Related
This is the following question after appyling comments from: Using GEKKO for Moving Horizon Estimation online
I have studied example from estimation iterative example on the Dynamic Optimization course website and revised my code as follows:
from gekko import GEKKO
import numpy as np
import matplotlib.pyplot as plt
import matplotlib; matplotlib.use('TkAgg')
class Observer():
def __init__(self, window_size, r_init, alpha_init):
self.m = GEKKO(remote=False)
self.dt = 0.05
self.m.time = [i*self.dt for i in range(window_size)]
#Parameters
self.m.u = self.m.MV()
#Variables
self.m.r = self.m.CV(lb=0) # value=r_init) #ub=20 can be over 20
self.m.alpha = self.m.CV() # value=alpha_init) #ub lb for angle?
#Equations
self.m.Equation(self.m.r.dt()== -self.m.cos(self.m.alpha))
self.m.Equation(self.m.alpha.dt()== self.m.sin(self.m.alpha)/self.m.r - self.m.u) # differential equation
#Options
self.m.options.MV_STEP_HOR = 2
self.m.options.IMODE = 5 # dynamic estimation
self.m.options.EV_TYPE = 2 #Default 1: absolute error form 2: squared error form
self.m.options.DIAGLEVEL = 0 #diagnostic level
self.m.options.NODES = 5 #nodes # collocation nodes default:2
self.m.options.SOLVER = 3 #solver_num
# STATUS = 0, optimizer doesn't adjust value
# STATUS = 1, optimizer can adjust
self.m.u.STATUS = 0
self.m.r.STATUS = 1
self.m.alpha.STATUS = 1
# FSTATUS = 0, no measurement
# FSTATUS = 1, measurement used to update model
self.m.u.FSTATUS = 1 #default
self.m.r.FSTATUS = 1
self.m.alpha.FSTATUS = 1
self.m.r.TR_INIT = 0
self.m.alpha.TR_INIT = 0
self.count = 0
def MHE(self, observed_state, u_data):
self.count =+ 1
self.m.u.MEAS = u_data
self.m.r.MEAS = observed_state[0]
self.m.alpha.MEAS = observed_state[1]
self.m.solve(disp=False)
return self.m.r.MODEL, self.m.alpha.MODEL
if __name__=="__main__":
FILE_PATH00 = '/home/shane16/Project/model_guard/uav_paper/adversarial/SA_PPO/src/DATA/4end_estimation_results_r.csv'
FILE_PATH01 = '/home/shane16/Project/model_guard/uav_paper/adversarial/SA_PPO/src/DATA/4end_estimation_results_alpha.csv'
FILE_PATH02 = '/home/shane16/Project/model_guard/uav_paper/adversarial/SA_PPO/src/DATA/4end_action_buffer_eps0.0_sig0.0.csv'
cycles = 55
x = np.arange(cycles) # 1...300
matrix00 = np.loadtxt(FILE_PATH00, delimiter=',')
matrix01 = np.loadtxt(FILE_PATH01, delimiter=',')
matrix02 = np.loadtxt(FILE_PATH02, delimiter=',')
vanilla_action_sigma_0 = matrix02
vanilla_estimation_matrix_r = np.zeros(cycles)
vanilla_estimation_matrix_alpha = np.zeros(cycles)
# sigma = 0.0
# vanilla model true/observed states
r_vanilla_sigma_0_true = matrix00[0, 3:] # from step 1
r_vanilla_sigma_0_observed = matrix00[1, 3:] # from step1
alpha_vanilla_sigma_0_true = matrix01[0, 3:]
alpha_vanilla_sigma_0_observed = matrix01[1, 3:]
# initialize estimator
sigma = 0.0 #1.0
solver_num = 3
nodes = 5
# for window_size in [5, 10, 20, 30, 40, 50]:
window_size = 5
observer = Observer(window_size, r_vanilla_sigma_0_observed[0], alpha_vanilla_sigma_0_observed[0])
for i in range(cycles):
if i % 100 == 0:
print('cylcle: {}'.format(i))
vanilla_observed_states = np.hstack((r_vanilla_sigma_0_observed[i], alpha_vanilla_sigma_0_observed[i])) # from current observed state
r_hat, alpha_hat = observer.MHE(vanilla_observed_states, vanilla_action_sigma_0[i]) # and current action -> estimate current state
vanilla_estimation_matrix_r[i] = r_hat
vanilla_estimation_matrix_alpha[i] = alpha_hat
#plot vanilla
plt.figure()
plt.subplot(3,1,1)
plt.title('Vanilla model_sig{}'.format(sigma))
plt.plot(x, vanilla_action_sigma_0[:cycles],'b:',label='action (w)')
plt.legend()
plt.subplot(3,1,2)
plt.ylabel('r')
plt.plot(x, r_vanilla_sigma_0_true[:cycles], 'k-', label='true_r')
plt.plot(x, r_vanilla_sigma_0_observed[:cycles], 'gx', label='observed_r')
plt.plot(x, vanilla_estimation_matrix_r, 'r--', label='time window: 10')
# plt.legend()
plt.subplot(3,1,3)
plt.xlabel('time steps')
plt.ylabel('alpha')
plt.plot(x, alpha_vanilla_sigma_0_true[:cycles], 'k-', label='true_alpha')
plt.plot(x, alpha_vanilla_sigma_0_observed[:cycles], 'gx', label='observed_alpha')
plt.plot(x, vanilla_estimation_matrix_alpha, 'r--', label='time window: {}'.format(window_size))
plt.legend()
plt.savefig('plot/revision/4estimated_STATES_vanilla_sig{}_window{}_cycles{}_solver{}_nodes{}.png'.format(sigma, window_size,cycles, solver_num, nodes))
plt.show()
csv files: https://drive.google.com/drive/folders/1jW_6zBCdbJHB7yU3HmCIhamEyOT1LJqD?usp=sharing
The code works when initialized with values specified at line 15,16 (m.r, m.alpha).
However, if I try with no initial value,(as same condition in example), solution is not found.
terminal output:
cylcle: 0 Traceback (most recent call last): File
"4observer_mhe.py", line 86, in
r_hat, alpha_hat = observer.MHE(vanilla_observed_states, vanilla_action_sigma_0[i]) # and current action -> estimate current
state File "4observer_mhe.py", line 49, in MHE
self.m.solve(disp=False) File "/home/shane16/Project/model_guard/LipSDP/lipenv/lib/python3.7/site-packages/gekko/gekko.py",
line 2140, in solve
raise Exception(apm_error) Exception: #error: Solution Not Found
What could be the solution to this problem?
I have tried below strategies, but couldn't find the solution.
Reduce the number of decision variables by using m.FV() or m.MV() with m.options.MV_STEP_HOR=2+ to reduce the degrees of freedom for the solver for the unknown parameters.
Try other solvers with m.options.SOLVER=1 or m.options.SOLVER=2.
I expect to see estimation results that follow the true state well.
But I guess I'm doing something wrong.
Could anyone help me please?
Thank you.
Solvers sometimes need good initial guess values or constraints (lower and upper bounds) on the degrees of freedom (MV or FV) to find the optimal solution.
One of the equations may be the source of the problem:
self.m.alpha.dt() == self.m.sin(self.m.alpha)/self.m.r - self.m.u
The initial value of r is zero (default) because no initial value is provided when it is declared as self.m.r = self.m.CV(lb=0). A comment suggests that it was formerly initialized with value r_init. The zero value creates a divide-by-zero for that equation. Try rearranging the equation into an equivalent form that avoids the potential for divide-by-zero either with the initial guess or when the solver is iterating.
self.m.r*self.m.alpha.dt() == self.m.sin(self.m.alpha) - self.m.r*self.m.u
There may be other things that are also causing the model to not converge. When the solution does not converge then the infeasibilities.txt file can be a source to troubleshoot the specific equations that are having trouble. Here are instructions to retrieve the infeasibilities.txt file: How to retrieve the 'infeasibilities.txt' from the gekko
As a civil engineer, I am working on a program to find the equilibrium of a concrete reinforced section submitted to a Flexural Moment.
Reinforced Concrete Cross Section Equilibrium:
Basically, I have 2 unknowns, which are eps_sup and eps_inf
I have a constant that is M
I have some variables that depend only on the values of (eps_sup,eps_inf). The functions are non-linear, no need to go into this.
When I have the right couple of values, the following equations are verified :
Fc + Fs = 0 (Forces Equilibrium)
M/z = Fc = -Fs (Moment Equilibrium)
My algorithm, as it is today, consists in finding the minimal value of : abs(Fc+Fs)/Fc + abs(M_calc-M)/M
To do this I iterate on Both e eps_sup and eps_inf between given limits, with a given step, and the step needs to be small enough to find a solution.
It is working, but it is very (very) slow since it goes through a very wide range of values without trying to reduce the number of iterations.
Surely I can find an optimized solution, and that is were I need your help.
'Constants :
M
'Variables :
delta = 10000000000000
eps_sup = 0
eps_inf = 0
M_calc = 0
Fc = 0
Fs = 0
z = 0
eps_sup_candidate = 0
eps_inf_candidate = 0
For eps_sup = 0 to 0,005 step = 0,000001
For eps_inf = -0,05 to 0 step = 0,000001
Fc = f(eps_sup,eps_inf)
Fs = g(eps_sup,eps_inf)
z = h(eps_sup,eps_inf)
M_calc = Fc * z
If (abs(Fc+Fs)/Fc + abs(M_calc-M)/M) < delta Then
delta = abs(Fc+Fs)/Fc + abs(M_calc-M)/M
eps_sup_candidate = eps_sup
eps_inf_candidate = eps_inf
End If
Next
Next
I built 5-layer neural network by using tensorflow.
I have a problem to get reproducible results (or stable results).
I found similar questions regarding reproducibility of tensorflow and the corresponding answers, such as How to get stable results with TensorFlow, setting random seed
But the problem is not solved yet.
I also set random seed like the following
tf.set_random_seed(1)
Furthermore, I added seed options to every random function such as
b1 = tf.Variable(tf.random_normal([nHidden1], seed=1234))
I confirmed that the first epoch shows the identical results, but not identical from the second epoch little by little.
How can I get the reproducible results?
Am I missing something?
Here is a code block I use.
def xavier_init(n_inputs, n_outputs, uniform=True):
if uniform:
init_range = tf.sqrt(6.0 / (n_inputs + n_outputs))
return tf.random_uniform_initializer(-init_range, init_range, seed=1234)
else:
stddev = tf.sqrt(3.0 / (n_inputs + n_outputs))
return tf.truncated_normal_initializer(stddev=stddev, seed=1234)
import numpy as np
import tensorflow as tf
import dataSetup
from scipy.stats.stats import pearsonr
tf.set_random_seed(1)
x_train, y_train, x_test, y_test = dataSetup.input_data()
# Parameters
learningRate = 0.01
trainingEpochs = 1000000
batchSize = 64
displayStep = 100
thresholdReduce = 1e-6
thresholdNow = 0.6
#dropoutRate = tf.constant(0.7)
# Network Parameter
nHidden1 = 128 # number of 1st layer nodes
nHidden2 = 64 # number of 2nd layer nodes
nInput = 24 #
nOutput = 1 # Predicted score: 1 output for regression
# save parameter
modelPath = 'model/model_layer5_%d_%d_mini%d_lr%.3f_noDrop_rollBack.ckpt' %(nHidden1, nHidden2, batchSize, learningRate)
# tf Graph input
X = tf.placeholder("float", [None, nInput])
Y = tf.placeholder("float", [None, nOutput])
# Weight
W1 = tf.get_variable("W1", shape=[nInput, nHidden1], initializer=xavier_init(nInput, nHidden1))
W2 = tf.get_variable("W2", shape=[nHidden1, nHidden2], initializer=xavier_init(nHidden1, nHidden2))
W3 = tf.get_variable("W3", shape=[nHidden2, nHidden2], initializer=xavier_init(nHidden2, nHidden2))
W4 = tf.get_variable("W4", shape=[nHidden2, nHidden2], initializer=xavier_init(nHidden2, nHidden2))
WFinal = tf.get_variable("WFinal", shape=[nHidden2, nOutput], initializer=xavier_init(nHidden2, nOutput))
# biases
b1 = tf.Variable(tf.random_normal([nHidden1], seed=1234))
b2 = tf.Variable(tf.random_normal([nHidden2], seed=1234))
b3 = tf.Variable(tf.random_normal([nHidden2], seed=1234))
b4 = tf.Variable(tf.random_normal([nHidden2], seed=1234))
bFinal = tf.Variable(tf.random_normal([nOutput], seed=1234))
# Layers for dropout
L1 = tf.nn.relu(tf.add(tf.matmul(X, W1), b1))
L2 = tf.nn.relu(tf.add(tf.matmul(L1, W2), b2))
L3 = tf.nn.relu(tf.add(tf.matmul(L2, W3), b3))
L4 = tf.nn.relu(tf.add(tf.matmul(L3, W4), b4))
hypothesis = tf.add(tf.matmul(L4, WFinal), bFinal)
print "Layer setting DONE..."
# define loss and optimizer
cost = tf.reduce_mean(tf.square(hypothesis - Y))
optimizer = tf.train.AdamOptimizer(learning_rate=learningRate).minimize(cost)
# Initialize the variable
init = tf.initialize_all_variables()
# save op to save and restore all the variables
saver = tf.train.Saver()
with tf.Session() as sess:
# initialize
sess.run(init)
print "Initialize DONE..."
# Training
costPrevious = 100000000000000.0
best = float("INF")
totalBatch = int(len(x_train)/batchSize)
print "Total Batch: %d" %totalBatch
for epoch in range(trainingEpochs):
#print "EPOCH: %04d" %epoch
avgCost = 0.
for i in range(totalBatch):
np.random.seed(i+epoch)
randidx = np.random.randint(len(x_train), size=batchSize)
batch_xs = x_train[randidx,:]
batch_ys = y_train[randidx,:]
# Fit traiing using batch data
sess.run(optimizer, feed_dict={X:batch_xs, Y:batch_ys})
# compute average loss
avgCost += sess.run(cost, feed_dict={X:batch_xs, Y:batch_ys})/totalBatch
# compare the current cost and the previous
# if current cost > the previous
# just continue and make the learning rate half
#print "Cost: %1.8f --> %1.8f at epoch %05d" %(costPrevious, avgCost, epoch+1)
if avgCost > costPrevious + .5:
#sess.run(init)
load_path = saver.restore(sess, modelPath)
print "Cost increases at the epoch %05d" %(epoch+1)
print "Cost: %1.8f --> %1.8f" %(costPrevious, avgCost)
continue
costNow = avgCost
reduceCost = abs(costPrevious - costNow)
costPrevious = costNow
#Display logs per epoch step
if costNow < best:
best = costNow
bestMatch = sess.run(hypothesis, feed_dict={X:x_test})
# model save
save_path = saver.save(sess, modelPath)
if epoch % displayStep == 0:
print "step {}".format(epoch)
pearson = np.corrcoef(bestMatch.flatten(), y_test.flatten())
print 'train loss = {}, current loss = {}, test corrcoef={}'.format(best, costNow, pearson[0][1])
if reduceCost < thresholdReduce or costNow < thresholdNow:
print "Epoch: %04d, Cost: %.9f, Prev: %.9f, Reduce: %.9f" %(epoch+1, costNow, costPrevious, reduceCost)
break
print "Optimization Finished"
It seems that your results are perhaps not reproducible because you are using Saver to write/restore from checkpoint each time? (i.e. the second time that you run the code, the variable values aren't initialized using your random seed -- they are restored from your previous checkpoint)
Please trim down your code example to just the code necessary to reproduce irreproducibility.
When I run the code shown below, the tic/toc pair inside the function shows it takes very short time (<< 1sec) to go through all the lines. However, it actually takes around 2.3secs to get the outputs!!! I use the tic/toc pair to measure the time.
tic
rnn.v = 11;
rnn.h = 101;
rnn.o = 7;
rnn.h_init = randn(1,rnn.h,'gpuArray');
rnn.W_vh = randn(rnn.v,rnn.h,'gpuArray');
rnn.W_hh = randn(rnn.h,rnn.h,'gpuArray');
rnn.W_ho = randn(rnn.h,rnn.o,'gpuArray');
inData.V = randn(10000,11,100,'gpuArray');
inData.TimeSteps =100;
inData.BatchSize = 10000;
[H,OX] = forward_pass(rnn, inData)
toc
All the matrices in rnn, and inData are gpuArray, so all the calculation are carried out in GPU. The outputs are also gpuArray.
function [H,OX] = forward_pass(rnn, inData)
tic;
%initial hidden state values
H_init = gpuArray(repmat(rnn.h_init,[inData.BatchSize,1]));
%initialize state H
H = zeros(inData.BatchSize, rnn.h, inData.TimeSteps,'gpuArray');
%initialize OX (which is H * Who)
OX = zeros(inData.BatchSize, rnn.o, inData.TimeSteps,'gpuArray');
for t = 1 : inData.TimeSteps
if t == 1
HX_t = H_init * rnn.W_hh...
+ inData.V(:,:,t) * rnn.W_vh;
else
HX_t = H(:,:,(t-1)) * rnn.W_hh...
+ inData.V(:,:,t) * rnn.W_vh;
end
H(:,:,t) = tanh(HX_t);
OX(:,:,t) = H(:,:,t) * rnn.W_ho;
end
toc;
end
Normally, if you use gather() function, it will be slow. I didn't use the gather() function to transfer the outputs to workspace, I don't know why it is still so slow. It looks like the last line "end" takes more than 2secs.
Anyone knows how to accelerate the function call?
First off, for proper benchmarking you do need to use gather either inside the function call or afterwards. In the former case, you would have a non-gpu output from the function call and in the latter case, a gpu-based datatype would be the output. Now, back to your problem, you are using very few TimeSteps and as such any optimization that you might try out won't reflect in a huge manner. Here's an optimized version that will show increased performance as you increase Timesteps -
function [H,OX] = forward_pass(rnn, inData)
H = zeros(inData.BatchSize, rnn.h, inData.TimeSteps,'gpuArray');
T = reshape(permute(inData.V,[1 3 2]),[],size(inData.V,2))*rnn.W_vh;
H(:,:,1) = tanh(bsxfun(#plus,rnn.h_init * rnn.W_hh,T(1:size(inData.V,1),:)));
for t = 2 : inData.TimeSteps
H(:,:,t) = tanh( H(:,:,(t-1))*rnn.W_hh + ...
T((t-1)*size(inData.V,1)+1: t*size(inData.V,1),:));
end
A = reshape(permute(H,[1 3 2]),[],size(H,2))*rnn.W_ho;
OX = permute(reshape(A,size(H,1),size(A,1)/size(H,1),[]),[1 3 2]);
return;
Benchmarking
Test Case #1
Parameters
rnn.v = 11;
rnn.h = 5;
rnn.o = 7;
inData.TimeSteps = 10000;
inData.BatchSize = 10;
Results
---- Original Code :
Elapsed time is 5.678876 seconds.
---- Modified Code :
Elapsed time is 3.821059 seconds.
Test Case #2
Parameters
inData.TimeSteps = 50000; (rest are same as in Test Case #1)
Results
---- Original Code :
Elapsed time is 28.392290 seconds.
---- Modified Code :
Elapsed time is 19.031776 seconds.
Please note that these are tested on GTX 750 Ti.
Recently I found this in some code I wrote a few years ago. It was used to rationalize a real value (within a tolerance) by determining a suitable denominator and then checking if the difference between the original real and the rational was small enough.
Edit to clarify : I actually don't want to convert all real values. For instance I could choose a max denominator of 14, and a real value that equals 7/15 would stay as-is. It's not as clear that as it's an outside variable in the algorithms I wrote here.
The algorithm to get the denominator was this (pseudocode):
denominator(x)
frac = fractional part of x
recip = 1/frac
if (frac < tol)
return 1
else
return recip * denominator(recip)
end
end
Seems to be based on continued fractions although it became clear on looking at it again that it was wrong. (It worked for me because it would eventually just spit out infinity, which I handled outside, but it would be often really slow.) The value for tol doesn't really do anything except in the case of termination or for numbers that end up close. I don't think it's relatable to the tolerance for the real - rational conversion.
I've replaced it with an iterative version that is not only faster but I'm pretty sure it won't fail theoretically (d = 1 to start with and fractional part returns a positive, so recip is always >= 1) :
denom_iter(x d)
return d if d > maxd
frac = fractional part of x
recip = 1/frac
if (frac = 0)
return d
else
return denom_iter(recip d*recip)
end
end
What I'm curious to know if there's a way to pick the maxd that will ensure that it converts all values that are possible for a given tolerance. I'm assuming 1/tol but don't want to miss something. I'm also wondering if there's an way in this approach to actually limit the denominator size - this allows some denominators larger than maxd.
This can be considered a 2D minimization problem on error:
ArgMin ( r - q / p ), where r is real, q and p are integers
I suggest the use of Gradient Descent algorithm . The gradient in this objective function is:
f'(q, p) = (-1/p, q/p^2)
The initial guess r_o can be q being the closest integer to r, and p being 1.
The stopping condition can be thresholding of the error.
The pseudo-code of GD can be found in wiki: http://en.wikipedia.org/wiki/Gradient_descent
If the initial guess is close enough, the objective function should be convex.
As Jacob suggested, this problem can be better solved by minimizing the following error function:
ArgMin ( p * r - q ), where r is real, q and p are integers
This is linear programming, which can be efficiently solved by any ILP (Integer Linear Programming) solvers. GD works on non-linear cases, but lack efficiency in linear problems.
Initial guesses and stopping condition can be similar to stated above. Better choice can be obtained for individual choice of solver.
I suggest you should still assume convexity near the local minimum, which can greatly reduce cost. You can also try Simplex method, which is great on linear programming problem.
I give credit to Jacob on this.
A problem similar to this is solved in the Approximations section beginning ca. page 28 of Bill Gosper's Continued Fraction Arithmetic document. (Ref: postscript file; also see text version, from line 1984.) The general idea is to compute continued-fraction approximations of the low-end and high-end range limiting numbers, until the two fractions differ, and then choose a value in the range of those two approximations. This is guaranteed to give a simplest fraction, using Gosper's terminology.
The python code below (program "simpleden") implements a similar process. (It probably is not as good as Gosper's suggested implementation, but is good enough that you can see what kind of results the method produces.) The amount of work done is similar to that for Euclid's algorithm, ie O(n) for numbers with n bits, so the program is reasonably fast. Some example test cases (ie the program's output) are shown after the code itself. Note, function simpleratio(vlo, vhi) as shown here returns -1 if vhi is smaller than vlo.
#!/usr/bin/env python
def simpleratio(vlo, vhi):
rlo, rhi, eps = vlo, vhi, 0.0000001
if vhi < vlo: return -1
num = denp = 1
nump = den = 0
while 1:
klo, khi = int(rlo), int(rhi)
if klo != khi or rlo-klo < eps or rhi-khi < eps:
tlo = denp + klo * den
thi = denp + khi * den
if tlo < thi:
return tlo + (rlo-klo > eps)*den
elif thi < tlo:
return thi + (rhi-khi > eps)*den
else:
return tlo
nump, num = num, nump + klo * num
denp, den = den, denp + klo * den
rlo, rhi = 1/(rlo-klo), 1/(rhi-khi)
def test(vlo, vhi):
den = simpleratio(vlo, vhi);
fden = float(den)
ilo, ihi = int(vlo*den), int(vhi*den)
rlo, rhi = ilo/fden, ihi/fden;
izok = 'ok' if rlo <= vlo <= rhi <= vhi else 'wrong'
print '{:4d}/{:4d} = {:0.8f} vlo:{:0.8f} {:4d}/{:4d} = {:0.8f} vhi:{:0.8f} {}'.format(ilo,den,rlo,vlo, ihi,den,rhi,vhi, izok)
test (0.685, 0.695)
test (0.685, 0.7)
test (0.685, 0.71)
test (0.685, 0.75)
test (0.685, 0.76)
test (0.75, 0.76)
test (2.173, 2.177)
test (2.373, 2.377)
test (3.484, 3.487)
test (4.0, 4.87)
test (4.0, 8.0)
test (5.5, 5.6)
test (5.5, 6.5)
test (7.5, 7.3)
test (7.5, 7.5)
test (8.534537, 8.534538)
test (9.343221, 9.343222)
Output from program:
> ./simpleden
8/ 13 = 0.61538462 vlo:0.68500000 9/ 13 = 0.69230769 vhi:0.69500000 ok
6/ 10 = 0.60000000 vlo:0.68500000 7/ 10 = 0.70000000 vhi:0.70000000 ok
6/ 10 = 0.60000000 vlo:0.68500000 7/ 10 = 0.70000000 vhi:0.71000000 ok
2/ 4 = 0.50000000 vlo:0.68500000 3/ 4 = 0.75000000 vhi:0.75000000 ok
2/ 4 = 0.50000000 vlo:0.68500000 3/ 4 = 0.75000000 vhi:0.76000000 ok
3/ 4 = 0.75000000 vlo:0.75000000 3/ 4 = 0.75000000 vhi:0.76000000 ok
36/ 17 = 2.11764706 vlo:2.17300000 37/ 17 = 2.17647059 vhi:2.17700000 ok
18/ 8 = 2.25000000 vlo:2.37300000 19/ 8 = 2.37500000 vhi:2.37700000 ok
114/ 33 = 3.45454545 vlo:3.48400000 115/ 33 = 3.48484848 vhi:3.48700000 ok
4/ 1 = 4.00000000 vlo:4.00000000 4/ 1 = 4.00000000 vhi:4.87000000 ok
4/ 1 = 4.00000000 vlo:4.00000000 8/ 1 = 8.00000000 vhi:8.00000000 ok
11/ 2 = 5.50000000 vlo:5.50000000 11/ 2 = 5.50000000 vhi:5.60000000 ok
5/ 1 = 5.00000000 vlo:5.50000000 6/ 1 = 6.00000000 vhi:6.50000000 ok
-7/ -1 = 7.00000000 vlo:7.50000000 -7/ -1 = 7.00000000 vhi:7.30000000 wrong
15/ 2 = 7.50000000 vlo:7.50000000 15/ 2 = 7.50000000 vhi:7.50000000 ok
8030/ 941 = 8.53347503 vlo:8.53453700 8031/ 941 = 8.53453773 vhi:8.53453800 ok
24880/2663 = 9.34284641 vlo:9.34322100 24881/2663 = 9.34322193 vhi:9.34322200 ok
If, rather than the simplest fraction in a range, you seek the best approximation given some upper limit on denominator size, consider code like the following, which replaces all the code from def test(vlo, vhi) forward.
def smallden(target, maxden):
global pas
pas = 0
tol = 1/float(maxden)**2
while 1:
den = simpleratio(target-tol, target+tol);
if den <= maxden: return den
tol *= 2
pas += 1
# Test driver for smallden(target, maxden) routine
import random
totalpass, trials, passes = 0, 20, [0 for i in range(20)]
print 'Maxden Num Den Num/Den Target Error Passes'
for i in range(trials):
target = random.random()
maxden = 10 + round(10000*random.random())
den = smallden(target, maxden)
num = int(round(target*den))
got = float(num)/den
print '{:4d} {:4d}/{:4d} = {:10.8f} = {:10.8f} + {:12.9f} {:2}'.format(
int(maxden), num, den, got, target, got - target, pas)
totalpass += pas
passes[pas-1] += 1
print 'Average pass count: {:0.3}\nPass histo: {}'.format(
float(totalpass)/trials, passes)
In production code, drop out all the references to pas (etc.), ie, drop out pass-counting code.
The routine smallden is given a target value and a maximum value for allowed denominators. Given maxden possible choices of denominators, it's reasonable to suppose that a tolerance on the order of 1/maxden² can be achieved. The pass-counts shown in the following typical output (where target and maxden were set via random numbers) illustrate that such a tolerance was reached immediately more than half the time, but in other cases tolerances 2 or 4 or 8 times as large were used, requiring extra calls to simpleratio. Note, the last two lines of output from a 10000-number test run are shown following the complete output of a 20-number test run.
Maxden Num Den Num/Den Target Error Passes
1198 32/ 509 = 0.06286837 = 0.06286798 + 0.000000392 1
2136 115/ 427 = 0.26932084 = 0.26932103 + -0.000000185 1
4257 839/2670 = 0.31423221 = 0.31423223 + -0.000000025 1
2680 449/ 509 = 0.88212181 = 0.88212132 + 0.000000486 3
2935 440/1853 = 0.23745278 = 0.23745287 + -0.000000095 1
6128 347/1285 = 0.27003891 = 0.27003899 + -0.000000077 3
8041 1780/4243 = 0.41951449 = 0.41951447 + 0.000000020 2
7637 3926/7127 = 0.55086292 = 0.55086293 + -0.000000010 1
3422 27/ 469 = 0.05756930 = 0.05756918 + 0.000000113 2
1616 168/1507 = 0.11147976 = 0.11147982 + -0.000000061 1
260 62/ 123 = 0.50406504 = 0.50406378 + 0.000001264 1
3775 52/3327 = 0.01562970 = 0.01562750 + 0.000002195 6
233 6/ 13 = 0.46153846 = 0.46172772 + -0.000189254 5
3650 3151/3514 = 0.89669892 = 0.89669890 + 0.000000020 1
9307 2943/7528 = 0.39094049 = 0.39094048 + 0.000000013 2
962 206/ 225 = 0.91555556 = 0.91555496 + 0.000000594 1
2080 564/1975 = 0.28556962 = 0.28556943 + 0.000000190 1
6505 1971/2347 = 0.83979548 = 0.83979551 + -0.000000022 1
1944 472/ 833 = 0.56662665 = 0.56662696 + -0.000000305 2
3244 291/1447 = 0.20110574 = 0.20110579 + -0.000000051 1
Average pass count: 1.85
Pass histo: [12, 4, 2, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
The last two lines of output from a 10000-number test run:
Average pass count: 1.77
Pass histo: [56659, 25227, 10020, 4146, 2072, 931, 497, 233, 125, 39, 33, 17, 1, 0, 0, 0, 0, 0, 0, 0]