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How define a graph where edges are defined using a condition?

asked 2024-02-29 11:46:13 +0100

sunilpasupulati gravatar image

I would like to define a graph where I want my vertices set to $\mathbb{F}_5^2$. I won't define there is an edge between $(A_0,B_0)$ and $(A,B)$ if $A\neq A_0$ and $F_{B-B_0}(A,B))=0$ Where $F_{A,B}(X)=X^3+AX+B$. Any help or reference is appreciated.

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What is $F_{B-B_0}(A,B)$? Do you mean $F_{A,B}(B-B_0)$?

rburing gravatar imagerburing ( 2024-02-29 14:53:34 +0100 )edit

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answered 2024-02-29 17:43:02 +0100

updated 2024-02-29 17:43:32 +0100

Here is a simple example: the vertices are elements of $\mathbb{F}_5$, and there is an edge from $a$ to $b$ if $a^2 = b^2$.

sage: from itertools import product
sage: vertices = list(GF(5))
sage: edges = [(a,b) for (a,b) in product(vertices, vertices) if a**2 == b**2 and a != b]

In the previous line, product(vertices, vertices) is an iterator consisting of all pairs (v,w) of vertices. I impose the condition a != b to avoid loops in the graph. Now construct the graph:

sage: Graph([vertices, edges])
Graph on 5 vertices
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You can use for a,b in Combinations(vertices,2) instead. No itertools is needed. Also, using product you are essentially adding each edge twice.

Max Alekseyev gravatar imageMax Alekseyev ( 2024-02-29 18:10:53 +0100 )edit

Sure, even better!

John Palmieri gravatar imageJohn Palmieri ( 2024-02-29 19:57:04 +0100 )edit
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answered 2024-02-29 19:46:03 +0100

Max Alekseyev gravatar image

updated 2024-02-29 19:50:53 +0100

This is a straighforward way to contruct the same graph as in John's code:

Graph( [ GF(5), lambda a,b: a^2 == b^2 ] )

Correspondingly, the graph in question can be constructed as

F = lambda X, A, B: X^3 + A*X + B
Graph( [ Tuples(GF(5),2), lambda AB0, AB: AB[0]!=AB0[0] and F(AB[1]-AB0[1],AB[0],AB[1])==0] )
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Asked: 2024-02-29 11:46:13 +0100

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Last updated: Feb 29