# Do Symmetric problems have Symmetric solutions?

There is a general notion of Symmetry in mathematics, that of an object being constant under some transformation.

If we think of our object as being a "mathematical problem" we can see certain mathemtical problems have symmetries in them, in terms of how the problem is stated.

This thought came to be when I was trying to solve a combinatorics problem on this website. The problem statement was "Given 11 people and one safe, how many locks do we need on the safe to ensure a subset of 5 people cannot open the safe but a susbet of 6 people can open the safe. For each lock we can get as many copies of its key as we like.".
I merely want to note that in this problem the group of 11 people all have the same constraints put on them. There is no mentioning of a property fulfilled by a certain "distinguished" subset of these 11 people, if we were to write the problem as: "Given a set of 11 people $\{p_1,\dots ,p_{11}\}$, how many locks do we need ..." (the rest of the problem statement is the same as before), then we would have a symmetry in the sense that if we permute the set $\{p_1,\dots ,p_{11}\}$ in the problem statement, and relabel the people accordingly, the problem statement does not change.

Intuitively, I expect a solution to such a problem to be symmetric in some way, for example I would expect each person do have the same number of keys. Intuition is not a proof.

The question is this: Is there a general principle in mathematics which relates symmetries in problem statemenets to symmetries in the solution?

• What is a "symmetry in problems"? Are you dealing with group actions on sets, or something more general? I have problems to understand what you mean with solution of a given problem... – Avitus Nov 26 '13 at 13:01
• – lhf Jan 4 '14 at 21:32

It is not in general true that a solution to a symmetric problem must itself be symmetric. For example:

Minimize $x^4-2x^2$.

This is an even function, so it is symmetric in reflection about the $y$ axis. However, the solutions are $1$ and $-1$, and neither of these solutions is itself symmetric. (Physicists speak about "symmetry breaking" in such cases, and it more complex variants it plays a significant role in modern particle physics).

What we can say is that if there's a symmetry of the problem, we can apply that same symmetry to the solution, and get a (possibly different) solution back. That's more or less the generic definition of what it means for a problem to have a symmetry. So the set of all solutions will be a symmetric as the problem is.

If we know somehow that the solution is unique, it will need to have all of the symmetries the problem has. This is sometimes a useful shortcut for finding it. (But be careful to check whether "unique" really means "unique modulo such-and-such symmetries which are present in the problem too").

In the vein of your example we could say

Given a box and two people, what is the minimum number of locks and keys we need to buy such that (a) the two people can open the box together, and (b) the box can't be opened if neither of the two people is present?

That's kind of a silly problem, but it's certainly symmetric under permutation of the two people. But the solution does not have the kind of symmetry you're intuitively expecting: One lock with one key, and hand the single key to one of the trusted people.

• $x=1$ and $x=-1$ are symmetric about the y-axis as well... – BlueRaja - Danny Pflughoeft Nov 26 '13 at 17:20
• @BlueRaja: The point is that $x=1$ is not symmetric about the $y$-axis, and $x=-1$ is not symmetric about the $y$-axis. – Henning Makholm Nov 27 '13 at 4:52
• I don't see how that statement is meaningful or relevant. The statement "The problem has symmetry but the solutions don't" is false, so though I'm sure there are valid counter-examples, this is not one of them. – BlueRaja - Danny Pflughoeft Nov 27 '13 at 6:56
• That statement is true because neither of the solutions has that symmetry. As I write, the set of solutions is symmetric, but that is different from saying that the solutions are (individually) symmetric. – Henning Makholm Nov 27 '13 at 11:27
• By that logic, you could say the problem is also non-symmetric because $x^4-2x^2$ is not symmetric about the x-axis. You can't just point out one way in which the solution is non-symmetric and conclude "therefore the whole thing has no symmetry" – BlueRaja - Danny Pflughoeft Nov 27 '13 at 16:38