Almost every problem in machine learning and data science starts with a dataset \( X \), a model \( g(\beta) \), which is a function of the parameters \( \beta \) and a cost function \( C(X, g(\beta)) \) that allows us to judge how well the model \( g(\beta) \) explains the observations \( X \). The model is fit by finding the values of \( \beta \) that minimize the cost function. Ideally we would be able to solve for \( \beta \) analytically, however this is not possible in general and we must use some approximative/numerical method to compute the minimum.
In our discussion on Logistic Regression we studied the case of two classes, with \( y_i \) either \( 0 \) or \( 1 \). Furthermore we assumed also that we have only two parameters \( \beta \) in our fitting, that is we defined probabilities $$ \begin{align*} p(y_i=1|x_i,\hat{\beta}) &= \frac{\exp{(\beta_0+\beta_1x_i)}}{1+\exp{(\beta_0+\beta_1x_i)}},\nonumber\\ p(y_i=0|x_i,\hat{\beta}) &= 1 - p(y_i=1|x_i,\hat{\beta}), \end{align*} $$ where \( \hat{\beta} \) are the weights we wish to extract from data, in our case \( \beta_0 \) and \( \beta_1 \).
Our compact equations used a definition of a vector \( \hat{y} \) with \( n \) elements \( y_i \), an \( n\times p \) matrix \( \hat{X} \) which contains the \( x_i \) values and a vector \( \hat{p} \) of fitted probabilities \( p(y_i\vert x_i,\hat{\beta}) \). We rewrote in a more compact form the first derivative of the cost function as $$ \frac{\partial \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}} = -\hat{X}^T\left(\hat{y}-\hat{p}\right). $$
If we in addition define a diagonal matrix \( \hat{W} \) with elements \( p(y_i\vert x_i,\hat{\beta})(1-p(y_i\vert x_i,\hat{\beta}) \), we can obtain a compact expression of the second derivative as $$ \frac{\partial^2 \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}\partial \hat{\beta}^T} = \hat{X}^T\hat{W}\hat{X}. $$ This defines what is called the Hessian matrix.
If we can set up these equations, Newton-Raphson's iterative method is normally the method of choice. It requires however that we can compute in an efficient way the matrices that define the first and second derivatives.
Our iterative scheme is then given by $$ \hat{\beta}^{\mathrm{new}} = \hat{\beta}^{\mathrm{old}}-\left(\frac{\partial^2 \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}\partial \hat{\beta}^T}\right)^{-1}_{\hat{\beta}^{\mathrm{old}}}\times \left(\frac{\partial \mathcal{C}(\hat{\beta})}{\partial \hat{\beta}}\right)_{\hat{\beta}^{\mathrm{old}}}, $$ or in matrix form as $$ \hat{\beta}^{\mathrm{new}} = \hat{\beta}^{\mathrm{old}}-\left(\hat{X}^T\hat{W}\hat{X} \right)^{-1}\times \left(-\hat{X}^T(\hat{y}-\hat{p}) \right)_{\hat{\beta}^{\mathrm{old}}}. $$ The right-hand side is computed with the old values of \( \beta \).
If we can compute these matrices, in particular the Hessian, the above is often the easiest method to implement.
Let us quickly remind ourselves how we derive the above method.
Perhaps the most celebrated of all one-dimensional root-finding routines is Newton's method, also called the Newton-Raphson method. This method requires the evaluation of both the function \( f \) and its derivative \( f' \) at arbitrary points. If you can only calculate the derivative numerically and/or your function is not of the smooth type, we normally discourage the use of this method.
The Newton-Raphson formula consists geometrically of extending the tangent line at a current point until it crosses zero, then setting the next guess to the abscissa of that zero-crossing. The mathematics behind this method is rather simple. Employing a Taylor expansion for \( x \) sufficiently close to the solution \( s \), we have $$ f(s)=0=f(x)+(s-x)f'(x)+\frac{(s-x)^2}{2}f''(x) +\dots. \label{eq:taylornr} $$
For small enough values of the function and for well-behaved functions, the terms beyond linear are unimportant, hence we obtain $$ f(x)+(s-x)f'(x)\approx 0, $$ yielding $$ s\approx x-\frac{f(x)}{f'(x)}. $$
Having in mind an iterative procedure, it is natural to start iterating with $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. $$
The above is Newton-Raphson's method. It has a simple geometric interpretation, namely \( x_{n+1} \) is the point where the tangent from \( (x_n,f(x_n)) \) crosses the \( x \)-axis. Close to the solution, Newton-Raphson converges fast to the desired result. However, if we are far from a root, where the higher-order terms in the series are important, the Newton-Raphson formula can give grossly inaccurate results. For instance, the initial guess for the root might be so far from the true root as to let the search interval include a local maximum or minimum of the function. If an iteration places a trial guess near such a local extremum, so that the first derivative nearly vanishes, then Newton-Raphson may fail totally
Newton's method can be generalized to systems of several non-linear equations and variables. Consider the case with two equations $$ \begin{array}{cc} f_1(x_1,x_2) &=0\\ f_2(x_1,x_2) &=0,\end{array} $$ which we Taylor expand to obtain $$ \begin{array}{cc} 0=f_1(x_1+h_1,x_2+h_2)=&f_1(x_1,x_2)+h_1 \partial f_1/\partial x_1+h_2 \partial f_1/\partial x_2+\dots\\ 0=f_2(x_1+h_1,x_2+h_2)=&f_2(x_1,x_2)+h_1 \partial f_2/\partial x_1+h_2 \partial f_2/\partial x_2+\dots \end{array}. $$ Defining the Jacobian matrix \( {\bf \hat{J}} \) we have $$ {\bf \hat{J}}=\left( \begin{array}{cc} \partial f_1/\partial x_1 & \partial f_1/\partial x_2 \\ \partial f_2/\partial x_1 &\partial f_2/\partial x_2 \end{array} \right), $$ we can rephrase Newton's method as $$ \left(\begin{array}{c} x_1^{n+1} \\ x_2^{n+1} \end{array} \right)= \left(\begin{array}{c} x_1^{n} \\ x_2^{n} \end{array} \right)+ \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right), $$ where we have defined $$ \left(\begin{array}{c} h_1^{n} \\ h_2^{n} \end{array} \right)= -{\bf \hat{J}}^{-1} \left(\begin{array}{c} f_1(x_1^{n},x_2^{n}) \\ f_2(x_1^{n},x_2^{n}) \end{array} \right). $$ We need thus to compute the inverse of the Jacobian matrix and it is to understand that difficulties may arise in case \( {\bf \hat{J}} \) is nearly singular.
It is rather straightforward to extend the above scheme to systems of more than two non-linear equations. In our case, the Jacobian matrix is given by the Hessian that represents the second derivative of cost function.
The basic idea of gradient descent is that a function \( F(\mathbf{x}) \), \( \mathbf{x} \equiv (x_1,\cdots,x_n) \), decreases fastest if one goes from \( \bf {x} \) in the direction of the negative gradient \( -\nabla F(\mathbf{x}) \).
It can be shown that if $$ \mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), $$ with \( \gamma_k > 0 \).
For \( \gamma_k \) small enough, then \( F(\mathbf{x}_{k+1}) \leq F(\mathbf{x}_k) \). This means that for a sufficiently small \( \gamma_k \) we are always moving towards smaller function values, i.e a minimum.
The previous observation is the basis of the method of steepest descent, which is also referred to as just gradient descent (GD). One starts with an initial guess \( \mathbf{x}_0 \) for a minimum of \( F \) and computes new approximations according to $$ \mathbf{x}_{k+1} = \mathbf{x}_k - \gamma_k \nabla F(\mathbf{x}_k), \ \ k \geq 0. $$
The parameter \( \gamma_k \) is often referred to as the step length or the learning rate within the context of Machine Learning.
Ideally the sequence \( \{\mathbf{x}_k \}_{k=0} \) converges to a global minimum of the function \( F \). In general we do not know if we are in a global or local minimum. In the special case when \( F \) is a convex function, all local minima are also global minima, so in this case gradient descent can converge to the global solution. The advantage of this scheme is that it is conceptually simple and straightforward to implement. However the method in this form has some severe limitations:
In machine learing we are often faced with non-convex high dimensional cost functions with many local minima. Since GD is deterministic we will get stuck in a local minimum, if the method converges, unless we have a very good intial guess. This also implies that the scheme is sensitive to the chosen initial condition.
Note that the gradient is a function of \( \mathbf{x} = (x_1,\cdots,x_n) \) which makes it expensive to compute numerically.
The gradient descent method is sensitive to the choice of learning rate \( \gamma_k \). This is due to the fact that we are only guaranteed that \( F(\mathbf{x}_{k+1}) \leq F(\mathbf{x}_k) \) for sufficiently small \( \gamma_k \). The problem is to determine an optimal learning rate. If the learning rate is chosen too small the method will take a long time to converge and if it is too large we can experience erratic behavior.
Many of these shortcomings can be alleviated by introducing randomness. One such method is that of Stochastic Gradient Descent (SGD), see below.
Ideally we want our cost/loss function to be convex(concave).
First we give the definition of a convex set: A set \( C \) in \( \mathbb{R}^n \) is said to be convex if, for all \( x \) and \( y \) in \( C \) and all \( t \in (0,1) \) , the point \( (1 − t)x + ty \) also belongs to C. Geometrically this means that every point on the line segment connecting \( x \) and \( y \) is in \( C \) as discussed below.
The convex subsets of \( \mathbb{R} \) are the intervals of \( \mathbb{R} \). Examples of convex sets of \( \mathbb{R}^2 \) are the regular polygons (triangles, rectangles, pentagons, etc...).
Convex function: Let \( X \subset \mathbb{R}^n \) be a convex set. Assume that the function \( f: X \rightarrow \mathbb{R} \) is continuous, then \( f \) is said to be convex if $$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2) $$ for all \( x_1, x_2 \in X \) and for all \( t \in [0,1] \). If \( \leq \) is replaced with a strict inequaltiy in the definition, we demand \( x_1 \neq x_2 \) and \( t\in(0,1) \) then \( f \) is said to be strictly convex. For a single variable function, convexity means that if you draw a straight line connecting \( f(x_1) \) and \( f(x_2) \), the value of the function on the interval \( [x_1,x_2] \) is always below the line as illustrated below.
In the following we state first and second-order conditions which ensures convexity of a function \( f \). We write \( D_f \) to denote the domain of \( f \), i.e the subset of \( R^n \) where \( f \) is defined. For more details and proofs we refer to: S. Boyd and L. Vandenberghe. Convex Optimization. Cambridge University Press.
Suppose \( f \) is differentiable (i.e \( \nabla f(x) \) is well defined for all \( x \) in the domain of \( f \)). Then \( f \) is convex if and only if \( D_f \) is a convex set and $$f(y) \geq f(x) + \nabla f(x)^T (y-x) $$ holds for all \( x,y \in D_f \). This condition means that for a convex function the first order Taylor expansion (right hand side above) at any point a global under estimator of the function. To convince yourself you can make a drawing of \( f(x) = x^2+1 \) and draw the tangent line to \( f(x) \) and note that it is always below the graph.
Assume that \( f \) is twice differentiable, i.e the Hessian matrix exists at each point in \( D_f \). Then \( f \) is convex if and only if \( D_f \) is a convex set and its Hessian is positive semi-definite for all \( x\in D_f \). For a single-variable function this reduces to \( f''(x) \geq 0 \). Geometrically this means that \( f \) has nonnegative curvature everywhere.
This condition is particularly useful since it gives us an procedure for determining if the function under consideration is convex, apart from using the definition.
The next result is of great importance to us and the reason why we are going on about convex functions. In machine learning we frequently have to minimize a loss/cost function in order to find the best parameters for the model we are considering.
Ideally we want the global minimum (for high-dimensional models it is hard to know if we have local or global minimum). However, if the cost/loss function is convex the following result provides invaluable information:
Consider the problem of finding \( x \in \mathbb{R}^n \) such that \( f(x) \) is minimal, where \( f \) is convex and differentiable. Then, any point \( x^* \) that satisfies \( \nabla f(x^*) = 0 \) is a global minimum.
This result means that if we know that the cost/loss function is convex and we are able to find a minimum, we are guaranteed that it is a global minimum.
We will use linear regression as a case study for the gradient descent methods. Linear regression is a great test case for the gradient descent methods discussed in the lectures since it has several desirable properties such as:
Let \( \mathbf{y} = (y_1,\cdots,y_n)^T \), \( \mathbf{\hat{y}} = (\hat{y}_1,\cdots,\hat{y}_n)^T \) and \( \beta = (\beta_0, \beta_1)^T \)
It is convenient to write \( \mathbf{\hat{y}} = X\beta \) where \( X \in \mathbb{R}^{100 \times 2} \) is the design matrix given by $$ X \equiv \begin{bmatrix} 1 & x_1 \\ \vdots & \vdots \\ 1 & x_{100} & \\ \end{bmatrix}. $$ The loss function is given by $$ C(\beta) = ||X\beta-\mathbf{y}||^2 = ||X\beta||^2 - 2 \mathbf{y}^T X\beta + ||\mathbf{y}||^2 = \sum_{i=1}^{100} (\beta_0 + \beta_1 x_i)^2 - 2 y_i (\beta_0 + \beta_1 x_i) + y_i^2 $$ and we want to find \( \beta \) such that \( C(\beta) \) is minimized.
Computing \( \partial C(\beta) / \partial \beta_0 \) and \( \partial C(\beta) / \partial \beta_1 \) we can show that the gradient can be written as $$ \nabla_{\beta} C(\beta) = (\partial C(\beta) / \partial \beta_0, \partial C(\beta) / \partial \beta_1)^T = 2\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\ \sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\ \end{bmatrix} = 2X^T(X\beta - \mathbf{y}), $$ where \( X \) is the design matrix defined above.
We can now write a program that minimizes \( C(\beta) \) using the gradient descent method with a constant learning rate \( \gamma \) according to $$ \beta_{k+1} = \beta_k - \gamma \nabla_\beta C(\beta_k), \ k=0,1,\cdots $$
We can use the expression we computed for the gradient and let use a \( \beta_0 \) be chosen randomly and let \( \gamma = 0.001 \). Stop iterating when \( ||\nabla_\beta C(\beta_k) || \leq \epsilon = 10^{-8} \).
And finally we can compare our solution for \( \beta \) with the analytic result given by \( \beta= (X^TX)^{-1} X^T \mathbf{y} \).
Here our simple example
# Importing various packages
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import sys
# the number of datapoints
m = 100
x = 2*np.random.rand(m,1)
y = 4+3*x+np.random.randn(m,1)
xb = np.c_[np.ones((m,1)), x]
beta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
print(beta_linreg)
beta = np.random.randn(2,1)
eta = 0.1
Niterations = 1000
for iter in range(Niterations):
gradients = 2.0/m*xb.T.dot(xb.dot(beta)-y)
beta -= eta*gradients
print(beta)
xnew = np.array([[0],[2]])
xbnew = np.c_[np.ones((2,1)), xnew]
ypredict = xbnew.dot(beta)
ypredict2 = xbnew.dot(beta_linreg)
plt.plot(xnew, ypredict, "r-")
plt.plot(xnew, ypredict2, "b-")
plt.plot(x, y ,'ro')
plt.axis([0,2.0,0, 15.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Gradient descent example')
plt.show()
# Importing various packages
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import SGDRegressor
x = 2*np.random.rand(100,1)
y = 4+3*x+np.random.randn(100,1)
xb = np.c_[np.ones((100,1)), x]
beta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
print(beta_linreg)
sgdreg = SGDRegressor(n_iter = 50, penalty=None, eta0=0.1)
sgdreg.fit(x,y.ravel())
print(sgdreg.intercept_, sgdreg.coef_)
We have also discussed Ridge regression where the loss function contains a regularized term given by the \( L_2 \) norm of \( \beta \), $$ C_{\text{ridge}}(\beta) = ||X\beta -\mathbf{y}||^2 + \lambda ||\beta||^2, \ \lambda \geq 0. $$
In order to minimize \( C_{\text{ridge}}(\beta) \) using GD we only have adjust the gradient as follows $$ \nabla_\beta C_{\text{ridge}}(\beta) = 2\begin{bmatrix} \sum_{i=1}^{100} \left(\beta_0+\beta_1x_i-y_i\right) \\ \sum_{i=1}^{100}\left( x_i (\beta_0+\beta_1x_i)-y_ix_i\right) \\ \end{bmatrix} + 2\lambda\begin{bmatrix} \beta_0 \\ \beta_1\end{bmatrix} = 2 (X^T(X\beta - \mathbf{y})+\lambda \beta). $$
We can easily extend our program to minimize \( C_{\text{ridge}}(\beta) \) using gradient descent and compare with the analytical solution given by $$ \beta_{\text{ridge}} = \left(X^T X + \lambda I_{2 \times 2} \right)^{-1} X^T \mathbf{y}. $$
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import cm
from matplotlib.ticker import LinearLocator, FormatStrFormatter
import sys
# the number of datapoints
m = 100
x = 2*np.random.rand(m,1)
y = 4+3*x+np.random.randn(m,1)
xb = np.c_[np.ones((m,1)), x]
XT_X = xb.T @ xb
#Ridge parameter lambda
lmbda = 0.001
Id = lmbda* np.eye(XT_X.shape[0])
beta_linreg = np.linalg.inv(XT_X+Id) @ xb.T @ y
print(beta_linreg)
# Start plain gradient descent
beta = np.random.randn(2,1)
eta = 0.1
Niterations = 100
for iter in range(Niterations):
gradients = 2.0/m*xb.T @ (xb @ (beta)-y)+2*lmbda*beta
beta -= eta*gradients
print(beta)
ypredict = xb @ beta
ypredict2 = xb @ beta_linreg
plt.plot(x, ypredict, "r-")
plt.plot(x, ypredict2, "b-")
plt.plot(x, y ,'ro')
plt.axis([0,2.0,0, 15.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Gradient descent example for Ridge')
plt.show()
Stochastic gradient descent (SGD) and variants thereof address some of the shortcomings of the Gradient descent method discussed above.
The underlying idea of SGD comes from the observation that the cost function, which we want to minimize, can almost always be written as a sum over \( n \) data points \( \{\mathbf{x}_i\}_{i=1}^n \), $$ C(\mathbf{\beta}) = \sum_{i=1}^n c_i(\mathbf{x}_i, \mathbf{\beta}). $$
This in turn means that the gradient can be computed as a sum over \( i \)-gradients $$ \nabla_\beta C(\mathbf{\beta}) = \sum_i^n \nabla_\beta c_i(\mathbf{x}_i, \mathbf{\beta}). $$
Stochasticity/randomness is introduced by only taking the gradient on a subset of the data called minibatches. If there are \( n \) data points and the size of each minibatch is \( M \), there will be \( n/M \) minibatches. We denote these minibatches by \( B_k \) where \( k=1,\cdots,n/M \).
The idea is now to approximate the gradient by replacing the sum over all data points with a sum over the data points in one the minibatches picked at random in each gradient descent step $$ \nabla_{\beta} C(\mathbf{\beta}) = \sum_{i=1}^n \nabla_\beta c_i(\mathbf{x}_i, \mathbf{\beta}) \rightarrow \sum_{i \in B_k}^n \nabla_\beta c_i(\mathbf{x}_i, \mathbf{\beta}). $$
Thus a gradient descent step now looks like $$ \beta_{j+1} = \beta_j - \gamma_j \sum_{i \in B_k}^n \nabla_\beta c_i(\mathbf{x}_i, \mathbf{\beta}) $$
where \( k \) is picked at random with equal probability from \( [1,n/M] \). An iteration over the number of minibathces (n/M) is commonly referred to as an epoch. Thus it is typical to choose a number of epochs and for each epoch iterate over the number of minibatches, as exemplified in the code below.
import numpy as np
n = 100 #100 datapoints
M = 5 #size of each minibatch
m = int(n/M) #number of minibatches
n_epochs = 10 #number of epochs
j = 0
for epoch in range(1,n_epochs+1):
for i in range(m):
k = np.random.randint(m) #Pick the k-th minibatch at random
#Compute the gradient using the data in minibatch Bk
#Compute new suggestion for
j += 1
Taking the gradient only on a subset of the data has two important benefits. First, it introduces randomness which decreases the chance that our opmization scheme gets stuck in a local minima. Second, if the size of the minibatches are small relative to the number of datapoints (\( M < n \)), the computation of the gradient is much cheaper since we sum over the datapoints in the \( k-th \) minibatch and not all \( n \) datapoints.
A natural question is when do we stop the search for a new minimum? One possibility is to compute the full gradient after a given number of epochs and check if the norm of the gradient is smaller than some threshold and stop if true. However, the condition that the gradient is zero is valid also for local minima, so this would only tell us that we are close to a local/global minimum. However, we could also evaluate the cost function at this point, store the result and continue the search. If the test kicks in at a later stage we can compare the values of the cost function and keep the \( \beta \) that gave the lowest value.
Another approach is to let the step length \( \gamma_j \) depend on the number of epochs in such a way that it becomes very small after a reasonable time such that we do not move at all.
As an example, let \( e = 0,1,2,3,\cdots \) denote the current epoch and let \( t_0, t_1 > 0 \) be two fixed numbers. Furthermore, let \( t = e \cdot m + i \) where \( m \) is the number of minibatches and \( i=0,\cdots,m-1 \). Then the function $$\gamma_j(t; t_0, t_1) = \frac{t_0}{t+t_1} $$ goes to zero as the number of epochs gets large. I.e. we start with a step length \( \gamma_j (0; t_0, t_1) = t_0/t_1 \) which decays in time \( t \).
In this way we can fix the number of epochs, compute \( \beta \) and evaluate the cost function at the end. Repeating the computation will give a different result since the scheme is random by design. Then we pick the final \( \beta \) that gives the lowest value of the cost function.
import numpy as np
def step_length(t,t0,t1):
return t0/(t+t1)
n = 100 #100 datapoints
M = 5 #size of each minibatch
m = int(n/M) #number of minibatches
n_epochs = 500 #number of epochs
t0 = 1.0
t1 = 10
gamma_j = t0/t1
j = 0
for epoch in range(1,n_epochs+1):
for i in range(m):
k = np.random.randint(m) #Pick the k-th minibatch at random
#Compute the gradient using the data in minibatch Bk
#Compute new suggestion for beta
t = epoch*m+i
gamma_j = step_length(t,t0,t1)
j += 1
print("gamma_j after %d epochs: %g" % (n_epochs,gamma_j))
# Importing various packages
from math import exp, sqrt
from random import random, seed
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import SGDRegressor
m = 100
x = 2*np.random.rand(m,1)
y = 4+3*x+np.random.randn(m,1)
xb = np.c_[np.ones((m,1)), x]
theta_linreg = np.linalg.inv(xb.T.dot(xb)).dot(xb.T).dot(y)
print("Own inversion")
print(theta_linreg)
sgdreg = SGDRegressor(max_iter = 50, penalty=None, eta0=0.1)
sgdreg.fit(x,y.ravel())
print("sgdreg from scikit")
print(sgdreg.intercept_, sgdreg.coef_)
theta = np.random.randn(2,1)
eta = 0.1
Niterations = 1000
for iter in range(Niterations):
gradients = 2.0/m*xb.T @ ((xb @ theta)-y)
theta -= eta*gradients
print("theta frm own gd")
print(theta)
xnew = np.array([[0],[2]])
xbnew = np.c_[np.ones((2,1)), xnew]
ypredict = xbnew.dot(theta)
ypredict2 = xbnew.dot(theta_linreg)
n_epochs = 50
t0, t1 = 5, 50
def learning_schedule(t):
return t0/(t+t1)
theta = np.random.randn(2,1)
for epoch in range(n_epochs):
for i in range(m):
random_index = np.random.randint(m)
xi = xb[random_index:random_index+1]
yi = y[random_index:random_index+1]
gradients = 2 * xi.T @ ((xi @ theta)-yi)
eta = learning_schedule(epoch*m+i)
theta = theta - eta*gradients
print("theta from own sdg")
print(theta)
plt.plot(xnew, ypredict, "r-")
plt.plot(xnew, ypredict2, "b-")
plt.plot(x, y ,'ro')
plt.axis([0,2.0,0, 15.0])
plt.xlabel(r'$x$')
plt.ylabel(r'$y$')
plt.title(r'Random numbers ')
plt.show()
Challenge: try to write a similar code for a Logistic Regression case.
The stochastic gradient descent (SGD) is almost always used with a momentum or inertia term that serves as a memory of the direction we are moving in parameter space. This is typically implemented as follows $$ \begin{align} \mathbf{v}_{t}&=\gamma \mathbf{v}_{t-1}+\eta_{t}\nabla_\theta E(\boldsymbol{\theta}_t) \nonumber \\ \boldsymbol{\theta}_{t+1}&= \boldsymbol{\theta}_t -\mathbf{v}_{t}, \label{_auto1} \end{align} $$
where we have introduced a momentum parameter \( \gamma \), with \( 0\le\gamma\le 1 \), and for brevity we dropped the explicit notation to indicate the gradient is to be taken over a different mini-batch at each step. We call this algorithm gradient descent with momentum (GDM). From these equations, it is clear that \( \mathbf{v}_t \) is a running average of recently encountered gradients and \( (1-\gamma)^{-1} \) sets the characteristic time scale for the memory used in the averaging procedure. Consistent with this, when \( \gamma=0 \), this just reduces down to ordinary SGD as discussed earlier. An equivalent way of writing the updates is $$ \Delta \boldsymbol{\theta}_{t+1} = \gamma \Delta \boldsymbol{\theta}_t -\ \eta_{t}\nabla_\theta E(\boldsymbol{\theta}_t), $$ where we have defined \( \Delta \boldsymbol{\theta}_{t}= \boldsymbol{\theta}_t-\boldsymbol{\theta}_{t-1} \).
Let us try to get more intuition from these equations. It is helpful to consider a simple physical analogy with a particle of mass \( m \) moving in a viscous medium with drag coefficient \( \mu \) and potential \( E(\mathbf{w}) \). If we denote the particle's position by \( \mathbf{w} \), then its motion is described by $$ m {d^2 \mathbf{w} \over dt^2} + \mu {d \mathbf{w} \over dt }= -\nabla_w E(\mathbf{w}). $$
We can discretize this equation in the usual way to get $$ m { \mathbf{w}_{t+\Delta t}-2 \mathbf{w}_{t} +\mathbf{w}_{t-\Delta t} \over (\Delta t)^2}+\mu {\mathbf{w}_{t+\Delta t}- \mathbf{w}_{t} \over \Delta t} = -\nabla_w E(\mathbf{w}). $$
Rearranging this equation, we can rewrite this as $$ \Delta \mathbf{w}_{t +\Delta t}= - { (\Delta t)^2 \over m +\mu \Delta t} \nabla_w E(\mathbf{w})+ {m \over m +\mu \Delta t} \Delta \mathbf{w}_t. $$
Notice that this equation is identical to previous one if we identify the position of the particle, \( \mathbf{w} \), with the parameters \( \boldsymbol{\theta} \). This allows us to identify the momentum parameter and learning rate with the mass of the particle and the viscous drag as: $$ \gamma= {m \over m +\mu \Delta t }, \qquad \eta = {(\Delta t)^2 \over m +\mu \Delta t}. $$
Thus, as the name suggests, the momentum parameter is proportional to the mass of the particle and effectively provides inertia. Furthermore, in the large viscosity/small learning rate limit, our memory time scales as \( (1-\gamma)^{-1} \approx m/(\mu \Delta t) \).
Why is momentum useful? SGD momentum helps the gradient descent algorithm gain speed in directions with persistent but small gradients even in the presence of stochasticity, while suppressing oscillations in high-curvature directions. This becomes especially important in situations where the landscape is shallow and flat in some directions and narrow and steep in others. It has been argued that first-order methods (with appropriate initial conditions) can perform comparable to more expensive second order methods, especially in the context of complex deep learning models.
These beneficial properties of momentum can sometimes become even more pronounced by using a slight modification of the classical momentum algorithm called Nesterov Accelerated Gradient (NAG).
In the NAG algorithm, rather than calculating the gradient at the current parameters, \( \nabla_\theta E(\boldsymbol{\theta}_t) \), one calculates the gradient at the expected value of the parameters given our current momentum, \( \nabla_\theta E(\boldsymbol{\theta}_t +\gamma \mathbf{v}_{t-1}) \). This yields the NAG update rule $$ \begin{align} \mathbf{v}_{t}&=\gamma \mathbf{v}_{t-1}+\eta_{t}\nabla_\theta E(\boldsymbol{\theta}_t +\gamma \mathbf{v}_{t-1}) \nonumber \\ \boldsymbol{\theta}_{t+1}&= \boldsymbol{\theta}_t -\mathbf{v}_{t}. \label{_auto2} \end{align} $$
One of the major advantages of NAG is that it allows for the use of a larger learning rate than GDM for the same choice of \( \gamma \).
In stochastic gradient descent, with and without momentum, we still have to specify a schedule for tuning the learning rates \( \eta_t \) as a function of time. As discussed in the context of Newton's method, this presents a number of dilemmas. The learning rate is limited by the steepest direction which can change depending on the current position in the landscape. To circumvent this problem, ideally our algorithm would keep track of curvature and take large steps in shallow, flat directions and small steps in steep, narrow directions. Second-order methods accomplish this by calculating or approximating the Hessian and normalizing the learning rate by the curvature. However, this is very computationally expensive for extremely large models. Ideally, we would like to be able to adaptively change the step size to match the landscape without paying the steep computational price of calculating or approximating Hessians.
Recently, a number of methods have been introduced that accomplish this by tracking not only the gradient, but also the second moment of the gradient. These methods include AdaGrad, AdaDelta, RMS-Prop, and ADAM.
In RMS prop, in addition to keeping a running average of the first moment of the gradient, we also keep track of the second moment denoted by \( \mathbf{s}_t=\mathbb{E}[\mathbf{g}_t^2] \). The update rule for RMS prop is given by $$ \begin{align} \mathbf{g}_t &= \nabla_\theta E(\boldsymbol{\theta}) \label{_auto3}\\ \mathbf{s}_t &=\beta \mathbf{s}_{t-1} +(1-\beta)\mathbf{g}_t^2 \nonumber \\ \boldsymbol{\theta}_{t+1}&=&\boldsymbol{\theta}_t - \eta_t { \mathbf{g}_t \over \sqrt{\mathbf{s}_t +\epsilon}}, \nonumber \end{align} $$
where \( \beta \) controls the averaging time of the second moment and is typically taken to be about \( \beta=0.9 \), \( \eta_t \) is a learning rate typically chosen to be \( 10^{-3} \), and \( \epsilon\sim 10^{-8} \) is a small regularization constant to prevent divergences. Multiplication and division by vectors is understood as an element-wise operation. It is clear from this formula that the learning rate is reduced in directions where the norm of the gradient is consistently large. This greatly speeds up the convergence by allowing us to use a larger learning rate for flat directions.
A related algorithm is the ADAM optimizer. In ADAM, we keep a running average of both the first and second moment of the gradient and use this information to adaptively change the learning rate for different parameters. In addition to keeping a running average of the first and second moments of the gradient (i.e. \( \mathbf{m}_t=\mathbb{E}[\mathbf{g}_t] \) and \( \mathbf{s}_t=\mathbb{E}[\mathbf{g}^2_t] \), respectively), ADAM performs an additional bias correction to account for the fact that we are estimating the first two moments of the gradient using a running average (denoted by the hats in the update rule below). The update rule for ADAM is given by (where multiplication and division are once again understood to be element-wise operations below) $$ \begin{align} \mathbf{g}_t &= \nabla_\theta E(\boldsymbol{\theta}) \label{_auto4}\\ \mathbf{m}_t &= \beta_1 \mathbf{m}_{t-1} + (1-\beta_1) \mathbf{g}_t \nonumber \\ \mathbf{s}_t &=\beta_2 \mathbf{s}_{t-1} +(1-\beta_2)\mathbf{g}_t^2 \nonumber \\ \hat{\mathbf{m}}_t&={\mathbf{m}_t \over 1-\beta_1^t} \nonumber \\ \hat{\mathbf{s}}_t &={\mathbf{s}_t \over1-\beta_2^t} \nonumber \\ \boldsymbol{\theta}_{t+1}&=\boldsymbol{\theta}_t - \eta_t { \hat{\mathbf{m}}_t \over \sqrt{\hat{\mathbf{s}}_t} +\epsilon}, \nonumber \\ \label{_auto5} \end{align} $$
where \( \beta_1 \) and \( \beta_2 \) set the memory lifetime of the first and second moment and are typically taken to be \( 0.9 \) and \( 0.99 \) respectively, and \( \eta \) and \( \epsilon \) are identical to RMSprop.
Like in RMSprop, the effective step size of a parameter depends on the magnitude of its gradient squared. To understand this better, let us rewrite this expression in terms of the variance \( \boldsymbol{\sigma}_t^2 = \hat{\mathbf{s}}_t - (\hat{\mathbf{m}}_t)^2 \). Consider a single parameter \( \theta_t \). The update rule for this parameter is given by $$ \Delta \theta_{t+1}= -\eta_t { \hat{m}_t \over \sqrt{\sigma_t^2 + m_t^2 }+\epsilon}. $$
Automatic differentiation (AD), also called algorithmic differentiation or computational differentiation,is a set of techniques to numerically evaluate the derivative of a function specified by a computer program. AD exploits the fact that every computer program, no matter how complicated, executes a sequence of elementary arithmetic operations (addition, subtraction, multiplication, division, etc.) and elementary functions (exp, log, sin, cos, etc.). By applying the chain rule repeatedly to these operations, derivatives of arbitrary order can be computed automatically, accurately to working precision, and using at most a small constant factor more arithmetic operations than the original program.
Automatic differentiation is neither:
Python has tools for so-called automatic differentiation. Consider the following example $$ f(x) = \sin\left(2\pi x + x^2\right) $$ which has the following derivative $$ f'(x) = \cos\left(2\pi x + x^2\right)\left(2\pi + 2x\right) $$ Using autograd we have
import autograd.numpy as np
# To do elementwise differentiation:
from autograd import elementwise_grad as egrad
# To plot:
import matplotlib.pyplot as plt
def f(x):
return np.sin(2*np.pi*x + x**2)
def f_grad_analytic(x):
return np.cos(2*np.pi*x + x**2)*(2*np.pi + 2*x)
# Do the comparison:
x = np.linspace(0,1,1000)
f_grad = egrad(f)
computed = f_grad(x)
analytic = f_grad_analytic(x)
plt.title('Derivative computed from Autograd compared with the analytical derivative')
plt.plot(x,computed,label='autograd')
plt.plot(x,analytic,label='analytic')
plt.xlabel('x')
plt.ylabel('y')
plt.legend()
plt.show()
print("The max absolute difference is: %g"%(np.max(np.abs(computed - analytic))))
Here we experiment with what kind of functions Autograd is capable of finding the gradient of. The following Python functions are just meant to illustrate what Autograd can do, but please feel free to experiment with other, possibly more complicated, functions as well.
import autograd.numpy as np
from autograd import grad
def f1(x):
return x**3 + 1
f1_grad = grad(f1)
# Remember to send in float as argument to the computed gradient from Autograd!
a = 1.0
# See the evaluated gradient at a using autograd:
print("The gradient of f1 evaluated at a = %g using autograd is: %g"%(a,f1_grad(a)))
# Compare with the analytical derivative, that is f1'(x) = 3*x**2
grad_analytical = 3*a**2
print("The gradient of f1 evaluated at a = %g by finding the analytic expression is: %g"%(a,grad_analytical))
To differentiate with respect to two (or more) arguments of a Python function, Autograd need to know at which variable the function if being differentiated with respect to.
import autograd.numpy as np
from autograd import grad
def f2(x1,x2):
return 3*x1**3 + x2*(x1 - 5) + 1
# By sending the argument 0, Autograd will compute the derivative w.r.t the first variable, in this case x1
f2_grad_x1 = grad(f2,0)
# ... and differentiate w.r.t x2 by sending 1 as an additional arugment to grad
f2_grad_x2 = grad(f2,1)
x1 = 1.0
x2 = 3.0
print("Evaluating at x1 = %g, x2 = %g"%(x1,x2))
print("-"*30)
# Compare with the analytical derivatives:
# Derivative of f2 w.r.t x1 is: 9*x1**2 + x2:
f2_grad_x1_analytical = 9*x1**2 + x2
# Derivative of f2 w.r.t x2 is: x1 - 5:
f2_grad_x2_analytical = x1 - 5
# See the evaluated derivations:
print("The derivative of f2 w.r.t x1: %g"%( f2_grad_x1(x1,x2) ))
print("The analytical derivative of f2 w.r.t x1: %g"%( f2_grad_x1(x1,x2) ))
print()
print("The derivative of f2 w.r.t x2: %g"%( f2_grad_x2(x1,x2) ))
print("The analytical derivative of f2 w.r.t x2: %g"%( f2_grad_x2(x1,x2) ))
Note that the grad function will not produce the true gradient of the function. The true gradient of a function with two or more variables will produce a vector, where each element is the function differentiated w.r.t a variable.
import autograd.numpy as np
from autograd import grad
def f3(x): # Assumes x is an array of length 5 or higher
return 2*x[0] + 3*x[1] + 5*x[2] + 7*x[3] + 11*x[4]**2
f3_grad = grad(f3)
x = np.linspace(0,4,5)
# Print the computed gradient:
print("The computed gradient of f3 is: ", f3_grad(x))
# The analytical gradient is: (2, 3, 5, 7, 22*x[4])
f3_grad_analytical = np.array([2, 3, 5, 7, 22*x[4]])
# Print the analytical gradient:
print("The analytical gradient of f3 is: ", f3_grad_analytical)
Note that in this case, when sending an array as input argument, the output from Autograd is another array. This is the true gradient of the function, as opposed to the function in the previous example. By using arrays to represent the variables, the output from Autograd might be easier to work with, as the output is closer to what one could expect form a gradient-evaluting function.
import autograd.numpy as np
from autograd import grad
def f4(x):
return np.sqrt(1+x**2) + np.exp(x) + np.sin(2*np.pi*x)
f4_grad = grad(f4)
x = 2.7
# Print the computed derivative:
print("The computed derivative of f4 at x = %g is: %g"%(x,f4_grad(x)))
# The analytical derivative is: x/sqrt(1 + x**2) + exp(x) + cos(2*pi*x)*2*pi
f4_grad_analytical = x/np.sqrt(1 + x**2) + np.exp(x) + np.cos(2*np.pi*x)*2*np.pi
# Print the analytical gradient:
print("The analytical gradient of f4 at x = %g is: %g"%(x,f4_grad_analytical))
import autograd.numpy as np
from autograd import grad
def f5(x):
if x >= 0:
return x**2
else:
return -3*x + 1
f5_grad = grad(f5)
x = 2.7
# Print the computed derivative:
print("The computed derivative of f5 at x = %g is: %g"%(x,f5_grad(x)))
import autograd.numpy as np
from autograd import grad
def f6_for(x):
val = 0
for i in range(10):
val = val + x**i
return val
def f6_while(x):
val = 0
i = 0
while i < 10:
val = val + x**i
i = i + 1
return val
f6_for_grad = grad(f6_for)
f6_while_grad = grad(f6_while)
x = 0.5
# Print the computed derivaties of f6_for and f6_while
print("The computed derivative of f6_for at x = %g is: %g"%(x,f6_for_grad(x)))
print("The computed derivative of f6_while at x = %g is: %g"%(x,f6_while_grad(x)))
import autograd.numpy as np
from autograd import grad
# Both of the functions are implementation of the sum: sum(x**i) for i = 0, ..., 9
# The analytical derivative is: sum(i*x**(i-1))
f6_grad_analytical = 0
for i in range(10):
f6_grad_analytical += i*x**(i-1)
print("The analytical derivative of f6 at x = %g is: %g"%(x,f6_grad_analytical))
import autograd.numpy as np
from autograd import grad
def f7(n): # Assume that n is an integer
if n == 1 or n == 0:
return 1
else:
return n*f7(n-1)
f7_grad = grad(f7)
n = 2.0
print("The computed derivative of f7 at n = %d is: %g"%(n,f7_grad(n)))
# The function f7 is an implementation of the factorial of n.
# By using the product rule, one can find that the derivative is:
f7_grad_analytical = 0
for i in range(int(n)-1):
tmp = 1
for k in range(int(n)-1):
if k != i:
tmp *= (n - k)
f7_grad_analytical += tmp
print("The analytical derivative of f7 at n = %d is: %g"%(n,f7_grad_analytical))
Note that if n is equal to zero or one, Autograd will give an error message. This message appears when the output is independent on input.
Assigning a value to the variable being differentiated with respect to
import autograd.numpy as np
from autograd import grad
def f8(x): # Assume x is an array
x[2] = 3
return x*2
f8_grad = grad(f8)
x = 8.4
print("The derivative of f8 is:",f8_grad(x))
Here, Autograd tells us that an 'ArrayBox' does not support item assignment. The item assignment is done when the program tries to assign x[2] to the value 3. However, Autograd has implemented the computation of the derivative such that this assignment is not possible.
import autograd.numpy as np
from autograd import grad
def f9(a): # Assume a is an array with 2 elements
b = np.array([1.0,2.0])
return a.dot(b)
f9_grad = grad(f9)
x = np.array([1.0,0.0])
print("The derivative of f9 is:",f9_grad(x))
Here we are told that the 'dot' function does not belong to Autograd's version of a Numpy array. To overcome this, an alternative syntax which also computed the dot product can be used:
import autograd.numpy as np
from autograd import grad
def f9_alternative(x): # Assume a is an array with 2 elements
b = np.array([1.0,2.0])
return np.dot(x,b) # The same as x_1*b_1 + x_2*b_2
f9_alternative_grad = grad(f9_alternative)
x = np.array([3.0,0.0])
print("The gradient of f9 is:",f9_alternative_grad(x))
# The analytical gradient of the dot product of vectors x and b with two elements (x_1,x_2) and (b_1, b_2) respectively
# w.r.t x is (b_1, b_2).
a += b
a -= b
a*= b
a /=b
Before we proceed, we would like to discuss the approach called the standard Steepest descent, which again leads to us having to be able to compute a matrix. It belongs to the class of Conjugate Gradient methods (CG).
The success of the CG method for finding solutions of non-linear problems is based on the theory of conjugate gradients for linear systems of equations. It belongs to the class of iterative methods for solving problems from linear algebra of the type $$ \begin{equation*} \hat{A}\hat{x} = \hat{b}. \end{equation*} $$
In the iterative process we end up with a problem like $$ \begin{equation*} \hat{r}= \hat{b}-\hat{A}\hat{x}, \end{equation*} $$ where \( \hat{r} \) is the so-called residual or error in the iterative process.
When we have found the exact solution, \( \hat{r}=0 \).
The residual is zero when we reach the minimum of the quadratic equation $$ \begin{equation*} P(\hat{x})=\frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T\hat{b}, \end{equation*} $$
with the constraint that the matrix \( \hat{A} \) is positive definite and symmetric. This defines also the Hessian and we want it to be positive definite.
We denote the initial guess for \( \hat{x} \) as \( \hat{x}_0 \). We can assume without loss of generality that $$ \begin{equation*} \hat{x}_0=0, \end{equation*} $$ or consider the system $$ \begin{equation*} \hat{A}\hat{z} = \hat{b}-\hat{A}\hat{x}_0, \end{equation*} $$ instead.
One can show that the solution \( \hat{x} \) is also the unique minimizer of the quadratic form $$ \begin{equation*} f(\hat{x}) = \frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T \hat{x} , \quad \hat{x}\in\mathbf{R}^n. \end{equation*} $$ This suggests taking the first basis vector \( \hat{r}_1 \) (see below for definition) to be the gradient of \( f \) at \( \hat{x}=\hat{x}_0 \), which equals $$ \begin{equation*} \hat{A}\hat{x}_0-\hat{b}, \end{equation*} $$ and \( \hat{x}_0=0 \) it is equal \( -\hat{b} \).
We can compute the residual iteratively as $$ \begin{equation*} \hat{r}_{k+1}=\hat{b}-\hat{A}\hat{x}_{k+1}, \end{equation*} $$ which equals $$ \begin{equation*} \hat{b}-\hat{A}(\hat{x}_k+\alpha_k\hat{r}_k), \end{equation*} $$ or $$ \begin{equation*} (\hat{b}-\hat{A}\hat{x}_k)-\alpha_k\hat{A}\hat{r}_k, \end{equation*} $$ which gives $$ \alpha_k = \frac{\hat{r}_k^T\hat{r}_k}{\hat{r}_k^T\hat{A}\hat{r}_k} $$ leading to the iterative scheme $$ \begin{equation*} \hat{x}_{k+1}=\hat{x}_k-\alpha_k\hat{r}_{k}, \end{equation*} $$
#include <cmath>
#include <iostream>
#include <fstream>
#include <iomanip>
#include "vectormatrixclass.h"
using namespace std;
// Main function begins here
int main(int argc, char * argv[]){
int dim = 2;
Vector x(dim),xsd(dim), b(dim),x0(dim);
Matrix A(dim,dim);
// Set our initial guess
x0(0) = x0(1) = 0;
// Set the matrix
A(0,0) = 3; A(1,0) = 2; A(0,1) = 2; A(1,1) = 6;
b(0) = 2; b(1) = -8;
cout << "The Matrix A that we are using: " << endl;
A.Print();
cout << endl;
xsd = SteepestDescent(A,b,x0);
cout << "The approximate solution using Steepest Descent is: " << endl;
xsd.Print();
cout << endl;
}
Vector SteepestDescent(Matrix A, Vector b, Vector x0){
int IterMax, i;
int dim = x0.Dimension();
const double tolerance = 1.0e-14;
Vector x(dim),f(dim),z(dim);
double c,alpha,d;
IterMax = 30;
x = x0;
r = A*x-b;
i = 0;
while (i <= IterMax){
z = A*r;
c = dot(r,r);
alpha = c/dot(r,z);
x = x - alpha*r;
r = A*x-b;
if(sqrt(dot(r,r)) < tolerance) break;
i++;
}
return x;
}
import numpy as np
import numpy.linalg as la
import scipy.optimize as sopt
import matplotlib.pyplot as pt
from mpl_toolkits.mplot3d import axes3d
def f(x):
return 0.5*x[0]**2 + 2.5*x[1]**2
def df(x):
return np.array([x[0], 5*x[1]])
fig = pt.figure()
ax = fig.gca(projection="3d")
xmesh, ymesh = np.mgrid[-2:2:50j,-2:2:50j]
fmesh = f(np.array([xmesh, ymesh]))
ax.plot_surface(xmesh, ymesh, fmesh)
And then as countor plot
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh)
guesses = [np.array([2, 2./5])]
Find guesses
x = guesses[-1]
s = -df(x)
Run it!
def f1d(alpha):
return f(x + alpha*s)
alpha_opt = sopt.golden(f1d)
next_guess = x + alpha_opt * s
guesses.append(next_guess)
print(next_guess)
What happened?
pt.axis("equal")
pt.contour(xmesh, ymesh, fmesh, 50)
it_array = np.array(guesses)
pt.plot(it_array.T[0], it_array.T[1], "x-")
In the CG method we define so-called conjugate directions and two vectors \( \hat{s} \) and \( \hat{t} \) are said to be conjugate if $$ \begin{equation*} \hat{s}^T\hat{A}\hat{t}= 0. \end{equation*} $$ The philosophy of the CG method is to perform searches in various conjugate directions of our vectors \( \hat{x}_i \) obeying the above criterion, namely $$ \begin{equation*} \hat{x}_i^T\hat{A}\hat{x}_j= 0. \end{equation*} $$ Two vectors are conjugate if they are orthogonal with respect to this inner product. Being conjugate is a symmetric relation: if \( \hat{s} \) is conjugate to \( \hat{t} \), then \( \hat{t} \) is conjugate to \( \hat{s} \).
An example is given by the eigenvectors of the matrix $$ \begin{equation*} \hat{v}_i^T\hat{A}\hat{v}_j= \lambda\hat{v}_i^T\hat{v}_j, \end{equation*} $$ which is zero unless \( i=j \).
Assume now that we have a symmetric positive-definite matrix \( \hat{A} \) of size \( n\times n \). At each iteration \( i+1 \) we obtain the conjugate direction of a vector $$ \begin{equation*} \hat{x}_{i+1}=\hat{x}_{i}+\alpha_i\hat{p}_{i}. \end{equation*} $$ We assume that \( \hat{p}_{i} \) is a sequence of \( n \) mutually conjugate directions. Then the \( \hat{p}_{i} \) form a basis of \( R^n \) and we can expand the solution $ \hat{A}\hat{x} = \hat{b}$ in this basis, namely $$ \begin{equation*} \hat{x} = \sum^{n}_{i=1} \alpha_i \hat{p}_i. \end{equation*} $$
The coefficients are given by $$ \begin{equation*} \mathbf{A}\mathbf{x} = \sum^{n}_{i=1} \alpha_i \mathbf{A} \mathbf{p}_i = \mathbf{b}. \end{equation*} $$ Multiplying with \( \hat{p}_k^T \) from the left gives $$ \begin{equation*} \hat{p}_k^T \hat{A}\hat{x} = \sum^{n}_{i=1} \alpha_i\hat{p}_k^T \hat{A}\hat{p}_i= \hat{p}_k^T \hat{b}, \end{equation*} $$ and we can define the coefficients \( \alpha_k \) as $$ \begin{equation*} \alpha_k = \frac{\hat{p}_k^T \hat{b}}{\hat{p}_k^T \hat{A} \hat{p}_k} \end{equation*} $$
If we choose the conjugate vectors \( \hat{p}_k \) carefully, then we may not need all of them to obtain a good approximation to the solution \( \hat{x} \). We want to regard the conjugate gradient method as an iterative method. This will us to solve systems where \( n \) is so large that the direct method would take too much time.
We denote the initial guess for \( \hat{x} \) as \( \hat{x}_0 \). We can assume without loss of generality that $$ \begin{equation*} \hat{x}_0=0, \end{equation*} $$ or consider the system $$ \begin{equation*} \hat{A}\hat{z} = \hat{b}-\hat{A}\hat{x}_0, \end{equation*} $$ instead.
One can show that the solution \( \hat{x} \) is also the unique minimizer of the quadratic form $$ \begin{equation*} f(\hat{x}) = \frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T \hat{x} , \quad \hat{x}\in\mathbf{R}^n. \end{equation*} $$ This suggests taking the first basis vector \( \hat{p}_1 \) to be the gradient of \( f \) at \( \hat{x}=\hat{x}_0 \), which equals $$ \begin{equation*} \hat{A}\hat{x}_0-\hat{b}, \end{equation*} $$ and \( \hat{x}_0=0 \) it is equal \( -\hat{b} \). The other vectors in the basis will be conjugate to the gradient, hence the name conjugate gradient method.
Let \( \hat{r}_k \) be the residual at the \( k \)-th step: $$ \begin{equation*} \hat{r}_k=\hat{b}-\hat{A}\hat{x}_k. \end{equation*} $$ Note that \( \hat{r}_k \) is the negative gradient of \( f \) at \( \hat{x}=\hat{x}_k \), so the gradient descent method would be to move in the direction \( \hat{r}_k \). Here, we insist that the directions \( \hat{p}_k \) are conjugate to each other, so we take the direction closest to the gradient \( \hat{r}_k \) under the conjugacy constraint. This gives the following expression $$ \begin{equation*} \hat{p}_{k+1}=\hat{r}_k-\frac{\hat{p}_k^T \hat{A}\hat{r}_k}{\hat{p}_k^T\hat{A}\hat{p}_k} \hat{p}_k. \end{equation*} $$
We can also compute the residual iteratively as $$ \begin{equation*} \hat{r}_{k+1}=\hat{b}-\hat{A}\hat{x}_{k+1}, \end{equation*} $$ which equals $$ \begin{equation*} \hat{b}-\hat{A}(\hat{x}_k+\alpha_k\hat{p}_k), \end{equation*} $$ or $$ \begin{equation*} (\hat{b}-\hat{A}\hat{x}_k)-\alpha_k\hat{A}\hat{p}_k, \end{equation*} $$ which gives $$ \begin{equation*} \hat{r}_{k+1}=\hat{r}_k-\hat{A}\hat{p}_{k}, \end{equation*} $$
Vector ConjugateGradient(Matrix A, Vector b, Vector x0){
int dim = x0.Dimension();
const double tolerance = 1.0e-14;
Vector x(dim),r(dim),v(dim),z(dim);
double c,t,d;
x = x0;
r = b - A*x;
v = r;
c = dot(r,r);
int i = 0; IterMax = dim;
while(i <= IterMax){
z = A*v;
t = c/dot(v,z);
x = x + t*v;
r = r - t*z;
d = dot(r,r);
if(sqrt(d) < tolerance)
break;
v = r + (d/c)*v;
c = d; i++;
}
return x;
}
The optimization problem is to minimize \( f(\mathbf {x} ) \) where \( \mathbf {x} \) is a vector in \( R^{n} \), and \( f \) is a differentiable scalar function. There are no constraints on the values that \( \mathbf {x} \) can take.
The algorithm begins at an initial estimate for the optimal value \( \mathbf {x}_{0} \) and proceeds iteratively to get a better estimate at each stage.
The search direction \( p_k \) at stage \( k \) is given by the solution of the analogue of the Newton equation $$ B_{k}\mathbf {p} _{k}=-\nabla f(\mathbf {x}_{k}), $$
where \( B_{k} \) is an approximation to the Hessian matrix, which is updated iteratively at each stage, and \( \nabla f(\mathbf {x} _{k}) \) is the gradient of the function evaluated at \( x_k \). A line search in the direction \( p_k \) is then used to find the next point \( x_{k+1} \) by minimising $$ f(\mathbf {x}_{k}+\alpha \mathbf {p}_{k}), $$ over the scalar \( \alpha > 0 \).


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Bundle The Amazon Fire TV Cube & Ring Video Doorbell 2 For $219
Amazon has a great bundle right now, which includes the Amazon Fire TV Cube and the Ring Video Doorbell 2, for just $218.99. Normally this would cost you $318.99, so you're saving $100 on this bundle.
The Amazon Fire TV Cube is basically a Fire TV and an Echo all-in-one. It can control your TV, and play Netflix, Hulu, Amazon Prime Video and much more. But it also has always listening Alexa, so you can literally use your voice to control your TV. Now that is a really cool feature and definitely one that we'd like to see more of.
The Fire TV Cube is able to control your TV because it is plugged into the HDMI port. So much like the Sonos Beam can control your TV, so can the Fire TV Cube.
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Then there's the Ring Video Doorbell 2. This is a video doorbell that you can install outside of your home. It'll be great for the holiday season when there are lots of packages coming to your door. This way you're able to see when they get there, and who might be trying to grab them. Since that is a real problem these days.
With the Ring Video Doorbell 2, you're also going to get notifications when someone is at your door. This works even if they don't ring the doorbell. So you'll know exactly when someone is at your door, even if they are trying to steal a package or two.
Ring Video Doorbell 2 does also work with Amazon Alexa, allowing you to bring up the video feed onto the big screen, and it can also ring through your Fire TV Cube and other Echo devices, like a traditional doorbell would.
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You can pick up this bundle of the Amazon Fire TV Cube and the Ring Video Doorbell 2 from Amazon by clicking here.


Brownie in a Cup
The following recipe is for brownie in a cup. This is a delicious, and very healthy brownie for university students to enjoy eating.
Ingredients
- 2 TBS Melted Butter
- 2 TBS Milk (2%)
- 4 TBS Sugar
- 2 TBS Flour
- 2 TBS Cocoa Powder
- 1 TSP Vanilla
Steps:
Step 1:
Melt the butter in a mug, add milk and sugar. Stir until consistent.
Step 2:
Add Cocoa powder and flower into the mug.
Step 3:
Add vanilla into the mug.
Step 4:
Stir ingredients in the mug until no dry chunks are left.
Step 5:
Microwave the mug with ingredients for 1 minute.
Step 6:
Take out of microwave and let stand for 3 minutes to cool down.
Finish!
Stir and enjoy the brownie!


This city is often considered the center of the universe in terms of finance, culture, fashion and more.
I'd love to go to New York city during the Christmas period to see the Rockefeller Christmas Tree, the decorated stores along Fifth Avenue, skating in Central Park, or the Rockettes at Radio City Music Hall.
It’s the type of holiday experience often immortalized in the movies, and yet I could experience it myself.
Japan
I'd like to experience Japan's 'bullet train'. It's the best, easiest and most cost effective way to travel, traveling more than 300km an hour.
I'd also like to visit Japan during cherry blossom season
Watch the timelapse video of blossoming trees in Tokyo, Kyoto and around Mt Fuji below


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