# An Anchor-Free Location Algorithm Based on Transition Coordinates
## 3 Algorithm Description  
This paper proposes an algorithm based on transition coordinates \(\mathbf{p}_{i}^{k},1\leq i\leq N\), \(0\leq k\leq K\) to achieve an anchor-free location. As can be seen from the previous section, we set the changes in the transition coordinates \(\mathbf{p}_{i}^{k},1\leq i\leq N,0\leq k\leq K-1\) of each member on its horizontal and vertical axes \(\Delta x_{i}^{k+1}\) and \(\Delta y_{i}^{k+1}\) at time \(k+1\), then they are written in the form of vectors \(\Delta\mathbf{p}_{i}^{k+1}=\left(\Delta x_{i}^{k+1},\Delta y_{i}^{k+1}\right)\), thus obtaining the transition coordinates of \(N\) members at time \(k+1\):  
\[\mathbf{p}_{i}^{k+1} =\mathbf{p}_{i}^{k}+\Delta\mathbf{p}_{i}^{k+1}\] \[=\left(x_{i}^{k},y_{i}^{k}\right)+\left(\Delta x_{i}^{k+1},\Delta y _{i}^{k+1}\right)\] \[=\left(x_{i}^{k}+\Delta x_{i}^{k+1},y_{i}^{k}+\Delta y_{i}^{k+1}\right) \tag{1}\]  
In formula (1), the left side of parentheses represents the horizontal coordinate \(x_{i}\) of node \(i\) at time \(k\), adding the increment along the \(x\)-axis, and the right side of parentheses  
Figure 1: Schematic diagram of anchor-free positioning. represents the vertical coordinate \(y_{i}\) of node \(y\) at time \(k\) adding the increment along the \(y\)-axis. The formula for \(\Delta x_{i}^{k+1}\) and \(\Delta y_{i}^{k+1}\) is:  
\[\left\{ \begin{array}{l}\Delta x_{i}^{k+1}=\frac{1}{(N-1)}\sum\limits_{ \begin{subarray}{c}j=1\text{to}N\\ \text{and}j\
eq i\end{subarray}}\Big{(}d_{ij}^{\prime}-d_{ij}^{k}\Big{)}cos \alpha_{ij}\\ \Delta y_{i}^{k+1}=\frac{1}{(N-1)}\sum\limits_{\begin{subarray}{c}j=1\text{to}N \\ \text{and}j\
eq i\end{subarray}}\Big{(}d_{ij}^{\prime}-d_{ij}^{k}\Big{)}sin \alpha_{ij}\\ \end{array} \right. \tag{2}\]  
In formula (2), \(\alpha_{ij}\) is the angle formed between vector \(\textbf{p}_{i}^{k},1\leq i\leq N,0\leq k\leq K\) and the positive direction of the \(x\)-axis, which ranges from 0 degrees to 360 degrees. So, formula (2) can be converted to:  
\[\left\{ \begin{array}{l}\Delta x_{i}^{k+1}=\frac{1}{(N-1)}\sum\limits_{ \begin{subarray}{c}j=1\text{to}N\\ \text{and}j\
eq i\end{subarray}}\Big{(}d_{ij}^{\prime}-d_{ij}^{k}\Big{)}\frac{x _{i}^{k}-x_{j}^{k}}{\sqrt{\left(x_{i}^{k}-x_{j}^{k}\right)^{2}+\left(y_{i}^{k }-y_{j}^{k}\right)^{2}}}\\ \Delta y_{i}^{k+1}=\frac{1}{(N-1)}\sum\limits_{\begin{subarray}{c}j=1\text{to}N \\ \text{and}j\
eq i\end{subarray}}\Big{(}d_{ij}^{\prime}-d_{ij}^{k}\Big{)}\frac{y _{i}^{k}-y_{j}^{k}}{\sqrt{\left(x_{i}^{k}-x_{j}^{k}\right)^{2}+\left(y_{i}^{k }-y_{j}^{k}\right)^{2}}}\\ \end{array} \right. \tag{3}\]  
In Equation (3), the \(d_{ij}^{\prime}-d_{ij}^{k}\) formula is \(d_{ij}^{\prime}-d_{ij}^{k}=d_{ij}^{\prime}-\sqrt{\left(x_{i}^{k}-x_{j}^{k} \right)^{2}+\left(y_{i}^{k}-y_{j}^{k}\right)^{2}}\). The coefficient of the formula is \(1/(N-1)\), which is because in addition to the member \(i\), there are \(N-1\) members, and the increment of \(d_{ij}^{\prime}\) of each member relative to \(d_{ij}^{k}\) can cause the change in the \(x\)-axis and \(y\)-axis coordinate increment of member \(i\). Let us assume that they have the identical effect, namely, their weighted average has the equivalent weight coefficient, so they are all equal to \(1/N-1\). From the above formula, we can calculate the distance between the transition coordinates \(\textbf{p}_{i}^{k},1\leq i\leq N,0\leq k\leq K-1\) of \(N\) members at time \(k+1\) and the expression \(d_{ij}^{k+1},i,j=1,2,\cdots,N,j\
eq i\) is as follows:  
\[d_{ij}^{k+1}=\sqrt{\left(x_{i}^{k}-x_{j}^{k}+\Delta x_{i}^{k+1}\right)^{2}+ \left(y_{i}^{k}-y_{j}^{k}+\Delta y_{i}^{k+1}\right)^{2}} \tag{4}\]  
In formula (not move along the direction perpendicular to \(\mathbf{e}_{ij}\).