**Victor Sanches Portella**

Abril 2024

cs.ubc.ca/~victorsp

junto de **Frederik Kunstner**, **Nick Harvey**, e **Mark Schmidt**

** , IME - USP **

**SNAIL **🐌

Training/Fitting a ML model is often cast a **(uncontrained) optimization problem**

Usually in ML, models tend to be BIG

\displaystyle \min_{x \in \mathbb{R}^{d}}~f(x)

**\(d\) is BIG**

First-order (i.e., **gradient based**) methods fit the bill

(stochastic even more so)

\(O(d)\) **time and space per iteration **is preferable

**\(f\) is convex**

Not the case with Neural Networks

Still quite useful in theory and practice

\displaystyle \Bigg\{

More conditions on \(f\) for rates of convergence

**\(L\)-smooth**

\displaystyle \min_{x \in \mathbb{R}^{d}}~f(x)

**\(\mu\)-strongly convex**

\preceq \nabla^2 f(x) \preceq

L \cdot I

\mu \cdot I

"Easy to optimize"

\displaystyle x_{t+1} = x_t - \alpha \nabla f(x_t)

Which step-size \(\alpha\) should we pick?

\displaystyle \implies

\displaystyle f(x_t) - f(x_*) \leq \left( 1 - \frac{\mu}{L} \right)^t (f(x_0) - f(x_*))

Condition number

\displaystyle \alpha = \frac{1}{L}

\displaystyle \kappa = \frac{L}{\mu}

\(\kappa\) Big \(\implies\) hard function

If we know \(L\), picking \(1/L\) always works

**and is optimal**

What if we do not know \(L\)?

\displaystyle x_{t+1} = x_t - \tfrac{1}{L} \nabla f(x_t)

\displaystyle f(x_{t+1}) \leq f(x_t) - \tfrac{1}{L} \tfrac{1}{2} \lVert \nabla f(x_t) \rVert_2^2

**"Descent Lemma"**

**Idea:** Pick \(\eta\) big and see if the "descent condition" holds

(Locally \(1/\eta\)-smooth)

**worst-case**

**Line-search**: test if step-size \(\alpha_{\max}/2\) makes enough progress:

\displaystyle f(x_{t+1}) \leq f(x_t) - \alpha_{\max} \tfrac{1}{2} \lVert \nabla f(x_t) \rVert_2^2

**Armijo condition**

If this fails, **cut out** everything bigger than \(\alpha_{\max}/2\)

0

\alpha_{0}

\tfrac{\alpha_{0}}{2}

\tfrac{1}{L}

\tfrac{\alpha_{0}}{4}

**Backtracking Line-Search**

Start with \(\alpha_{\max} > 2 L\)

\(\alpha \gets \alpha_{\max}/2\)

**If**

\displaystyle f(x_{t+1}) \leq f(x_t) - \alpha \tfrac{1}{2} \lVert \nabla f(x_t) \rVert_2^2

\(t \gets t+1\)

**Else**

**While** \(t \leq T\)

\displaystyle \alpha_{\max} \gets \alpha_{\max}/2

Halve candidate space

**Guarantee**: step-size will be at least \(\tfrac{1}{2} \cdot \tfrac{1}{L}\)

Makes enough progress?

\displaystyle f(x) = x^T A x

\displaystyle A =
\begin{pmatrix}
1000 & 0 \\
0 & 0.001
\end{pmatrix}

\displaystyle \kappa = 10^{-6}

\displaystyle x_{t+1} = x_t -
\begin{pmatrix}
0.001 & 0 \\
0 & 1000
\end{pmatrix}
\nabla f(x_t)

**Converges in 1 step**

\(P\)

Can we find a good \(P\) automatically?

**"Adapt to \(f\)"**

**Preconditioer \(P\)**

P_t = \nabla^2 f(x_t)

Newton's method

is usually a great preconditioner

**Superlinear** convergence

...when \(\lVert x_t - x_*\rVert\) small

**Newton **may diverge otherwise

Using step-size with Newton and QN method ensures convergence away from \(x_*\)

**Worse than GD**

\displaystyle f(x_t) - f(x_*) \leq \left( 1 - \frac{1}{\kappa^2} \right)^t (f(x_0) - f(x_*))

\displaystyle \phantom{\kappa}^2

\(\nabla^2 f(x)\) is usually expensive to compute

P_t \approx \nabla^2 f(x_t)

should also help

Quasi-Newton Methods, e.g. BFGS

**Toy Problem:** Minimizing the **absolute value function**

\displaystyle x_{t+1} = x_t - P_t \cdot \nabla f(x_t)

Preconditioner at round \(t\)

**AdaGrad from Online Learning**

\displaystyle P_t = \mathrm{Diag}\Big( \sum_{i \leq t} \nabla f(x_i) \nabla f(x_i)^T \Big)^{-1/2}

Convergence guarantees

"adapt" to the function itself

Attains** linear rate in classical convex opt** (proved later)

But... Online Learning is** too adversarial**, AdaGrad is **"conservative"**

Also... "approximates the Hessian" is not quite true

But... Online Learning is** too adversarial**, AdaGrad is **"conservative"**

"**Fixes**": Adam, RMSProp, and other workarounds

\displaystyle G_t = (1 - \rho) \cdot \nabla f(x_i) \nabla f(x_i)^T + \rho \cdot G_{t-1}

\displaystyle P_t = G_t^{-1/2}

**RMSProp**

**Adam**

Uses **"momentum"** (weighted sum of gradients)

Similar preconditioner to the above

"AdaGrad inspired anincredible number of clones, most of them withsimilar, worse, or no regret guarantees.(...) Nowadays, [adaptive] seems to denoteany kind of coordinate-wise learning rates that does not guarantee anything in particular."

**Francesco Orabona** in "A Modern Introduction to Online Learning", Sec. 4.3

**Idea: **look at step-size/preconditioner choice as an optimization problem

Gradient descent on the hyperparameters

How to pick the step-size of this? Well...

Little/ No theory

Unpredictable

... and popular?!

**What does it mean for a method to be adaptive?**

**(Quasi-)Newton Methods**

\displaystyle \Bigg \{

Super-linear convergence close to opt

May need 2nd-order information.

**Hypergradient Methods**

Hyperparameter tuning as an opt problem

Unstable and no theory/guarantees

\displaystyle \Bigg \{

**Online Learning **

Formally adapts to adversarial and changing inputs

\displaystyle \Bigg \{

Too conservative in this case (e.g., AdaGrad)

"Fixes" (e.g., Adam) have few guarantees

adaptive methods

f(x_t) - f(x_*) \displaystyle \lesssim \left( 1 - O\Big(\frac{1}{\kappa}\Big) \right)^t

only guarantee (globally)

In **Smooth**

and **Strongly Convex** optimization,

**Should be better** if there is a good Preconditioner \(P\)

**Online Learning**

**Smooth Optimization**

**1 step-size**

**\(d\) step-sizes**

(diagonal preconditioner )

Backtracking Line-search

Diagonal AdaGrad

**Multidimensional Backtracking**

Scalar AdaGrad

(and others)

(and others)

(non-smooth optmization)

\displaystyle \frac{1}{\kappa} P^{-1} \preceq \nabla^2 f(x) \preceq P^{-1}

**Optimal step-size**: biggest that guarantees progress

**Optimal preconditioner**: **biggest (??)** that guarantees progress

\displaystyle P = \tfrac{1}{L} I

\displaystyle \kappa = \tfrac{L}{\mu}

**\(L\)-smooth**

**\(\mu\)-strongly convex**

\(f\) is

and

\preceq \nabla^2 f(x) \preceq

L \cdot I

\mu \cdot I

**Optimal Diagonal Preconditioner**

\(\kappa_* \leq \kappa\), hopefully \(\kappa_* \ll \kappa\)

**Over diagonal matrices**

\displaystyle P_*

minimizes \(\kappa_*\) such that

\displaystyle \frac{1}{\kappa_*} P^{-1} \preceq \nabla^2 f(x) \preceq P^{-1}

\displaystyle P_*, \kappa_*

**attain**

**Line-search**

\displaystyle \Bigg\{

Worth it if \(\sqrt{2d} \kappa_* \ll 2 \kappa\)

\displaystyle f(x_t) - f(x_*) \leq\Big(1 - \phantom{\frac{1}{2}} \cdot \frac{1}{\kappa}\Big)^t (f(x_0) - f(x_*))

\displaystyle
\frac{1}{2}

**Multidimensional Backtracking**

\displaystyle \Bigg\{

(our algorithm)

# backtracks \(\lesssim\)

\displaystyle d \cdot \log\Big( \alpha_0 \cdot L\Big)

\displaystyle f(x_t) - f(x_*) \leq\Big(1 - \phantom{\frac{1}{\sqrt{2d}}} \cdot \frac{1}{\phantom{\kappa_*}}\Big)^t (f(x_0) - f(x_*))

\displaystyle\frac{1}{\sqrt{2d}}

\displaystyle \kappa_*

# backtracks \(\leq\)

\displaystyle \log\Big(\alpha_{0} \cdot L \Big)

**Line-search**: test if step-size \(\alpha_{\max}/2\) makes enough progress:

\displaystyle f(x_{t+1}) \leq f(x_t) - \alpha_{\max} \tfrac{1}{2} \lVert \nabla f(x_t) \rVert_2^2

**Armijo condition**

If this fails, **cut out** everything bigger than \(\alpha_{\max}/2\)

**Preconditioner search:**

0

\alpha_{0}

\tfrac{\alpha_{0}}{2}

\tfrac{1}{L}

\tfrac{\alpha_{0}}{4}

Test if preconditioner \(P\) makes enough progress:

Candidate preconditioners \(\mathcal{S}\): diagonals in a box

\displaystyle f(x_{t+1}) \leq f(x_t)

- \tfrac{1}{2} \lVert \nabla f(x_t) \rVert_P^2

If this fails, **cut out** everything bigger than \(P\)

\langle \nabla f(x_t), P \nabla f(x_t) \rangle

**Preconditioner search:**

Test if preconditioner \(P\) makes enough progress:

Candidate preconditioners \(\mathcal{S}\): diagonals in a box

\displaystyle f(x_{t+1}) \leq f(x_t)

\langle \nabla f(x_t), P \nabla f(x_t) \rangle

- \tfrac{1}{2} \lVert \nabla f(x_t) \rVert_P^2

If this fails, **cut out** everything bigger than \(P\)

\(P\) does not yield **sufficient progress**

Which preconditioners can be thrown out?

All \(Q\) such that \(P \preceq Q\) works, but it is **too weak**

\displaystyle
h(P) \coloneqq f(x - P^{-1} \nabla f(x)) - f(x) + \tfrac{1}{2} \lVert \nabla f(x) \rVert_P^2

\(P \) does not yield sufficient progress \(\iff\) \(h(P) > 0\)

\displaystyle
h(Q) \geq h(P) + \langle \nabla h(P), Q - P \rangle

Convexity \(\implies\)

\displaystyle
h(P) + \langle \nabla h(P), Q - P \rangle > 0

\(\implies\) \(Q\) **is invalid**

A separating hyperplane!

\(P\) in this half-space

\displaystyle \Bigg\{

that maximizes

that maximizes

We want to use the **Ellipsoid method** as our cutting plane method

\(\Omega(d^3)\) time per iteration

We can exploit symmetry!

\(O(d)\) time per iteration

**Constant volume decrease** on each CUT

\displaystyle f(x_t) - f(x_*) \lesssim\Big(1 - \phantom{\frac{1}{\sqrt{2d}}} \cdot \frac{1}{\kappa_*}\Big)^t

\displaystyle
\frac{1}{\sqrt{2d}}

**Query point** \(1/\sqrt{2d}\) away from boundary

\kappa \approx 10^{13}

\kappa_* \approx 10^{2}

Theoretically principled adaptive optimization method for strongly convex smooth optimization

A theoretically-informed use of "hypergradients"

ML Optimization meets Cutting Plane methods

Stochastic case?

Heuristics for non-convex case?

Other cutting-plane methods?