I have a target of target = 11.82 with two variables
x = 9
y = 15
How do I find the percentage weighting that would blend x & y to equal my target? i.e. 55% of x and 45% of y - what function is most efficient way to calc a weighting to obtain my target?
Looking at it again, what I think you want is really two equations:
9x + 15y = 11.82
x + y = 1
Solving that system of equations is pretty fast on pen and paper (just do linear combination). Or you could use sympy to solve the system of linear equations:
>>> from sympy import *
>>> from sympy.solvers.solveset import linsolve
>>> x, y = symbols('x, y')
>>> linsolve([x + y - 1, 9 * x + 15 * y - 11.82], (x, y)) # make 0 on right by subtraction
FiniteSet((0.53, 0.47))
We can confirm this by substitution:
>>> 9 * 0.53 + 15 * 0.47
11.82
Related
I have a nonconvex optimization problem for which I am calculating a lower bound using the McCormick envelope. Each bilinear term is replaced with an auxiliary variable which has the following constraints defined:
w_{ij} >= x_i^L * x_j + x_i * x_j^L - x_i^L * x_j^L
w_{ij} >= x_i^U * x_j + x_i * x_j^U - x_i^U * x_j^U
w_{ij} <= x_i^U * x_j + x_i * x_j^L - x_i^U * x_j^L
w_{ij} <= x_i^L * x_j + x_i * x_j^U - x_i^L * x_j^U
where
x_U <= x <= x_L
I am given a function taking in several arguments:
def convex_bounds(n,m,c,H,Q,A,b,lb,ub):
# n is the number of optimization variables
# m is the number of eq constraints
# H = positive, semidefinite matrix from objetcive function (n x n)
# Q is (mxn) x n
# A is m x n
# b is RHS of non linear eq constraints (m x 1)
# c,lb,ub are vectors size (n x 1)
......................................
# Create matrix B & b_ineq for inequality constraints
# where B*x <= b_ineq
B = np.eye(3)
b_ineq = np.array((10,10,10))
## these values would work in a scenario with no bilinear terms
My problem is that I don't know how to specify the inequality constraints matrix B and vector b_ineq. For this particular exercise my variables are x1, x2 and x3 with bounds 0 (x_L) and 10 (x_U). My bilinear terms are x_12 and x_23 (which will lead to auxiliary variables w_12 and w_23). How can I specify the known bounds (0 and 10) for x1,x2 and x3 and the calculated ones (as in the theory pasted above) in B and b_ineq?
I don't actually know how to proceed with this.
I'm struggling a bit finding a fast algorithm that's suitable.
I just want to minimize:
norm2(x-s)
st
G.x <= h
x >= 0
sum(x) = R
G is sparse and contains only 1s (and zeros obviously).
In the case of iterative algorithms, it would be nice to get the interim solutions to show to the user.
The context is that s is a vector of current results, and the user is saying "well the sum of these few entries (entries indicated by a few 1.0's in a row in G) should be less than this value (a row in h). So we have to remove quantities from the entries the user specified (indicated by 1.0 entries in G) in a least-squares optimal way, but since we have a global constraint on the total (R) the values removed need to be allocated in a least-squares optimal way amongst the other entries. The entries can't go negative.
All the algorithms I'm looking at are much more general, and as a result are much more complex. Also, they seem quite slow. I don't see this as a complex problem, although mixes of equality and inequality constraints always seem to make things more complex.
This has to be called from Python, so I'm looking at Python libraries like qpsolvers and scipy.optimize. But I suppose Java or C++ libraries could be used and called from Python, which might be good since multithreading is better in Java and C++.
Any thoughts on what library/package/approach to use to best solve this problem?
The size of the problem is about 150,000 rows in s, and a few dozen rows in G.
Thanks!
Your problem is a linear least squares:
minimize_x norm2(x-s)
such that G x <= h
x >= 0
1^T x = R
Thus it fits the bill of the solve_ls function in qpsolvers.
Here is an instance of how I imagine your problem matrices would look like, given what you specified. Since it is sparse we should use SciPy CSC matrices, and regular NumPy arrays for vectors:
import numpy as np
import scipy.sparse as spa
n = 150_000
# minimize 1/2 || x - s ||^2
R = spa.eye(n, format="csc")
s = np.array(range(n), dtype=float)
# such that G * x <= h
G = spa.diags(
diagonals=[
[1.0 if i % 2 == 0 else 0.0 for i in range(n)],
[1.0 if i % 3 == 0 else 0.0 for i in range(n - 1)],
[1.0 if i % 5 == 0 else 0.0 for i in range(n - 1)],
],
offsets=[0, 1, -1],
)
a_dozen_rows = np.linspace(0, n - 1, 12, dtype=int)
G = G[a_dozen_rows]
h = np.ones(12)
# such that sum(x) == 42
A = spa.csc_matrix(np.ones((1, n)))
b = np.array([42.0]).reshape((1,))
# such that x >= 0
lb = np.zeros(n)
Next, we can solve this problem with:
from qpsolvers import solve_ls
x = solve_ls(R, s, G, h, A, b, lb, solver="osqp", verbose=True)
Here I picked CVXOPT but there are other open-source solvers you can install such as ProxQP, OSQP or SCS. You can install a set of open-source solvers by: pip install qpsolvers[open_source_solvers]. After some solvers are installed, you can list those for sparse matrices by:
print(qpsolvers.sparse_solvers)
Finally, here is some code to check that the solution returned by the solver satisfies our constraints:
tol = 1e-6 # tolerance for checks
print(f"- Objective: {0.5 * (x - s).dot(x - s):.1f}")
print(f"- G * x <= h: {(G.dot(x) <= h + tol).all()}")
print(f"- x >= 0: {(x + tol >= 0.0).all()}")
print(f"- sum(x) = {x.sum():.1f}")
I just tried it with OSQP (adding the eps_rel=1e-5 keyword argument when calling solve_ls, otherwise the returned solution would be less accurate than the tol = 1e-6 tolerance) and it found a solution is 737 milliseconds on my (rather old) CPU with:
- Objective: 562494373088866.8
- G * x <= h: True
- x >= 0: True
- sum(x) = 42.0
Hoping this helps. Happy solving!
In particle physics, we have to compute the invariant mass a lot, which is for a two-body decay
When the momenta (p1, p2) are sometimes very large (up to a factor 1000 or more) compared to the masses (m1, m2). In that case, there is large cancellation happening between the last two terms when the calculation is carried out with floating point numbers on a computer.
What kind of numerical tricks can be used to compute this accurately for any inputs?
The question is about suitable numerical tricks to improve the accuracy of the calculation with floating point numbers, so the solution should be language-agnostic. For demonstration purposes, implementations in Python are preferred. Solutions which reformulate the problem and increase the amount of elementary operations are acceptable, but solutions which suggest to use other number types like decimal or multi-precision floating point numbers are not.
Note: The original question presented a simplified 1D dimensional problem in form of a Python expression, but the question is for the general case where the momenta are given in 3D dimensions. The question was reformulated in this way.
With a few tricks listed on Stackoverflow and the transformation described by Jakob Stark in his answer, it is possible to rewrite the equation into a form that does not suffer anymore from catastrophic cancellation.
The original question asked for a solution in 1D, which has a simple solution, but in practice, we need the formula in 3D and then the solution is more complicated. See this notebook for a full derivation.
Example implementation of numerically stable calculation in 3D in Python:
import numpy as np
# numerically stable implementation
#np.vectorize
def msq2(px1, py1, pz1, px2, py2, pz2, m1, m2):
p1_sq = px1 ** 2 + py1 ** 2 + pz1 ** 2
p2_sq = px2 ** 2 + py2 ** 2 + pz2 ** 2
m1_sq = m1 ** 2
m2_sq = m2 ** 2
x1 = m1_sq / p1_sq
x2 = m2_sq / p2_sq
x = x1 + x2 + x1 * x2
a = angle(px1, py1, pz1, px2, py2, pz2)
cos_a = np.cos(a)
if cos_a >= 0:
y1 = (x + np.sin(a) ** 2) / (np.sqrt(x + 1) + cos_a)
else:
y1 = -cos_a + np.sqrt(x + 1)
y2 = 2 * np.sqrt(p1_sq * p2_sq)
return m1_sq + m2_sq + y1 * y2
# numerically stable calculation of angle
def angle(x1, y1, z1, x2, y2, z2):
# cross product
cx = y1 * z2 - y2 * z1
cy = x1 * z2 - x2 * z1
cz = x1 * y2 - x2 * y1
# norm of cross product
c = np.sqrt(cx * cx + cy * cy + cz * cz)
# dot product
d = x1 * x2 + y1 * y2 + z1 * z2
return np.arctan2(c, d)
The numerically stable implementation can never produce a negative result, which is a commonly occurring problem with naive implementations, even in double precision.
Let's compare the numerically stable function with a naive implementation.
# naive implementation
def msq1(px1, py1, pz1, px2, py2, pz2, m1, m2):
p1_sq = px1 ** 2 + py1 ** 2 + pz1 ** 2
p2_sq = px2 ** 2 + py2 ** 2 + pz2 ** 2
m1_sq = m1 ** 2
m2_sq = m2 ** 2
# energies of particles 1 and 2
e1 = np.sqrt(p1_sq + m1_sq)
e2 = np.sqrt(p2_sq + m2_sq)
# dangerous cancelation in third term
return m1_sq + m2_sq + 2 * (e1 * e2 - (px1 * px2 + py1 * py2 + pz1 * pz2))
For the following image, the momenta p1 and p2 are randomly picked from 1 to 1e5, the values m1 and m2 are randomly picked from 1e-5 to 1e5. All implementations get the input values in single precision. The reference in both cases is calculated with mpmath using the naive formula with 100 decimal places.
The naive implementation loses all accuracy for some inputs, while the numerically stable implementation does not.
If you put e.g. m1 = 1e-4, m2 = 1e-4, p1 = 1 and p2 = 1 in the expression, you get about 4e-8 with double precision but 0.0 with single precision calculation. I assume, that your question is about how one can get the 4e-8 as well with single precision calculation.
What you can do is a taylor expansion (around m1 = 0 and m2 = 0) of the expression above.
e ~ e|(m1=0,m2=0) + de/dm1|(m1=0,m2=0) * m1 + de/dm2|(m1=0,m2=0) * m2 + ...
If I calculated correctly, the zeroth and first order terms are 0 and the second order expansion would be
e ~ (p1+p2)/p1 * m1**2 + (p1+p2)/p2 * m2**2
This yields exactly 4e-8 even with single precision calculation. You can of course do more terms in the expansion if you need, until you hit the precision limit of a single float.
Edit
If the mi are not always much smaller than the pi you could further massage the equation to get
The complicated part is now the one in the square brackets. It essentially is sqrt(x+1)-1 for a wide range of x values. If x is very small, we can use the taylor expansion of the square root (e.g. like here). If the x value is larger, the formula works just fine, because the addition and subtraction of 1 are no longer discarding the value of x due to floating point precision. So one threshold for x must be choosen below one switches to the taylor expansion.
import numpy as np
import matplotlib.pyplot as plt
n = 500
T = 10
dw = 2 * np.pi / T
t = np.linspace(0, T, n)
x = 5 * np.sin(20 * t + np.pi) + 10 * np.sin( 40 * t + np.pi/2)
fftx = np.fft.rfft(x)
freq = np.fft.rfftfreq(n) * n * dw
amps = np.abs(fftx) * 2 / n
angs = np.angle(fftx)
_, ax = plt.subplots(3, 1)
ax[0].plot(t, x)
ax[1].plot(freq, amps)
ax[2].plot(freq, angs)
I get correct values for frequency and amplitude. But as seen from the plot the phase values are not correct. How to extract correct values for phase from fft? What exactly am I looking at in the phase plot?
I am expecting approx 3.14 and 3.14/2 for frequencies 20 and 40 respectively.
There are two issues with computing the phase:
Your input signal is not an integer number of periods. If you replicate the signal repeatedly, you'll see you actually have a different set of frequency components than you assume when you construct the signal (the DFT can the thought of as using an infinite repetition of your signal as input). This causes the peaks to have some width to them, it also causes the phase to shift a bit.
This issue you can fix by either windowing your signal, or creating it so it has an integer number of periods. The latter is:
T = 3 * np.pi
t = np.linspace(0, T, n, endpoint=False)
The frequencies where there is no signal (which after the fix above is all except for two frequencies), the phase will be given by noise. You can set the phase here to zero:
angs[amps < 1] = 0
Now your plot looks like this:
The phases are not as you expected, because the sine has a phase of -π/2. Repeat the experiment with cos instead of sin and you get the phases you were expecting.
I am trying to implement a loss function which tries to minimize the negative log likelihood of obtaining ground truth values (x,y) from predicted bivariate gaussian distribution parameters. I am implementing this in tensorflow -
Here is the code -
def tf_2d_normal(self, x, y, mux, muy, sx, sy, rho):
'''
Function that implements the PDF of a 2D normal distribution
params:
x : input x points
y : input y points
mux : mean of the distribution in x
muy : mean of the distribution in y
sx : std dev of the distribution in x
sy : std dev of the distribution in y
rho : Correlation factor of the distribution
'''
# eq 3 in the paper
# and eq 24 & 25 in Graves (2013)
# Calculate (x - mux) and (y-muy)
normx = tf.sub(x, mux)
normy = tf.sub(y, muy)
# Calculate sx*sy
sxsy = tf.mul(sx, sy)
# Calculate the exponential factor
z = tf.square(tf.div(normx, sx)) + tf.square(tf.div(normy, sy)) - 2*tf.div(tf.mul(rho, tf.mul(normx, normy)), sxsy)
negRho = 1 - tf.square(rho)
# Numerator
result = tf.exp(tf.div(-z, 2*negRho))
# Normalization constant
denom = 2 * np.pi * tf.mul(sxsy, tf.sqrt(negRho))
# Final PDF calculation
result = -tf.log(tf.div(result, denom))
return result
When I am doing the training, I can see the loss value decreasing but it goes well past below 0. I can understand that should be because, we are minimizing the 'negative' likelihood. Even the loss values are decreasing, I can't get my results accurate. Can someone help in verifying, if the code that I have written for the loss function is correct or not.
Also is such a nature of loss desirable for training Neural Nets(specifically RNN)?
Thankss
I see you've found the sketch-rnn code from magenta, I'm working on something similar. I found this piece of code not to be stable by itself. You'll need to stabilize it using constraints, so the tf_2d_normal code can't be used or interpreted in isolation. NaNs and Infs will start appearing all over the place if your data isn't normalized properly in advance or in your loss function.
Below is a more stable loss function version I'm building with Keras. There may be some redundancy in here, it may not be perfect for your needs but I found it to be working and you can test/adapt it. I included some inline comments on how large negative log values can arise:
def r3_bivariate_gaussian_loss(true, pred):
"""
Rank 3 bivariate gaussian loss function
Returns results of eq # 24 of http://arxiv.org/abs/1308.0850
:param true: truth values with at least [mu1, mu2, sigma1, sigma2, rho]
:param pred: values predicted from a model with the same shape requirements as truth values
:return: the log of the summed max likelihood
"""
x_coord = true[:, :, 0]
y_coord = true[:, :, 1]
mu_x = pred[:, :, 0]
mu_y = pred[:, :, 1]
# exponentiate the sigmas and also make correlative rho between -1 and 1.
# eq. # 21 and 22 of http://arxiv.org/abs/1308.0850
# analogous to https://github.com/tensorflow/magenta/blob/master/magenta/models/sketch_rnn/model.py#L326
sigma_x = K.exp(K.abs(pred[:, :, 2]))
sigma_y = K.exp(K.abs(pred[:, :, 3]))
rho = K.tanh(pred[:, :, 4]) * 0.1 # avoid drifting to -1 or 1 to prevent NaN, you will have to tweak this multiplier value to suit the shape of your data
norm1 = K.log(1 + K.abs(x_coord - mu_x))
norm2 = K.log(1 + K.abs(y_coord - mu_y))
variance_x = K.softplus(K.square(sigma_x))
variance_y = K.softplus(K.square(sigma_y))
s1s2 = K.softplus(sigma_x * sigma_y) # very large if sigma_x and/or sigma_y are very large
# eq 25 of http://arxiv.org/abs/1308.0850
z = ((K.square(norm1) / variance_x) +
(K.square(norm2) / variance_y) -
(2 * rho * norm1 * norm2 / s1s2)) # z → -∞ if rho * norm1 * norm2 → ∞ and/or s1s2 → 0
neg_rho = 1 - K.square(rho) # → 0 if rho → {1, -1}
numerator = K.exp(-z / (2 * neg_rho)) # → ∞ if z → -∞ and/or neg_rho → 0
denominator = (2 * np.pi * s1s2 * K.sqrt(neg_rho)) + epsilon() # → 0 if s1s2 → 0 and/or neg_rho → 0
pdf = numerator / denominator # → ∞ if denominator → 0 and/or if numerator → ∞
return K.log(K.sum(-K.log(pdf + epsilon()))) # → -∞ if pdf → ∞
Hope you find this of value.