#undirected graph

[positively-knotted]

The polycule category

So obviously a polycule is just an undirected graph with no double edges or self-loops. Properly we should probably only consider connected polycules, but for technical convenience we'll say that an object of Polycule isn't necessarily connected.

The big question is what is a morphism of polycules. The natural idea is that a morphism should roughly be a cobordism, reflecting the evolution of a polycule.

Inspired by this, we define the following elementary morphisms:

A(v): Addition/deletion of a disconnected vertex v;

E(v,w): Addition/deletion of an edge between v and w.

We probably want to define a morphism to just be sequence of elementary morphisms (that is, a word). But to make things slightly interesting, we add two equivalences:

(Independence). If W, X are words involving disjoint sets of vertices, then WX = XW.

(Antibubbling). A(v)A(v) = E(v,w)E(v,w) = id.

Informally, changes in separated parts of the polycule commute, and if you form a relationship and break up (or vice versa) too fast it doesn't count.

So a morphism in Polycule(G,H) is an equivalence class of words taking G to H.

Interesting properties?

Idk but it probably has some. I wanted the empty polycule to be initial but there are two maps from the empty set to u—v—w depending on the order that you add the edges.

So yeah. I assume it has properties but I'll leave that as an exercise to the category theorists.

[thenegoteator]

back to say that to express my hype for Andor s2 I did the normal thing and made a social network graph of the relationships in s1!

notes for fellow network nerds below the cut:

The labels are sized by the characters' degree. The 5 characters with the highest degree are: 1 - Cassian Andor (no surprise); 2 - Bix Caleen; [3 - Ferrix crowd - this was a bit of a cheat to express the collective action, but interesting to consider the crowd as an agent]; 4 - Mon Mothma; 5 - Kino Loy; 6 - Luthen Rael

The characters with the most betweenness (in this case, eigenvector centrality) are: 1 - Cassian (still no surprise); 2 - Vel Sartha; 3 - Cinta Kaz; 4 - Bix Caleen; 5 - Kino Loy

The data was not gathered automatically!!!! this is vibes based from my rewatch of s1 - it still only represents what we see onscreen, not things that are verbally reported, and only interactions which I could argue were somewhat mutual so i could make it undirected

it's undirected.

ignore any temptation to read into the edge weight. that's just how I imported the data into gephi bc I was feeling too lazy to change it

The colours roughly correspond to locations in which these characters were found - grey means that they travelled from two locations. Light green is Ferrix, orange is Narkina 5, pink is Aldhani, blue is Coruscant, yellow is Morlana One

[yarn-dragon]

So fun fact about me is I have a math minor and while taking a graph theory course last semester I realized something kinda fun (well, I think it's fun)

So as other people have pointed out most love triangles in media aren't actually triangles cause two of the points don't connect

Now there is a type of graph in graph theory called a tree, which is basically when any two vertices are only connected by one path

Now if you were to graph most love triangles, where the vertices are charatcers and the edges are who's in love with who you'd get a tree (assuming being in love is undirected which meaning it goes both ways)

So therefore, we should be calling it a love tree

[caprice-nisei-enjoyer]

Say I'm looking at a connected undirected graph and I want to uniquely identify a point. In the spirit of vagueposting, I decide to identify this point by listing other points in the graph and their distance from the unique point. Is it always possible to identify a point uniquely? And what's the best strategy?

[ao3demographicssurvey2024]

What exactly does it mean when you say a sexual or romantic orientation is "undirected"? Is that the ace and pan respondents, or something else?

It's not a label I would use for myself, and not actually one I've heard before, so that pie chart just isn't very meaningful.

Also, since the term wasn't used in the survey quiestion itself, it isn't very accurate to sort orientations into a chart based on data that was not actually gathered.

As discussed in the post linked under the graphs, there were three answers that specified a direction of orientation - "Heterosexual or Straight", "Homosexual or Gay or Lesbian", and "Bisexual or Pansexual" (or the equivalent romantic orientations).

For the "Direction of Orientation" charts, only the inclusion (or lack) of these three answers was counted, in order to help group together similar answers, e.g. "Bisexual or Pansexual" is grouped with "Bisexual or Pansexual + Queer", "Bisexual or Pansexual + Questioning", and "Bisexual or Pansexual + Gray-Asexual or Demisexual or Asexual-Spectrum".

"Undirected" was the term I used to indicate that an answer didn't include any of those options, i.e. it consisted of some combination of "Asexual", "Gray-Asexual or Demisexual or Asexual-Spectrum", "Queer", "Questioning", "I do not label my sexual orientation or I only identify by romantic orientation", and "Other" (or the corresponding answers for romantic orientation).

The chart in question was intended to give people a more wholistic overview of how sexual and romantic orientation related to each other within the dataset. I have in general been trying to avoid constructing groups like this in my analysis as it can give a misleading impression of how people identified (which is in part why I gave this constructed group a label which is not a commonly-used identity). However, these multi-answer questions can be difficult to get a feel for without some level of interpretation being applied, and in this case I judged it was necessary to aid understanding.

[bubbloquacious]

So defining directed multigraphs as presheaves over a category V ⇉ A is pretty cool. I think you can get undirected multigraphs very similarly. You add an endomorphism p: A -> A satisfying p ∘ p = id_A, p ∘ s = t, and p ∘ t = s (where s and t are the parallel morphisms V -> A). You can get graphs with labelled vertices by adding an object L and a morphism ℓ: L -> A. That one's interesting cause you get one more representable presheaf.

[dandelionmoss]

i was wondering about how the directed nature of following someone on sites like tumblr relates to the friendship paradox (where on average your friends probably have more friends than you do).

the original version is based on undirected graphs, but i found this paper that talks about directed social networks. it covers four variants of the "paradox" in directed networks, where friend refers to a person you're following:

"Random friend Y has more followers than a random node X, on average"

"Random follower Z has more friends than a random node X, on average"

"Random friend Y has more friends than a random node X, on average"

"Random follower Z has more followers than a random node X, on average"

the last two apply only if the number of followers and the number of people one follows are positively correlated, which probably is the case on tumblr? if so, then on average, someone's followers have more followers than that person does. this conclusion feels somewhat off, so maybe the assumption of correlation is wrong, or maybe it's just the counterintuitive nature of this phenomenon.

[lonelydipshit]

I mean, you can represent a polycule as an undirected graph where each person is a node and each relationship an edge, and similarly you represent a monogamous relationship with 2 nodes and an edge between them, they’re both just graphs from the mathematical perspective

I don't necessarily agree with this but it has such a mathematical quality to it somehow

[codingprolab]

CS 440: INTRODUCTION TO ARTIFICIAL INTELLIGENCE Exam

Problem 1 (30 points): Figure 1: Find the shortest path from A to Z Consider the graph in Figure 1 for a search problem from A to Z. All edges of the graph are undirected (bidirectional) and labelled with their cost. The labels of the nodes contain a heuristic approximation (in paranthesis) w.r.t. the remaining cost to the goal node Z. Whenever several nodes have identical priorities for…

[ankitcodinghub]

CSCI570HOMEWORK2 Solved

Q1. What is the tight bound on worst-case runtime performance of the procedure below? Give an explanation for your answer. (10 points) int c = 0; for(int k = 0; k <= log2n; k++) for(int j = 1; j <= 2^k; j++) c=c+1 return c Q2. Given an undirected graph G with n nodes and m edges, design an O(m+n) algorithm to detect whether G contains a cycle. Your algorithm should output a cycle if G contains…

[tpointtech1]

Bron-Kerbosch Algorithm in C++

The Bron-Kerbosch algorithm in C++ is an efficient recursive method for finding all maximal cliques in an undirected graph. It systematically explores subsets of vertices, expanding cliques while maintaining constraints to avoid unnecessary checks. The algorithm uses three sets: candidates, current clique, and excluded vertices. By recursively refining these sets, it identifies all maximal cliques without redundantly revisiting already discovered ones, making it useful for graph analysis and network research.

For more information visit our website: https://www.tpointtech.com/bron-kerbosch-algorithm-in-cpp

[21st-century-minutiae]

In graph theory, a tree is defined as a graph that contains no cycles. That is to say, no matter where you start, it is impossible to get back to that place without reusing an edge. An alternate definition is that there are no enclosed areas/volumes/etc in an (undirected) tree.

The above image depicts a tree shaped like a dog, and a non-tree graph that is shaped like a tree. This is humorously playing with the word "tree" in an unexpected manner.

In the early twenty-first century, not everyone knows graph theory, so many would not get the joke. But the barrier to entry is very low, so many would find this humorous.

From What Do Trees and Hypercubes Have in Common? by Henry Martyn Mulder.

[programmingandengineering]

Lab 08 CS 303 Algorithms and Data Structures Solved

Notes: Implement the algorithm and analyze the results using the give input files Deliverables: Report.pdf file and your code file (please do not send a zip file. If you have more than one class in your code, then submit each file separately through Canvas.) Homework report must follow the guidelines provided in the sample report uploaded in Canvas Objectives: Implement an undirected graph…

[tccicomputercoaching]

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Arrays

Linked Lists

Stacks

Queues

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Graphs (Directed, Undirected, Weighted, etc.)

Hash Tables

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[myprogrammingsolver]

Computer Science CSC373H Hw #5

Question 1. (10 marks) Describe an algorithm that, given any undirected graph G = (V; E), determines the minimum number of edges that must be removed to disconnect the graph. Your algorithm should work by running the maximum- ow algorithm on at most jV j ow networks, each having O(V ) vertices and O(E) edges. Prove the correctness of your algorithm. Hint: The basic idea is to relate the minimum…

[lorxus-is-a-fox]

as a mathematician (and even a topologist!): IKEA has nontrivial fundamental group and a key part of what it is to be a labyrinth is that it has trivial fundamental group[oid? look dw about it it's the deep magic] such that equivalently there's exactly one kind of path - the one where you don't get lost.

adding to the problem is that if you're a dumbass who doesn't follow floor arrows and thus has to treat a random walk on the IKEA room graph as being on the undirected rather than directed version of the graph, you get caught in basically all of the loops.

You might be tempted to get nitpicky and call IKEA a "labyrinth" , as there's only one path so you can't get lost. With no branches and no dead ends, surely this is a labyrinth, right?

Nope! While IKEA's design resembles a labyrinth, it's only actually one if you ignore the included "shortcuts". Since there's nothing stopping you from taking a shortcut backwards to an early section of the maze, you can indeed get stuck in an infinite loop and never make it to the end. You can get lost in an IKEA, making it a maze and not a labyrinth.

I'd draw some topological diagrams to explain this but I'm on mobile. Maybe later, if this is too confusing.