# rub your belly and pat your head

its either that or the other way around, i forget. but also, imagine if you cant do both on your own.
however, say you and your friend are each individually capable of one of the elements. that is, you can
rub any belly, and they can pat any head. now we're getting somewhere: the two of you can get into an
arrangement where both of you are getting your heads patted and your bellies rubbed as a closed system.

we'll label rubbers as R and patters as P, now see the below diagram

.--.   .--.
|  |   |  |
.->R-->P<-.
   ^   |
   |   |
   .---.
   
See? everyone's happy. One person failed, but two succeed together.
But you fail again if you have three, since there's always either one
extra rubbing arm and one too few patting arms, or the other way around.
See below for a single example:

   .---.
   |   |
   v   |
<--R-->R-->P--.
   ^   ^   |  |
   |   |   |  |
   |   .---.  |
   |          |
   .----------.
   
note how the first R's hand goes to waste and the P uses both hands on
its friends but has no P hand for itself? Tragic.

any 2n players can succeed by simply pairing up, and you can easily prove
that 2n+1 will fail, since for 2n+1 players you need 2n+1 R hands and 2n+1
P hands, but each player brings 2n hands to the table. it never lines up.

that being said, we could include some hypothetical hybrid who is able to
do both R and P, one on each hand, and we'll call them B for Both. or Bofa
this solves some issues in that now 2n+1 will always work while 2n fails:

.-----------.
|           |
|   .---.   |
|   |   |   |
|   v   |   |
.-->R-->B-->P<--.  (+ arbitrary number of R + P pairings)
    |   ^   |   |
    |   |   |   |
    |   .---.   |
    |           |
    .-----------.


.--.
|  |
.->R-->B--> (already used its P hand but doesnt need to use its R hand on itself)
   ^   |
   |   |
   .---.
   
again, easily proven: 2n homo-hands + 1 hetero-hand will always make 2n+1 of each
hand (2n for each of the homos' types + 1 P and 1 R for the single hetero), but
2n-1 homos + 1 het will result in *still* 2n+1 of each hand, since homos *always*
provide 2n of each, and 1 het always equals 1 of each. if you allow more than one
het it runs into the issue that any single het is self sufficient, as seen below:

.--.
|  |
.->B--.  (obviously)
   ^  |
   |  |
   .--.
   
which is obvious but my point is that if you don't cap the hybrids, any diagram n
has freedom to become this:

.--.
|  |
.->B--.  (+ arbirary number of identical copies)
   ^  |
   |  |
   .--.

so you could maybe then ask, how many hets does it take for an even total number of hets
and homos to all get along and rub bellies and pat heads? that's a reasonable question,
maybe we try and find the minimum, as a function of n. for 2n of (B, P, R), how many
Bs minimum does it take to satisfy everyone? it's 2, per our previous reasoning.
2n homos + 2n hets = 2n of each and we're done. example:

.--.     .--.     
|  |     |  |     
.->B--.  .->B--.  (+ arbitrary number of R + P pairings)
   ^  |     ^  |  
   |  |     |  | 
   .--.     .--.
   
and this is of course just assuming we cant have 0 hybrids...

the only problem that having 1 hybrid introduced is that there was no longer able to
be an even number of P and R hands. 1 more brings us back to parity. not interesting.
so then is there a version of this that is more interesting? maybe if we stop looking
at regular* humans and start looking at mutated ones with more than two arms. or if
we bring about more needs to fulfill, such as a need to scratch one's nose. sure!

*if a bit discoordinated

if everyone's got 3 arms but the same initial goal, 2n still always clears since we've got
no rule against having spare hands, but what 2n+1 cases succeed? maybe ones that are 3n?

   .---.
   |   | .---.
   v   |/    |
.--P-->R<----.
|  ^   |
|  |   v
|  .---P
|      ^
|      |
.------.

yeah this works. we should have expected this from the original 3 body problem, since this
just gives everyone an extra arm to work with. only the minority even ends up needing it,
in this case the R just needs an extra rubber, and having a new hand they can rub themself

this also means we can stick a (+ arbitrary number of R + P pairings) at the end and get that
all 2n+1 work above n=0, which will never work without a hybrid. since 2n and 2n+1 (barring n=0)
all work, we now know all n>1 work. yaay

restricting free hands always fails in this setup, since there are always more hands than needs,
and i wont explore varying number of hands yet, since i want to touch on the nose itching first.
we'll call scratchers S for short, or rather for scratchers, even though S could also be for short.

we obviously need at least 3 people, since everyone is capable of satisfying at most one need, but
it isn't enough to have 3 or more, since needs scale faster than satisfying hands, since for each
player there are 2 hands but 3 needs. let's give everyone a mutation, 3 arms. no need for another
diagram yet as this is trivially just a mapping of every element to every element, but i'll do it
anyway to show off my brand new double headed arrows:

.---.    .---.
|   |    |   |
.-->R<-->P<--.
    ^    ^  
    |    |      
    |    v
    .--->S<--.
         |   |
         .---.

so 3n works of course, but 3n+1 and 3n+2 could never work since each new
player after 3n requires there be 3 fresh hands of each type, which only
lines up on 3n. its the same logic from before. so lets see what hybrids
can do about it; of course if you add a triple hybrid who can do all the
actions they immediately cause the same issue as the bofa hybrids:

T (satisfied singlet)
TT (satisfied duplet)
RPS (satisfied triplet)
RPST (satisfied 3n+1)
RPSTT (satisfied 3n+2)

but what if certain hybrids can only satisfy two needs? say if we bring
the original bofa back and he can only pat and rub, and maybe he can do
any combination with his mutation but he cant ever scratch the itch, of
the nose that is. what can he do? what groups can he cooperate with?:

.---.    .---.
|   |    |   |
.-->B<-->B<--.
    ^    ^  
    |    |      
    |    v
    .--->S<--.
         |   |
         .---.

well that's unsurprising. i skipped a diagram but if you just have RBS
the B one just has 3 P hands which is obvious, but in this case we can
say theyre like an R and a P who swap one pair of hands to pat and rub
the S with. essentially this whole thing never gets you far since only
S can ever scratch, so you always need an S, and then you always need
an a P or B to pat it, and then you always need a third guy to handle
the combined needs that the S can't fulfill. im playing it a little
looser with the proofs but you should be able to figure it out.

so varying arm counts... uhh... probably also doesnt work, sorry if i
kept you in suspense. if you let a guy have more arms you need less of
that guy, over time, and if you let a guy have less arms you lose
because theres never enough of that guy to satisfy itself. everyone
needs to have at least k arms for k needs to be satisfiable, and more
just makes it easier to satisfy off-parity numbers of guys.

diagram of some boobies: ( . Y . )

#tits #RPS #multitasking