# Table 3 Calculation of the temporal correlation coefficient C for time series with identical unconnected components of equal size and isolated node

Snapshots 1st calculation step 2nd calculation step 3rd calculation step
$$t_{m} , t_{m + 1}$$ $$C_{i = 1} \left( {t_{m} , t_{m + 1} } \right) = \frac{1}{\sqrt 3 }$$ Method 1: $$C_{m} = \frac{1}{N}\mathop \sum \nolimits_{i = 1}^{N} C_{i} \left( {t_{m} , t_{m + 1} } \right) \approx 0.32$$
Method 2: $$C_{m} = \frac{1}{{\text{max} \left[ {N\left( {t_{m} } \right), N\left( {t_{m + 1} } \right)} \right]}}\mathop \sum \nolimits_{i = 1}^{N} C_{i} \left( {t_{m} , t_{m + 1} } \right) \approx 0.39$$
Method 3: $$C_{m} = \frac{1}{{\text{max} \left[ {A\left( {t_{m} } \right), A\left( {t_{m + 1} } \right)} \right]}}\mathop \sum \nolimits_{i = 1}^{N} C_{i} \left( {t_{m} , t_{m + 1} } \right) \approx 0.39$$
Method 1: $$C = \frac{1}{M - 1}\mathop \sum \nolimits_{m}^{M - 1} C_{m} \approx 0.56$$
Method 2: $$C = \frac{1}{M - 1}\mathop \sum \nolimits_{m}^{M - 1} C_{m} \approx 1.20$$
Method 3: $$C = \frac{1}{M - 1}\mathop \sum \nolimits_{m}^{M - 1} C_{m} \approx 0.70$$
$$C_{i = 2} \left( {t_{m} , t_{m + 1} } \right) = 1$$
$$C_{i = 3} \left( {t_{m} , t_{m + 1} } \right) = 0$$
$$C_{i = 4} \left( {t_{m} , t_{m + 1} } \right) = 0$$
$$C_{i = 5} \left( {t_{m} , t_{m + 1} } \right) = 0$$
$$t_{m + 1} , t_{m + 2}$$ $$C_{i = 1} \left( {t_{m + 1} , t_{m + 2} } \right) = 1$$ Method 1: $$C_{m + 1} = \frac{1}{N}\mathop \sum \nolimits_{i = 1}^{N} C_{i} \left( {t_{m + 1} , t_{m + 2} } \right) = 0.80$$
Method 2: $$C_{m + 1} = \frac{1}{{\text{max} \left[ {N\left( {t_{m + 1} } \right), N\left( {t_{m + 2} } \right)} \right]}}\mathop \sum \nolimits_{i = 1}^{N} C_{i} \left( {t_{m + 1} , t_{m + 2} } \right) = 2$$
Method 3: $$C_{m + 1} = \frac{1}{{\text{max} \left[ {A\left( {t_{m + 1} } \right), A\left( {t_{m + 2} } \right)} \right]}}\mathop \sum \nolimits_{i = 1}^{N} C_{i} \left( {t_{m + 1} , t_{m + 2} } \right) = 1$$
$$C_{i = 2} \left( {t_{m + 1} , t_{m + 2} } \right) = 1$$
$$C_{i = 3} \left( {t_{m + 1} , t_{m + 2} } \right) = 1$$
$$C_{i = 4} \left( {t_{m + 1} , t_{m + 2} } \right) = 1$$
$$C_{i = 5} \left( {t_{m + 1} , t_{m + 2} } \right) = 0$$