While using support vector machines (SVM), we encounter 3 types of lines (for a 2D case). One is the decision boundary and the other 2 are margins:

decision boundary of SVM

Why do we use $+1$ and $-1$ as the values after the $=$ sign while writing the equations for the SVM margins? What's so special about $1$ in this case?

For example, if $x$ and $y$ are two features then the decision boundary is: $ax+by+c=0$. Why are the two marginal boundaries represented as $ax+by+c=+1$ and $ax+by+c=-1$?


2 Answers 2


It's important for the optimization formulation of the SVM that $y_i=\{-1,1\}$ which is why it makes sense to also output $y=\{-1,1\}$. If we look at the soft-margin linear SVM we want to minimize:

$\left[\frac{1}{n}\sum_{i=1}^n\max{(0,1-y_i(w\cdot x_i+b))}\right]+\lambda\| w\| ^2$

The $y_i$ is either +1 or -1 which flips the hyperplane in the soft-margin definition of the problem.

  • $\begingroup$ I think you didn't understand my question. I have given some more details in my question. Please see it $\endgroup$ Commented May 31, 2017 at 17:19
  • 1
    $\begingroup$ I think he understood it. $y$ is not a feature; it's the response, and it was chosen the way it is for the loss function to have the tractable form given above. Try substituting $y=0$ as you suggested and examining the effect of changing the prediction $w\cdot x + b$. $\endgroup$
    – Emre
    Commented May 31, 2017 at 17:28

This is just mathematical convenience.

Suppose we have $W*X+b \ge k$ for positive points and $W*X+b \le -k$ for negative points. If we scale the equation with $1 \over k$, $W'*X+b' \ge 1$, where $W'= {W \over k}$, $b'= {b \over k}$. This doesn't change the optimization target for $W$ and $b$.


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