1
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so that $H_0$ is equidistant from $H_1$ and $H_2$.

However, here the variable $\delta$ is not necessary. So we can set $\delta=1$ to simplify the problem.

$$w\cdot x+b=1 $$ and

$$w\cdot x+b=−1$$

Why is this assumption is taken? If it is taken, we can get the distance between two planes as $2$ directly because both are parallel and differ by $2$. how it is $\frac{2}{\left\|w\right\|}$ instead.

got equations from this https://www.svm-tutorial.com/2015/06/svm-understanding-math-part-3/

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After we have

$$w^Tx + b = \pm \delta$$

We can always divide everything by $\delta$,

$$\left( \frac{w}{\delta}\right)^Tx + \left( \frac{b}{\delta}\right)=\pm1$$

Now, we can set $\tilde{w}=\frac{w}{\delta}$ and $\tilde{b}=\frac{b}{\delta}$.

$$\tilde{w}^Tx+\tilde{b}=\pm1$$

This is as if we have set $\delta=1$ from the beginning.

The derivation of the distance formula has been given in equation $(19)$ in the article that you linked to and you might like to be more specific if you can't understand it. The distance should be $\frac{2\delta}{\|w\|}$ if $\delta$ is not set to be $1$.

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  • $\begingroup$ Thanks, i have checked that equation (19). what am not able to get intuitively is if both hyper planes are parallel, after replacing W and X with scalar values what you get is two lines differing by 2 from -1 to 1 so the margin will be 2. $\endgroup$ – Naveen Meka Sep 16 '19 at 3:52
  • $\begingroup$ Referring equation (3) we get y-ax+b = ±1 so both differ by 2. $\endgroup$ – Naveen Meka Sep 16 '19 at 3:59
  • $\begingroup$ Note that the distance between $y=x+1$ and $y=x-1$ is $\sqrt2$ rather than $2$. The distance is measured along the normal direction. $\endgroup$ – Siong Thye Goh Sep 16 '19 at 5:12

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