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In this post we describe several variations of normalized gradient descent, known generally as steepest descent algorithms. These variations arise out of a small but fundamental change to the gradient descent setup, a change that has significant impact on the form of a descent direction but importantly, not on the convergence of the methods (steepest descent methods converge under the same general condition as gradient descent, i.e., that the length of the steps eventually diminishes).
When we first discussed gradient descent the descent direction given by the negative gradient fell out naturally from a geometric understanding of hyperplanes. Here we begin our discussion of steepest descent algorithms by re-visiting gradient descent and deriving the descent direction using a more rigorous mathematical framework. This framework immediately suggests a way to generalize the gradient descent algorithm under different norms, which we then explore.
When first deriving the gradient descent direction of a multi-input function $g(\mathbf{w})$ at a point $\mathbf{v}$ we began by examining the tangent hyperplane to $g$ at this point
\begin{equation} h\left(\mathbf{w}\right)=g\left(\mathbf{v}\right)+\nabla g\left(\mathbf{v}\right)^{T}\left(\mathbf{w}-\mathbf{v}\right) \end{equation}We then reasoned out that since in general we know that the ascent direction on a hyperplane is given by its set of 'slopes' - stored in $\nabla g(\mathbf{v})$ - that intuitively therefore the descent direction is given by the negative gradient as $-\nabla g(\mathbf{v})$, or normalized to unit length as $-\frac{\nabla g(\mathbf{v})}{\left\Vert \nabla g(\mathbf{v}) \right\Vert_2 }$. We often used the latter unit-normalized version since after all we care only about the descent direction.
This descent direction can be derived more formally as follows. Note that $\mathbf{d} = \mathbf{w} - \mathbf{v}$ is a general search direction centered at the point $\mathbf{v}$. We want to find the unit-length direction $\mathbf{d}$ that provides the smallest evaluation on the hyperplane, i.e., the one that gives the smallest value $g(\mathbf{v}) + \nabla g(\mathbf{v})^T(\mathbf{w} - \mathbf{v}) = g(\mathbf{v}) + \nabla g(\mathbf{v})^T\mathbf{d}$. Since $\mathbf{d}$ is only present in the second term we only need care about finding the unit length direction $\mathbf{d}$ so that $\nabla g(\mathbf{v})^T\mathbf{d}$ is as small as possible (after all, $g(\mathbf{v})$ is constant).
Formally this is a simple constrained minimization problem
Using the toy in the next Python cell we explore possible solutions to this problem in two dimensions for a particular choice of gradient vector, gaining valuable geometric intuition as to what the solution should be. There we plot an example gradient vector $\nabla g (\mathbf{v})$ as a red arrow, along with the $\ell_2$ unit ball. Moving the slider from left to right you can test various directions $\mathbf{d}$ (each shown as a black arrow) on this $\ell_2$ unit ball computing the inner product $\nabla g(\mathbf{v})^{T}\mathbf{d}$, whose value is simultaneously plotted in the right panel. As you move the slider right the direction providing the smallest value is shown as a green arrow in the left panel, and is highlighted in green in the plot in the right panel.
As can be seen it appears as if indeed $\mathbf{d} = -\frac{\nabla g(\mathbf{v})}{\left\Vert \nabla g(\mathbf{v}) \right\Vert_2 }$ is the direction producing the smallest inner product.
To prove this formally we can use the inner product rule which tells us almost immediately what the solution to this problem is. According to the inner product rule $\nabla g(\mathbf{v})^{T}\mathbf{d}$ can be written as
$$\nabla g(\mathbf{v})^{T}\mathbf{d} = \left \Vert \nabla g(\mathbf{v}) \right \Vert_2 \left \Vert \mathbf{d} \right \Vert_2 \text{cos}(\theta)$$where $\theta$ is the angle between $\nabla g(\mathbf{v})$ and $\mathbf{d}$.
Noting that both $\left \Vert \nabla g(\mathbf{v}) \right \Vert_2$ and $\left \Vert \mathbf{d} \right \Vert_2$ have constant values (the former is the length of the gradient at $\mathbf{v}$, and latter is just $1$), the value of $\nabla g(\mathbf{v})^{T}\mathbf{d}$ is smallest when $\text{cos}(\theta)$ is smallest, i.e., when $\theta = \pi$. In other words, $\mathbf{d}$ must point in the same direction as $-\nabla g(\mathbf{v})$, and also be unit length. Thus we have $\mathbf{d} = -\frac{\nabla g(\mathbf{v})}{\left \Vert \nabla g(\mathbf{v}) \right \Vert_2}$, that is indeed the normalized gradient descent direction. In deriving the gradient descent direction in this way, $\mathbf{d}$ is referred to more generally as the steepest descent direction.
With this setup we can actually choose any norm in the constraint that we wish to determine a descent direction. For example, replacing the $\ell_2$ with the $\ell_1$ norm we have a similar looking problem
whose solution defines a new steepest descent direction with respect to the $\ell_1$ norm. The direction we find here will certainly be different than the $\ell_2$ constrained version.
Using the toy in the next Python cell we explore possible solutions to this problem in two dimensions for a particular choice of gradient vector, gaining valuable geometric intuition as to what the solution should be. There we plot an example gradient vector $\nabla g (\mathbf{v})$ as a red arrow, along with the $\ell_1$ unit ball. Moving the slider from left to right you can test various directions $\mathbf{d}$ (each shown as a black arrow) on this $\ell_1$ unit ball computing the inner product $\nabla g(\mathbf{v})^{T}\mathbf{d}$, whose value is simultaneously plotted in the right panel. As you move the slider right the direction providing the smallest value is shown as a green arrow in the left panel, and is highlighted in green in the plot in the right panel.
Let us try a different a gradient vector $\nabla g (\mathbf{v})$, this time in the second quadrant.
Toying around with the slider above it looks as if the descent direction here is defined along a coordinate axis as the negative direction of only the largest (in magnitude) partial derivative of the gradient. Denoting by $j$ the index of the largest (in magnitude) entry in $\nabla g (\mathbf{v})$
$$j=\underset{i}{\text{argmax}}\left | \frac{\partial}{\partial w_i}g(\mathbf{v}) \right |$$the descent direction $\mathbf{d}$ will be along the vector
$$\begin{array}{cc} \left[\begin{array}{c} 0\\ \vdots\\ 0\\ -\frac{\partial}{\partial w_{j}}g(\mathbf{v})\\ 0\\ \vdots\\ 0 \end{array}\right] & \rightarrow j^{th}\text{ position}\end{array}$$which can be written, using the standard basis notation, as $ - \frac{\partial}{\partial w_j}g(\mathbf{v})\, \mathbf{e}_j$, where $\mathbf{e}_j$ is a standard basis vector with a $1$ in its $j^{th}$ position and zeros elsewhere. Dividing this vector by its norm (so that it becomes unit length) we have
$$\mathbf{d}=\frac{-\frac{\partial}{\partial w_{j}}g(\mathbf{v})}{\left\Vert -\frac{\partial}{\partial w_{j}}g(\mathbf{v})\,\mathbf{e}_{j}\right\Vert _{2}}\mathbf{e}_{j}=\frac{-\frac{\partial}{\partial w_{j}}g(\mathbf{v})}{\left|\frac{\partial}{\partial w_{j}}g(\mathbf{v})\right|}\mathbf{e}_{j}=-\text{sign}\left(\frac{\partial}{\partial w_{j}}g(\mathbf{v})\right)\mathbf{e}_{j}$$It is certainly possible for more than one partial derivative to be maximal (in magnitude), e.g., when $\nabla g (\mathbf{v})=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ in the setup above. In this case not only both $-\text{sign}\left(\frac{\partial}{\partial w_{1}}g(\mathbf{v})\right)\mathbf{e}_{1}$ and $-\text{sign}\left(\frac{\partial}{\partial w_{2}}g(\mathbf{v})\right)\mathbf{e}_{2}$ minimize the inner product, but so will their $\ell_1$ length normalized sum $-\frac{\text{sign}\left(\frac{\partial}{\partial w_{1}}g(\mathbf{v})\right)\mathbf{e}_{1} + \text{sign}\left(\frac{\partial}{\partial w_{2}}g(\mathbf{v})\right)\mathbf{e}_{2}}{2}$.
More generally speaking, if we denote $\mathcal{S}$ the set of indices of those maximal partial derivatives the descent direction that combines all indices in $\mathcal{S}$ is given by
\begin{equation} \mathbf{d} = - \frac{1}{\left| \mathcal{S}\right|}\sum_{j \in \mathcal{S} } \,\,\text{sign}\left(\frac{\partial}{\partial w_j}g(\mathbf{v})\right)^{\,} \mathbf{e}_j^{\,} \end{equation}This formula can be proven to be true rigorously, as we show in the Appendix Section of this post.
The steepest descent step at $\mathbf{w}^{k-1}$ in this direction looks like
\begin{equation} \mathbf{w}^{k} = \mathbf{w}^{k-1} - \alpha \sum_{j \in \mathcal{S} } \,\,\text{sign}\left(\frac{\partial}{\partial w_j}g(\mathbf{w}^{k-1})\right)^{\,} \mathbf{e}_j^{\,} \end{equation}where the index set $\mathcal{S}$ changes from step to step depending on the full gradient $\nabla g\left(\mathbf{w}^{k-1} \right)$, and the constant $\frac{1}{\left| \mathcal{S}\right|}$ has been absorbed into $\alpha$. Given how each direction is formed these steps tend to go along the coordinate directions, like a coordinate descent approach. This descent step appears less useful then the $\ell_2$ analog in general, since we must still compute the entire gradient in order to use it and then we 'throw away' all but its largest values.
In the next Python cell we use the $\ell_1$ steepest descent approach to find the minimum of
\begin{equation} g\left(w_1,w_2 \right) = 0.26\left(w_1^2 + w_2^2\right) - 0.48w_1w_2 \end{equation}located at the origin. We draw the path taken by this approach directly on the function contours - which are colored green as the method starts and gradually change red as it halts.
Replacing the $\ell_2$ norm with the $\ell_{\infty}$ norm in our original constrained optimization problem we have a similar looking problem whose solution defines a new kind of steepest descent direction
Using the toy in the next Python cell we explore possible solutions to this problem in two dimensions for a particular choice of gradient vector, gaining valuable geometric intuition as to what the solution should be. Here once again we plot an example gradient vector $\nabla g (\mathbf{v})$ as a red arrow, along with the $\ell_{\infty}$ unit ball. Moving the slider from left to right you can test various directions $\mathbf{d}$ (each shown as a black arrow) on this $\ell_{\infty}$ unit ball computing the inner product $\nabla g(\mathbf{v})^{T}\mathbf{d}$, whose value is simultaneously plotted in the right panel. As you move the slider right once more the direction providing the smallest value is shown as a green arrow in the left panel, and is highlighted in green in the plot in the right panel.
Based on the example above you may now have a hunch about what $\mathbf{d}$ should be. But let's make sure by examining a different gradient vector $\nabla g (\mathbf{v})$.
Toying around with the slider above it looks as if each entry of the descent direction here is the negative unit length partial derivative along its respective coordinate. That is, the $j^{th}$ entry is given as
$$d_j = \frac{ \frac{\partial}{\partial w_j}g(\mathbf{v}) }{\left| \frac{\partial}{\partial w_j}g(\mathbf{v}) \right|} = \text{sign}\left(\frac{\partial}{\partial w_j}g(\mathbf{v})\right)$$and hence the entire descent direction can be written simply as
\begin{equation} \mathbf{d} = -\, \text{sign}\left(\nabla g(\mathbf{v}) \right) \end{equation}These directions tend to most often be constrainted to the corners of the unit square or - in higher dimensions - the unit hypercube (the exceptions being when one partial derivative is exactly equal to zero).
This intuitied formula is in fact correct, and can be proven more rigorously as show in the Appendix Section of this post.
The steepest descent step at $\mathbf{w}^{k-1}$ in this direction looks like
\begin{equation} \mathbf{w}^{k} = \mathbf{w}^{k-1} - \alpha\, \text{sign}\left(\nabla g(\mathbf{w}^{k-1}) \right) \end{equation}Due to the $\text{sign} (\cdot)$ function these steps tend to be at 'diagonal' traveling largely in directions defined by the corners of the unit square. Because the set of directions is restricted one often sees the use of coordinate-wise steplength parameters with this method (e.g., Rprop).
In the next Python cell we use the $\ell_{\infty}$ steepest descent approach to find the minimum of
\begin{equation} g\left(w_1,w_2 \right) = w_1^2 + 5w_2^2 \end{equation}located at the origin. We draw the path taken by this approach directly on the function contours - which are colored green as the method starts and gradually change red as it halts.
Here we provide formal proofs of the $\ell_1$ and $\ell_\infty$ descent directions. Note that throughout for ease of reading we denote $\mathbf{a} = -\nabla g(\mathbf{v})$.
In what follows we find a closed form solution to
\begin{equation} \begin{array}{cc} \underset{\mathbf{d}}{\text{maximize}} & \mathbf{a}^{T}\mathbf{d}\\ \text{subject to} & \left\Vert \mathbf{d}\right\Vert _{1}=1\\ \end{array} \end{equation}written entry-wise as
\begin{equation} \begin{array}{cc} \underset{d_1,\,d_2,\,\ldots,\, d_n}{\text{maximize}} & a_1d_1+a_2d_2+\cdots+a_nd_n\\ \text{subject to} & \left\vert d_1\right\vert + \left\vert d_2\right\vert + \cdots + \left\vert d_n\right\vert = 1\\ \end{array} \end{equation}Firstly notice that if the mathematical sign of $d_i$ differs from that of $a_i$, then replacing $d_i$ with $-d_i$ would increase (or at least not decrease) the objective value without violating the equality constraint. Therefore by writing $d_i$ as
$$d_i=\text{sign}(a_i)\,y_i$$where $y_i$'s are now all nonnegative, our original problem can be reformulated equivalently as
\begin{equation} \begin{array}{cc} \underset{y_1,\,y_2,\,\ldots,\, y_n}{\text{maximize}} & a_1\,\text{sign}(a_1)\,y_1+a_2\,\text{sign}(a_2)\,y_2+\cdots+a_n\,\text{sign}(a_n)\,y_n\\ \text{subject to} & \left\vert \text{sign}(a_1)\,y_1\right\vert + \left\vert \text{sign}(a_2)\,y_2\right\vert + \cdots + \left\vert \text{sign}(a_n)\,y_n\right\vert =1\\ & y_1, y_2, \ldots, y_n \geq 0\\ \end{array} \end{equation}Noticing that we have $a_i\,\text{sign}(a_i)\,y_i=\left\vert a_i\right\vert \,y_i$ (in the objective), and $\left\vert \text{sign}(a_i)\,y_i\right\vert = \left\vert \text{sign}(a_i)\right\vert \left\vert y_i\right\vert = y_i$, we can rewrite the optimization problem as
\begin{equation} \begin{array}{cc} \underset{y_1,\,y_2,\,\ldots,\, y_n}{\text{maximize}} & \left\vert a_1\right\vert y_1+\left\vert a_2\right\vert y_2+\cdots+\left\vert a_n\right\vert y_n\\ \text{subject to} & y_1 + y_2 + \cdots + y_n =1\\ & y_1, y_2, \ldots, y_n \geq 0\\ \end{array} \end{equation}Finally, denoting by $j$ an index of the largest entry in $\mathbf{a}$ (in magnitude)
$$j=\underset{i=1,\,2,\,\ldots,\, n}{\text{argmax}}\left|a_{i}\right|$$we can find an upperbound to the objective
$$\left|a_{1}\right|y_{1}+\left|a_{2}\right|y_{2}+\ldots+\left|a_{n}\right|y_{n}\leq\left|a_{j}\right|y_{1}+\left|a_{j}\right|y_{2}+\ldots+\left|a_{j}\right|y_{n}=\left|a_{j}\right|$$Notice that the $j^{th}$ standard basis vector (having $1$ in its $j^{th}$ position and zero elsewhere) gives the same value as this upperbound, and hence $\mathbf{y}=\mathbf{e}_{j}$ is a solution to the reformulated optimization problem. Moreover if there are several entries of $\mathbf{a}$ that are equally maximal in magnitude the above is true for each of their corresponding indices, the set of which we denote as $\mathcal{S}$.
All together then a solution is given as
$$ \mathbf{d}^{\star}= \frac{1}{\left| \mathcal{S}\right|}\sum_{j\in \mathcal{S}}\text{sgn}(a_j)\,\mathbf{e}_{j} $$where $\left| \mathcal{S}\right|$ denotes the number of elements in $\mathcal{S}$.
Here we find a closed form solution to
\begin{equation} \begin{array}{cc} \underset{\mathbf{d}}{\text{maximize}} & \mathbf{a}^{T}\mathbf{d}\\ \text{subject to} & \left\Vert \mathbf{d}\right\Vert _{\infty}=1\\ \end{array} \end{equation}written entry-wise as
\begin{equation} \begin{array}{cc} \underset{d_1,\,d_2,\,\ldots,\, d_n}{\text{maximize}} & a_1d_1+a_2d_2+\cdots+a_nd_n\\ \text{subject to} & \text{max}\left\{ \left|d_{1}\right|,\,\left|d_{2}\right|,\,\ldots,\,\left|d_{n}\right|\right\} =1\\ \end{array} \end{equation}Notice that the constraint set
$$\mathcal{S}=\left\{ \mathbf{d}\left|\text{ max}\left(\left|d_{1}\right|,\,\left|d_{2}\right|,\,\ldots,\,\left|d_{n}\right|\right)=1\right.\right\}$$is a subset of the set
$$\mathcal{T}=\left\{ \mathbf{d}\left|\, -1\leq d_{1},\,d_{2},\,\ldots,\,d_{n}\leq1\right.\right\}$$Therefore the maximum objective value subject to $\mathbf{d}\in\mathcal{S}$ is upperbounded by the maximum objective value subject to $\mathbf{d}\in\mathcal{T}$. The latter problem can be then written as
Now note that this problem can be broken down into $n$ independent optimization problems of the form
with the easily calculable solution given as $d_i^{\star}=\text{sign}(a_i)$. Taken all together
$$\mathbf{d}^{\star}=\text{sign}(\mathbf{a})$$maximizes the objective subject to $\mathbf{d}\in\mathcal{T}$. Given we have that $\mathbf{d}^{\star}\in\mathcal{S}$ as well, $\mathbf{d}^{\star}$ also maximizes the objective subject to $\mathbf{d}\in\mathcal{S}$.