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Home/ Questions/Q 865799
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T09:41:43+00:00 2026-05-15T09:41:43+00:00

I have a toroidal-ish Euclidean-ish map. That is the surface is a flat, Euclidean

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I have a toroidal-ish Euclidean-ish map. That is the surface is a flat, Euclidean rectangle, but when a point moves to the right boundary, it will appear at the left boundary (at the same y value), given by x_new = x_old % width

Basically, points are plotted based on: * see edit

(x_new, y_new) = ( x_old % width, y_old % height)

Think Pac Man — walking off one edge of the screen will make you appear on the opposite edge.

What’s the best way to calculate the shortest distance between two points? The typical implementation suggests a large distance for points on opposite corners of the map, when in reality, the real wrapped distance is very close.

The best way I can think of is calculating Classical Delta X and Wrapped Delta X, and Classical Delta Y and Wrapped Delta Y, and using the lower of each pair in the Sqrt(x^2+y^2) distance formula.

But that would involve many checks, calculations, operations — some that I feel might be unnecessary.

Is there a better way?


edit

When an object moves, it moves to position (x_old,y_old), runs it through the above formula, and stores (x_new, y_new) as its position. The above formula was only added to clarify what happens when objects move across the boundary; in reality, only one (x,y) pair is stored in each object at a time.

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  1. Editorial Team
    Editorial Team
    2026-05-15T09:41:43+00:00Added an answer on May 15, 2026 at 9:41 am

    The best way I can think of is calculating Classical Delta X and Wrapped Delta X, and Classical Delta Y and Wrapped Delta Y, and using the lower of each pair in the Sqrt(x^2+y^2) distance formula.

    That’s it, I don’t think there is any quicker way. But it’s not too hard of a computation; you could do something like

    dx = abs(x1 - x2);
    if (dx > width/2)
      dx = width - dx;
    // again with x -> y and width -> height
    

    (I trust you can translate that into your preferred language)

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