A075464
https://oeis.org/A075464。复杂度取决于解空间的维数。现在这个序列已经算到帝第60项，是因为下一个达到40维。我们需要穷举2＾40个解，这个问题为关灯问题，对于n*n的电灯阵列，每关开关一个灯都会改变上下左右相邻灯的状态。如果开始所有灯开着，目标是关掉所有的灯，解必然存在，但是可能不唯一，这个序列就是要找到开关次数最小的解。
而对于n*n问题存在一些变开关方案保持所有灯状态不变，它们构成二阶域上线性空间。所以对于本题，求出任意一个特解，以及这个线性空间一组基后，就可以穷举所有解找出最优解了 现在序列以已经算到第60项，因为下一个是40维，也就是需要穷举2＾40个解，每个需要算40个长为三千多比特向量的异或并数结果1的数目。如果采用递归估计可以减少为三平均三次，由于一个时钟可操作32比特，平均300时钟可处理一个数，所以应该还属于可以忍受的计算量 我的csdn博客里面有穷举算法，但是效率很低，可以作为参考
http://blog.csdn.net/mathe/article/details/1143634
/*The code to turn off all bulbs in a matrix of bulbs
*/
#include <stdio.h>
#include <intrin.h>
#include <stdlib.h>
#include <string.h>
#define N 10000
typedef char ** MATRIX;
typedef char * VECTOR;
typedef char *const* CONST_MATRIX;
typedef const char * CONST_VECTOR;
MATRIX matrix_alloc2(int n, int m){
MATRIX x= (MATRIX)malloc(sizeof(char *)*n+sizeof(char)*n*m);
int i;
x=(char *)(x+n);
for(i=1;i<n;i++)x=x+m;
return x;
}
int maxtrix_count_one(MATRIX x,int n, int m){
int r=0;
int i,j;
for(i=0;i<n;i++)for(j=0;j<n;j++)if(x!=0)r++;
return r;
}
MATRIX matrix_alloc(int n){
MATRIX x = (MATRIX)malloc(sizeof(char *)*n+sizeof(char)*n*n);
int i;
x=(char *)(x+n);
for(i=1;i<n;i++)x=x+n;
return x;
}
void matrix_free(MATRIX x){
free(x);
}
VECTOR vector_alloc(int n){
VECTOR x = (VECTOR)malloc(sizeof(char)*n);
return x;
}
void vector_free(VECTOR x){
free(x);
}
void matrix_sum(MATRIX A, CONST_MATRIX B, int n){
int i,j;
for(i=0;i<n;i++){
for(j=0;j<n;j++){
A ^= B;
}
}
}
void vector_sum(VECTOR a, CONST_VECTOR b, int n){
int i;
for(i=0;i<n;i++)a^=b;
}
void matrix_mul(MATRIX out, CONST_MATRIX in1, CONST_MATRIX in2, int n){
int i,j,k;
for(i=0;i<n;i++)for(j=0;j<n;j++){
int sum=0;
for(k=0;k<n;k++){
if(in1)
sum^=in2;
}
out=sum;
}
}
void matrix_mul_H(MATRIX out, CONST_MATRIX in, int n){
int i,j;
for(i=0;i<n;i++)for(j=0;j<n;j++){
int sum=in;
if(i>0)sum^=in;
if(i<n-1)sum^=in;
out=sum;
}
}
void matrix_mul_vector(VECTOR out, CONST_MATRIX m, CONST_VECTOR in, int n){
int i,j;
for(i=0;i<n;i++){
int sum=0;
for(j=0;j<n;j++){
sum^=m∈
}
out=sum;
}
}
void H_mul_vector(VECTOR out, CONST_VECTOR in, int n){
int i;
for(i=0;i<n;i++){
int sum=in;
if(i>0)sum^=in;
if(i<n-1)sum^=in;
out=sum;
}
}
void vector_init_const(VECTOR v, char c, int n){
int i;
for(i=0;i<n;i++)v=c;
}
void matrix_init_O(MATRIX o, int n){
int i,j;
for(i=0;i<n;i++)for(j=0;j<n;j++)o=0;
}
void matrix_init_const(MATRIX m, char c, int n1, int n2){
int i,j;
for(i=0;i<n1;i++)for(j=0;j<n2;j++)m=c;
}
void matrix_init_E(MATRIX e, int n){
int j;
matrix_init_O(e,n);
for(j=0;j<n;j++)e=1;
}
void matrix_init_H(MATRIX h, int n){
int i;
matrix_init_O(h,n);
for(i=0;i<n;i++){
if(i>=1)h=1;
h=1;
if(i<n-1)h=1;
}
}
void matrix_copy(MATRIX m, CONST_MATRIX k, int n){
int i,j;
for(i=0;i<n;i++)for(j=0;j<n;j++)m=k;
}
void matrix_copy2(MATRIX m, CONST_MATRIX k, int n1, int n2){
int i,j;
for(i=0;i<n1;i++)for(j=0;j<n2;j++)m=k;
}
void vector_copy(VECTOR v, CONST_VECTOR w, int n){
int i;
for(i=0;i<n;i++)v=w;
}
void matrix_output2(CONST_MATRIX m, int n1, int n2){
int i,j;
for(i=0;i<n1;i++){
for(j=0;j<n2;j++){
printf("%d",m);
}
printf("\n");
}
}
void matrix_output(CONST_MATRIX m, int n){
int i,j;
for(i=0;i<n;i++){
for(j=0;j<n;j++){
printf("%d",m);
}
printf("\n");
}
}
void vector_output(VECTOR v, int n){
int i;
for(i=0;i<n;i++)printf("%d",v);
printf("\n");
}
void Usage(const char *program_name){
printf("Usage:\n");
printf("\t$%s n\n",program_name);
printf("\t\twhere n is a positive integer no more than %d\n", N);
printf("Or\n");
printf("\t$%s * * ... *\n", program_name);
printf("\t\tIn this format, there're total n strings of 0 and 1 and length of all 01 strings are same\n");
printf("The program assumes there're an n*m array. Each cell in the array has a switch "
"and a light. On touches of the switch in a cell will changes the state of the "
"light in the cell as well as the lights in the neighbors of the cell, so that "
"one switch could change states of 5 lights at most\n");
printf("The program will try to find a solution to turn all lights off\n");
printf("The first format of input gives an initial n*n array with all lights on\n");
printf("The second format requires n strings of 0 and 1, and the length of all strings"
" are m. Each string is correspondent to state of lights in one line of the "
"array. A digit 0 means the correspondent light is off and 1 means the light is "
"on.\n");
printf("The program will output an n*m matrix of 0 and 1 when there's a solution which "
"tells whether we need to touch a switch to turn off all the lights\nOr the "
"program will output \"NO SOLUTION\" when there's no solution\n");
exit(-1);
}
MATRIX P0,P1,H;
VECTOR ME;
MATRIX init;
static void copy_matrix_to_bit_string(MATRIX x, int m, int n, unsigned long long *bitstrings, int ll_len)
{
int i,j;
memset(bitstrings, 0, ll_len*sizeof(unsigned long long));
for(i=0;i<n;i++){
for(j=0;j<m;j++){
if(x!=0){
int index=i*m+j;
int word = index/64;
int off = index%64;
bitstrings|=1ll<<off;
}
}
}
}
int count_one(const unsigned long long *x, int ll_len)
{
int r=0;
int i;
const unsigned int *p=(const unsigned int*)x;
for(i=0;i<2*ll_len;i++)r+=_mm_popcnt_u32(p);
return r;
}
static void llxor(unsigned long long *r,const unsigned long long *x, int ll_len)
{
int i;
for(i=0;i<ll_len;i++)r^=x;
}
void search_base_result(const unsigned long long *any_root, const unsigned long long *base, int base_count, int ll_len)
{
long long x;
int j;
unsigned long long *best_r = (unsigned long long *)malloc(ll_len*sizeof(unsigned long long));
unsigned long long *cur_r = (unsigned long long *)malloc(ll_len*sizeof(unsigned long long));
int best_count = count_one(any_root, ll_len);
memcpy(best_r, any_root, ll_len*sizeof(unsigned long long));
memcpy(cur_r, any_root, ll_len*sizeof(unsigned long long));
for(x=1LL;x<=(1LL<<base_count)-1;x++){
for(j=0;j<base_count&&(x&(1LL<<j))==0;j++){
llxor(cur_r, base+j*ll_len, ll_len);
}
llxor(cur_r, base+j*ll_len, ll_len);
int cur_count = count_one(cur_r, ll_len);
if(cur_count<best_count){
best_count = cur_count;
memcpy(best_r, cur_r, ll_len*sizeof(unsigned long long));
}
}
free(best_r);
free(cur_r);
printf("best count:%d\n",best_count);
}
//First we need to solve equation P*X(n) = ME;
//There could be multiple solutions
//For each solution X(n), try
//X(n-1) = INIT+ H*X(n);
//X(K) = INIT + H*X(k+1)+X(k+2);//for k<n-1
void Solve(MATRIX P, VECTOR ME, int n, int m)
{
int *freedom_index;
int freedom_count=0;
int i,j,k,t;
int failed=0;
int longlong_len = (n*m+63)/64;
freedom_index = (int *)malloc(sizeof(int)*m);
//First solve equition P*X(n) = ME
for(i=0,t=0;i<m;i++){
for(j=t;j<m;j++)if(P==1)break;
if(j==m){
//find a freedom factor
freedom_index=i;
}else{
if(j!=t){
//Swap line j and t
int temp;
for(k=i;k<m;k++){
temp=P;
P=P;
P=temp;
}
temp = ME;
ME = ME;
ME = temp;
}
for(j=0;j<m;j++){
if(j!=t&&P){
//If P is 1, add Line i into Line j
for(k=i;k<m;k++){
P^=P;
}
mathe 发表于 2014-3-21 14:17
ME^=ME;
}
}
t++;
}
}
for(;t<m;t++){
if(ME!=0){
failed=1;
break;
}
}
fprintf(stderr,"Freedom factor = %d\n", freedom_count);
if(failed){
printf("NO SOLUTION\n");
free(freedom_index);
return;
}
if(freedom_count>0){
int delta=freedom_count;
for(k=m-1,t=freedom_count-1;k>=0&&delta>0;k--){
if(k==freedom_index){
int i;
for(i=0;i<m;i++)P=0;
ME=0;
delta--;t--;
}else if(delta>0){
for(i=0;i<m;i++)P=P;
ME=ME;
}
}
}
//Now ME hold's one solution for X,
//And random reset index inside freedom_index of ME to 0, 1
// will result in another solution for X;
//Output one solution
#ifdef _DEBUG
printf("ME:");vector_output(ME, m);printf("\n");
printf("H:\n");matrix_output(H,m);
printf("init:\n");
matrix_output(init,m);
#endif
unsigned long long *any_root = (unsigned long long *)malloc(longlong_len*sizeof(unsigned long long));
unsigned long long *base = (unsigned long long *)malloc(longlong_len *sizeof(unsigned long long) *freedom_count);
// if(freedom_count>=20)freedom_count=20;//do not search too much, so if the freedom_count is more than 15, the output may not be optimal
long long u;
int best_one_count=n*m+1;
MATRIX bm=matrix_alloc2(n,m);
for(u=0;u<1LL;u++)//calculate only one roots
{
MATRIX x=matrix_alloc2(n,m);
vector_copy(x,ME,m);
for(k=0;k<freedom_count;k++){
if(u&(1<<k)){
x]=1;
for(i=0;i<m;i++){
if(P]!=0){
x^=1;
}
}
}else{
x]=0;
}
}
matrix_mul_vector(x,H,x,m);
vector_sum(x,init,m);
for(k=n-3;k>=0;k--){
H_mul_vector(x,x,m);
#ifdef _DEBUG
printf("x[%d] 1:",k);vector_output(x,m);printf("\n");
#endif
vector_sum(x,init,m);
#ifdef _DEBUG
printf("x[%d] 2:",k);vector_output(x,m);printf("\n");
#endif
vector_sum(x,x,m);
#ifdef _DEBUG
printf("x[%d] 3:",k);vector_output(x,m);printf("\n");
#endif
}
copy_matrix_to_bit_string(x, m, n, any_root, longlong_len);
matrix_free(x);
}
for(u=0;u<freedom_count;u++){
MATRIX x=matrix_alloc2(n,m);
for(k=0;k<m;k++)x=0;
x]=1;//only set one value to bit 1
for(i=0;i<m;i++){
if(P]!=0){
x^=1;
}
}
matrix_mul_vector(x,H,x,m);
for(k=n-3;k>=0;k--){
H_mul_vector(x,x,m);
vector_sum(x,x,m);
}
copy_matrix_to_bit_string(x, m, n, base+u*longlong_len, longlong_len);
matrix_free(x);
}
search_base_result(any_root, base, freedom_count, longlong_len);
matrix_free(bm);
free(freedom_index);
free(any_root);
free(base);
}
void parse(int argc, char **argv, int *pn,int *pm){
if(argc == 2){
int n = atoi(argv);
if(n<0||n>N){
Usage(argv);
}else if(n<=1){
printf("%d\n",n);
exit(0);
}else{
init=matrix_alloc(n);
matrix_init_const(init,1,n,n);
*pn=*pm=n;
return;
}
}else if(argc==3){
int n = atoi(argv);
int m = atoi(argv);
if(n<0||n>N||m<0||m>N){
Usage(argv);
}else if(n<=1||m<=1){
printf("%d\n",n);
exit(0);
}else{
init=matrix_alloc2(n,m);
matrix_init_const(init,1,n,m);
*pn=n;*pm=m;
return;
}
}else{
int n=argc-1;
int i,j;
int m=strlen(argv);
init=matrix_alloc2(n,m);
for(i=0;i<n;i++){
char *s=argv;
if(strlen(s)!=m){
matrix_free(init);
Usage(argv);
}
for(j=0;j<m;j++){
if(s!='0'&&s!='1'){
matrix_free(init);
Usage(argv);
}
init=s-'0';
}
}
*pm=m;*pn=n;
return;
}
}
int main(int argc, char **argv){
int n,m;
int i;
MATRIX temp_matrix;
VECTOR temp_vector;
if(argc<2){
Usage(argv);
}
parse(argc,argv,&n,&m);
#ifdef _DEBUG
printf("Input:\n");
matrix_output2(init,n,m);
#endif
ME = vector_alloc(m);
temp_vector = vector_alloc(m);
P0 = matrix_alloc(m);
P1 = matrix_alloc(m);
H= matrix_alloc(m);
temp_matrix = matrix_alloc(m);
matrix_init_H(H,m);
matrix_init_E(P0,m);
matrix_init_H(P1,m);
#ifdef _DEBUG
printf("P(0):\n");matrix_output(P0,m);
printf("P(1):\n");matrix_output(P1,m);
#endif
matrix_mul_vector(ME, P0, init, m);
matrix_mul_vector(temp_vector, P1, init, m);
#ifdef _DEBUG
printf("M(0):");vector_output(ME,m);printf("\n");
printf("M(1):");vector_output(temp_vector,m);printf("\n");
#endif
vector_sum(ME,temp_vector,m);
#ifdef _DEBUG
printf("S(1):");vector_output(ME,m);printf("\n");
#endif
for(i=2;i<=n;i++){
matrix_mul_H(temp_matrix,P1,m);
matrix_sum(temp_matrix,P0,m);//P(k)= H*P(k-1) + P(k-2)
matrix_copy(P0,P1,m);
matrix_copy(P1,temp_matrix,m);
#ifdef _DEBUG
printf("P(%d):\n",i);matrix_output(P1,m);
#endif
if(i<n){
matrix_mul_vector(temp_vector, P1, init, m);
vector_sum(ME,temp_vector,m);
#ifdef _DEBUG
printf("M(%d):",i);vector_output(temp_vector,m);printf("\n");
printf("S(%d):",i);vector_output(ME,m);printf("\n");
#endif
}
}
Solve(P1, ME, n,m);
matrix_free(init);
vector_free(ME);
vector_free(temp_vector);
matrix_free(P0);
matrix_free(P1);
matrix_free(H);
matrix_free(temp_matrix);
return 0;
}
android上载代码不是很方便。上面代码计算n=39大概花4分钟，估计n=61大概两天半，n=65大概10天
如果能优化数倍就可以舒服很多了，主要是那个 search_开头的函数 发现A075463只计算了29项，其实其计算难度应该不高。上面有个mathematica代码，有了能介绍一下其算法吗？我看不懂。 mathe 发表于 2014-3-22 20:43
发现A075463只计算了29项，其实其计算难度应该不高。上面有个mathematica代码，有了能介绍一下其算法吗？我 ...
mathe都看不明白,我就不去尝试了.
只觉得代码的效率并不高,把16换成19就超级慢了. 我是对mathematica不熟悉 其实复杂度不比 https://oeis.org/A075462 多多少