我可以使用Processing一步一步地绘制回溯吗?
我想使用processing编写一个回溯-8 Queens-可视化代码 所以我尝试在安装程序中使用noLoop,并在每个更新步骤中调用redraw和delay100,但没有成功 这是我的职责我可以使用Processing一步一步地绘制回溯吗?,processing,backtracking,Processing,Backtracking,我想使用processing编写一个回溯-8 Queens-可视化代码 所以我尝试在安装程序中使用noLoop,并在每个更新步骤中调用redraw和delay100,但没有成功 这是我的职责 int cellH = 38, cellW = 38, n = 8; PImage img; boolean [][] grid; boolean [] visC, visMD, visSD; boolean firstTime = true; void drawQueen(int r, int c){
int cellH = 38, cellW = 38, n = 8;
PImage img;
boolean [][] grid;
boolean [] visC, visMD, visSD;
boolean firstTime = true;
void drawQueen(int r, int c){
image(img, c*cellW, r*cellH, cellW, cellH);
}
void drawGrid(){
background(255);
for(int r = 0 ; r < n ; ++r){
for(int c = 0 ; c < n ; ++c){
if((r&1) != (c&1)) fill(0);
else fill(255);
rect(c*cellW, r*cellH, (c+1)*cellW, (r+1)*cellH);
}
}
}
void updateQueens(){
for(int r = 0 ; r < n ; ++r)
for(int c = 0 ; c < n ; ++c)
if(grid[r][c] == true)
drawQueen(r, c);
}
boolean backTrack(int r){
if(r == n) return true;
else{
for(int c = 0 ; c < n ; ++c){
if(!visC[c] && !visMD[n+r-c] && !visSD[r+c]){
//Do
grid[r][c] = visC[c] = visMD[n+r-c] = visSD[r+c] = true;
redraw();
delay(100);
//Recurse
if(backTrack(r+1)) return true;
//Undo
grid[r][c] = visC[c] = visMD[n+r-c] = visSD[r+c] = false;
}
}
}
return false;
}
void setup(){
size(280, 280);
cellH = 280/n;
cellW = 280/n;
grid = new boolean[n][n];
visC = new boolean[n];
visMD = new boolean[2*n];
visSD = new boolean[2*n];
noLoop();
img = loadImage("queen.png");
backTrack(0);
}
void draw(){
drawGrid();
updateQueens();
}
运行草图时,仅显示最终状态
还有其他的想法吗?根据:
在构建程序时,只有在内部调用redraw才有意义
事件,例如鼠标按下。这是因为重画不会运行
立即绘制它仅设置一个标志,指示正在进行更新
需要
重画不会导致绘制屏幕。它设置了一个需要调用draw的标志,该标志发生在循环结束时。一种解决方案是将draw重命名为drawScreen,并调用它,而不是重新绘制。根据:
在构建程序时,只有在内部调用redraw才有意义
事件,例如鼠标按下。这是因为重画不会运行
立即绘制它仅设置一个标志,指示正在进行更新
需要
重画不会导致绘制屏幕。它设置了一个需要调用draw的标志,该标志发生在循环结束时。一种解决方案是将draw重命名为drawScreen并调用它,而不是重新绘制。处理的方式是,通过将draw函数转到该循环的主体来模拟循环,并在setup函数中执行所有初始化 为了模拟一个递归,你可以把它变成一个循环,然后做上面的事情,通常这可以用一个堆栈来完成,你基本上用你的堆栈替换系统的堆栈;我对一些想法做了一些阅读检查,我发现如果递归调用位于函数体的末尾,那么将递归转换为带有堆栈的循环将非常容易
现在的问题是,在调用返回后应该执行的递归调用之后,您有一些代码,但是查看它,它只是撤销对全局变量所做的更改,如果我们认为这些变量作为状态的一部分不是非常有效的,并且不能很好地扩展,那么我们可以克服它。但是在你的例子中,撤销部分是不必要的,递归调用将是函数体中的最后一件事,让我们暂时离开内部for循环
为此,我定义了一个名为State的类,如下所示 class State {
private final int SIZE = 8;
private boolean grid[][], visC[], visR[], visMD[], visSD[];
int r, c;
State() {
visC = new boolean[SIZE];
visR = new boolean[SIZE];
visMD = new boolean[2*SIZE];
visSD = new boolean[2*SIZE];
grid = new boolean[SIZE][SIZE];
}
State(State other) {
this();
cpyArr(other.visMD, this.visMD);
cpyArr(other.visSD, this.visSD);
cpyArr(other.visC, this.visC);
cpyArr(other.visR, this.visR);
for (int i = 0 ; i < other.grid.length ; ++i)
for (int j = 0 ; j < other.grid[i].length ; ++j)
this.grid[i][j] = other.grid[i][j];
this.r = other.r;
this.c = other.c;
}
void cpyArr(boolean from[], boolean to[]) {
for (int i = 0 ; i < from.length ; ++i) to[i] = from[i];
}
boolean isValid(int r, int c) {
return (r < SIZE && c < SIZE && !visR[r] && !visC[c] && !visMD[SIZE + r - c] && !visSD[r + c]);
}
// actually update this sate with r and c
void affect() {
grid[r][c] = visC[c] = visMD[SIZE + r - c] = visSD[r + c] = true;
}
PVector[] getPositions() {
ArrayList<PVector> ret = new ArrayList<PVector>();
for (int i = 0; i < SIZE; ++i)
for (int j = 0; j < SIZE; ++j)
if (grid[i][j]) ret.add(new PVector(j, i));
return ret.toArray(new PVector[0]);
}
}
我们可以将绘制函数的主体看作是while循环的主体,在栈为空时使用NOLoop停止它,所以最后的代码将类似于…
。stack.push(initialState);
while(stack.size() != 0) {
State currentState = stack.pop();
// do stuff ...
stack.push(nextState);
}
import java.util.Stack;
final int GRID_SIZE = 8;
float cellH, cellW;
PImage img;
Stack<State> stack;
void setup() {
size(500, 500);
frameRate(5);
cellH = (float) height / GRID_SIZE;
cellW = (float) width / GRID_SIZE;
img = loadImage("queen.png");
stack = new Stack<State>();
State state = new State();
state.r = -1;
state.c = -1;
stack.push(state);
noLoop();
}
void draw() {
// stop if the stack is empty
if (stack.size() == 0) {
noLoop();
return;
}
State current = stack.pop();
drawGrid(current);
// stop when a solution is found
if (current.r == GRID_SIZE - 1) {
noLoop();
return;
}
for (int c = GRID_SIZE - 1; c >= 0; --c) {
State next = new State(current);
if (!next.isValid(current.r+1, c)) continue;
next.c = c;
next.r = current.r + 1;
next.affect();
stack.push(next);
}
}
void drawGrid(State state) {
float cellH = height / GRID_SIZE;
float cellW = width / GRID_SIZE;
background(255);
for (int r = 0; r < GRID_SIZE; ++r) {
for (int c = 0; c < GRID_SIZE; ++c) {
if ((r&1) != (c&1)) fill(0);
else fill(255);
rect(c*cellW, r*cellH, (c+1)*cellW, (r+1)*cellH);
}
}
PVector pos[] = state.getPositions();
for (PVector vec : pos) {
image(img, vec.x * cellW + cellW * 0.1, vec.y * cellH + cellH * 0.1, cellW * 0.8, cellH * 0.8);
}
}
// to resume the search after a solution is found
void keyPressed() {
if (key == ' ') loop();
}
请注意,我们稍后留下的内部for循环部分是反向的,因此要执行的第一个状态与回溯将探索的第一个状态相同
现在在数据文件中为queen.png资源放置一些漂亮的图像,结果非常好
处理的方式是,通过将draw函数转到循环体来模拟循环,并在setup函数中执行所有初始化 为了模拟一个递归,你可以把它变成一个循环,然后做上面的事情,通常这可以用一个堆栈来完成,你基本上用你的堆栈替换系统的堆栈;我对一些想法做了一些阅读检查,我发现如果递归调用位于函数体的末尾,那么将递归转换为带有堆栈的循环将非常容易
现在的问题是,在调用返回后应该执行的递归调用之后,您有一些代码,但是查看它,它只是撤销对全局变量所做的更改,如果我们认为这些变量作为状态的一部分不是非常有效的,并且不能很好地扩展,那么我们可以克服它。但是在你的例子中,撤销部分是不必要的,递归调用将是函数体中的最后一件事,让我们暂时离开内部for循环
为此,我定义了一个名为State的类,如下所示 class State {
private final int SIZE = 8;
private boolean grid[][], visC[], visR[], visMD[], visSD[];
int r, c;
State() {
visC = new boolean[SIZE];
visR = new boolean[SIZE];
visMD = new boolean[2*SIZE];
visSD = new boolean[2*SIZE];
grid = new boolean[SIZE][SIZE];
}
State(State other) {
this();
cpyArr(other.visMD, this.visMD);
cpyArr(other.visSD, this.visSD);
cpyArr(other.visC, this.visC);
cpyArr(other.visR, this.visR);
for (int i = 0 ; i < other.grid.length ; ++i)
for (int j = 0 ; j < other.grid[i].length ; ++j)
this.grid[i][j] = other.grid[i][j];
this.r = other.r;
this.c = other.c;
}
void cpyArr(boolean from[], boolean to[]) {
for (int i = 0 ; i < from.length ; ++i) to[i] = from[i];
}
boolean isValid(int r, int c) {
return (r < SIZE && c < SIZE && !visR[r] && !visC[c] && !visMD[SIZE + r - c] && !visSD[r + c]);
}
// actually update this sate with r and c
void affect() {
grid[r][c] = visC[c] = visMD[SIZE + r - c] = visSD[r + c] = true;
}
PVector[] getPositions() {
ArrayList<PVector> ret = new ArrayList<PVector>();
for (int i = 0; i < SIZE; ++i)
for (int j = 0; j < SIZE; ++j)
if (grid[i][j]) ret.add(new PVector(j, i));
return ret.toArray(new PVector[0]);
}
}
我们可以将绘制函数的主体看作是while循环的主体,在栈为空时使用NOLoop停止它,所以最后的代码将类似于…
。stack.push(initialState);
while(stack.size() != 0) {
State currentState = stack.pop();
// do stuff ...
stack.push(nextState);
}
import java.util.Stack;
final int GRID_SIZE = 8;
float cellH, cellW;
PImage img;
Stack<State> stack;
void setup() {
size(500, 500);
frameRate(5);
cellH = (float) height / GRID_SIZE;
cellW = (float) width / GRID_SIZE;
img = loadImage("queen.png");
stack = new Stack<State>();
State state = new State();
state.r = -1;
state.c = -1;
stack.push(state);
noLoop();
}
void draw() {
// stop if the stack is empty
if (stack.size() == 0) {
noLoop();
return;
}
State current = stack.pop();
drawGrid(current);
// stop when a solution is found
if (current.r == GRID_SIZE - 1) {
noLoop();
return;
}
for (int c = GRID_SIZE - 1; c >= 0; --c) {
State next = new State(current);
if (!next.isValid(current.r+1, c)) continue;
next.c = c;
next.r = current.r + 1;
next.affect();
stack.push(next);
}
}
void drawGrid(State state) {
float cellH = height / GRID_SIZE;
float cellW = width / GRID_SIZE;
background(255);
for (int r = 0; r < GRID_SIZE; ++r) {
for (int c = 0; c < GRID_SIZE; ++c) {
if ((r&1) != (c&1)) fill(0);
else fill(255);
rect(c*cellW, r*cellH, (c+1)*cellW, (r+1)*cellH);
}
}
PVector pos[] = state.getPositions();
for (PVector vec : pos) {
image(img, vec.x * cellW + cellW * 0.1, vec.y * cellH + cellH * 0.1, cellW * 0.8, cellH * 0.8);
}
}
// to resume the search after a solution is found
void keyPressed() {
if (key == ' ') loop();
}
请注意,我们稍后留下的内部for循环部分是反向的,因此要执行的第一个状态与回溯将探索的第一个状态相同
现在在数据文件中为queen.png资源放置一些漂亮的图像,结果非常好
我试图用线程解决它,它给了我一个很好的输出,下面是我的代码:
int cellH = 38, cellW = 38, n = 8;
PImage img;
boolean [][] grid;
boolean [] visC, visMD, visSD;
boolean firstTime = true;
Thread thread;
void setup(){
size(560, 560);
cellH = 560/n;
cellW = 560/n;
grid = new boolean[n][n];
visC = new boolean[n];
visMD = new boolean[2*n];
visSD = new boolean[2*n];
img = loadImage("queen.png");
thread = new Thread(new MyThread());
thread.start();
}
void draw(){
if(thread.isAlive())
drawGrid();
else{
noLoop();
endRecord();
return;
}
}
void drawGrid(){
background(255);
for(int r = 0 ; r < n ; ++r){
for(int c = 0 ; c < n ; ++c){
if((r&1) != (c&1)) fill(0);
else fill(255);
rect(c*cellW, r*cellH, (c+1)*cellW, (r+1)*cellH);
if(grid[r][c] == true)
image(img, c*cellW, r*cellH, cellW, cellH);
}
}
}
boolean backTrack(int r){
if(r == n) return true;
else{
for(int c = 0 ; c < n ; ++c){
if(!visC[c] && !visMD[n+r-c] && !visSD[r+c]){
//Do
grid[r][c] = visC[c] = visMD[n+r-c] = visSD[r+c] = true;
try{
Thread.sleep(200);
}catch(InterruptedException e){System.out.println(e);}
//Recurse
if(backTrack(r+1)) return true;
//Undo
grid[r][c] = visC[c] = visMD[n+r-c] = visSD[r+c] = false;
try{
Thread.sleep(200);
}catch(InterruptedException e){System.out.println(e);}
}
}
}
return false;
}
class MyThread implements Runnable{
public void run(){
backTrack(0);
}
}
以下是输出:
我试图用线程解决它,它给了我一个很好的输出,下面是我的代码:
int cellH = 38, cellW = 38, n = 8;
PImage img;
boolean [][] grid;
boolean [] visC, visMD, visSD;
boolean firstTime = true;
Thread thread;
void setup(){
size(560, 560);
cellH = 560/n;
cellW = 560/n;
grid = new boolean[n][n];
visC = new boolean[n];
visMD = new boolean[2*n];
visSD = new boolean[2*n];
img = loadImage("queen.png");
thread = new Thread(new MyThread());
thread.start();
}
void draw(){
if(thread.isAlive())
drawGrid();
else{
noLoop();
endRecord();
return;
}
}
void drawGrid(){
background(255);
for(int r = 0 ; r < n ; ++r){
for(int c = 0 ; c < n ; ++c){
if((r&1) != (c&1)) fill(0);
else fill(255);
rect(c*cellW, r*cellH, (c+1)*cellW, (r+1)*cellH);
if(grid[r][c] == true)
image(img, c*cellW, r*cellH, cellW, cellH);
}
}
}
boolean backTrack(int r){
if(r == n) return true;
else{
for(int c = 0 ; c < n ; ++c){
if(!visC[c] && !visMD[n+r-c] && !visSD[r+c]){
//Do
grid[r][c] = visC[c] = visMD[n+r-c] = visSD[r+c] = true;
try{
Thread.sleep(200);
}catch(InterruptedException e){System.out.println(e);}
//Recurse
if(backTrack(r+1)) return true;
//Undo
grid[r][c] = visC[c] = visMD[n+r-c] = visSD[r+c] = false;
try{
Thread.sleep(200);
}catch(InterruptedException e){System.out.println(e);}
}
}
}
return false;
}
class MyThread implements Runnable{
public void run(){
backTrack(0);
}
}
以下是输出:
添加打印重绘时;在重新绘制之前;,它被打印了很多次吗?你能给我们看看你的drawGrid函数吗?@J.D.是的,它被打印了。@DWuest添加了完整的代码。@Rabbid76有什么诀窍吗?当你添加printlnRedrawing时;在重新绘制之前;,它印了很多次了吗?可以吗
请给我们看一下drawGrid函数?@J.D.是的,它会打印。@DWuest添加了完整的代码。@Rabbid76有什么诀窍吗?我不工作,文档上说:所有处理程序都会在绘图结束时更新屏幕,决不会更早。如果添加PrintLN调用绘图;以上重画和打印图纸;你会明白我的意思的。”“调用绘图”打印多次,最后“绘图”打印一次。现在将redraw重命名为draw-将其转换为直接函数调用,print语句将交替进行,这意味着每次更新都会被绘制。是的,单词会被打印出来,但不会更新场景。我不工作,文档中说:所有处理程序都会在绘制结束时更新屏幕,而不会更早。如果添加PrintlCalling draw;以上重画和打印图纸;你会明白我的意思的。”“调用绘图”打印多次,最后“绘图”打印一次。现在将redraw重命名为draw-将其转换为直接函数调用print语句将交替进行,这意味着每次更新都会被绘制。是的,单词会被打印,但不会更新场景。非常好的方法。它具有很好的通用性,可以用很少的编辑来可视化几乎任何算法。非常好的方法。它具有很好的通用性,并且可以用于可视化几乎任何只需很少编辑的算法。