An introduction to relational complexity: …An introduction to relational complexity: background,...

An introduction to relational complexity: background,questions, and a few answers

Joshua Wiscons

California State University, Sacramento

Workshop on the Model Theory of Finite and PseudofiniteStructures

Joshua Wiscons Relational complexity

Relational Complexity

Let X = (X,R1, . . . ,Rm) be a finite relational structure.

Let L(X) be the language obtained by naming the definable subsets of allfinite powers of X.

Let Lk(X) be the language obtained by considering only the definablesubsets of Xi for i ≤ k.

DefinitionThe relational complexity of X is defined to be the least k such that for everyrelation R in L(X) there exists a quantifier-free formula ϕ in Lk(X) withX |= ∀x (R(x)↔ ϕ(x)).

Definition (via Homogeneity)The relational complexity of X is the least k such that X is equivalent to ahomogeneous structure (X, S1, . . . , Sn) with every Si of arity at most k.

1 rc(X) is really an invariant of X, hence of Aut(X) = Aut(X).2 The definable subsets of Xk are unions of orbits of Aut(X) (acting

diagonally) on Xk.

Definition (Fuzzy Reformulation)The relational complexity of X is defined to be the least k such that the orbitsof Aut(X) on Xk “determine” the orbits of Aut(X) on Xn for all finite n.

1 rc(X) is really an invariant of X, hence of Aut(X) = Aut(X).

2 The definable subsets of Xk are unions of orbits of Aut(X) (actingdiagonally) on Xk.

diagonally) on Xk.

Terminology

Let (X,G) be a permutation group.

1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in

the same orbit as the corresponding k elements from y.

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Note: 3-equivalenceimplies 2-equivalence

Terminology

Let (X,G) be a permutation group.

1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).

2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.

3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are inthe same orbit as the corresponding k elements from y.

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

Let (X,G) be a permutation group.1 Every g ∈ G acts diagonally on Xn via (x1, . . . , xn)g = (x1g, . . . , xng).2 The orbits of G on Xn are called n-types.3 Two tuples x, y ∈ Xn are k-equivalent if every k elements from x are in

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

(x1, x2, x3)

(y1, y2, y3)

(3-)equivalence

(x1, x2, x3)

(y1, y2, y3)

2-equivalence

Terminology

DefinitionThe relational complexity of a permutation group (X,G) is the smallest k suchthat for all n ≥ k, k-equivalence of n-tuples implies equivalence.

A First Example

Example (Petersen Graph)

Let G be the automorphism group of the graph.

Question: (1, 2, 3) ∼ (1, 5, 4)? Yes.(1, 2, 3)

(1, 5, 4)

?reflection: (25)(34)(7 10)(89)

Thus, 2-equivalence does not imply equivalence.

Thus, rc(G) > 2.

It turns out that rc(G) = 3.

RemarkBe aware that, for example, 2-equivalence may imply 3-equivalence withoutimplying 4-equivalence.

A First Example

Let G be the automorphism group of the graph.Question: (1, 2, 3) ∼ (1, 5, 4)?

(1, 2, 3)

(1, 5, 4)

reflection: (25)(34)(7 10)(89)

Thus, rc(G) > 2.

A First Example

Let G be the automorphism group of the graph.Question: (1, 2, 3) ∼ (1, 5, 4)? Yes.

(1, 2, 3)

(1, 5, 4)

reflection: (25)(34)(7 10)(89)

Thus, rc(G) > 2.

A First Example

Let G be the automorphism group of the graph.

Question: (1, 3, 7) ∼ (1, 3, 9)?(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼ (1, 3, 9)?

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼ (1, 3, 9)? No.

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)?

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

(1, 3, 7)

(1, 3, 9)

reflection

Thus, rc(G) > 2.

A First Example

(1, 3, 7)

(1, 3, 9)

reflection

Thus, rc(G) > 2.

A First Example

(1, 3, 7)

(1, 3, 9)

it works . . . but hard to see

Thus, rc(G) > 2.

A First Example

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

Let G be the automorphism group of the graph.Question: (1, 3, 7) ∼2 (1, 3, 9)? Yes!

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2.

A First Example

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2. It turns out that rc(G) = 3.

A First Example

(1, 3, 7)

(1, 3, 9)

Thus, rc(G) > 2. It turns out that rc(G) = 3.

Remarks And More Examples

RemarkLet (X,G) be a nontrivial permutation group.

1 rc(X,G) ≥ 22 rc(X,G) = max(2, rc′(X,G))

where rc′(X,G) is computed only for tuples with no repeated entries

3 rc(X,G) ≤ max(2, |X| − 1)

1 rc(X,G) ≥ 2

2 rc(X,G) = max(2, rc′(X,G))

3 rc(X,G) ≤ max(2, |X| − 1)

1 rc(X,G) ≥ 2

2 rc(X,G) = max(2, rc′(X,G))

3 rc(X,G) ≤ max(2, |X| − 1)

Proof.

Choose a nontrivial g ∈ G.

Choose distinct x, y ∈ X with xg = y.

Then (x, x) and (x, y) are 1-equivalent but not 2-equivalent.

1 rc(X,G) ≥ 2

2 rc(X,G) = max(2, rc′(X,G))

3 rc(X,G) ≤ max(2, |X| − 1)

Proof.

1 rc(X,G) ≥ 2

2 rc(X,G) = max(2, rc′(X,G))

3 rc(X,G) ≤ max(2, |X| − 1)

Proof.

1 rc(X,G) ≥ 2

2 rc(X,G) = max(2, rc′(X,G))

3 rc(X,G) ≤ max(2, |X| − 1)

Proof.

1 rc(X,G) ≥ 2

2 rc(X,G) = max(2, rc′(X,G))

3 rc(X,G) ≤ max(2, |X| − 1)

where rc′(X,G) is computed only for tuples with no repeated entries3 rc(X,G) ≤ max(2, |X| − 1)

3 rc(X,G) ≤ max(2, |X| − 1)

Proof.

Let x and y be (|X| − 1)-equivalent.

We may assume that neither tuple has repeated entries, so

x and y areenumerations of X.

If g ∈ G takes the first (|X| − 1) entries of x to the first (|X| − 1) entriesy, it must take the remaining entry of x to the remaining entry of y.

Thus, x and y are |X|-equivalent.

Proof.

We may assume that neither tuple has repeated entries, so

x and y areenumerations of X.

Proof.

We may assume that neither tuple has repeated entries, so x and y areenumerations of X.

Proof.

Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) =

22 rc(X,Alt(n)) =

Assume x and y are 2-equivalent.

x1 x2 · · · xn

y1 y2 · · · yn

· · ·

all distinct

Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) =

22 rc(X,Alt(n)) =

x1 x2 · · · xn

y1 y2 · · · yn

· · ·

all distinct

Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 2

2 rc(X,Alt(n)) =

x1 x2 · · · xn

y1 y2 · · · yn

· · ·

all distinct

2 rc(X,Alt(n)) =

x1 x2 · · · xn

y1 y2 · · · yn

· · ·

all distinct

2 rc(X,Alt(n)) =

x1 x2 · · · xn

y1 y2 · · · yn

· · ·

all distinct

2 rc(X,Alt(n)) =

x1 x2 · · · xn

y1 y2 · · · yn

· · ·

all distinct

2 rc(X,Alt(n)) =

x1 x2 · · · xn

y1 y2 · · · yn

?· · ·

all distinct

2 rc(X,Alt(n)) =

x1 x2 · · · xn

y1 y2 · · · yn

Of Course!· · ·

all distinct

2 rc(X,Alt(n)) =

Assume rc(X,Alt(n)) = r with 2 ≤ r ≤ n− 2.

Let x, y be any two enumerations of X.

Key point: (X,Alt(n)) is (n− 2)-transitive.

Thus, x and y are (n− 2)-equivalent.

So, if r ≤ n− 2, x and y are equivalent.

Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 22 rc(X,Alt(n)) =

Example (Let X = {1, . . . , n}.)1 rc(X,Sym(n)) = 22 rc(X,Alt(n)) = n−1

Some Problems (posed by Cherlin)

Problems

1 Gather data: determine the relational complexities of natural permutationgroups.

E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).

2 Classify the finite permutation groups of relational complexity at most r.

Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?

3 Classify the finite permutation groups of relational complexity at least rwhen r is “big” compared to |X|.

Problems1 Gather data: determine the relational complexities of natural permutation

groups.

E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).

groups.E.g. study the various natural actions of classical groups.

E.g, study the various natural actions of Sn (and An).2 Classify the finite permutation groups of relational complexity at most r.

groups.E.g. study the various natural actions of classical groups.E.g, study the various natural actions of Sn (and An).

2 Classify the finite permutation groups of relational complexity at most r.Need to restrict to so-called primitive groups.

E.g, what are the finite primitive permutation groups of relationalcomplexity 2?

2 Classify the finite permutation groups of relational complexity at most r.Need to restrict to so-called primitive groups.E.g, what are the finite primitive permutation groups of relationalcomplexity 2?

An easy, natural example

Example (GL(V))

Let V = Fd with char(F) > 2.

Then rc(V,GL(V)) = ?1 Lower bound:

rc(V,GL(V)) ≥ d + 1Let x = (e1, . . . , ed,

∑ei) and y = (e1, . . . , ed,

∑2ei).

Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.

2 Upper bound:

rc(V,GL(V)) ≤ d + 1

Example (GL(V))

Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?

1 Lower bound:

rc(V,GL(V)) ≥ d + 1Let x = (e1, . . . , ed,

∑ei) and y = (e1, . . . , ed,

∑2ei).

2 Upper bound:

rc(V,GL(V)) ≤ d + 1

Example (GL(V))

Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound:

rc(V,GL(V)) ≥ d + 1Let x = (e1, . . . , ed,

∑ei) and y = (e1, . . . , ed,

∑2ei).

2 Upper bound:

rc(V,GL(V)) ≤ d + 1

Example (GL(V))

rc(V,GL(V)) ≥ d + 1

Let x = (e1, . . . , ed,∑

ei) and y = (e1, . . . , ed,∑

2 Upper bound:

rc(V,GL(V)) ≤ d + 1

Example (GL(V))

rc(V,GL(V)) ≥ d + 1

Let x = (e1, . . . , ed,∑

ei) and y = (e1, . . . , ed,∑

2ei).Key point: GL(V) is transitive on the set of bases.

Thus, x and y are d-equivalent.But, x and y are not equivalent.

2 Upper bound:

rc(V,GL(V)) ≤ d + 1

Example (GL(V))

rc(V,GL(V)) ≥ d + 1

Let x = (e1, . . . , ed,∑

ei) and y = (e1, . . . , ed,∑

2ei).Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.

But, x and y are not equivalent.2 Upper bound:

rc(V,GL(V)) ≤ d + 1

Example (GL(V))

rc(V,GL(V)) ≥ d + 1

Let x = (e1, . . . , ed,∑

ei) and y = (e1, . . . , ed,∑

2ei).Key point: GL(V) is transitive on the set of bases.Thus, x and y are d-equivalent.But, x and y are not equivalent.

2 Upper bound:

rc(V,GL(V)) ≤ d + 1

Example (GL(V))

Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 1

Let x = (e1, . . . , ed,∑

ei) and y = (e1, . . . , ed,∑

2 Upper bound:

rc(V,GL(V)) ≤ d + 1

Example (GL(V))

Let x = (e1, . . . , ed,∑

ei) and y = (e1, . . . , ed,∑

2 Upper bound:

rc(V,GL(V)) ≤ d + 1

Example (GL(V))

Let x = (e1, . . . , ed,∑

ei) and y = (e1, . . . , ed,∑

2 Upper bound: rc(V,GL(V)) ≤ d + 1

Example (GL(V))

Let x = (e1, . . . , ed,∑

ei) and y = (e1, . . . , ed,∑

2 Upper bound: rc(V,GL(V)) ≤ d + 1

Example (GL(V))

Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = ?1 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1

Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that

- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to y

m + 1 equivalence implies that

- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym

=⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1

Repeating, we find that x′i = yi for all i ≥ mThus, x ∼ (y1, . . . , ym, x′m+1, . . . , x

′n) = y

Example (GL(V))

Want to show that (d + 1)-equivalence implies equivalence

Assume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that

′n) = y

Example (GL(V))

Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalent

Let m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independentm equivalence implies that

′n) = y

Example (GL(V))

Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)

Without loss, assume that x1, . . . , xm are independentm equivalence implies that

′n) = y

Example (GL(V))

Want to show that (d + 1)-equivalence implies equivalenceAssume that x = (x1, . . . , xn) and y = (y1, . . . , yn) are (d + 1)-equivalentLet m be the size of a maximal independent subset of x. (m ≤ d)Without loss, assume that x1, . . . , xm are independent

m equivalence implies that

′n) = y

Example (GL(V))

′n) = y

Example (GL(V))

- x = (x1, . . . , xm, xm+1, . . . , xn) ∼ (y1, . . . , ym, x′m+1, . . . , x′n) and

- (y1, . . . , ym, x′m+1, . . . , x′n) is (d + 1)-equivalent to ym + 1 equivalence implies that

′n) = y

Example (GL(V))

′n) = y

Example (GL(V))

′n) = y

Example (GL(V))

m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so

- g fixes y1, . . . , ym

′n) = y

Example (GL(V))

m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym

′n) = y

Example (GL(V))

m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym =⇒ g fixes span{y1, . . . , ym}

=⇒ x′m+1 = ym+1

′n) = y

Example (GL(V))

m + 1 equivalence implies that- there exists g ∈ GL(V) taking (y1, . . . , ym, x′m+1) to (y1, . . . , ym, ym+1), so- g fixes y1, . . . , ym =⇒ g fixes span{y1, . . . , ym} =⇒ x′m+1 = ym+1

′n) = y

Example (GL(V))

Repeating, we find that x′i = yi for all i ≥ m

Thus, x ∼ (y1, . . . , ym, x′m+1, . . . , x′n) = y

Example (GL(V))

′n) = y

Example (GL(V))

Let V = Fd with char(F) > 2. Then rc(V,GL(V)) = d + 11 Lower bound: rc(V,GL(V)) ≥ d + 12 Upper bound: rc(V,GL(V)) ≤ d + 1

′n) = y

A harder, natural example

DefinitionPm(n) is the set of all partitions of {1, . . . , n} into subsets of (equal) size m.

Example

P4(8) :

[1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) :

[12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

Of course, there are many partitions of a given set, e.g. |P3(9)| = 280.

Definition (Action on Partitions)

Every σ ∈ Sn can be naturally regarded as a permutation of Pm(n) via the rule

σ ([x1 · · · xm | y1 · · · ym | · · · ]) = [σ(x1) · · ·σ(xm) | σ(y1) · · ·σ(ym) | · · · ].

Example

P4(8) :

[1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) :

[12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

ExampleP4(8) :

[1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) :

[12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678],

[2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) :

[12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357],

. . . note: [1234 | 5678] = [8765 | 4321]

P2(8) :

[12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . .

note: [1234 | 5678] = [8765 | 4321]

P2(8) :

[12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) :

[12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) :

[12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) : [12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) : [12 | 34 | 56 | 78], . . .

P3(9) :

[147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) : [12 | 34 | 56 | 78], . . .

P3(9) : [147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) : [12 | 34 | 56 | 78], . . .

P3(9) : [147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) : [12 | 34 | 56 | 78], . . .

P3(9) : [147 | 258 | 369], . . .

ExampleP4(8) : [1234 | 5678], [2468 | 1357], . . . note: [1234 | 5678] = [8765 | 4321]

P2(8) : [12 | 34 | 56 | 78], . . .

P3(9) : [147 | 258 | 369], . . .

ProblemDetermine the relational complexity of Sn acting on Pm(n).

Some answers are known, but the reasons why aren’t so unclear.

P2(4) : [∗ ∗ | ∗ ∗] =⇒ rc(S4) = 2P2(6) : [∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S6) = 3P2(8) : [∗ ∗ | ∗ ∗ | ∗ ∗ | ∗ ∗] =⇒ rc(S8) = 4