Summary of Polay theorem

Resources:

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Feel the recent has been easy to meet such a topic .... A little more complicated, it's not going to be.

First a summed up theorem:

Use one of the simplest examples to illustrate

How many different images can I get from the 2*2 square, which is colored in black and white? The two schemes, which are matched by rotation, are considered the same.

Set G={P1,P2,..., PG} is a permutation group on Ω, such as permutation group g={to 0°. Turn 90°, turn 180°, turn 270°}

C (PK) is the number of cycles that displace the PK

The loop section of G1 permutation {to 0°} is 4. {(1), (2), (3), (4)}

G2 permutation {to 90°}, the Loop section is 1, {(4,3,2,1)}

G3 permutation {to 180°}, the Loop section is 2, {(1,3), (2,4)}

The loop section of G4 permutation {270°} is 1. {(1,2,3,4)}

The elements in Ω are colored with the colors in M,
Number of shading schemes is L = 1/| G|*[c1 (p1) +c1 (p2) +c1 (p3) +...c1 (P[g])

= 1/| G|*[m^c (p1) +m^c (p2) +m^c (p3) +...m^c (P[g])

| G| is the total number of permutations, the number of M colors

C1 (PI) refers to the number of fixed points of the displacement pi (both 1 points of the cyclic section)

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Obvious four numbers are 16 2 4 2, respectively.

L = 1/| G| * [16 + 2 + 4 + 2] = 6

C (PI) refers to the number of cycles of the permutation pi.

L = 1/| G| *[2^4 + 2^1 + 2^2 + 2^1] = 6

Let's start with a simple topic:

POJ 2409:http://poj.org/problem?

id=2409

Main topic:

A necklace company produces bracelets. n beads form a ring, with the M color to the n beads dyed, you get a variety of bracelets. However, the same scheme is calculated by rotating and flipping the match.

For example, when 5 beads are dyed with 2 colors, the number of solutions is 8.

Solution of the problem:

One: rotate (for example, there are n beads.) The angle that can be rotated every time is 360/n)

Two: Flip (consider the axis of symmetry. Odd number of beads, that each axis of symmetry can pass through a bead. With a total of n axis of symmetry)

An even number of beads, each axis of symmetry through two beads, a common n/2 axis of symmetry, or every axis of symmetry does not pass through the beads, this symmetry axis is also a N/2

So all in all, either odd or even. The way of change is that there are 2*n kinds of flipping.

It can be proved that each flip of its cycle section is GCD (i,n) 0<i<=n

#include <iostream> #include <stdio.h> #include <string.h> #include <math.h>using namespace std ; int gcd (int a,int b) {    return b==0?a:gcd (b,a%b);} Long long rotate (int c,int n) {  //rotation  rotation (360/n) * I-degree    long long sum=0;    for (int i=1;i<=n;i++)        Sum+=pow (C*1.0,GCD (n,i));   The circular section of each rotation is gcd (n,i)    return sum; Long long Turn (int c,int n) {    //flip    long long sum=0;    if (n%2)        Sum+=n*pow (c*1.0, (n+2)/2);//odd the axis of symmetry is crossed by a bead a total  of n each permutation of the cyclic section is n/2+1    else    sum+=n/2* (POW ( C*1.0,N/2) +pow (c*1.0, (n+2)/2));  The even number is N/2 through the beads or not through the beads, and the n/2+1 and N/2 are   able to calculate the return sum with the following formula    ;} void Polya (int c,int n) {    int i,j;    Long long sum=0;    Sum+=rotate (c,n);    Sum+=turn (c,n);    printf ("%lld\n", sum/(2*n));} int main () {    int n,m;    while (scanf ("%d%d", &n,&m), n| | m) {        Polya (n,m);    }    return 0;}

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#include <cstring> #include <string> #include <fstream> #include <iostream> #include < iomanip> #include <cstdio> #include <cctype> #include <algorithm> #include <queue> #include <map> #include <set> #include <vector> #include <stack> #include <ctime> #include < cstdlib> #include <functional> #include <cmath>using namespace std #define PI acos ( -1.0) #define EPS 1e-7# Define INF 0x7fffffff#define llinf 0x7fffffffffffffff#define seed 1313131#define MOD 1000000007#define ll Long Long#defin E ull unsigned ll#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1//to replace the circulation section, Polya principle//perm[0..n-1] is 0. N-1 a permutation (permutation)//return displacement minimum period, NUM returns the number of circular nodes # define MAXN 1000int gcd (int a,int b) {return B?GCD (b,a%b): A;} int Polya (int* perm,int n,int& num) {int i,j,p,v[maxn]={0},ret=1;for (num=i=0;i<n;i++) if (!v[i]) {for (num++,j=0, p=i;! v[p=perm[p]];j++) V[P]=1;RET*=J/GCD (ret,j);} return ret;}    int main () {int perm1[6]={0,5,4,3,2,1}; int perm2[6]={1,0,5,4,3,2};    int num;    Cout<<polya (Perm2,6,num) <<endl; Cout<<num<<endl;}

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Summary of Polay theorem