I'm trying to find the minimal value of the Parabola y=(x+2)**2-3, apparently, the answer should be y==-3, when x ==-2.
But z3 gives the answer [x = 0, y = 1], which doesn't meet the ForAll assertion.
Am I assuming wrongly with something?
Here is the python code:
from z3 import *
x, y, z = Reals('x y z')
print(Tactic('qe').apply(And(y == (x + 2) ** 2 - 3,
ForAll([z], y <= (z + 2) ** 2 - 3))))
solve(y == x * x + 4 * x +1,
ForAll([z], y <= z * z + 4 * z +1))
And the result:
[[y == (x + 2)**2 - 3, True]]
[x = 0, y = 1]
The result shows that 'qe' tactic eliminated that ForAll assertion into True, although it's NOT always true.
Is that the reason that solver gives a wrong answer?
What should I code to find the minimal (or maximal) value of such an expression?
BTW, the Z3 version is 4.3.2 for Mac.
I refered
How does Z3 handle non-linear integer arithmetic?
and found a partial solution, using 'qfnra-nlsat' and 'smt' tactics.
from z3 import *
x, y, z = Reals('x y z')
s1 = Then('qfnra-nlsat','smt').solver()
print s1.check(And(y == (x + 2) ** 2 - 3,
ForAll([z], y <= (z + 2) ** 2 - 3)))
print s1.model()
s2 = Then('qe', 'qfnra-nlsat','smt').solver()
print s2.check(And(y == (x + 2) ** 2 - 3,
ForAll([z], y <= (z + 2) ** 2 - 3)))
print s2.model()
And the result:
sat
[x = -2, y = -3]
sat
[x = 0, y = 1]
Still the 'qe' tactic and the default solver seem buggy. They don't give the correct result.
Further comments and discussions are needed.
Related
I am new to CVXPY, I learn by running examples and I found this post: How to construct a SOCP problem and solve using cvxpy and cvxpylayers #ben
The author provided the codes (Below I've corrected the line that caused an error in OP's EDIT 2):
import cvxpy as cp
import numpy as np
N = 5
Q_sqrt = cp.Parameter((N, N))
Q = cp.Parameter((N, N))
x = cp.Variable(N)
z = cp.Variable(N)
p = cp.Variable()
t = cp.Variable()
objective = cp.Minimize(p - t)
constraint_soc = [z == Q # x, x.value * z >= t ** 2, z >= 0, x >= 0] # <-- my question
constraint_other = [cp.quad_over_lin(Q_sqrt # x, p) <= N * p, cp.sum(x) == 1, p >= 0, t >= 0]
constraint_all = constraint_other + constraint_soc
matrix = np.random.random((N, N))
a_matrix = matrix.T # matrix
Q.value = a_matrix
Q_sqrt.value = np.sqrt(a_matrix)
prob = cp.Problem(objective, constraint_all)
prob.solve(verbose=True)
print("status:", prob.status)
print("optimal value", prob.value)
My question is here: x.value * z >= t ** 2
why only x.value while z not?
Actually I tried x * z, x.value * z.value, they both throw out errors, only the one in the original post works, which is x.value * z.
I googled and found this page and this, looks most relevant, but still failed to find an answer.
But both x and z are variables and defined as such
x = cp.Variable(N)
z = cp.Variable(N)
why only x needs a .value after?? Maybe it's a trivial question to experienced users, but I really don't get it. Thank you.
Update: two follow-up questions
Thanks to #MichalAdamaszek the first question above is clear: x.value didn't make sense here. A suggestion is using constraint_soc = [z == Q # x] + [ z[i]>=cp.quad_over_lin(t, x[i]) for i in range(N) ]
I have two following questions:
is it better to write the second of the soc constraint as : [ x[i]>=cp.quad_over_lin(t, z[i]) for i in range(N) ]? because in the question we only know that z[i] > 0 for sure. Or it doesn't matter at all in cvxpy? I tired both, both returned a similar optimal value.
I noticed that for the two constraints:
$x^TQx <= Np^2 $ and $ x_i z_i >= t^2 $
the quadratic terms were always intentionally split into two linear terms:
cp.quad_over_lin(Q_sqrt # x, p) <= N * p and z[i]>=cp.quad_over_lin(t, x[i]) respectively.
Is it because that in cvxpy, it is not allowed to have quadratic terms in (in)equality constraints? Is there an documentation to those basic rules?
Thank you.
I'm setting up a linear programming optimization model using CPLEX and am wondering if it's possible to accomplish a modification of the cost function dependent upon which binary decision variables are 'active' in an arbitrary solution. This is mostly a question about how to formulate the LP model (if it's even possible), but responses in the context of CPLEX are welcome or even preferred.
Say I have an LP problem in canonical form:
minimize cTx
s.t. Ax <= b
With cost function:
c = [c_1, c_2,...,c_100]
All variables are binary. I have this basic setup modeled and running effectively in CPLEX.
Now say I have a subset of variables:
efficiency_set = [x_1, x_2,...,x_5]
With the condition:
if any x_n in efficiency_set == 1
then c_n for all other x_n in the set = 0.9 * c_n
Essentially there is a dependency where if any x_n in the efficiency set is 'active', it becomes 10% less expensive for other variables in the set to appear in the solution.
I thought that CPLEX indicator constraints were what I was looking for, but after reading through documentation, I don't think I can enforce an on-the-fly change to cost function with them (I could be wrong). So I feel like it needs to be done through formulation of the LP, but I can't reason how to accomplish it. Any ideas?. Thanks.
In CPLEX you have many APIs, let me answer you with the easiest one OPL.
Your canonical form can be written
int n=3;
int m=4;
range N=1..n;
range M=1..m;
float A[N][M]=[[1,4,9,6],[8,5,0,8],[2,9,0,2]];
float B[M]=[3,1,3,0];
float C[N]=[1,1,1];
dvar boolean x[N];
minimize sum(i in N) C[i]*x[i];
subject to
{
forall(j in M) sum(i in N) A[i,j]*x[i]>=B[j];
}
and then you can you write logical constraints:
int n=3;
int m=4;
range N=1..n;
range M=1..m;
float A[N][M]=[[1,4,9,6],[8,5,0,8],[2,9,0,2]];
float B[M]=[3,1,3,0];
float C[N]=[1,1,1];
{int} efficiencySet={1,2};
dvar boolean activeEfficiencySet;
dvar boolean x[N];
minimize sum(i in N) C[i]*x[i]*(1-0.1*activeEfficiencySet*(i not in efficiencySet));
subject to
{
forall(j in M) sum(i in N) A[i,j]*x[i]>=B[j];
activeEfficiencySet==(1<=sum(i in efficiencySet) x[i]);
}
Using Alex's data, I have written the program in docplex (cplex python API)
from docplex.mp.model import Model
n = 3
m = 4
A = {}
A[0, 0] = 1
A[0, 1] = 4
A[0, 2] = 9
A[0, 3] = 6
A[1, 0] = 8
A[1, 1] = 5
A[1, 2] = 0
A[1, 3] = 8
A[2, 0] = 2
A[2, 1] = 9
A[2, 2] = 0
A[2, 3] = 2
B = {}
B[0] = 3
B[1] = 1
B[2] = 3
B[3] = 0
C = {}
C[0] = 1
C[1] = 1
C[2] = 1
efficiencySet = [0, 1]
mdl = Model(name="")
activeEfficiencySet = mdl.binary_var()
x = mdl.binary_var_dict(range(n), name="x")
# constraint 1:
for j in range(m):
mdl.add_constraint(mdl.sum(A[i, j] * x[i] for i in range(n)) >= B[j])
# constraint 2:
mdl.add(activeEfficiencySet == (mdl.sum(x) >= 1))
# objective function:
# expr = mdl.linear_expr()
lst = []
for i in range(n):
if i not in efficiencySet:
lst.append((C[i] * x[i] * (1 - 0.1 * activeEfficiencySet)))
else:
lst.append(C[i] * x[i])
mdl.minimize(mdl.sum(lst))
mdl.solve()
for i in range(n):
print(str(x[i]) + " : " + str(x[i].solution_value))
activeEfficiencySet.solution_value
Using Z3Py, once a model has been checked for an optimization problem, is there a way to convert ArithRef expressions into values?
Such as
y = If(x > 5, 0, 0.5 * x)
Once values have been found for x, can I get the evaluated value for y, without having to calculate again based on the given values for x?
Many thanks.
You need to evaluate, but it can be done by the model for you automatically:
from z3 import *
x = Real('x')
y = If(x > 5, 0, 0.5 * x)
s = Solver()
r = s.check()
if r == sat:
m = s.model();
print("x =", m.eval(x, model_completion=True))
print("y =", m.eval(y, model_completion=True))
else:
print("Solver said:", r)
This prints:
x = 0
y = 0
Note that we used the parameter model_completion=True since there are no constraints to force x (and consequently y) to any value in this model. If you have sufficient constraints added, you wouldn't need that parameter. (Of course, having it does not hurt.)
I have the following problem:
I want to sum a smaller matrix M to a bigger one , N, starting from i,j in N.
Here is the code:
PutMintoN[M_, Q_, i_, j_] := Module[{Mrow, Mcol},
{Mrow, Mcol} = Dimensions[M];
For[k = 1, k <= Mrow, k++,
For[q = 1, q <= Mcol, q++,
Q[[i + k - 1, j + q - 1]] =
Q[[i + k - 1, j + q - 1]] + M[[k, q]]]];
Q
];
The problem seems not to be in the algorithm, but in the module because if i copy the inner code outside , it works.
Thanks in advance.
Great, that you found your error on your own.
I produced a more elegant and robust Module for the same purpose, by the use of ArrayPad, to bring M to the same Dimension as N and than just Add M to N. It even works, if i or j runns out of the dimensions of N, wich is a problem for your original module.
putMintoN[M_, N_, i_, j_] := Module[{Mrow, Mcol, Nrow, Ncol, mn},
{Mrow, Mcol} = Dimensions[M]; {Nrow, Ncol} = Dimensions[N];
mn = {{Min[i - 1, Nrow], Min[(Nrow - Mrow) - i + 1, Nrow]},
{Min[j - 1, Ncol], Min[(Ncol - Mcol) - j + 1, Ncol]}};
ArrayPad[M, mn] + N]
test:
IN: putMintoN[{{x, y, p}, {z, w, q}}, {{a, b, c}, {d, e, f}, {g, h, i}, {j, k, l}}, 2, 1]
OUT: {{a, b, c}, {d + x, e + y, f + p}, {g + z, h + w, i + q}, {j, k, l}}
In Mathematica it is often possible to avoid the use of for-lopes, with Listable functions, map, apply, and so on.
Hope this inspires you.
Best regards.
I need to write a program to generate and display a piecewise quadratic Bezier curve that interpolates each set of data points (I have a txt file contains data points). The curve should have continuous tangent directions, the tangent direction at each data point being a convex combination of the two adjacent chord directions.
0.1 0,
0 0,
0 5,
0.25 5,
0.25 0,
5 0,
5 5,
10 5,
10 0,
9.5 0
The above are the data points I have, does anyone know what formula I can use to calculate control points?
You will need to go with a cubic Bezier to nicely handle multiple slope changes such as occurs in your data set. With quadratic Beziers there is only one control point between data points and so each curve segment much be all on one side of the connecting line segment.
Hard to explain, so here's a quick sketch of your data (black points) and quadratic control points (red) and the curve (blue). (Pretend the curve is smooth!)
Look into Cubic Hermite curves for a general solution.
From here: http://blog.mackerron.com/2011/01/01/javascript-cubic-splines/
To produce interpolated curves like these:
You can use this coffee-script class (which compiles to javascript)
class MonotonicCubicSpline
# by George MacKerron, mackerron.com
# adapted from:
# http://sourceforge.net/mailarchive/forum.php?thread_name=
# EC90C5C6-C982-4F49-8D46-A64F270C5247%40gmail.com&forum_name=matplotlib-users
# (easier to read at http://old.nabble.com/%22Piecewise-Cubic-Hermite-Interpolating-
# Polynomial%22-in-python-td25204843.html)
# with help from:
# F N Fritsch & R E Carlson (1980) 'Monotone Piecewise Cubic Interpolation',
# SIAM Journal of Numerical Analysis 17(2), 238 - 246.
# http://en.wikipedia.org/wiki/Monotone_cubic_interpolation
# http://en.wikipedia.org/wiki/Cubic_Hermite_spline
constructor: (x, y) ->
n = x.length
delta = []; m = []; alpha = []; beta = []; dist = []; tau = []
for i in [0...(n - 1)]
delta[i] = (y[i + 1] - y[i]) / (x[i + 1] - x[i])
m[i] = (delta[i - 1] + delta[i]) / 2 if i > 0
m[0] = delta[0]
m[n - 1] = delta[n - 2]
to_fix = []
for i in [0...(n - 1)]
to_fix.push(i) if delta[i] == 0
for i in to_fix
m[i] = m[i + 1] = 0
for i in [0...(n - 1)]
alpha[i] = m[i] / delta[i]
beta[i] = m[i + 1] / delta[i]
dist[i] = Math.pow(alpha[i], 2) + Math.pow(beta[i], 2)
tau[i] = 3 / Math.sqrt(dist[i])
to_fix = []
for i in [0...(n - 1)]
to_fix.push(i) if dist[i] > 9
for i in to_fix
m[i] = tau[i] * alpha[i] * delta[i]
m[i + 1] = tau[i] * beta[i] * delta[i]
#x = x[0...n] # copy
#y = y[0...n] # copy
#m = m
interpolate: (x) ->
for i in [(#x.length - 2)..0]
break if #x[i] <= x
h = #x[i + 1] - #x[i]
t = (x - #x[i]) / h
t2 = Math.pow(t, 2)
t3 = Math.pow(t, 3)
h00 = 2 * t3 - 3 * t2 + 1
h10 = t3 - 2 * t2 + t
h01 = -2 * t3 + 3 * t2
h11 = t3 - t2
y = h00 * #y[i] +
h10 * h * #m[i] +
h01 * #y[i + 1] +
h11 * h * #m[i + 1]
y