# Exact Line Search

The gradient descent I learnt uses $x^{k+1} = x^k + t\triangledown f(x)$ and we learnt to set $t$ heuristically. Am I right to say that exact line search simply computes the optimal value of $t$ that minimizes the $f(x) ?$

Wouldn't I be able to look for the global minima in 1 iteration in that case ? I can't see the negatives of this algorithm eg being stuck in a local minima. Can someone give an example ?

• What if the global minimum isn't on the line you're searching along?
– user856
Mar 9, 2018 at 12:27
• I am finding it extremely hard to vizualize an example in 3D. All I can do is vizualize one in 2D
– Kong
Mar 9, 2018 at 12:48
• do you have a link to a webpage with a 3d example
– Kong
Mar 9, 2018 at 12:49
• please tell me what is the heuristic method of setting $t$, I'm not sure how to get the $t = argmin_{s\ge0} f(x+s\Delta x)$ while implementing the descent
– ALEX
Jul 16, 2019 at 16:20

Exact line search computes the global solution of the one-dimensional problem $$\min_{t\in \mathbb R} f(x_k - t\nabla f(x_k)).$$ This might be easy if $f$ has nice structure (quadratic).
If the global minima of $f$ are not on the line $x_k - t\nabla f(x_k)$, then you will not find the global minimum in the current step.