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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:

JadeMeenah
 
JakeCD

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, Jade ♦ CD and Meenah ♦ Jake).

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:

 
KankriRoseMeenahRufioh
RoseKankriRufiohMeenah
MeenahRufiohKankriRose
RufiohMeenahRoseKankri

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:

 
KankriRoseMeenahRufioh