Coupling Problem Equation

[Edo4169]

Before you read this thread, I had the idea from this manga.

With 2 people you'll get 4 ways of story with genderswap think, either it straight, reverse, yaoi or yuri. But the question is how many wil you get when the love interest is 2 or more. Some case I think of:

1. MC is male and the love interest is man and woman.
2. What if there's 5 side love interest.

What kind of equations you use to get answer?

 

[Blocky_Mudkip]

If there's 3 people( A,B,C), there's 12 possible couples. There's 3 possible combos of people (AB AC BC) and each couple could be straight, reverse, Yuri, or yaoi. 3x4=12

 

[Blocky_Mudkip]

Wait does the MC have to be involved

 

[UnnamedPlayer]

I'm pretty sure you can find an equation by using graph theory. For each person, create two points : one for the female and another for the male. Then separate the male and female into two different sets. Deal with the straight/reverse couples by treating it as a bipartite graph, where only relations allowed are bijectives, identity rejected.I just looked on the net, but there is n! bijections between two sets, so it should be n! if only straight relations are allowed.

Deal with the yuri/yaoi couples by taking each set and calculating the number of pair it can contain, by divising its cardinal by two. So it should be close of (n! - 1) + n for n persons. EDIT : And it don't work. It works if you allow only straight couples, but blending yuri/yaoi couples with straight mess up something. But i think you can solve the problem with the binomial coefficient and some combinatorials.

One detail : if you have a yuri relationship, it must imply a yaoi realationship to compensate.

 

[UnnamedPlayer]

If there's 3 people( A,B,C), there's 12 possible couples. There's 3 possible combos of people (AB AC BC) and each couple could be straight, reverse, Yuri, or yaoi. 3x4=12

For 4 people, the logic don't work like that, since if a pair form, it force the other pair. If A choose B, C or D, everything is done. You have either AB + CD, AC + BD, AD + BC. Each couple/pair can be Straight, Reverse, Yuri or Yaoi, which mean each pair of pair can lead to 4^2 combinations possibles. Multiply by 3, and you have : 48 possibilites.

I'm pretty sure the equation will be different if the number of persons is odd or even... One case will lead to singles, not the other.

 

[Data_Gold]

i'm kinda shit with math so it could be very wrong, but from what i can work out, the formula
2^((n*(n-1)/2)+1) with n=the number of people
should give you your number, but from what i can tell, it does give you pairings with excluded members, for example a pairing with 3 people, it will include things like AB pairings ignoring C
with it throwing out;
1 person: 2
2 people: 4
3 people: 16
4 people: 128
5 people: 2048

 

[UnnamedPlayer]

I'm pretty sure it's 2 * n!, it fit for 2, 3 and 4 persons. I'm pretty sure you can proove it by recurrence, btw. Let's try it.

EDIT : and it give me another equation.

Imagine you already have n persons. For n persons, you have X_n possibilites. And now, you add one more love interest, what will happen ?

If n is odd, then there is someone single (may be the MC), and the new love interest will og with him/her and make a new pair. Since the pair can lead to either a straight, reverse, yuri or yaoi relationship, you have :

X_n+1 = 4 * X_n-1 for n odd.

If n is even, the new love interest can be left single, or can pair with someone else. There is at least X_n possibilities where he/she don't pair with anyone and is left single, because adding him/her don't change anything. If he/she pair, there is n pairs possibles, multiply by 4 to take the 4 possibilities in a pair, and it will leave n-1 person so X_n-1 possibilites for each pair. So it lead to :

X_n+1 = X_n + 4 * n * X_n-1 = (n+1) X_n for n even

You have X(1) = 1, X(2) = 4. The recurrence give 12 for n=3, 48 for n=4, and so on.

 

[Edo4169]

Wait does the MC have to be involved

Yes, that's the MC for.

I'm more confuse reading franchie explanation

i'm kinda shit with math so it could be very wrong, but from what i can work out, the formula
2^((n*(n-1)/2)+1) with n=the number of people
should give you your number, but from what i can tell, it does give you pairings with excluded members, for example a pairing with 3 people, it will include things like AB pairings ignoring C
with it throwing out;
1 person: 2
2 people: 4
3 people: 16
4 people: 128
5 people: 2048

Me too, I'm not good in math either.

I'm pretty sure it's 2 * n!, it fit for 2, 3 and 4 persons. I'm pretty sure you can proove it by recurrence, btw. Imagine you already have N persons, and add one more love interest, what will happen ?

Can you proof it? This equation is mor make sense but I still can't grasp it.

 

[Angry_Panda]

New rule: no math after bedtime.

For n individuals:

Sex assignment is easy - 2^n.

Coupling - assuming the maximum number of pairs must be formed (so there is never more than one single individual), this works out to the product of all odd natural numbers equal to or less than n. So for 1 or 2 individuals, 1 possible pairing; for 3 or 4, 3×1; for 5 or 6, 5×3×1.... (Too tired to get out of bed and go to the computer to type the derivation of this.)

Hence, for 4 individuals, 2^4×(3×1) = 16×3 = 48 possible combinations.

For 5 individuals, 2^5×(5×3×1) = 32×15 = 480.

For 10 individuals, 2^10×(9×7×5×3×1) = 1024×945 = 967680.

 

[Edo4169]

New rule: no math after bedtime.

For n individuals:

Sex assignment is easy - 2^n.

Coupling - assuming the maximum number of pairs must be formed (so there is never more than one single individual), this works out to the product of all odd natural numbers equal to or less than n. So for 1 or 2 individuals, 1 possible pairing; for 3 or 4, 3×1; for 5 or 6, 5×3×1.... (Too tired to get out of bed and go to the computer to type the derivation of this.)

Hence, for 4 individuals, 2^4×(3×1) = 16×3 = 48 possible combinations.

For 5 individuals, 2^5×(5×3×1) = 32×15 = 480.

For 10 individuals, 2^10×(9×7×5×3×1) = 1024×945 = 967680.

Doesn't this equation the same with @UnnamedPlayer did?