#nonconvex

Question 1: Single choice

A lightweight Newton method preconditions with the Hessian diagonal, $\mathbf{x} \leftarrow \mathbf{x} - \eta\, \textrm{diag}(\mathbf{H})^{-1} \nabla f(\mathbf{x})$ . On nonconvex functions, why divide by $|H_{ii}|$ rather than by $H_{ii}$ ?

Where $H_{ii} < 0$ the step would go uphill along that coordinate; $|H_{ii}|$ keeps it a descent step1
Negative entries make $\textrm{diag}(\mathbf{H})$ singular, so it cannot be inverted2
The absolute value is cheaper to obtain by backpropagation than the signed entry3
It makes the method converge in one step on every quadratic, convex or not4

Timed test