张麟的博客

一个sympy的例子

张麟 2010年11月03日星期三 09:19 | 2189次浏览 | 1条评论

it is from http://stochastix.wordpress.com/2008/07/05/playing-with-sympy/#comment-70270

An example

Let’s consider the system of polynomial equations in $\mathbb{R}[x_1, x_2, x_3]$

$\begin{array}{rl}x_1 + x_2 + x_3 = y_1 + y_2 + y_3 \\x_1^2 + x_2^2 + x_3^2 = y_1^2 + y_2^2 + y_3^2\\x_1^3 + x_2^3 + x_3^3 = y_1^3 + y_2^3 + y_3^3\end{array}$ ,

where $y = (y_1, y_2, y_3) \in \mathbb{R}^3$ is a known vector (whose entries are distinct). We will denote x = (x_1, x_2, x_3) . Let function $g_k: \mathbb{R}^3 \rightarrow \mathbb{R}$ be defined as

$g_k(z) = \displaystyle\sum_{i=1}^3 z_i^k$ .

The system of polynomial equations can thus be rewritten as

$\begin{array}{rl}p_1(x) = 0\\p_2(x) = 0\\p_3(x) = 0\end{array}$

where p_k(x) = g_k(x) - g_k(y) for all $k \in \{1,2,3\}$ . According to my conjecture , this system of equations should have exactly 3! = 6 solutions. Solving the system is not very easy because the equations p_k(x) = 0 are nonlinear in variables x_1, x_2, x_3 . A CAS would be useful.

For example, we could implement function $g_k: \mathbb{R}^3 \rightarrow \mathbb{R}$ with the following Python script:

def g(z,k):
    """Computes g_k(z), where z is a list of reals and k is a positive integer"""

    # checks function arguments for errors
    if len(z)==0:
        return "ERROR: First argument must be a non-empty list of symbols!"
    if (type(k) != int) or (k < 1):
        return "ERROR: Second argument must be a positive integer!"

# computes g_k(z) and returns it
    acc = 0
    for i in range(0,len(z)):
    acc += (z[i])**k
    return acc

We could then compute the Gröbner basis of the set of polynomials $\{p_1(x), p_2(x), p_3(x)\}$ , and the solutions of the system of polynomial equations for a given , say, y = (1,2,3) :

       from sympy import *

# number of equations and symbolic variables
n = 3

# defines set S = {1, 2,..., n}
S = range(1,n+1)

# declares symbolic variables
x = [Symbol('x%d' % i) for i in S]

# initializes y vector
#y = [Symbol('y%d' % i) for i in S]
y = S

# defines system of polynomial equations in variables [x1, x2, x3]
P = [g(x,k) - g(y,k) for k in S]

# computes Groebner basis and solutions of the system of polynomials
GB = groebner(P,x, order='lex')
Sols = solve_system(P,x)

# prints results
print "Groebner basis: %s" % GB
print "Solutions: %s" % Sols
print "There are %d solutions." % len(Sols)

The output is:

Gröbner basis: the Gröbner basis is a set of three polynomials $\{b_1(x), b_2(x), b_3(x)\}$ , where

$\begin{array}{l} b_1(x) = -6 + x_1 + x_2 + x_3\\ b_2(x) = 11 - 6 x_2 - 6 x_3 + x_2 x_3 + x_2^2 + x_3^2\\ b_3(x) = -6 + 11 x_3 - 6 x_3^2 + x_3^3\end{array}$ .

Solutions: there are solutions

$\{(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)\}$ .

As expected, the solutions of the system of polynomial equations are the 3! = 6 permutations of the elements of y = (1,2,3) . Of course, this does not prove the conjecture . All it proves is that my conjecture works for the particular case where n=3 and .

分享添加到桌面

回复张麟 2010年11月03日星期三 10:38

晕死，怎么不显示。

0条回复

一个sympy的例子

评论

回复 张麟 2010年11月03日 星期三 10:38

回复张麟 2010年11月03日星期三 10:38