java 需要帮助解决一些已知值的最小稀疏线性

brccelvz 于 2024-01-05 发布在 Java

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我有下面描述的问题
x1c 0d1x的数据
我需要找到x1'，x2'，x3 '，x4'，x5'的值，
（x1-x1'）^2+（x2-x2'）^2+（x3-x3 '）^2+（x4-x4'）^2+（x5-x5 '）^2 =最小值
和
x1' + x2' + x3' + x4' + x5' = 1
x1+ x2 + x3 + x4 + x5 = 1
注：我们知道a，B，c，d，e，x1，x2，x3，x4，x5的值
在这种情况下，有人会帮助我吗？
我尝试了google/or-tools库，但无法添加条件来查找最小值。

MPSolver solver = createSolver(solverType);
    double infinity = MPSolver.infinity();
    MPVariable x1 = solver.makeNumVar(0.0, infinity, "x1");
    MPVariable x2 = solver.makeNumVar(0.0, infinity, "x2");
    MPVariable x3 = solver.makeNumVar(0.0, infinity, "x3");
    MPVariable x4 = solver.makeNumVar(0.0, infinity, "x4");
    MPVariable x5 = solver.makeNumVar(0.0, infinity, "x5");
    // 0.15 <= x1 <= 0.35
    MPConstraint c1 = solver.makeConstraint(-infinity, 0.35);
    c1.setCoefficient(x1, 1);   
    MPConstraint c2 = solver.makeConstraint(0.15, infinity);
    c2.setCoefficient(x1, 1);
    // 0.1 <= x2 <= 0.3
    MPConstraint c3 = solver.makeConstraint(-infinity, 0.3);
    c3.setCoefficient(x2, 1);   
    MPConstraint c4 = solver.makeConstraint(0.1, infinity);
    c4.setCoefficient(x2, 1);
    // 0.0 <= x3 <= 0.2
    MPConstraint c5 = solver.makeConstraint(-infinity, 0.2);
    c5.setCoefficient(x3, 1);   
    MPConstraint c6 = solver.makeConstraint(0.0, infinity);
    c6.setCoefficient(x3, 1);
    // 0.15 <= x4 <= 0.35
    MPConstraint c7 = solver.makeConstraint(-infinity, 0.35);
    c7.setCoefficient(x4, 1);   
    MPConstraint c8 = solver.makeConstraint(0.15, infinity);
    c8.setCoefficient(x4, 1);
    // 0.1 <= x5 <= 0.3
    MPConstraint c9 = solver.makeConstraint(-infinity, 0.3);
    c9.setCoefficient(x5, 1);   
    MPConstraint c10 = solver.makeConstraint(0.1, infinity);
    c10.setCoefficient(x5, 1);
    // x1 + x2 + x3 + x4 + x5 = 1
    MPConstraint c11 = solver.makeConstraint(-infinity, 1.0);
    c11.setCoefficient(x1, 1);
    c11.setCoefficient(x2, 1);
    c11.setCoefficient(x3, 1);
    c11.setCoefficient(x4, 1);
    c11.setCoefficient(x5, 1);
    MPConstraint c12 = solver.makeConstraint(1.0, infinity);
    c12.setCoefficient(x1, 1);
    c12.setCoefficient(x2, 1);
    c12.setCoefficient(x3, 1);
    c12.setCoefficient(x4, 1);
    c12.setCoefficient(x5, 1);
    MPObjective objective = solver.objective();
    objective.setCoefficient(x1, 1);
    objective.setCoefficient(x2, 1);
    objective.setCoefficient(x3, 1);
    objective.setCoefficient(x4, 1);
    objective.setCoefficient(x5, 1);
    objective.setMinimization();

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Java

来源：https://stackoverflow.com/questions/37175905/need-to-help-solving-least-sparse-linear-with-some-known-values

2条答案

按热度按时间

4urapxun1#

这是一个带有凸目标函数的基本约束优化问题。https://en.wikipedia.org/wiki/Constrained_optimization有很多软件可以帮助你做到这一点。
http://cvxopt.org/documentation/index.html的

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vwkv1x7d2#

我想为Long Duong先生感谢你。
下面是他的解决方案，帮助我解决我的问题：

#@author: Long Duong
#Using this : http://cvxopt.org/userguide/coneprog.html#quadratic-programming
#Need to install cvxopt using (pip install cvxopt --user)
from cvxopt import matrix, solvers
# Need to MODIFY the value here 
# Hold the value of x1,x2,x3,x4,x5 
xi = matrix([0.5,0.6,0.7,0.8,0.9])
# Hold the value for a,b,c,d,e 
cons = matrix([0.2,0.2,0.2,0.2,0.2])
### Main part #### 
# Ensure the contrain: x1' + x2' + x3' + x4' + x5' = 1
A = matrix([1.0,1.0,1.0,1.0,1.0], (1,5)) 
b = matrix(1.0)
# Ensure the contrain:  cons[i] -0.1 < x'[i] < cons[i] + 0.1
G = matrix([[1.0,0.0,0.0,0.0,0.0],
        [-1.0,0.0,0.0,0.0,0.0],
        [0.0,1.0,0.0,0.0,0.0],
        [0.0,-1.0,0.0,0.0,0.0],
        [0.0,0.0,1.0,0.0,0.0],
        [0.0,0.0,-1.0,0.0,0.0],
        [0.0,0.0,0.0,1.0,0.0],
        [0.0,0.0,0.0,-1.0,0.0],
        [0.0,0.0,0.0,0.0,1.0],
        [0.0,0.0,0.0,0.0,-1.0]]).T
temp = []
for i in range(5):
    temp.append(cons[i] + 0.1)
    temp.append(-1 * (cons[i] - 0.1))
h = matrix(temp)
# Now need to solve the main function to minimize sum((x'[i]-xi[i])^2)
# P is kind of identity matrix since (x-a)^2 = x^2 - 2ax + a^2  
P = 2 * matrix([[1.0,0.0,0.0,0.0,0.0],
           [0.0,1.0,0.0,0.0,0.0],
           [0.0,0.0,1.0,0.0,0.0],
           [0.0,0.0,0.0,1.0,0.0],
           [0.0,0.0,0.0,0.0,1.0]])
q = -2 * xi #  
# All done 
sol=solvers.qp(P, q, G, h, A, b)
print "[RESULT] :"
print sol['x']

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java 需要帮助解决一些已知值的最小稀疏线性

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