## 3 Chaotic Activity Filtering Approach Based on Direct Following Relationship
### Direct Chaotic Activity Filtering
First of all, for activity x, find the activities that have both a direct following relationship and direct preceding relationship with \(x\) in the relation table, and find that #(\(<\)_x,b\(>\),L_), #(\(<\)_b,x\(>\),L_) are all greater than 0, #(\(<x,b>,L)-\#(<b,x>,L)|<\frac{\#(<x,b>,L)+\#(<x,b>,L)}{2}\) #(\(<\)_x,c\(>\),L_), #(\(<\)_c,x\(>\),L_) are greater than 0, \(\left|\#(<x,c>,L)-\#(<c,x>,L)\right|<\frac{\#(<x,c>,L)+\#(<c,x>,L)}{2}\) #(\(<\)_x,d\(>\),L_), #(\(<\)_d,x\(>\),L_) are all greater than 0, and \(\left|\#(<x,d>,L)-\#(<d,x>,L)\right|<\frac{\#(<x,c>,L)+\#(<d,x>,L)}{2}\). So for activity \(x\), the collection of activities that satisfies the conditions is \(\{b,c,d\}\). This calculation is performed for all activities in the log, and the corresponding results of each activity are obtained, and the average value of all results is obtained. In this log, the average value is 1.2. The result of activity \(x\) is greater than the average, so it is judged that the candidate chaotic activity is \(x\). **Condition (4):** By dividing the value obtained from condition (3) with the value obtained from condition (2), the set of activities that have both a direct following relationship and direct preceding relationship with an activity in the log is obtained through condition (2), and through condition (3), we obtain the set of activities with a small difference between the direct following frequency and the direct preceding frequency between the activities in the log. Calculate the proportion of the latter to the former; this is the chaos degree of the activity. If the value calculated for an activity is 0 for condition (2), the value calculated for that activity is 0 for condition (4). A threshold is set, and if the chaos degree of activity x exceeds this threshold, it is judged that x is a candidate chaotic activity. Take the log \(L=[\)_ca,b,c,d,x\(>\)\({}^{10}\), _ca,b,x,c,d\(>\)\({}^{10}\), _ca,x,b,c,d\(>\)\({}^{10}\)_, _ca,x,b,c,d\(>\)\({}^{10}]\)_ as an example, and use Table 1 to calculate activity \(x\). Firstly, for activity \(x\) in the event log, condition (2) is calculated, and \(CH_{2}\) (\(x\)) = 3 is obtained. For condition (3), \(CH_{3}\) (\(x\)) = 3, so for condition (4), the calculated result is \(CH_{4}\) (\(x\)) = 1. This calculation is performed for all of the activities in the log, and the corresponding results of each activity are obtained. For all of the results, the average value is 0.8. The results of activities \(b\), \(c\), and \(x\) are greater than the average, so it is judged that the candidate chaotic activities are \(b\), \(c\), and \(x\). Take the event log \(L=[\)_ca,b,c,d,x\(>\)\({}^{10}\), _ca,b,x,c,d\(>\)\({}^{10}\), _ca,x,b,c,d\(>\)\({}^{10}\)_, _ca,x,b,c,d\(>\)\({}^{10}]\)_ as an example. For condition (1), we obtain \(\{CH_{1}(a)=2\), \(CH_{1}(b)=4\), \(CH_{1}(c)=4\), \(CH_{1}(d)=3\), \(CH_{1}(x)=7\}\), with an average of 5, so the candidate chaotic activities are \(b\), \(x\). For condition (2), it is found that \(\{CH_{2}(a)=0\), \(CH_{2}(b)=1\), \(CH_{2}(c)=1\), \(CH_{2}(d)=1\), \(CH_{2}(x)=3\}\) and the average value is 1.2, so the candidate chaotic activity is \(x\). For condition (3), we find that \(\{CH_{3}(a)=0\), \(CH_{3}(b)=1\), \(CH_{3}(c)=1\)\(CH_{3}(d)=1\), \(CH_{3}(x)=3\}\) and the average value is 1.2, so the candidate chaotic activity is x. For condition (4), it is found that \(CH_{4}(a)=0\), \(CH_{4}(b)=1\), \(CH_{4}(c)=1\), \(CH_{4}(d)=1\), \(CH_{4}(x)=1\), with an average value of 0.8, and the candidate chaotic activities are \(b\), \(c\), \(d\), and \(x\). Finally, the chaotic activity is \(x\).