do you reset the master apprentice relationship each incarnation?
Trying to work out the amount of permutations in a number set that behaves in this manner:
"This is a reincarnation problem. There are six incarnations of artists. Three are alive and three are waiting in limbo to be reborn. There is always a Child, an Apprentice, and a Old Master. As the Old Master dies the Apprentice takes his place as Master, the Child becomes the Apprentice, and one of the three souls in limbo is reborn as the Child. The Old Master cannot be reborn until his Apprentice has passed away too because after spending all that time together, they would recognize each other and all of their conversations would be super awkward. So, how many different variations of Child, Apprentice, and Old Master are possible? How would you express the equation if there were more than three souls waiting in limbo? Four, five, six, etc.?"
The only formula I remember from my days in math class is:
but I think this is a little too simple to factor in the weird rule about consecutive persons being removed from the pool until they both are "dead". Any thoughts?
Yes. The way I'm looking at it appears to be like this [a,b,c,] are in limbo and {d,e,f} are alive with d the child and f the master. When f dies there is one space to fill from a pool of [a,b,c]. So say after one permutation of that, it would then look like this: [b,c] in limbo {a,d,e} alive in the respective roles, and (f) ruled out until e has also died. This is where I start to get lost in the math.
are you only interested in the number of possible sets of {c,a,m} given an initial condition and infinite running of the cycle?
If you can arbitrarily set the initial conditions I'm pretty sure you arrive at all of the possible combinations.
I'm mostly interested in the minimum amount I'd need in the [a,b,c] limbo pool to arrive at or above 50 possible unique combinations of {c,a,m} without repeating an a or m in the same set
so any of the letters within the limbo pool [a,b,c] can become the child essentially...the only real rule is that the master can't be immediately reincarnated, but has to sit out a round.
Ok. I ran this thing for various size populations long enough until I was certain of seeing all the permutations for a given size. These are the numbers. The first number is the population size, the 2nd the number of permutations given your scheme.
$\begin{array}{cc}&4 &4 \\ &5 &60 \\ &6 &120 \\ &7 &210 \\ &8 &336 \\ &9 &504 \\ &10 &720 \\&11 &990 \\ &12 &1320\end{array}$
for population size N > 4 this is $N(N-1)(N-2)$
which is just what you'd expect selecting 3 unique items from N.
Your scheme does affect a population of 4 though reducing the number of valid permutations from 24 to 4.
Thank you so much for working this out for me! My mind tends to get hung up on things like this, and I could've been stuck here for months. If I ever get this damn book written I'll be sure to include you in the thanks!
All the best
--Jim
Here's the code if you're interested. It's in Mathematica.
(* Initialize *)
ClearAll[Evaluate[Context[]<>"*"]];
maxRuns=2000;
nPeople=6;
people =Table[i,{i,0,nPeople-1}];
initpool=people;
k=RandomInteger[{1,nPeople}];
om= initpool[[k]];
initpool=Delete[initpool,k];
k=RandomInteger[{1,nPeople-1}];
a= initpool[[k]];
initpool=Delete[initpool,k];
k=RandomInteger[{1,nPeople-2}];
c= initpool[[k]];
initpool=Delete[initpool,k];
limbo=initpool;
holding=-1;
snaps={};
(* start the cycles *)
For[i=1,i<=maxRuns,i++,
AppendTo[snaps,{c,a,om}];
If[holding != -1,AppendTo[limbo,holding]];
holding=om;
om=a;
a=c;
k=RandomInteger[{1,Length[limbo]}];
c=limbo[[k]];
limbo=Delete[limbo,k];
];
Length[DeleteDuplicates[snaps]]