The (In)Complete List of Shippers - V8 - Page 14

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Thread: The (In)Complete List of Shippers - V8

  1. #196
    追放されたバカ Transfinite's Avatar
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    Hee hee hee.

    Found my list! More ships!


    SudoSnagShipper: Miror B. & Wes
    (Wes snagged Miror B.'s Sudowoodo.)

    EnteiSnagShipper: Dakim & Wes
    SuicuneSnagShipper: Venus & Wes
    RaiSnagShipper: Ein & Wes
    (These should be apparent.)

    SuicuneShipper: Eusine & Venus
    (Yay for Suicune!)

    EnteiShipper: Dakim & Molly Hale
    (Yay for Entei!)

    SudocoloShipper: Miror B. & Brock
    (They both have Ludicolo, and Brock fell in love with a Sudowoodo once if I'm not mistaken.)

    CelebiShipper: Vicious & a Time Flute
    (Don't kill me, Blackjack.)

    TimeFluteShipper: Vander & Wes
    (Vander gave Wes a Time Flute. I used mine on Entei -_-)

    IdentityShipper: Rui & any Shadow Pokemon
    (Rui can see the auras of Shadow Pokemon.)

    PlusleShipper: Duking & Wes
    (Duking gives Wes his Plusle.)

    NetShipper: Lanette & Nett
    (*shrug*)

    SnagShipper: Gonzap & Wes
    (Gonzap is the leader of Team Snagem, and Wes defected from it because of a lover's quarrel.

    ...hey, it never SAYS why he blew their hideout up and stole that machine!)

    WescourShipper: Wes & Nascour
    (I couldn't think of anything better...)


    And some random, non-Colosseum ships:

    OhayoNekoShipper: Skitty & Hello Kitty

    GorgeousShipper: Painter Celina & Lady Gillian
    (They're two random trainers at Resort Gorgeous in FR/LG. I swear there's something between them.)

    PlushyShipper: Lorelei and the G/S/C character's mom
    BoxShipper: Lanette & Bridgette

    ShoujoShipper: Hazel & Haruka
    (That's Haruka from Chamo-Chamo Pretty, not May-Haruka.)
    Last edited by Transfinite; 29th November 2004 at 03:39 PM.

  2. #197
    Pie
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    Quote Originally Posted by Ketsuban
    SudocoloShipper: Miror B. & Brock
    (They both have Ludicolo, and Brock fell in love with a Sudowoodo once if I'm not mistaken.)
    ...

    *DIES*

    That ship just made my day. No, really. Though I admit part of it's just from wanting to be in the middle of it...

    Now that I think about it, there's more you can draw on, too...such as the fact that Miror B has the awesome Pokéfro, and then there's that episode that started with everyone getting their hair turned poofy and 'fro like and Brock was the only one who seemed pleased with his...

    CelebiShipper: Vicious & a Time Flute
    You know, now I'm very tempted to write a fic based on this ship...

    I heart your Colosseum ships. Seriously.

    Edit - Huh, hundredth post.
    Last edited by Pie; 29th November 2004 at 10:50 PM.
    By reading this, you have given me temporary control of your mind.

  3. #198
    MondoTR's Avatar
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    Quote Originally Posted by Ketsuban
    GorgeousShipper: Painter Celina & Lady Gillian
    That name's already taken:
    Quote Originally Posted by Archaic
    GorgeousShipper - Pistachio & Ginger

  4. #199
    A black and white world Blackjack Gabbiani's Avatar
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    Quote Originally Posted by Pie
    You know, now I'm very tempted to write a fic based on this ship...
    You do, and I kill you. And Ketsuban for coming up with it.

  5. #200
    Pie
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    Can't kill me, Blackjack. Remember?

    Quote Originally Posted by Pie
    *DIES*
    I'm already dead.
    By reading this, you have given me temporary control of your mind.

  6. #201
    MondoTR's Avatar
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    *revives Pie* Now Blackjack can kill you.

  7. #202
    φιλομαθής Zhen Lin's Avatar Vice-Webmaster
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    Quote Originally Posted by Murgatroyd
    And it turns out that one-sided ships don't make it as complicated as I thought. Including all one-sided relationship components among all characters, including whether each character loves him/herself, the maximum number of ships is 2^(n^2), where n is the number of characters.
    This does not quite compute for me. How is it 2^(n^2)? 2^n is simple enough to understand... but 2^(n^2) sounds more like the possible number of combinations of presence/non-presence of a particular ordered pairing. Or is that what is meant by one-sided?

    If that is the case, it seems to be inconsistent with the definition of ship used to compute 2^n...

    Hmm. Now that would be an interesting project, to solidly define what a ship is, so that we can theoretically count them all.

  8. #203
    Cheers to the Freeze Luna Tiger's Avatar Bulbapedia Staff
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    ....Right. So stepping away from the death and math as I amend something I said about, oh, a hundred pages back during my rampage:

    Redthreadshipping - Gary/Drew
    Avatar artwork by アカネ
    PRIA is my haunt, 45500 is my identity,
    the black stronghold is my fortress.

  9. #204
    追放されたバカ Transfinite's Avatar
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    That name's already taken:
    :(

    Okay then, ResortGorgeousShipping.

  10. #205
    Unseen Watcher Murgatroyd's Avatar
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    Quote Originally Posted by Zhen Lin
    This does not quite compute for me. How is it 2^(n^2)? 2^n is simple enough to understand... but 2^(n^2) sounds more like the possible number of combinations of presence/non-presence of a particular ordered pairing. Or is that what is meant by one-sided?

    If that is the case, it seems to be inconsistent with the definition of ship used to compute 2^n...

    Hmm. Now that would be an interesting project, to solidly define what a ship is, so that we can theoretically count them all.

    For each ordered pair of characters (A,B), either A loves B (call this "1") or A does not love B ("0").
    Let S denote the space of all characters, of cardinality n. S^2 is the space of all ordered pairs of characters, and has cardinality n^2.
    Let f be a function from S^2 to {0,1}, that is, for each ordered pair of characters, it tells whether the first loves the second. Every ship, whether one-sided, two-sided, love triangle, or even more complicated, can be represented by a unique function of this type. There are 2^(n^2) such functions. Thus, 2^(n^2) is an upper bound on the number of Ships.
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  11. #206
    φιλομαθής Zhen Lin's Avatar Vice-Webmaster
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    I see. Amazing what one night of sleep can do in understanding a problem. I didn't realise that any graph could be reduced to a list of (ordered) pairs.

  12. #207
    Unseen Watcher Murgatroyd's Avatar
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    The definition of "Ship" used in my proof, however, seems a bit broad. It considers the union of two unrelated Ships to be another distinct Ship (i.e. "A loves B", "C loves D", and ""A loves B and C loves D" are all Ships). It seems to me that for our purposes, we need a stronger definition, so that "PalletShipping and Gymshipping", for example, is not a Ship of its own. I think the key to the improved definition would be that all characters in a Ship must be connected somehow, either directly or in a chain of characters loving/being loved by others. This gives a new problem which, when solved, should give an exact count of Ships:
    How many weakly connected subgraphs are there of a complete digraph on n vertices?
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  13. #208
    φιλομαθής Zhen Lin's Avatar Vice-Webmaster
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    Better question: can a purely directed graph distinguish between a mutual A love B and a cyclic, directed A love B love A? Should those be even distinguished?

  14. #209
    Unseen Watcher Murgatroyd's Avatar
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    No, it cannot make the distinction. I don't quite understand the distinction myself, either.
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  15. #210
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    Maybe in the former case they know they love each other, and in the latter they don't?


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