table of Contents
@ 0 - @ References
@ 1-- problem introduced @
A classic question:
A square into 4 squares, 2 colored, how many kinds of programs?
Wherein rotation of the same image after operator the same program.
If the rotation is not considered, the various programs as follows.
First of all can be found, only four rotation angle: 0 °, 90 °, 180 °, 270 °.
Can then be obtained per one rotation can be a unique mapping scheme to another scheme (which may be itself).
So we can use a permutation p effect described in the first rotation, wherein the p [i] denotes the i-th program became the first mapped p [i] kinds of programs.
In this example, four kinds of substitution (0 °, 90 °, 180 °, 270 °) form a permutation group .
(What is the permutation group ... emmm can be left on their own Baidu)
(The following notation is not rigorous, because it is based on my own memory write)
If the displacement map to element i such that f j, we denoted by \ (f: i -> J \) .
For the permutation group G, if there is \ (f \ in G \) , and \ (f: the X--> the y-\) , we say x, y is the equivalence class.
Obviously, this transitive equivalence relation, (corresponding to group several properties) reflexive symmetry.
Then we seek equal to the number of equivalence classes under the action of the group G substitution .
@ 2 - burnside lemma @
First cast conclusion:
We call for f, meet f: x -> x x for fixed point under f role. Note C (f) indicates the number of the fixed point f.
The number of equivalence classes under the action of G:
\ [CNT = \ FRAC {\ sum_ {F \ G} in C (F)} {| G |} \]
BB is proven to be my own, quite different from the mainstream Proof (I can not remember the mainstream Proof), it may not be easy.
Introducing (a very strict) notation: Write \ (S_ {i, j} \) shows a j i is mapped to a set of replacement thereof.
First prove a few lemmas:
Lemma. 1: \ (| of S_ {I, J} | = | of S_ {J, I} | \)
Proof: equivalent to prove \ (| S_ {i, j } | \ le | S_ {j, i} | \) and \ (| of S_ {J, I} | \ Le | of S_ {I, J} | \) .
Just consider \ (| of S_ {I, J} | \ Le | of S_ {J, I} | \) , the other half same reason.
For \ (F \ in of S_ {I, J} \) , its inverse \ (F ^ {-}. 1 \ in of S_ {J, I} \) . Depending on the nature and the permutation group, the inverse of each element is unique and different inverse different elements.
So \ (S_ {j, i} \) comprising at least \ (| S_ {i, j } | \) elements - i.e. their inverse element.
Lemma 2: If \ (S_ {i, j} \) is not empty, then \ (| S_ {i, j
} | = | S_ {i, i} | \) Proof: just proof \ (| of S_ {i, j} | \ le | S_ {i, i} | \) and \ (| of S_ {I, I} | \ Le | of S_ {I, J} | \) .
From \ (S_ {j, i} \) optionally an element \ (F \) , \ (F \) and \ (S_ {i, j} \) Each \ (G \) obtained by multiplying \ (f * G \) .
Apparently \ (F * g \ in of S_ {I, I} \) , and g are not the same, the result must also different, so \ (| S_ {i, j } | \ le | S_ {i, i } | \) .
Then remains from \ (S_ {j, i} \) optionally an element \ (F \) , \ (F \) and \ (S_ {i, i} \) Each \ (G \) multiplied\ (f * G \) .
There \ (F * g \ in of S_ {J, I} \) , and g are not the same, the results are different. It is also to give \ (| of S_ {I, I} | \ Le | of S_ {J, I} | = | of S_ {I, J} | \) (Lemma 1).
Fixed point number of the sum of the equivalence class = | G |: Lemma 3.
Proof: provided that the equivalence class \ (E = {x_1, x_2, ..., x_k} \) .
There \ (\ sum_. 1 = {I} ^ {K} | of S_ {x_1, x_i} | = | G | \) (each permutation in G must put \ (x_1 \) mapped into equivalence classes in a number).
Lemma Lemma 1 + 2 can be obtained \ (\ sum_ I = {1} ^ {K} | {x_i of S_, x_i} | = | G | \) , thus proved.
With Lemma 3, we can deduce the correctness of burnside lemma, because the total number of fixed points = the number of equivalence classes * | G |.
@ 3 - pólya Theorem @
(Note that pólya theorems rather than polya theorem, although this is not important but still want to look professional point)
burnside lemma general method of statistical equivalence classes, and proposed a pólya Theorem dyeing problems Fixed Point specific calculation method.
We started a problem given, if we are to honestly counted, to calculate the 2 ^ (2 + 2) th program in whether replacing fixed point.
When the data range increases, the sharp increase in the number of programs, burnside lemma would be meaningless.
We might as well put 2 * 2 points out, build permutation group. The following assumed that the number of available colors is m.
We will break down each of the replacement cycle, fixed point under the effect of this displacement is certainly an element within the same cycle painted the same color.
After setting the total cycle decomposition transducer f C (f) cycles, then f does not move under the effect of the number of points of \ (m ^ {C (f)} \) .
Fixed point number under the action of G:
\ [CNT = \ FRAC {\ sum_ {F \ ^ {m} G in C (F)}} {| G |} \]
This allows you to exponential complexity into polynomial complexity.
@ 4 - pólya form generating function theorem @
Can be found, pólya theorem is very obvious.But at the same pólya theorems it is also very popular application.
but! If the combination of universal generating function, we can get some of the more interesting things.
A typical example is the beginning, if we want to know how many there are fixed black plaid, essentially different program number.
This time we need to make use of the generating function.
If cyclic permutation decomposed f k cycles, each cycle of the loop section size are x1, x2, ..., xk.
This displacement corresponding to the generated function f \ (F. (F) = \ prod_ {I} = ^ {K}. 1 (B x_i ^ {^} + {W} x_i) \) . Wherein b, w is a formal variable.
Permutation function to generate the entire group G is:
\ [F. (G) = \ {FRAC \ sum_ {F \ in G} F. (F)} {| G |} \]
Of the G \ (b ^ i * w ^ j \) coefficients of terms equal to the number of black and white painted j program grid cell i.
Such as for example, we start the generating function:
\ [F. (G) = \ FRAC {. 1} {. 4} ((B + W) ^. 4 + (B ^. 4 + W ^. 4) + (B ^ 2 + w ^ 2) ^ 2 + (b ^ 4 + w ^ 4)) = w ^ 4 + w ^ 3b + 2w ^ 2b ^ 2 + wb ^ 3 + b ^ 4 \]
Verify its correctness can be found.