How do we find weight and b value of SVM classifier for XNOR operation by hand? Problem statement

Show that the Boolean function (x1 ∧ x2) ∨ (¬x1 ∧ ¬x2) is not linearly separable (i.e. there is no linear classifier sign(w1 x1 + w2 x2 + b) that classifies all 4 possible input points correctly). Assume that “true” is represented by 1 and “false” is represented by −1. Show that there is a linear separator for this Boolean function when we use the kernel K(x, y) = (x · y)2 (x.y denotes the ordinary inner product) . Give the weights and the value of b for one such separator

I was trying to solve the above problem in SVM by hand. The input dataset is XNOR truth table (here -1 is used as false value)




x1
x2
x1 XNOR x2




-1
-1
1


-1
1
-1


1
-1
-1


1
1
1




It was given to prove that this can not be classified using any linear classifier. Which I already did. Now the second part of the problem is to use the kernel function
$K(X,Y) = (X.Y)^{2}$
and prove that there is linear separator for this data when this kernel function is applied.
I proved it following way:
Applying transformation using that kernel, I could see
$\phi(X) = [x_1^{2}  ,\sqrt{2}x_1x_2,x_2^{2}]$
And I transformed entire dataset to this
$$\begin{bmatrix}
&1&\sqrt{2}&1\\
&1&-\sqrt{2}&1\\
&1&-\sqrt{2}&1\\
&1&\sqrt{2}&1\\
\end{bmatrix}
$$
Which can be easily separable.
But the problem wants me to find the weight and value of b variable.
This is so far I have done, but I'm really stuck thinking around
I know
$\alpha_i(y_i(W.X+b)-1)=0 $
and for non support vectors $\alpha_i$ is zero
But how can I know which data points are support vectors? I know these are points lying closer to "gutter" or hyperplane. But how can I know it mathematically? as this can't separated using a linear classifier and therefore I can't know which points lies closer to the imaginative plane. Also how will I find $\alpha_i$ for these support vectors? Don't I need weight and b value for it? But I want to find $\alpha_i$ first inorder to find weight and b value.
This question probably is a naïve one, which came up due to not having deeper understanding of SVM. But I would appreciate giving me some pointers to correct my thinking and guide me to solve this problem.
Edit: added my original problem statement, in case my explanation wasn't that clear.
 A: Separable case
If you want to get the weights for the separable case, look at the transformed dataset:
\begin{bmatrix}
&1&\sqrt{2}&1\\
&1&-\sqrt{2}&1\\
&1&-\sqrt{2}&1\\
&1&\sqrt{2}&1\\
\end{bmatrix}
and your y values are basically $[1, -1, -1, 1]$.
You have to apply the SVM method in the transformed space, i.e., your equation would be: $$\alpha_i(y_i(W.\phi(x_i)+b)-1)=0.$$
Notice that if you take the second coordinate of the $\phi(x_i),$ you can have a direct relationship with the y values. So take W to be $(0, 1, 0)$ which picks the second coordinate. Now your $W.\phi(x_i)$ values are $\sqrt{2}\ [1,-1,-1,1].$
If you if you scale the $W$ to be $(0, 1/\sqrt{2}, 0).$ and look carefully, the $y_i * (W.\phi(x_i))$ values are all $1.$
Hence, $b=0$ and any real values of $\alpha_i>0$ would work in this case.
Non-Separable case
Now you are in two-dimensional space, and your quadrant 1 and 3 basically have 1's, and the rest have -1's. First, take any line visually and check that it can not divide them. If the first and the fourth points are on the same side, at least one other point would be in the same side too.
Mathematically, why?
Take any line in the form $f(x_1,x_2)=x_2 - m x_1 - c = 0.$ Now if a point $(x,y)$ satisfies $f(x,y)>0,$ it will lie on one side, and if the value is $<0,$ it will lie on the other side. The values of $f$ for your four points are, respectively: $-1+m-c, 1+m-c, -1-m-c, 1-m-c.$
If $-1+m-c<0,$ and $1-m-c<0,$ both happen (i.e., first and fourth points are on the same side), then $1-c<m<c+1.$ If $c>0,$ the point $(1,-1)$ will be less than 0 too, i.e., that is on the same side of $(1,1) (-1,-1).$ If $c<0,$ the other point will be on the same side of them. Similarly, you can show that if $-1+m-c>0,$ and $1-m-c>0,$ both happen, one of the other points will be on the same side.
Also, note that the y-axis can not separate them.
You can translate this whole argument in the form of hyperplanes too.
