[英]finding the optimized location for a list of coordinates x and y
我是編程新手,尤其是python,但我正在努力學習它,到目前為止我發現它非常迷人。
我有一個30個固定坐標x和y的列表。
x = np.array([[13,10,12,13,11,12,11,13,12,13,14,15,15,16,18,2,3,4,6,9,1,3,6,7,8,10,12,11,10,30]])
y = np.array([[12,11,10,9,8,7,6,6,7,8,11,12,13,15,14,18,12,11,10,13,15,16,18,17,16,15,14,13,12,3]])
我想找到一個優化的(集中式)位置,通過找到最小距離可以連接最多10個固定坐標。
我厭倦了使用優化算法,但我只能得到一個坐標列表的最佳位置(即我不能將位置約束到10個坐標)。
任何幫助,將不勝感激。
def objective(x):
px = x[0]
py = x[1]
wdistance = sum(((px - wx)**2 + (py - wy)**2)**0.5)
tdistance = (((px - tx)**2 + (py - ty)**2)**0.5)
cost = wdistance * 2.5 + tdistance * 5
return cost
# initial estimate of the centralized location
x0 = [10,10]
# boundary values for the centralized locations
bnds = ((0,15), (0,15))
sol = minimize(objective, x0, bounds=bnds, method='SLSQP')
print(sol)
正如我的評論中所指出的,我認為這個問題是NP難的 ,因此解決它並不容易。
這是某種基數約束的問題,這意味着,通常很難/不可能使用連續優化(假設scipy.optimize中的幾乎所有內容)。
此外,最小化最小距離並不是特別平滑,幾乎所有內容都假設在scipy內。 但重新制定是可能的(至少在支持約束的情況下!)。
如果沒有基數約束(這個問題的離散性的核心),它就會崩潰成最小的封閉圓問題,很容易在線性時間內解決。 更一般的維基文章 。 但這對給定的問題沒有多大幫助。
這是一個演示,可能不適合膽小的人(取決於背景知識):
使用cvxpy作為建模工具
以下代碼將解決20個問題中的10個問題(前20個問題;為了更好的可視化,丟棄了一個復制品!)
它會因完整問題而失敗(很可能會返回近似值;但不是最佳值!)。 我責備ECOS_BB(但同時非常感謝Han Wang的代碼!)。 通過更好的替代方案,您的問題肯定會很容易解決。 但是不要指望任何解決方案能夠做到1000分:-)。
import numpy as np
import cvxpy as cvx
import matplotlib.pyplot as plt
from scipy.spatial.distance import pdist
""" (Reduced) Data """
# removed a duplicate point (for less strange visualization)
x = np.array([13,10,12,13,11,12,11,13,13,14,15,15,16,18,2,3,4,6,9,1])#,3,6,7,8,10,12,11,10,30])
y = np.array([12,11,10,9,8,7,6,6,8,11,12,13,15,14,18,12,11,10,13,15])#,16,18,17,16,15,14,13,12,3])
N = x.shape[0]
M = 10
""" Mixed-integer Second-order cone problem """
c = cvx.Variable(2) # center position to optimize
d = cvx.Variable(N) # helper-var: lower-bound on distance to center
d_prod = cvx.Variable(N) # helper-var: lower-bound on distance * binary/selected
b = cvx.Bool(N) # binary-variables used for >= M points
dists = pdist(np.vstack((x,y)).T)
U = np.amax(dists) # upper distance-bound (not tight!) for bigM-linearization
helper_expr_c_0 = cvx.vec(c[0] - x)
helper_expr_c_1 = cvx.vec(c[1] - y)
helper_expr_norm = cvx.norm(cvx.hstack(helper_expr_c_0, helper_expr_c_1), axis=1)
constraints = []
constraints.append(cvx.sum_entries(b) >= M) # M or more points covered
constraints.append(d >= helper_expr_norm) # lower-bound of distances
constraints.append(d_prod <= U * b) # linearization of product
constraints.append(d_prod >= 0) # """
constraints.append(d_prod <= d) # """
constraints.append(d_prod >= d - U * (1-b)) # """
objective = cvx.Minimize(cvx.max_entries(d_prod))
problem = cvx.Problem(objective, constraints)
problem.solve(verbose=False)
print(problem.status)
# Visualization
center = np.array(c.value.flat)
tmp = np.array(np.round(b.value.T.flat), dtype=int)
selected_points = np.where(tmp == 1)[0]
non_selected_points = np.where(tmp == 0)[0]
ax = plt.subplot(aspect='equal')
ax.set_title('Optimal solution radius: {0:.2f}'.format(problem.value))
ax.scatter(x[non_selected_points], y[non_selected_points], c='blue', s=70, alpha=0.9)
ax.scatter(x[selected_points], y[selected_points], c='magenta', s=70, alpha=0.9)
ax.scatter(c[0].value, c[1].value, c='red', s=70, alpha=0.9)
circ = plt.Circle((center[0], center[1]), radius=problem.value, color='g', fill=True, alpha=0.1)
ax.add_patch(circ)
plt.tight_layout()
plt.show()
optimal
編輯
我將上面的代碼移植到julia,以便能夠使用上面提到的求解器Pajarito。 盡管過去對我來說沒有太多的julia編程,但它現在正在運行:
這種方法正在解決您的完整問題(我再次刪除了重復點)以達到最佳狀態!
它輸出一些數值不穩定警告,但聲稱結果是最佳的! (Pajarito仍然是相當新的研究軟件,我沒有調整眾多選項;我們使用的開源解算器與商業替代品相比並不那么強大)
我對結果非常滿意,我很高興我試試了Julia / Convex.jl / Pajarito!
using Convex, Pajarito, ECOS, GLPKMathProgInterface
# DATA
x = [13 10 12 13 11 12 11 13 13 14 15 15 16 18 2 3 4 6 9 1 3 6 7 8 10 12 11 10 30]
y = [12 11 10 9 8 7 6 6 8 11 12 13 15 14 18 12 11 10 13 15 16 18 17 16 15 14 13 12 3]
N = size(x)[2]
M = 10
# MISOCP
c = Variable(2)
d = Variable(N)
d_prod = Variable(N)
b = Variable(N, :Bin)
U = 100 # TODO
# Objective
problem = minimize(maximum(d_prod))
# Constraints
problem.constraints += sum(b) >= M
problem.constraints += d_prod <= U*b
problem.constraints += d_prod >= 0
problem.constraints += d_prod <= d
problem.constraints += d_prod >= d - U * (1-b)
for i = 1:N
problem.constraints += d[i] >= norm([c[1] - x[i], c[2] - y[i]]) # ugly
end
# Solve
mip_solver_drives = false
log_level = 2
rel_gap = 1e-8
mip_solver = GLPKSolverMIP()
cont_solver = ECOSSolver(verbose=false)
solver = PajaritoSolver(
mip_solver_drives=mip_solver_drives,
log_level=log_level,
rel_gap=rel_gap,
mip_solver=mip_solver,
cont_solver=cont_solver,
init_exp=true,
)
# option taken from https://github.com/JuliaOpt/Pajarito.jl/blob/master/examples/gatesizing.jl
# not necessarily optimal
solve!(problem, solver)
# Print out results
println(problem.status)
println(problem.optval)
println(b.value)
println(c.value)
Problem dimensions:
variables | 120
constraints | 263
nonzeros in A | 466
Cones summary:
Cone | Count | Min dim. | Max dim.
Second order | 29 | 3 | 3
Variable types:
continuous | 91
binary | 29
Transforming data... 1.10s
Creating conic subproblem... 0.52s
Building MIP model... 2.35s
Solving conic relaxation... 1.38s
Starting iterative algorithm
0 | +Inf | +3.174473e-11 | Inf | 6.434e+00
Warning: numerical instability (primal simplex, phase I)
Iter. | Best feasible | Best bound | Rel. gap | Time (s)
1 | +2.713537e+00 | +2.648653e+00 | 2.391e-02 | 1.192e+01
Warning: numerical instability (primal simplex, phase I)
Warning: numerical instability (primal simplex, phase I)
Warning: numerical instability (dual simplex, phase II)
... some more ...
2 | +2.713537e+00 | +2.713537e+00 | 5.134e-12 | 1.341e+01
Iterative algorithm summary:
- Status = Optimal
- Best feasible = +2.713537e+00
- Best bound = +2.713537e+00
- Relative opt. gap = 5.134e-12
- Total time (s) = 1.34e+01
Outer-approximation cuts added:
Cone | Relax. | Violated | Nonviol.
Second order | 58 | 32 | 24
0 numerically unstable cone duals encountered
Distance to feasibility (negative indicates strict feasibility):
Cone | Variable | Constraint
Linear | NA | 5.26e-13
Second order | NA | 2.20e-11
Distance to integrality of integer/binary variables:
binary | 0.00e+00
Optimal
2.7135366682753825
[1.0; 1.0; 1.0; 1.0; 0.0; 0.0; 0.0; 0.0; 0.0; 1.0; 1.0; 1.0; 0.0; 0.0; 0.0; 0.0; 0.0; 0.0; 0.0; 0.0; 0.0; 0.0; 0.0; 0.0; 0.0; 1.0; 1.0; 1.0; 0.0]
[12.625; 11.6875]
在所有點中還有一個選擇-23的情節:
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