x = matrix_NN.colPivHouseholderQr().solve(v_Nd);
时间: 2024-04-06 13:31:55 浏览: 15
这段代码是什么意思?
这段代码是使用列主元的 Householder QR 分解方法求解线性方程组 Ax = v,其中 matrix_NN 是一个 N × N 的矩阵,v_Nd 是一个 N 维列向量,x 是一个 N 维列向量。这个方法比较稳定,能够处理矩阵的秩不满的情况。函数 colPivHouseholderQr() 返回一个对象,它有一个 solve() 方法,用于求解线性方程组。
相关问题
clc syms a10,a11,a12,a13,a20,a21,a22,a23,a30,a31,a32,a33,tc1,tc2,tcf,theta1,theta2,theta3,theta4; equ1=a11+2*a12*tc1+3*a13*(tc1)^2-a21-2*a22*tc1-3*a23*(tc1)^2==0; equ2=2*a12+6*a13*tc1-2*a22-6*a23*tc1==0; equ3=a10==0; equ4=a11==0; equ5=a10+a11*tc1+a12*(tc1)^2+a13*(tc1)^3-theta1==0; equ6=a20+a21*tc1+a22*(tc1)^2+a23*(tc1)^3-theta1==0; equ7=a20+a21*tc2+a22*(tc2)^2+a23*(tc2)^3-theta2==0; equ8=a21+2*a22*tc2+3*a23*(tc2)^2-a31-2*a32*tc2-3*a33(tc2)^2==0; equ9=2*a22+6*a23*tc2-2*a32-6*a33*tc2==0; equ10=a30+a31*tc2+a32*(tc2)^2+a33*(tc2)^3-theta3==0; equ11=a30+a31*tcf+a32*(tcf)^2+a33*(tcf)^3-theta4==0; equ12=a31+2*a32*tcf+3*a33*(tcf)^2==0; bottom_t = solve([equ1,equ2,equ3,equ4,equ5,equ6,equ7,equ8,equ9,equ10,equ11,equ12], [a10,a11,a12,a13,a20,a21,a22,a23,a30,a31,a32,a33]); bottom_a10=bottom_t.a10 bottom_a11=bottom_t.a11 bottom_a12=bottom_t.a12 bottom_a13=bottom_t.a13 bottom_a20=bottom_t.a20 bottom_a21=bottom_t.a21 bottom_a22=bottom_t.a22 bottom_a23=bottom_t.a23 bottom_a30=bottom_t.a30 bottom_a31=bottom_t.a31 bottom_a32=bottom_t.a32 bottom_a33=bottom_t.a33 % simplify(bottom_a10) % simplify(bottom_t3)
这是一个关于多项式的求解问题,看起来是使用 MATLAB 中的符号计算工具箱来解决的。大致的步骤是建立方程组,然后使用 solve 函数求解。通过对方程组的求解,得到了一系列的解。
其中,bottom_t 是求解得到的解集,bottom_a10 到 bottom_a33 分别是解集中对应的变量的值。最后,代码中被注释掉的两行 simplify 函数可以对求解结果进行简化。
将以下代码转化为伪代码def get_bays(self): bays = {} for n in irange(1, self.N): conf = [[None] * self.T for _ in range(self.S)] for c in irange(n, self.C): stack = next(s - 1 for s in irange(1, self.S) if round(self.a[n, self.C + s, c].solution_value) == 1) tier = sum(round(self.a[n, d, c].solution_value) for d in irange(n, self.C)) conf[stack][tier] = c bays[n] = Bay(self.S, self.T, conf) return bays def get_n_relos(self): return self.model.objective_value def test(): conf = [[4], [3, 1], [2, 5, 6]] bay = Bay(3, 3, conf) print('bay') print(bay) crp_i = CRP_I(bay) if crp_i.model.solve(): print() print('n_relos = {}'.format(crp_i.get_n_relos())) bays = crp_i.get_bays() for n in irange(1, crp_i.N): print('n = {}'.format(n)) print(bays[n]) if __name__ == '__main__': test()
以下是伪代码的转化:
```
function get_bays():
bays = {}
for n in range(1, N):
conf = initialize 2D array with None of size S x T
for c in range(n, C):
stack = None
for s in range(1, S+1):
if round(a[n, C+s, c].solution_value) == 1:
stack = s-1
break
tier = 0
for d in range(n, C+1):
if round(a[n, d, c].solution_value) == 1:
tier += 1
conf[stack][tier] = c
bays[n] = Bay(S, T, conf)
return bays
function get_n_relos():
return model.objective_value
function test():
conf = [[4], [3, 1], [2, 5, 6]]
bay = new Bay(3, 3, conf)
print('bay')
print(bay)
crp_i = new CRP_I(bay)
if crp_i.model.solve():
print()
print('n_relos = {}'.format(crp_i.get_n_relos()))
bays = crp_i.get_bays()
for n in range(1, crp_i.N):
print('n = {}'.format(n))
print(bays[n])
if __name__ == '__main__':
test()
```