Circular and Reflectional Symmetry

Question

If you have two types of objects, $a$, and $b$. You have $n_1$ of the first object, $n_2$ of the second object. How many distinct ways can you rearrange these objects on a ring. Arrangements are only unique if the ring can't be rotated or flipped to form another arrangement.

Answer 1

Consulting the following fact sheet on necklaces and bracelets we get for the cycle index of the cyclic group

$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$

We will apply PET so we need with $k$ objects of type $A$ and $m$ objects of type $B$ for the cyclic component:

$$[A^k B^m] Z(C_{k+m}; A+B) = [A^k B^m] \frac{1}{k+m} \sum_{d|k+m} \varphi(d) (A^d + B^d)^{(k+m)/d} \\ = [A^k B^m] \frac{1}{k+m} \sum_{d|\gcd(k,m)} \varphi(d) (A^d + B^d)^{(k+m)/d} \\ = \frac{1}{k+m} \sum_{d|\gcd(k,m)} \varphi(d) [A^k B^m] (A^d + B^d)^{(k+m)/d} \\ = \frac{1}{k+m} \sum_{d|\gcd(k,m)} \varphi(d) [A^{k/d} B^{m/d}] (A + B)^{(k+m)/d}.$$

This is

$$\bbox[5px,border:2px solid #00A000]{P_{k,m} = \frac{1}{k+m} \sum_{d|\gcd(k,m)} \varphi(d) {(k+m)/d\choose k/d}.}$$

Now for reflectional i.e. dihedral symmetry we get for $k+m$ odd the extra contributiom

$$[A^k B^m] \frac{1}{2} (A+B) (A^2 + B^2)^{(k+m-1)/2}.$$

If $k$ is the odd one this is

$$[A^{k-1} B^m] \frac{1}{2} (A^2 + B^2)^{(k+m-1)/2} \\ = [A^{(k-1)/2} B^{m/2}] \frac{1}{2} (A + B)^{(k+m-1)/2}.$$

This is

$$\bbox[5px,border:2px solid #00A000]{ Q_{1,k,m} = \frac{1}{2} P_{k,m} + \frac{1}{2} {(k+m-1)/2 \choose m/2}.}$$

On the other hand if $m$ is the odd one we get

$$\bbox[5px,border:2px solid #00A000]{ Q_{2,k,m} = \frac{1}{2} P_{k,m} + \frac{1}{2} {(k+m-1)/2 \choose k/2}.}$$

With $k+m$ even they can both be odd or both be even. The contribution is

$$[A^k B^m] \frac{1}{4} (A+B)^2 (A^2 + B^2)^{(k+m)/2-1} + [A^k B^m] \frac{1}{4} (A^2 + B^2)^{(k+m)/2}.$$

If they are both odd we are left with just

$$[A^k B^m] \frac{1}{2} AB (A^2 + B^2)^{(k+m)/2-1} \\ = [A^{k-1} B^{m-1}] \frac{1}{2} (A^2 + B^2)^{(k+m)/2-1} \\ = [A^{(k-1)/2} B^{(m-1)/2}] \frac{1}{2} (A + B)^{(k+m)/2-1}.$$

We obtain

$$\bbox[5px,border:2px solid #00A000]{ Q_{3,k,m} = \frac{1}{2} P_{k,m} + \frac{1}{2} {(k+m)/2-1 \choose (k-1)/2}.}$$

The last one is when both are even and we obtain

$$[A^k B^m] \frac{1}{4} (A^2 + B^2) (A^2 + B^2)^{(k+m)/2-1} + [A^{k/2} B^{m/2}] \frac{1}{4} (A + B)^{(k+m)/2} \\ = [A^{k-2} B^m] \frac{1}{4} (A^2 + B^2)^{(k+m)/2-1} + [A^k B^{m-2}] \frac{1}{4} (A^2 + B^2)^{(k+m)/2-1} \\ + \frac{1}{4} {(k+m)/2\choose k/2} \\ = [A^{k/2-1} B^{m/2}] \frac{1}{4} (A + B)^{(k+m)/2-1} + [A^{k/2} B^{m/2-1}] \frac{1}{4} (A + B)^{(k+m)/2-1} \\ + \frac{1}{4} {(k+m)/2\choose k/2}.$$

This is

$$\bbox[5px,border:2px solid #00A000]{ \begin{array}{l} Q_{4,k,m} & = \frac{1}{2} P_{k,m} + \frac{1}{4} {(k+m)/2-1\choose m/2} + \frac{1}{4} {(k+m)/2-1\choose k/2} \\ & + \frac{1}{4} {(k+m)/2\choose k/2}. \end{array}}$$

There is some Maple code to verify this which computes the desired statistic from PET and from the closed form.

with(numtheory);
with(combinat);


pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res, k;

    res := ind;

    polyvars := indets(poly);
    indvars := indets(ind);

    for v in indvars do
        pot := op(1, v);

        subs1 :=
        [seq(polyvars[k]=polyvars[k]^pot,
             k=1..nops(polyvars))];

        subs2 := [v=subs(subs1, poly)];

        res := subs(subs2, res);
    od;

    res;
end;

pet_cycleind_cyclic :=
proc(n)
local d;

    add(phi(d)*a[d]^(n/d), d in divisors(n))/n;
end;

pet_cycleind_dihedral :=
proc(n)
local s;

    s := 1/2*pet_cycleind_cyclic(n);

    if(type(n, odd)) then
        s := s + 1/2*a[1]*a[2]^((n-1)/2);
    else
        s := s + 1/4*(a[1]^2*a[2]^((n-2)/2) + a[2]^(n/2));
    fi;

    s;
end;

CYCLIC_CIND :=
proc(k,m)
option remember;
local cind, sind;

    cind := pet_cycleind_cyclic(k+m);
    sind := pet_varinto_cind(A+B, cind);

    coeff(coeff(expand(sind), A, k), B, m);
end;

CYCLIC_X :=
proc(k, m)
local d;

    1/(k+m)*add(phi(d)*binomial((k+m)/d, k/d),
                d in divisors(gcd(k,m)));
end;


DIHEDRAL_CIND :=
proc(k,m)
option remember;
local cind, sind;

    cind := pet_cycleind_dihedral(k+m);
    sind := pet_varinto_cind(A+B, cind);

    coeff(coeff(expand(sind), A, k), B, m);
end;

DIHEDRAL_X :=
proc(k, m)
local d;
    if type(k+m, odd) then
        if type(k, odd) then
            1/2*CYCLIC_X(k,m)
            + 1/2*binomial((k+m-1)/2, m/2)
        else
            1/2*CYCLIC_X(k,m)
            + 1/2*binomial((k+m-1)/2, k/2)
        fi;
    else
        if type(k, odd) then
            1/2*CYCLIC_X(k,m)
            + 1/2*binomial((k+m)/2-1, (k-1)/2);
        else
            1/2*CYCLIC_X(k,m)
            + 1/4*binomial((k+m)/2-1, m/2)
            + 1/4*binomial((k+m)/2-1, k/2)
            + 1/4*binomial((k+m)/2, k/2);
        fi;
    fi;
end;