lcz

Ask a Question

Create a Post

11 Badges

Member for: 6 years, 131 days

Location: changsha, China

Contact lcz

MaplePrimes Activity

These are questions asked by lcz

Two colors dye the same Vertex of graph...

Asked: lcz 994 Product: Maple 2020

September 17 2020

2 3

I'm thinking of better demonstrating the cartesian product of a graph.
With the help documentation, we can easily find the cartesian product of two graphs.

with(GraphTheory):
G := CycleGraph([v__1,v__2,v__3,v__4]);
H:=Graph({{u__1,u__2}}):
DrawGraph(G,size=[250,250],stylesheet=[vertexborder=false,vertexpadding=10,edgecolor = "Red",
vertexcolor="navy",edgethickness=3]);
DrawGraph(H,size=[250,250],stylesheet=[vertexborder=false,vertexpadding=10,edgecolor = "Blue",
vertexcolor="Gold",edgethickness=3]);

GH:=CartesianProduct(G,H)
DrawGraph(GH,style=spring)

When I saw Wikipedia's demo diagram, https://en.wikipedia.org/wiki/Cartesian_product_of_graphs

I was fascinated,and I also wanted to visually reflect the nature of Cartesian product by doing different staining of vertices.
It is easy for me to dye the vertices in one color, but it is difficult for two different colors .

count the total number of Eulerian trail...

Asked: lcz 994 Product: Maple 2020

August 19 2020

2 1

I'm thinking about a question about counting the total number of Eulerian trails of following graphs .

The Seven Bridges in Gonisburg lost one bridge due to war.

G-e₇

We looked up on Wikipedia to add some details..

For the existence of Eulerian trails it is necessary that zero or two vertices have an odd degree; this means the Königsberg graph is not Eulerian. If there are no vertices of odd degree, all Eulerian trails are circuits. If there are exactly two vertices of odd degree, all Eulerian trails start at one of them and end at the other. A graph that has an Eulerian trail but not an Eulerian circuit is called semi-Eulerian.

It's easy to know that G-e₇ is semi-Eulerian. So I want find all Eulerian trails. A simple attempt, found the following one in graph (Pink label).

I feel a little tricky dealing with this through Maple. The first reason is that the Maple graph theory package does not support parallel edges. Unfortunately， e₅and e₆ are parallel edges of G-e₇.

To take a step back, what if there are no parallel edges? For considering G-e₇-e₆.

There are some built-in commands related like FindEulerianPath , but can only find a Euler circuits.

Can this formula be simplified?...

Asked: lcz 994 Product: Maple

August 04 2020

1 2

I tried to simplify it, but It didn't work that well.

simplify(%)

I always feel that this formula can be further simplified, but there is no way to start. Of course, My thoughts maybe incorrect. maybe this is the simplest form.

Expression with products in both numerat...

Asked: lcz 994 Product: Maple 2020

August 03 2020

1 4

I have an expression

product(q^(n -2)- q^i, i = 0 .. r )/product(q^(n-2) - q^(i+1), i = 0 .. r)

Obviously, it can be further simplified.

But in maple I can't do that, I use simplify and expand , all failed.

simplify(expand(%))

What should I do to get the simplified result I want？

convert every row of matrix to list...

Asked: lcz 994

August 01 2020

3 2

I'd like to convert every row of matrix to list, but failed. I want to use batch operations map(~) not for-loop.

>	data:=Matrix(3, 5, [[2, -6, 3, 0, 0], [5, -2, 4, 1, 2], [17, -4, 10, 20, 99]])

Matrix(3, 5, {(1, 1) = 2, (1, 2) = -6, (1, 3) = 3, (1, 4) = 0, (1, 5) = 0, (2, 1) = 5, (2, 2) = -2, (2, 3) = 4, (2, 4) = 1, (2, 5) = 2, (3, 1) = 17, (3, 2) = -4, (3, 3) = 10, (3, 4) = 20, (3, 5) = 99})

(2)

>	convert~(data,list)

Matrix(3, 5, {(1, 1) = [2], (1, 2) = [-6], (1, 3) = [3], (1, 4) = [0], (1, 5) = [0], (2, 1) = [5], (2, 2) = [-2], (2, 3) = [4], (2, 4) = [1], (2, 5) = [2], (3, 1) = [17], (3, 2) = [-4], (3, 3) = [10], (3, 4) = [20], (3, 5) = [99]})