我正在开发名为Lights Out的游戏。因此，为了解决这个问题，我必须在模块2中计算 AX = B 的答案。因此，我选择jscience库。在这个游戏中，A的大小是25x25矩阵，X和B都是25x1矩阵。我编写了如下代码：
AllLightOut.java上课：
public class AllLightOut {
public static final int SIZE = 5;
public static double[] Action(int i, int j) {
double[] change = new double[SIZE * SIZE];
int count = 0;
for (double[] d : Switch(new double[SIZE][SIZE], i, j))
for (double e : d)
change[count++] = e;
return change;
}
public static double[][] MatrixA() {
double[][] mat = new double[SIZE * SIZE][SIZE * SIZE];
for (int i = 0; i < SIZE; i++)
for (int j = 0; j < SIZE; j++)
mat[i * SIZE + j] = Action(i, j);
return mat;
}
public static SparseVector<ModuloInteger> ArrayToDenseVectorModule2(
double[] array) {
List<ModuloInteger> list = new ArrayList<ModuloInteger>();
for (int i = 0; i < array.length; i++) {
if (array[i] == 0)
list.add(ModuloInteger.ZERO);
else
list.add(ModuloInteger.ONE);
}
return SparseVector.valueOf(DenseVector.valueOf(list),
ModuloInteger.ZERO);
}
public static SparseMatrix<ModuloInteger> MatrixAModule2() {
double[][] mat = MatrixA();
List<DenseVector<ModuloInteger>> list = new ArrayList<DenseVector<ModuloInteger>>();
for (int i = 0; i < mat.length; i++) {
List<ModuloInteger> l = new ArrayList<ModuloInteger>();
for (int j = 0; j < mat[i].length; j++) {
if (mat[i][j] == 0)
l.add(ModuloInteger.ZERO);
else
l.add(ModuloInteger.ONE);
}
list.add(DenseVector.valueOf(l));
}
return SparseMatrix.valueOf(DenseMatrix.valueOf(list),
ModuloInteger.ZERO);
}
public static double[][] Switch(double[][] action, int i, int j) {
action[i][j] = action[i][j] == 1 ? 0 : 1;
if (i > 0)
action[i - 1][j] = action[i - 1][j] == 1 ? 0 : 1;
if (i < action.length - 1)
action[i + 1][j] = action[i + 1][j] == 1 ? 0 : 1;
if (j > 0)
action[i][j - 1] = action[i][j - 1] == 1 ? 0 : 1;
if (j < action.length - 1)
action[i][j + 1] = action[i][j + 1] == 1 ? 0 : 1;
return action;
}
}
主要课程如下：
public class Main {
public static void main(String[] args) {
double[] bVec = new double[] { 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0,
1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0 };
SparseMatrix<ModuloInteger> matA = AllLightOut.MatrixAModule2();
SparseVector<ModuloInteger> matB = AllLightOut
.ArrayToDenseVectorModule2(bVec);
ModuloInteger.setModulus(LargeInteger.valueOf(2));
Vector<ModuloInteger> matX = matA.solve(matB);
System.out.println(matX);
}
}
我跑了这个程序大约30分钟，但没有结果。我的代码是否包含致命错误或错误？为什么需要太长时间？
感谢您的关注：）
修改
这一行Matrix<ModuloInteger> matX = matA.inverse();发生了减速。请注意，JScience benchmark result，此库的速度非常快，但我不知道为什么我的程序运行得太慢了！
EDIT2
请注意，当我尝试SIZE = 3时，我会真正得到答案。例如：
马特：
{{1, 1, 0, 1, 0, 0, 0, 0, 0}, {1, 1, 1, 0, 1, 0, 0, 0, 0}, {0, 1, 1, 0, 0, 1, 0, 0, 0}, {1, 0, 0, 1, 1, 0, 1, 0, 0}, {0, 1, 0, 1, 1, 1, 0, 1, 0}, {0, 0, 1, 0, 1, 1, 0, 0, 1}, {0, 0, 0, 1, 0, 0, 1, 1, 0}, {0, 0, 0, 0, 1, 0, 1, 1, 1}, {0, 0, 0, 0, 0, 1, 0, 1, 1}}
MatB：
{1,1,1,1,1,1,1,0,0}
MatC：
{0,0,1,1,0,0,0,0,0}
但是当我尝试SIZE = 5时，发生了减速。
答案 0 :(得分：4)
此行
发生了减速Matrix<ModuloInteger> matX = matA.inverse();
那是因为系数矩阵matA对于SIZE == 5（或4,9,11,14,16，......？）是不可逆的。
我有点惊讶图书馆没有检测到并抛出异常。如果库试图在solve()中反转矩阵，则会产生相同的后果。
对于某些尺寸的系数矩阵的奇异性的结果是，并非所有这些尺寸的谜题都是可解的，而其他谜题都有多种解决方案。
由于我们正在计算模2，我们可以使用位或boolean来模拟状态/切换，使用XOR进行加法，使用&进行乘法。我已经使用高斯消除法编写了一个简单的求解器，也许它可以帮助你（我没有花太多时间考虑设计，所以它并不漂亮）：
public class Lights{
private static final int SIZE = 5;
private static boolean[] toggle(int i, int j) {
boolean[] action = new boolean[SIZE*SIZE];
int idx = i*SIZE+j;
action[idx] = true;
if (j > 0) action[idx-1] = true;
if (j < SIZE-1) action[idx+1] = true;
if (i > 0) action[idx-SIZE] = true;
if (i < SIZE-1) action[idx+SIZE] = true;
return action;
}
private static boolean[][] matrixA() {
boolean[][] mat = new boolean[SIZE*SIZE][];
for(int i = 0; i < SIZE; ++i) {
for(int j = 0; j < SIZE; ++j) {
mat[i*SIZE+j] = toggle(i,j);
}
}
return mat;
}
private static void rotateR(boolean[] a, int r) {
r %= a.length;
if (r < 0) r += a.length;
if (r == 0) return;
boolean[] tmp = new boolean[r];
for(int i = 0; i < r; ++i) {
tmp[i] = a[i];
}
for(int i = 0; i < a.length - r; ++i) {
a[i] = a[i+r];
}
for(int i = 0; i < r; ++i) {
a[i + a.length - r] = tmp[i];
}
}
private static void rotateR(boolean[][] a, int r) {
r %= a.length;
if (r < 0) r += a.length;
if (r == 0) return;
boolean[][] tmp = new boolean[r][];
for(int i = 0; i < r; ++i) {
tmp[i] = a[i];
}
for(int i = 0; i < a.length - r; ++i) {
a[i] = a[i+r];
}
for(int i = 0; i < r; ++i) {
a[i + a.length - r] = tmp[i];
}
}
private static int count(boolean[] a) {
int c = 0;
for(int i = 0; i < a.length; ++i) {
if (a[i]) ++c;
}
return c;
}
private static void swapBits(boolean[] a, int i, int j) {
boolean tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
private static void addBit(boolean[] a, int i, int j) {
a[j] ^= a[i];
}
private static void swapRows(boolean[][] a, int i, int j) {
boolean[] tmp = a[i];
a[i] = a[j];
a[j] = tmp;
}
private static void xorb(boolean[] a, boolean[] b) {
for(int i = 0; i < a.length; ++i) {
a[i] ^= b[i];
}
}
private static boolean[] boolBits(int bits, long param) {
boolean[] bitArr = new boolean[bits];
for(int i = 0; i < bits; ++i) {
if (((param >> i) & 1L) != 0) {
bitArr[i] = true;
}
}
return bitArr;
}
private static boolean[] solve(boolean[][] m, boolean[] b) {
// Move first SIZE rows to bottom, so that on the diagonal
// above the lowest SIZE rows, there are unit matrices
rotateR(m, SIZE);
// modify right hand side accordingly
rotateR(b,SIZE);
// clean first SIZE*(SIZE-1) columns
for(int i = 0; i < SIZE*(SIZE-1); ++i) {
for(int k = 0; k < SIZE*SIZE; ++k) {
if (k == i) continue;
if (m[k][i]) {
xorb(m[k], m[i]);
b[k] ^= b[i];
}
}
}
// Now we have a block matrix
/*
* E 0 0 ... 0 X
* 0 E 0 ... 0 X
* 0 0 E ... 0 X
* ...
* 0 0 ... E 0 X
* 0 0 ... 0 E X
* 0 0 ... 0 0 Y
*
*/
// Bring Y to row-echelon form
int i = SIZE*(SIZE-1), j, k, mi = i;
while(mi < SIZE*SIZE){
// Try to find a row with mi-th bit set
for(j = i; j < SIZE*SIZE; ++j) {
if (m[j][mi]) break;
}
if (j < SIZE*SIZE) {
// Found one
if (j > i) {
swapRows(m,i,j);
swapBits(b,i,j);
}
for(k = 0; k < SIZE*SIZE; ++k) {
if (k == i) continue;
if (m[k][mi]) {
xorb(m[k], m[i]);
b[k] ^= b[i];
}
}
// cleaned up column, good row, next
++i;
}
// Look at next column
++mi;
}
printMat(m,b);
boolean[] best = b;
if (i < SIZE*SIZE) {
// We have zero-rows in the matrix,
// check whether the puzzle is solvable at all,
// i.e. all corresponding bits in the rhs are 0
for(j = i; j < SIZE*SIZE; ++j) {
if (b[j]) {
System.out.println("Puzzle not solvable, some lights must remain lit.");
break;
// throw new IllegalArgumentException("Puzzle is not solvable!");
}
}
// Pretending it were solvable if not
if (j < SIZE*SIZE) {
System.out.println("Pretending the puzzle were solvable...");
for(; j < SIZE*SIZE; ++j) {
b[j] = false;
}
}
// Okay, puzzle is solvable, but there are several solutions
// Let's try to find the one with the least toggles.
// We have the canonical solution with last bits all zero
int toggles = count(b);
System.out.println(toggles + " toggles in canonical solution");
int freeBits = SIZE*SIZE - i;
long max = 1L << freeBits;
System.out.println(freeBits + " free bits");
// Check all combinations of free bits whether they produce
// something better
for(long param = 1; param < max; ++param) {
boolean[] base = boolBits(freeBits,param);
boolean[] c = new boolean[SIZE*SIZE];
for(k = 0; k < freeBits; ++k) {
c[i+k] = base[k];
}
for(k = 0; k < i; ++k) {
for(j = 0; j < freeBits; ++j) {
c[k] ^= base[j] && m[k][j+i];
}
}
xorb(c,b);
int t = count(c);
if (t < toggles) {
System.out.printf("Found new best for param %x, %d toggles\n",param,t);
printMat(m,c,b);
toggles = t;
best = c;
} else {
System.out.printf("%d toggles for parameter %x\n", t, param);
}
}
}
return best;
}
private static boolean[] parseLights(int[] lights) {
int lim = lights.length;
if (SIZE*SIZE < lim) lim = SIZE*SIZE;
boolean[] b = new boolean[SIZE*SIZE];
for(int i = 0; i < lim; ++i) {
b[i] = (lights[i] != 0);
}
return b;
}
private static void printToggles(boolean[] s) {
for(int i = 0; i < s.length; ++i) {
if (s[i]) {
System.out.print("(" + (i/SIZE + 1) + ", " + (i%SIZE + 1) + "); ");
}
}
System.out.println();
}
private static void printMat(boolean[][] a, boolean[] rhs) {
for(int i = 0; i < SIZE*SIZE; ++i) {
for(int j = 0; j < SIZE*SIZE; ++j) {
System.out.print((a[i][j] ? "1 " : "0 "));
}
System.out.println("| " + (rhs[i] ? "1" : "0"));
}
}
private static void printMat(boolean[][] a, boolean[] sol, boolean[] rhs) {
for(int i = 0; i < SIZE*SIZE; ++i) {
for(int j = 0; j < SIZE*SIZE; ++j) {
System.out.print((a[i][j] ? "1 " : "0 "));
}
System.out.println("| " + (sol[i] ? "1" : "0") + " | " + (rhs[i] ? "1" : "0"));
}
}
private static void printGrid(boolean[] g) {
for(int i = 0; i < SIZE; ++i) {
for(int j = 0; j < SIZE; ++j) {
System.out.print(g[i*SIZE+j] ? "1" : "0");
}
System.out.println();
}
}
public static void main(String[] args) {
int[] initialLights = new int[] { 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0,
1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0 };
boolean[] b = parseLights(initialLights);
boolean[] b2 = b.clone();
boolean[][] coefficients = matrixA();
boolean[] toggles = solve(coefficients, b);
printGrid(b2);
System.out.println("--------");
boolean[][] check = matrixA();
boolean[] verify = new boolean[SIZE*SIZE];
for(int i = 0; i < SIZE*SIZE; ++i) {
if (toggles[i]) {
xorb(verify, check[i]);
}
}
printGrid(verify);
xorb(b2,verify);
if (count(b2) > 0) {
System.out.println("Aww, shuck, screwed up!");
printGrid(b2);
}
printToggles(toggles);
}
}
答案 1 :(得分：2)
如果可以避免，你几乎不想计算矩阵的实际逆。这样的操作是有问题的并且非常耗时。查看JScience的文档，您是否考虑过使用solve方法？ matX = matA.solve（matB）中的某些东西应该会给你你正在寻找的东西，我怀疑他们正在使用逆来计算它，虽然我还没有挖到JScience，所以这并非不可能。