I am trying to put together a simple code without any python libraries.
the aim is to find the centerline of a rectangle (for the longest side specifically).
This is my code so far. (it is looking for the center coordinates).
I want to be able to fine the centerline nonmatter the orientation of the rectangle. how do you recommend I do it. I'm open to any advice.
import matplotlib.pyplot as plt
x = [4,4,-4,-4]
y = [8,-5,-5,8]
x1,y1 = x[0],y[0]
x2,y2 = x[1],y[1]
x3,y3 = x[2],y[2]
x4,y4 = x[3],y[3]
ch_y1 = y1-y2
ch_y2 = y1-y4
ch_x1 = x1-x2
ch_x2 = x1-x4
l1 = ((x1-x2)**2+(y1-y2)**2)**0.5
l4 = ((x1-x4)**2+(y1-y4)**2)**0.5
l2 = ((x2-x3)**2+(y2-y3)**2)**0.5
l3 = ((x3-x4)**2+(y3-y4)**2)**0.5
if l2 > l1:
if abs(1-l1/l3) < 0.3 :
if x1>x2:
x11 = x1 - l2/2
if x1<x2:
x11 = x1 + l2/2
if x1 == x2:
x11 = x1
if y1>y2:
y11 = y1 - l2/2
if y1<y2:
y11 = y1 + l2/2
if y1 == y2:
y11 = y1
if x2>x3:
x12 = x2 - l3/2
if x2<x3:
x12 = x2 + l3/2
if x2 == x3:
x12 = x2
if y2>y3:
y12 = y2 - l3/2
if y2<y3:
y12 = y2 + l3/2
if y2 == y3:
y12 = y2
if l1 >l2:
x1 = x1
x2 = x3
x3 = x4
y1 = y1
y2 = y3
y3 = y4
if abs(1-l2/l4) < 0.3 :
if x1>x2:
x11 = x1 - l2/2
if x1<x2:
x11 = x1 + l2/2
if x1 == x2:
x11 = x1
if y1>y2:
y11 = y1 - l2/2
if y1<y2:
y11 = y1 + l2/2
if y1 == y2:
y11 = y1
if x2>x3:
x12 = x2 - l4/2
if x2<x3:
x12 = x2 + l4/2
if x2 == x3:
x12 = x2
if y2>y3:
y12 = y2 - l4/2
if y2<y3:
y12 = y2 + l4/2
if y2 == y3:
y12 = y2
plt.plot(x,y)
plt.plot([x11,x12],[y11,y12])
That is way too complicated. You can just draw the lines going through the center of the segments like so (note that I added a copy of x[0] and y[0] as x[4] and y[4] for simplicity):
import matplotlib.pyplot as plt
x = [4,4,-4,-4]
y = [8,-5,-5,8]
x = x + [x[0]]
y = y + [y[0]]
plt.plot(x,y)
for i in range(2):
plt.plot([(x[i]+x[i+1])/2, (x[i+2]+x[i+3])/2], [(y[i]+y[i+1])/2, (y[i+2]+y[i+3])/2])
Related
If i have 2 z3 Ints for examaple x1 and x2, and a 2d array of numbers for example:
list = [[1,2],[12,13],[45,7]]
i need to right a rule so that x1 and x2 are any of the pairs of numbers in the list for example x1 would be 1 and x2 would be 2 or x1 is 12 and x2 is 13
im guessing it would be something like:
solver = Solver()
for i in range(o,len(list)):
solver.add(And((x1==list[i][0]),(x2==list[i][1])))
but this would obviously just always be unsat, so i need to right it so that x1 and x2 can be any of the pairs in the list. It's worth noting that the number of pairs in the list could be anything not just 3 pairs.
You're on the right track. Simply iterate and form the disjunction instead. Something like:
from z3 import *
list = [[1,2],[12,13],[45,7]]
s = Solver()
x1, x2 = Ints('x1 x2')
s.add(Or([And(x1 == p[0], x2 == p[1]) for p in list]))
while s.check() == sat:
m = s.model()
print("x1 = %2d, x2 = %2d" % (m[x1].as_long(), m[x2].as_long()))
s.add(Or(x1 != m[x1], x2 != m[x2]))
When run, this prints:
x1 = 1, x2 = 2
x1 = 12, x2 = 13
x1 = 45, x2 = 7
Let's say I have two ranges (x1,x2) and (z1,z2). What is fastest way to check if they intersect?
if
#case: (2,4), (3,6)
x1 > z1 < x2 and (z2 < x1 or z2 > x2)
or
#case: (2,4) , (1,3)
x1 > z2 < x2 and (z1 < x1 or z1 > x2)
or
#case: (3,6), (2,4)
............. reverse(1) x <--> z
or
#case: (1,3), (2,4)
............. reverse(2) x <--> z
BTW containing one range in the other is OK, for example
(2,5) , (3,4)
(2,5) , (3,5)
(2,5) , (2,4)
(this makes the question not a duplicate. In addition the conditions are <, > rather than >=, <=. And one more I expect input ranges to be also unordered, for example (x2,x1),(z2,z1) ; (z1,z2),(x1,x2) )
But this, for example, does not count as intersection:
(2,5), (2,5)
What is a faster way with less checks?
NOT a duplicate ... here is the solution :
def olap(x1,x2,z1,z2):
if x1 == x2 or z1 == z2 : return True
if x1 == z1 and x2 == z2 : return True
if x1 > x2 : x1,x2 = x2,x1
if z1 > z2 : z1,z2 = z2,z1
if x1 > z1 or x2 > z2 : z1,z2,x1,x2 = x1,x2,z1,z2
return x1 < z1 < x2 and (z2 < x1 or z2 > x2)
Simplest check is to verify a property of intersecting intervals: None of the leftmost endpoints (smaller) can be higher than a rightmost endpoint (higher).
x1 x2
a ------------ if x1 > y2: a would be to the right of b
b ---------------- if y1 > x2: b would be to the right of a
y1 y2
def overlap(x1, x2, y1, y2):
""" Overlap means neither of the intervals
is "to the right or left" of the other.
"""
return x1 <= y2 and y1 <= x2
I am trying to use http://www.philipkalinda.com/ds9.html to set up a constrained optimisation.
prob = pulp.LpProblem('FantasyTeam', pulp.LpMaximize)
decision_variables = []
res = self.team_df
# Set up the LP
for rownum, row in res.iterrows():
variable = str('x' + str(rownum))
variable = pulp.LpVariable(str(variable), lowBound = 0, upBound = 1, cat= 'Integer') #make variables binary
decision_variables.append(variable)
print ("Total number of decision_variables: " + str(len(decision_variables)))
total_points = ""
for rownum, row in res.iterrows():
for i, player in enumerate(decision_variables):
if rownum == i:
formula = row['TotalPoint']* player
total_points += formula
prob += total_points
print ("Optimization function: " + str(total_points))
The above, however, creates an optimisation where if points earned by x1 = X1, x2=X2.... and xn=Xn, it maximises
x1*X1 + x2*X2 +..... + xn*XN. Here xi is the points earned by the XI variable. However, in my case, I need to double the points for the variable that earns the most points. How do I set this up?
Maximize
OBJ: 38.1 x0 + 52.5 x1 + 31.3 x10 + 7.8 x11 + 42.7 x12 + 42.3 x13 + 4.7 x14
+ 49.5 x15 + 21.2 x16 + 11.8 x17 + 1.4 x18 + 3.2 x2 + 20.8 x3 + 1.2 x4
+ 24 x5 + 25.9 x6 + 27.8 x7 + 6.2 x8 + 41 x9
When I maximise the sum x1 gets dropped but when I maximise with the top guy taking double points, it should be there
Here are the constraints I am using:-
Subject To
_C1: 10.5 x0 + 21.5 x1 + 17 x10 + 7.5 x11 + 11.5 x12 + 12 x13 + 7 x14 + 19 x15
+ 10.5 x16 + 5.5 x17 + 6.5 x18 + 6.5 x2 + 9.5 x3 + 9 x4 + 12 x5 + 12 x6
+ 9.5 x7 + 7 x8 + 14 x9 <= 100
_C10: x12 + x2 + x6 >= 1
_C11: x10 + x11 + x17 + x3 <= 4
_C12: x10 + x11 + x17 + x3 >= 1
_C13: x0 + x10 + x11 + x12 + x13 + x14 + x15 + x18 + x2 <= 5
_C14: x0 + x10 + x11 + x12 + x13 + x14 + x15 + x18 + x2 >= 3
_C15: x1 + x16 + x17 + x3 + x4 + x5 + x6 + x7 + x8 + x9 <= 5
_C16: x1 + x16 + x17 + x3 + x4 + x5 + x6 + x7 + x8 + x9 >= 3
_C2: x0 + x1 + x10 + x11 + x12 + x13 + x14 + x15 + x16 + x17 + x18 + x2 + x3
+ x4 + x5 + x6 + x7 + x8 + x9 = 8
_C3: x0 + x14 + x16 + x5 <= 4
_C4: x0 + x14 + x16 + x5 >= 1
_C5: x15 + x18 + x4 + x7 + x8 <= 4
_C6: x15 + x18 + x4 + x7 + x8 >= 1
_C7: x1 + x13 + x9 <= 4
_C8: x1 + x13 + x9 >= 1
_C9: x12 + x2 + x6 <= 4
Naturally, maximising A + B + C + D doesn't maximise max(2A+B+C+D, A+2B+C+D, A+B+2C+D, A+B+C+2D)
I'm going to answer the question I think you're asking and you can correct me if I'm wrong. My understanding of your question is:
I have a series of binary variables x0...xN, and if a variable is included is receives some points. If it is not included it receives no points.
There are some constraints which apply to the selection
If (and only if) a variable is selected and if (and only if) it is the chosen variable which receives the highest number of points, then that particular variable gets double points
The objective is to maximize the total points including the doubling of the highest scoring one.
Assuming that's your question here is a dummy example that does that. Basically we add an auxiliary binary variable for each variable which is true iff (if and only if) that variable scores the most number of points:
from pulp import *
n_vars = 4
idxs = range(n_vars)
points = [2.0, 3.0, 4.0, 5.0]
prob = pulp.LpProblem('FantasyTeam', pulp.LpMaximize)
# Variables
x = LpVariable.dicts('x', idxs, cat='Binary')
x_highest_score = LpVariable.dicts('x_highest_score', idxs, cat='Binary')
# Objective
prob += lpSum([points[i]*(x[i] + x_highest_score[i]) for i in idxs])
# Constraints
# Exactly one item has highest score:
prob += lpSum([x_highest_score[i] for i in idxs]) == 1
# If a score is to be highest, it has to be chosen
for i in idxs:
prob += x_highest_score[i] <= x[i]
# And some selection constraints:
prob += x[0] + x[1] + x[2] + 1.5*x[3] <= 3
prob += x[0] + x[2] + 3*x[3] <= 3
prob += x[0] + x[1] + x[2] + 2*x[3] <= 3
# etc...
# Solve problem
prob.solve()
# Get soln
x_soln = [x[i].varValue for i in idxs]
x_highest_soln = [x_highest_score[i].varValue for i in idxs]
# And print the outputs
print (("Status: "), LpStatus[prob.status])
print ("Total points: ", value(prob.objective))
print ("x = ", x_soln)
print ("x_highest_soln = ", x_highest_soln)
This should return the following:
Status: Optimal
Total points: 13.0
x = [0.0, 1.0, 0.0, 1.0]
x_highest_soln = [0.0, 0.0, 0.0, 1.0]
If you turn off the double-points option, by changing the constraint to the following:
prob += lpSum([x_highest_score[i] for i in idxs]) == 1
I.E. none scores highest, you'll find a different set of choices is made.
I am using the following Integer Programming Model for solving the Two Dimensional Bin Packing Problem. The following model illustrates the one dimensional version. The code I have written incorporates the constraints for the additional dimension.
I am using Python PuLP for solving the optimization problem. The code is as follows :
from pulp import *
#knapsack problem
def knapsolve(item):
prob = LpProblem('BinPacking', LpMinimize)
ys = [LpVariable("y{0}".format(i+1), cat="Binary") for i in range(item.bins)]
xs = [LpVariable("x{0}{1}".format(i+1, j+1), cat="Binary")
for i in range(item.items) for j in range(item.bins)]
#minimize objective
nbins = sum(ys)
prob += nbins
print(nbins)
#constraints
t = nbins >= 1
print(t)
prob += t
for i in range(item.items):
con1 = sum(xs[(i + j*item.bins)] for j in range(item.bins))
t = con1 == 1
prob += t
print(t)
for k in range(item.bins):
x = xs[k*item.bins : (k+1)*item.bins]
con1 = sum([x1*w for x1, w in zip(x, item.itemweight)])
t = con1 <= item.binweight[k] * ys[k]
#t = con1 <= item.binweight[k]
prob += t
print(t)
for k in range(item.bins):
x = xs[k*item.bins : (k+1)*item.bins]
con1 = sum([x1*w for x1, w in zip(x, item.itemheight)])
t = con1 <= item.binheight[k] * ys[k]
#t = con1 <= item.binheight[k]
prob += t
print(t)
status = prob.solve()
print(LpStatus[status])
print("Objective value:", value(prob.objective))
print ('\nThe values of the variables : \n')
for v in prob.variables():
print(v.name, "=", v.varValue)
return
class Item:
#bins, binweight, items, weight, itemheight, binheight
bins = 5
items = 5
binweight = [2,3,2,5,3]
itemweight = [1,2,2,1,3]
itemheight = [2,1,4,5,3]
binheight = [4,9,10,8,10]
item = Item()
knapsolve(item)
It produces the following output :
y1 + y2 + y3 + y4 + y5
y1 + y2 + y3 + y4 + y5 >= 1
x11 + x21 + x31 + x41 + x51 = 1
x12 + x22 + x32 + x42 + x52 = 1
x13 + x23 + x33 + x43 + x53 = 1
x14 + x24 + x34 + x44 + x54 = 1
x15 + x25 + x35 + x45 + x55 = 1
x11 + 2*x12 + 2*x13 + x14 + 3*x15 - 2*y1 <= 0
x21 + 2*x22 + 2*x23 + x24 + 3*x25 - 3*y2 <= 0
x31 + 2*x32 + 2*x33 + x34 + 3*x35 - 2*y3 <= 0
x41 + 2*x42 + 2*x43 + x44 + 3*x45 - 5*y4 <= 0
x51 + 2*x52 + 2*x53 + x54 + 3*x55 - 3*y5 <= 0
2*x11 + x12 + 4*x13 + 5*x14 + 3*x15 - 4*y1 <= 0
2*x21 + x22 + 4*x23 + 5*x24 + 3*x25 - 9*y2 <= 0
2*x31 + x32 + 4*x33 + 5*x34 + 3*x35 - 10*y3 <= 0
2*x41 + x42 + 4*x43 + 5*x44 + 3*x45 - 8*y4 <= 0
2*x51 + x52 + 4*x53 + 5*x54 + 3*x55 - 10*y5 <= 0
Optimal
Objective value: 3.0
The values of the variables :
x11 = 0.0
x12 = 0.0
x13 = 0.0
x14 = 0.0
x15 = 0.0
x21 = 0.0
x22 = 0.0
x23 = 1.0
x24 = 0.0
x25 = 0.0
x31 = 0.0
x32 = 0.0
x33 = 0.0
x34 = 0.0
x35 = 0.0
x41 = 0.0
x42 = 1.0
x43 = 0.0
x44 = 0.0
x45 = 1.0
x51 = 1.0
x52 = 0.0
x53 = 0.0
x54 = 1.0
x55 = 0.0
y1 = 0.0
y2 = 1.0
y3 = 0.0
y4 = 1.0
y5 = 1.0
The sample input data that has been hard coded, should produce 1 bin as the output, that is one y variable should have the value 1. However this is not the case. Are the equations modeled properly? Is there another way to specify the constraints?
The mathematical model for the standard bin-packing problem uses x(bins,items) while in the Python model you seem to use a mix of of x(bins,items) and x(items,bins). The assignment to xs uses x(items,bins) but the construct xs[(i + j*item.bins)] implies x(bins,items). This is easily seen by inspecting the output: x11 + x21 + x31 + x41 + x51 = 1 which indicates x(bins,items). This type of modeling with explicit index calculations is rather unreliable in practice. This is a toy model, but for real models the lack of type checking can be very dangerous.
Different bin weights and heights should be no problem.
Also given your data
binweight = [2,3,2,5,3]
itemweight = [1,2,2,1,3]
itemheight = [2,1,4,5,3]
binheight = [4,9,10,8,10]
I don't believe this can be handled by just 1 bin as you claim. You need 3 bins for this (bins 2,4 and 5). (You are lucky here because although there are actually bugs in the Python code you get good solutions).
I'd like to minimize a set of equations where the variables are known with their uncertainties. In essence I'd like to test the hypothesis that the given measured variables conform to the formula constraints given by the equations. This seems like something I should be able to do with scipy-optimize. For example I have three equations:
8 = 0.5 * x1 + 1.0 * x2 + 1.5 * x3 + 2.0 * x4
4 = 0.0 * x1 + 0.0 * x2 + 1.0 * x3 + 1.0 * x4
1 = 1.0 * x1 + 1.0 * x2 + 0.0 * x3 + 0.0 * x4
And four measured unknowns with their 1-sigma uncertainty:
x1 = 0.246 ± 0.007
x2 = 0.749 ± 0.010
x3 = 1.738 ± 0.009
x4 = 2.248 ± 0.007
Looking for any pointers in the right direction.
This is my approach. Assuming x1-x4 are approximately normally distributed around each mean (1-sigma uncertainty), the problem is turning into one of minimizing the sum of square of errors, with 3 linear constrain functions. Therefore, we can attack it using scipy.optimize.fmin_slsqp()
In [19]:
def eq_f1(x):
return (x*np.array([0.5, 1.0, 1.5, 2.0])).sum()-8
def eq_f2(x):
return (x*np.array([0.0, 0.0, 1.0, 1.0])).sum()-4
def eq_f3(x):
return (x*np.array([1.0, 1.0, 0.0, 0.0])).sum()-1
def error_f(x):
error=(x-np.array([0.246, 0.749, 1.738, 2.248]))/np.array([0.007, 0.010, 0.009, 0.007])
return (error*error).sum()
In [20]:
so.fmin_slsqp(error_f, np.array([0.246, 0.749, 1.738, 2.248]), eqcons=[eq_f1, eq_f2, eq_f3])
Optimization terminated successfully. (Exit mode 0)
Current function value: 2.17576389592
Iterations: 4
Function evaluations: 32
Gradient evaluations: 4
Out[20]:
array([ 0.25056582, 0.74943418, 1.74943418, 2.25056582])
I appear to me that I have a very similar problem. I am relatively new to py and used it mostly to sort and reduce data with pandas.
I have a set of linear equations, where I want to find the best fit parameters. However, the dataset has known uncertainties that need to be considered given in parentheses).
x1*99(1)+x2*45(1)=52(0.2)
x1*1(0.5)+x2*16(1)=15(0.1)
Moreover there are constraints:
x1>=0
x2>=0
x1+x2=1
My approach would be to treat the equations as constraints and solve the sum of the residues as it has been shown in the example above.
Solving this without uncertainties is not the issue. I ask to get a hint on how to account for the uncertainties while finding the best fit parameters.
As given, the problem has no solution. This is because if the inputs x1, x2, x3 and x4 are gaussian, then the outputs:
y1 = 0.5 * x1 + 1.0 * x2 + 1.5 * x3 + 2.0 * x4 - 8.0
y2 = 0.0 * x1 + 0.0 * x2 + 1.0 * x3 + 1.0 * x4 - 4.0
y3 = 1.0 * x1 + 1.0 * x2 + 0.0 * x3 + 0.0 * x4 - 1.0
are also gaussian.
Assuming that x1, x2, x3 and x4 are independent random variables, this is easy to see with OpenTURNS:
import openturns as ot
x1 = ot.Normal(0.246, 0.007)
x2 = ot.Normal(0.749, 0.010)
x3 = ot.Normal(1.738, 0.009)
x4 = ot.Normal(2.248, 0.007)
y1 = 0.5 * x1 + 1.0 * x2 + 1.5 * x3 + 2.0 * x4 - 8.0
y2 = 0.0 * x1 + 0.0 * x2 + 1.0 * x3 + 1.0 * x4 - 4.0
y3 = 1.0 * x1 + 1.0 * x2 + 0.0 * x3 + 0.0 * x4 - 1.0
The following script produces the graph:
graph1 = y1.drawPDF()
graph1.setLegends(["y1"])
graph2 = y2.drawPDF()
graph2.setLegends(["y2"])
graph3 = y3.drawPDF()
graph3.setLegends(["y3"])
graph1.add(graph2)
graph1.add(graph3)
graph1.setColors(["dodgerblue3",
"darkorange1",
"forestgreen"])
graph1.setXTitle("Y")
The previous script produces the following output.
Given the location of the 0.0 in this distribution, I would say that solving the equations is mathematically impossible, but physically consistent with the data.
Actually, I guess that the gaussian distributions you gave for x1, ..., x4 are estimated from data. So I would rather reformulate the problem as follows:
Given a sample of observed values of x1, x2, x3, x4, what is the value of e1, e2, e3 which is so that :
y1 = 0.5 * x1 + 1.0 * x2 + 1.5 * x3 + 2.0 * x4 - 8 + e1 = 0
y2 = 0.0 * x1 + 0.0 * x2 + 1.0 * x3 + 1.0 * x4 - 4 + e2 = 0
y3 = 1.0 * x1 + 1.0 * x2 + 0.0 * x3 + 0.0 * x4 - 1 + e3 = 0
This turns the problem into an inversion problem, which can be solved by calibrating e1, e2, e3. Furthermore, given the finite sample size of x1, ..., x4, we might want to produce the distribution of e1, e2, e3. This can be done by bootstraping the input / output pairs (x, y): the distribution of e1, e2, e3 then reflects the variability of these parameters depending on the sample at hand.
First, we have to generate a sample from the distribution (I suppose that you have this sample, but did not publish it so far):
distribution = ot.ComposedDistribution([x1, x2, x3, x4])
sampleSize = 10
xobs = distribution.getSample(sampleSize)
Then we define the model:
formulas = [
"y1 := 0.5 * x1 + 1.0 * x2 + 1.5 * x3 + 2.0 * x4 + e1 - 8.0",
"y2 := 0.0 * x1 + 0.0 * x2 + 1.0 * x3 + 1.0 * x4 + e2 - 4.0",
"y3 := 1.0 * x1 + 1.0 * x2 + 0.0 * x3 + 0.0 * x4 + e3 - 1.0"
]
program = ";".join(formulas)
g = ot.SymbolicFunction(["x1", "x2", "x3", "x4", "e1", "e2", "e3"],
["y1", "y2", "y3"],
program)
And set the observed outputs, which is a sample of zeros:
yobs = ot.Sample(sampleSize, 3)
We start with initial values equal to zero, and define the function to calibrate:
e1Initial = 0.0
e2Initial = 0.0
e3Initial = 0.0
thetaPrior = ot.Point([e1Initial,e2Initial,e3Initial])
calibratedIndices = [4, 5, 6]
mycf = ot.ParametricFunction(g, calibratedIndices, thetaPrior)
Then we can calibrate the model:
algo = ot.NonLinearLeastSquaresCalibration(mycf, xobs, yobs, thetaPrior)
algo.run()
calibrationResult = algo.getResult()
print(calibrationResult.getParameterMAP())
This prints:
[0.0265988,0.0153057,0.00495758]
This means that the errors e1, e2, e3 are rather small.
We can compute a confidence interval:
thetaPosterior = calibrationResult.getParameterPosterior()
print(thetaPosterior.computeBilateralConfidenceIntervalWithMarginalProbability(0.95)[0])
This prints:
[0.0110046, 0.0404756]
[0.00921992, 0.0210059]
[-0.00601084, 0.0156665]
The third parameter e3 might be zero, but neither e1 nor e2.
Finally, we can get the distribution of the errors:
thetaPosterior = calibrationResult.getParameterPosterior()
and draw it:
graph1 = thetaPosterior.getMarginal(0).drawPDF()
graph2 = thetaPosterior.getMarginal(1).drawPDF()
graph3 = thetaPosterior.getMarginal(2).drawPDF()
graph1.add(graph2)
graph1.add(graph3)
graph1.setColors(["dodgerblue3",
"darkorange1",
"forestgreen"])
graph1
This produces:
This shows that e3 might be zero given the variability in the observed inputs x1, ..., x4. But e1 and e2 cannot be zero. The conclusion for this sample is that the third equation is approximately solved by the observed values of x1, ..., x4, but not the two first equations.