I am trying to multiply 2 matrices of 2*2 order. One of the matrix contains an unknown parameter "k1". I want to check a satisfiable solution that for which value of k1. The product of two matrices will be equal to the third one.
Note: I dont want to convert the multiplication into a linear relation or set of equation I want to manipulate it as matrices.
Here is where I am stuck.
k1 = Int ('k1')
x = [ [ Int("x_%s_%s" % (i+1, j+1)) for j in range(2) ]
for i in range(2) ]
a =((1,k1),(3,4))
b =((1,1),(1,1))
c= ((3,3),(7,7))
s = Solver()
s.add(a[0][1]>0)
s.add(a*b==c)
if s.check() == sat:
m = s.model()
r = [ [ m.evaluate(x[i][j]) for j in range(2) ]
for i in range(2) ]
print_matrix(r)
else:
print "failed to solve"
Any way Out?
One possible solution is
k1 = Int ('k1')
x = [ [ Int("x_%s_%s" % (i+1, j+1)) for j in range(2) ]
for i in range(2) ]
a =((1,k1),(3,4))
b =((1,1),(1,1))
c= ((3,3),(7,7))
s = Solver()
eq1= a[0][1]>0
eq2 =[[sum(a[i][k]*b[k][j] for k in range(2)) == c[i][j] for i in range(2) ]
for j in range(2) ]
s.add(eq1)
s.add(eq2[0][0])
s.add(eq2[0][1])
s.add(eq2[1][0])
s.add(eq2[1][1])
print s
print s.check()
m = s.model()
print m
and the corresponding output is
[k1 > 0, 1 + k1*1 == 3, True, 1 + k1*1 == 3, True]
sat
[k1 = 2]
Please run this example online here
Related
I have a convex optimization problem I am trying to solve with cvxpy. Given a 1 x n row vector y and an m x n matrix C, I want to find a scalar b and a 1 x m row vector a such that the sum of squares of y - (aC + b(aC # aC)) is as small as possible (the # denotes element wise multiplication). In addition, all entires in a must be nonnegative and sum to 1 and -100 <= b <= 100. Below is my attempt to solve this using cvxpy.
import numpy as np
import cvxpy as cvx
def find_a(y, C, b_min=-100, b_max=100):
b = cvx.Variable()
a = cvx.Variable( (1,C.shape[0]) )
aC = a * C # this should be matrix multiplication
x = (aC + cvx.multiply(b, cvx.square(aC)))
objective = cvx.Minimize ( cvx.sum_squares(y - x) )
constraints = [0. <= a,
a <= 1.,
b_min <= b,
b <= b_max,
cvx.sum(a) == 1.]
prob = cvx.Problem(objective, constraints)
result = prob.solve()
print a.value
print result
y = np.asarray([[0.10394265, 0.25867508, 0.31258457, 0.36452763, 0.36608997]])
C = np.asarray([
[0., 0.00169811, 0.01679245, 0.04075472, 0.03773585],
[0., 0.00892802, 0.03154158, 0.06091544, 0.07315024],
[0., 0.00962264, 0.03245283, 0.06245283, 0.07283019],
[0.04396226, 0.05245283, 0.12245283, 0.18358491, 0.23886792]])
find_a(y, C)
I keep getting a DCPError: Problem does not follow DCP rules. error when I try to solve for a. I am thinking that either my function is not really convex, or I do not understand how to construct the proper cvxpy Problem. Any help would be greatly appreciated.
In Z3 I can call (get-objectives) to have a dump of the resulting weights.
(e.g. here)
It prints something like this:
(objectives
(aaa 1)
(bbb 0)
)
In z3py however Optimize.objectives() prints a dump of the calculation for the objectives, not however the calculated weights, as seen here:
[If(a == 3, 0, 1), If(b == 3, 0, 1)]
Is there a way how I can get the calculated weights? or the weight of a specific objective as in the standard z3?
Here is my example code:
from z3 import *
a, b = Ints('a b')
s = Optimize()
s.add(3 <= a, a <= 10)
s.add(3 <= b, b <= 10)
s.add(a >= 2*b)
s.add_soft(a == 3, weight=1, id="aaa")
s.add_soft(b == 3, weight=1, id="bbb")
print(s.check())
print(s.model())
print(s.objectives())
You can use the model to evaluate the objectives:
m = s.model()
print [m.evaluate(o) for o in s.objectives()]
This yields:
sat
[1, 0]
I'm trying to implement a function that computes the Relu derivative for each element in a matrix, and then return the result in a matrix. I'm using Python and Numpy.
Based on other Cross Validation posts, the Relu derivative for x is
1 when x > 0, 0 when x < 0, undefined or 0 when x == 0
Currently, I have the following code so far:
def reluDerivative(self, x):
return np.array([self.reluDerivativeSingleElement(xi) for xi in x])
def reluDerivativeSingleElement(self, xi):
if xi > 0:
return 1
elif xi <= 0:
return 0
Unfortunately, xi is an array because x is an matrix. reluDerivativeSingleElement function doesn't work on array. So I'm wondering is there a way to map values in a matrix to another matrix using numpy, like the exp function in numpy?
Thanks a lot in advance.
That's an exercise in vectorization.
This code
if x > 0:
y = 1
elif xi <= 0:
y = 0
Can be reformulated into
y = (x > 0) * 1
This is something that will work for numpy arrays, since boolean expressions involving them are turned into arrays of values of these expressions for elements in said array.
I guess this is what you are looking for:
>>> def reluDerivative(x):
... x[x<=0] = 0
... x[x>0] = 1
... return x
>>> z = np.random.uniform(-1, 1, (3,3))
>>> z
array([[ 0.41287266, -0.73082379, 0.78215209],
[ 0.76983443, 0.46052273, 0.4283139 ],
[-0.18905708, 0.57197116, 0.53226954]])
>>> reluDerivative(z)
array([[ 1., 0., 1.],
[ 1., 1., 1.],
[ 0., 1., 1.]])
Basic function to return derivative of relu could be summarized as follows:
f'(x) = x > 0
So, with numpy that would be:
def relu_derivative(z):
return np.greater(z, 0).astype(int)
def dRelu(z):
return np.where(z <= 0, 0, 1)
Here z is a ndarray in my case.
def reluDerivative(self, x):
return 1 * (x > 0)
You are on a good track: thinking on vectorized operation. Where we define a function, and we apply this function to a matrix, instead of writing a for loop.
This threads answers your question, where it replace all the elements satisfy the condition. You can modify it into ReLU derivative.
https://stackoverflow.com/questions/19766757/replacing-numpy-elements-if-condition-is-met
In addition, python supports functional programming very well, try to use lambda function.
https://www.python-course.eu/lambda.php
This works:
def dReLU(x):
return 1. * (x > 0)
As mentioned by Neil in the comments, you can use heaviside function of numpy.
def reluDerivative(self, x):
return np.heaviside(x, 0)
If you want to use pure Python:
def relu_derivative(x):
return max(sign(x), 0)
If you want it with the derivative you can use:
def relu(neta):
relu = neta * (neta > 0)
d_relu = (neta > 0)
return relu, d_relu
When x is larger than 0, the slope is 1.
When x is smaller than or equal to 0, the slope is 0.
if (x > 0):
return 1
if (x <= 0):
return 0
This can be written more compact:
return 1 * (x > 0)
I was on the site rise4fun a few weeks ago and they had a python code that converted a sudoku puzzle input file to z3. I checked again today and the file is gone, and was wondering if anyone had this code or could explain to me how to implement it. Thanks!!
# 9x9 matrix of integer variables
X = [ [ Int("x_%s_%s" % (i+1, j+1)) for j in range(9) ]
for i in range(9) ]
# each cell contains a value in {1, ..., 9}
cells_c = [ And(1 <= X[i][j], X[i][j] <= 9)
for i in range(9) for j in range(9) ]
# each row contains a digit at most once
rows_c = [ Distinct(X[i]) for i in range(9) ]
# each column contains a digit at most once
cols_c = [ Distinct([ X[i][j] for i in range(9) ])
for j in range(9) ]
# each 3x3 square contains a digit at most once
sq_c = [ Distinct([ X[3*i0 + i][3*j0 + j]
for i in range(3) for j in range(3) ])
for i0 in range(3) for j0 in range(3) ]
sudoku_c = cells_c + rows_c + cols_c + sq_c
# sudoku instance, we use '0' for empty cells
instance = ((5,3,0,0,7,0,0,0,0),
(6,0,0,1,9,5,0,0,0),
(0,9,8,0,0,0,0,6,0),
(8,0,0,0,6,0,0,0,3),
(4,0,0,8,0,3,0,0,1),
(7,0,0,0,2,0,0,0,6),
(0,6,0,0,0,0,2,8,0),
(0,0,0,4,1,9,0,0,5),
(0,0,0,0,8,0,0,7,9))
instance_c = [ If(instance[i][j] == 0,
True,
X[i][j] == instance[i][j])
for i in range(9) for j in range(9) ]
s = Solver()
s.add(sudoku_c + instance_c)
if s.check() == sat:
m = s.model()
r = [ [ m.evaluate(X[i][j]) for j in range(9) ]
for i in range(9) ]
print_matrix(r)
else:
print "failed to solve"
Certain problem in transport Phenomena is solved using the following code:
T_max, T_0, S, R, k, I, k_e, L, R, E, a = Reals('T_max T_0 S R k I k_e L R E a')
k = a*k_e*T_0
I = k_e*E/L
S = (I**2)/k_e
eq = T_0 + S* R**2/(4*k)
print eq
equations = [
T_max == eq,
]
print "Temperature equations:"
print equations
problem = [
R == 2, L == 5000,
T_0 == 20 + 273,
T_max == 30 + 273, k_e == 1,
a == 2.23*10**(-8), E > 0
]
print "Problem:"
print problem
print "Solution:"
solve(equations + problem)
using this code online we obtain
This output gives the correct answer but there are two issues in the code: a) the expresion named "eq" is not fully simplified and then it is necessary to give an arbitrary value for k_e . My question is: How to simplify the expression "eq" in such way that k_e be eliminated from "eq"?
Other example: To determine the radius of a tube
Code:
def inte(n,a,b):
return (b**(n+1))/(n+1)-(a**(n+1))/(n+1)
P_0, P_1, L, R, mu, q, C = Reals('P_0 P_1 L R mu q C')
k = (P_0 - P_1)/(2*mu*L)
equations = [0 == -k*inte(1,0,R) +C,
q == 2*3.1416*(-(k/2)*inte(3,0,R) + C*inte(1,0,R))]
print "Fluid equations:"
print equations
problem = [
L == 50.02/100, mu == (4.03*10**(-5)),
P_0 == 4.829*10**5, P_1==0,
q == 2.997*10**(-3), R >0
]
print "Problem:"
print problem
print "Solution:"
solve(equations + problem)
Output:
Fluid equations:
[-((P_0 - P_1)/(2·mu·L))·(R2/2 - 0) + C = 0, q =
3927/625·
(-(((P_0 - P_1)/(2·mu·L))/2)·(R4/4 - 0) + C·(R2/2 - 0))]
Problem:
[L = 2501/5000, mu = 403/10000000, P_0 = 482900, P_1 = 0, q = 2997/1000000, R > 0]
Solution:
[R = 0.0007512843?,
q = 2997/1000000,
P_1 = 0,
P_0 = 482900,
mu = 403/10000000,
L = 2501/5000,
C = 3380.3149444289?]