Network Mathematics and Rival Factions | Infinite Series

Where can I learn more about this?

Structural balance goes out of the window if we talk about the Syrian civil war. The complexity of enemy friend relations there is insane…

Combine this with the Cops and Robbers graph – who would be the cops, who are the robbers, and who are the dead-ends?

6:43 What if the 6 people split into 2-4 or 1-5?

i love your videos, your channel, and you. you are so pretty and smart – which is irrelevant but doesn't hurt lol

that is why there is only ios and android…

Is this video spoiler safe?

"No one could be persuaded to switch sides" cough cough italy cough cough

So you chose to go from a deeply interesting complex drama to the garbage of D&D … sorry WW1 but mathematicians fear history and love HBO ruining a an author's life work.

No idea what I just watched… BUT MATH! lol

Hair is amazeballs

Answer to the challenge: define a faction as a set X such that if a and b are both in X, then (1) for each c in the complete graph the relationship between a and c and that between b and c is the same, and (2) a and b are friends. Then for each x, the set of all individuals with whom x is friendly forms a faction. This is because if x is friendly with a and b, a and b are friends with each other. If x is friendly with a and an enemy to c, a must also be hostile towards c. Thus, a weakly structurally balanced complete graph can be divided into factions, each of which is hostile to all others. Note that a faction may contain only one member (with x always being friendly to x).

Better way of thinking about it: according to the definition of weak balance, friendliness is an equivalence relation. Thus the graph may be divided into equivalence classes (aka rival factions).

Could It be possible to have three different groups and be a balanced network or graph?

I think the number of nodes in the Romeo and Juliet chart should have dropped by 2 by the ending.

Thanks for color blind friendly visualisation

In the 6-dot case with weak structural balance, you can have three 2-cliques. If only A-B, C-D, and E-F are green, and all others red then any triangle either has 0 green edges or 1 green edge because every point is only on one green segment and to have the disallowed arrangement you must have a vertex with two green edges connecting to it.

A couple notes/observations/ questions
This model seems to operate on perfect information and a binary choice, what about ambivalence or ignorance as options?
I noticed all the examples you showed had either 3 or 6 people, and two factions, are there other possibilities, or is that a limitation of this?
this seems to operate on a symmetry of opinion. having gone through high school, I can tell you from personal experience that feelings aren't always reciprocated, it's possible for you to be rather fond of somebody while they hate your guts, or vice versa. is there a version of this that accounts for that potential added level of complexity?

I would visit a soothsayer, if she used this to predict my future. I mean, this must be taught at Hogwarts.

Where is this week's video?

I like this channel.

With weak structural balance, there can now be more than two rival groups, where everyone within a group is friends, but people in different rival groups are enemies. To prove this, take one person and consider their set of friends. They must form a group, where everyone within the group is friends, otherwise there would be disallowed triangles. They must also all be enemies with everyone outside the group for the same reason. But now two people outside the group need not be friends with each other. So we can pick someone outside the group, form their group of friends, and continue this way until everyone belongs to a group.

I have the solution for the weak case. Define a clique as a set of
completely connected vertices where the connections are all friends or
all enemies or whatever. The vertices will be divided into sets of
disjoint friend cliques. They must be disjoint because partial connections are not possible. If A and B are in a friend clique, then they are friends. If X is friends with A, then the only way that that is allowable is if X is also friends with B.

This means that the possible friend-clique sizes are, for these total sizes:
2: 2
3: 2, 3
4: 2, 22, 3, 4
5: 2, 22, 23, 3, 4, 5
6: 2, 22, 222, 23, 24, 3, 33, 4, 5, 6
in addition to the trivial case of no friend cliques

Imposing the strong condition means that there are no enemy cliques that are separate from the friend cliques. This gives us
2: 2
3: 2, 3
4: 22, 3, 4
5: 22, 23, 4, 5
6: 222, 23, 24, 33, 5, 6

The paper with the graphs for late-19th-cy European alliances is http://physics.bu.edu/~redner/pubs/pdf/dresden.pdf

Let's say there were more kinds of relationship than just friend or enemy. Suppose member a can have a positive attitude toward member b, but member b has negative attitude toward member a. This introduces 2 new line types, which are inherently unstable, I think. Those line types would have to change to full enemy or full friend for the map to become stable. I think. Thoughts?

I mean, narrative wise, it's completely possible. Father loves son, but the son hates the father. Inherently unstable. But Add in mom. Son loves mom, but mom hates son. Mom loves father but father hates mom.

Is this stable or unstable? It's balanced, but I don't think that makes it stable.

Hi, could you recommend any materials on sociocentric models where there are more than two types of relations and/or graphs are directed (both people might perceive the relation differently)?

She still doesn't know what to do with her hands.

Been away from my computer for a bit, but first thoughts on watching for the question is allowing all enemy triangles means that you can have n groups of allies that hate everyone not in their friend group.

7:00 it can be said that the group of all friends is simply one in which the rival faction has population zero, thereby demanding two factions always be present (even if one be of 0 population) or that a maximum of two factions can have a non-zero population, and removing the concept of "either", the removal being desirable in mathematics (especially in proofs and graphs)

I think in the US you have a voting system that have some votes matter slightly more since the amount of presidential electors doesnt exactly match the percentage of the represented population.

My answer to the challenge question:

The weak definition of structural balance demands that a person can never have allies that aren't allied with each other, because that would form a triangle with 2 alliances and 1 animosity.
That is the same as in the strong structural balance and demands that allies be grouped together in full alliances.

Unlike the strong balance, in the weak balance it is allowed for a person's enemy be enemy with his/hers enemy. That means that there can be a greater number of full allied groups than 2. There can be any number of allied groups up to the maximum number of the total number of people.

The strong structural balance only accepts structures with full alliances in the number of 1 (everyone is allied) and 2 (2 opposing groups), in each case each alliance can have any number of people (including a single one) as long as the sum is the total number of people. Weak structural balance increases the number of alliances to the maximum without putting restrictions on the number of people in each one.

What happens if there is a third type of relationship status: indifferent/neutral. In this case we'd assume a connection of pure neutrality is stable, but what would happen if any relationship switches to either "friend" or "enemy"? I'd assume the peers would need some sort of emotional connection to someone else in order to sway another peer, so an indifferent relationship would not influence any change, but only a change in relationships between their friends or enemies will cause a shift. So what would this mean for the graph's status and factions as a whole? How would a graph look if it were either balanced or imbalanced?