Creating adversarial sample against logistic regreession

In logistic regression, you're assuming $p(text{class } y!=!1|x) = sigma(w^top!x+b)$ for some $w,b$, which can be estimated from the data. Typically, a point $x$ is assigned to class $y!=!1$ If $sigma(w^top!x+b)>0.5$, otherwise $y!=!0$. The decision boundary are those $x$ where $sigma(w^top!x+b)=0.5$, or simply $w^top!x+b =0$ (a hyper-plane orthogonal to $w$, with $w^top!x=-b$).

Given some $x$ that the model labels $y!=!1$ (i.e. $w^top!x+b >0$), it is projected to the closest point $x'$ on the decision boundary, by subtracting some amount ($lambda$) of the vector $w$, since that is orthogonal to the decision boundary. That is, $lambda$ satisfies $0=w^top!x'+b =w^top!(x!-!lambda w)+b$.

Rearranging, gives $lambda = tfrac{w^top!x+b}{w^top w} $, so $ x' = x!-!tfrac{w^top!x+b}{w^top w} w $ is the projection of $x$ on the decision boundary. Picking a larger $lambda$ gives a point past the decision boundary, which is assigned class $y!=!0$. (Note that $lambda = tfrac{w^top!x+b}{w^top w}$ must be positive since $w^top!x+b$ is positive for the given $x$ and $w^top!w$ is always positive).

So, for $x$ labelled $y!=!1$, the closest point $x^*$ labelled $y!=!0$ (by Euclidean distance) is $ x^* = x-lambda^* w $, where $ lambda^*=tfrac{w^top!x+b}{w^top w}+delta $, and $delta!>!0$ is small.