# Network Mathematics and Rival Factions | Infinite Series

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# Network Mathematics and Rival Factions | Infinite Series

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Career In Forensic Science

please do more on this sort of thing! really liked this video

CHALLENGE RESPONSE

6:45

Not exactly correct – It isn't just TWO groups.

You can split the graph into any number of distinct groups and the graph remains balanced. For example, if you label your hexagon vertices A through F, then three distinct groups can form, EG: AB, CD, and EF.

You can add or remove new vertices freely into any distinct group by simply inserting the vertex, drawing green edges to the other group members, and red to all other vertices. In this sense, the new vertex inherits the groups relationship (red) to all other vertices. Put simply, the new vertex has joined a gang.

As long as each group has no Romeo-Juliet-style rogue relationships, then the graph remains completely stable regardless of the number of groups and the variance in the number of members in each group. EG: ABC, DE, F, GHIJ, KLMN.

Finally, you could simplify each group into a single vertex for a simplified version of the graph as long as all vertices are in conformance with their group's relationships.

Thing is, with this in mind, each vertex triangle is only balanced in the simplified graph of each vertex representing a whole group under the weak structure definition, because you have a graph that is totally red.

Now this brings us to the whole argument about the two-party system, how smaller groups naturally ally to form a larger, more powerful group. (Or don't)

Really there's a point where you have to decide which structural definition is more realistic: The strong or weak. Personally, I think the weak structure is more realistic, as it is common in politics for parties to oppose other parties based on the fear of losing voters by allying with a rival that their voters do not approve of, thus funneling voters to the other overall common rival.

At this point I'm just rambling, because what I described above deviates from the red-green model, opening up the possibility for other types of relationship.

This is starting to look like a problem way beyond exponential complexity, and way outside of my humble grasp.

Any number of factions all against each other.

At 10:14 didn't you forbid the triangle with

onegreen edge rather thantwo?there was no "Russia" back then. There was the Soviet Union

Of course, the two groups aren't necessarily symmetrical. You can have everyone against the same person, and it would still be stable.

Challange Problem: there are n groups where all are friends with each other, and all the groups are against one another.

Proof: if there are n such groups, they are clearly weakly structurally balanced – any three people are either in the same group (stable), in two separate groups (stable) or in three separate groups (weakly stable).

If the graph is weakly stable, beginning with some person x, its friends must be friends with each other to be stable, so they are in the same group. His enemies must be enemies of his friends to be stable. However, his enemies can be either friends or enemies of each other. applying the same logic to any node in the graph, if it is stable, every person's friends form a group of friends, who are all enemies with his enemies, making the graph a collection of groups against each other.

The answer is 333.

Why is it so? It is a kind of combinatorics question, but hard to explain the generalization in a simple or elegant way.

1st we have 6 nodes choose 6 friends, then 6 choose 5 friends, then 6 choose 4, but the last term has 2 ways to go either the remaining 2 are enemies with each other and everyone else, or they can be friends so we have 6_C_4 *2. This expands even further for 6 choose 2, but here we already had a 6 choose 4, so we have 6 choose 2 *[ 4 choose 2 * 2]. I hope someone will help simplify this, or I'll have to wait until next week.

Re: challenge question

The solution for weak stability is the same for strong stability, only you can have an arbitrary number of cliques/factions instead of just one or two.

We can use a constructive proof: assume we have a weakly stable graph and add another vertex. This new vertex can either be friends with some of the old ones and enemies with the rest, be enemies with everyone, or be friends with everyone.

If he is friends with some and enemies with the other, the analysis is similar to the strongly stable case: every vertex the new vertex is friends with is must be friends among themselves, but the the vertices the new one is enemy with can have any relationship between themselves, as long as the subgraph is weakly stable. So the graph has the same number of factions and one of the factions grew by one.

By a similar logic, if he is friends with everyone, that means that everyone is friends with everyone else, so the graph only has one faction.

The case where the new vertex is enemies with everyone is were things are different. For strong stability, this can only happen if the original graph only has one faction (everyone is friends), because other wise there is a forbidden triangle between the new vertex and one from each faction. When you remove the restriction on the triangle were no one is friends, we can now add a new vertex that hates everyone in the original graph, without breaking weak stability regardless of the state of the old graph. The result is an graph with one more (one member) faction.

The weaker structural balance allows more than two rivaling factions. All edges within the factions are green, all edges between the factions are red.

For example, in a graph with 6 vertices, you can have 3 rivaling factions of 2 vertices each.

This way,

o) all triangles between 3 vertices of the same faction are all-green (stable)

o) all triangles with 2 vertices of the same faction and 1 vertex of another faction are green-red-red (stable)

and

o) all triangles with 3 vertices which are all from different factions are red-red-red (weakly stable)

-> the requirements for weak structural balance are fulfilled

The third case shows why you can only have 2 factions when you want to have strong structural balance. You can't have an all-red triangle where each of the 3 vertices is from different factions. Thus, you can only have 2 factions to avoid this case.

(edited for grammar and clarity)

I'm pretty sure the weak structural balance allows for multiple factions, but that's it.

To clarify, instead of the only possibilities being either "everyone friendly" or "two rival groups," weak structural balance allows for multiple rival groups.

I use the term "group," but I mean, you could just have trivial groups with one person each. You could even have a sort of "hell" where every person hates everyone else and is his or her own group.

However, within those groups, every person must like every other person to be balanced. To see why this is the case, consider a group of three or more people (otherwise, everyone in the group already likes each other). If not everyone likes each other, there's at least one red line between two people. Mark the two vertices, and consider the vertices that have a green line between them and one of the two marked vertices. Notice that each of those vertices connecting to one of the marked vertices cannot be connected to the other marked vertex, otherwise the forbidden triangle pops up, nor can they be connected to a vertex that is connected to the other marked vertex, or again, the forbidden triangle pops up. So we see that if there is one red edge between any two vertices in a group, then the group can be split into at least two subgroups. Assuming we have finitely many points, we can continue this splitting process until everyone in their subgroup likes each other (again, recognizing the possibility of groups having size 1 or 2).

Oh, I just realized that this is basically equivalence classes. The groups sort of form a partition of the set of vertices, and each vertex "likes" itself, obviously reflexivity is demanded by the fact that the graph is directionless, and I basically just showed transitivity above.

So I guess this could work without assuming finitely many points, I just wanted to show that this process terminates. Besides, there is only finite amounts of matter in the universe (at least, to my knowledge), so an infinite graph doesn't seem like it would model reality very well (but then again, it might, so maybe someone else can give some examples).

Sorry if that was confusing, I can't really include pictures to aid intuition in a comment.

But hey, if you're still reading – here's an interesting graph to consider. Let each positive integer be its own vertex, and let these vertices form a complete graph (in perhaps an abstract sense). Two integers "like" each other if:

(1) their greatest common divisor is nonzero, and

(2) for the smallest prime that divides them both, any power of that prime divides both numbers or divides neither.

If two integers do not meet the above conditions, they dislike each other.

Can you describe how splitting might never terminate using my method?

Challenge answer: A graph is weak structurally balanced iff it splits into n rival groups where n is an integer. The proof goes analogously to the one in the video.

First, the graph nodes are then colored in successively in the following fashion: Pick one yet uncolored node adjoint to a red line and give it and all its neighbors connected by a green line a new color. Repeat until no nodes are left uncolored.

The rationale is now the same as in the video for the following node combinations: 3 nodes of the same color, 2 nodes of the same and 1 of a different color. For 3 pairwise different colored nodes A, B ,C consider wlog a node D of the same color as A. The triangle ABD has red edges between A and B and D and B, with the same logic we can deduce that the edge between A and C has to be red. Since the graph is weakly stable, the triangle ABC must have 3 red edges.

Concluding, the coloring chosen in the first step yields n rival groups. Conversely, every graph already split into n rival groups forms a weakly stable graph.

Before I clicked I thought this was about rival factions in mathematics 🙂

Weak Structural Balance allows for all the same configurations before in structural balance. In addition to these it allows for a type of factional system.

Assume that some graph has Weak Structural Balance. In the weak structural balance if two vertices, vertex A and B, are friendly, then if some other vertex C is friendly with A, then vertex C must also be friendly with vertex B, otherwise if some vertex is not friendly with at least 1 other vertex then it must be hostile to all other vertices. Lets call this type of grouping "Allies" where all vertex in an ally group are friendly with each other and unfriendly with all outside their group. Note that the ally group must contain at least 2 vertices, but can contain much more as well.

Additionally, there may be several ally groups in one weak structural balanced graph. For instance if you had 10 vertices, a weak structural balanced system could be that one ally group of 5 vertices, then we have two other ally groups consisting of 2 vertices each, then one vertex which is hostile to everything. Note for a graph with m vertices, there will be at most floor(m/2) colorings. Trivial cases being that everyone is friendly or hostile.

To this week question. I figured out that what is possible with the weak definition is that there are more separate factions, each one with the property that all its members are friends with each other and enemy with everyone that is not int that given faction. The proof is kind simple with the elements shown in the video. To start i want to separate three cases all are friends with each other, all are enemies with each other o the one with different factions. Lets suppose that there is a person, here we name A, that is friend with another B and C, therefor B and C must be friends with each other so we do not contradict the definition of weak balance, therefor all the people that are friends with A must be friends with each other and make a faction all together with A, the rest of the people can have three options, one is that every one is friend with each other, all are enemies with each other, or there are factions in this remaining, lets take a person D, in this case the situation for D is the same that was for A, we can select the group of D and continue in this way, removing factions, until all the remaining people are friends or enemies within each other. This is actually a equivalence relation, so we can make classes of equivalences and each class is a faction.

The solution to the weak structural balance problem can be summed up in one sentence.

"A complete graph has weak structural balance if there is any number of rival factions, where a faction can consist of any number of people."

This includes everything from everyone being friends (one faction of size N), everyone being enemies (0 factions), to multiple factions of varying sizes.

god i love maths, why did i choose engineering?

how are tyrion and the queen´s mother enemies?

For the weak problem, it's any number of factions.

Challenge problem proof:

Assume there is at least one pair of enemies. Choose one vertex of the graph and call it A. All vertices that are friends with A must be friends with each other or else an unstable triangle will occur. Then, choose a vertex enemies with A and call it B. All friends of B must be friends of each other or there will be an unstable triangle. Then choose an enemy of B that is not a member of an already established faction and call them C. All their friends must be friends or there will be an unstable triangle. Choose an enemy of C that isn't a member of a faction that is already established. All C's friends are friends. Continue choosing vertices that aren't in factions until there are none left. Then, all vertices are part of factions that are enemies with all other factions.

Wanted to double check my brain meat. The number of permutations shown for the 6 vertex figure was 32768. So that's 2^15th, but can it be correctly thought of as states^(instances)?

We can generalize this to more than two colors. We can consider the green edges to be 0 and the red edges to be 1; a balanced graph has the sum of the values in every triangle equaling 0 mod 2. What if we have three colors, labeled 0, 1, and 2, and require every triangle's sum to equal 0 mod 3? (This would mean every triangle either has three of one color, or one of each color.)

Needs complex analysis (an additional dimension) to transition between green and red conservatively—without a big bang….

Ayyy that's some dank background music

Weak structurally balanced graphs have n disjoint groups where n is a natural number. Members of a group are friends with each other and enemies with everyone else. n can be 0, where all members are enemies. Likewise, n can be 1 where all are friends.

Prove: the cases where all are enemies and all are friends are trivial. Otherwise, we iterate over all friendships in the graph. Pick the first 2 freinds. They need to be either both be firends or both be enemies with any other member. Any 2 members who are friends with the 2 initial picked friends, must also be friends with each other. This inevitably forms a group within which everyone is friends and within which everyone is enemies with everyone outside the group. Pick the next freindship relation in the remaining graph – with the same reasoning it forms another group where all are friends within and enemies with everyone else. This process can be repeated until all friendships are traversed.

Inside the groups of friends, the triangle relations are always 3-friendly releations. Triangle releations between 2 members of one group of friends and 1 member from another group of friends are always 1-friendly-2-hostile releations.

Then any remaining relations must be from members who are not part of any group of friends and therefore enemies with everyone else. They always form triangles relations with at least 2 hostilities, so either 3-hostile or 1-friendly-2-hostile triangles.

q.e.d. 🙂

That's it, I'm done! GoT and Infinity Series!!!! Hype! Don't even know what to write. Subtle avoiding spoilers too, thought it was clever. yes, Robert and Rhaegar get into a fight7 hehe.

Man, I can't get enough of this series.

.

Q What are all the possible colourings (hey im Canadian) of complete graphs that have weak structural balance:

A: Our guess is that there are 8 possible colorings including the examples given in the video(2 + 6 others). No proof yet.

This is to much speculation and not based on facts. Why would 3 green be ballanced but 3 red not be?

Challenge Problem:

You can pretty much use the same proof for the challenge question as the original question if you add the condition that there can be more than two rivals factions. If you color in the vertices with three factions and force people on rivals factions to have a negative relationship while forcing people on the same faction to have a positive relationships, you necessarily end up with at least a weak structural balance. So all the possible configurations are simply just the number of ways to split up the vertices into factions of 1, 2, 3, 4, 5 , or even 6, though that is fairly trivial. The factions also do not have to be of equal size, so for example you could have 3 people supporting the blue faction, while having 2 support the red faction, and having one guy without any allies on his own faction.

Side Note:

I believe the last point is also true of complete structural balance, the factions must be clearly split in two, but I didn't find any issue with splitting them unevenly, such as a team of four against a team of 2. The proof in the video filled out the purple vertices relationships in such a way that it has two friends and three enemies, but if this step was done instead by drawing three friends and two enemies, then the network would still be structurally balanced. Correct me if I'm wrong please.

Under the weak balance system, any non-negative number of factions of any positive size is allowed, and no partial connections between factions, because exclusive factions (all friendly with each other and not friendly with anyone else) can be effectively merged and treated as one vertex that is hostile to all others, until you've collapsed it all into all hostile vertices or, in the case of universal friendliness, a single vertex. Any mixed relations where part of a could-be-faction is friendly with an "outsider" makes an unstable configuration.

Brace yourself. Spoilers are coming.

To the challenge question:

Since we only use complete graphs with either red or green edges, we can model each graph as a relation F over the Nodes:

for two nodes a and b we cay (a, b) in F, if and only if a there is a green edge between them. Since the graph is not ordered, F is symmetric.

The forbidden configuration in the WSB can be stated as "If (there is a green edge between a and b) and (there is a green edge between b and c), there must be a green edge between a and c". This can be translated to the language of relations: (a, b) in F and (b, c) in F => (a, c) in F.

So the WSB condition is the same as saying F is transitive.

This makes F almost an equivalence relation. Only reflexivity is missing. If F were reflexive, F would fall apart in equivalence classes. So the graph is divides into several parties, where each party is connected with green edges.

Now lets add in the second condition for SB: This condition says, that for every a, b and c in F: (a, b),(b, a) in F or (b, c),(c, b) in F or (a, c),(c, a) in F. So lets assume we have a Graph with WSB and at least three parties. We will take one representative of each party. All three representatives are connected with red edges. making (a, b),(b, a), (b, c),(c, b), (a, c) and (c, a) not in F.

The second condition for SB forbids this. so at least two of the tree parties must be connected with a green edge. Since corresponding relation must fall apart into equivalence classes, The graph itself must fall apart into 2 or less parties.

a network is weakly balanced if it can be divided into 3 or more groups, where the members of the same group are mutual friends and members of different groups and always enemies.

if there are only two groups, the network is strongly balanced, as shown in the video (two factions). if there is only one group, the network is strongly balanced (all friends)

edit: it is also weakly balanced if all members of the network are mutual enemies.

can't you have more than two groups and then apply the rules of the individuals to the groups and then make groups of groups and so on and so forth, to make more complicated balanced graphs?

Challenge: You can split the graph into an arbitrary number of groups. Say we start at a "neutral" graph where everyone is enemies. Adding any friend-edges to that forces you to also add friend-edges in between friends of friends, so you always end up with a set of groups where within each group everyone is friends with each other and in between groups everyone is enemies with each other.

np complete. more then seven nodes becomes problematic in terms of vertex weight. NSA figured this out.

In th question at the end of the video, you only need one bad triangle to see that the network isn't balanced to see the whole thing isn't balanced, that leaves you with 12 lines can be coloured green or red striped and there is only one unbalanced triangle so you don't need to multiply it by anything meaning 2^12 out of the 2^15 total are not solutions so there would be 28672 possible combinations. I think this is right, but I don't know much about combinations and permutations yet.

Yay! I won last episode's challenge! How do I collect my T-shirt?

I'm not really sure what classrooms

you'vebeen in where everyone knows everyone else well enough to be friends or enemies.You can probably prove that all stable configurations can be reduced to smaller graphs, so stable configurations could be minimized to sets of vertices enemies with each other, including the 1-point network.

I like how you've done symbols for colour-blind viewers 🙂

The switching between "vertices" and "vertexes" freaked me out

I think everyone should just get along 😛 I'm more used to doing optimising problems (designer). I remember this one problem at uni where we had to find the optimum number of ball bearings to fit round a shaft – the problem was simple enough that you could just punch numbers into a calculator until you came up with the best answer. But there was also a mathematical solution (that required imaginery numbers the way I did it). My solution hadn't been done before on my course, and I made erros and got the wrong answer – so I was initially given a low grade (almost a fail), but I appealled to let my solution be part of the grading process (in-which a selection of students work seen to fit a certain grade is compared and given second opinions and everyone else's grades are re-adjusted based on the outcome – I forget the name of the process). A lecturer noticed that even though I had made mistakes and got the wrong answer, my process produced a general solution that would work for any set-up, and that this required a higher level of reasoning than anyone else had used – if only I had checked for errors properly I would've got the right answer… and so they gave me a top grade.

Is this the latest science/Mathematics of Graph Theory Analysis? Idk i feel like there's been more contributed to the subject matter

anyway, thanks for the videos!

By allowing 3x enemy triangles, you obtain roughly the same result, except more than two factions can form. This can be proven by proving that the defined relation exactly satisfies the criteria for an equivalence relation (and therefore partitions the set of people):

1. A is friends with A. While technically this is not true according to our graph, I think we can safely say that the "A is friends with A" edges are only omitted because it is implicitly assumed.2. A is friends with B if and only if B is friends with A. Our edges are not directed, so this is trivially the case for any complete relationship graph.3. If A is friends with B and B is friends with C, then C is friends with A. If this is not the case, then C is enemies with A, and we have a forbidden friends/friends/enemies triangle.

Similarly, if there is a friends/friends/enemies triangle, then that also means that means that condition 3 is not satisfied. Therefore, not only is every weakly balanced complete graph a partition into groups of friends, but every partition into groups of friends is weakly balanced.

One of the possible answers if you changed the definition of balanced networks, would be this:

https://i.imgur.com/kuyBexN.png

Could you have more than 2 factions if you use hypergraphs with the "strong" definition of stability?

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?

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.

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.

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)

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.

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

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)?

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.

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

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

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 like this channel.

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

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?

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.

Thanks for color blind friendly visualisation

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

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

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).

Hair is amazeballs

No idea what I just watched… BUT MATH! lol

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 one could be persuaded to switch sides" cough cough

italycough coughIs this video spoiler safe?

that is why there is only ios and android…

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

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

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

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

Where can I learn more about this?

Eigenchris also

Cool. Now Riot please try to apply this to all LoL champions on Summoner's Rift to see which configuration will make the game balance.

N groups of people, be friend to each other and hates others.

Math is coming… Maybe…

How a graph has balanced labelings-32768, whether it is for every complete graph or for the complete graph with 6 edges

I noticed that they no longer use specifically coloured lines! although it's red and green, red line is dashed. I hadn't realized why until I suddenly remembered the comment about some people watching may be colour blind. nice to see that

opting out of red is less stable than opting out of green, for completely connected population

+or-

This theorem only points out to two states of relationship, rival or friend. Is there a theorem where there are three states of relationship, neutral is the third one? It could actually help some writers to logically group their characters into rival parties.

Take that Machiavelli. Did you predict all this in the Prince?

The network node/vertices have to be composed of axial holes/numerical frequency of e-Pi-i resonance and 1st law of Thermodynamics.., and the network structural spacing/distribution on 2nd law of Thermodynamics and conduction-duration periods. (Rough "algerbraic" equivalents, and a kind of inherent "Rigour" of eternity-now temporal superposition probabilities)

Smart

What math classes teach this type of thing?

This is very interesting!

Indeviduels with kingdoms being representet with a court thers the King and his relation and thers the Council members that Can agree with the King or disagree with the King if a mayority agree with the King then the Kings realasion is the contrys (country = kingdom) BUT if a mayority disagree the disagreing Council members Will resolve it amongst themself

What about non-comPLETE graths?

Il send you a power graph representing a kingdom with King gurd and citecin try explaning it and a litel historyafter

I manet history after

Thers not seven kingdoms

Not seven houses atlaest

What if tow group buomp in to each other

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please come back Kelsey, we miss you …