A novel parameterized neutrosophic score function and its application in genetic algorithm

GSF on SvNSs

Inspired by Guo, Şengür & Ye (2014), a kind of GSF on SvNSs is introduced as follows.

Definition 5: Assume that a is a single-valued neutrosophic components on Y, it is denoted as (8)${\displaystyle a=〈{\mathrm{T}}_A\left(y\right),{\mathrm{I}}_A\left(y\right),F_A\left(y\right)〉.}$

Then, a kind of GSF on a is denoted as (9)${\displaystyle S\left(a\right)=f\left({\mathrm{T}}_A\left(y\right),\gamma \right),}$

where $f_{{\mathrm{T}}_A}^{\prime }>0,$ $f_{\gamma }^{\prime }>0,$ $f\left({\mathrm{T}}_A\left(y\right),\gamma \right)\in \left[0,1\right],$ whereas (10)${\displaystyle \gamma =cos〈\left({\mathrm{T}}_A\left(y\right),{\mathrm{I}}_A\left(y\right),F_A\left(Y\right)\right),\left(1,0,0\right)〉.}$

It is noteworthy that $S\left(\cdot \right)$ is composed of two independent variables, where one is ${\mathrm{T}}_A\left(y\right)$ which indicates the modulus, whereas the other is γ which indicates the degree of similarity between a and $〈1,0,0〉$.

Definition 6: Suppose a is a single-valued neutrosophic number on Y, it can be denoted as (11)${\displaystyle a=〈{\mathrm{T}}_A\left(y\right),{\mathrm{I}}_A\left(y\right),F_A\left(y\right)〉.}$

Then, a specific GSF on a is denoted as (12)${\displaystyle S_F\left(a\right)={\mathrm{T}}_A\left(y\right)\cdot \gamma ={\mathrm{T}}_A\left(y\right)\cdot \frac{{\mathrm{T}}_{{}_A}^{\lambda }\left(y\right)}{\sqrt{{\mathrm{T}}_{{}_A}^{2\lambda }\left(y\right)+{\mathrm{I}}_{{}_A}^{2\lambda }\left(y\right)+F_{{}_A}^{2\lambda }\left(y\right)}},}$

where λ is a positive integer which represents the decision-maker’s subjective attitude towards extremely large or small values. Obviously, $S_{F,{\mathrm{T}}_A}^{\prime }>0,$ $S_{F,\gamma }^{\prime }>0,$ $S_F\left(\cdot ,\cdot \right)\in \left[0,1\right]$. Especially, when the weight vector for $〈\mathrm{T},\mathrm{I},F〉$ is given as $〈w_1,w_2,w_3〉$, $S_F\left(a\right)$ can be expressed as (13)${\displaystyle S_F\left(a\right)==\frac{{\mathrm{T}}_A\left(y\right)\cdot {\left(w_1{\mathrm{T}}_A\left(y\right)\right)}^{\lambda }}{\sqrt{{\left(w_1{\mathrm{T}}_A\left(y\right)\right)}^{2\lambda }+{\left(w_2{I}_A\left(y\right)\right)}^{2\lambda }+{\left(w_3{F}_A\left(y\right)\right)}^{2\lambda }}}.}$

PNO-PGA

On the basis of combining GA and NS theories, a novel heuristic algorithm is proposed. For convenience, the studied problem will be introduced firstly.

(iii) Parameterized PNO–PGA

Based on the problem analysis and by referring Definition 2, the introduced MOO problem (Eq. (17)) can be solved, and the detailed steps are shown as follows.

Step 1: For any given feasible solution x for Eq. (17), by using C in SGA, code x and convert it into a vector x′ which is the representation of a feasible solution.

Step 2: Generate the initial population P₀ by using x′, and denote the population size of P₀ as M. Denote $X_0=\left\{x_{01},x_{02},\dots ,x_{0M}\right\},$ where $x_{0m}\left(1\le m\le M\right)$ represents an individual. Denote $P_0=\left\{x_{01}^{\prime },x_{02}^{\prime },\dots ,x_{0M}^{\prime }\right\},$ where $x_{0m}^{\prime }\left(1\le m\le M\right)$ is a vector which is corresponding to x_0m. The novel PNO–PGA will start the iteration with P₀. Set the iteration variable as n, and set the number of iterations as N.

Step 3: For any given $x_{0m}\left(1\le m\le M\right)$ in X₀, its corresponding $E\left(x_{0m}\right),$ $S\left(x_{0m}\right),$ and $C\left(x_{0m}\right)$ according to the actual condition in production practice are calculated. For the dangerous goods transport scenario mentioned in this article, $E\left(x_{0m}\right),$ $S\left(x_{0m}\right),$ and $C\left(x_{0m}\right)$ according to Eqs. (39), (40), and (41) are calculated. Then, it gets the vector (20)${\displaystyle V_{x_{0m}}=\left(E\left(x_{0m}\right),S\left(x_{0m}\right),C\left(x_{0m}\right)\right).}$

For the whole initial population P₀, it gets a set of G_P0. For convenience, the set is denoted as (21)${\displaystyle V_{P_0}=\left\{V_{x_{01}},V_{x_{02}},\dots ,V_{x_{0M}}\right\}.}$

Denote (22)${\displaystyle \left\{\begin{array}{c}{E}_{MAX,0}=\left\{E\left(x_{01}\right),E\left(x_{02}\right),\dots ,E\left(x_{0M}\right)\right\},\hfill \\ S_{MAX,0}=\left\{S\left(x_{01}\right),S\left(x_{02}\right),\dots ,S\left(x_{0M}\right)\right\},\hfill \\ C_{MAX,0}=\left\{C\left(x_{01}\right),C\left(x_{02}\right),\dots ,C\left(x_{0M}\right)\right\}.\hfill \end{array}\right.}$

For any given $m\left(1\le m\le M\right)$. Denote (23)${\displaystyle \left\{\begin{array}{c}{E}_N\left(x_{0m}\right)=E\left(x_{0m}\right)\cdot {\left(E_{MAX,0}\right)}^{-1},\hfill \\ S_N\left(x_{0m}\right)=S\left(x_{0m}\right)\cdot {\left(S_{MAX,0}\right)}^{-1},\hfill \\ C_N\left(x_{0m}\right)=C\left(x_{0m}\right)\cdot {\left(C_{MAX,0}\right)}^{-1}.\hfill \end{array}\right.}$

Then, it gets a NS $G_0=〈E_N,1-S_N,C_N〉.$ For any given $x_{0m}\left(1\le m\le M\right),$ by using Eq. (23), it gets a neutrosophic number (24)${\displaystyle G_{x_{0m}}=〈E_N\left(x_{0m}\right),1-S_N\left(x_{0m}\right),C_N\left(x_{0m}\right)〉}$

and a set (25)${\displaystyle X_G=\left\{\left(x,〈E_N\left(x\right),1-S_N\left(x\right),C_N\left(x\right)〉\right)\left|x\in X\right.\right\}.}$

Step 4: For any given $G_{x_{0m}}\left(1\le m\le M\right)$, by using Eq. (13), its corresponding GSF value can be obtained as $S_F\left(G_{x_{0m}}\right)$. Thereafter, by using $S_F\left(\cdot \right),$ a monotonic descending sequence according to $S_F\left(\cdot \right)$ is obtained as (26)${\displaystyle l_{0F}=S_F\left(G_{x_{0m_1}}\right),S_F\left(G_{x_{0m_2}}\right),\dots ,S_F\left(G_{x_{0m_M}}\right),}$

where m₁, m₂, …, m_M isa permutation of 1, 2, …, M.

Set a positive integer $M^{\ast }\left(M^{\ast }<M\right)$. By using M^∗, X₀ is divided into two parts, where one is (27)${\displaystyle X_{01}=\left\{x_{0m_1},x_{0m_2},\dots ,x_{0m_{M^{\ast }}}\right\},}$

and the other is (28)${\displaystyle X_{02}=\left\{x_{0m_{M^{\ast }+1}},x_{0m_{M^{\ast }+2}},\dots ,x_{0m_M}\right\}.}$

Step 5: By using certain crossover operator Γ,  X₀₁ is transferred to $X_{01}^{\prime }$, where $X_{01}^{\prime }=\left\{x_{0m_1}^{\prime },x_{0m_2}^{\prime },\dots ,x_{0m_{M^{\ast }}}^{\prime }\right\}$. By using mutation operator Ψ,  X₀₂ is transferred to $X_{02}^{\prime }$, where $X_{02}^{\prime }=\left\{x_{0m_{M^{\ast }+1}}^{\prime },x_{0m_{M^{\ast }+2}}^{\prime },\dots ,x_{0m_M}^{\prime }\right\}$. Denote $X_0^{\prime }=\left\{X_{01}^{\prime },X_{02}^{\prime }\right\}$ by using $S_F\left(\cdot \right),$ a monotonic descending sequence is obtained as (29)${\displaystyle l_{0F}^{\prime }=S_F\left(G_{x_{0m_1}^{\prime }}\right),S_F\left(G_{x_{0m_2}^{\prime }}\right),\dots ,S_F\left(G_{x_{0m_M}^{\prime }}\right).}$

For any given $m\left(1\le m\le M\right),$ denote (30)${\displaystyle S_F\left(G_{x_{1k}}\right)=max\left\{S_F\left(G_{x_{0m_k}}\right),S_F\left(G_{x_{0m_k}^{\prime }}\right)\right\},}$

denote the corresponding set of feasible solution X₁, where $X_1=\left\{x_{11},x_{12},\dots ,x_{1M}\right\},$ and denote the 1 st optimal solution as x₁₁.

Step 6: If N ≠ 1,  then, turn to Step 3; and denote n = 2,  denote (31)${\displaystyle \left\{\begin{array}{c}{E}_{MAX,1}^{\prime }=max\left\{E_{MAX,0},E_{MAX,1}\right\},\hfill \\ S_{MAX,1}^{\prime }=max\left\{S_{MAX,0},S_{MAX,1}\right\},\hfill \\ C_{MAX,1}^{\prime }=max\left\{C_{MAX,0},C_{MAX,1}\right\}.\hfill \end{array}\right.}$

Denote (32)${\displaystyle \left\{\begin{array}{c}{E}_N\left(x_{1,m}\right)=E\left(x_{1,m}\right)\cdot {\left(E_{MAX,1}^{\prime }\right)}^{-1},\hfill \\ S_N\left(x_{1,m}\right)=S\left(x_{1,m}\right)\cdot {\left(S_{MAX,1}^{\prime }\right)}^{-1},\hfill \\ C_N\left(x_{1,m}\right)=C\left(x_{1,m}\right)\cdot {\left(C_{MAX,1}^{\prime }\right)}^{-1}.\hfill \end{array}\right.}$

Then, turn to Step 4, Step 5, Step 6 in turn. If n < N,  turn to Step 3; and denote n = n + 1,  (33)${\displaystyle \left\{\begin{array}{c}{E}_{MAX,n}^{\prime }=max\left\{E_{MAX,n-1}^{\prime },E_{MAX,n}\right\},\hfill \\ S_{MAX,n}^{\prime }=max\left\{S_{MAX,n-1}^{\prime },S_{MAX,n}\right\},\hfill \\ C_{MAX,n}^{\prime }=max\left\{C_{MAX,n-1}^{\prime },E_{MAX,n}\right\}.\hfill \end{array}\right.}$

Denote (34)${\displaystyle \left\{\begin{array}{c}{E}_N\left(x_{n,m}\right)=E\left(x_{n,m}\right)\cdot {\left(E_{MAX,n}^{\prime }\right)}^{-1},\hfill \\ S_N\left(x_{n,m}\right)=S\left(x_{n,m}\right)\cdot {\left(S_{MAX,n}^{\prime }\right)}^{-1},\hfill \\ C_N\left(x_{n,m}\right)=C\left(x_{n,m}\right)\cdot {\left(C_{MAX,n}^{\prime }\right)}^{-1}.\hfill \end{array}\right.}$

Then, turn to Step 4, Step 5, Step 6 in turn. If t < N,  turn to Step 3; if else, select the final optimal solution as x_N1. It is noteworthy that there is a series of virtual NSs which is structured by Eq. (22), and is denoted as $V_n=\left\{〈E_n,1-S_n,C_n〉\left|n=1,2,\dots ,N\right.\right\}.$ For any given individual x_n,m,  the neutrosophic number corresponding to it is denoted as $V_{n,m}=〈E_N\left(x_{n,m}\right),1-S_N\left(x_{n,m}\right),C_N\left(x_{n,m}\right)〉$. For ease of understanding, please refer to Fig. 2.

Exemplification introduction

To illustrate a MOO problem in warehouse operation, we cited the data which were collected as previously described in Zhang et al. (2023). Assume that the length, width, and entrance width of the warehouse are 64 m, 17.5 m, and 4 m respectively. And there are 14 points inbound and points outbound. Not only should a distance of 12 m between two container vehicles but also a distance from the warehouse to the container truck be 2 m. For convenience, the inbound point is denoted as ${\Delta }_i\left(1\le i\le 14\right)$, the outbound point is denoted as ${\nabla }_j\left(1\le j\le 14\right)$. In general, the work flow of the forklift is as follows. Firstly, the hazardous goods are transported from containers by the operating forklifts. Secondly, the goods are unloaded by forklifts to the inbound point. Thirdly, the outbound goods should be discharged onto the outbound container by an empty forklift traveling to the outbound port. Assuming that there are three forklifts responsible for both inbound and outbound hazardous materials transfer. The forklift turns back to the inbound container and prepares the next inbound activity. For convenience, the chain of the above operations is named as a closed-loop storage chain. The forklifts work on successive chains until operations are over. According to Sun et al. (2021), the original warehouse layout structure diagram is shown in Fig. 3.

The distance for forklifts to travel to finish a closed-loop storage chain depends on the location of the point incoming and the point outbound. For any given Δ_i and ∇_j, there is a corresponding closed-loop storage chain whose distance is denoted as l_ij. Denote the distance of all pairs of $\left({\Delta }_i,{\nabla }_j\right)\left(1\le i,j\le 14\right)$ as $L={\left(l_{ij}\right)}_{14\times 14}$ where the incoming point is represented by the row in L, while the outgoing point is represented by the column in L. To simplify the calculation, the original distances of the closed-loop storage chains are normalized by using the function (37)${\displaystyle l_{{p_i}_{{}_{k1}},q_{j_{k2}}}^{\prime }=\frac{l_{{p_i}_{{}_{k1}},q_{j_{k2}}}-\underset{S1,S2\in M}{min}{l}_{{p_i}_{{}_{S1}},{q_j}_{{}_{S2}}}}{\underset{S1,S2\in M}{max}{l}_{{p_i}_{{}_{S1}},{q_j}_{{}_{S2}}}-\underset{S1,S2\in M}{min}{l}_{{p_i}_{{}_{S1}},{q_j}_{{}_{S2}}}}\vee 0.1.}$

For any given $l_{{p_i}_{{}_1},q_{j_1}}^{\prime },l_{{p_i}_{{}_2},q_{j_2}}^{\prime },l_{{p_i}_{{}_3},q_{j_3}}^{\prime },$ the similarity degree r′ on them can be obtained as (38)${\displaystyle r^{\prime }=1-\frac{\left|\left(l^{\prime }{p_i}_{{}_1},q_{j_1}-l^{\prime }{p_i}_{{}_2},q_{j_2}\right)\cdot \left(l^{\prime }{p_i}_{{}_1},q_{j_1}-l^{\prime }{p_i}_{{}_3},q_{j_3}\right)\cdot \left(l^{\prime }{p_i}_{{}_2},q_{j_2}-l^{\prime }{p_i}_{{}_3},q_{j_3}\right)\right|}{{\left(max\left\{l^{\prime }{p_i}_{{}_1},q_{j_1}\cdot l^{\prime }{p_i}_{{}_2},q_{j_2}\cdot l^{\prime }{p_i}_{{}_3},q_{j_3}\right\}\right)}^3}.}$

Denote the fixed cost for operation management per hour as C_r,  denote the normal speed index of the forklifts as τ_v,  denote the time cost of the operation as $C_t=C_r\cdot {\tau }_v^{-1}\cdot l^{\prime },$ denote the influence factor of variable speed on operation time as τ_α,  denote the influence factor of variable speed on fuel cost as τ_β,  anddenote the normal fuel cost index for forklift to travel as C_f. Then, according to field survey, the efficiency optimization model can be expressed as (39)${\displaystyle \left\{\begin{array}{c}{\gamma }_1={\left({\tau }_v\right)}^{-1}\sum _{i_s=1}^{14}\sum _{j_t=1}^{14}{x}_{i_s{j}_t}\cdot l_{i_s{j}_t}^{\prime }-{\tau }_{\alpha }\cdot {\left(C_{14}^3\right)}^{-1}\cdot \sum _{i_s,j_t=1}^{14}\left(x_{i_s,j_t,1}\cdot x_{i_s,j_t,2}\cdot x_{i_s,j_t,3}\cdot r_{\left(p_{i_1},q_{j_1}\right),\left(p_{i_2},q_{j_2}\right),\left(p_{i_3},q_{j_3}\right)}^{\prime }\right),\hfill \\ s.t.\sum _{i_s=1}^{14}{x}_{i_s{j}_t}=1,\sum _{j_t=1}^{14}{x}_{i_s{j}_t}=1,\sum _{i_s=1}^{14}{x}_{i_s,j_t,1}=1,\sum _{j_t=1}^{14}{x}_{i_s,j_t,1}=1,\hfill \\ \sum _{i_s=1}^{14}{x}_{i_s,j_t,2}=1,\sum _{j_t=1}^{14}{x}_{i_s{j}_t,2}=1,\sum _{i_s=1}^{14}{x}_{i_s,j_t,3}=1,\sum _{j_t=1}^{14}{x}_{i_s{j}_t,3}=1,\hfill \\ x_{i_s,j_t,1}\ne x_{i_s,j_t,2}\ne x_{i_s,j_t,3},x_{i_s{j}_t}\in \left\{0,1\right\},x_{i_s,j_t,1}\in \left\{0,1\right\},x_{i_s,j_t,2}\in \left\{0,1\right\},x_{i_s,j_t,3}\in \left\{0,1\right\}.\hfill \end{array}\right.}$

The safety optimization model can be expressed as (40)${\displaystyle \left\{\begin{array}{c}{\gamma }_2={\left(C_{14}^3\right)}^{-1}\cdot \sum _{i_s,j_t=1}^{14}\left(x_{i_s,j_t,1}\cdot x_{i_s,j_t,2}\cdot x_{i_s,j_t,3}\cdot r_{\left({p_i}_{{}_1},q_{j_1}\right),\left({p_i}_{{}_2},q_{j_2}\right),\left({p_i}_{{}_3},q_{j_3}\right)}^{\prime }\right),\hfill \\ s.t.\sum _{i_s=1}^{14}{x}_{i_s{j}_t}=1,\sum _{j_t=1}^{14}{x}_{i_s{j}_t}=1,\sum _{i_s=1}^{14}{x}_{i_s,j_t,1}=1,\sum _{j_t=1}^{14}{x}_{i_s,j_t,1}=1,\hfill \\ \sum _{i_s=1}^{14}{x}_{i_s,j_t,2}=1,\sum _{j_t=1}^{14}{x}_{i_s{j}_t,2}=1,\sum _{i_s=1}^{14}{x}_{i_s,j_t,3}=1,\sum _{j_t=1}^{14}{x}_{i_s{j}_t,3}=1,\hfill \\ x_{i_s,j_t,1}\ne x_{i_s,j_t,2}\ne x_{i_s,j_t,3},x_{i_s{j}_t}\in \left\{0,1\right\},x_{i_s,j_t,1}\in \left\{0,1\right\},x_{i_s,j_t,2}\in \left\{0,1\right\},x_{i_s,j_t,3}\in \left\{0,1\right\}.\hfill \end{array}\right.}$

The cost optimization model can be expressed as (41)${\displaystyle \left\{\begin{array}{c}{\gamma }_3=C_f\sum _{i_s=1}^{14}\sum _{j_t=1}^{14}{x}_{i_s{j}_t}\cdot l_{i_s{j}_t}^{\prime }+{\tau }_{\beta }\cdot {\left(C_{14}^3\right)}^{-1}\cdot \sum _{i_s,j_t=1}^{14}\left(x_{i_s,j_t,1}\cdot x_{i_s,j_t,2}\cdot x_{i_s,j_t,3}\cdot r_{\left({p_i}_{{}_1},q_{j_1}\right),\left({p_i}_{{}_2},q_{j_2}\right),\left({p_i}_{{}_3},q_{j_3}\right)}^{\prime }\right),\hfill \\ s.t.\sum _{i_s=1}^{14}{x}_{i_s{j}_t}=1,\sum _{j_t=1}^{14}{x}_{i_s{j}_t}=1,\sum _{i_s=1}^{14}{x}_{i_s,j_t,1}=1,\sum _{j_t=1}^{14}{x}_{i_s,j_t,1}=1,\hfill \\ \sum _{i_s=1}^{14}{x}_{i_s,j_t,2}=1,\sum _{j_t=1}^{14}{x}_{i_s{j}_t,2}=1,\sum _{i_s=1}^{14}{x}_{i_s,j_t,3}=1,\sum _{j_t=1}^{14}{x}_{i_s{j}_t,3}=1,\hfill \\ x_{i_s,j_t,1}\ne x_{i_s,j_t,2}\ne x_{i_s,j_t,3},x_{i_s{j}_t}\in \left\{0,1\right\},x_{i_s,j_t,1}\in \left\{0,1\right\},x_{i_s,j_t,2}\in \left\{0,1\right\},x_{i_s,j_t,3}\in \left\{0,1\right\}.\hfill \end{array}\right.}$

Optimization process using PNO–PGA

In this subsection, the introduced operation optimization problem would be solved by using the proposed novel PNO–PGA model. First of all, a series of parameters are obtained according to production practice. Specifically, in Eq. (21), it values that τ_v = 3.5,  τ_α = 0.6; in Eq. (41), it values that C_f = 2.4,  τ_β = 0.3. Then, by using the proposed PNO–PGA model, the problem is solved in the following manner.

Step 1: For any given feasible solution xfor Eq. (17), code x and convert it into a vector x′ whose size is 14 and is a permutation from 1 to 14.

Step 2: Generate the initial population P₀ by using x′, and denote the population size of P₀ as M. Denote $X_0=\left\{x_{0,1},x_{0,2},\dots ,x_{0,200}\right\},$ where $x_{0m}\left(1\le m\le 200\right)$ represents an individual. Denote $P_0=\left\{x_{0,1}^{\prime },x_{0,2}^{\prime },\dots ,x_{0,200}^{\prime }\right\},$ where $x_{0m}^{\prime }\left(1\le m\le 200\right)$ is a vector which is corresponding to x_0m. Denote Group number = 200, cross rate = 0.8 and mutation rate = 0.2. The novel PNO–PGA will start the iteration with P₀. Set the iteration variable as n, and set the number of iterations as N = 500. Here L′ is a distance between 14 inbound and 14 outbound warehouses. ${\displaystyle L^{\prime }=\left[\begin{array}{cccccccccccccc}0.13\hfill & 0.18\hfill & 0.13\hfill & 0.24\hfill & 0.28\hfill & 0.28\hfill & 0.32\hfill & 0.34\hfill & 0.54\hfill & 0.51\hfill & 0.57\hfill & 0.54\hfill & 0.77\hfill & 0.77\hfill \\ 0.09\hfill & 0.18\hfill & 0.13\hfill & 0.21\hfill & 0.25\hfill & 0.25\hfill & 0.28\hfill & 0.31\hfill & 0.51\hfill & 0.47\hfill & 0.54\hfill & 0.51\hfill & 0.77\hfill & 0.73\hfill \\ 0.09\hfill & 0.18\hfill & 0.13\hfill & 0.18\hfill & 0.21\hfill & 0.21\hfill & 0.25\hfill & 0.28\hfill & 0.47\hfill & 0.43\hfill & 0.51\hfill & 0.47\hfill & 0.77\hfill & 0.73\hfill \\ 0.62\hfill & 0.59\hfill & 0.60\hfill & 0.80\hfill & 0.62\hfill & 0.19\hfill & 0.19\hfill & 0.19\hfill & 0.44\hfill & 0.41\hfill & 0.47\hfill & 0.41\hfill & 0.44\hfill & 0.41\hfill \\ 0.09\hfill & 0.18\hfill & 0.13\hfill & 0.18\hfill & 0.21\hfill & 0.21\hfill & 0.25\hfill & 0.28\hfill & 0.47\hfill & 0.43\hfill & 0.51\hfill & 0.47\hfill & 0.77\hfill & 0.73\hfill \\ 0.58\hfill & 0.55\hfill & 0.58\hfill & 0.77\hfill & 0.13\hfill & 0.15\hfill & 0.13\hfill & 0.19\hfill & 0.41\hfill & 0.38\hfill & 0.44\hfill & 0.38\hfill & 0.40\hfill & 0.38\hfill \\ 0.55\hfill & 0.52\hfill & 0.55\hfill & 0.74\hfill & 0.09\hfill & 0.12\hfill & 0.13\hfill & 0.19\hfill & 0.38\hfill & 0.34\hfill & 0.41\hfill & 0.34\hfill & 0.37\hfill & 0.34\hfill \\ 0.55\hfill & 0.52\hfill & 0.55\hfill & 0.74\hfill & 0.09\hfill & 0.12\hfill & 0.13\hfill & 0.19\hfill & 0.38\hfill & 0.34\hfill & 0.41\hfill & 0.34\hfill & 0.37\hfill & 0.34\hfill \\ 0.44\hfill & 0.41\hfill & 0.41\hfill & 0.44\hfill & 0.06\hfill & 0.06\hfill & 0.09\hfill & 0.03\hfill & 0.19\hfill & 0.19\hfill & 0.19\hfill & 0.22\hfill & 0.23\hfill & 0.20\hfill \\ 0.41\hfill & 0.38\hfill & 0.38\hfill & 0.41\hfill & 0.03\hfill & 0.03\hfill & 0.03\hfill & 0.01\hfill & 0.13\hfill & 0.15\hfill & 0.19\hfill & 0.19\hfill & 0.20\hfill & 0.17\hfill \\ 0.31\hfill & 0.28\hfill & 0.28\hfill & 0.31\hfill & 0.01\hfill & 0.01\hfill & 0.03\hfill & 0.01\hfill & 0.09\hfill & 0.12\hfill & 0.19\hfill & 0.15\hfill & 0.17\hfill & 0.13\hfill \\ 0.81\hfill & 0.78\hfill & 0.81\hfill & 1\hfill & 0.35\hfill & 0.38\hfill & 0.35\hfill & 0.41\hfill & 0.55\hfill & 0.58\hfill & 0.55\hfill & 0.15\hfill & 0.13\hfill & 0.19\hfill \\ 0.75\hfill & 0.72\hfill & 0.75\hfill & 0.93\hfill & 0.29\hfill & 0.32\hfill & 0.29\hfill & 0.34\hfill & 0.49\hfill & 0.52\hfill & 0.49\hfill & 0.09\hfill & 0.06\hfill & 0.12\hfill \\ 0.78\hfill & 0.75\hfill & 0.78\hfill & 0.96\hfill & 0.32\hfill & 0.35\hfill & 0.32\hfill & 0.39\hfill & 0.52\hfill & 0.55\hfill & 0.52\hfill & 0.12\hfill & 0.09\hfill & 0.15\hfill \end{array}\right].}$