## Appendix A  
For subdomain \(I_{a}\) in Figure 11, the variable \(y\) ranges from \(-u\) to \(n-u\). When \(y\) is fixed, \(x\) ranges from \(-y\) to \(u\). Hence, the integral for subdomain \(I_{a}\) is  
\[I_{a}=\int_{-u}^{n-u}\int_{-y}^{u}T_{\text{ss}}(x,y)f(x,y)dxdy. \tag{11}\]  
From Equations (33) and (34), we know that \(T_{\text{ss}}(x,y)=\frac{L_{\text{ss}}(x,y)}{n}\cdot\Delta t\). Additionally, from Equation (30), \(S_{M}=8n\frac{M(M+1)}{2}\), and from Equation (31), \(M=\left\lfloor\frac{x}{n}\right\rfloor\). Therefore, the value of \(L_{a}\) on subdomain \(I_{a}\) is  
\[L_{a}(x,y)=S_{M}+x-Mn. \tag{12}\]  
Substituting for \(S_{M}\) and \(M\):  
\[L_{a}(x,y)=8n\frac{\left\lfloor\frac{x}{n}\right\rfloor\left(\left\lfloor\frac {x}{n}\right\rfloor+1\right)}{2}+x-\left\lfloor\frac{x}{n}\right\rfloor n. \tag{13}\]  
Thus, the expression for \(L_{a}(x,y)\) becomes \[L_{a}(x,y)=4n\left[\frac{x}{n}\right]^{2}+3n\left[\frac{x}{n}\right]+x. \tag{124}\]  
Therefore,  
\[I_{a}=\int_{-u}^{n-u}\int_{-y}^{u}4x\Big{(}\frac{x}{n}+1\Big{)}\frac{\Delta t}{2 \pi n\sigma^{2}}\exp\left(-\frac{1}{2}\frac{x^{2}+y^{2}}{\sigma^{2}}\right)dxdy. \tag{125}\]  
For subdomain \(I_{b}\) in Figure 11, the variable \(y\) ranges from \(n-u\) to \(0\). When \(y\) is fixed, \(x\) ranges from \(-y\) to \(n-y\). This can be seen by considering the Cartesian equations of both diagonals, which have slopes of \(-1\) and intercepts of \(0\) and \(n\), respectively. Since the value of \(L_{b}\) remains the same as before, we obtain  
\[I_{b}=\int_{n-u}^{0}\int_{-y}^{n-y}4x\Big{(}\frac{x}{n}+1\Big{)}\frac{\Delta t }{2\pi n\sigma^{2}}\exp\left(-\frac{1}{2}\frac{x^{2}+y^{2}}{\sigma^{2}} \right)dxdy. \tag{126}\]  
For subdomain \(I_{c}\) in Figure 12, the variable \(y\) ranges from \(0\) to \(\frac{u}{2}\). When \(y\) is fixed, \(x\) ranges from \(y\) to \(n-y\). Since the value of \(L_{c}\) remains the same as before, we obtain  
\[I_{c}=\int_{0}^{\frac{u}{2}}\int_{y}^{n-y}4x\Big{(}\frac{x}{n}+1\Big{)}\frac{ \Delta t}{2\pi n\sigma^{2}}\exp\left(-\frac{1}{2}\frac{x^{2}+y^{2}}{\sigma^{2} }\right)dxdy. \tag{127}\]For domain \(I_{2}\) in Figure A4, the variable \(x\) ranges from \(\frac{u}{2}\) to \(u\). When \(x\) is fixed, \(y\) ranges from \(n-x\) to \(x\). Hence, the integral for domain \(I_{2}\) is  
\[I_{2}=\int_{\frac{u}{2}}^{u}\int_{n-x}^{x}T_{\rm ss}(x,y)f(x,y)dydx.\] (A8)  
In domain \(I_{2}\), we have \(x\geq 0\) and \(|x|\geq|y|\), so \(M=\left\lfloor\frac{x}{n}\right\rfloor\). Therefore, the value of \(L_{2}\) on domain \(I_{2}\) is  
\[L_{2}(x,y)=S_{M-1}+Mn+y.\] (A9)  
Substituting for \(S_{M-1}\) and \(M\):  
\[L_{2}(x,y)=4n\left\lfloor\frac{x}{n}\right\rfloor\left(\left\lfloor\frac{x}{n} \right\rfloor-1\right)+\left\lfloor\frac{x}{n}\right\rfloor n+y.\] (A10)  
Thus, the expression for \(L_{2}(x,y)\) becomes  
\[L_{a}(x,y)=S_{M}+x-Mn. \tag{12}\]0 (A11)  
Finally, we obtain  
\[L_{a}(x,y)=S_{M}+x-Mn. \tag{12}\]1 (A12)  
For domain \(I_{3}\) in Figure A5, the variable \(y\) ranges from \(0\) to \(u\). When \(y\) is fixed, \(x\) ranges from \(-y\) to \(y\). Hence, the integral for domain \(I_{3}\) is  
\[L_{a}(x,y)=S_{M}+x-Mn. \tag{12}\]2  
In domain \(I_{3}\), we have \(y\geq 0\) and \(|y|\geq|x|\), so \(M=\left\lfloor\frac{y}{n}\right\rfloor\). Therefore, the value of \(L_{3}\) on domain \(I_{3}\) is  
\[L_{a}(x,y)=S_{M}+x-Mn. \tag{12}\]3  
Substituting for \(S_{M-1}\) and \(M\):  
\[L_{a}(x,y)=S_{M}+x-Mn. \tag{12}\]4  
Thus, the expression for \(L_{3}(x,y)\) becomes  
\[L_{a}(x,y)=S_{M}+x-Mn. \tag{12}\]5  
Finally, we obtain  
\[L_{a}(x,y)=S_{M}+x-Mn. \tag{12}\]6  
For domain \(I_{4}\) in Figure 10, as \(x\) varies from \(-u\) to \(0\), \(y\) varies from \(x\) to \(-x\). Hence, the integral for domain \(I_{4}\) is  
\[L_{a}(x,y)=S_{M}+x-Mn. \tag{12}\]7  
In domain \(I_{4}\), we have \(x\leq 0\) and \(|x|\geq|y|\). Therefore, \(M=\left\lfloor-\frac{x}{n}\right\rfloor\).