I'm trying to a Steepest descent for a function with 2 variables. It works fine with known step size which = 0.3. But I want to find a way to optimize step size and create a function to find a good step size. I found something called Armijo–Goldstein condition but I didn't understand it and the formula was kind of confusing for me. So I ask your help, if you have an idea how to balance this 'cause everything that connected to the step size at my code is wrong, I think. It have to calculate step size deepening on x and y we're on, I guess.
x, y = f.optimal_range() ##getting random start
step = 0.3 ## <--- This number have to be random between 0 to 1. But my step size calculations are wrong so I can't do it.
while f.fprime_x(x) != 0: ##loop to find 0 point for derivative of a function on x
fx = -f.fprime_x(x)
x = x + (step * fx)
print(x)
if not f.delta_check(step, x, y): <--- Here's the problem. By the defenition the step have to be smaller if it doesn't pass the check, but if I make it smaller - it enters the eternal loop around the mi
step = step * 1.001
while f.fprime_y(y) != 0: ##loop to find 0 point for derivative of a function on x
fy = -f.fprime_y(y)
y = y + (step * fy)
print(x, y)
if not f.delta_check(step, x, y):
step = step * 1.001
print(f"\n\nMIN ({x}, {y})")
Here's the function of delta / step size checking:
def delta_check(delta, x, y):
ux = -fprime_x(x)
uy = -fprime_y(y)
f_del_xy = func(x + (delta * ux), y + (delta * uy))
return f_del_xy <= func(delta * ux, delta * uy) + delta
Here's a notional Armijo–Goldstein implementation. Can't test it without a data+function example, though.
# both should be less than, but usually close to 1
c = 0.8 # how much imperfection in function improvement we'll settle up with
tau = 0.8 # how much the step will be decreased at each iteration
x = np.array(f.optimal_range()) # assume everything is a vector; x is an n-dimensional coordinate
# NOTE: the part below is repeated for every X update
step = 0.3 # alpha in Armijo–Goldstein terms
gradient = np.array(f.fprime_x(x[0]), f.fprime_y(x[1]), ...)
# in the simplest case (SGD) p can point in the direction of gradient,
# but in general they don't have to be the same, e.g. because of added momentum
p = -gradient / ((gradient**2).sum() **0.5)
m = gradient.dot(p) # "happy case" improvement per unit step
t = - c * m # improvement we'll consider good enough
# func(*x) might be worth precomputing
while func(*x) - func(*(x + step*p)) < step * t: # good enough step size found
step *= tau
# update X and repeat
I do have a function, for example , but this can be something else as well, like a quadratic or logarithmic function. I am only interested in the domain of . The parameters of the function (a and k in this case) are known as well.
My goal is to fit a continuous piece-wise function to this, which contains alternating segments of linear functions (i.e. sloped straight segments, each with intercept of 0) and constants (i.e. horizontal segments joining the sloped segments together). The first and last segments are both sloped. And the number of segments should be pre-selected between around 9-29 (that is 5-15 linear steps + 4-14 constant plateaus).
Formally
The input function:
The fitted piecewise function:
I am looking for the optimal resulting parameters (c,r,b) (in terms of least squares) if the segment numbers (n) are specified beforehand.
The resulting constants (c) and the breakpoints (r) should be whole natural numbers, and the slopes (b) round two decimal point values.
I have tried to do the fitting numerically using the pwlf package using a segmented constant models, and further processed the resulting constant model with some graphical intuition to "slice" the constant steps with the slopes. It works to some extent, but I am sure this is suboptimal from both fitting perspective and computational efficiency. It takes multiple minutes to generate a fitting with 8 slopes on the range of 1-50000. I am sure there must be a better way to do this.
My idea would be to instead using only numerical methods/ML, the fact that we have the algebraic form of the input function could be exploited in some way to at least to use algebraic transforms (integrals) to get to a simpler optimization problem.
import numpy as np
import matplotlib.pyplot as plt
import pwlf
# The input function
def input_func(x,k,a):
return np.power(x,1/a)*k
x = np.arange(1,5e4)
y = input_func(x, 1.8, 1.3)
plt.plot(x,y);
def pw_fit(func, x_r, no_seg, *fparams):
# working on the specified range
x = np.arange(1,x_r)
y_input = func(x, *fparams)
my_pwlf = pwlf.PiecewiseLinFit(x, y_input, degree=0)
res = my_pwlf.fit(no_seg)
yHat = my_pwlf.predict(x)
# Function values at the breakpoints
y_isec = func(res, *fparams)
# Slope values at the breakpoints
slopes = np.round(y_isec / res, decimals=2)
slopes = slopes[1:]
# For the first slope value, I use the intersection of the first constant plateau and the input function
slopes = np.insert(slopes,0,np.round(y_input[np.argwhere(np.diff(np.sign(y_input - yHat))).flatten()[0]] / np.argwhere(np.diff(np.sign(y_input - yHat))).flatten()[0], decimals=2))
plateaus = np.unique(np.round(yHat))
# If due to rounding slope values (to two decimals), there is no change in a subsequent step, I just remove those segments
to_del = np.argwhere(np.diff(slopes) == 0).flatten()
slopes = np.delete(slopes,to_del + 1)
plateaus = np.delete(plateaus,to_del)
breakpoints = [np.ceil(plateaus[0]/slopes[0])]
for idx, j in enumerate(slopes[1:-1]):
breakpoints.append(np.floor(plateaus[idx]/j))
breakpoints.append(np.ceil(plateaus[idx+1]/j))
breakpoints.append(np.floor(plateaus[-1]/slopes[-1]))
return slopes, plateaus, breakpoints
slo, plat, breaks = pw_fit(input_func, 50000, 8, 1.8, 1.3)
# The piecewise function itself
def pw_calc(x, slopes, plateaus, breaks):
x = x.astype('float')
cond_list = [x < breaks[0]]
for idx, j in enumerate(breaks[:-1]):
cond_list.append((j <= x) & (x < breaks[idx+1]))
cond_list.append(breaks[-1] <= x)
func_list = [lambda x: x * slopes[0]]
for idx, j in enumerate(slopes[1:]):
func_list.append(plateaus[idx])
func_list.append(lambda x, j=j: x * j)
return np.piecewise(x, cond_list, func_list)
y_output = pw_calc(x, slo, plat, breaks)
plt.plot(x,y,y_output);
(Not important, but I think the fitted piecewise function is not continuous as it is. Intervals should be x<=r1; r1<x<=r2; ....)
As Anatolyg has pointed out, it looks to me that in the optimal solution (for the function posted at least, and probably for any where the derivative is different from zero), the horizantal segments will collapse to a point or the minimum segment length (in this case 1).
EDIT---------------------------------------------
The behavior above could only be valid if the slopes could have an intercept. If the intercepts are zero, as posted in the question, one consideration must be taken into account: Is the initial parabolic function defined in zero or nearby? Imagine the function y=0.001 *sqrt(x-1000), then the segments defined as b*x will have a slope close to zero and will be so similar to the constant segments that the best fit will be just the line that without intercept that fits better all the function.
Provided that the function is defined in zero or nearby, you can start by approximating the curve just by linear segments (with intercepts):
divide the function domain in N intervals(equal intervals or whose size is a function of the average curvature (or second derivative) of the function along the domain).
linear fit/regression in each intervals
for each interval, if a point (or bunch of points) in the extreme of any interval is better fitted by the line of the neighbor interval than the line of its interval, this point is assigned to the neighbor interval.
Repeat from 2) until no extreme points are moved.
Linear regressions might be optimized not to calculate all the covariance matrixes from scratch on each iteration, but just adding the contributions of the moved points to the previous covariance matrixes.
Then each linear segment (LSi) is replaced by a combination of a small constant segment at the beginning (Cbi), a linear segment without intercept (Si), and another constant segment at the end (Cei). This segments are easy to calculate as Si will contain the middle point of LSi, and Cbi and Cei will have respectively the begin and end values of the segment LSi. Then the intervals of each segment has to be calculated as an intersection between lines.
With this, the constant end segment will be collinear with the constant begin segment from the next interval so they will merge, resulting in a series of constant and linear segments interleaved.
But this would be a floating point start solution. Next, you will have to apply all the roundings which will mess up quite a lot all the segments as the conditions integer intervals and linear segments without slope can be very confronting. In fact, b,c,r are not totally independent. If ci and ri+1 are known, then bi+1 is already fixed
If nothing is broken so far, the final task will be to minimize the error/cost function (I assume that it will be the integral of the error between the parabolic function and the segments). My guess is that gradients here will be quite a pain, as if you change for example one ci, all the rest of the bj and cj will have to adapt as well due to the integer intervals restriction. However, if you can generalize the derivatives between parameters ( how much do I have to adapt bi+1 if ci changes a unit), you can propagate the change of one parameter to all other parameters and have kind of a gradient. Then for each interval, you can estimate what would be the ideal parameter and averaging all intervals calculate the best gradient step. Let me illustrate this:
Assuming first that r parameters are fixed, if I change c1 by one unit, b2 changes by 0.1, c2 changes by -0.2 and b3 changes by 0.2. This would be the gradient.
Then I estimate, comparing with the parabolic curve, that c1 should increase 0.5 (to reduce the cost by 10 points), b2 should increase 0.2 (to reduce the cost by 5 points), c2 should increase 0.2 (to reduce the cost by 6 points) and b3 should increase 0.1 (to reduce the cost by 9 points).
Finally, the gradient step would be (0.5/1·10 + 0.2/0.1·5 - 0.2/(-0.2)·6 + 0.1/0.2·9)/(10 + 5 + 6 + 9)~= 0.45. Thus, c1 would increase 0.45 units, b2 would increase 0.45·0.1, and so on.
When you add the r parameters to the pot, as integer intervals do not have an proper derivative, calculation is not straightforward. However, you can consider r parameters as floating points, calculate and apply the gradient step and then apply the roundings.
We can integrate the squared error function for linear and constant pieces and let SciPy optimize it. Python 3:
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize
xl = 1
xh = 50000
a = 1.3
p = 1 / a
n = 8
def split_b_and_c(bc):
return bc[::2], bc[1::2]
def solve_for_r(b, c):
r = np.empty(2 * n)
r[0] = xl
r[1:-1:2] = c / b[:-1]
r[2::2] = c / b[1:]
r[-1] = xh
return r
def linear_residual_integral(b, x):
return (
(x ** (2 * p + 1)) / (2 * p + 1)
- 2 * b * x ** (p + 2) / (p + 2)
+ b ** 2 * x ** 3 / 3
)
def constant_residual_integral(c, x):
return x ** (2 * p + 1) / (2 * p + 1) - 2 * c * x ** (p + 1) / (p + 1) + c ** 2 * x
def squared_error(bc):
b, c = split_b_and_c(bc)
r = solve_for_r(b, c)
linear = np.sum(
linear_residual_integral(b, r[1::2]) - linear_residual_integral(b, r[::2])
)
constant = np.sum(
constant_residual_integral(c, r[2::2])
- constant_residual_integral(c, r[1:-1:2])
)
return linear + constant
def evaluate(x, b, c, r):
i = 0
while x > r[i + 1]:
i += 1
return b[i // 2] * x if i % 2 == 0 else c[i // 2]
def main():
bc0 = (xl + (xh - xl) * np.arange(1, 4 * n - 2, 2) / (4 * n - 2)) ** (
p - 1 + np.arange(2 * n - 1) % 2
)
bc = scipy.optimize.minimize(
squared_error, bc0, bounds=[(1e-06, None) for i in range(2 * n - 1)]
).x
b, c = split_b_and_c(bc)
r = solve_for_r(b, c)
X = np.linspace(xl, xh, 1000)
Y = [evaluate(x, b, c, r) for x in X]
plt.plot(X, X ** p)
plt.plot(X, Y)
plt.show()
if __name__ == "__main__":
main()
I have tried to come up with a new solution myself, based on the idea of #Amo Robb, where I have partitioned the domain, and curve fitted a dual - constant and linear - piece together (with the help of np.maximum). I have used the 1 / f(x)' as the function to designate the breakpoints, but I know this is arbitrary and does not provide a global optimum. Maybe there is some optimal function for these breakpoints. But this solution is OK for me, as it might be appropriate to have a better fit at the first segments, at the expense of the error for the later segments. (The task itself is actually a cost based retail margin calculation {supply price -> added margin}, as the retail POS software can only work with such piecewise margin function).
The answer from #David Eisenstat is correct optimal solution if the parameters are allowed to be floats. Unfortunately the POS software can not use floats. It is OK to round up c-s and r-s afterwards. But the b-s should be rounded to two decimals, as those are inputted as percents, and this constraint would ruin the optimal solution with long floats. I will try to further improve my solution with both Amo's and David's valuable input. Thank You for that!
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
# The input function f(x)
def input_func(x,k,a):
return np.power(x,1/a) * k
# 1 / f(x)'
def one_per_der(x,k,a):
return a / (k * np.power(x, 1/a-1))
# 1 / f(x)' inverted
def one_per_der_inv(x,k,a):
return np.power(a / (x*k), a / (1-a))
def segment_fit(start,end,y,first_val):
b, _ = curve_fit(lambda x,b: np.maximum(first_val, b*x), np.arange(start,end), y[start-1:end-1])
b = float(np.round(b, decimals=2))
bp = np.round(first_val / b)
last_val = np.round(b * end)
return b, bp, last_val
def pw_fit(end_range, no_seg, **fparams):
y_bps = np.linspace(one_per_der(1, **fparams), one_per_der(end_range,**fparams) , no_seg+1)[1:]
x_bps = np.round(one_per_der_inv(y_bps, **fparams))
y = input_func(x, **fparams)
slopes = [np.round(float(curve_fit(lambda x,b: x * b, np.arange(1,x_bps[0]), y[:int(x_bps[0])-1])[0]), decimals = 2)]
plats = [np.round(x_bps[0] * slopes[0])]
bps = []
for i, xbp in enumerate(x_bps[1:]):
b, bp, last_val = segment_fit(int(x_bps[i]+1), int(xbp), y, plats[i])
slopes.append(b); bps.append(bp); plats.append(last_val)
breaks = sorted(list(x_bps) + bps)[:-1]
# If due to rounding slope values (to two decimals), there is no change in a subsequent step, I just remove those segments
to_del = np.argwhere(np.diff(slopes) == 0).flatten()
breaks_to_del = np.concatenate((to_del * 2, to_del * 2 + 1))
slopes = np.delete(slopes,to_del + 1)
plats = np.delete(plats[:-1],to_del)
breaks = np.delete(breaks,breaks_to_del)
return slopes, plats, breaks
def pw_calc(x, slopes, plateaus, breaks):
x = x.astype('float')
cond_list = [x < breaks[0]]
for idx, j in enumerate(breaks[:-1]):
cond_list.append((j <= x) & (x < breaks[idx+1]))
cond_list.append(breaks[-1] <= x)
func_list = [lambda x: x * slopes[0]]
for idx, j in enumerate(slopes[1:]):
func_list.append(plateaus[idx])
func_list.append(lambda x, j=j: x * j)
return np.piecewise(x, cond_list, func_list)
fparams = {'k':1.8, 'a':1.2}
end_range = 5e4
no_steps = 10
x = np.arange(1, end_range)
y = input_func(x, **fparams)
slopes, plats, breaks = pw_fit(end_range, no_steps, **fparams)
y_output = pw_calc(x, slopes, plats, breaks)
plt.plot(x,y_output,y);
THIS PART IS JUST BACKGROUND IF YOU NEED IT
I am developing a numerical solver for the Second-Order Kuramoto Model. The functions I use to find the derivatives of theta and omega are given below.
# n-dimensional change in omega
def d_theta(omega):
return omega
# n-dimensional change in omega
def d_omega(K,A,P,alpha,mask,n):
def layer1(theta,omega):
T = theta[:,None] - theta
A[mask] = K[mask] * np.sin(T[mask])
return - alpha*omega + P - A.sum(1)
return layer1
These equations return vectors.
QUESTION 1
I know how to use odeint for two dimensions, (y,t). for my research I want to use a built-in Python function that works for higher dimensions.
QUESTION 2
I do not necessarily want to stop after a predetermined amount of time. I have other stopping conditions in the code below that will indicate whether the system of equations converges to the steady state. How do I incorporate these into a built-in Python solver?
WHAT I CURRENTLY HAVE
This is the code I am currently using to solve the system. I just implemented RK4 with constant time stepping in a loop.
# This function randomly samples initial values in the domain and returns whether the solution converged
# Inputs:
# f change in theta (d_theta)
# g change in omega (d_omega)
# tol when step size is lower than tolerance, the solution is said to converge
# h size of the time step
# max_iter maximum number of steps Runge-Kutta will perform before giving up
# max_laps maximum number of laps the solution can do before giving up
# fixed_t vector of fixed points of theta
# fixed_o vector of fixed points of omega
# n number of dimensions
# theta initial theta vector
# omega initial omega vector
# Outputs:
# converges true if it nodes restabilizes, false otherwise
def kuramoto_rk4_wss(f,g,tol_ss,tol_step,h,max_iter,max_laps,fixed_o,fixed_t,n):
def layer1(theta,omega):
lap = np.zeros(n, dtype = int)
converges = False
i = 0
tau = 2 * np.pi
while(i < max_iter): # perform RK4 with constant time step
p_omega = omega
p_theta = theta
T1 = h*f(omega)
O1 = h*g(theta,omega)
T2 = h*f(omega + O1/2)
O2 = h*g(theta + T1/2,omega + O1/2)
T3 = h*f(omega + O2/2)
O3 = h*g(theta + T2/2,omega + O2/2)
T4 = h*f(omega + O3)
O4 = h*g(theta + T3,omega + O3)
theta = theta + (T1 + 2*T2 + 2*T3 + T4)/6 # take theta time step
mask2 = np.array(np.where(np.logical_or(theta > tau, theta < 0))) # find which nodes left [0, 2pi]
lap[mask2] = lap[mask2] + 1 # increment the mask
theta[mask2] = np.mod(theta[mask2], tau) # take the modulus
omega = omega + (O1 + 2*O2 + 2*O3 + O4)/6
if(max_laps in lap): # if any generator rotates this many times it probably won't converge
break
elif(np.any(omega > 12)): # if any of the generators is rotating this fast, it probably won't converge
break
elif(np.linalg.norm(omega) < tol_ss and # assert the nodes are sufficiently close to the equilibrium
np.linalg.norm(omega - p_omega) < tol_step and # assert change in omega is small
np.linalg.norm(theta - p_theta) < tol_step): # assert change in theta is small
converges = True
break
i = i + 1
return converges
return layer1
Thanks for your help!
You can wrap your existing functions into a function accepted by odeint (option tfirst=True) and solve_ivp as
def odesys(t,u):
theta,omega = u[:n],u[n:]; # or = u.reshape(2,-1);
return [ *f(omega), *g(theta,omega) ]; # or np.concatenate([f(omega), g(theta,omega)])
u0 = [*theta0, *omega0]
t = linspan(t0, tf, timesteps+1);
u = odeint(odesys, u0, t, tfirst=True);
#or
res = solve_ivp(odesys, [t0,tf], u0, t_eval=t)
The scipy methods pass numpy arrays and convert the return value into same, so that you do not have to care in the ODE function. The variant in comments is using explicit numpy functions.
While solve_ivp does have event handling, using it for a systematic collection of events is rather cumbersome. It would be easier to advance some fixed step, do the normalization and termination detection, and then repeat this.
If you want to later increase efficiency somewhat, use directly the stepper classes behind solve_ivp.
Currently my convergence criteria for SGD checks whether the MSE error ratio is within a specific boundary.
def compute_mse(data, labels, weights):
m = len(labels)
hypothesis = np.dot(data,weights)
sq_errors = (hypothesis - labels) ** 2
mse = np.sum(sq_errors)/(2.0*m)
return mse
cur_mse = 1.0
prev_mse = 100.0
m = len(labels)
while cur_mse/prev_mse < 0.99999:
prev_mse = cur_mse
for i in range(m):
d = np.array(data[i])
hypothesis = np.dot(d, weights)
gradient = np.dot((labels[i] - hypothesis), d)/m
weights = weights + (alpha * gradient)
cur_mse = compute_mse(data, labels, weights)
if cur_mse > prev_mse:
return
The weights are update w.r.t. to a single data point in the training set.
With an alpha of 0.001, the model is supposed to have converged within a few iterations however I get no convergence. Is this convergence criteria too strict?
I'll try to answer the question. First, the pseudocode of stochastic gradient descent looks something like this:
input: f(x), alpha, initial x (guess or random)
output: min_x f(x) # x that minimizes f(x)
while True:
shuffle data # good practice, not completely needed
for d in data:
x -= alpha * grad(f(x)) # df/dx
if <stopping criterion>:
break
There can be other regularization parameters added to the function that you want to minimize, such as the l1 penalty to avoid overfitting.
Going back to your problem, looking at your data and definition of the gradient, looks like you want to solve a simple linear system of equations of the form:
Ax = b
which yields the objevtive function:
f(x) = ||Ax - b||^2
stochastic gradient descent uses one row data at a time:
||A_i x - b||
where || o || is the euclidean norm and _i means index of a row.
Here, A is your data, x is your weights and b is your labels.
The gradient of the function is then computed as a:
grad(f(x)) = 2 * A.T (Ax - b)
Or in the case of the stochastic gradient descent:
2 * A_i.T (A_i x - b)
where .T means transpose.
Putting everything back into your code... first I will setup a synthetic data:
A = np.random.randn(100, 2) # 100x2 data
x = np.random.randn(2, 1) # 2x1 weights
b = np.random.randint(0, 2, 100).reshape(100, 1) # 100x1 labels
b[b == 0] = -1 # labels in {-1, 1}
Then, define the parameters:
alpha = 0.001
cur_mse = 100.
prev_mse = np.inf
it = 0
max_iter = 100
m = A.shape[0]
idx = range(m)
And loop!
while cur_mse/prev_mse < 0.99999 and it < max_iter:
prev_mse = cur_mse
shuffle(idx)
for i in idx:
d = A[i:i+1]
y = b[i:i+1]
h = np.dot(d, x)
dx = 2 * np.dot(d.T, (h - y))
x -= (alpha * dx)
cur_mse = np.mean((A.dot(x) - b)**2)
if cur_mse > prev_mse:
raise Exception("Not converging")
it += 1
This code is pretty much the same as yours, with a couple of additions:
Another stopping criterion based on the number of iterations (to avoid looping forever if the system doesn't converge or does too slowly)
Redefinition of the gradient dx (still similar to yours). You have the sign inverted and therefore the weight update is positive + since in my example is negative - (makes sense since you are going down in a gradient).
Indexing of data and labels. While data[i] gives a tuple of size (2,) (in this case for a 100x2 data), using fancy indexing data[i:i+1] will return a view of the data without reshaping it (e.g with shape (1, 2)) and therefore will allow you to perform the proper matrix multiplications.
You can add a 3rd stopping criterion based on acceptable mse error, i.e: if cur_mse < 1e-3: break.
This algorithm, with random data, converges in 20-40 iterations for me (depending on the generated random data).
So... assuming that this is the function you want to minimize, if this method doesn't work for you, it might mean that your system is underdeterminated (you have less training data than features, which means A is more wide than high).
Hope it helps!
So I am writing a program that handles gradient descent. Im using this method to solve equations of the form
Ax = b
where A is a random 10x10 matrix and b is a random 10x1 matrix
Here is my code:
import numpy as np
import math
import random
def steepestDistance(A,b,xO, e):
xPrev = xO
dPrev = -((A * xPrev) - b)
magdPrev = np.linalg.norm(dPrev)
danger = np.asscalar(((magdPrev * magdPrev)/(np.dot(dPrev.T,A * dPrev))))
xNext = xPrev + (danger * dPrev)
step = 1
while (np.linalg.norm((A * xNext) - b) >= e and np.linalg.norm((A * xNext) - b) < math.pow(10,4)):
xPrev = xNext
dPrev = -((A * xPrev) - b)
magdPrev = np.linalg.norm(dPrev)
danger = np.asscalar((math.pow(magdPrev,2))/(np.dot(dPrev.T,A * dPrev)))
xNext = xPrev + (danger * dPrev)
step = step + 1
return xNext
##print(steepestDistance(np.matrix([[5,2],[2,1]]),np.matrix([[1],[1]]),np.matrix([[0.5],[0]]), math.pow(10,-5)))
def chooseRandMatrix():
matrix = np.zeros(shape = (10,10))
for i in range(10):
for a in range(10):
matrix[i][a] = random.randint(0,100)
return matrix.T * matrix
def chooseRandColArray():
arra = np.zeros(shape = (10,1))
for i in range(10):
arra[i][0] = random.randint(0,100)
return arra
for i in range(4):
matrix = np.asmatrix(chooseRandMatrix())
array = np.asmatrix(chooseRandColArray())
print(steepestDistance(matrix, array, np.asmatrix(chooseRandColArray()),math.pow(10,-5)))
When I run the method steepestDistance on the random matrix and column, I keep getting an infinite loop. It works fine when simple 2x2 matrices are used for A, but it loops indefinitely for 10x10 matrices. The problem is in np.linalg.norm((A * xNext) - b); it keeps growing indefinitely. Thats why I put an upper bound on it; Im not supposed to do it for the algorithm however. Can someone tell me what the problem is?
Solving a linear system Ax=b with gradient descent means to minimize the quadratic function
f(x) = 0.5*x^t*A*x - b^t*x.
This only works if the matrix A is symmetric, A=A^t, since the derivative or gradient of f is
f'(x)^t = 0.5*(A+A^t)*x - b,
and additionally A must be positive definite. If there are negative eigenvalues,then the descent will proceed to minus infinity, there is no minimum to be found.
One work-around is to replace b by A^tb and A by a^t*A, that is to minimize the function
f(x) = 0.5*||A*x-b||^2
= 0.5*x^t*A^t*A*x - b^t*A*x + 0.5*b^t*b
with gradient
f'(x)^t = A^t*A*x - A^t*b
But for large matrices A this is not recommended since the condition number of A^t*A is about the square of the condition number of A.