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norm of differences of variables in cvxpy

stackoverflow.com › questions › 71632766 › norm-of-differences-of-variables-in-cvxpy

Your problem is solved exactly because you don't have any constraint. I skipped the function to make you a shorter script.

import numpy as np
import cvxpy as cp
pts = np.array([(-2, 3), (-3, 5), (-1, 4), (-3, 7), (0, 3),
                (-2, -2), (-3, 8), (3, 1), (1, -2), (2, 6),
                (-2, 6), (-2, 5), (-4, 3), (-4, 6), (-3, 10),
                (2, -3)])

n = pts.shape[0]
x = cp.Variable((n, 2))
objective = cp.Minimize(cp.sum(cp.norm(pts - x, 2, axis=1)))
prob = cp.Problem(objective)
prob.solve()

print(x.value)
# result
# [[-2.  3.]
#  [-3.  5.]
#  [-1.  4.]
#  [-3.  7.]
#  [ 0.  3.]
#  [-2. -2.]
#  [-3.  8.]
#  [ 3.  1.]
#  [ 1. -2.]
#  [ 2.  6.]
#  [-2.  6.]
#  [-2.  5.]
#  [-4.  3.]
#  [-4.  6.]
#  [-3. 10.]
#  [ 2. -3.]]

P.S. maybe this is not your correct answer. The problem is that from what I see in your equation. I'm not able to differentiate the parameters and your values.

Does it mean that you have two functions to minimize? one approaching the points and the other one approaching all the previous points to the current one?

If that's the case, you can replace the code with:

cost1 = cp.sum(cp.norm(pts - x, 2, axis=1))
cost2 = cp.sum([cp.norm(x[j] - x[i],2) for i in range(1, n) for j in range(0, i)])
objective = cp.Minimize(cost1+cost2)
prob = cp.Problem(objective)
prob.solve()
print(x.value)
#result
# [[-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]]

But as you can see, the result is not that interesting.

Answer from silgon on Stack Overflow

Cvxpy

cvxpy.org › tutorial › functions › index.html

Atomic Functions -

The functions max and min give ... and minimum to find the max or min of a list of scalar expressions. The CVXPY function sum sums all the entries in a single expression....

Cvxpy

cvxpy.org › _modules › cvxpy › atoms › norm.html

cvxpy.atoms.norm -

""" x = Expression.cast_to_const(x) # matrix norms take precedence num_nontrivial_idxs = sum([d > 1 for d in x.shape]) if axis is None and x.ndim == 2: if p == 1: # matrix 1-norm return cvxpy.atoms.max(norm1(x, axis=0)) # Frobenius norm elif p == 'fro' or (p == 2 and num_nontrivial_idxs == 1): return pnorm(vec(x, order='F'), 2) elif p == 2: # matrix 2-norm is largest singular value return sigma_max(x) elif p == 'nuc': # the nuclear norm (sum of singular values) return normNuc(x) elif p in [np.inf, "inf", "Inf"]: # the matrix infinity-norm return cvxpy.atoms.max(norm1(x, axis=1)) else: raise RuntimeError('Unsupported matrix norm.') else: if p == 1 or x.is_scalar(): return norm1(x, axis=axis, keepdims=keepdims) elif str(p).lower() == "inf": return norm_inf(x, axis=axis, keepdims=keepdims) elif str(p).lower() == "fro": # TODO should not work for vectors.

Cvxpy

cvxpy.org › _modules › cvxpy › atoms › pnorm.html

cvxpy.atoms.pnorm -

[docs] def pnorm(x, p: Union[int, str] = 2, axis=None, keepdims: bool = False, max_denom: int = 1024, approx: bool = True): """Factory function for a mathematical p-norm. Parameters ---------- p : numeric type or string The type of norm to construct; set this to np.inf or 'inf' to construct an infinity norm.

Cvxpy

cvxpy.org › api_reference › cvxpy.atoms.other_atoms.html

Other Atoms -

An Expression representing the mixed norm. Return type:¶ · Expression · cvxpy.atoms.norm.norm(x, p: int | str = 2, axis=None, keepdims: bool = False)[source]¶ · Wrapper on the different norm atoms. Parameters:¶ · x : Expression or numeric constant¶ ·

Readthedocs

ajfriendcvxpy.readthedocs.io › en › latest › tutorial › functions › pnorm.html

p-norm — CVXPY 0.2.25 documentation - Read the Docs

If given a matrix variable, pnorm will treat it as a vector, and compute the p-norm of the concatenated columns.

Cvxpy

cvxpy.org › version › 1.2 › _modules › cvxpy › atoms › norm.html

cvxpy.atoms.norm — CVXPY 1.2 documentation

Parameters ---------- x : Expression or numeric constant The value to take the norm of. If `x` is 2D and `axis` is None, this function constructs a matrix norm. Returns ------- Expression An Expression representing the norm. """ return norm(x, p=2, axis=axis) Install · User Guide · Examples · API Documentation · FAQ · Citing CVXPY ·

Cvxpy

cvxpy.org › _modules › cvxpy › atoms › norm1.html

cvxpy.atoms.norm1 -

""" if self.axis is None: values = np.array(values[0]).flatten() else: values = np.array(values[0]) return np.linalg.norm(values, 1, axis=self.axis, keepdims=self.keepdims) def sign_from_args(self) -> Tuple[bool, bool]: """Returns sign (is positive, is negative) of the expression.

Cvxpy

cvxpy.org › examples › basic › least_squares.html

Least-squares -

# Import packages. import cvxpy as cp import numpy as np # Generate data. m = 20 n = 15 np.random.seed(1) A = np.random.randn(m, n) b = np.random.randn(m) # Define and solve the CVXPY problem. x = cp.Variable(n) cost = cp.sum_squares(A @ x - b) prob = cp.Problem(cp.Minimize(cost)) prob.solve() # Print result. print("\nThe optimal value is", prob.value) print("The optimal x is") print(x.value) print("The norm of the residual is ", cp.norm(A @ x - b, p=2).value) The optimal value is 7.005909828287484 The optimal x is [ 0.17492418 -0.38102551 0.34732251 0.0173098 -0.0845784 -0.08134019 0.293119 0.27019762 0.17493179 -0.23953449 0.64097935 -0.41633637 0.12799688 0.1063942 -0.32158411] The norm of the residual is 2.6468679280023557 Back to top

Stack Overflow

stackoverflow.com › questions › 71632766 › norm-of-differences-of-variables-in-cvxpy

python - norm of differences of variables in cvxpy - Stack Overflow

Top answer

1 of 1

Your problem is solved exactly because you don't have any constraint. I skipped the function to make you a shorter script.

import numpy as np
import cvxpy as cp
pts = np.array([(-2, 3), (-3, 5), (-1, 4), (-3, 7), (0, 3),
                (-2, -2), (-3, 8), (3, 1), (1, -2), (2, 6),
                (-2, 6), (-2, 5), (-4, 3), (-4, 6), (-3, 10),
                (2, -3)])

n = pts.shape[0]
x = cp.Variable((n, 2))
objective = cp.Minimize(cp.sum(cp.norm(pts - x, 2, axis=1)))
prob = cp.Problem(objective)
prob.solve()

print(x.value)
# result
# [[-2.  3.]
#  [-3.  5.]
#  [-1.  4.]
#  [-3.  7.]
#  [ 0.  3.]
#  [-2. -2.]
#  [-3.  8.]
#  [ 3.  1.]
#  [ 1. -2.]
#  [ 2.  6.]
#  [-2.  6.]
#  [-2.  5.]
#  [-4.  3.]
#  [-4.  6.]
#  [-3. 10.]
#  [ 2. -3.]]

P.S. maybe this is not your correct answer. The problem is that from what I see in your equation. I'm not able to differentiate the parameters and your values.

Does it mean that you have two functions to minimize? one approaching the points and the other one approaching all the previous points to the current one?

If that's the case, you can replace the code with:

cost1 = cp.sum(cp.norm(pts - x, 2, axis=1))
cost2 = cp.sum([cp.norm(x[j] - x[i],2) for i in range(1, n) for j in range(0, i)])
objective = cp.Minimize(cost1+cost2)
prob = cp.Problem(objective)
prob.solve()
print(x.value)
#result
# [[-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]
#  [-1.7909687  4.4720621]]

But as you can see, the result is not that interesting.

Find elsewhere

Google Bing Mojeek

Byu

acme.byu.edu › 00000181-a756-d9e0-a789-bf7665e10001 › convex-opt-with-cvxpy-pdf pdf

15 CVXPY Lab Objective:

CVXPY includes the l1 norm and many other useful functions.

Readthedocs

ajfriendcvxpy.readthedocs.io › en › latest › tutorial › functions

Functions — CVXPY 0.2.25 documentation - Read the Docs

This section of the tutorial describes the functions that can be applied to CVXPY expressions.

Snyk

snyk.io › advisor › cvxpy › functions › cvxpy.norm

How to use the cvxpy.norm function in cvxpy | Snyk

def accelerate(self): """ Perform acceleration """ print("Accelerating...") import cvxpy as cvx a = cvx.Variable() constraints = [cvx.sum(a) == 1, a >= 0] objective = cvx.norm(self.work.acceleration.F * a) problem = cvx.Problem(cvx.Minimize(objective), constraints) problem.solve() u = self.work.acceleration.G * a.value print("u = ", u) print("Updating work variables...") x = u[:self.work.data.n] v = u[self.work.data.n:] z = self.project(v) y = self.work.rho_vec * (v - z) self.work.x = x self.work.z = z self.work.y = y

Cvxpy

cvxpy.org › api_reference › cvxpy.atoms.html

Atoms -

cvxpy.atoms.min.min · mixed_norm · cvxpy.atoms.mixed_norm.mixed_norm · Parameters · X · p · q · Returns · Return type · norm · cvxpy.atoms.norm.norm · Parameters · x · p · axis · keepdims · Returns · Return type · norm1 · cvxpy.atoms.norm1.norm1 ·

Byu

labs.acme.byu.edu › Volume2 › CVXPY_Intro › CVXPY_Intro.html

Intro to CVXPY — ACME Labs

To specify a norm in CVXPY, use the syntax cp.norm(x, a) where \(a\) represents your choice of norm (1 in this case).

GitHub

github.com › cvxpy › cvxpy › issues › 443

Infinity norm for matrices different than convention · Issue #443 · cvxpy/cvxpy

February 23, 2018 - According to the documentation, the function norm(X, 'inf') when X is a matrix computes \sum_{i,j}|X_{i,j}|, which is different than the convention, \max_i \sum_j |a_{i,j}|. This could lead to some confusion · source: http://mathworld.wolfram.com/MatrixNorm.html

Author sbarratt

Ee227c

ee227c.github.io › code › lecture22.html

Non-convex constraints part 2: projected gradient descent

def l1min(y, A): x_hat = cvxpy.Variable(A.shape[1]) objective = cvxpy.Minimize(cvxpy.norm(x_hat, 1)) constraints = [A * x_hat == y] prob = cvxpy.Problem(objective, constraints) prob.solve() return np.squeeze(np.array(x_hat.value)), (prob.value, prob.status) Let's see if we get an accurate solution, e.g., in terms of relative error.

Stack Exchange

scicomp.stackexchange.com › questions › 36704 › why-is-cvxpy-throwing-a-dcp-error-with-cp-sqrt-but-no-error-with-cp-norm

python - Why is CVXPY throwing a DCP error with cp.sqrt but no error with cp.norm - Computational Science Stack Exchange