基于符号的Sympy与平凡解的方程组的符号解

Question

我有一个如下的方程式系统:

对于这个特定的系统,我知道只有当p1 == p2时,才存在一个非平凡的解决方案

.

但是,如何使用Sympy在一般情况下确定这一点？

对于此示例,我的实现如下:

from sympy import Matrix, symbols, pprint, lcm, latex
from sympy.solvers import solve_linear_system

top_matrix = Matrix.zeros(8,7)
p1 = symbols("p1")
p2 = symbols("p2")

top_matrix[0,0] = 1
top_matrix[0,1] = -1

top_matrix[1,1] = (1-p1)
top_matrix[1,2] = -1

top_matrix[2,2] = 1
top_matrix[2,4] = p2-1

top_matrix[3,1] = p1
top_matrix[3,3] = -1

top_matrix[4,3] = 1
top_matrix[4,4] = -p2

top_matrix[5,4] = 1
top_matrix[5,5] = -1

top_matrix[6,1] = -1
top_matrix[6,6] = 1

top_matrix[7,4] = -1
top_matrix[7,6] = 1

pprint(top_matrix)
vars = symbols("a1, a2, a3, a4, a5, a6, a7, a8")
print solve_linear_system(top_matrix, *vars)

结果是

None

如果我订

p2 = p1

结果是

{a1: -1, a5: -1, a2: -1, a6: -1, a3: p1 - 1, a4: -p1}

有没有办法自动发现这个要求？

Answer 1

在您的示例代码中，solve_linear_system需要一个增强系统，即，如果右侧为零，则矩阵应声明为Matrix.zeros(8,8)。通过此修改，您的代码将产生

{a3: 0, a1: 0, a5: 0, a7: 0, a6: 0, a2: 0, a4: 0}

这确实是一个解决方案，尽管不是有趣的......

为了解决这个问题，可以明确请求将解决方案的一个组件标准化为 1。因此，如果您执行以下操作：

from sympy import Matrix, symbols, pprint, lcm, latex, solve

top_matrix = Matrix.zeros(8,7)
p1,p2 = symbols("p1, p2")

top_matrix[0,0] = 1
top_matrix[0,1] = -1

top_matrix[1,1] = (1-p1)
top_matrix[1,2] = -1

top_matrix[2,2] = 1
top_matrix[2,4] = p2-1

top_matrix[3,1] = p1
top_matrix[3,3] = -1

top_matrix[4,3] = 1
top_matrix[4,4] = -p2

top_matrix[5,4] = 1
top_matrix[5,5] = -1

top_matrix[6,1] = -1
top_matrix[6,6] = -1

top_matrix[7,4] = 1
top_matrix[7,6] = 1

pprint(top_matrix)

a1,a2,a3,a4,a5,a6,a7 = list(symbols("a1, a2, a3, a4, a5, a6, a7"))

B = Matrix([[1],[a2],[a3],[a4],[a5],[a6],[a7]])

C = top_matrix * B

print(solve(C, (a2,a3,a4,a5,a6,a7,p1,p2)))

并求解剩余变量以及参数p1,p2，结果为：

[{a2: 1, a7: -1, a4: p2, a6: 1, a5: 1, p1: p2, a3: -p2 + 1}]

这确实是所需的解决方案。