The Homestuck Troll Theorem
Created: / Modified:
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.
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!
- 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:
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, WV ♦ Black King and Tavros ♦ Equius).
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:
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:
|Eridan||Feferi||Horuss||CD||Latula & Dirk|
|Feferi||Eridan||CD||Horuss||Jack & Jane|
|Horuss||CD||Eridan||Feferi||Roxy & White King|
|CD||Horuss||Feferi||Eridan||Jake & Kurloz|
|Latula||Jack||Roxy||Jake||Eridan & Dirk|
|Jack||Latula||Jake||Roxy||Feferi & Jane|
|Roxy||Jake||Latula||Jack||Horuss & White King|
|Jake||Roxy||Jack||Latula||CD & Kurloz|
|Dirk||Jane||White King||Kurloz||Eridan & Latula|
|Jane||Dirk||Kurloz||White King||Feferi & Jack|
|White King||Kurloz||Dirk||Jane||Horuss & Roxy|
|Kurloz||White King||Jane||Dirk||CD & Jake|
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.
|DD||Black King||Equius||Aradia||Tavros & Cronus|
|Tavros||Aradia||Black King||Equius||DD & Cronus|
|Cronus||Equius||Aradia||Black King||DD & Tavros|
|Black King||DD||Tavros||Cronus||Aradia & Equius|
|Aradia||Tavros||Cronus||DD||Black King & Equius|
|Equius||Cronus||DD||Tavros||Black King & Aradia|
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 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.
1There's at least a billion reasons why this wouldn't work to begin with, but still!
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