## 3 Reconstruction Algorithms
### Algorithm
Also among these points, we have one point at \(x=a\) and another one at \(x=b\). Moreover, we will assume that the initial dataset is consistent. Since Algorithm 1 recomputes the inconsistent points at all steps, we can also consider in the following that any new numerical observation is actually consistent. Also, we need to guarantee that the weighted area \(\operatorname{WA}^{(n)}\) will permanently oscillate about \(\mathcal{E}\) as the iteration step \(n\) is increasing; this is the purpose of Assumption 3 below as shown in the subsequent Proposition 1. From these properties it will then be shown that Algorithm 1 is convergent, as stated in Theorem 1. **Assumption 2**. : _Any new numerical value obtained by Algorithm 1 is consistent._

**Assumption 3**. : \(q_{-}^{(n)}\to+\infty\) _as \(n\to\infty\)._

Within Assumption 2 all points have a consistency of 1, and therefore \(q=r>0\) the reliability. Besides, one has \(G(x_{i}^{(n)},q_{i}^{(n)})\leq G(x_{i+1}^{(n)},q_{i+1}^{(n)})\), i.e., \(y_{i}^{(n)}\leq y_{i+1}^{(n)}\) for all points \(i\) and steps \(n\). We finally define the sequence of piecewise constant reconstruction functions \(F^{(n)}\) as follows. **Definition 1**. : _For each \(x\in[a,b]\), we define the reconstructing function \(F^{(n)}\) at step \(n\) as:_

\[F^{(n)}(x)=\sum_{i=1}^{I^{(n)}-1}y_{i}^{(n)}\mathbbm{1}_{x_{i}^{(n)}}(x)\,,\]

_and \(F^{(n)}(x_{I^{(n)}}^{(n)})=F^{(n)}(b)=y_{I^{(n)}}^{(n)}\)._

Now let

\[E^{+}:=\left\{n\in\mathbb{N}\,;\,\operatorname{WA}^{(n)}\geq\mathcal{E}\right\}, E^{-}:=\left\{n\in\mathbb{N}\,;\,\operatorname{WA}^{(n)}<\mathcal{E} \right\}, \tag{12}\]

which are such that \(E^{+}\cup E^{-}=\mathbb{N}\) and \(E^{+}\cap E^{-}=\varnothing\). In order to prove the convergence (in a sense to be given) of Algorithm 1, we first need to establish the following intermediate results, Proposition 1, Proposition 2, and Proposition 3. They clarify the behaviour of the sequence \(\operatorname{WA}^{(n)}\) when points are added to the dataset and the largest area \(a_{+}^{(n)}\) is divided into four parts at each iteration step \(n\); see \[q_{-}^{(n)}=\min_{1\leq i\leq I^{(n)}}\{q_{i}^{(n)}\}\,,\quad\text{A}^{(n)}= \sum_{i=1}^{I^{(n)}-1}a_{i}^{(n)}\,, \tag{6}\]0

**Proposition 1**. : \(E^{+}\) _is infinite._

Proof. : Let us assume that \(E^{+}\) is finite: \(\exists N\) such that \(\forall n\geq N\), \(n\in E^{-}\). Therefore we are in the situation \(\operatorname{WA}^{(n)}<\mathcal{E}\), the minimum quality \(q_{-}^{(n)}\) of the data goes to infinity, and the total area \(\operatorname{A}^{(n)}\) is

\[q_{-}^{(n)}=\min_{1\leq i\leq I^{(n)}}\{q_{i}^{(n)}\}\,,\quad\text{A}^{(n)}= \sum_{i=1}^{I^{(n)}-1}a_{i}^{(n)}\,, \tag{6}\]1 New area when one adds a point at the middle of the biggest rectangle. modified although the evaluation points \(\{x_{i}^{(n)}\}_{i=1}^{I^{(n)}}\) and their number \(I^{(n)}\) are unchanged; thus they are independent of \(n\). Repeating this step yields

\[q_{-}^{(n)}=\min_{1\leq i\leq I^{(n)}}\{q_{i}^{(n)}\}\,,\quad\text{A}^{(n)}= \sum_{i=1}^{I^{(n)}-1}a_{i}^{(n)}\,, \tag{6}\]2

since \(F^{\dagger}\) is monotonically increasing and non-constant on \([a,b]\), and Assumption 1 is used. Consequently \(\text{WA}^{(n)}\to+\infty\) as \(n\to\infty\), that is \(\text{WA}^{(n)}\geq\mathcal{E}\ \forall n\geq N_{1}\) for some \(N_{1}\), which is a contradiction. The set \(E^{+}\) is therefore of the form

\[q_{-}^{(n)}=\min_{1\leq i\leq I^{(n)}}\{q_{i}^{(n)}\}\,,\quad\text{A}^{(n)}= \sum_{i=1}^{I^{(n)}-1}a_{i}^{(n)}\,, \tag{6}\]3

where

\[q_{-}^{(n)}=\min_{1\leq i\leq I^{(n)}}\{q_{i}^{(n)}\}\,,\quad\text{A}^{(n)}= \sum_{i=1}^{I^{(n)}-1}a_{i}^{(n)}\,, \tag{6}\]4

Let us introduce the strictly increasing application \(\varphi:\mathbb{N}\to\mathbb{N}\) such that \(\varphi(p)\) is the \(p^{\text{th}}\) element of \(E^{+}\) (in increasing order), and \(\left[m_{k},n_{k}\right]=\varphi(\left[p_{k}+1,p_{k+1}\right])\). \(p\) is the counter of the elements of \(E^{+}\), and \(n\) is the corresponding iteration number. Proposition 2: _Let \(I^{(\varphi(p))}=I^{(\varphi(0))}+p\).