14 Agnieszka Patyk

_Pat = c00100, male = c00111, 66 = c11000, name = c00010,  sex = c11100,  age = c10001, }_
$$\mathbf{\tau}_{i}$$
$$(S4)$$
$$)\otimes1,$$
$$(\mathbf{55})$$
$\eqref{eq:walpha}$
$\epsilon$ ). 
$\overline{\overline{\phantom{\rule{0.000pt}{0ex}}}}$ . 
$\sigma_1\subset\sigma_2\subset\sigma_3$
$\left(57\right)$. 
$$\tau_{1}\otimes\sigma_{1}\otimes\sigma_{2})$$

$$(58)$$
$$(59)$$
$\eqref{eq:walpha}$. 

![0_image_0.png](0_image_0.png)

as described in Section 2. Their explicit matrix representations are

 $\rho_2=(\sigma_1\otimes\sigma_1\otimes\sigma_1\otimes\sigma_1\otimes\sigma_3)$  $1\otimes(-i\sigma_1)$. 
$$=1\otimes1\otimes1\otimes(-i\sigma_{1}),$$
$$\sigma_{1}\otimes\sigma_{1}\otimes\sigma_{2}\otimes1,$$
Pat = c00100 = b3 = σ1 ⊗σ1 ⊗σ1 ⊗σ3 ⊗1, (56)
male = c00111 = b3b4b5
= (σ1 ⊗σ1 ⊗σ1 ⊗σ3 ⊗1)(σ1 ⊗σ1 ⊗σ1 ⊗σ2 ⊗1)(σ1 ⊗σ1 ⊗σ3 ⊗1⊗1) = σ1 ⊗σ1 ⊗σ3 ⊗(−iσ1)⊗1, (57)
66 = c11000 = b1b2 = (σ1 ⊗σ1 ⊗σ1 ⊗σ1 ⊗σ3)(σ1 ⊗σ1 ⊗σ1 ⊗σ1 ⊗σ2)
= 1⊗1⊗1⊗1⊗(−iσ1), (58)
name = c00010 = b4 = σ1 ⊗σ1 ⊗σ1 ⊗σ2 ⊗1, (59)
sex = c11100 = b1b2b3
= (σ1 ⊗σ1 ⊗σ1 ⊗σ1 ⊗σ3)(σ1 ⊗σ1 ⊗σ1 ⊗σ1 ⊗σ2)(σ1 ⊗σ1 ⊗σ1 ⊗σ3 ⊗1)
= σ1 ⊗σ1 ⊗σ3 ⊗1⊗(−iσ1), (60)
age = c10001 = b1b5 = (σ1 ⊗σ1 ⊗σ1 ⊗σ1 ⊗σ3)(σ1 ⊗σ1 ⊗σ3 ⊗1⊗1)
= 1⊗1⊗(−iσ2)⊗σ1 ⊗σ3. (61)

$\mathbf{j}(\mathbf{o}_1\otimes\mathbf{o}_1\otimes\mathbf{o}_1\otimes\mathbf{o}_1\otimes\mathbf{o}_1)$  $\mathbf{r}_1$). 
$$|\rangle_{*}$$
$$=1\otimes1\otimes(-i\sigma_{2})\otimes\sigma_{1}\otimes\sigma_{3}.$$

$\mathbf{a}$
Figure 8 shows six blades making up *PSmith* for n = 5 and Figure 9 shows the matrix representation of *PSmith* for n ∈ {6,7}, black dots indicate nonzero matrix entries.

The regularity of patterns placed along the diagonals is not an accident. Consider Cartan representation of blades b2k and b2k−1 - the shortest sequence of n − k σ1's will occur for k = d n 2 e (in other words, in blade bn). Therefore each blade b1*,...,*bn has at least b n 2 c of σ1's placed at the beginning of the formula describing its representation. Hence, there are exactly 2b n 2 c"boxes" of patterns placed along one of the diagonals, each one of dimensions 2d n 2 e × 2 d n 2 e. To extract a part individual for a given blade, one needs to consider only the last d n 2 e+1 of σ's or unit matrices belonging to its representation - the extra σ1 bearing the number n− d n2 e is needed to preserve the direction of the diagonal - either "top left to bottom right" or "top