Notes on Species

Introduction
The "transport of structures"is introduced in the very first page of the very first article on species.

An example of transport is given using endofunctions. Let A be a set of three elements. Then there are 27 endofunctions of A, grouped in seven types:


 * 27 = 1 + 2 + 3 + 6 + 6 + 6 + 3

The associated species is:


 * 3.X3 + 2.X·E2 + C3   + E3

and the cycle index for 3-endofunctins is


 * (27.x3 + 9.x.t + 6.c )/6

The question
If we look at the above species formula we can read ( in the 2.X·E2 term) that bijections like:

may be transported in functions like:

because they have the very same subjacent species, namely X·E2. The question is: What the transport of structures is transporting if the endofunctions are NOT transported ?

The answer
Normally bijections transport cardinality. A chain of bijections A->B->C-> ... ensures that all A, B, C,... have the same cardinality.

If the chain eventually hits back the set A like A->B->C->...->A, the chain of bijections act as a permutation on A. This permutation on A is then copied on every other set B, C, ...

The transport of structure transports permutations and nothing else, one at a time. If we consider a group of permutations, this group may not act transitively nor on A neither on F[A]; in this case a multi-sort species that preserve a partitional structure arise.

Note on the transport of structures
Since three basic operations are defined by splitting the set A with n elements like n = (n-1) + 1 (differentiation), n = m1 + m2 (product) or n = m1 + m2 +...+ mk (partitional composition) one should understand the transport of structure as a transport of partitional stuctures. boicu|Nicolae-boicu 2012 aug

Integration of species
In terms of permutation groups, primitives of molecular species are called transitive point extensions. This does not close the issue, since (UV)'= U'V+UV'

The name of species
The combinatorial species Lin does not necessarily stands for a Linear Order, in the very same way that Ens does not stands for a Boolean Algebra.

A word xyzt has the very same combinatorial structure as { βx, 3y, Az, Πt } where β, 3, A, Π are specific "slots".

Ens are cardinal numbers, Lin are sets
- a mathematical set is rock, a solid rigid chunk of members where there is no way to confuse a member with other. Eg, in {b,c,d} the element b has no copies, no clones, no conjugates or whatever cousins. When writing down a set we say by mouth "order does not matter" just to correct the imprecision of the written notation. A mathematical set is, in fact, a Lin species, an X.X.....X

- a combinatorial set is a cardinal number. We could write Ens3(X) = {X,X,X} = {X}^3, and Ens = N where N are all natural numbers and Ens is the combinatorial species.

2808845 - A question about the definition of species : Sudoku
https://math.stackexchange.com/questions/172401/a-question-about-the-definition-of-species/2808845#2808845 Q A species F is defined as an endofunctor of the category of finite sets. What if our combinatorial structure is not defined for sets of arbitrary size. More precisely, can we define a species from a subcategory of finite sets? For example, the species of Sudoko puzzles.

A The combinatoric point of view is that the theory is about labelled structures. In the case of Sudoku, unfortunately, we only have 9 labels to label something (even they are repeating).

If we consider that by relabeling eg we interchange 1 and 2 in a Sudoku puzzle we got the same Sudoku, the species is somebignumber.Ens9 i.e. a large number of copies of the set {1,2,3,4,5,6,7,8,9} with 9 elements - and nothing else.

If we consider that relabeling produces different Sudoku's, the species is samebignumber×9!×Lin9, i.e. a large number of copies of all 123456789 like sequences.

the Roger Blazey's problem
"Consider a well shuffled pack of playing cards from a 52 card standard pack. Now on drawing out 1 card at a time and exposing each card at the moment of withdrawal....ON AVERAGE how many cards would you need to draw from the pack in order to obtain at least 1 card from each suit?

the David Radcliff's solution
The expected number of cards drawn is 52*p, where p is the probability that any particular card is drawn. To be definite, let's say that p is the probability that the Ace of Spaces is drawn.

There are four events such that the Ace of Spades (AS) is drawn:
 * E1 = AS is the first spade in the deck.
 * E2 = AS comes before all hearts in the deck.
 * E3 = AS comes before all clubs in the deck.
 * E4 = AS comes before all clubs in the deck.

We use the law of inclusion-exclusion to find the probability. This requires finding the probability of each event E_i and the probabilities of the intersections of the E_i.


 * P(E1) = 1/13
 * P(E2) = P(E3) = P(E4) = 1/14
 * P(E1 E2) = P(E1 E3) = P(E1 E4) = 1/26
 * P(E2 E3) = P(E2 E4) = P(E3 E4) = 1/27
 * P(E1 E2 E3) = P(E1 E2 E4) = P(E1 E3 E4) = 1/39
 * P(E2 E3 E4) = 1/40
 * P(E1 E2 E3 E4) = 1/52

$$ p = ( \frac 1 {13} + \frac 3 {14} - \frac 3 {26} -  \frac 3 {27} + \frac 3 {39} +  \frac 1 {40} - \frac 1 {52}) = \frac {4829} {32760} $$

So the expected number of cards is 52 * p = 4829 / 630 = 7.665 (approx).

If there are n cards in each suit then the expected number of draws is (5n+1)(10n^2+5n+1) / [(n+1)(2n+1)(3n+1)] which approaches 25/3 in the limit. Conjecture: Suppose that a deck of cards has n*k cards in k suits, with n cards in each suit. The deck is shuffled and the cards are turned over, one by one, until all k suits have appeared at least once. Then the expected number of cards that must be turned over is given by the following formula.

$$(k.n+1) - \frac {k!.n^k} {(n+1)(2n+1)...((k-1)n+1)}$$

Using Species
The describing species of a deck is

$$[lin.lin (Y+Z+T) - lin.lin(Y+Z) - lin.lin(Y+T) - lin.lin(Z+T) $$

$$ + lin.lin(Y) + lin.lin(Z) + lin.lin(T) -1]$$

$$.X.lin(X+Y+Z+T)$$

We have to compare these structures to

$$[lin (Y+Z+T) - lin(Y+Z) - lin(Y+T) - lin(Z+T) $$

$$ + lin(Y) + lin(Z) + lin(T) -1]$$

$$.X.lin(X+Y+Z+T)$$

For each structure of the second kind, there are L structures of the first kind, where L is the length of the initial segment, encoded in Lin.Lin part of formula. By dividing the coefficients of x13.y13.z13.t13 we get the average length.

note on electrical interpretation
File:Interpretation electrique.png|thumbnail|right|Electrical interpretation By interrupting one of L1 and L2 signals the output COM gets L2 or L1.

note on Cuantum Mechanics interpretation
Schrödinger's cat: a cat, a flask of poison, and a source are placed in a sealed box. If an internal monitor detects radioactivity (i.e. a single atom decaying), the flask is shattered, releasing the poison that kills the cat. The Copenhagen interpretation of quantum mechanics implies that after a while, the cat is simultaneously alive and dead. Yet, when one looks in the box, one sees the cat either alive or dead, not both alive and dead. This poses the question of when exactly quantum superposition ends and reality collapses into one possibility or the other.

Let's take the Schrödinger's cat. Then one has :

( alive and dead )' = dead or alive