poj 1222 EXTENDED LIGHTS OUT 增广矩阵消元法

这篇博客介绍了如何使用增广矩阵消元法解决POJ 1222题——EXTENDED LIGHTS OUT问题。题目要求找到一个开关布局,使得5x6矩阵中的所有灯最终熄灭。通过构建包含30个未知数的方程组,利用异或运算,将题目转化为线性代数的消元法求解。文章提到,虽然ACM中的消元法扩展了运算,但建立方程组仍然具有挑战性。代码实现部分未展示。

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EXTENDED LIGHTS OUT
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 8682 Accepted: 5624

Description

In an extended version of the game Lights Out, is a puzzle with 5 rows of 6 buttons each (the actual puzzle has 5 rows of 5 buttons each). Each button has a light. When a button is pressed, that button and each of its (up to four) neighbors above, below, right and left, has the state of its light reversed. (If on, the light is turned off; if off, the light is turned on.) Buttons in the corners change the state of 3 buttons; buttons on an edge change the state of 4 buttons and other buttons change the state of 5. For example, if the buttons marked X on the left below were to be pressed,the display would change to the image on the right. 

The aim of the game is, starting from any initial set of lights on in the display, to press buttons to get the display to a state where all lights are off. When adjacent buttons are pressed, the action of one button can undo the effect of another. For instance, in the display below, pressing buttons marked X in the left display results in the right display.Note that the buttons in row 2 column 3 and row 2 column 5 both change the state of the button in row 2 column 4,so that, in the end, its state is unchanged. 

Note: 
1. It does not matter what order the buttons are pressed. 
2. If a button is pressed a second time, it exactly cancels the effect of the first press, so no button ever need be pressed more than once. 
3. As illustrated in the second diagram, all the lights in the first row may be turned off, by pressing the corresponding buttons in the second row. By repeating this process in each row, all the lights in the first 
four rows may be turned out. Similarly, by pressing buttons in columns 2, 3 ?, all lights in the first 5 columns may be turned off. 
Write a program to solve the puzzle.

Input

The first line of the input is a positive integer n which is the number of puzzles that follow. Each puzzle will be five lines, each of which has six 0 or 1 separated by one or more spaces. A 0 indicates that the light is off, while a 1 indicates that the light is on initially.

Output

For each puzzle, the output consists of a line with the string: "PUZZLE #m", where m is the index of the puzzle in the input file. Following that line, is a puzzle-like display (in the same format as the input) . In this case, 1's indicate buttons that must be pressed to solve the puzzle, while 0 indicate buttons, which are not pressed. There should be exactly one space between each 0 or 1 in the output puzzle-like display.

Sample Input

2
0 1 1 0 1 0
1 0 0 1 1 1
0 0 1 0 0 1
1 0 0 1 0 1
0 1 1 1 0 0
0 0 1 0 1 0
1 0 1 0 1 1
0 0 1 0 1 1
1 0 1 1 0 0
0 1 0 1 0 0

Sample Output

PUZZLE #1
1 0 1 0 0 1
1 1 0 1 0 1
0 0 1 0 1 1
1 0 0 1 0 0
0 1 0 0 0 0
PUZZLE #2
1 0 0 1 1 1
1 1 0 0 0 0
0 0 0 1 0 0
1 1 0 1 0 1
1 0 1 1 0 1

题意:一个5*6的矩阵,其中1代表灯亮,0代表灯灭,对于每一个开关布局,若b[i][j]==1,则a[i][j]与其相邻四个元素的值变化(0变1,1变0),现让你求出一个开关布局,使得每一个灯最后都是灭的。

记得以前写高代课作业的时候曾经写过一个消元法求答案的程序,没用结构体,一遇到分数就gg……这个题目作为消元法的入门题感觉着实有点难啊,它的运算不是加减乘除,是异或(虽然加法取摸也行)…没想到acm中将消元法的运算扩展了,更容易编码,但是构建方程组还是有点难想的……


这个题目有30个未知量,倒着想,最后状态全是0,对于答案矩阵c,0 xor c[i][j]  xor c[i-1][j] xor c[i+1][j] xor c[i][j-1] xor c[i][j+1] ==a[i][j],其中a[i][j]为题目给出的输入。所以,每个方程中,只有5项的系数为1,其余全为零。设b为增广矩阵,a[i][j]的系数离散化到矩阵中的第6*i-6+j行,将a[i][j]放到b[6*i-6+j][31]中,这样就能构建出增广矩阵了。系数矩阵是不变的,如下:


至于消元法,高等代数中解释的很详细了,虽然不知道线性代数中是怎么讲的,但是

肯定也是讲过方法的,这里就不赘述了。


代码:

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int a[35],x[40],map[10][10],b[35][35];  //b:系数矩阵的增广矩阵
    int i,j,k,m,n,t;

void gauss()
{
 int i,j,k;
 for(k=1;k<=30;k++)
   {for(i=k;i<=30;i++)
      if(b[i][k]==1)break;
    if(i!=k)
      for(j=1;j<=31;j++)swap(b[i][j],b[k][j]);
    for(i=1;i<=30;i++)
     if(i!=k && b[i][k]==1)
      for(j=1;j<=31;j++)b[i][j]=b[i][j]^b[k][j];    
   }
}


int main(int argc, char const *argv[])
{   
	
    scanf("%d",&t);
    for(int t1=1;t1<=t;t1++)
    { memset(a,0,sizeof(a));
      memset(x,0,sizeof(x));
      memset(b,0,sizeof(b));
      for(i=1;i<=5;i++)       //构建30*31的矩阵
      	for(j=1;j<=6;j++)
      		{scanf("%d",&map[i][j]);
      		 b[6*i-6+j][31]=map[i][j];
             b[6*i-6+j][6*i-6+j]=1;
             if(i>1)b[6*i-6+j][6*i-12+j]=1;
             if(i<=4)b[6*i-6+j][6*i+j]=1;
             if(j>1)b[6*i-6+j][6*i-6+j-1]=1;
             if(j<=5)b[6*i-6+j][6*i-6+j+1]=1; 
              }
      printf("PUZZLE #%d\n",t1);        
      gauss();
      for(i=1;i<=30;i++)
       if(i%6==0)printf("%d\n",b[i][31]);
        else printf("%d ",b[i][31]); 
     

    }
 	return 0;
}


### 矩阵在扩展版 Lights Out 游戏中的实现 #### 背景介绍 关灯游戏(Lights Out)是一种基于逻辑的益智游戏,其核心机制涉及通过一系列操作使所有灯光熄灭。该游戏可以通过线性代数的方法建模并求解,其中矩阵扮演了重要角色。具体来说,游戏的状态可以用二进制向量表示,而每次点击的操作可以看作是对当前状态施加的一个变换。 对于扩展版本的游戏(如 POJ 1222),通常是一个 \( M \times N \) 的网格,初始状态下某些位置可能亮起或熄灭。目标是找到一种最小化翻转次数的方式使得整个网格变为全零状态(即所有灯均关闭)。这一过程可通过构建一个对应的布尔方程组来描述,并利用高斯消元法或其他数值技术求解[^3]。 #### 数学模型建立 假设我们有一个大小为 \( m=5, n=6 \) 的棋盘,则总共有 30 个独立变量代表各个格子是否被按压过。设这些未知数构成列向量 \( x=[x_1,x_2,...,x_{30}]^T \),如果某位 i 对应的位置需要按下一次就令 xi=1 否则等于 0 。接着定义另一个长度相同的响应矢量 y ,它记录的是最终期望达到的目标配置 —— 这里就是全是 'off' 或者说都是 0 值的情况下的布局图样 [y₁,y₂,…,yn]=₀₆₀⁰[^4]. 为了表达上述关系,我们需要创建一个系数矩阵 A 来捕捉每一个按钮动作如何影响周围邻居节点的变化情况: \[ Ax = b \] 这里, - **A** 是一个稀疏矩阵,每一行对应于某个特定单格及其邻域的影响模式; - **b** 表示初始条件下的光源分布状况; - 解决这个系统意味着寻找合适的输入序列 x 以满足给定的要求 b. 由于涉及到异或运算特性(xor operation), 实际上是在有限字段 GF(2)=Z/2Z 上执行计算而不是普通的实数空间 R^n 中进行处理. 因此标准算法比如 Gaussian elimination 需要做适当调整以便适应这种特殊的算术环境. #### 使用高斯去法解决问题 一旦建立了这样的数学框架之后,就可以采用经典的 Gauss-Jordan Elimination 方法逐步简化增广矩阵直到得到唯一可行解或者是证明无解存在为止。特别注意当面对大规模实例时效率问题变得至关重要因此也可能考虑其他更高效的替代策略例如 Bitmasking Techniques 结合快速傅立叶变换 FFT 加速乘法步骤等等优化手段提升性能表现水平. 以下是用 Python 编写的简单例子展示如何运用 NumPy 库完成基本功能演示: ```python import numpy as np def create_matrix(m,n): """Create coefficient matrix for lights out game.""" size=m*n mat=np.zeros((size,size)) # Fill diagonal elements and neighbors based on grid structure. for r in range(m): for c in range(n): idx=r*n+c # Current cell itself always toggles. mat[idx,idx]=1 # Toggle top neighbor if exists. if r>0: mat[(r-1)*n+c,idx]=mat[idx,(r-1)*n+c]=1 # Bottom neighbor similarly handled... ... return mat.astype(int) # Example usage creating small test case setup. if __name__=="__main__": M,N=(5,6) initial_state=[[int(c)for c in line.strip().split()]for _in range(M)] target_vector=sum(initial_state,[]) coeff_mat=create_matrix(M,N) augmented=np.hstack([coeff_mat,np.array(target_vector).reshape(-1,1)]) rank_before_elim=np.linalg.matrix_rank(coeff_mat) reduced_form,rref_rows,_=gauss_jordan(augmented) solution_exists=len(rref_rows)==len(set(map(tuple,reduced_form[:,-1]))) print("Solution Exists:",solution_exists) ``` 以上脚本片段仅作为概念验证用途并未完全覆盖边界情形检测等功能完善需求实际部署前还需进一步测试改进确保鲁棒性和兼容性良好。 ---
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