## The Homestuck Troll Theorem

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##### Central Hypothesis: For a given group individuals, in order for *every* member of the group to fill *all* of their Homestuck Troll romantic quadrants, the minimum necessary size of the group is 6.

### Preface

This started when a friend of mine said that I'd somehow given him the idea of Flandre Scarlet using her "Four Of A Kind" Spell Card, and all of them filling the Homestuck Troll quadrants with each other. I quickly realized that this was impossible, due to the nature of the Ashen Quadrant, which involves *three* individuals, rather than a pair: you'd also need to get Remilia involved.^{1}

But this got me started on wondering how many trolls you'd need in order to have *all* of them to fill *all* of their quadrants! I decided I'd conclusively figure it out, and report my findings. So here it is!

### The Proof

**Axiom #1:**There are four quadrants. The Flushed, Pale, and Caliginous Quadrants each involve precisely two individuals; the Ashen Quadrant involves precisely three.**Corollary #1 to Axiom 1:**the group-size must be a multiple of 2, for every member of the group to be part of a pairing, and it must also be a multiple of 3, for every member to be part of a trio.**Axiom #2:**All relationships are mutual: if someone is your Matesprit, you are also their Matesprit. If two people are in your Ashen Quadrant, they are also in yours, and in each other's.- For the purpose of this theorem, I will not distinguish between individual members of the Conciliatory relationships: it doesn't matter if you're the Auspistice or one of the two enemies.

**Axiom #3:**No individual may occupy more than one of another individual's quadrants.**Corollary #2 (to all axioms)**: in order for an*individual*person to completely fill all of their quadrants, they need to have five other people (for a total of six).

I was having difficulty figuring it out, when I came across:

##### Sub-Hypothesis #1: In order for *every* member of a group to fill *both* Concupiescent relationships, the minimum group-size is 4.

This one is relatively simple: four people, two sets of pairings both ways. If you want a visual demonstration:

Eridan | ♥ | Sollux |

♠ | ♠ | |

John | ♥ | Ms. Paint |

Note: In this table and others, all names are taken from Homestuck; they are randomly selected every time you load the page, and do not necessarily reflect the author's shipping preferences.

I wondered if you could fit moirallegiance into that, and you can: connect them *diagonally* in the above table (so, Eridan ♦ Ms. Paint and Sollux ♦ John).

##### Sub-Hypothesis #2: In order for every member of a group to fill both Concupiescent relationships *and* the Pale Quadrant, the minimum group-size is 4.

And here's another table, with an arrangement that will work nicer with the main hypothesis, and with a new set of names:

♥ | ♠ | ♦ | |
---|---|---|---|

Vriska | Equius | AR | White King |

Equius | Vriska | White King | AR |

AR | White King | Vriska | Equius |

White King | AR | Equius | Vriska |

This is an easy arrangement to work with, because in order for relationships to be properly unique, each name must appear in every precisely once (except for auspisticism, in which it appears twice), and each name must appear in every row no more than once. Another useful aspect is that it gives you nice little self-contained groups of four, which you can add to the table as needed. Which leads us to:

##### Sub-Hypothesis #3: For a given group individuals, in order for *every* member of the group to fill *all* of their Homestuck Troll romantic quadrants, the minimum necessary size of the group is *not greater than* 12.

Taking the above arrangement to the next logical step, using three groups of four, we can put each member of each group into the Ashen Quadrants of the corresponding members of the other groups, like so:

♥ | ♠ | ♦ | ♣ | |
---|---|---|---|---|

Vriska | Equius | AR | White King | HB & Jane |

Equius | Vriska | White King | AR | Jake & WV |

AR | White King | Vriska | Equius | Aranea & Meulin |

White King | AR | Equius | Vriska | Black King & Nepeta |

♥ | ♠ | ♦ | ♣ | |

HB | Jake | Aranea | Black King | Vriska & Jane |

Jake | HB | Black King | Aranea | Equius & WV |

Aranea | Black King | HB | Jake | AR & Meulin |

Black King | Aranea | Jake | HB | White King & Nepeta |

♥ | ♠ | ♦ | ♣ | |

Jane | WV | Meulin | Nepeta | Vriska & HB |

WV | Jane | Nepeta | Meulin | Equius & Jake |

Meulin | Nepeta | Jane | WV | AR & Aranea |

Nepeta | Meulin | WV | Jane | White King & Black King |

#### Proof of the Central Hypothesis

Now, this neatly proves that you don't need more than 12. However, it does *not* prove that you can't have *fewer* than 12, i.e. 6 like the central hypothesis says; nothing says that the Concupiescent and Pale quadrants all have to be in a group of four. So let's shuffle the names again, and try it with a group of six (the only lower number which is a multiple of both 2 and 3, as per Corollary #1). I'll arrange them by the Ashen Quadrants: each group of three is part of a single Auspisticism. All of their other relationships must therefore come from the other group. Matespritship is the corresponding member of the other Auspisticism, Kismesissitude is the corresponding member offset by one, and Moirallegience is the corresponding member offset in the other direction.

♥ | ♠ | ♦ | ♣ | |
---|---|---|---|---|

Nepeta | Cronus | Feferi | Damara | HB & AR |

HB | Damara | Cronus | Feferi | Nepeta & AR |

AR | Feferi | Damara | Cronus | Nepeta & HB |

♥ | ♠ | ♦ | ♣ | |

Cronus | Nepeta | HB | AR | Damara & Feferi |

Damara | HB | AR | Nepeta | Cronus & Feferi |

Feferi | AR | Nepeta | HB | Cronus & Damara |

So there you have it. There exists a configuration of 6 individuals such that they can all fill each other's Quadrants.

Conclusion: The central hypothesis is correct.

### Proof of the Flandre Hypothesis

And here's this just for kicks.

##### Flandre Hypothesis #1: For a given group of individuals to *collectively* fill all four quadrants at least once between them, the minimum size of the group is 4.

♥ | ♦ | ♠ | ♣ | |
---|---|---|---|---|

Flandre #1 | Flandre #2 | ? | ||

Flandre #2 | Flandre #1 | Flandre #3 | ? | |

Flandre #3 | Flandre #2 | Flandre #4 | ? | |

Flandre #4 | Flandre #3 | ? |

Flandre #1 may have #3 or #4 (but not both) in her Ashen Quadrant, #4 may have #1 or #2 (but not both), #2 may only have #4, and #3 may only have #1. All other possibilities result in relationship-conflict. Any other similar configurations of Flandres and pairings would produce the same problem.

♥ | ♦ | ♠ | ♣ | |
---|---|---|---|---|

Flandre #1 | Flandre #2 | Flandre #3 | ? | |

Flandre #2 | Flandre #1 | Flandre #3 | ? | |

Flandre #3 | Flandre #2 | Flandre #1 | ? | |

Flandre #4 | ? | ? | ? | ? |

Flandre #4 can have any *one* of Flandres #1, #2, and #3 in her Ashen Quadrant; however, she cannot have more than one, because they are all in pre-existing relationships with each other. Again, any other similar configurations would produce the same problem.

Flandre Hypothesis #1 is thus *disproven*.

##### Flandre Hypothesis #2: For a given group of individuals to collectively fill all four quadrants at least once, the minimum size of the group is 5.

♥ | ♦ | ♠ | ♣ | |
---|---|---|---|---|

Flandre #1 | Flandre #2 | Flandre #4/Remilia | ||

Flandre #2 | Flandre #1 | Flandre #3 | ||

Flandre #3 | Flandre #2 | Flandre #4 | ||

Flandre #4 | Flandre #3 | Flandre #1/Remilia | ||

Remilia | Flandre #4/Flandre #1 |

This arrangement fulfills all the requirements; any combination of individuals and pairings will work.

^{1}There's at least a billion reasons why this wouldn't work to begin with, but still!

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#### 19 Comments *(auto-closed)*

*Dizzy H. Muffin*

Well, I had it down as Flandre 4 auspisticizing Remilia/Flandre 1 before I made the tables into something more comprehensible.

###### Tarquin

@Muffin: So now you have the Remilia / Flandre 1 / Flandre 4 triad, where each of the three is auspisticizing the other two. I have to assume that this is either:

(a) some sort of delicately-balanced situation where all three of them are aware that it has the potential to blow up into a horrific mansion-destroying disaster if anything goes wrong, or:

(b) the platonic auspistice version of a stable polyamorous relationship, where everyone gets on suitably badly (but not *too* badly) and this is just good practice to keep the relationship healthy. (But then the work to maintain the relationship *is* the relationship, which makes it easier to keep working but is also a very confusing idea...)

###### Wymar

Somewhere, somebody is trying to ship the 6 main inhabitants of the SDM with each other in that way.

###### DiMastine

They dont already?

Fanon either has Sakuya stabbing Meiling or Meiling dragging Sakuya into bed.

Sakuya nosebleeds over Remi.

Koakuma is thought to be a succubus summoned by Patchouli for kinky things.

Flandre is often paired with Remilia for cheesecake.

###### Pawel

Objection!

Axiom #3 has a fundamental logic flaw which renders all of the following proof wrong. Oops...

Explanation: The ashen quadrant's (auspisticing) purpose is to stabilize a pair of trolls in their concupiscent relationship.

Even in the auspisticized relationship the trolls still have the black/red feelings for each other.

So, this leads to the fact that every troll has 3 concupiscent relationships - flushed, caliginous and the relationship that is a part of ashen quadrant.

This is obviously wrong - as each troll should have exactly 2 concupiscent relationships - a matesprit and a kismesis.

Axiom #3 is in fact proven wrong and should also have "No individual may occupy more than one of another individual's quadrants, with an exception of ashen quadrant which involves either a matesprit or kismesis pair and another troll."

Corollary #2 should then be "every troll is related to FOUR other trolls" - his matesprit and his kismesis (with one of the two he is being auspisticized with) and his moirail.

Sorry if there is any mistake.

###### Pawel

Yeah, great this is probably the quickest change of mind.

After reading some more it was me that was wrong. Disregard my previous comment. :(

###### Zeph

I'm surprised you objected to that rather than how he hasn't actually proved his hypothesis, unless I'm missing something.

That is, the "minimum necessary" part. You've proved that there exists a combination of 6, but where does that show it's the minimum?

###### Wymar

It must be at least 6. Every Troll must have 5 romantic interests, one for ♥, ♦, and ♠ respectively, and two for ♣. Add the troll themselves and you have 6.

###### Zeph

Oh, I'm not arguing that the result is true or false. I'm just saying that as far as I can see, he hasn't formally proven his hypothesis.

*Dizzy H. Muffin*

The axioms already state that it must be a multiple of both 2 and 3. 6 is the lowest positive integer which is a multiple of both 2 and 3. Proof of the latter, and the logical connect between it and the conclusions, is an exercise left to the reader.

###### Zeph

Fair enough, I figured you were going for a formal proof since you were using axioms and corollaries to begin with, and those generally spell everything out pretty explicitly.

###### Henix Aurorus

I love how you put the troll romance thing in perspective using Flan and Remilia. It's a great way of connecting Homestuck and Touhou.

## Bocaj

For some reason, the first thing that springs to mind is to wonder whether Remilia is auspistizing for two Flandres or whether a Flandre is auspistizing between Remilia and a Flandre.