This is discussion archived from a time before the current discussion method was installed.
Madacaek: Why is this a JPEG? It looks horrible.
Dr Dedman: You left one out. B and C are attracted to each other but are also attracted to/loyal to A which messes with their relationship. Berserk is the canatonical example (Gutts and Caska are B and C Griffin is A).
Fast Eddie: That would be #8, no? Ah. Okay. That's in. //Later: Hmm. It occurs to me that the arrow heads could also represent antipathy, a graph of who-hates-who. That should probably be a separate entry, if at all.
Sniffnoy: Actually, assuming we stick to just love relations, there are in fact 4 possibilities not shown. These are (numbering them arbitrarily):
10.
a↔b
b→c
[Could be like a partial #7 or inverted #4, with the "affair" being purely imaginary, or could be a case of b having settled for #2. Er, not triangle #2... you know what I mean.]
11.
a↔b
a↔c
b→c
[???]
Fast Eddie: One of the people having an affair with 'a' ('b') becomes interested in another person ('c') who is also having an affair with 'a'.
12.
a→b
b→c
a↔c
[Like #2 but 2 of the characters are already in a relationship? Alternatively an extended #4 or an extended #10.]
13.
a→b
b→c
a→c
[Like a one-sided #9?]
I don't know what all these mean storywise, or how frequently they occur, but combinatorially, that's it.
Seth: I fleshed out a few by imagining situations where they might happen. Which leads me to 11, i honestly can't think of a situation like that. It would be like Ross attracted to his wife's girlfriend or something.
osh: You know it's math when it has a bunch of letters that don't spell stuff.
Fast Eddie: pulled ....
- Dude, it's equilateral.
- The Defenestrator: What joke? It's not at all set up as one.
- TJ Devil 02: It's in response to asking "Whether a is Vyse or not is open to interpretation." The author did not mean equilateral, but symmetric; it doesn't matter whether Vyse is at a, b, or c — the relationships are all the same.
//Later: Huh. Twin Bird backed some some comments here that seemed like good discussion. What's up with that?. A Lso, pulled ...
- It's not unusual for both A and B to realize they're better off together.
Rob Mandeville: I remember seeing a variation on #8 in a webcomic I don't recall the name of. B and C were a married couple, each of which having an affair with bisexual A. The arrows all look the same, but the triangle is isosceles rather than equilateral, with BC being the long side.
Fast Eddie: Axed some natter.
Tsundere Lightning: I was under the impression that Tedd was the Author Avatar in El Goonish Shive, not (or at least in addition to) Elliott? Also, the name Mary Supermarket is Made Of Win.
Daibhid C: I'm honestly struggling to work out where the arrows should go in the tangled web Arthur King Of Time And Space has made of the King Arthur triangle. It seems like it should be an ll, with unfaithful-but-guilt-ridden Guenevere at A, loyal-but-conflicted Lancelot at B, and clueless-or-is-he Arthur at C. But that fails to take into account Arthur's loyalty to his best knight. Adding that in gives us an 8, but the text calls that a "straight-up threesome" which isn't right either (although it may become so in some of the arcs).
Daniel LC: I just want to make sure all of the combinations are here. I assume orientation doesn't matter, and all there must be somehow involved. 1) There are 4^3=64 gross combinations. That counts most, but not all, six times. 2) To find the flippable ones, leg A must be symmentric, so it can be reciprocated or nonexistant, there can't be two nonexistant, which leaves three for the other two legs. Naturally, they must be identical. 2*3=6 flippable ones. They're counted three times, and we want to count them six, so we count them another three times. 64+6*3=82 3) The only one that rotates but doesn't flip is what's currently type 2. It's counted twice, so we count it four more times. 82+4=86 4) There is one (currently number eight) that rotates and flips. I've counted it once in step 1 and three more times in step 2, so I need to count it two more times. 86+2=88 5) This counts the one where there are no relations. 88-1=87 6) There also shouldn't be ones where there are two empty relations. The actual relation can be on any side, and there are three actual relations. 87-3*3=78 7) Now every one is counted six times. 78/6=13 Good. As a side note, should the actual rules for distinct relationships be put on the page?
Anyone seen one of the following in anything, or should I put it on my list of (non-Rule 34-connected) ideas to fiddle with? (Or is it on the list already and I shouldn't read trope pages while tired?) A and B are together. C is aware of this. A is cheating on B with C. B is cheating on A with C. Neither A or B is aware of the other's infidelity. C knows everything, and can be either a romantic Chessmaster and Jerkass or can Angst/Wangst about the entire situation. Obviously requires Ho Yay somewhere.
Twin Bird: This sounds like Type 9, although the way it's written skirts the Ho Yay.
Question: Why does the Type Seven HAVE to be a cheating one? One way a non-cheating Type Seven would be that A is attracted to both B and C. B and C are attracted to A, but not to each other. Both B and C are willing to Let The Best Girl Win decide who gets A. A decides to take both of them. B and C agree, as long as A doesn't try and force them together. Basically
A<->B
A<->C
B<-/->C\
