我想求解一个矩形系统(解中具有任意参数)。如果失败,我想向矩阵添加行,直到它成为正方形。
print matrix_a
print vector_b
print len(matrix_a),len(matrix_a[0])
给出:
[[0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[1, 1, 1, 1, 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, 1, 1, 1, 1, 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, 1, 1, 1, 1, 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, 1, 1, 1, 1, 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, 1, 1, 1, 1, 1, 1]]
[2, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1]
11 26
我的完整代码位于 http://codepad.org/tSgstpYe
如你所见,我有一个系统 Ax=b。 我知道每个 x 值 x1,x2.. 必须为 1 或 0,并且我希望在这个限制下系统应该只有一个解决方案。
事实上我期望的答案是 x=[0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0 ,0,1,0,0,0,0]
我看了看 http://docs.scipy.org/doc/numpy/reference/ generated/numpy.linalg.solve.html#numpy.linalg.solve 但它似乎只需要方阵。
任何解决这个系统的帮助都会很棒!
这是一个简单的实现(具有硬编码阈值),但它提供了您正在寻找的测试数据解决方案。
它基于迭代重新加权最小二乘法。
from numpy import abs, diag, dot, zeros
from numpy.linalg import lstsq, inv, norm
def irls(A, b, p= 1):
"""Solve least squares problem min ||x||_p s.t. Ax= b."""
x, p, e= zeros((A.shape[1], 1)), p/ 2.- 1, 1.
xp= lstsq(A, b)[0]
for k in xrange(1000):
if e< 1e-6: break
Q= dot(diag(1./ (xp** 2+ e)** p), A.T)
x= dot(dot(Q, inv(dot(A, Q))), b)
x[abs(x)< 1e-1]= 0
if norm(x- xp)< 1e-2* e** .5: e*= 1e-1
xp= x
return k, x.round()
根据您期望的输入,使用简单的树搜索算法可能会更好。您的结果向量包含相对较低的数字,可以尽早切断大多数树枝。我尝试实现这个算法在 0.2 秒后产生了预期的结果:
def testSolution(a, b, x):
result = 0
for i in range(len(b)):
n = 0
for j in range(len(a[i])):
n += a[i][j] * x[j]
if n < b[i]:
result = -1
elif n > b[i]:
return 1
return result
def solve(a, b):
def solveStep(a, b, result, step):
if step >= len(result):
return False
result[step] = 1
areWeThere = testSolution(a, b, result)
if areWeThere == 0:
return True
elif areWeThere < 0 and solveStep(a, b, result, step + 1):
return True
result[step] = 0
return solveStep(a, b, result, step + 1)
result = map(lambda x: 0, range(len(a[0])))
if solveStep(a, b, result, 0):
return result
else:
return None
matrix_a = [[0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[1, 1, 1, 1, 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, 1, 1, 1, 1, 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, 1, 1, 1, 1, 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, 1, 1, 1, 1, 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, 1, 1, 1, 1, 1, 1]]
vector_b = [2, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1]
print solve(matrix_a, vector_b)
这必须使用您的输入仅测试 1325 个可能的向量,这比所有可能的结果(6700 万)要少得多。当然,最坏的情况仍然是 6700 万次测试。
设
Ax = b
为系统,A|b
为增广矩阵,则
有3种可能
rank(A) < rank(A|b)
rank(A) = rank(A|b) = n
rank(A) = rank(A|b) < n
其中
n
是未知数。
对于矩形矩阵,最好使用最小二乘法 - IRLS 看起来像正态方程 - 具有矩阵梯度下降:
import numpy as np
from numpy.linalg import lstsq, inv, norm
A = np.array([[0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0],
[0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0],
[1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0],
[1, 1, 1, 1, 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, 1, 1, 1, 1, 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, 1, 1, 1, 1, 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, 1, 1, 1, 1, 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, 1, 1, 1, 1, 1, 1]])
b = np.array([2, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1])
######################
##https://people.duke.edu/~ccc14/sta-663-2020/notebooks/S09F_Least_Squares_Optimization.html
# Matrix form of gradient
def grad_m(X, y, b):
return X.T@X@b- X.T@y
max_iter = 1000
a1 = 0.01 # learning rate
beta2 = np.zeros(26)
for i in range(max_iter):
beta2 -= a1 * grad_m(A, b, beta2)
print(np.round(beta2,0))
结果仍然
[ 0. 1. 0. -0. 0. 1. 0. 0. 0. 0. 0. 0. 0. 1. -0. 0. 0. 1. 0. -0. 0. 0. -0. 0. 0. 0.]
与你的不同。
也许,原因是你的线性系统不确定(许多x,更少的方程)-没有单一的解决方案,因此我认为不同的方法会导致不同的解决方案