在javascript中从圆圈中排除点

Question

to get a random point ona circle i use this code :

const position = randomCirclePoint(circleRadius, circleX, circleY);

const x = position.x;
const y = position.y;

function randomCirclePoint(circleRadius, circleX, circleY) {
    let ang = Math.random() * 2 * Math.PI,
        hyp = Math.sqrt(Math.random()) * circleRadius,
        adj = Math.cos(ang) * hyp,
        opp = Math.sin(ang) * hyp

        const x = circleX + adj;
        const y = circleY + opp;
        
    return {x, y}
}

But how can I exclude some points from where the loop can take the random point ?

Answer 1

If you want to exclude certain angles from the circle, you could do something like this:

const randomChoice = (list) => {
    return list[Math.floor(Math.random() * list.length)];
};

const randomCirclePoint = (circleRadius, circleX, circleY) => {
    const ang = 2 * Math.PI * Math.random();
    // for example, only in the upper half
    const angUpper = Math.PI * Math.random();
    // or only in the lower half
    const angLower = Math.PI + Math.PI * Math.random();
    // or only in quadrants 1 and 3
    const angQ1Q3 = randomChoice([
        (Math.PI / 2) * Math.random(),
        Math.PI + (Math.PI / 2) * Math.random(),
    ]);
    // or only in certain small slices of the circle
    const angSpecial = randomChoice([
        (Math.PI / 4) * Math.random(),
        (Math.PI * 8) / 7 + (Math.PI / 6) * Math.random(),
        (Math.PI * 2) / 3 + (Math.PI / 6) * 4 * Math.random(),
    ]);
    const hyp = Math.sqrt(Math.random()) * circleRadius;
    const adj = Math.cos(ang) * hyp;
    const opp = Math.sin(ang) * hyp;

    const x = circleX + adj;
    const y = circleY + opp;

    return { x, y };
};

console.log(randomCirclePoint(1, 0, 0));

Just change the way you generate the random angle and restrict it to a certain set of angles.

Since we don't know your specific application, it's a bit difficult to tell what you actually want. Hopefully this helps.