我如何生成到原点的距离顺序的3d整数坐标

Question

我想按距原点的距离顺序生成一系列坐标。序列显然是无限的，因此仅生成它们并按距离排序对我不起作用。

对于那些距离相同的点，我不在乎顺序。

例如，这是一些点，它们与原点的距离最多为两步。

# d² = 0
(0,0,0)
# d² = 1
(0,0,-1)
(0,-1,0)
(-1,0,0)
(1,0,0)
(0,1,0)
(0,0,1)
# d² = 2
(0,-1,-1)
(-1,0,-1)
(1,0,-1)
(0,1,-1)
(-1,-1,0)
(1,-1,0)
(-1,1,0)
(1,1,0)
(0,-1,1)
(-1,0,1)
(1,0,1)
(0,1,1)
# d² = 3
(-1,-1,-1)
(1,-1,-1)
(-1,1,-1)
(1,1,-1)
(-1,-1,1)
(1,-1,1)
(-1,1,1)
(1,1,1)
# d² = 4
(0,0,-2)
(0,-2,0)
(-2,0,0)
(2,0,0)
(0,2,0)
(0,0,2)
# d² = 5
(0,-1,-2)
(-1,0,-2)
(1,0,-2)
(0,1,-2)
(0,-2,-1)
(-2,0,-1)
(2,0,-1)
(0,2,-1)
(-1,-2,0)
(1,-2,0)
(-2,-1,0)
(2,-1,0)
(-2,1,0)
(2,1,0)
(-1,2,0)
(1,2,0)
(0,-2,1)
(-2,0,1)
(2,0,1)
(0,2,1)
(0,-1,2)
(-1,0,2)
(1,0,2)
(0,1,2)
# d² = 6
(-1,-1,-2)
(1,-1,-2)
(-1,1,-2)
(1,1,-2)
(-1,-2,-1)
(1,-2,-1)
(-2,-1,-1)
(2,-1,-1)
(-2,1,-1)
(2,1,-1)
(-1,2,-1)
(1,2,-1)
(-1,-2,1)
(1,-2,1)
(-2,-1,1)
(2,-1,1)
(-2,1,1)
(2,1,1)
(-1,2,1)
(1,2,1)
(-1,-1,2)
(1,-1,2)
(-1,1,2)
(1,1,2)
# d² = 8
(0,-2,-2)
(-2,0,-2)
(2,0,-2)
(0,2,-2)
(-2,-2,0)
(2,-2,0)
(-2,2,0)
(2,2,0)
(0,-2,2)
(-2,0,2)
(2,0,2)
(0,2,2)
# d² = 9
(-1,-2,-2)
(1,-2,-2)
(-2,-1,-2)
(2,-1,-2)
(-2,1,-2)
(2,1,-2)
(-1,2,-2)
(1,2,-2)
(-2,-2,-1)
(2,-2,-1)
(-2,2,-1)
(2,2,-1)
(-2,-2,1)
(2,-2,1)
(-2,2,1)
(2,2,1)
(-1,-2,2)
(1,-2,2)
(-2,-1,2)
(2,-1,2)
(-2,1,2)
(2,1,2)
(-1,2,2)
(1,2,2)
# d² = 12
(-2,-2,-2)
(2,-2,-2)
(-2,2,-2)
(2,2,-2)
(-2,-2,2)
(2,-2,2)
(-2,2,2)
(2,2,2)

Answer 1

[从解决方案aaa开始，然后是aab，然后是abc，您可以检索所有其他“排列”，例如aba，baa，-a-a-b等。因此，您可以保留a

[在几何上：地球仪上的第八个三角形。对于一个圆x²+y²，一个人将在x的四分之一处进行迭代，并让y从r²-x²中获得。此处是给定的a。

不幸的是，对于我在一个阳光明媚的星期日回答来说，编码太过分了。（足够硬）。用伪代码示意性地表示：

int distance = -1;
int a;
int b;
int c;
PermutationIterator perm = ...

Point next() {
    if (perm.atEnd()) { // Initially true.
        perm.nextDistance();
        ++distance;
        a = distance;
        b = a;
        c = a;
        // Will return Point(a, a, a);
    }
    return perm.nextPerm();
}