Consistency is a popular feature subset evaluation measure. The measure is motived by the idea that a good feature subset should have high consistency or low inconsistency among matching instances. Two instances are considered matching instances if their values of all features $($except the class$)$ match. For $2$ matching instances, an inconsistency occurs if their class values are different. For n matching instances, the number of inconsistencies $($i$.$e$.,$ inconsistency count$)$ is determined by n $$ max$\{$m$0,$ m$1\},$ where m$0+$m$1=$n$,$ and m$0$ and m$1$ are the number instances for class $0$ and class $1$ respectively. In other words, it is the count of the instances in the minority class. For example, if we only consider the data in the two columns of F$1$ and C $($ignoring features F$2,$ F$3,$ and F$4),$ there are $4$ matching instances with F$1$ being $1.$ The inconsistency count for this set of matching instances is $4-$max$\{2,2\}=2.$ Similarly, there are $4$ matching instances with F$1$ being $0,$ and the same inconsistency count $2.$ The total inconsistency count for F$1$ is $2+2=4.$ Consider the following data with four binary features $($F$1,$ F$2,$ F$3,$ and F$4)$ and binary class label C$.$ Perform feature selection using Sequential Backward Selection $($SBS$)$ search coupled with total inconsistency count as the subset evaluation measure $($smaller inconsistency is better$).$ What will be the selected subset of two features $($illustrate your solution by intermediate steps$)?$ F$1(1,1,1,0,1,0,0,0)$ F$2(1,1,0,1,0,0,0,1)$ F$3(1,0,0,0,1,1,0,1)$ F$4(1,1,0,0,0,1,0,1)$ C $(1,1,0,0,0,1,1,0)(20$ points$)$ For the same data set above, if Sequential forward Selection $($SFS$)$ is used with the same subset evaluation measure, what will be the selected subset of two features $($illustrate your solution$)?3.(10$ points$)$ Discuss which method, SBS or SFS$,$ is better for the above data and why?