https://arxiv.org/pdf/hep-lat/0005018.pdf

1. The larger Wilson loops have less signal to noise ratio (mean values are lost in error).
2. Using HYP smearing did helped but not for desired accuracy. We can make use of 'Bresenham Algorithm'.
3. We were taking rectangular Wilson loops, but it limited the number of Wilson loops for given 'R(separation of heavy Quark and anti-quark pair)', so signal to noise ratio was still very small.

The main idea is to make use of rotational symmetry, and use 'R' not only limited to along any spatial direction (x,y,z), but to any arbitrary spatial direction.

The use of this Algorithm is used to draw any angular lines in display, note that the pixels are square boxes, so in order to draw a diagonal line, Bresenham algorithm is used.

The paper referenced above only takes R from \(R_{min} \le R \le R_{max}\), where \(R_{min}=10\) is and \(R_{max}=12 \sqrt(3)\) , they did it as they wanted to study string breaking effect(for our case we are interested in spatial string tension(so we are not restricted to use this range of R)).

For our case :

\[ \begin{eqnarray} R_{max}=N_{\sigma}-1 \\ R_{min}=0\\ \end{eqnarray} \]

We shall have to find the solutions of :

\[ \begin{eqnarray} 0\le R_{x}^2+R_{y}^2+R_{z}^2\le R^2\\ \end{eqnarray} \]

Since our Wilson loop is spatial, so our R is in 2d rather than 3d, and one spatial axis is our fictitious time. So for us :

\[ \begin{eqnarray} 0\le R_{x}^2+R_{y}^2 \le R^2\\ \end{eqnarray} \]

We need to find all these points (set of values of (\(R_{x}, R{y}\)) which will satisfy above equation. R can take continuous values from 0 to \(R_{max}\).