Suppose we have some probability distribution over objects of a given order, and a way of randomly selecting sub-objects of a given object of some fixed order. Suppose we also have some equivalence relation on these objects. Then we say that an object O is n-symmetric if the distribution over equivalence classes given by choosing a random order-n subobject of O is the same as the one given by choosing a random order-n object.

So in the case of permutations, our order is length, our distributions are the uniform ones, our sub-object relation is to choose a uniformly random subset of n elements in the partition, and our equivalence relation is that of having the same order type. In the case of graphs, our order is number of vertices, our distributions are the ones given by choosing edges with probability 0.5, our sub-object relation is to choose the subgraph induced by a uniformly random order-n subset of the vertices, and our equivalence relation is having the same number of vertices and edges.

Proposition: Suppose that for all k≤n≤m the following 2 conditions hold:

(1) The distribution over equivalence classes given by choosing a random order-k subobject of a random order-n object is the same as the one given by choosing a random element of K.

(2) The distribution over equivalence classes given by choosing a random order-k subobject of a random order-n subobject of a given order-m object O is the same as the one given by choosing a random order-k subobject of O.

Then every n-symmetric object is also a k-symmetric object for all k≤n.

Proof: Let k≤n≤m, and let K,N,M be the sets of all objects of orders k,n,m respectively. Let S be an n-symmetric element of M. Then we have equality between the following distributions (with respect to the chosen equivalence):

[random order-k subobject in S] = [random order-k subobject of (random order-n subobject of S)] = [random order-k subobject of (random element of N)] = [random element of K]

which is exactly the condition for S to be k-symmetric.

Some other objects which can have symmetry defined in this way and satisfy the given conditions under the natural choices for order/equivalence/random object:

* Posets (order = number of elements)

* Directed graphs

* Graphs under other probability distributions (e.g. where a random graph is chosen with edge probability 2/3)

* Words in a finite language with random substrings (not necessarily contiguous). (This doesn’t work if our distribution is contiguous substrings; see below.)

The argument doesn’t hold for every such object – e.g. groups – but it does in most cases where the sub-object relation is “nice” (and in particular when an object’s order counts something, and we can take sub-objects by choosing random subsets of those somethings). (Open question: is there some nontrivial interpretation of random groups and random subgroups which DOES obey these conditions?)

The consequences in each specific case here are nontrivial. For one thing, it restricts our possible orders substantially: the possible orders of 3-symmetric permutations go from 9, 10, 18, 20, … to 9, 20, … and restrict even further for higher orders (the minimal possible orders for 4-symmetric and 5-symmetric permutations go up to 64 and 128, respectively).

An example which is kind of interesting but for which this proposition doesn’t hold: words in a finite language with contiguous substrings. Indeed, BAABB is a 2-symmetric word in the language {A,B} which is not 1-symmetric. The number of 2-symmetric objects has a closed-form solution here: it’s 0 for orders not congruent to 1 mod 4, and at 4k+1 is (2k choose k)^2.

An obvious question for generating n-symmetric graphs: is there any sort of natural operation on a hypercube which yields such graphs? (The motivation being that the complete graph on 2^n vertices always has potential to be n-symmetric.)

I’m not particularly knowledgeable about this, but a friend of mine suggests that this may have connections to the theory of flag algebras (https://arxiv.org/pdf/1607.04741.pdf).

]]>The reason that the number of permutations of 5 is not a multiple of 4 is because the pair 25314 41352 is closed under reversal AND element-wise inversion, so it only generates 2 permutations from the seed instead of 4. (This happens for 9 too, but the number of such pairs is 150, an even number.)

Iterating through all permutations for large values is computationally intractable, so I checked some things probabilistically: none of 54,000,000 random permutations of 18 elements were 3-balanced, though with only one nontrivial datapoint it’s hard to tell what our expected density should be – there could easily be millions and I wouldn’t expect to have encountered any.

At least one 2-balanced permutation exists for every multiple of 4: if n=4k, we take (1,2,…k,4k-k+1,4k-k+2,…,4k,4k-k,4k-k-1,….k+2,k+1). I’m less sure about things which are 1 mod 4; the similar pattern I tried breaks down in the 20s.

The 3-balanced patterns for 9 have this nice structure if you plot them; you’ve got this symmetric structure about 5, with alternatingly increasing and decreasing runs of equal length. Sadly, this fails to be 3-balanced for 29, the next 3-valid number congruent to 1 mod 4.

]]>Now everyone knows A is Mafia and B is detective. So C is telling truth, hence innocent. Now if D is Mafia, since D knows there are only two Mafia, D would immediately knows that E is innocent. But D says D knows who E is, so D is not Mafia.

So E is Mafia.

A – Mafia

B – Detective

C – Innocent

D – Innocent

E – Mafia

Quite interesting, but once you write it down, it would be quite easy.

]]>I fixed the typo.

]]>The sequence of the number of 2-balanced permutations is submitted, but is not approved yet. I was planning to submit more sequences on the subject.

]]>I would expect these numbers to be a result of some simple computation, but they’re weird: 3836 has a factor of 137. Not sure what’s going on here, and it suggests a basic combinatorial argument probably won’t cut it (I was thinking something about viewing the permutation as a chosen breakpoint in a circular rearrangement of 1,…,n, since from that perspective everything is nice with respect to 2-balancing or its analogue, but this doesn’t seem to lead anywhere nice).

There’s also something going on with the allowable cycle decompositions in these things ((a,b,c) means cycles of lengths a, b, c with multiplicity):

* 4 only has (4) and (1,1,2)

* 5 has (1,4) (2,3) (1,1,1,2)

* 8 has (1,1,1,1,2,2) (1,1,1,5) (1,1,2,4) (1,1,3,3) (1,2,2,3) (2,2,2,2) (1,7) (2,6) (3,5) (4,4)

I haven’t thought about this to tell whether the ones that don’t occur are trivially rules out by some simple consideration.

Curiously enough, this sequence (both with added zeros and without) isn’t in the OEIS, nor is your list of nontrivial sizes; I would have expected something this simple to show up at least in the case of 2.

I can also confirm your results for 3-balanced permutations of sizes 9 and 10.

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