[英]Fractal Tree - branches not drawn
目前,我正在嘗試通過IFS(迭代函數系統)繪制對稱二叉樹:
但結果總是只有分支提示:
。
我無法弄清楚我做錯了什么或我錯過了什么。
這是IFS:
這是我的代碼:
RenderWindow window(VideoMode(480, 640), "fractals everywhere");
CircleShape point(1);
int chance;
float x, y, w, h, nx, ny, px, py;
void SymmetricBinaryTrees()
{
float r = 0.57f;
float o = 0.785f;
chance = rand() % 3;
switch (chance)
{
case 0:
nx = r * cos(o) * x + (-1 * r * sin(o) * y);
ny = r * sin(o) * x + r * cos(o) * y + 1;
break;
case 1:
nx = r * cos(o) * x + r * sin(o) * y;
ny = -1 * r * sin(o) * x + r * cos(o) * y + 1;
break;
case 2:
nx = x;
ny = y;
break;
}
}
void nextPoint()
{
SymmetricBinaryTrees();
x = nx; y = ny;
}
void drawPoint()
{
px = _map(x, -1.078, 1.078f, 0, w); py = _map(y, 0.f, 2.078f, h, 0); // maps the position accordingly
point.setPosition(px, py);
window.draw(point);
}
int main()
{
srand(time(NULL));
w = window.getSize().x * 1.f;
h = window.getSize().y * 1.f;
x = 0.f; y = 0.f;
window.setFramerateLimit(60);
while (window.isOpen())
{
Event e;
while (window.pollEvent(e))
if (e.type == Event::Closed) window.close();
for (int i = 1; i <= 500; i++)
{
drawPoint();
nextPoint();
}
window.display();
}
return 0;
}
這是我用於我的代碼的網站。
如果有人能幫助我或有任何想法我會非常感激,謝謝。
我贊同@beyond意見,我認為你太復雜了。 使用不同的方法會更容易。 讓我們更輕松。
使用遞歸函數,我們可以輕松了解每一步應該做什么。 考慮我們從一個初始點開始,然后在一個給定長度的角度上追蹤一條線 ,所以我們需要一個像下面這樣的函數:
void createTreeRecursive(sf::VertexArray &tree, sf::Vector2f point, float angle, float lenght)
tree
將成為我們的行集,它組成了樹本身。
我們能做的第一件事就是設置第一點,這是已知的:
// Add first point
tree.append(sf::Vertex(point, treeColor));
現在我們需要計算下一個點,形成一條線。 使用簡單的三角函數,我們可以確定這一點:
float newX = point.x + (cos((2.f * PI / 360.f) * angle) * lenght);
float newY = point.y - (sin((2.f * PI / 360.f) * angle) * lenght); // Caution here! Minus(-) sign because we're drawing upwards
所以我們添加第二個點,然后將樹分成2個新的分支,每個分支旋轉一定程度:
// Add second point
tree.append(sf::Vertex(nextPoint, treeColor));
// Create sub-tree from 2nd point, rotating +45 degrees (i.e. counterclockwise), reducing lenght of the new branch by 0.6 factor
createTreeRecursive(tree, nextPoint, angle + O, lenght * R);
// Same with the other sub-tree, but rotating -45 (i.e. clockwise)
createTreeRecursive(tree, nextPoint, angle - O, lenght * R);
我們需要一個基本情況來處理遞歸函數,在這種情況下,我選擇3作為最小長度:
if (lenght < 3)
// End condition, can be modified
return;
這必須首先檢查。
所以我們完成了,我們只需要初始調用:
sf::VertexArray createTree(){
// Our tree will be made out of lines
sf::VertexArray ret(sf::PrimitiveType::Lines);
// Initial point at botton-center(250, 450), with a 90 degrees rotation, first branch lenght 200
createTreeRecursive(ret, sf::Vector2f(250, 450), 90, 200);
return ret;
}
結果是:
#include <SFML/Graphics.hpp>
const double PI = 3.141592;
const double R = 0.57; // Reduction factor
const double O = 45; // Degree rotation each time
sf::Color treeColor = sf::Color::Blue;
void createTreeRecursive(sf::VertexArray &tree, sf::Vector2f point, float angle, float lenght){
if (lenght < 3)
// End condition, can be modified
return;
// Add first point
tree.append(sf::Vertex(point, treeColor));
float newX = point.x + (cos((2.f * PI / 360.f) * angle) * lenght);
float newY = point.y - (sin((2.f * PI / 360.f) * angle) * lenght); // Caution here! Minus(-) sign because we're drawing upwards
sf::Vector2f nextPoint(newX, newY);
// Add second point
tree.append(sf::Vertex(nextPoint, treeColor));
// Create sub-tree from 2nd point, rotating +45 degrees (i.e. counterclockwise), reducing lenght of the new branch by 0.6 factor
createTreeRecursive(tree, nextPoint, angle + O, lenght * R);
// Same with the other sub-tree, but rotating -45 (i.e. clockwise)
createTreeRecursive(tree, nextPoint, angle - O, lenght * R);
}
sf::VertexArray createTree(){
// Our tree will be made out of lines
sf::VertexArray ret(sf::PrimitiveType::Lines);
// Initial point at botton-center(250, 450), with a 90 degrees rotation, first branch lenght 200
createTreeRecursive(ret, sf::Vector2f(250, 450), 90, 200);
return ret;
}
int main()
{
RenderWindow window({ 500, 500 }, "SFML Tree", Style::Close);
auto tree = createTree();
while (window.isOpen())
{
for (Event event; window.pollEvent(event);){
if (event.type == Event::Closed)
window.close();
}
window.clear();
window.draw(tree);
window.display();
}
return EXIT_SUCCESS;
}
我建議你使用一個函數遞歸,1)繪制當前分支(作為一行),然后2)從當前分支創建兩個新的分支。 使用全局變量也無濟於事。 看起來你應該重新考慮你的方法。
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