python - 遍历同心圆中的每个像素

Question

我试图遍历每个像素坐标，从 (0, 0) 开始，以便在它们不重叠的最近偏移处融合两个像素化形状。
到现在为止，我一直在使用同心正方形，这确实很容易做到，但最终会使嫁接图像放置得更远。然后我实现了 Bresenham 圆算法如下:

def generate_offsets(maxRadius : int):
    """Generate x and y coordinates in concentric circles around the origin
    Uses Bresenham's Circle Drawing Algorithm
    """
    
    for radius in range(maxRadius):
        x = 0
        y = radius
        d = 3 - (2 * radius)
        while x < y:
        
            yield x, y
            yield y, x
            yield y, -x
            yield x, -y
            yield -x, -y
            yield -y, -x
            yield -y, x
            yield -x, y
            
            if d < 0:
                d += (4 * x) + 6
            else:
                d += (4 * (x-y)) + 10
                y -= 1
            
            x += 1

然而，这样做的缺点是不检查某些像素偏移。我找到的所有填充孔的解决方案都建议跟踪从 0,0 到像素的整条线，这在这里会非常浪费。
如何在不重新访问任何像素的情况下修复漏洞？

这是显示所述孔的示例，这表示每个圆或半径 1-9。探索的像素为 # ，而未探索的像素是 . :

....................
....................
........#####.......
......#########.....
.....###########....
....#..#######..#...
...##..#.###.#..##..
...####.#####.####..
..####.#.###.#.####.
..#######.#.#######.
..########.########.
..#######.#.#######.
..####.#.###.#.####.
...####.#####.####..
...##..#.###.#..##..
....#..#######..#...
.....###########....
......#########.....
........#####.......
....................

更新:这是我当前的解决方案，它确实填满了整个圆圈，但存储的状态比我想要的多得多:

import itertools
def generate_offsets(minRadius : int = 0, maxRadius : int = 3_750_000):
    """Generate x and z coordinates in concentric circles around the origin
    Uses Bresenham's Circle Drawing Algorithm
    """
    def yield_points(x, y):
        
            yield x, y
            yield x, -y
            yield -x, -y
            yield -x, y
            
            if x != y:
                yield y, x
                yield y, -x
                yield -y, -x
                yield -y, x
    
    def yield_circle(radius, previousCircle):
        x = 0
        y = radius
        d = 3 - (2 * radius)
        while x < y:
        
            for point in yield_points(x, y):
                if point not in previousCircle:
                    yield point
            
            if d < 0:
                d += (4 * x) + 6
            else:
                d += (4 * (x-y)) + 10
                for point in itertools.chain(yield_points(x + 1, y), yield_points(x, y - 1)):
                    if point not in previousCircle:
                        yield point
                y -= 1
            
            x += 1
    
    previousCircle = [(0,0)]
    for radius in range(minRadius, maxRadius):
    
        circle = set()
        for point in yield_circle(radius, previousCircle):
            if point not in circle:
                yield point
                circle.add(point)
        
        previousCircle = circle

这是迄今为止我在内存和处理方面找到的最平衡的解决方案。它只记住前一个圆圈，这将冗余率(访问两次像素的比率)从没有任何内存的大约 50% 降低到大约 1.5%

Answer 1

离开我的头顶......
生成 集 坐标一次。在探索的同时，保留一个 集 访问的坐标。集合之间的差异将是未访问的坐标。如果您不想处理圆外的像素，也许可以跟踪 x 和 y 极值以进行比较 - 也许像字典一样:{each_row_visited:max_and_min_col_for that row,} .

I would prefer a solution that doesn't expand in memory as time progresses !

而不是制作越来越大的圆圈希望填满圆盘:

使用 Bresenham 算法确定具有所需半径的点

找到每个 x 的最小和最大 y 值(反之亦然)

使用这些极值产生极值之间的所有点
从 pprint 导入 pprint
from 操作符导入 itemgetter
从 itertools 导入 groupby
X = itemgetter(0)
Y = itemgetter(1)

此函数修改自 question in a different forum

def circle(radius):
    '''Yield (x,y) points of a disc
    
    Uses Bresenham complete circle algorithm
    '''
    # init vars
    switch = 3 - (2 * radius)
    # points --> {x:(minY,maxY),...}
    points = set()
    x = 0
    y = radius
    # first quarter/octant starts clockwise at 12 o'clock
    while x <= y:
        # first quarter first octant
        points.add((x,-y))
        # first quarter 2nd octant
        points.add((y,-x))
        # second quarter 3rd octant
        points.add((y,x))
        # second quarter 4.octant
        points.add((x,y))
        # third quarter 5.octant
        points.add((-x,y))
        # third quarter 6.octant
        points.add((-y,x))
        # fourth quarter 7.octant
        points.add((-y,-x))
        # fourth quarter 8.octant
        points.add((-x,-y))
        if switch < 0:
            switch = switch + (4 * x) + 6
        else:
            switch = switch + (4 * (x - y)) + 10
            y = y - 1
        x = x + 1
    circle = sorted(points)
    for x,points in groupby(circle,key=X):
        points = list(points)
        miny = Y(points[0])
        maxy = Y(points[-1])
        for y in range(miny,maxy+1):
            yield (x,y)

那应该最小化状态。从圆圈创建光盘时会有一些重复/重新访问 - 我没有尝试量化这一点。

结果...

def display(points,radius):
    ''' point: sequence of (x,y) tuples, radius: int
    '''
    not_visited, visited = '-','█'
    
    # sort on y
    points = sorted(points,key=Y)

    nrows = ncols = radius * 2 + 1 + 2

    empty_row = [not_visited for _ in range(ncols)]    # ['-','-',...]
   
    # grid has an empty frame around the circle
    grid = [empty_row[:] for _ in range(nrows)]   # list of lists
    # iterate over visited points and substitute symbols
    for (x,y) in points:
        # add one for the empty row on top and colun on left
        # add offset to address negative coordinates
        y = y + radius + 1
        x = x + radius + 1
        grid[y][x] = visited

    grid = '\n'.join(' '.join(row) for row in grid)

    print(grid)
    return grid

for r in (3,8):
    points = circle(r)  # generator/iterator
    grid = display(points,r)

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