生成三角形/六边形坐标(xyz)

Question

我正在尝试提出一个迭代函数,为六边形网格生成xyz坐标.使用起始十六进制位置(简单来说就是0,0,0),我想计算六边形的每个连续"环"的坐标,如下所示:

到目前为止,我已经设法提出的是这个(例如在javascript中):

var radius = 3
var xyz = [0,0,0];

// for each ring
for (var i = 0; i < radius; i++) {
    var tpRing = i*6;
    var tpVect = tpRing/3;
    // for each vector of ring
    for (var j = 0; j < 3; j++) {
        // for each tile in vector
        for(var k = 0; k < tpVect; k++) {
            xyz[0] = ???;
            xyz[1] = ???;
            xyz[2] = ???;
            console.log(xyz);
        }
    }
}

我知道每个环包含比前一个多6个点,每个120°向量包含从中心开始的每个步骤的一个附加点.我也知道x + y + z = 0所有瓷砖.但是,如何生成遵循以下顺序的坐标列表？

    0, 0, 0

    0,-1, 1
    1,-1, 0
    1, 0,-1
    0, 1,-1
   -1, 1, 0
   -1, 0, 1

    0,-2, 2
    1,-2, 1
    2,-2, 0
    2,-1,-1
    2, 0,-2
    1, 1,-2
    0, 2,-2
   -1, 2,-1
   -2, 2, 0
   -2, 1, 1
   -2, 0, 2
   -1,-1, 2

Answer 1

不仅如此x + y + z = 0,而且x,y和z的绝对值等于环的半径的两倍.这足以识别每个连续环上的每个六边形:

var radius = 4;
for(var i = 0; i < radius; i++)
{
    for(var j = -i; j <= i; j++)
    for(var k = -i; k <= i; k++)
    for(var l = -i; l <= i; l++)
        if(Math.abs(j) + Math.abs(k) + Math.abs(l) == i*2 && j + k + l == 0)
            console.log(j + "," + k + "," + l);
    console.log("");
}

Answer 2

另一种可能的解决方案,在O(半径²)中运行,与tehMick解决方案的O(半径⁴)不同(以很多样式为代价)是:

radius = 4
for r in range(radius):
    print "radius %d" % r
    x = 0
    y = -r
    z = +r
    print x,y,z
    for i in range(r):
        x = x+1
        z = z-1
        print x,y,z
    for i in range(r):
        y = y+1
        z = z-1
        print x,y,z
    for i in range(r):
        x = x-1
        y = y+1
        print x,y,z
    for i in range(r):
        x = x-1
        z = z+1
        print x,y,z
    for i in range(r):
        y = y-1
        z = z+1
        print x,y,z
    for i in range(r-1):
        x = x+1
        y = y-1
        print x,y,z

或者写得更简洁:

radius = 4
deltas = [[1,0,-1],[0,1,-1],[-1,1,0],[-1,0,1],[0,-1,1],[1,-1,0]]
for r in range(radius):
    print "radius %d" % r
    x = 0
    y = -r
    z = +r
    print x,y,z
    for j in range(6):
        if j==5:
            num_of_hexas_in_edge = r-1
        else:
            num_of_hexas_in_edge = r
        for i in range(num_of_hexas_in_edge):
            x = x+deltas[j][0]
            y = y+deltas[j][1]
            z = z+deltas[j][2]            
            print x,y,z

它的灵感来自于六边形实际上位于六边形的外部,因此您可以找到其中1个点的坐标,然后通过移动其6个边来计算其他坐标.

Answer 3

这是一个有趣的谜题.

O(半径²)但是(希望)比Ofri的解决方案更具风格. 在我看来,可以生成坐标,就像你使用方向(移动)向量在环上"行走"一样,转弯相当于在移动向量周围移动零点.

这个版本也比Eric的解决方案更有优势,因为它永远不会触及无效坐标(Eric拒绝它们,但这个甚至不必测试它们).

# enumerate coords in rings 1..n-1; this doesn't work for the origin
for ring in range(1,4):
    # start in the upper right corner ...
    (x,y,z) = (0,-ring,ring)
    # ... moving clockwise (south-east, or +x,-z)
    move = [1,0,-1]         

    # each ring has six more coordinates than the last
    for i in range(6*ring):
        # print first to get the starting hex for this ring
        print "%d/%d: (%d,%d,%d) " % (ring,i,x,y,z)
        # then move to the next hex
        (x,y,z) = map(sum, zip((x,y,z), move))

        # when a coordinate has a zero in it, we're in a corner of
        # the ring, so we need to turn right
        if 0 in (x,y,z):
            # left shift the zero through the move vector for a
            # right turn
            i = move.index(0)
            (move[i-1],move[i]) = (move[i],move[i-1])

    print # blank line between rings

三个欢呼为python的序列切片.