Counting votes in coupled decisions

Abstract

We consider scenarios with distributed decision processes, e.g., coupled majorities and personal union in parliament chambers, supranational decisions and supervisory boards. When computing the adoption rate for reaching a decision in these scenarios, multiple linear inequality restrictions in combinatorial countings are present. These rates cannot be computed in closed form. We introduce a general method for incorporating multiple inequality conditions in multiple majority decisions, which significantly reduces the number of involved summations and removes restrictions on the summation indices. Exact solutions are provided through (a) integral representations which can be evaluated numerically, and (b) unrestricted, contracted sums over discrete events. Further, we provide methods to reduce the number of necessary summations by splitting or recurring the original problem to easier sub-problems. For five dedicated scenarios, full results are given which indeed require a single unrestricted summation only.

Appendices

Appendix 1: Proof of Theorem 3

In this section, we prove Theorem 3 (“Chamber”) and the formulae for the special cases. A proof for the special case R(u, K, K) has previously been presented in (Wendemuth and Simonelli 2013).

The following analysis transforms the triple-indexed sum 2.1 —with dependencies of the individual indexes amongst each other—into a double indexed sum with arbitrary bounds, enabling transforming into a double integral as well. The obtained double integral can be evaluated, resulting in a single unrestricted sum. This constitutes a novel, and the maximum possible, contraction of the triple-indexed sum 2.1.

[5] Let r and Q be positive integer numbers. Values for feasible and suitable r and Q are given later. Define complex variables

$$\begin{aligned} z_j= & {} \exp {(i 2 \pi j / r) } \end{aligned}$$

(7.1)

$$\begin{aligned} v_k= & {} \exp {(i 2 \pi k / r) } \end{aligned}$$

(7.2)

[6,7] We show that, for suitable r and Q,

$$\begin{aligned} R(u,K,L)= & {} {\frac{1}{r^2}} \sum _{j=0}^{r-1} \sum _{k=0}^{r-1} I_{j,k} \end{aligned}$$

(7.3)

$$\begin{aligned} I_{j,k}= & {} \left\{ \frac{{z_j}^{1/2}+ {z_j}^{-1/2}}{2} \right\} ^{K-u} \left\{ \frac{{v_k}^{1/2}+ {v_k}^{-1/2}}{2} \right\} ^{L-u} \nonumber \\&\times \left\{ \frac{(z_j v_k)^{1/2}+ (z_j v_k)^{-1/2}}{2} \right\} ^{u} \frac{1 - z_j^{Q+1}}{1 - z_j} \frac{1 - v_k^{Q+1}}{1 - v_k} (z_j v_k)^{1/2}\nonumber \\ \end{aligned}$$

(7.4)

Proof

[8,9] Expanding the powers with binomial factors, and interpreting the fractions as finite geometric series, results in the following expression. Note that sums without index ranges are unrestricted sums, where this is justified since in general ${N \atopwithdelims ()j} = 0$ for $j<0$ or $j > N$.

$$\begin{aligned} I_{j,k}= & {} 2^{(-L-K+u)} \sum _{l= 0}^{K-u}\sum _{m = 0}^{L-u}\sum _{n=0}^u \sum _{p=0}^Q \sum _{q=0}^Q {K-u \atopwithdelims ()l} {L-u \atopwithdelims ()m} {u \atopwithdelims ()n} \nonumber \\&\times z_j^{ -l-n+p+ (K+1)/2} \quad v_k^{-m-n+q+ (L+1)/2} \end{aligned}$$

(7.5)

[10] Performing in expression 7.5 the sum over j as expressed in Eq. (7.3), and inserting Eq. (7.1), results in the term

$$\begin{aligned} \sum _{j=0}^{r-1} z_j^{ -l-n+p+ (K+1)/2} = \sum _{j=0}^{r-1} \exp {\left\{ i (2 \pi / r) \, j \, (-l-n+p+ (K+1)/2) \right\} }\qquad \end{aligned}$$

(7.6)

Now, generally, for $M \ne 0$ and $r>| M|$, the identities hold:

$$\begin{aligned} \sum _{j=0}^{r-1} \exp {(i (2 \pi / r) \, j M)} = 0 \end{aligned}$$

(7.7)

whereas for $M = 0$

$$\begin{aligned} \sum _{j=0}^{r-1} \exp {(i (2 \pi / r) \, j M)} = r \end{aligned}$$

(7.8)

Using that, in Eq. (7.6), $M = -l-n+p+ (K+1)/2 \in \{- (K-1)/2, \ldots , (K+1)/2 +Q\}$, we have $|M| \le (K+1)/2 +Q$. Acting equivalently for $v_k$, $|M| \le (L+1)/2 +Q$. Hence, since $L \ge M$, if we choose

$$\begin{aligned} r > (L+1)/2 +Q \end{aligned}$$

(7.9)

we obtain for Eq. (7.6):

$$\begin{aligned} \sum _{j=0}^{r-1} z_j^{ -l-n+p+ (K+1)/2} = r \delta (-l-n+p+ (K+1)/2) \end{aligned}$$

(7.10)

with the $\delta$-series where $\delta (n) = 1$ for $n=0$, and $\delta (n) = 0$ for $n\ne 0$.

[12] Hence, using Eq. (7.10) (equivalently for $v_k$) for the removing of l and m in Eqs. (7.3) and (7.4) gives

$$\begin{aligned} R(u,K,L) = 2^{(-L-K+u)} \sum _{n=0}^u \sum _{p=0}^Q \sum _{q=0}^Q {\textstyle {K-u \atopwithdelims ()-n+p+ (K+1)/2} {L-u \atopwithdelims ()-n+q+ (L+1)/2} {u \atopwithdelims ()n} } \end{aligned}$$

[11] This is identical to Eq. (2.1) if one observes that the sum of the two committee parts is $p+ (K+1)/2$, respectively, $q+ (L+1)/2$ which, since $p \ge 0$ and $q \ge 0$, constitutes the required majorities. Hence, Q should be chosen large enough (we will see later that the actual value is not important) to cover all terms in the binomials, $Q \ge (L-1)/2$.

This completes the first part of the proof, showing that Eq. (7.3) is identical to Eq. (2.1), if one takes the latter finding and Eq. (7.9) into account:

$$\begin{aligned} L \le Q + (L+1)/2< r \end{aligned}$$

(7.11)

[14] Having found this equivalence, Eq. (2.1) can be immediately written, using Euler identities in Eq. (7.3) , and observing Eq. (7.11):

$$\begin{aligned} R(u,K,L)= & {} {\frac{1}{r^2}} \sum _{j=0}^{r-1} \sum _{k=0}^{r-1} \left\{ \cos (\pi j/r) \right\} ^{K-u} \left\{ \cos (\pi k /r) \right\} ^{L-u} \left\{ \cos (\pi (j+k)/r) \right\} ^{u} \nonumber \\&\times \frac{(1 - z_j^{Q+1}) \; (1 - v_k^{Q+1} )}{4 \sin (\pi j/r)\sin (\pi k/r) } \; (-1) \end{aligned}$$

(7.12)

[15] We can now choose r suitably such that as many terms as possible disappear. The choice which we follow here is

$$\begin{aligned} r = 2 (Q+1) > L \end{aligned}$$

(7.13)

Then, the last fractions become:

$$\begin{aligned}&\frac{(1 - z_j^{Q+1}) }{ \sin (\pi j/r)} = \frac{(1 - \exp (i 2 \pi j (Q+1)/r)) }{\sin (\pi j/r)} = \frac{(1 - \exp (i\pi j)) }{\sin (\pi j/r)} = i r \;(j=0) \\&\quad 1-\exp (i\pi j) = 0 \quad (j> 0, \mathrm{even}) \\&\quad 1-\exp (i\pi j) = 2 \quad (j> 0, \mathrm{odd}) \end{aligned}$$

Then, Eq. (7.12) is (only real terms must be kept):

$$\begin{aligned} R(u,K,L)=\frac{1}{4} - {\frac{1}{r^2}} \sum _{j,k=1, \mathrm{odd}}^{r-1} \frac{\scriptstyle \left\{ \cos (\pi j/r) \right\} ^{K-u} \left\{ \cos (\pi k /r) \right\} ^{L-u} \left\{ \cos (\pi (j+k)/r) \right\} ^{u} }{\sin (\pi j/r)\sin (\pi k/r) }\qquad \end{aligned}$$

(7.14)

Equation 7.14 is the desired result where sums decouple and the upper limit r can be freely chosen, only observing $r>L$ (Eq. (7.13)).

[13] Since r has no upper bound, one can let $r \rightarrow \infty$ and obtain an integral representation (note only every second term is taken in the sums):

$$\begin{aligned} R(u,K,L)=\frac{1}{4}-\frac{1}{4} \int _0^{1}\mathrm{d}x \int _0^{1}\mathrm{d}y \frac{\scriptstyle \left\{ \cos (\pi x) \right\} ^{K-u}\left\{ \cos (\pi y) \right\} ^{L-u}\left\{ \cos (\pi (x+y))\right\} ^{u}}{\sin (\pi x)\sin (\pi y)}\qquad \end{aligned}$$

(7.15)

This is already an integral representation, containing no sums, as required in Theorem 1. However, due to the denominator, the integrand diverges at the integration limits, which makes numerical computation difficult. We will shortly give an integral representation in Eq. (7.18) which does not suffer from this problem.

We now factorize the coupled cosine term in Eq. (7.15), using the triangular identity:

$$\begin{aligned} \cos (\pi (x+y)) = \cos (\pi x)\cos (\pi y) - \sin (\pi x)\sin (\pi y) \end{aligned}$$

and obtain upon expansion of the powers of this term, a factorized double integral:

$$\begin{aligned} R(u,K,L)= & {} \frac{1}{4}- \frac{1}{4} \sum _j{u\atopwithdelims ()j} (-1)^j \left\{ \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x)\right\} ^{K-j} \left\{ \sin (\pi x) \right\} ^{j-1}\right\} \nonumber \\&\times \left\{ \int _0^{1} \mathrm{d}y \left\{ \cos (\pi y) \right\} ^{L-j} \left\{ \sin (\pi y) \right\} ^{j-1} \right\} \end{aligned}$$

(7.16)

[16b] For even j, the integrals in Eq. (7.16) vanish since K and L are odd and since the integrands are antisymmetric about 1 / 2. Hence, we can write $j= 1+ 2\, q$ and obtain

$$\begin{aligned} R(u,K,L)= & {} \frac{1}{4} + \frac{1}{4} \sum _q {u \atopwithdelims (){1 + 2q}} \left\{ \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x) \right\} ^{K-1-2 q} \left\{ \sin (\pi x) \right\} ^{2 q} \right\} \nonumber \\&\times \left\{ \int _0^{1} \mathrm{d}y \left\{ \cos (\pi y) \right\} ^{L-1-2 q} \left\{ \sin (\pi y) \right\} ^{2 q} \right\} \end{aligned}$$

(7.17)

[16a] This allows us to globally symmetrize the integral representation 7.15, resulting in

$$\begin{aligned}&R(u,K,L)=\frac{1}{4} + \frac{1}{8} \int _0^{1}\mathrm{d}x \int _0^{1}\mathrm{d}y \frac{ \cos (\pi x)^{K-u} \cos (\pi y)^{L-u} \left\{ \cos (\pi (x-y))^{u} - \cos (\pi (x+y))^{u} \right\} }{\sin (\pi x)\sin (\pi y) }\nonumber \\ \end{aligned}$$

(7.18)

Performing the same expansions as before again results in Eq. (7.17). Beneficially, the integrand in Eq. (7.18) now does not diverge at the integration limits, and numerical integration can be performed without problems.

An example plot of the integrand in Eq. (7.18) is given in the main text in Fig. 4, for $u=3$ and $K=L=7$. Equation (7.18) can therefore be seen as an example of Theorem 1.

[17] We now continue by evaluating Eq. (7.17), where again discrete expressions will result.

The definite integrals in Eq. (7.17) can be evaluated in closed form. Several options are possible to achieve this. Euler’s Beta functions can be used (Abramowitz and Stegun 1972) since the definite integrals represent the general trigonometric form of Beta functions. The closed-form solution used here, without Beta functions, is, e.g., given in the manuscript notebooks of Gould (2010). The result is

$$\begin{aligned} B(K,q)= & {} \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x) \right\} ^{K-1-2 q} \left\{ \sin (\pi x) \right\} ^{2 q} \nonumber \\= & {} (2)^{(1-K)} \frac{ {K-1 \atopwithdelims ()(K-1)/2} {(K-1)/2 \atopwithdelims ()q}}{ {K-1 \atopwithdelims (){2q}}} \end{aligned}$$

(7.19)

Inserting this result (equivalently for the second integral) into Eq. (7.17) finally gives

$$\begin{aligned} R(u,K,L) \!=\! \frac{1}{4} \!+\! \frac{ {K-1 \atopwithdelims ()(K-1)/2} {L-1 \atopwithdelims ()(L-1)/2} }{2^{K+L}}\sum _q {u \atopwithdelims (){1 + 2q}} \frac{{(K-1)/2 \atopwithdelims ()q}}{ {K-1 \atopwithdelims (){2q}}} \frac{{(L-1)/2 \atopwithdelims ()q}}{ {L-1 \atopwithdelims (){2q}}}\qquad \end{aligned}$$

(7.20)

which equals the first theorem Eq. (3.3). $\square$

Appendix 2: Proof of theorem 4

[3] For the Proof of Theorem 4 (“Union”) as in Eq. (3.4), we recall the conditions given in 2.2: At least two out of three States with L, M and N citizens have to approve with simple majority, and there has to be a majority from all $K = L+M+N$ citizens. Let us symbolically denote occurence rates with calligraphic letters which correspond to the given population sizes, e.g., ${\mathcal {L}}$ will be the ratio of the number of all events which give the desired result (example: a majority in the State with L citizens) to the number of all possible events associated with the decision (in the example, $2^L$). Simultaneous observance of several such conditions is denoted by a product, e.g., ${\mathcal {L}} {\mathcal {K}}$ denotes majority in the State with L citizens and overall majority amongst the K citizens. As usual, ${{\bar{\mathcal {A}}}}$ denotes the ratio for the inverse of ${\mathcal {A}}$, with ${\mathcal {A}} + {\bar{\mathcal {A}}}= 1$.

Proof

With this notation, we carefully write all four conditions to obtain mutually exclusive products which can simply be added. We are to compute the ratio

$$\begin{aligned} S=\mathcal {K} \mathcal {L} \mathcal {M}{\bar{\mathcal {N}}} + \mathcal {K}\mathcal {L}{\bar{\mathcal {M}}}\mathcal {N} + \mathcal {K}{\bar{\mathcal {L}}} \mathcal {M}\mathcal {N} +\mathcal {K}\mathcal {L}\mathcal {M}\mathcal {N} \end{aligned}$$

(8.1)

Note ${\mathcal {K}\mathcal {L}\mathcal {M}\mathcal {N}=\mathcal {L}\mathcal {M}\mathcal {N}} = 1/8$ since if majorities in all three States with L,M and N citizens are obtained, majority in the total population with K citizens is satisfied automatically. The other terms, involving 4 conditions, can be reduced to terms involving three conditions as follows, by adding and subtracting $\mathcal {K}\mathcal {L}\mathcal {M}\mathcal {N}$ to the three first terms:

$$\begin{aligned} S = {{\mathcal {K}}{\mathcal {L}}{\mathcal {M}} + {\mathcal {K}}{\mathcal {L}}{\mathcal {N}} + {\mathcal {K}}{\mathcal {M}}{\mathcal {N}}} - 1/4 \end{aligned}$$

(8.2)

Now, we introduce a technique to reduce the terms with three majority conditions to terms with two conditions. Observe the following identity, which is obtained by marginalizing the terms:

$$\begin{aligned} {\mathcal {A}}{\mathcal {B}} + {\mathcal {B}}{\mathcal {C}} + {\mathcal {A}}{\mathcal {C}}= & {} 3 {\mathcal {A}}{\mathcal {B}}{\mathcal {C}} + {\mathcal {A}}{\mathcal {B}}{\bar{\mathcal {C}}} + {\bar{\mathcal {A}}}{\mathcal {B}}{\mathcal {C}} + {\mathcal {A}}{{\bar{\mathcal {B}}}}{\mathcal {C}}\nonumber \\= & {} -{\mathcal {A}}{\bar{\mathcal {B}}}{\bar{\mathcal {C}}} + {\bar{\mathcal {A}}}{\mathcal {B}}{\mathcal {C}} + 2 {\mathcal {A}}{\mathcal {B}}{\mathcal {C}} + ({\mathcal {A}}{\mathcal {B}}{\mathcal {C}} + {\mathcal {A}}{\mathcal {B}}{\bar{\mathcal {C}}} + {\mathcal {A}}{{\bar{\mathcal {B}}}}{\mathcal {C}} + {\mathcal {A}}{\bar{\mathcal {B}}}{\bar{\mathcal {C}}})\nonumber \\= & {} (-{\mathcal {A}}{\bar{\mathcal {B}}}{\bar{\mathcal {C}}} + {\bar{\mathcal {A}}}{\mathcal {B}}{\mathcal {C}}) + 2 {\mathcal {A}}{\mathcal {B}}{\mathcal {C}} + {\mathcal {A}} \end{aligned}$$

(8.3)

Now, note that the ratio for any product of majority conditions is identical to that of the product of the inverse conditions. For example, if this ratio is ${\mathcal {K}}{\mathcal {L}}$, then ${\mathcal {K}}{\mathcal {L}}= {\bar{\mathcal {K}}}{\bar{\mathcal {L}}}$. The reason is that each individual majority condition forms a halfspace in the space of indices of the product of binomial or multinomial conditions. Since the maximum index (e.g., population size N) is odd, the inverse condition forms the inverse halfspace. This has the same total ratio, since all binomials or multinomials are point-symmetric to N/2. Hence, a product of conditions will be the intersection of halfspaces which has identical total ratio than the intersection of the inverse halfspaces. Further, note that for a single majority condition ${\mathcal {A}}$, always ${\mathcal {A}} = 1/2$.

Hence, the first two terms cancel, and we obtain for majority conditions:

$$\begin{aligned} 2 {\mathcal {A}}{\mathcal {B}}{\mathcal {C}} = {\mathcal {A}}{\mathcal {B}} + {\mathcal {B}}{\mathcal {C}} + {\mathcal {A}}{\mathcal {C}} - 1/2 \end{aligned}$$

(8.4)

Applying Eqs. (8.4) to (8.2) gives:

$$\begin{aligned} 2 S = 2( {\mathcal {K}} {\mathcal {L}} + {\mathcal {K}}{\mathcal {M}} + {\mathcal {K}}{\mathcal {N}}) + {\mathcal {L}}{\mathcal {M}} + {\mathcal {L}}{\mathcal {N}} + {\mathcal {M}}{\mathcal {N}} -2 \end{aligned}$$

and, since the uncoupled terms give ${\mathcal {L}}{\mathcal {M}} ={\mathcal {L}}{\mathcal {N}} ={\mathcal {M}}{\mathcal {N}} = 1/4$,

$$\begin{aligned} S = {\mathcal {K}}{\mathcal {L}} + {\mathcal {K}}{\mathcal {M}} + {\mathcal {K}}{\mathcal {N}} -5/8. \end{aligned}$$

(8.5)

[4–17] We can now make use of Theorem 3.3, in the special case $u=L$ where u is a true subset of K. We have

$$\begin{aligned} {\mathcal {K}}{\mathcal {L}}= & {} R(L,K,L) = \frac{1}{4} \nonumber \\&+ \frac{ {K-1 \atopwithdelims ()(K-1)/2}}{2^K} \frac{ {L-1 \atopwithdelims ()(L-1)/2}}{2^{L}} \sum _q {L \atopwithdelims (){1 + 2q}} \frac{{(K-1)/2 \atopwithdelims ()q}}{ {K-1 \atopwithdelims (){2q}}} \frac{{(L-1)/2 \atopwithdelims ()q}}{ {L-1 \atopwithdelims (){2q}}} \end{aligned}$$

(8.6)

Inserting Eq. (8.6) (likewise for ${\mathcal {K}}{\mathcal {M}}$ and ${\mathcal {K}}{\mathcal {N}}$) into 8.5 directly gives Theorem 4 as in Eq. (3.4). $\square$

Appendix 3: Proof of Theorem 5

In this section, we prove Theorem 5 (“Parity”). We shall keep this proof short where the methods are identical to those in the proof of Theorem 3.

Proof

[5] Let r and Q be positive integer numbers as before, where values for feasible and suitable r and Q are also selected as before. As introduced in Eqs. (7.1) and (7.2), define complex variables $z_j$ and $v_k$.

[6,7] We show that, for suitable r and Q,

$$\begin{aligned} T(N)= & {} {\frac{1}{r^2}} \sum _{j=0}^{r-1} \sum _{k=0}^{r-1} I_{j,k}\end{aligned}$$

(9.1)

$$\begin{aligned} I_{j,k}= & {} \left\{ {z_j}^{3/2} {v_k}^{-1/2}+ 3 {z_j}^{1/2} {v_k}^{1/2}+ 3 {z_j}^{-1/2} {v_k}^{-1/2}+ {z_j}^{-3/2} {v_k}^{1/2} \right\} ^{N} \nonumber \\&\times 2^{(-3 N)} \frac{1 - z_j^{Q+1}}{1 - z_j} \frac{1 - v_k^{Q+1}}{1 - v_k} (z_j v_k)^{1/2} \end{aligned}$$

(9.2)

[8,9] Expanding the power with multinomial factors, and interpreting the fractions as finite geometric series, results in the following expression.

$$\begin{aligned} I_{j,k}= & {} 2^{(- 3 N)} \sum _{d, t, o = 0}^{N} \sum _{p=0}^Q \sum _{q=0}^Q {N \atopwithdelims (){d\; t \; o}} \; 3^{t+o} \nonumber \\&\times z_j^{-(3d + 2t + o) +p + (3N+1)/2} \quad v_k^{d+o+q -(N-1)/2} \end{aligned}$$

(9.3)

[10] Performing in expression 9.3 the sum over j as expressed in Eq. (9.1), and inserting Eq. (7.1), results in the term

$$\begin{aligned}&\sum _{j=0}^{r-1}z_j^{-(3d+2t+o)+p+(3N+1)/2}\nonumber \\&\quad =\sum _{j=0}^{r-1}\exp {\left\{ i(2\pi /r)\,j\,(-(3d+ 2t+o)+p+(3N+1)/2)\right\} } \end{aligned}$$

(9.4)

Now, generally, for $M \ne 0$ and $r>| M|$, the $\delta$-identities 7.7 and 7.8 hold. Using them for large enough r, we obtain for Eq. (9.4):

$$\begin{aligned} \sum _{j=0}^{r-1} z_j^{-(3d + 2t + o) +p + (3N+1)/2} = r \delta (-(3d + 2t + o) +p + (3N+1)/2) \end{aligned}$$

(9.5)

[11] Performing now also the sum over p gives

$$\begin{aligned} \sum _{p=0}^{Q} \sum _{j=0}^{r-1} z_j^{-(3d + 2t + o) +p + (3N+1)/2}= & {} r \sum _{p=0}^{Q} \delta (-(3d + 2t + o) +p + (3N+1)/2)\nonumber \\= & {} r \; \varTheta ((3d + 2t + o) - (3N+1)/2) \end{aligned}$$

(9.6)

with the usual unit step series $\varTheta (n) = 1$ if $n >= 0$, and 0 otherwise.

Likewise, performing the sum over q and k gives

$$\begin{aligned} \sum _{q=0}^{Q} \sum _{k=0}^{r-1} v_k^{d + o +q - (N-1)/2}= & {} r \sum _{q=0}^{Q} \delta (-(d + o) -q + (N-1)/2)\nonumber \\= & {} r \; \varTheta (-(d + o) + (N-1)/2) \end{aligned}$$

(9.7)

which expresses minority for odd parity, i.e., it exactly satisfies the condition for majority with even parity. Hence, Q should be chosen large enough (we will see later that the actual value is not important) to cover all terms in the binomials, $Q \ge (N-1)/2$.

[12] This completes the first part of the proof, showing that Eq. (9.1) is identical to Eq. (2.3), if one takes into account the latter finding, and chooses large enough r, expressed equivalently in Eq. (7.11).

[14,15] Having found this equivalence, Eq. (2.3) can be immediately written, using Euler identities in Eq. (9.1):

$$\begin{aligned} T(N)= & {} {\frac{1}{4^N r^2}} \sum _{j=0}^{r-1} \sum _{k=0}^{r-1} \left\{ \cos ((3 j - k) \pi /r) + 3 \cos ((j + k) \pi /r) \right\} ^{N} \nonumber \\&\times \frac{\left( 1 - z_j^{Q+1}\right) \; \left( 1 - v_k^{Q+1}\right) }{4 \sin (\pi j/r)\sin (\pi k/r) } \; (-1) \end{aligned}$$

(9.8)

We can now choose r suitably such that as many terms as possible disappear. The choice follows the same rationale which was leading to Eq. (7.13). Then, Eq. (9.8) is:

$$\begin{aligned} T(N)=\frac{1}{4} - {\frac{1}{4^N r^2}}\sum _{j=1, \mathrm{odd}}^{r-1} \sum _{k=1, \mathrm{odd}}^{r-1} \frac{\left\{ \cos ((3j-k)\pi /r) + 3 \cos ((j + k) \pi /r) \right\} ^{N} }{\sin (\pi j/r)\sin (\pi k/r)}\nonumber \\ \end{aligned}$$

(9.9)

[13] Since r has no upper bound, one can let $r \rightarrow \infty$ and obtain an integral representation (note only every second term is taken in the sums):

$$\begin{aligned} T(N)=\frac{1}{4}- \frac{1}{4^{N+1}} \int _0^{1} \mathrm{d}x \int _0^{1} \mathrm{d}y \frac{ \left\{ \cos ((3 x - y) \pi ) + 3 \cos ((x + y) \pi ) \right\} ^{N} }{\sin (\pi x)\sin (\pi y) }\nonumber \\ \end{aligned}$$

(9.10)

We rewrite the coupled cosine terms with the triangular identity:

$$\begin{aligned}&\textstyle \cos ((3 x - y) \pi ) + 3 \cos ((x + y) \pi ) \nonumber \\&\quad =4 (\cos (\pi x))^3 \cos (\pi y) - 4 (\sin (\pi x))^3\sin (\pi y) \end{aligned}$$

(9.11)

and obtain upon expansion of the powers of this term, a factorized double integral:

$$\begin{aligned} T(N)= & {} \frac{1}{4}- \frac{1}{4} \sum _j {N \atopwithdelims ()j} (-1)^j \left\{ \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x) \right\} ^{3(N-j)} \left\{ \sin (\pi x) \right\} ^{3j-1} \right\} \nonumber \\&\times \left\{ \int _0^{1} \mathrm{d}y \left\{ \cos (\pi y) \right\} ^{N-j} \left\{ \sin (\pi y) \right\} ^{j-1} \right\} \end{aligned}$$

(9.12)

[16b] For even j, the second integral in Eq. (9.12) vanishes since N is odd and since the integrand is antisymmetric about 1 / 2. Hence, we can write $j= 1+ 2\, q$ and obtain

$$\begin{aligned} T(N)= & {} \frac{1}{4}+ \frac{1}{4} \sum _q {N \atopwithdelims (){1 + 2q}} \left\{ \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x) \right\} ^{3(N-1-2q)} \left\{ \sin (\pi x) \right\} ^{6q+2} \right\} \nonumber \\&\times \left\{ \int _0^{1} \mathrm{d}y \left\{ \cos (\pi y) \right\} ^{N-1-2q} \left\{ \sin (\pi y) \right\} ^{2q} \right\} \end{aligned}$$

(9.13)

[16a] With these symmetry considerations, we can again obtain from Eq. (9.10), where the integrand diverges at the integration limits, the following symmetrized version which does not suffer from divergence problems and which can be seen as another example for Theorem 1.

$$\begin{aligned}&T(N)=\frac{1}{4} + \frac{1}{4^{N+1}} \int _0^{1} \mathrm{d}x \int _0^{1} \mathrm{d}y \nonumber \\&\quad \times \frac{\left\{ \cos ((3 x + y) \pi ) + 3 \cos ((x - y) \pi ) \right\} ^{N}-\left\{ \cos ((3 x - y) \pi ) + 3 \cos ((x + y) \pi ) \right\} ^{N}}{2 \sin (\pi x)\sin (\pi y)}\nonumber \\ \end{aligned}$$

(9.14)

[17] We now continue by evaluating Eq. (9.13), where again discrete expressions will result. The definite integrals in Eq. (9.13) can again be evaluated in closed form, as already known from Eq. (7.19). The result is

$$\begin{aligned} \int _0^{1} \mathrm{d}y \left\{ \cos (\pi y) \right\} ^{N-1-2 q} \left\{ \sin (\pi y) \right\} ^{2 q} = \frac{ {N-1 \atopwithdelims ()(N-1)/2} { (N-1)/2 \atopwithdelims ()q}}{ 2^{\, N-1} \; {N-1 \atopwithdelims (){2q}}} \end{aligned}$$

(9.15)

and likewise, identifying the exponents,

$$\begin{aligned} \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x) \right\} ^{3(N-1-2q)} \left\{ \sin (\pi x) \right\} ^{6q+2} = \frac{ {3N -1 \atopwithdelims ()(3N-1)/2} {(3N-1)/2 \atopwithdelims ()3q+1}}{ 2^{\, 3N-1} \; {3N -1 \atopwithdelims (){6q+2}}} \end{aligned}$$

(9.16)

Inserting these results into Eq. (9.13) finally gives the Theorem 5, Eq. (3.5). $\square$

In Theorem 5, we have that $T(N=1) = 1/8$, monotonically rising to $T(N\rightarrow \infty ) \rightarrow 1/4$. The latter can be shown by applying the central limit theorem for multinomial distributions directly in Eq. (2.3).

Note that Theorem 5 is a very comfortable result with only one unrestricted sum, which was made possible using the triangular identity Eq. (9.11). As a remark, it is not clear whether such identities can be found in other interesting cases. Effort should be invested in exploiting such identities. In the next section, we will give a corresponding example.

Appendix 4: Proof of Theorem 6

In this section, we prove Theorem 6 (“Families”).

In contrast to Theorem 5 (“Parity”), families take majority decisions where 2 or 3 family members must approve. Then, following the same steps as in the derivation of Theorem 5, the coupled cosine terms—analogous to Eq. (9.11)—would read, with a plus-sign (instead of a minus-sign) in the first term:

$$\begin{aligned} \cos ((3 x + y) \pi ) + 3 \cos ((x + y) \pi ) \end{aligned}$$

Decoupling the variables in this expression leads to 3 terms as follows

$$\begin{aligned}&\cos ((3 x + y) \pi ) + 3 \cos ((x + y) \pi ) \\&\quad = 4 (\cos (\pi x))^3 \cos (\pi y) + 4 (\sin (\pi x))^3\sin (\pi y) - 6 \sin (\pi x) \sin (\pi y) \end{aligned}$$

Performing the expansion of the Nth power of this term (analogously to Eq. (9.10)) gives

$$\begin{aligned} \sum _j \sum _k {N \atopwithdelims (){j \; k}} 4^{j+k} (-6)^{N-j-k} (\cos (\pi x))^{3 j} (\sin (\pi x))^{N - j + 2 k} (\cos (\pi y))^j (\sin (\pi y))^{N - j}\nonumber \\ \end{aligned}$$

(10.1)

and consequently the result will contain, after x- and y-integrations as in Eq. (9.12), a double (!) sum which we want to avoid. We will therefore introduce now a method to overcome this problem and to re-instate the desired single summation again. This method consists in splitting the (o, t, d)-summations in two parts, both of which can be computed using a single sum only.

Proof

[4] For the proof, let us recall the two conditions that we must meet in the summations:

$$\begin{aligned} \begin{aligned}&o+2t+3d\ge (3N +1)/2 \\&\quad t +d \ge (N+1)/2 \end{aligned} \end{aligned}$$

(10.2)

It is now favorable to rewrite these conditions by introducing vector notation. To avoid special treatment of the boundary (the equality conditions), we can take advantage of the fact that N is odd and introduce boundaries that will never be met. Let ${\mathbf x} = (o,t,d)$, then

$$\begin{aligned} \begin{aligned}&{\mathbf a}{\mathbf x} > 3 N/2 \quad {\text{ with }} \quad {\mathbf a} = (1,2,3) \\&\quad {\mathbf b}{\mathbf x} > N/2 \quad {\text{ with }} \quad {\mathbf b} = (0,1,1) \end{aligned} \end{aligned}$$

(10.3)

Here, we generalize the line of arguments, introducing a lemma:

Lemma 1

(Subspace splitting) In a real-valued vector space ${\mathbf x}$, let a subspace be defined by two given inequalities

$$\begin{aligned} \begin{aligned} {\mathbf a}{\mathbf x}&\ge A \\ {\mathbf b}{\mathbf x}&\ge B \end{aligned} \end{aligned}$$

(10.4)

with any vectors ${\mathbf a}$ and ${\mathbf b}$ and real-valued constants A and B. Now, we can construct any new vector ${\mathbf c} = q_a {\mathbf a} - q_b {\mathbf b} \ne {\mathbf 0}$ as a weighted difference of ${\mathbf a}$ and ${\mathbf b}$, with real-valued weights $q_a > 0$ and $q_b > 0$. Then, the subspace defined by the Eq. (10.4) can be split into two subspaces as follows:

$$\begin{aligned} {\mathbf b}{\mathbf x}\ge & {} B\nonumber \\ {\mathbf c}{\mathbf x}\ge & {} q_a A - q_b B \end{aligned}$$

(10.5)

and

$$\begin{aligned} {\mathbf a}{\mathbf x}\ge & {} A\nonumber \\ {\mathbf c}{\mathbf x}\le & {} q_a A - q_b B \end{aligned}$$

(10.6)

Special provision must be taken that the boundary case ${\mathbf c}{\mathbf x} = q_a A - q_b B$, if not of measure 0, is not counted twice.

Proof

It is obvious that from the Eq. (10.4) two subspaces can be constructed by marginalizing these inequalities with a third inequality ${\mathbf c}{\mathbf x} \ge q_a A - q_b B$, or, respectively, ${\mathbf c}{\mathbf x} \le q_a A - q_b B$. These subspaces, if united, again construct the subspace given by the Eq. (10.4). The boundary case ${\mathbf c}{\mathbf x} = q_a A - q_b B$ needs special treatment to avoid double counting. The following proof shows that in these two cases, one of the three inequalities indeed follows from the two others, resulting in Eqs. (10.5) and (10.6).

In the first case 10.5, we have ${\mathbf c}{\mathbf x} \ge q_a A - q_b B$ and ${\mathbf b}{\mathbf x} \ge B$. Hence, from the definition of ${\mathbf c}$,

$$\begin{aligned} q_a {\mathbf a}{\mathbf x}= & {} {\mathbf c}{\mathbf x} + q_b {\mathbf b}{\mathbf x} \\\ge & {} q_a A - q_b B + q_b B = q_a A \end{aligned}$$

Hence, the first condition in 10.4, that is ${\mathbf a}{\mathbf x} \ge A$, is automatically satisfied.

In the second case 10.6, we have $- {\mathbf c}{\mathbf x} \ge - q_a A + q_b B$ and ${\mathbf a}{\mathbf x} \ge A$. Hence, from the definition of ${\mathbf c}$,

$$\begin{aligned} q_b {\mathbf b}{\mathbf x}= & {} - {\mathbf c}{\mathbf x} + q_a {\mathbf a}{\mathbf x}\\\ge & {} q_b B - q_a A + q_a A = q_b B \end{aligned}$$

Hence, the second condition in 10.4, that is ${\mathbf b}{\mathbf x} \ge B$, is automatically satisfied. This proves the lemma. $\square$

We now return to the proof of Theorem 6. Applying Lemma 1 to the Eq. (10.3), we choose $q_a=1$ and $q_b=2$, resulting in ${\mathbf c}=(1,0,1)$. The newly introduced boundary becomes ${\mathbf c}{\mathbf x} = d + o = (3N)/2- 2 N/2 = N/2$. Since N is odd, this boundary will never be met, hence no double counting problem arises. Then, the Eqs. (10.5) and 10.6 give

$$\begin{aligned} \text{ subspace } \text{ A: }&3d + 2t + o \ge (3N +1)/2 \\&d + o \le (N-1)/2\\ \text{ subspace } \text{ B: }&d + t \ge (N+1)/2 \\&d + o \ge (N+1)/2 \end{aligned}$$

Inspecting subspace A, we see that it is exactly what is occupied by the “Parity” majority decisions, which sums to a portion of T(N) of all states. Inspecting subspace B, we must perform the computation.

[5–7] Proceeding as before, we obtain the corresponding fraction of states B(N):

$$\begin{aligned} B(N)= & {} {\frac{1}{r^2}} \sum _{j=0}^{r-1} \sum _{k=0}^{r-1} I_{j,k} \end{aligned}$$

(10.7)

$$\begin{aligned} I_{j,k}= & {} \left\{ {z_j}^{1/2} {v_k}^{1/2}+ 3 {z_j}^{1/2} {v_k}^{-1/2}+ 3 {z_j}^{-1/2} {v_k}^{1/2}+ {z_j}^{-1/2} {v_k}^{-1/2} \right\} ^{N} \nonumber \\&\times 2^{(-3 N)} \frac{1 - z_j^{Q+1}}{1 - z_j} \frac{1 - v_k^{Q+1}}{1 - v_k} {z_j}^{1/2} {v_k}^{1/2} \end{aligned}$$

(10.8)

[8–12] With the same methods as used before after Eq. (9.2), it is easy to see that this expression counts the fraction of states in subspace B.

[14,15] Using Euler identities in Eq. (10.7) gives

$$\begin{aligned} B(N)= & {} {\frac{1}{4^{N} r^2}} \sum _{j=0}^{r-1} \sum _{k=0}^{r-1} \left\{ \cos ((j + k) \pi /r) + 3 \cos ((j - k) \pi /r) \right\} ^{N} \nonumber \\&\times \frac{(1 - z_j^{Q+1}) \; (1 - v_k^{Q+1} )}{4 \sin (\pi j/r)\sin (\pi k/r) } \; (-1) \end{aligned}$$

(10.9)

We can now choose r suitably such that as many terms as possible disappear. The choice follows the same rationale which was leading to Eq. (7.13). Then, Eq. (10.9) is (only real terms must be kept):

$$\begin{aligned} B(N)=\frac{1}{4} - {\frac{1}{4^{N} r^2}} \sum _{j=1, \mathrm{odd}}^{r-1} \sum _{k=1, \mathrm{odd}}^{r-1} \frac{ \left\{ \cos ((j + k) \pi /r) + 3 \cos ((j - k) \pi /r) \right\} ^{N} }{\sin (\pi j/r)\sin (\pi k/r)}\nonumber \\ \end{aligned}$$

(10.10)

[13] Since r has no upper bound, one can let $r \rightarrow \infty$ and obtain an integral representation (note only every second term is taken in the sums):

$$\begin{aligned} B(N) = \frac{1}{4} - \frac{1}{4^{N+1}} \int _0^{1}\mathrm{d}x \int _0^{1}\mathrm{d}y \frac{ \left\{ \cos ((x + y) \pi ) + 3 \cos ((x - y) \pi ) \right\} ^{N} }{\sin (\pi x)\sin (\pi y) }\nonumber \\ \end{aligned}$$

(10.11)

The coupled cosine terms in Eq. (10.11) have to be factorized to factorize the double sum/double integral. We rewrite the coupled cosine terms with the triangular identity:

$$\begin{aligned} \cos ((x + y) \pi ) + 3 \cos ((x - y) \pi ) = 4 \cos (\pi x) \cos (\pi y) + 2 \sin (\pi x) \sin (\pi y)\nonumber \\ \end{aligned}$$

(10.12)

and obtain upon expansion of the powers of this term, a factorized double integral:

$$\begin{aligned} B(N)= & {} \frac{1}{4} - \frac{1}{4} \sum _j {N \atopwithdelims ()j} 2^{-j} \left\{ \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x) \right\} ^{(N-j)} \left\{ \sin (\pi x) \right\} ^{j-1} \right\} \nonumber \\&\times \left\{ \int _0^{1} \mathrm{d}y \left\{ \cos (\pi y) \right\} ^{N-j} \left\{ \sin (\pi y) \right\} ^{j-1} \right\} \end{aligned}$$

(10.13)

[16b] For even j, the integrals in Eq. (10.13) vanish since N is odd and since the integrand is antisymmetric about 1 / 2. Hence, we can write $j= 1+ 2\, q$ and obtain

$$\begin{aligned} B(N)= & {} \frac{1}{4} - \frac{1}{8} \sum _q \frac{ {N \atopwithdelims (){1 + 2q}} }{ 4^{q} } \left\{ \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x) \right\} ^{N-1-2q} \left\{ \sin (\pi x) \right\} ^{2q} \right\} ^2 \end{aligned}$$

(10.14)

[16a] With these symmetry considerations, we can again obtain from Eq. (10.11), where the integrand diverges at the integration limits, the following symmetrized version which does not suffer from divergence problems and which can be seen as another example for 1.

$$\begin{aligned} B(N)= & {} \frac{1}{4} + \frac{1}{4^{N+1}} \int _0^{1} \mathrm{d}x \int _0^{1} \mathrm{d}y \nonumber \\&\times \frac{\left\{ \cos ((x - y) \pi ) + 3 \cos ((x + y) \pi ) \right\} ^{N} - \left\{ \cos ((x + y) \pi ) + 3 \cos ((x - y) \pi ) \right\} ^{N} }{2 \sin (\pi x)\sin (\pi y) }\nonumber \\ \end{aligned}$$

(10.15)

[17] The definite integrals in Eq. (10.14) can again be evaluated in closed form, as already known from Eq. (7.19). Inserting this into Eq. (10.14) finally gives the result for subspace B,

$$\begin{aligned} B(N) = \frac{1}{4} - \frac{1}{2^{2 N +1}} {N-1 \atopwithdelims ()(N-1)/2}^2 \sum _q \frac{1}{4^q} {N \atopwithdelims (){1 + 2q}} \frac{{(N-1)/2 \atopwithdelims ()q}^2}{ {N-1 \atopwithdelims (){2q}}^2}\qquad \end{aligned}$$

(10.16)

From this, we can now compile the terms for U(N)

$$\begin{aligned} U(N)= & {} T(N) + B(N)\nonumber \\= & {} \frac{1}{2} + \frac{1}{2^{4 N}} {N-1 \atopwithdelims ()(N-1)/2} {3N-1 \atopwithdelims ()(3N-1)/2} \sum _q {N \atopwithdelims (){1 + 2q}} \frac{{(N-1)/2 \atopwithdelims ()q}}{ {N-1 \atopwithdelims (){2q}}} \frac{{(3N-1)/2 \atopwithdelims ()3q+1}}{ {3N-1 \atopwithdelims (){6q+2}}} \nonumber \\&-\frac{1}{2^{2 N +1}} {N-1 \atopwithdelims ()(N-1)/2}^2 \sum _q \frac{1}{4^q} {N \atopwithdelims (){1 + 2q}} \frac{{(N-1)/2 \atopwithdelims ()q}^2}{ {N-1 \atopwithdelims (){2q}}^2} \end{aligned}$$

(10.17)

This can be favorably computed with one sum, giving the final result for Theorem 6, Eq. (3.6). $\square$

In Theorem 6, $U(N=1)=1/2$, monotonically decreasing to $U(N\rightarrow \infty )\rightarrow 5/12\simeq 0.417$. The latter can be shown by applying the central limit theorem for multinomial distributions directly in Eq. (2.4).

Appendix 5: Proof of Theorem 7

[3] For the proof of Theorem 7 (“Egality”) as in Eq. (3.7), we observe that our general transformation method, applied directly to the sevenfold sum with four conditions, would result in multiple (actually four) summations. Hence, we aim at recurring the problem into a sequence of sub-problems which will result in single summations.

The motivation for the generation of sub-problems is as follows. Considering the conditions for the summations in Eq. (2.5), we introduce a vector notation for the summation variables and write ${\mathbf n} = (n_0, n_1, \ldots , n_7)$. Then, the four conditions read

$$\begin{aligned} \begin{aligned} {\mathbf n} \cdot {\mathbf m} > 0 ; \quad {\mathbf m}&= (-1,-1,-1,-1,+1,+1,+1,+1) \\ {\mathbf n} \cdot {\mathbf w} > 0 ; \quad {\mathbf w}&= (-1,-1,+1,+1,-1,-1,+1,+1)\\ {\mathbf n} \cdot {\mathbf c} > 0 ; \quad {\mathbf c}&= (-1,+1,-1,+1,-1,+1,-1,+1) \\ {\mathbf n} \cdot {\mathbf f} > 0 ; \quad {\mathbf f}&=(-1,-1,-1,+1,-1,+1,+1,+1) \end{aligned} \end{aligned}$$

(11.1)

where the notation of the vectors ${\mathbf m}$, ${\mathbf w}$, ${\mathbf c}$, ${\mathbf f}$ stands for the conditions for men, women, children, families. Now, we note that ${\mathbf m}$, ${\mathbf w}$, ${\mathbf c}$ are pairwise normal whereas ${\mathbf f}$ is not. The vector ${\mathbf f}$ can be decomposed as

$$\begin{aligned} {\mathbf f} = \frac{1}{2} ( {\mathbf m} + {\mathbf w} + {\mathbf c} + {\mathbf p} ) ;\quad {\mathbf p} = (+1,-1,-1,+1,-1,+1,+1,-1) \end{aligned}$$

where now the four vectors ${\mathbf m}$, ${\mathbf w}$, ${\mathbf c}$, ${\mathbf p}$ are pairwise normal; i.e., ${\mathbf p}$ is the component of ${\mathbf f}$ normal to the space spanned by ${\mathbf m}$, ${\mathbf w}$, ${\mathbf c}$. Symbolically, the notation ${\mathbf p}$ stands for “parity” since the condition ${\mathbf n} \cdot {\mathbf p} > 0$ selects exactly those tuples $n_i$ where 2 or 0 family members will approve. The family condition in Eq. (11.1) can now be written as

$$\begin{aligned} 2 {\mathbf n} \cdot {\mathbf f} = {\mathbf n} \cdot ({\mathbf m} + {\mathbf w} + {\mathbf c} + {\mathbf p} ) > 0 \end{aligned}$$

(11.2)

We have now expressed the required conditions, Eq. (11.1), in terms of pairwise orthogonal vectors. Note the extraction of the “parity” condition would have been hard to see without the vector notation. Pairwise orthogonality is beneficial since it will lead to a decoupling in the counting of the occurence ratios. To see this, consider the case if we had only the three conditions for men, women and children: then it were obvious that the occurence ratios could be counted independent of each other. With the fourth family condition, we have now moved the problem into the eight-dimensional ${\mathbf n}$-space where the dimensions represent events which occur independently of each other, except for the sum of all events being N.

To capture the occurence rates, we again symbolically denote them with calligraphic letters as in Appendix 2. For example, ${\mathcal {M}}$ will be the ratio of the number of all events which give the desired result (here: at least half of the men approving, i.e., ${\mathbf n} \cdot {\mathbf m} > 0$ ) to the number of all possible events associated with the decision (in the example, $2^N$). Simultaneous observance of several such conditions is denoted by a product. ${{\bar{\mathcal {A}}}}$ denotes the ratio for the inverse of ${\mathcal {A}}$, with ${\mathcal {A}} + {\bar{\mathcal {A}}} = 1$.

Proof

With the introduced notations, we start the proof. We are now to compute the occurence rate

$$\begin{aligned} V = {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal F \end{aligned}$$

(11.3)

Now, we can apply a method, also demonstrated in Appendix 2, to decompose this occurence rate into rates which can actually be computed with single summations. We shall do so in the following lemma:

Lemma 2

(Subspace recurrence) The relation holds:

$$\begin{aligned} V = {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal F = \frac{3}{16} - \frac{1}{2} {{\bar{\mathcal {M}}}} \mathcal F + \frac{1}{4} {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P \end{aligned}$$

(11.4)

Hence, computing the occurence rate V can be recurred to computing two simpler relations which, as we shall see, can be done with single summations.

Proof

We marginalize the condition 11.3 with respect to the parity condition $\mathcal P$:

$$\begin{aligned} V = {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P \mathcal F + {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} {\bar{\mathcal P}} \mathcal F = {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P + {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} {\bar{\mathcal P}} \mathcal F \end{aligned}$$

(11.5)

where we used that if the fourfold condition ${\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P$ is obeyed, than the condition $\mathcal F$ is obeyed automatically by Eq. (11.2). Further, we consider $W = {\mathcal {M}} {\mathcal {W}} \mathcal F$ since we are able to compute this occurence ratio with three conditions, using Eq. (8.4). We marginalize W with respect to ${\mathcal {C}}$ and $\mathcal P$:

$$\begin{aligned} W = {\mathcal {M}} {\mathcal {W}} \mathcal F = {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P + {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} {\bar{\mathcal P}} \mathcal F + {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} \mathcal P \mathcal F + {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F \end{aligned}$$

(11.6)

Combining Eqs. (11.5) and (11.6) results in

$$\begin{aligned} 2\,V=W + {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P + ({\mathcal {M}} {\mathcal {W}} {\mathcal {C}} {\bar{\mathcal P}} \mathcal F - {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} \mathcal P \mathcal F) - {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F \end{aligned}$$

(11.7)

With Eq. (11.7), we now have found an expression in which we can show, by symmetry arguments, the following relations for the individual terms:

$$\begin{aligned}&{\mathcal {M}} {\mathcal {W}} {\mathcal {C}} {\bar{\mathcal P}} \mathcal F = {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} \mathcal P \mathcal F \end{aligned}$$

(11.8)

$$\begin{aligned}&{\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F = \frac{1}{2} {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P \end{aligned}$$

(11.9)

This can be seen as follows:

In Eq. (11.8), we can exchange variables ${\mathcal {C}}$ and $\mathcal P$ and obtain the equality. In Eq. (11.9), we can exchange variables $({\mathcal {C}}, \mathcal P)$ and $({\mathcal {M}}, {\mathcal {W}})$ and obtain ${\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F = {{\bar{\mathcal {M}}}} {{\bar{\mathcal {W}}}} {\mathcal {C}} \mathcal P \mathcal F$. Then, we use that for majority conditions, identical values will be obtained by inverting all conditions, hence ${{\bar{\mathcal {M}}}} {{\bar{\mathcal {W}}}} {\mathcal {C}} \mathcal P \mathcal F = {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} {\bar{\mathcal F}}$. Therefore, $2 {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F = {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F + {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} {\bar{\mathcal F}}$ or $2 {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F = {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} (\mathcal F + {\bar{\mathcal F}}) = {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}}$. Finally, inverting two variables yields identical results than the non-inverted case, hence Eq. (11.9) follows.

A detailed treatment of these symmetry considerations is given in Appendix 6.

Using Eq. (8.4) we can further conclude that

$$\begin{aligned} W = {\mathcal {M}} {\mathcal {W}} \mathcal F = \frac{1}{2} {\left( {{\mathcal {M}} {\mathcal {W}} + {\mathcal {M}} \mathcal F + {\mathcal {W}} \mathcal F}\right) } - \frac{1}{4} \end{aligned}$$

(11.10)

With ${\mathcal {M}} {\mathcal {W}} = 1/4$ and ${\mathcal {M}} \mathcal F = {\mathcal {W}} \mathcal F = 1/2 - {{\bar{\mathcal {M}}}} \mathcal F$ we can rewrite Eq. (11.10):

$$\begin{aligned} W = {\mathcal {M}} {\mathcal {W}} \mathcal F = \frac{3}{8} - {{\bar{\mathcal {M}}}} \mathcal F \end{aligned}$$

(11.11)

Combining Eqs. (11.7), (11.8), (11.9) and (11.11) gives

$$\begin{aligned} V = \frac{3}{16} - \frac{1}{2} {{\bar{\mathcal {M}}}} \mathcal F + \frac{1}{4} {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P \end{aligned}$$

(11.12)

which completes the proof of Lemma 2, Eq. (11.4). $\square$

[4–17] We continue with the proof of Theorem 7. We are now left with calculating the terms ${{\bar{\mathcal {M}}}} \mathcal F$ and ${\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P$.

[5–7] For calculating the term ${{\bar{\mathcal {M}}}} \mathcal F$, we obtain the corresponding fraction of states $\tilde{B}(N)$ by inspecting the conditions in Eq. (11.1). We see that in 1 case, namely $n_3$, both conditions are obeyed, whereas in three cases $n_5$, $n_6$ and $n_7$ only $\mathcal F$ is obeyed. The conditions invert for the cases $n_4$ and $n_0, n_1, n_2$. Using $z_j$ for the condition ${{\bar{\mathcal {M}}}}$ and $v_k$ for the condition $\mathcal F$, we obtain, proceeding as before:

$$\begin{aligned} \tilde{B}(N)= & {} {\frac{1}{r^2}} \sum _{j=0}^{r-1} \sum _{k=0}^{r-1} I_{j,k} \end{aligned}$$

(11.13)

$$\begin{aligned} I_{j,k}= & {} \left\{ {z_j}^{1/2} {v_k}^{1/2}+ 3 {z_j}^{1/2} {v_k}^{-1/2}+ 3 {z_j}^{-1/2} {v_k}^{1/2}+ {z_j}^{-1/2} {v_k}^{-1/2} \right\} ^{N} \nonumber \\&\times \, 2^{(-3 N)} \frac{1 - z_j^{Q+1}}{1 - z_j} \frac{1 - v_k^{Q+1}}{1 - v_k} {z_j}^{1/2} {v_k}^{1/2} \end{aligned}$$

(11.14)

We see immediately that this is identical to B(N) as in Eq. (10.8). Hence, we can directly recall the result $\tilde{B}(N) = B(N)$ from Eq. (10.16).

The term ${\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P$ can be directly calculated, giving the corresponding fraction of states $\tilde{C}(N)$. For the four conditions, we introduce complex variables with the same letters, i.e., $m_i$, $w_j$, $c_k$, and $p_l$. Then, by inspecting the conditions in Eq. (11.1) and Eq. (11.2), we obtain, proceeding as before (the terms are ordered for the conditions $0 \ldots 7$):

$$\begin{aligned} \tilde{C}(N)= & {} {\frac{1}{r^4}} \sum _{i=0}^{r-1} \sum _{j=0}^{r-1} \sum _{k=0}^{r-1} \sum _{l=0}^{r-1} I_{i,j,k,l} \end{aligned}$$

(11.15)

$$\begin{aligned} I_{i,j,k,l}= & {} 2^{(-3 N)} \left\{ {m_i}^{-1/2} {w_j}^{-1/2} {c_k}^{-1/2} {p_l}^{1/2} + {m_i}^{-1/2} {w_j}^{-1/2} {c_k}^{1/2} {p_l}^{-1/2} \right. \nonumber \\&\left. +\,{m_i}^{-1/2} {w_j}^{1/2} {c_k}^{-1/2} {p_l}^{-1/2} + {m_i}^{-1/2} {w_j}^{1/2} {c_k}^{1/2} {p_l}^{1/2} \right. \nonumber \\&\left. +\,{m_i}^{1/2} {w_j}^{-1/2} {c_k}^{-1/2} {p_l}^{-1/2} + {m_i}^{1/2} {w_j}^{-1/2} {c_k}^{1/2} {p_l}^{1/2}\right. \nonumber \\&\left. +\,{m_i}^{1/2} {w_j}^{1/2} {c_k}^{-1/2} {p_l}^{1/2} + {m_i}^{1/2} {w_j}^{1/2} {c_k}^{1/2} {p_l}^{-1/2} \right\} ^{N}\nonumber \\&\times \, \frac{1 - m_i^{Q+1}}{1 - m_i} \frac{1 - w_j^{Q+1}}{1 - w_j} \frac{1 - c_k^{Q+1}}{1 - c_k} \frac{1 - p_l^{Q+1}}{1 - p_l} {m_i}^{1/2} {w_j}^{1/2} {c_k}^{1/2} {p_l}^{1/2}\nonumber \\ \end{aligned}$$

(11.16)

[8–12] With the same methods as used before after Eq. (9.2), it is easy to see that this expression counts the fraction of states for ${\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P$.

[14,15] Using Euler identities in Eq. (11.16) gives

$$\begin{aligned} \tilde{C}(N)= & {} {\frac{1}{4^N r^4}} \sum _{i=0}^{r-1} \sum _{j=0}^{r-1} \sum _{k=0}^{r-1} \sum _{l=0}^{r-1} \nonumber \\&\{\cos ((i+j + k-l) \pi /r) +\cos ((i+j - k+l) \pi /r) \nonumber \\&+\cos ((i - j + k +l) \pi /r) +\cos ((-i+j + k+l) \pi /r) \}^{N} \nonumber \\&\times \frac{\left( 1 - m_i^{Q+1}\right) \; \left( 1 - w_j^{Q+1}\right) \; \left( 1 - c_k^{Q+1}\right) \; \left( 1 - p_l^{Q+1}\right) }{16 \sin (\pi i/r)\sin (\pi j/r)\sin (\pi k/r)\sin (\pi l/r) }\qquad \end{aligned}$$

(11.17)

We can now choose r suitably such that as many terms as possible disappear. The choice follows the same rationale which was leading to Eq. (7.13). Since only real terms must be kept, Eq. (11.17) will result in terms with all indices = 0, two indices = 0 and no indices = 0:

$$\begin{aligned}&\tilde{C}(N)=\frac{1}{16} + {\frac{1}{4^{N} r^4}} \sum _{i=1, \mathrm{odd}}^{r-1} \sum _{j=1, \mathrm{odd}}^{r-1} \sum _{k=1, \mathrm{odd}}^{r-1} \sum _{l=1, \mathrm{odd}}^{r-1}\nonumber \\&\quad \frac{\left\{ \cos ((i+j + k{-}l) \pi /r) + \cos ((i+j {-} k+l) \pi /r) + \cos ((i {-} j + k +l) \pi /r) + \cos (({-}i+j + k+l) \pi /r) \right\} ^{N} }{\sin (\pi i/r)\sin (\pi j/r)\sin (\pi k/r)\sin (\pi l/r) }\nonumber \\&\quad +\;{\text{ terms } \text{ with } \text{ two } \text{ indices }}= 0 \end{aligned}$$

(11.18)

We start with discussing the terms with two indices = 0. They will result in double sums

$$\begin{aligned} \sum _{k=1, \mathrm{odd}}^{r-1} \sum _{l=1, \mathrm{odd}}^{r-1} \frac{ \left\{ \cos ((k-l) \pi /r) + \cos ((k +l) \pi /r) \right\} ^{N} }{\sin (\pi k/r)\sin (\pi l/r) } \end{aligned}$$

which, by factorizing the sum of cosine terms, leads to terms

$$\begin{aligned} \left\{ \sum _{k=1, \mathrm{odd}}^{r-1} \frac{\left\{ \cos (k \pi /r) \right\} ^{N} }{\sin (\pi k/r) } \right\} ^2 \end{aligned}$$

The sums vanish since N is odd and since the summand is antisymmetric about r / 2. Hence, the terms with two indices = 0 in Eq. (11.18) need not be considered.

[13] Continuing with the discussion of Eq. (11.18), since r has no upper bound, one can let $r \rightarrow \infty$ and obtain an integral representation (note only every second term is taken in the sums):

$$\begin{aligned}&\tilde{C}(N)=\frac{1}{16} + \frac{1}{16 \, 4^{N}} \int _0^{1} \mathrm{d}w \int _0^{1} \mathrm{d}x \int _0^{1} \mathrm{d}y \int _0^{1} \mathrm{d}z\nonumber \\&\quad \frac{ \left\{ \cos ((w + x + y {-}z) \pi ) + \cos ((w + x - y +z) \pi ) + \cos ((w {-} x + y +z) \pi ) + \cos ((-w + x + y + z) \pi ) \right\} ^{N} }{\sin (\pi w)\sin (\pi x)\sin (\pi y)\sin (\pi z)}\nonumber \\ \end{aligned}$$

(11.19)

The coupled cosine terms in Eq. (11.19) have to be factorized to factorize the double sum/double integral. We rewrite the coupled cosine terms with the triangular identity:

$$\begin{aligned}&{\cos ((w + x + y {-}z) \pi ) + \cos ((w + x {-} y +z) \pi ) + \cos ((w - x + y +z) \pi ) + \cos ((-w + x + y +z) \pi )}\nonumber \\&\quad = 4 \cos (\pi w)\cos (\pi x) \cos (\pi y)\cos (\pi z) - 4 \sin (\pi w)\sin (\pi x) \sin (\pi y) \sin (\pi z) \end{aligned}$$

(11.20)

and obtain upon expansion of the powers of this term, factorized integrals:

$$\begin{aligned} \tilde{C}(N)=\frac{1}{16}+ \frac{1}{16} \sum _j {N \atopwithdelims ()j} (-1)^j \left\{ \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x) \right\} ^{(N-j)} \left\{ \sin (\pi x) \right\} ^{j-1} \right\} ^4\nonumber \\ \end{aligned}$$

(11.21)

[16a] We refrain from symmetrizing Eq. (11.19) with the help of Eq. (11.20) at this point, which would result in another example for Theorem 1.

[16b] For even j, the integrals in Eq. (11.21) vanish since N is odd and since the integrand is antisymmetric about 1 / 2. Hence, we can write $j= 1+ 2\, q$ and obtain

$$\begin{aligned} \tilde{C}(N)= \frac{1}{16} - \frac{1}{16} \sum _q {N \atopwithdelims (){1 + 2q}} \left\{ \int _0^{1} \mathrm{d}x \left\{ \cos (\pi x) \right\} ^{N-1-2q} \left\{ \sin (\pi x) \right\} ^{2q} \right\} ^4\nonumber \\ \end{aligned}$$

(11.22)

[17] The definite integrals in Eq. (11.22) can again be evaluated in closed form, as already known from Eq. (7.19). Inserting this into Eq. (11.22) finally gives the result for C(N),

$$\begin{aligned} \tilde{C}(N) = \frac{1}{16} - \frac{1}{16} \sum _q C(N,q) \left\{ B(N,q) \right\} ^3 \end{aligned}$$

(11.23)

From this, we can now compile the terms for V(N)

$$\begin{aligned} V(N)= & {} \frac{3}{16} - \frac{1}{2} \tilde{B}(N) + \frac{1}{4} \tilde{C}(N) \end{aligned}$$

(11.24)

$$\begin{aligned}= & {} \frac{5}{64}+ \frac{1}{16} \sum _q C(N,q) \, B(N,q) \left\{ \frac{1}{4^{q}} - \frac{1}{4\texttt {}} (B(N, q))^2 \right\} \end{aligned}$$

(11.25)

as in Theorem 7, Eq. (3.7). This completes the proof. $\square$

For the egality conditions, we can easily deduct bounds for $N\rightarrow \infty$ from previous results. Due to orthogonality, ${\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P = \tilde{C}(N) \rightarrow 1/16$ (for a full discussion see Appendix 7). Further, as previously computed, $\tilde{B}(N) = B(N) = U(N) - T(N) \rightarrow 5/12 - 1/4 = 1/6$. Hence, with Eq. (11.24 11.25), ${\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal F = V(N) \rightarrow 3/16 - 1/12 + 1/64 = 23/192$.

An interpretation of the egality conditions in the light of these results is given in the main text.

We can now move to the proof of Corollary 1 (“Reduced Egality”), which can easily be performed using already existing results.

Proof

We are to compute

$$\begin{aligned} W(N) = {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal F + {{\bar{\mathcal {M}}}} {\mathcal {W}} {\mathcal {C}} \mathcal F + {\mathcal {M}} {{\bar{\mathcal {W}}}} {\mathcal {C}} \mathcal F + {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} \mathcal F \end{aligned}$$

(11.26)

This can easily be transformed to

$$\begin{aligned} W(N)= & {} - 2 {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal F + {\mathcal {W}} {\mathcal {C}} \mathcal F + {\mathcal {M}} {\mathcal {C}} \mathcal F + {\mathcal {M}} {\mathcal {W}} \mathcal F\nonumber \\= & {} - 2 {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal F + 3 {\mathcal {M}} {\mathcal {W}} \mathcal F \end{aligned}$$

(11.27)

Using Eqs. (11.11) and (11.12), this transforms into

$$\begin{aligned} W(N)= & {} \frac{3}{4} - 2 {{\bar{\mathcal {M}}}} \mathcal F - \frac{1}{2} {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P\nonumber \\= & {} \frac{3}{4} - 2 \tilde{B}(N) - \frac{1}{2} \tilde{C}(N) \end{aligned}$$

(11.28)

Recalling that $\tilde{B}(N) = B(N)$ is given by Eq. (10.16), and $\tilde{C}(N)$ is given by Eq. (11.23), the Corollary 1 (Eq. 3.8) directly follows. $\square$

As stated above, we can now use the limiting values $B(N) \rightarrow 1/6$ and $\tilde{C}(N) \rightarrow 1/16$ directly in Eq. (11.28) to obtain

$$\begin{aligned} W(N\rightarrow \infty ) = 37/96 \simeq 0.3854. \end{aligned}$$

(11.29)

Appendix 6: Symmetry arguments in subspace recurrence

Here, we technically explicate the symmetry arguments which prove Eqs. (11.8) and (11.9), which are repeated here:

$$\begin{aligned}&{\mathcal {M}} {\mathcal {W}} {\mathcal {C}} {\bar{\mathcal P}} \mathcal F = {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} \mathcal P \mathcal F \end{aligned}$$

(12.1)

$$\begin{aligned}&{\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F = \frac{1}{2} {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P \end{aligned}$$

(12.2)

We will show that the relevant steps, already indicated in the discussion following Eqs. (11.8) and (11.9), can be derived from mappings which directly transform the right-hand side into the left-hand side of the relevant equations, and which preserve the number of counted configurations.

Here, to prove Eq. (12.1), we first wish to calculate the left- hand side ${\mathcal {M}} {\mathcal {W}} {\mathcal {C}} {\bar{\mathcal P}} \mathcal F$. To this end, we recall which of the eight family conditions (decimals $0 \cdots 7$, corresponding to family triples), already introduced in Eq. (11.1), are counted when the relevant majorities are obeyed. The following truth Table 1 holds (zeroes are omitted):

Table 1 $\mathcal {M}\mathcal {W}\mathcal {C} {\bar{\mathcal {P}}}\mathcal {F}$

Now, we can apply a mapping by exchanging the family condition columns $0 \leftrightarrow 1$ and $6 \leftrightarrow 7$, and by renumbering the columns. This mapping preserves the number of counted configurations for all five conditions to be observed. The truth Table 2 results.

Table 2 $\mathcal {M}\mathcal {W}{\bar{\mathcal {C}}} \mathcal {P}\mathcal {F}$

Note that after the column exchange, one can re-identify the majorities which are obtained. Where ${\mathcal {M}}$, ${\mathcal {W}}$, and $\mathcal F$ remain unchanged under the mapping, we identify two new majorities ${\mathcal {C}}$ and ${\bar{\mathcal P}}$, depicted in the two unseparated rows. Hence, the truth table represents ${\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} \mathcal P \mathcal F$ and it has been shown that the number of counted configurations, as demanded Eq. (12.1), equals.

To prove Eq. (11.9), respectively, Eq. (12.2), we first wish to exchange variables $({\mathcal {C}}, \mathcal P)$ and $({\mathcal {M}}, {\mathcal {W}})$ to obtain ${\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F = {{\bar{\mathcal {M}}}} {{\bar{\mathcal {W}}}} {\mathcal {C}} \mathcal P \mathcal F$. To see that this is feasible, we again proceed as above, obtaining the truth Table 3.

Table 3 $\mathcal {M}\mathcal {W} {\bar{\mathcal {C}}}{\bar{\mathcal {P}}}\mathcal {F}$

Now, we can apply a mapping by exchanging the family condition columns $0 \leftrightarrow 2$, $1 \leftrightarrow 4$, $3 \leftrightarrow 6$, $5 \leftrightarrow 7$, and by renumbering the columns. This mapping preserves the number of counted configurations for all five conditions to be observed. The following truth Table 4 results.

Table 4 ${\bar{\mathcal {M}}} {\bar{\mathcal {W}}} {\mathcal {C}} {\mathcal {P}} {\mathcal {F}}$

After the column exchange, one can re-identify the majorities which are obtained. Where $\mathcal F$ remained unchanged under the mapping, we identify four new majorities, depicted in the four unseparated rows. Hence, the truth table represents ${{\bar{\mathcal {M}}}} {{\bar{\mathcal {W}}}} {\mathcal {C}} \mathcal P \mathcal F$ and it has been shown that the number of counted configurations equals in ${\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F = {{\bar{\mathcal {M}}}} {{\bar{\mathcal {W}}}} {\mathcal {C}} \mathcal P \mathcal F$.

Then, we use that for majority conditions, identical values will be obtained by inverting all conditions, hence ${{\bar{\mathcal {M}}}} {{\bar{\mathcal {W}}}} {\mathcal {C}} \mathcal P \mathcal F = {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} {\bar{\mathcal F}}$. Therefore, $2 {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F = {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F + {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} {\bar{\mathcal F}}$ or $2 {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} \mathcal F = {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} (\mathcal F + {\bar{\mathcal F}})= {\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}}$.

Finally, we show that ${\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} = {\mathcal {M}} {\mathcal {W}} {{{\mathcal {C}}}} {{\mathcal P}}$. It is easy to see this by applying a mapping which directly inverts all family conditions belonging to ${{\mathcal {C}}}$. Obviously, this will not affect ${\mathcal {M}}$ and ${\mathcal {W}}$. With all conditions for ${\mathcal {M}}$ and ${\mathcal {W}}$ unchanged, and with all conditions for ${{\mathcal {C}}}$ inverted, parity will be automatically inverted as well. This shows that the number of counted configurations equals in ${\mathcal {M}} {\mathcal {W}} {{\bar{\mathcal {C}}}} {\bar{\mathcal P}} = {\mathcal {M}} {\mathcal {W}} {{{\mathcal {C}}}} {{\mathcal P}}$. For completeness, note that working with truth tables as above, this mapping amounts to exchanging the family condition columns $0 \leftrightarrow 1$, $2 \leftrightarrow 3$, $4 \leftrightarrow 5$, and $6 \leftrightarrow 7$.

Hence, Eq. (11.9), respectively, Eq. (12.2) finally follows.

Appendix 7: Obtaining the limits in Theorem 7

We now wish to explicate the following bound for $N \rightarrow \infty$, ${\mathcal {M}} {\mathcal {W}} {\mathcal {C}} \mathcal P = \tilde{C}(N) \rightarrow 1/16$, which was used in the discussion of Theorem 7.

To do so, we use the original (counting) formulation and apply directly the limiting theorem for multinomial distributions.

We shall calculate ${\mathcal {M}} {\mathcal {W}} {\mathcal {C}} {\bar{\mathcal P}} = {\hat{C}(N)}$, since for this calculation $n_0$ plays no role and hence $n_0$ can be absorbed in the equality condition for N. Due to the four conditions, it is necessary to treat all eight configurations for each family separately. As in Eq. (2.5), direct counting gives a sum over sevenfold multinomials ($n_0$ is given by the last equality condition)

$$\begin{aligned} \hat{C}(N)=2^{(-3 N)} \sum _{{{\scriptstyle n_1=0}}}^N \cdots {\mathop {\mathop {\mathop {\mathop {\mathop {\mathop {\sum }\limits _{{\scriptstyle n_7 = 0}}}\limits _{\scriptstyle n_7 + n_6 + n_5 + n_4 > N/2}}\limits _{{\scriptstyle n_7 + n_6 + n_3 + n_2 > N/2}}}\limits _{{\scriptstyle n_7 + n_5 + n_3 + n_1 > N/2}}}\limits _{{\scriptstyle n_7 + n_4 + n_2 + n_1 > N/2}}}\limits _{{\scriptstyle \sum _{i=0}^7 n_i = N}}^{N}} {N\atopwithdelims (){n_1 \ldots n_7}} \end{aligned}$$

(13.1)

In the limit $N \rightarrow \infty$, we can make use of the limiting theorem for multinomials, which states that in the seven-dimensional space $\underline{\mathbf {n}}$, the multinomial will be asymptotically equal to a multivariate normal distribution with mean vector $\underline{\mathbf {\mu }} = N/8 \; [1,1,1,1,1,1,1]$ and covariance matrix $\underline{\underline{\mathbf {\Sigma }}} = \underline{\underline{\mathbf {\text{ diag }}}}(\underline{\mathbf {\mu }}) - 1/N \; \underline{\mathbf {\mu }}^T \, \underline{\mathbf {\mu }}$.

We shall show now that with the metric given by $\underline{\underline{\mathbf {\Sigma }}}$, the conditions in Eq. (13.1) are pairwise orthogonal. To do so, a first coordinate transformation will shift the origin of the vector space $\underline{\mathbf {n}}$ to $\underline{\mathbf {\mu }}$, resulting in $\underline{\mathbf {m}} = \underline{\mathbf {n}} - \underline{\mathbf {\mu }}$.

Since, e.g., the second condition can be written $(n_7 - N/8) + (n_6 -N/8) + (n_3 - N/8) + (n_2 - N/8)> 0$, we have $m_7 + m_6 + m_3 + m_2 > 0$ or $\underline{\mathbf {m}} \cdot {{\underline{\mathbf {m}}}_{2}}^T > 0$, with ${\underline{\mathbf {m}}}_{2} = [0,1,1,0,0,1,1]$. One proceeds in the same fashion with all conditions, which results in halfspaces in $\underline{\mathbf {m}}$-space, given by hyperplanes, passing the origin, with normal vectors ${\underline{\mathbf {m}}}_{i}$ and with conditions $\underline{\mathbf {m}} \cdot {{\underline{\mathbf {m}}}_{i}}^T > 0$. The normal vectors of the hyperplanes are:

$$\begin{aligned} \begin{aligned} {\underline{\mathbf {m}}}_{1}&=[0,0,0,1,1,1,1]\\ {\underline{\mathbf {m}}}_{2}&=[0,1,1,0,0,1,1]\\ {\underline{\mathbf {m}}}_{3}&=[1,0,1,0,1,0,1]\\ {\underline{\mathbf {m}}}_{4}&=[1,1,0,1,0,0,1] \end{aligned} \end{aligned}$$

(13.2)

Now, in the metric given by covariance matrix $\underline{\underline{\mathbf {\Sigma }}}$, indeed ${\underline{\mathbf {m}}}_{i} \, \underline{\underline{\mathbf {\Sigma }}} \, {{\underline{\mathbf {m}}}_{j}}^T = 0$ for all pairs $i \ne j$. Hence, after whitening, the total probability mass summed in Eq. (13.1) is asymptotically equal to integrating a multivariate normal distribution, centered a the origin, with unit covariance matrix, over a volume in Euclidian space, produced by the intersection of four halfspaces which are pairwise orthogonal. Since this amounts to 1/16 of the full space volume, we obtain 1/16 of the total probability mass 1, i.e., $\hat{C}(N) \rightarrow 1/16$. Now, $\tilde{C}(N) = {\mathcal {M}} {\mathcal {W}} {\mathcal {C}} - \hat{C}(N) = 1/8 - \hat{C}(N) \rightarrow 1/16$ which completes the discussion.