JavaScript 中使用递归的碰撞检测圆

Question

我想在距主圆特定距离的随机位置生成圆。圆圈不应相互交叉。我创建了一个方法来检查每个圆圈是否与所有其他圆圈相交。碰撞检测完成了它的工作，但我代码中的递归部分不起作用：

if (collision > 0) {
    console.log("GENERATE NEW RANDOM COORDINATE")
    generateRandomCoordinate(d, r)
}

结果

如您在输出中所见，我的应用生成了新坐标（158,290）当它检测到碰撞（138,272）时。然而，它仍然吸引在 (138,272) 处圈出。

知道为什么会这样吗？如果您知道更有效的方法，请告诉我...

生成随机坐标函数

function generateRandomCoordinate(d, r) {
    var phi = Math.random() * 2 * Math.PI
    var x = topicCoordinates[0].x + Math.round(d * Math.cos(phi))
    var y = topicCoordinates[0].y + Math.round(d * Math.sin(phi))
    var collision = 0
    for (var j = 0; j < topicCoordinates.length; j++) {
        var dx = x - topicCoordinates[j].x
        var dy = y - topicCoordinates[j].y
        var distance = Math.sqrt(dx * dx + dy * dy)

        if (distance < r + topicCoordinates[j].r) {
            collision++
            console.log("COLLISION DETECTED BETWEEN POINT(" + "x: " + x + ", y: " + y + ") AND  POINT(x1: " + topicCoordinates[j].x + ", y1: " + topicCoordinates[j].y + ")")
        }
    }
    if (collision > 0) {
        console.log("GENERATE NEW RANDOM COORDINATE")
        generateRandomCoordinate(d, r)
    }

    var topicCoordinate = {
        "x": x,
        "y": y,
        "r": r
    }
    return topicCoordinate
}

调用生成随机坐标函数

var randomCoordinate = generateRandomCoordinate(d, r)
x = randomCoordinate.x
y = randomCoordinate.y
r = randomCoordinate.r
}

console.log("ADDED POINT(x: " + x + ", y: " + y + ", r: " + r + ")")
var topicCoordinate = {
    "x": x,
    "y": y,
    "r": r
}
topicCoordinates.push(topicCoordinate)
drawCircle(x, y, r)

输出

更新工作代码

function generateRandomCoordinate(d, r) {
    var collision = true
    while (collision) {
        collision = false
        var phi = Math.random() * 2 * Math.PI
        var x = topicCoordinates[0].x + Math.round(d * Math.cos(phi))
        var y = topicCoordinates[0].y + Math.round(d * Math.sin(phi))
        for (var j = 0; j < topicCoordinates.length; j++) {
            var dx = x - topicCoordinates[j].x
            var dy = y - topicCoordinates[j].y
            var distance = Math.sqrt(dx * dx + dy * dy)

            if (distance < r + topicCoordinates[j].r) {
                collision = true
                console.log("COLLISION DETECTED BETWEEN POINT(" + "x: " + x + ", y: " + y + ") AND  POINT(x1: " + topicCoordinates[j].x + ", y1: " + topicCoordinates[j].y + ")")
            }
        }
    }

    var topicCoordinate = { "x" : x, "y" : y, "r" : r }
    return topicCoordinate
}

Answer 1

递归函数调用不需要。只是整理循环（伪代码）

 repeat
     iscollision = false 
     generate x,y
     for other circles check
         if collision occurs
             iscollision = true
             break  //no need to check later circles    
 until iscollision = false

关于你的递归混淆 - 因为每个递归调用级函数实例都包含自己的内部变量副本，所以外部函数不知道内部 x,y - 你在这段代码中不知道 return 它们件数：

  if (collision > 0) {
    vvvvvvvvvvvvvvvvvv
    generateRandomCoordinate(d, r)
    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
}

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