bjd2385 - 1 year ago 114

Python Question

I've found this answer, which seems to be somewhat related to this question, but I'm wondering if it's possible to generate the coordinates one by one without the additional ~22% (1 - pi / 4) loss of comparing each point to the radius of the circle (by computing the distance between the circle's center and that point).

So far I have the following function in Python. I know by Gauss' circle problem the *number* of coordinates I will end up with, but I'd like to generate those points one by one as well.

`from typing import Iterable`

from math import sqrt, floor

def circCoord(sigma: float =1.0, centroid: tuple =(0, 0)) -> Iterable[tuple]:

r""" Generate all coords within $3\vec{\sigma}$ of the centroid """

# The number of least iterations is given by Gauss' circle problem:

# http://mathworld.wolfram.com/GausssCircleProblem.html

maxiterations = 1 + 4 * floor(3 * sigma) + 4 * sum(\

floor(sqrt(9 * sigma**2 - i**2)) for i in range(1, floor(3 * sigma) + 1)

)

for it in range(maxiterations):

# `yield` points in image about `centroid` over which we loop

What I'm trying to do is iterate over only those pixels lying within 3 * sigma of a pixel (at

`centroid`

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Answer Source

What about simply something like this (for a circle at origin)?

```
X = int(R) # R is the radius
for x in range(-X,X+1):
Y = int((R*R-x*x)**0.5) # bound for y given x
for y in range(-Y,Y+1):
yield (x,y)
```

This can easily be adapted to the general case when the circle is not centred at the origin.

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