A Simple Combinatorial Approximation Problem

This is certainly a well-known problem. But I pose it here because there is a geometric solution, and geometric solutions are always nice.

Assume you have two sets of n points

\(x_1,\ldots,x_n, \quad y_1,\ldots,y_n\)

The problem is to find a permutation p such that

\(\sum_{k=1}^n (x_k-y_{p(k)})^2\)

is minimized. Of course, we can assume that the x’s are sorted. And it is easy to guess that the minimum is achieved if the permutation sorts the y’s. But how to prove that?

Here is a simple idea. We claim that swapping two of the y’s into the right order makes the sum of squares smaller. I.e.,

\(x_1 < x_2, \, y_1<y_2 \Longleftrightarrow (x_1-y_1)^2+(x_2-y_2)^2 < (x_1-y_2)^2+(x_2-y_1)^2\)

If you try that algebraically the right hand side is equivalent to

\(-x_1y_1-x_2y_2 < -x_1y_2-x_2y_1\)

Putting everything one side and factorizing, you end up with

\((x_2-x_1)(y_2-y_1) > 0\)

and the equivalence is proved.

But there is a geometric solution. We plot the points into the plane. Two points are on the same side of the symmetry line between the x- and y-axis if they have the same order. And the connection to the mirrored point will always be larger.

We can extend this result to more norms. E.g., the sorting of the y’s will also minimize

\(\sum_{k=1}^n |x_k-y_{p(k)}|, \, \max_k |x_k-y_{p(k)}|\)

As long as the underlying norm is symmetric with respect to the variables, i.e.,

\(\|(\ldots,t,\ldots,s,\ldots)\|=\|(\ldots,s,\ldots,t,\ldots)\|\)

The prove for this can be based on the same geometrical argument.

17. Juni 2020 von mga010
Kategorien: Uncategorized | Schreibe einen Kommentar

Schreibe einen Kommentar

Pflichtfelder sind mit * markiert


Diese Website verwendet Akismet, um Spam zu reduzieren. Erfahre mehr darüber, wie deine Kommentardaten verarbeitet werden.