确定图的色多项式的问题
对于作业图论，我被要求确定下图的色多项式
对于 < strong>色多项式分解定理。如果G=(V,E)，则为连通图，e属于E，
P (G, λ) = P (Ge, λ) -P(Ge', λ)
其中Ge表示从G中删除de边e得到的子图(Ge=Ge)，Ge'是通过识别顶点{a,b得到的子图} = e
当计算色多项式时，我应在图周围放置括号以指示其色多项式。去除原图的任意一条边，通过分解的方法计算色多项式。
P (G, λ) = P (Ge, λ)-P (Ge', λ) = λ (λ-1)^4 - [λ(λ-1)*(λ^2 - 3λ + 3)]
但是答案键和老师的回应是：
P (G, λ) = λ (λ-1)(λ-2)(λ^2-2λ-2)
我已经对多项式进行了运算，但我不能达到我问的解决方案..我做错了什么？
for a homework graph theory, I'm asked to determine the chromatic polynomial of the following graph
For the Descomposition Theorem of Chromatic Polynomials. if G=(V,E), is a connected graph and e belong E
P (G, λ) = P (Ge, λ) -P(Ge', λ)
where Ge denotes de subgraph obtained by deleting de edge e from G (Ge= G-e) and Ge' is the subgraph obtained by identifying the vertices {a,b} = e
When calculating chromatic Polynomials, i shall place brackets about a graph to indicate its chromatic polynomial. removes an edge any of the original graph to calculate the chromatic polynomial by the method of decomposition.
P (G, λ) = P (Ge, λ)-P (Ge', λ) = λ (λ-1)^4 - [λ(λ-1)*(λ^2 - 3λ + 3)]
But the response from the answer key and the teacher is:
P (G, λ) = λ (λ-1)(λ-2)(λ^2-2λ-2)
I have operated on the polynomial but I can not reach the solution that I ask .. what am I doing wrong?
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math.stackexchange.com 告诉我作为解决我的问题的一种方法。这是解决方案：
https://math.stackexchange。 com/questions/33946/问题确定图的色多项式
math.stackexchange.com told me as a way to solve my problem. Here's the solution:
https://math.stackexchange.com/questions/33946/problem-to-determine-the-chromatic-polynomial-of-a-graph
你的答案是正确的，老师的答案也是正确的——他们是平等的。
[顺便说一句，很好的图片和解释。]
奇数循环不能有 2 着色，因此 5 循环可以
没有 2 色，因此它的色多项式 f(x)，
必须有
x * [x - 1] * [x - 2]
作为除数。如果将 f(x) 的表达式与
除掉 ，
你会发现剩下的可以被 [x - 2]整除，并且
商是你老师写的。
——乔纳森·金
Your answer is correct, and so is the teacher's --they are equal.
[By the way, nice picture and explanation.]
An odd-cycle can have no 2-coloring, hence the 5-cycle can
have no 2-coloring, so its chromatic polynomial, f(x),
must have
x * [x - 1] * [x - 2]
as a divisor. If you combine your expression for f(x) and
divide out the
then you'll find that what remains is divisible by [x - 2], and the
quotient is what your teacher wrote.
-Jonathan King
在我关注的书（图论与应用 - Deo Prentice Hall）中，它的做法有所不同。它们不是排除边，而是连接两个不相邻的顶点。
使用这种技术，我得到
P (G, λ) = 2λ(λ-1)^2(λ-2) + 2λ(λ-1)(λ-2)(λ-3) + λ(λ -1)(λ-2)(λ-3)(λ-4) 这也不等于您的任何一个结果。
In the book I am following (Graph Theory with Applications - Deo Prentice Hall) it is done differently. Instead of excluding the edge they connect two non-adjacent vertices.
Using this technique I am getting
P (G, λ) = 2λ(λ-1)^2(λ-2) + 2λ(λ-1)(λ-2)(λ-3) + λ(λ-1)(λ-2)(λ-3)(λ-4)which is also not equal to either of your results.