# Regular Meeting (2020.3.2)

## Hough transform for curve

The curve eq. in x-y space, we can express with:

$$(x-h)^2 + (y-k)^2 = r^2 ...eq.1$$


If a center point of circuit at $(0,0)$,

then we can assume:

$$h = r cos(t)$$

$$k = r sin(t)$$

Substituting $r cos(t)$ and $r sin(t)$ into $(h,k)$ of $eq.1$,

we can obtain:

$$(x-r cos(t))^2 + (y-r sin(t))^2 = r^2 ...eq.2$$

Arranging $eq.2$:

$$r = \frac{1}{2} \frac{x^2 + y^2}{x cos(t) + y sin(t)} ...eq.3$$

[Demo](https://www.youtube.com/watch?v=Ltqt24SQQoI):


Now,

assume that we have the eqution of circit $$(x-0)^2 + (y-0)^2 = 5^2 ...eq.4$$

its chart and some points:


We mapping

$f(x,y) -> g(t,r)$ where $t = rad. degree,$ $r=eq.3$

Using these points(above chart)

[0,5]
[3,4]
[5,0]
[0,-5]
[-4,-3]

plot $g(t,r)$ of $eq.4$

We get **error**(python with matplotlib, numpy, math)

$ python ./hough-curve-space-2D.py
Traceback (most recent call last):
  File "./hough-curve-space-2D.py", line 35, in <module>
    theta
  File "./hough-curve-space-2D.py", line 24, in get_hough_spcae
    r_axis[i] = numerator / denominator
ZeroDivisionError: float division by zero

Because denominator of $eq.3$ is 0.

The denominator $x cos(t) + y sin(t)$ of $eq.3$ goes to **0** very easy,

for example:

{ `t = 0` **or** `x = 0` **or** `y = 0` }

But numerator not,

if the numerator $x^2 + y^2$ of $eq.3$ goes to **0**,

only the condition

{ `x = 0` **and** `y = 0` }

can be satisfied.

So we swap denominator and numerator from $eq.3$

$$\frac{1}{r} = 2 \frac{x cos(t) + y sin(t)} {x^2 + y^2} ...eq.5$$
[code](https://github.com/curly-wei/small_program_training/blob/master/python_training/hough/hough-curve-spce-2D-1r.py)

But the chart...

###### tags: `Regular Meeting` `DeWei`