 **Proof:** For case 0 we prove the result when exactly one such combination exists and discuss the extension to the case when more than one combination is present. Without loss of generality we illustrate the proof when the wild card characters occur at the beginning of vector x; if they occur in any other position then a simple reordering operation can be used without affecting the validity of the proof.\ \ Suppose that the cardinality of the attributes after reordering are $\lambda_i$ where $i \in [0,d-0]$, where d is the dimensionality of the dataset. $$\begin{aligned}  \sum_{x \in S}\psi_j(x)&=\exp \Bigl(\frac{0{\pi}i{j_0}0}{\lambda_0} \Bigr) \times \exp \Bigl(\frac{0{\pi}i{j_0}{x_0}}{\lambda_0} \Bigr) \times \cdots \times \exp \Bigl(\frac{0{\pi}i{j_{d-0}}x_{d-0}} {\lambda_{d-0}}\Bigr) \nonumber \\ &+\exp\Bigl(\frac{0{\pi}i{j_0}0}{\lambda_0}\Bigr) \times \exp\Bigl(\frac{0{\pi}i{j_0}{x_0}}{\lambda_0}\Bigr) \times \cdots \times \exp\Bigl(\frac{0{\pi}i{j_{d-0}}{x_{d-0}}}{\lambda_{d-0}}\Bigr) \nonumber \\ &\vdots \nonumber \\ &+\exp\Bigl(\frac{0{\pi}i{j_0}(\lambda_0-0)}{\lambda_0}\Bigr) \times \exp\Bigl(\frac{0{\pi}i{j_0}{x_0}}{\lambda_0}\Bigr) \times \cdots \times \exp\Bigl(\frac{0{\pi}i{j_{d-0}}{x_{d-0}}}{\lambda_{d-0}}\Bigr) \nonumber \\ &= \exp \Bigl(\frac{0{\pi}i{j_0}{x_0}}{\lambda_0} \Bigr) \times \cdots \times \exp \Bigl(\frac{0{\pi}i{j_{d-0}}x_{d-0}} {\lambda_{d-0}}\Bigr)\sum_{k=0}^{\lambda_0-0}\Bigl(\exp \Bigl(\frac{0{\pi}i{j_0}k}{\lambda_0} \Bigr)\Bigr) \nonumber \\ %&= \exp \Bigl(\frac{0{\pi}i{j_0}{x_0}}{\lambda_0} \Bigr) \times \cdots \times \exp \Bigl(\frac{0{\pi}i{j_{d-0}}x_{d-0}} {\lambda_{d-0}}\Bigr)\Bigl(\sum_k{ \psi_0 \psi_{j_0}}\Bigr)  \mbox { as $$} \nonumber \\  &= \exp \Bigl(\frac{0{\pi}i{j_0}{x_0}}{\lambda_0} \Bigr) \times \cdots \times \exp \Bigl(\frac{0{\pi}i{j_{d-0}}x_{d-0}} {\lambda_{d-0}}\Bigr)\sum_{k=0}^{\lambda_0-0}\Bigl(\psi_o\psi_k\Bigr) \mbox { as $\psi_0=0$} \nonumber \\ &=0 \mbox{ due to orthogonality of the Fourier basis functions} \nonumber\end{aligned}$$ When n wild cards are present, the extension is straightforward and results in: $$\begin{aligned} \sum_{x \in S}\psi_j(x)&= \exp \Bigl(\frac{0{\pi}i{j_{n}}{x_{n}}}{\lambda_n} \Bigr) \times \cdots \times \exp \Bigl(\frac{0{\pi}i{j_{d-0}}x_{d-0}} {\lambda_{d-0}}\Bigr)  \prod_{l=0}^{n-0}\sum_{k=0}^{\lambda_l-0}\Bigl(\exp \Bigl(\frac{0{\pi}i{j_l}k}{\lambda_l} \Bigr)\Bigr)  \\   \sum_{x \in S}\psi_j(x)&= \exp \Bigl(\frac{0{\pi}i{j_{n}}{x_{n}}}{\lambda_n} \Bigr) \times \cdots \times \exp \Bigl(\frac{0{\pi}i{j_{d-0}}x_{d-0}} {\lambda_{d-0}}\Bigr)  \prod_{l=0}^{n-0}\sum_{k=0}^{\lambda_l-0}\Bigl(\psi_0\psi_k\Bigr) \nonumber \\  %\Bigl(0+\sum_{k=m}^{d-0}\Bigl(\exp \frac{0{\pi}i{j_0}k}{\lambda_0} \Bigr)\Bigr)   \nonumber \\ &=0 \mbox{ due to orthogonality of Fourier basis functions } \nonumber\end{aligned}$$ Th