## 2 Theoretical Formulation  
A perfect conductor buried in one half-space is depicted in Figure 1. The permittivity and conductivity in regions 1 and 2 were \((\varepsilon_{1},\sigma_{1})\) and \((\varepsilon_{2},\sigma_{2})\), respectively. In each region, the permeability of the free space was set to \(\mu_{0}\), that is, only non-magnetic substances were considered. Let the scatterer be a cylindrical conductor extending infinitely in the z-axis. Its cross-sectional area in the _xy_-plane can be  
\[E(x,y)= \begin{cases}E_{1}(x,y)+E_{s}(x,y)&,y\leq-a\\ E_{2}(x,y)+E_{s}(x,y)&,y>-a\end{cases}  \tag{6}\]  
where \(E_{s}(x,y)\) is the scatterrd field. Since the size of the interested object in the resonance region is about one wavelength, the scattered field will have a severe diffraction effect. In order to search for the scattered field, [PERSON]'s function \(G(x,y;x^{\prime},y^{\prime})\) must be calculated carefully by first generating a line current source at \((x^{\prime},y^{\prime})\) and the corresponding scattered field at \((x,y)\). By the Fourier transformation technique, \(G(x,y;x^{\prime},y^{\prime})\) can be expressed as:  
\[G(x,y;x^{\prime},y^{\prime})= \begin{cases}G_{1}(x,y;x^{\prime},y^{\prime})\\ G_{2}(x,y;x^{\prime},y^{\prime})=G_{f}(x,y;x^{\prime},y^{\prime})+G_{s}(x,y;x^ {\prime},y^{\prime})\end{cases} ,y\leq-a \tag{7}\]  
\[G_{1}(x,y;x^{\prime},y^{\prime})=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{ j}{\gamma_{1}+Y_{2}}e^{j\gamma_{2}(y+a)}e^{-j\gamma_{2}(y+a)}e^{-ja(x-x^{ \prime})}d\alpha \tag{8}\]  
\[G_{f}(x,y;x^{\prime},y^{\prime})=\frac{j}{4}H_{0}^{(2)}\left[k_{2}\sqrt{(x-x^ {\prime})^{2}+(y-y^{\prime})^{2}}\right] \tag{9}\]  
\[G_{s}(x,y;x^{\prime},y^{\prime})=\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{ j}{2\gamma_{2}}\left(\frac{y_{2}-\gamma_{1}}{\gamma_{2}+\gamma_{1}}\right)e^{-j \gamma_{2}(y+2a+y^{\prime})}e^{-ja(x-x^{\prime})}d\alpha \tag{10}\]  
where \(\gamma_{l}^{2}=k_{l}^{2}-\alpha^{2},Im(\gamma_{l})\leq 0,i=1,2\). Conceptually the scattered field \(E_{S}(x,y)\) can be regarded as the surface-induced current \(J_{s}\) on the conductor radiating in half-space. By means of the two-dimensional Green's function, the scattered field outside the conductor can be expressed as:  
\[E_{s}= \begin{cases}-\int_{0}^{2\pi}G_{1}(\tilde{r};F(\theta^{\prime}),\theta )J(\theta^{\prime})d\theta^{\prime}&,\quad y\leq-a\\ -\int_{0}^{2\pi}G_{2}(\tilde{r};F(\theta^{\prime}),\theta)J(\theta^{\prime})d \theta^{\prime}&,\quad y>-a\end{cases}  \tag{11}\]  
where  
\[\tilde{r}=(x,y),\quad\quad J(\theta)=-j\omega\mu_{0}\sqrt{F^{2}(\theta)+F^{ \prime 2}(\theta)}J_{s}(\theta) \tag{12}\]  
The boundary condition for a perfect conductor is that the total electric field in the tangential direction on the surface of the conductor is zero. According to this boundary condition, we can obtain the integral equation of \(J(\theta)\). \[E_{2}(F(\theta),\theta)=\int_{0}^{2\pi}G_{2}(F(\theta),\theta;F(\theta^{\prime }),\theta^{\prime})J(\theta^{\prime})d\theta^{\prime} \tag{13}\]  
The scattered field in region 1 is:  
\[E_{s}(x,y)=-\int_{0}^{2\pi}G_{1}(x,y;F(\theta^{\prime}),\theta^{\prime})J( \theta^{\prime})d\theta^{\prime}\ \,y\leq-a \tag{14}\]  
For the direct scattering problem, \(E_{s}\) was calculated from \(J(\theta)\) using the given \(F(\theta)\) shape function in region 1. For inverse problem, any point within the scatterer was presumed. Under this circumstance, \(F(\theta)\) can be expressed as:  
\[F(\theta)=\sum_{n=0}^{N/2}B_{n}\cos(n\theta)+\sum_{n=1}^{N/2}C_{n}\sin(n\theta) \tag{15}\]  
where \(B_{n}\) and \(C_{n}\) denote the parameters to be computed, and N + 1 is an unknown number of the parameters. In the inverse problem, the following objective function, which was minimized using the WOA and SHO, was defined as:\[OF=\left\{\frac{1}{M_{t}}\sum_{m=1}^{M_{t}}\left|E_{s}^{exp}(\tilde{r}_{m})-E_{s}^ {cal}(\tilde{r}_{m})\right|^{2}/\left|E_{s}^{exp}(\tilde{r}_{m})\right|^{2} \right\}^{\frac{1}{2}} \tag{16}\]  
where \(M_{t}\) is the measured number.