The Geometric Median

Given a set of numbers on the real line, say \( a_1, a_2, a_3, …, a_n\), the median of these numbers is the middle number when you arrange these numbers in ascending/descending order. To understand more about the median for a set of numbers, go through this.

Let’s ask a different question altogether.

Given a set of numbers on the real line, say \( a_1, a_2, a_3, …, a_n\), what is the number \( x\) such that \( |x-a_1| + |x-a_2| + … |x-a_n|\) is minimized?

In other words, what is the point on the real number line, such that the sum of the distances of that point from the given set of points is minimum? In some sense, this point gives an idea about the central behavior of the set of points.

In fact, this turns out to be the median of the set of numbers { \( a_1, a_2, a_3, …, a_n\)}. To understand the proof, go through this.

But, I suggest you try this yourself, in inductive steps. Start with two points, any point in between them is the median. Now, add a point in between those two points, Observe that the new median is the new point added. Continue similarly in this inductive process.

My contentment is to ask questions and seek. Therefore, a lot of questions inpoured.

  • What happens if we take those n points in 2D space?
  • What happens if we take those n points in 3D space?
  • What happens if we take those n points in n-D space?
  • What happens if we take a different distance idea instead of the normal distance?

Well, the intelligent way is to answer the easiest of all and then generalize the idea – the first question.

Given a set of numbers in the 2D space, say \( a_1, a_2, a_3, …, a_n\), what is the point \( x\) such that \( |x-a_1| + |x-a_2| + … |x-a_n|\) is minimized?

Let’s take the easiest distance concept of all – The Euclidean Distance \(d(x,y) = |x-y|= \sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2} \).

This concept is known as Geometric Median in the literature.

Now, the problem seems difficult even in the case of three points.

Let’s explore it for a quadrilateral ( \( n = 4 \) ).

Let’s take the point G.

Step 1: Does there exist a better point like that?

Yes, there is! Let’s explore.

Observe, I is the point such that AG + FG = BI + EI. You can argue this by continuity. But, here I have explicitly constructed it by using the idea of an ellipse, where the intersection point of BE and the ellipse with focal points A and F and passing through G gives the required point.

Observe that (AI + IF) + (BI + IE) = (AG + GF) + (BE) < (AG + GF) + (BG + GE). We are just using the triangle inequality.

Hence, G is a better point than I.

Step 2: Does there exist a better point than I?

Albeit, there is!

Just observe that the intersection point of the diagonals J is a better point

(AI + IF) + (BI + IE) = (AI + IF) + (BE) = (AI + IF) + (BJ + JE) > (AG + GF) + (BG + GE). We are just using the triangle inequality.

Thus, this algorithm results in the converges to the intersection of the diagonals as a special point. This is easy.

But what about, \( n = 3\). It may be easy. No, it’s not.

The case for triangles is not at all easy. In the case of triangles, the point is called Fermat’s Point.

Construct an equilateral triangle on each of the two arbitrarily chosen sides of the given triangle. Draw a line from each new vertex to the opposite vertex of the original triangle.
The two lines intersect at the Fermat point.

I will not repeat the solution again. The following pic gives a cute solution. Using the same idea, that the distance between two points is minimum along a straight line.

Source: Cut the Knot

Now, the rest of the questions remain too difficult to solve. Even there hardly exists a good algorithm to solve it.

Now, this left to your imagination. Stay tuned!

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