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Subject: Character/Relationship Distribution Help Needed rss

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Michael M.
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Hello everyone! I'm currently designing a game that includes 40 characters, each of which has a relationship of some kind with 6 other characters. I'm trying to figure out a formula with which to evenly distribute all the characters. This is actually the second thread about this problem, since I feel I mishandled how I asked the question in the first one. I waited for a while and decided to start a new thread, but here's a link to the old one if you're so inclined to check it out.

A quick description of the theme: The characters are all attendees at a school dance, think Prom or Homecoming. Each character has their own "Dance Card" (incidentally, what I'm thinking will be the name of the game) that lists their desired dance partners, and their ultimate crush. It also lists their friends who can give them advice as well as a rival that will make it harder for them to score a dance with the partner of their choice. The first player to successfully dance with all three partners on their card (partner 1, partner 2 and their crush) win the game.

Alright, so here are the characters:



As you can see, there are 20 male characters and 20 female characters. Each gendered column is broken down into 4 "classes," or colors. Then, each character has to be assigned two friends and a rival, two partners and a crush. Here are the parameters I want to stay inside:

1. Friend 1 and Friend 2 must be a different class than the character in question, the Rival must be the same color as the character in question, and the crush must be the same class as Friend 1.

2. Characters who are friends in one character's row must be friends on the other character's row as well. For example, If Adam's Friend 1 is Eddie and his Friend 2 is Kent, then on Eddie's row Adam must be one of his friends and on Kent's row Adam must be one of his friends. However, friends of friends do not need to be friends, meaning Eddie and Kent would not have to be friends, even though they are both friends with Adam.

3. Partner 1, Partner 2 and the Crush should all be different classes from the given character. So, if the given character is Quintin (Yellow), then Partner 1, Partner 2 and the Crush should be some combination of Blue, Green and Pink.

4. The same character cannot show up multiple times in the same row. For example, Eddie cannot simultaneously be someone's Friend 2 and their Crush, or someone's Rival and their Partner 1. Each name that shows up in a given row must be unique.

Note: When I say row, I mean for a specific character, not merely the visual row in the excel sheet. For example, Adam's row and Alexis's row are two separate rows, even though, in Excel-speak, they are currently on the same row.

5. No character should show up any more or less times than any other across the grid. They should, in the end, all be evenly distributed.

Gender Specific distribution:

For the most part, characters will be friends and rivals with characters of the same gender, and partners and crushes with characters of the opposite gender. However, I would like to design a game that represents diversity and is inclusive of other orientations, so here is a layout of the gender distribution I see for each row. The "M" stands for male and the "F" stands for female.



Now, there is another layer I must mention that I think makes balancing everything I said above particularly difficult, and why I probably need the help so bad:

Each of the 40 characters will be represented by standees that get randomly placed on the dance floor at the beginning of the game, and there are four sections of the dance floor, so 10 per section. Then, AFTER all the characters have been randomly placed, you draw one of 40 "Dance Cards" (one corresponding to each character) which determines your character and your partners. The Dance Card looks like this:



The square in the upper left corner will have YOUR character's portrait. The two smaller circles to the right of that will represent your friends, and the last circle will represent your rival.

Then, the 3 tilted rectangles under your portrait are the portraits of, in descending order, Partner 1, Partner 2 and your crush. To the right of each of those will be their friends and their crush.

SO, on each card, there will be 16 character faces. None of these 16 can be overlapped. Meaning, on a single card, you can't have one character show up as your friend, and also the crush of one of your partners, or the friend of two of your partners, etc. This includes your own character (upper left box), which can't show up as the friend of one of your partners or anything like that. So this makes the balancing I outline above very important, and perhaps very difficult. Which is why I need the help!

Alright, I feel I've laid out what I need pretty clearly. That wasn't the case the first time I asked about this. I really appreciate anyone who helps me figure out this distribution, because this really isn't my strong point at all. Obviously if my parameters make it mathematically impossible to distribute the way I've laid out, let me know, and feel free to suggest solutions to such problems. Thanks so much!
 
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Matt D
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Part of this mathy stuff makes my head hurt, but is it requisite that the crush be in the same class as Friend 1, and not Friend 2?

Are you making a clear distinction between Friend 1 and Friend 2 from a mechanics standpoint?

I'm wondering (without doing the math) if it's easier to make a distribution where F1/F2/C are three different classes, and R is the same class, thus giving each person a relationship with one person of each class.

I can see if from a thematic standpoint you want the crush to be someone in the same class as one of the friends (that's how they get info, and such) but it may be easier to knock out that distinction.

Also, from a symmetry standpoint, I think you may want to consider the Crush being from the same class, and not the Rival. Unless you mean for the Rival to not be a pair -- IE, Adam's rival is Brandon, but Brandon's rival is Connor, and Connor's Rival is Derek, and Derek's rival is Adam. As compared to Adam and Brandon both being Rival's to each other.

Finally, what determines the other two partners? What's the difference between one Crush and two assigned partners?

Do you think it might make sense to just give each character two Friends, a Rival, and a Crush, and then let them choose their other two misc. partners?

Not sure what exactly you are doing from a mechanic standpoint, but if you do it that way you can introduce things where the choice of other partners could help (or hinder) the ultimate goal of the dance with the Crush.

And it also helps prevent limitation -- I mean, from a mechanics standpoint, unless each character has multiple cards with different partner 1/2 on it, what's the effective difference between a partner and a Crush?

P.S.: I do think this is way better laid out that your first thread on this, which I also followed.
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Michael M.
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hestiansun wrote:
Part of this mathy stuff makes my head hurt, but is it requisite that the crush be in the same class as Friend 1, and not Friend 2?


Not really, there's no meaningful class distinction between friends 1 and 2, I just thought that from a "plug-in-a-formula" standpoint it would just be easier if you knew that the crush and friend 1 always shared a class. But if it's easier not to think of it that way and that the crush just shares a class with ONE of the the friends, and the overall distribution across the grid doesn't favor any one class over all, then that's fine to.

hestiansun wrote:
Are you making a clear distinction between Friend 1 and Friend 2 from a mechanics standpoint?


See above!

hestiansun wrote:
I'm wondering (without doing the math) if it's easier to make a distribution where F1/F2/C are three different classes, and R is the same class, thus giving each person a relationship with one person of each class.

I can see if from a thematic standpoint you want the crush to be someone in the same class as one of the friends (that's how they get info, and such) but it may be easier to knock out that distinction.


Hmm. I haven't done the math yet either (hence the thread!) so I don't know. I'd like to stick to the way I've done it so far, but if after trying a formula once the way you're suggesting is better, then I'd consider doing that instead. It's early enough in the design that all this stuff is malleable.

hestiansun wrote:
Also, from a symmetry standpoint, I think you may want to consider the Crush being from the same class, and not the Rival. Unless you mean for the Rival to not be a pair -- IE, Adam's rival is Brandon, but Brandon's rival is Connor, and Connor's Rival is Derek, and Derek's rival is Adam. As compared to Adam and Brandon both being Rival's to each other.


Rivalries do not have to be mutually perceived. One character can view another character as their rival, but not the other way around. I think either way COULD work, but it has to be consistent. Either ALL Rivalries are shared, or ALL rivalries are one way. I don't really have a strong preference either way.

hestiansun wrote:
Finally, what determines the other two partners? What's the difference between one Crush and two assigned partners?


Basically increased difficulty. This will start to get into the mechanics of the game now, but successfully dancing with someone is the result of a D6 roll. Generally, Partner 1 will be easier than partner 2, and partner 2 will be easier than the crush. Consider it this way:

P1: 3
P2: 5
P3: 7

So if you're attempting to dance with your partner 1, you'll roll the die and as long as you roll a 3 or higher, you successfully convince them to dance with you. You're probably wondering "How in the world do you roll a 7 with a D6 then?" That's in the answer to your next question...

hestiansun wrote:
Do you think it might make sense to just give each character two Friends, a Rival, and a Crush, and then let them choose their other two misc. partners?

Not sure what exactly you are doing from a mechanic standpoint, but if you do it that way you can introduce things where the choice of other partners could help (or hinder) the ultimate goal of the dance with the Crush.


Not with the current setup I have. So, to make it easier to understand, here's some more explanation on the actual game play. First, during setup, you place all 40 character standees randomly on the dance floor, which is broken up into 4 sections. So there will be 10 randomly placed characters in each section of the board. Then, each place gets a random Dance Card:



The top left corner of the card the character you will be playing as. To the right it shows your two friends and your one rival. Being in a space with a friend gives you +1 to your rolls. Being in a space with your rival gives you -1. You don't have to interact with them or anything, this effect happens just by being in the same area. There will be game effects that allow you manipulate who is where on the board, allowing you to call your friends over or move your rival away.

Each turn you may perform two actions, and your choices are Move or Chat. Move means you move from your current section of the dance floor to another. You can do this once, then chat, or you can move twice, up to you. As for chat, if you are in a space containing a friend of a partner or your crush, you can spend an action to "chat" which gives you a token that counts as a +1 to rolls when you eventually try to dance with their friend. Chatting up a partner or a crush initiates the roll to see if you successfully convince them to dance with you or not.

So, it's a combination of being in the same space a your friends, not in a space containing your rival, having chatted up the friends of your partners ahead of time, and a few other factors, that tilt the game in your favor. You want to make sure you're really ready to make that roll, because if you fail, you put an X on the face of the person you attempted to dance with and you can't try again until you remove that X. The way you do that is by successfully dancing with someone else, or through some other special ability that may come up.

There are other things that allow you to manipulate the placement of characters on the dance floor and tilt the odds in your favor, but the gameplay is pretty straight forward. You also have this meta layer on top where players can end up playing as the friends/rivals/partners/crushes of other players, making you want to move away from, or into, spaces containing other players. What happens when you end up playing the crush of another player and you're constantly trying to get away from them while they're chasing after you on the dance floor? I hope that gives birth to some fun moments.

hestiansun wrote:
P.S.: I do think this is way better laid out that your first thread on this, which I also followed.


Thank you very much! I really appreciate the support!

 
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Michael M.
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I just added another section to my original post because it wasn't until I posted that image of the Dance Card I realized that I'd forgotten to mention that you can't have character repeats on any given dance card, they all need to be 16 unique faces. That was one of the big challenges to figuring out the distribution that I needed help with!
 
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Jeremy Lennert
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M_Strauss wrote:
As you can see, there are 20 male characters and 20 female characters.

Your picture actually shows 16 and 16.

My gut feeling is that it should be possible to meet all your requirements given a sufficiently large set of characters, but that 16+16 is probably not sufficiently large. Every dance card is going to have half your characters on it, with fairly complex requirements (e.g. for your partners+crush, you need 3 people who are all of different classes, aren't friends with each other, and also have no friends in common). I can't see any specific reason that it's definitely impossible, but I wouldn't hold my breath.

If I were developing this game, my next step would probably be to write a computer program to search the different permutations and see if it can find one that works.
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Michael M.
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You're absolutely right, it used to be 5 per class, but I thought that may have been too much and reduced it right before posting. I may have to bump it back up.

As for writing a program, I agree that would probably be best. I'm not a programmer though. Is there anyone who could help with this?
 
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Jeremy Lennert
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Do you know a programmer who owes you a favor?

This isn't amazingly hard, but it's beyond the complexity of what I'd typically code for the sake of a BGG thread. Probably several hours for an experienced programmer.
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Nathaniel Grisham

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I don't really have any answers for you, but I do have more questions for you.

I know that us analytic types like things to be nice and symmetrical and balanced. But will this work with your theme? I can see how your constraints make the game symmetrical, but not how they fit the theme.

How are the factions (can I say cliques? I think it works better here, but I obviously don't know everything about your game) separated? Is anyone part of more than one clique? Some friendships reach passed barriers, but not all.
Edit: Oh, you said classes, not factions. That makes a little more sense.

What makes someone my rival? Is it someone who has the same crush as you (actually, usually these people were my friends)? Or is it the crush of your crush?

What about those people who no one really has any kind of crush on? Is there anyone who nobody really wants to dance with, but they will just to be nice?

I'm not saying that you shouldn't have symmetry in your game, but I do think that there are other ways you can make it symmetric while still making the theme feel more immersive.
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Sturv Tafvherd
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I have a solution for the first part (Friend 1 and 2)

Take your list of X names onto individual cards.

Lay out the cards in a circle, such that each card has a neighbor on the left and a neighbor on the right. (Imagine them holding hands and standing in a circle)

So now, each card has 2 neighbors. The card on the left can be Friend 1. The card on the right is Friend 2.

Let's assign the Y colors. ideally, X needs to be a multiple of Y for this to work. And all I'd do is go around the circle and mark them in a pattern ... Red, Blue, Green, Yellow, Red, Blue, Green, Yellow... etc

note: X = n * Y ... where n is the number of people in a color.

...

For the second part: the Rival and the Crush, I was thinking you can just do something like:

Rival = the person Y cards to the right (which should be the next person of your color)

Crush = the person who is Y-1 cards to the right (which would be Y cards to the right of your Friend 1, placing Friend 1 and Crush in the same color)

(incidentally, with that method, your Rival and Crush are friends)


If you want to maintain gender pairs of Crush as much as possible, and assuming a continuous half of the circle is male and the other half is female... then replace Crush with:

Crush = X/2 + Y - 1 cards to the right of "you"
... = 3/2 * Y - 1


edit.. adding this:

Partner 1 = 1 card to the left of Crush
... = 3/2 * Y - 2 cards to the right of "you"

Partner 2 = 2 cards to the left of Crush
... = 3/2 * Y - 3 cards to the right of "you"

... so incidentally, Partner 1 is friends with Partner 2 and your Crush.
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Matt D
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Nice.

That was the math part of this that was hurting my brain.
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Jeremy Lennert
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Stormtower wrote:
(incidentally, with that method, your Rival and Crush are friends)
...
... so incidentally, Partner 1 is friends with Partner 2 and your Crush.

These are forbidden in the problem statement. Every face on your dance card is supposed to be unique, and your dance card includes the friends and crushes of your partners and crush. So your partners can't be friends with each other, or be friends with your friends, or even have the same friends as each other.
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Sturv Tafvherd
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Implementing that method with your group of 32 names:

Adam is friends with Vicky and Eddie
Rival with Brandon, Crush with Paula
Partner 1 and 2 are Katie and Elyse

Eddie is friends with Adam and Kent
Rival with Frank, Crush with Becca
Partner 1 and 2 are Paula and Katie

Kent is friends with Eddie and Quintin
Rival with Lance, Crush with Frannie
Partner 1 and 2 are Becca and Paula

Brandon is friends with Quintin and Frank
Rival is Connor, Crush with Reba
Partner 1 and 2 are Lisa and Frannie

Vicky is friends with Ophelia and Adam
Rival with Quintin, Crush with Katie
Partner 1 and 2 are Elyse and Alexis
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Sturv Tafvherd
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Antistone wrote:
Stormtower wrote:
(incidentally, with that method, your Rival and Crush are friends)
...
... so incidentally, Partner 1 is friends with Partner 2 and your Crush.

These are forbidden in the problem statement. Every face on your dance card is supposed to be unique, and your dance card includes the friends and crushes of your partners and crush. So your partners can't be friends with each other, or be friends with your friends, or even have the same friends as each other.


Ah! I missed that part. I was reading the numbered list, and didn't read the description of the dance card.

Lemme see if there is another formula for the Partners.
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Michael M.
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Thanks Sturv, I really appreciate it!
 
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Sturv Tafvherd
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Ok, re-formulated.

People still in a circle. colors alternating red, blue, green, yellow as you go around.

Number them 1 to X, counterclockwise. (so if we're all holding hands and looking in, and I am "1", the person to my left is "32" and the person to my right is "2")

My number is Z.

My friends are the people I'm holding hands with.
Friend 1 = Z-1
Friend 2 = Z+1

My Rival is the next person on my right that matches my color.
Rival = Z+Y = Z+4

My crush is about halfway around the circle, same color as my Friend 1
Crush = Z+ (X/2) +Y-1

And then...
Partner 1 = Z+ (X/2) + 2*Y -2 (3 down from crush)
Partner 1 = Z+ (X/2) + 3*Y -3 (3 down from partner 1)


Example 1:
Adam is friends with Vicky and Eddie
Rival with Brandon, Crush with Paula
Partners are Lisa and Gina

Paula's friends are Katie and Becca, and crush with Lance
Lisa's friends are Frannie and Reba, and crush with Gabe
Gina's friends are Celia and Michelle, and crush with Derek

(yay, 16 different names)


edit...
X is 32, the number of people, total.
Y is 4, the number of colors.
Z is the individual's index number, 1 thru 32
1 thru 16 are male
17 thru 32 are female

most will have friends of the same gender. Adam and Vicky, Zack and Alexis being at the "ends" of the two halves are the exceptions.

most will have crushes on the opposite gender. exceptions are Howie, Peter, and Zack, Irene, Ophelia, and Vicky.

most will have rivals of the same gender. exceptions are Derek, Howie, Peter, and Zack, Daphne, Irene, Ophelia, and Vicky.

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Michael M.
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Thanks so much! I'm gonna plug this formula in when I get home from work later today and post how it turns out. I really appreciate all the community support!
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Michael M.
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I'm plugging in the formula, and it seems to be working well until I get to around character 12 (I upped it from 32 to 40 characters, 5 instead of 4 in each class, and 1-20 are male). From that point forward, every male has at least one male among his partners or crushes. That ratio is a little off for my taste. While I don't personally have an issue with a high school where almost 50% of the students have an interest in the same sex, it does seem a little unrealistic, and may make parents a little (or a lot) uncomfortable. The Crush ratio is fine, but even with crushes being female, having so many male partners seems off. Of course, there can always be flavor text that says something like "Show me what you got?" "Wanna battle?" or something like that to indicate that sometimes you're doing dance battles as opposed to trying to dance with a romantic interest. Dunno, just thoughts.
 
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Michael M.
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Actually, what I'm thinking is that I will go ahead and continue plugging in this formula until I've filled out every character. THEN, there must be a formula to swapping out male partners for female partners (and vice versa) that doesn't cause overlaps on a given dance card.
 
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Sturv Tafvherd
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M_Strauss wrote:
Actually, what I'm thinking is that I will go ahead and continue plugging in this formula until I've filled out every character. THEN, there must be a formula to swapping out male partners for female partners (and vice versa) that doesn't cause overlaps on a given dance card.


It should just be a matter of swapping the positions of a few members of the same color in the circle. And then apply the formulas

like, Adam switches with Alexis, and Kent swaps with Katie.
 
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Sturv Tafvherd
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I tried out my suggestion above... it doesn't solve the issue. So ... I studied the pattern.

Confession, I didn't test the formulas with different number of colors. So... Ok, re-formulated again

People still in a circle. colors alternating red, blue, green, yellow as you go around.

X = Number of people
Y = Number of colors

X/2 = number of males

Number them 1 to X, counterclockwise. (so if we're all holding hands and looking in, and I am "1", the person to my left is "32" and the person to my right is "2").

All the males on one side of the circle, females on the opposite side.


My number is Z.

My friends are the people I'm holding hands with.
Friend 1 = Z-1
Friend 2 = Z+1

My Rival is the next person on my right that matches my color.
Rival = Z+Y = Z+4

My crush is about halfway around the circle, same color as my Friend 1
Crush = Z+ (X/2) +Y-1

And then...
Partner 1 = Z+ (X/2) + Y +2 (3 down from crush)
Partner 1 = Z+ (X/2) + Y +5 (3 down from partner 1)

These formulas work for Y = 4,5,6 ... I haven't bothered with 7 ... and I know it won't work for 3 colors just because all your partners have to be a color different from yours.

...................

That said, the crush and two partners span an arc of the circle that is 7 people wide. And that's the controlling factor of how many people you'll have that have a "dance interest" with at least 1 person of the same gender.

Let's call them "curious", just for the sake of shortening the description.

For some reason, it's controlled by Y, the number of colors.

For 4 colors, you'll always get 9 curious of each gender.
... 32 people: 9 men of 16 will have at least 1 male partner
... 40 people: 9 men of 20 will have at least 1 male partner
... 48 people: 9 men of 24 will have at least 1 male partner

For 5 colors, you'll always get 10 curious of each gender
... 40 people: 10 men of 20 will have at least 1 male partner
... 50 people: 10 men of 25 will have at least 1 male partner

For 6 colors, you'll always get 11 curious of each gender
... 48 people: 11 men of 24 will have at least 1 male partner


So... if you want to reduce the ratio of curious people, I'd recommend staying to 4 colors, and having at least 48 people. Maybe even get to 72. Which would give you 9 men of 36 will have at least 1 male partner.

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Michael M.
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I don't know if I want to go over 40, the idea of randomly putting out 40 standees on a board may already be a tedious setup for some players. I'm thinking about maybe getting rid of the word "crush" and just having it be "Partner 1, Partner 2, Partner 3." That way its simply people you wish to dance with, which may be romantic but it could also be competitive to show off. I'll still keep the term crush for the part of your dance card that lists your partner's friends and their crush, because there will only be 3 same-sex crushes for each gender, which I'm fine with. The crush will still be determined by that player's partner 3 from their own dance card.

It's coming along great! Your formula helped immensely, I'll show you how it all breaks down once the game is completely plugged into your formula. It's time consuming, but worth it!
 
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Michael M.
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OR... and I haven't tried this yet, but it may work... instead of having characters 1-20 be male, and 21-40 be female, I can have characters 1-15 be male, 16-35 be female, and characters 36-40 be male. That would balance it, right?
 
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Eric Smith
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I like the idea of arranging the students in a circle, but I think mixing boys and girls is leading to your problem of balancing crushes on like genders. What if you made two rings and indexed along the rings?

To make things easier, instead of names, I'm going to use ring position in the format, Number-Color-Gender with number = 1...16; color = G(reen), B(lue), P(ink), Y(ellow); gender = M(ale), F(emale).

Male Pattern = 1PM, 2GM, 3YM, 4BM, 5PM, 6GM, 7YM, 8BM... 16BM
Female Pattern = 1PF, 2GF, 3YF, 4BF, 5PF, 6GF, 7YF, 8BF... 16BF

== for males ==
Dancer = 1PM
Friends = 16BM, 2GM (+15M,+1M)
Rival = 13PM (+12M)
Partners = 2GF, 7YF (+1F,+6F)
Crush = 12BF (+11F)

== for females ==
Dancer = 1PF
Friends = 16BF, 2GF (+15F,+1F)
Rival = 13PF (+12F)
Partners = 6GM, 11YF (+5M,+10M)
Crush = 4BM (+3M)


I think this works out but of course gives 100% like gendered friends and different gendered partners.

Edit: It doesn't work...
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Michael M.
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Bloomington-Normal
Illinois
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Alright, I think I somehow made it worse, BUT, there may be a way to fix it, with your guys's help!

So, here is the character ring I made. Characters start with the letter M are male, and characters that start with the letter F are female. For simplicity's sake its a diamond instead of a circle:



This was after I mixed it up so it wasn't all males 1-20 then all females 21-40. I thought this would bring the gender balance in partners and crushes down to something more reasonable. It didn't.



From left to right, the columns represent:

Number/Name/Friend1/Friend2/Rival/Parner1/Partner2/Crush. Just for simplicity's sake I added three small columns at the end, the circles representing the same gender in categories Partner1, Partner2 and Crush. As you can see now, its the minority of students who don't have any same sex partners or crushes.

However, all may not be lost. Perhaps with this out in front of us now, we can figure out a formula to redistribute based on the model the way it is now as opposed to starting from scratch all over again. What would the formula be for that though?

Edit: Here's another view of the grid,

 
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Michael M.
United States
Bloomington-Normal
Illinois
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And this needs to all be done in such a way where there won't be repeats of characters on any given dance card. I haven't tested that myself yet, I'm just trusting Struv's calculations were correct.
 
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