Add a 3rd dimension to each of the extra input dimension so that the data becomes linearly separable.
X Y
00 1
01 -1
10 -1
11 1
->Third value is equal to first value for each data
->Third value is 1 for one data point, and 0 for other three
->Third value is opposite of second value for each data point
->None of the above
Please explain the solution for this question
hey @samriddhijain2000 ,
as you know the given data is not linearly separable , so we need to introduce a third variable to make it linearly separable. There is no such perfect rule that states on what basis we need to introduce our third variable , so we need to try the given options and find the one which works.
we tried both 1 and 3 options but we weren’t able to plot a hyper plane that can divide them into different classes . So we tried option 2. where i got the respective hyper plane when i added a 1 for 1 point and 0 for others.
after adding them we get points as
A( 0,0,1 ) , B( 0,1,0 ) , C( 1,0,0 ) ,D( 1,1,0 )
You can have a look at the below plot to understand how this option is correct for the above question.
F ( 0,0,0.8 ) and G ( 1, 0.8, 0 ) and H( 0,1.5,0 )
point H is currently invisible in this plot below.
using these two points F and G i have shown your hyperplane distinguishing points A,D and B,C
The hyperplane can also be shifted to get some other point nearby , but for that you need to try various values. This hyperplane can be drawn with given equation - 0.56x - 0.8y - 1.5z = -1.2
I hope this would have resolved your doubt and we apologize for such late response.
Thank You
Happy Learning 
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