# Elements of Hyperstructure Theory in UWSN Design and Data Aggregation
## 3 Use of the Theory of Quasi-Automata
Every tier has an arbitrary number of clusters, i.e., an arbitrary number of cluster heads, each with the same memory capacity. Then by vector e.g., \(\vec{v}=(0.3;0.1;0;0.5)\), we describe such a state of the system that, at a certain point in time, \(30\%\) data has been collected (or rather, \(30\%\) memory capacity has been used) from tier 1 with the numbers being \(10\%\) from tier 2, \(0\%\) from tier 3, and \(50\%\) from tier 4. Now regard a matrix:_  
\[\mathbf{A}= \begin{bmatrix}0&0&0&0\\ 1&1&0&0\\ 0&0&0&0\\ 0&0&1&0\end{bmatrix} \in I_{M},\]  
_which we apply on \(\vec{v}\). Then \(\delta(\mathbf{A},\vec{v})=(0.1;0.1;0.5;0)\), which describes the new state of the system. Of course, the first \(30\%\) of data (represented by the first component of \(\vec{v}\)) did not get lost, which we can regard as output, i.e., to each quasi-automaton we can assign output, meaning that data from the upmost tier 1 have been processed, sent out of the system. Furthermore, in our example, data from tier 2 were transferred to tier 1 (and left in tier 2, i.e., in this case we have a backup copy). Also, data from tier 4 were transferred to tier 3, which had been empty, and simultaneously deleted in tier 4. It can be easily observed that the issue of suitable construction of matrices used for quasi-automata inputs is a topic for further research which will be closely linked to various aspects of optimization theory._  
When describing the state hypergroup of \((I_{M},S_{v},\delta)\), we must first of all decide whether or not \((S_{v},\circ)\) defined by Equation (3) is trivial. However, this task is rather simple. **Lemma 6**. : _The state hypergroup \((S_{v},\circ)\) of \((I_{M},S_{v},\delta)\) is not trivial, i.e., there exist states \(s,t\in S_{v}\) such that \(\emptyset\
eq\delta(I_{M}^{*},s)\cup\delta(I_{M}^{*},t)\
eq S_{v}\)._  
Proof. : The transition function \(\delta\) is defined on \((I_{M},S_{v},\delta)\) by \(\delta(\mathbf{A},\vec{v})=\vec{v}\mathbf{A}\) for all \(\mathbf{A}\in I_{M}\) and all \(\vec{v}\in S_{v}\). Moreover, by Equation (5), components of vectors from \(S_{v}\) are real numbers from interval \(\langle 0;1\rangle\), and by (6) matrices from \(I_{M}\) are upper triangular with entries taken from the same interval. Finally, in a text preceding Theorem 1 we have already mentioned that \(I_{M}^{*}=I_{M}\). Thus, for all \(s,t\in I_{M}\) and all \(\mathbf{A}\in I_{M}\), we have that:  
\[\delta(\mathbf{A},\vec{s})=\vec{s}\mathbf{A}=(s_{1},s_{2},\ldots,s_{n}) \begin{bmatrix}a_{11}&0&0&\ldots&0\\ a_{21}&a_{22}&0&\ldots&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ a_{n1}&a_{n2}&a_{n3}&\ldots&a_{nn}\end{bmatrix} =\left(\sum_{i=1}^{n}s_{i}a_ {i1},\sum_{i=2}^{n}s_{i}a_{i2},\ldots\sum_{i=n-1}^{n}s_{i}a_{in-1},s_{n}a_{nn }\right).\]  
Now suppose that \(s_{n}=0.1\). Obviously in this case there cannot be \(\delta(\mathbf{A},\vec{s})=S_{v}\) because we are not able to find such a matrix \(\mathbf{A}\in I_{M}^{*}=I_{M}\) that \(s_{n}a_{nn}=0.9\) as this would require \(a_{nn}=9\) which is out of the interval \(\langle 0;1\rangle\) in which all entries of \(\mathbf{A}\) should be. Naturally, there are infinitely many of these choices. The fact that \(\delta(I_{M}^{*},s)\cup\delta(I_{M}^{*},t)\
eq\emptyset\) is obvious. Regard now the set \(\hat{I}_{M}\) which will be different from \(I_{M}\) because the condition of the column norm will be dropped and we will assume that \(b_{ii}\geq b_{ij}\) for all \(i\in\{1,2,\ldots,n\}\) and \(j\in\{1,\ldots,i\}\), i.e., that each diagonal element will be greater than all other elements in the given row. On \(\hat{I}_{M}\) we define a hyperoperation \"\(*\)\" by:  
\[\mathbf{A}*\mathbf{B}=\left\{ \begin{bmatrix}c_{11}&0&0&\ldots&0\\ c_{21}&c_{22}&0&\ldots&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ c_{n1}&c_{n2}&c_{n3}&\ldots&c_{nn}\end{bmatrix}