation of #closest() from game tiles to a cost function to get the actual closest. It will impact scripts in many contexts, for example Varrock West Bank where 2/3 Bankers are all 2 tiles away, therefore OSBot picks one which could mean that the player moves around between bankers.

Let's take this square as an example:

Its width and height are both 2 arbitrary units.

Now, let's cut it into 2 triangles: (one removed to see easier)

With the help of our greek pal Pythagoras, we can conclude that  c != 2.

Why is this relevant?

OSBot will count #closest() in the form of game tiles. Now, if we label our own position as x, and 2 positions both 2 game tiles away (one horizontal, and one diagonal) x and y respectively, OSBot decides that dist(y) = dist(z) where dist(n) = abs(n - x). However, this is wrong:

We can clearly see in this diagram that dist(z) > dist(y), however OSBot still concludes that they are of equal distance apart.

From this point, OSBot (seems) to choose a random entity from this list of those that are of the same distance apart.

Putting this into context: Varrock West Bank.

If you run getNpcs().closest("Banker") while standing in one of the stall-line things, you will switch between the bankers around you even though the one directly horizontal to you is the closest.

Solution: use a calculation like below.

If we set two cost variables, one for diagonal and one for non-diagonal. Their costs are 10 and 14 respectively (you could do 1 and sqrt(2) at the cost of more CPU). We can then perform a calculation like this: (pseudo)

horizontal = abs(n.x - x.x)
vertical = abs(n.y - x.y)

cost = max(horizontal, vertical) + (diagonalcost - nondiagonalcost) * min(horizontal, vertical)

Now, back to our pictures: (added the costs)

From this, we can conclude that dist(y) < dist(z), because our calculations:

dist(y) = max(2, 0) + (14 - 10) * min(2, 0)
dist(z) = max(2, 2) + (14 - 10) * min(2, 2)

This will return the true closest.

Thanks for taking the time to read this