Hi I'm using contour detection in Julia to detect contours. UPDATE: I need to find closest distance between two contours (made from broken lines).Is there function that finds the closest distance ? I'm not sure whats the best way of joining those two.
The case I'm looking at is
so I've used the code given at link example link
[https://juliaimages.org/v0.21/democards/examples/contours/contour_detection/] I've contours with n-element Vector{CartesianIndex} datatype. How do I join the line / two contours?
UPDATE: I've tried removing close contour points using the following
sorted_contours = sort!(contours, by=x -> x[1])
for i in eachindex(sorted_contours)
if check_adjacent(sorted_contours[i], sorted_contours) >= 1
sorted_contours[i] = CartesianIndex.(0, 0)
end
end
deleteat!(sorted_contours,findall(sorted_contours.==CartesianIndex.(0, 0))) but from this [![enter image description here][3]][3] it improved to this only [![enter image description here][3]][3]
I want to remove these points as these are incorrect contour points. Or how could I remove the spur which causes it?
You are vague in your question (as of 27 November) about the packages you are using. If you are using Contour, here is a nonoptimized way to find the minimum distance between two contours. You would need to select the two contours using contour() for the different x, y, z sets of values in each of your different contours.
using Contour
using Distances
const x = -3:0.01:3
const y = -4:0.02:5
const z = [Float64((xi^2 + yi^2)) for xi in x, yi in y]
x5, y5 = contour(x, y, z, 5) |> lines |> first |> coordinates
points5 = [[x5[i], y5[i]] for i in eachindex(x5)]
x10, y10 = contour(x, y, z, 10) |> lines |> first |> coordinates
points10 = [[x10[i], y10[i]] for i in eachindex(x10)]
distances = [(euclidean(p1, p2), p1, p2) for p1 in points5, p2 in points10]
d, idx = findmin(t -> t[1], distances)
#show d, idx #(d, idx) = (0.9261944913478685, CartesianIndex(918, 126))
#show distances[idx] # distances[idx] = (0.9261944913478685, [1.64, 1.52], [2.32, 2.1488372093023256])
#show points5[idx[1]] points10[idx[2]]
# points5[idx[1]] = [1.64, 1.52]
# points10[idx[2]] = [2.32, 2.1488372093023256]
It's the best approach to show your effort and explain where you have the problem, and you'll get the best answers to solve it. Since this topic interested me and made me curious, I found it worthwhile, although your question doesn't meet the requirements of a good question.
To better detect the contours, First, I perform thinning on the image to locate the contours easily. I use the following image as the input to see the results in a bigger size:
using Images, FileIO, ImageBinarization
img = load("img.png")
gimg = Gray.(img)
bin = binarize(gimg, UnimodalRosin()) .> 0.5
thinned = thinning(bin, algo=GuoAlgo())
The thinned variable represents the following picture:
julia> Gray.(thinned)
Now, I define my functions to find the contours:
"""
check_adjacent(loc::CartesianIndex{2}, all_locs::Vector{CartesianIndex{2}})
Return the number of adjacent locations that are in `all_locs` to `loc`.
This function is used to determine whether a location is a corner (contour) or not.
# Arguments
- `loc::CartesianIndex{2}`: the location to check.
- `all_locs::Vector{CartesianIndex{2}}`: all the locations that aren't black.
# Returns
- `Int`: the number of adjacent locations that are in `all_locs` to `loc`.
"""
function check_adjacent(loc::CartesianIndex{2}, all_locs::Vector{CartesianIndex{2}})
conditions = [
loc - CartesianIndex(0,1) ∈ all_locs,
loc + CartesianIndex(0,1) ∈ all_locs,
loc - CartesianIndex(1,0) ∈ all_locs,
loc + CartesianIndex(1,0) ∈ all_locs,
loc - CartesianIndex(1,1) ∈ all_locs,
loc + CartesianIndex(1,1) ∈ all_locs,
loc - CartesianIndex(1,-1) ∈ all_locs,
loc + CartesianIndex(1,-1) ∈ all_locs
]
return sum(conditions)
end
"""
find_the_contour(img::BitMatrix)
Return the contours of the image `img`.
# Arguments
- `img::BitMatrix`: the image to get the contours of.
# Returns
- `Vector{CartesianIndex{2}}`: the contours of the image `img`.
"""
function find_the_contour(img::BitMatrix)
img_matrix = convert(Array{Float64}, img)
not_black = findall(!=(0.0), img_matrix)
contours = Vector{CartesianIndex{2}}()
for nb∈not_black
t = check_adjacent(nb, not_black)
t==1 && push!(contours, nb)
end
return contours
end
Now I use the find_the_contour function:
julia> contours = find_the_contour(thinned)
4-element Vector{CartesianIndex{2}}:
CartesianIndex(161, 334)
CartesianIndex(256, 452)
CartesianIndex(69, 1213)
CartesianIndex(210, 1243)
How should I know this function locates the contours correctly? I define another function to highlight the contours:
"""
highlight_contours(img, contours)
Return the image `img` with the contours `contours` highlighted with red color.
# Arguments
- `img::BitMatrix`: the image to highlight the contours on.
- `contours::Vector{CartesianIndex{2}}`: the contours to highlight.
- `resize`::Bool=false: whether to resize the image to 512x512 or not.
- `scale`::Union{Int64, Float64}=1: the scale to resize the image to.
# Returns
- `img_matrix`: the image `img` with the contours `contours` highlighted with red color.
"""
function highlight_contours(img, contours; resize::Bool=false, scale::Union{Int64, Float64}=1)
img_matrix = convert(Array{RGB, 2}, img)
for c∈contours
img_matrix[c] = RGB(1,0,0)
end
if resize
img_matrix = typeof(scale)==Int64 ? imresize(img_matrix, size(img).*scale) : imresize(img_matrix, ratio=scale)
end
return img_matrix
end
Then I pass the thinned Matrix and contours to it:
julia> highlited = highlight_contours(thinned, contours)
If you zoom in a little bit, you find the contours with red color:
Now that I made sure the contours are located successfully, I go for calculating the distance between contours. I used the Euclidean distance between the CartesianIndexes of each contour! You can adjust it to match your taste:
"""
distance_between_contours(V::Vector{T}, v::T) where T<:Tuple{Int64, Int64}
Return the euclidean distance between the contour `v` and the contours in `V`.
# Arguments
- `V::Vector{T}`: the contours to check the distance to.
- `v::T`: the contour to check the distance from.
# Returns
- `Float64`: the euclidean distance between the contour `v` and the contours in `V`.
"""
function distance_between_contours(V::Vector{T}, v::T) where T<:Tuple{Int64, Int64}
return broadcast((x, y)->sqrt(sum((x.-y).^2)), V, Ref(v))
end
"""
distance_between_contours(all_contours::Vector{CartesianIndex{2}})
Return the euclidean distance between each pair of contours in `all_contours`.
# Arguments
- `all_contours::Vector{CartesianIndex{2}}`: the contours to calculate the distance between.
# Returns
- `Matrix{Float64}`: the euclidean distance between each pair of contours in `all_contours`.
"""
function distance_between_contours(all_contours::Vector{CartesianIndex{2}})
n = length(all_contours)
distances = Matrix{Float64}(undef, n, n)
for j in eachindex(all_contours)
distances[:, j] .= distance_between_contours(
Tuple.(all_contours),
Tuple(all_contours[j])
)
end
return distances
end
Now I pass it the contours and calculate the pairwise Euclidean distance between each:
julia> distance_between_contours(contours)
4×4 Matrix{Float64}:
0.0 151.489 883.801 910.32
151.489 0.0 783.639 792.336
883.801 783.639 0.0 144.156
910.32 792.336 144.156 0.0
In this matrix, for example, the 151.489 value shows the distance between the first contour and the second. Remember that the first and second contours are the followings:
CartesianIndex(161, 334)
CartesianIndex(256, 452)
I followed the same procedure for your image, and this the followings are the distances and the contours:
julia> distance_between_contours(contours)
6×6 Matrix{Float64}:
0.0 27.7308 88.6397 168.6 166.676 170.188
27.7308 0.0 60.9262 140.872 139.284 142.622
88.6397 60.9262 0.0 80.1311 79.2465 82.0061
168.6 140.872 80.1311 0.0 26.2488 19.2354
166.676 139.284 79.2465 26.2488 0.0 8.06226
170.188 142.622 82.0061 19.2354 8.06226 0.0
highlited = highlight_contours(thinned, contours, resize=true, scale=2)
TL;DR
Code:
using Images, FileIO, ImageBinarization
function check_adjacent(loc::CartesianIndex{2}, all_locs::Vector{CartesianIndex{2}})
conditions = [
loc - CartesianIndex(0,1) ∈ all_locs,
loc + CartesianIndex(0,1) ∈ all_locs,
loc - CartesianIndex(1,0) ∈ all_locs,
loc + CartesianIndex(1,0) ∈ all_locs,
loc - CartesianIndex(1,1) ∈ all_locs,
loc + CartesianIndex(1,1) ∈ all_locs,
loc - CartesianIndex(1,-1) ∈ all_locs,
loc + CartesianIndex(1,-1) ∈ all_locs
]
return sum(conditions)
end
function find_the_contour(img::BitMatrix)
img_matrix = convert(Array{Float64}, img)
not_black = findall(!=(0.0), img_matrix)
contours = Vector{CartesianIndex{2}}()
for nb∈not_black
t = check_adjacent(nb, not_black)
t==1 && push!(contours, nb)
end
return contours
end
function distance_between_contours(V::Vector{T}, v::T) where T<:Tuple{Int64, Int64}
return broadcast((x, y)->sqrt(sum((x.-y).^2)), V, Ref(v))
end
function distance_between_contours(all_contours::Vector{CartesianIndex{2}})
n = length(all_contours)
distances = Matrix{Float64}(undef, n, n)
for j in eachindex(all_contours)
distances[:, j] .= distance_between_contours(
Tuple.(all_contours),
Tuple(all_contours[j])
)
end
return distances
end
img = load("img.png")
gimg = Gray.(img)
bin = binarize(gimg, UnimodalRosin()) .> 0.5
thinned = thinning(bin, algo=GuoAlgo())
contours = find_the_contour(thinned)
distance_between_contours(contours)
Related
I am new to the Julia language and need to draw a circular sector on an image (2-dimensional UInt8 array for gray version or 3-dimensional UInt8 array for an RGB version). Afterwards this image is to be used as a mask to select data in other arrays, so I need the result, not as an image object, but as an array of booleans or integers.
There is the way to draw a circle by means of the ImageDraw package:
draw!(img, Ellipse(CirclePointRadius(350,200,100), fill = tue))
but found no way to provide a start and end angle.
You can use Luxor.jl's pie or sector function:
julia> begin
img = readpng("/path/Images/deepam.png")
Drawing(img.width, img.height, "sector-on-img.png")
placeimage(img)
origin()
sethue("orange")
pie(0, 0, 100, π/2, π, :fill)
sethue("olive")
sector(25, 125, 3π/2, 0, 15, :fill)
finish()
end
true
Result:
(Original png image scaled down, for comparison:
)
I think Julia is a great language, because (among other things) all libraries are implemented in the same language and you have ease acces to their sources.
And in this way, I have been able to modify the ellipse2d.jl script of the ImageDraw library.
The modification consits of adding another definition of the draw! funciton for ellipse objects (multiple dispatch of Julia is also great) that accepts a start and end angle.
I think the best way could be to define new objects, ellipse_sector and circle_sector, which would be the same as the ellipse and circle objects but with two more members: start_angle and end_angle. Then the correspondent drawing functions should be implemented. I would like to write to the ImageDraw package developers in order to make this suggestion or even offer me to make these changes, but I do not know the manage of github.
My solution, instead, does not modify any existing object, just adds a method to the draw! function that accpets two more arguments: startAngle and endAngle.
Here is the code, to be copied to the end of the ellipse2d.jl script:
function draw!(img::AbstractArray{T, 2}, ellipse::Ellipse, startAng::Real, endAng::Real, color::T) where T<:Colorant
# Solution to find out if an angle lies between two given ones, borrowed from:
# https://stackoverflow.com/questions/11406189/determine-if-angle-lies-between-2-other-angles/11412077#11412077
# Make all angles to lie in [0, 2π)
# rem2pi(ϕ, RoundNearest) returns the remainder of the division by 2π in the range [−π,π]
# mod2pi returns the remainder of the division by 2π in the range [0,2π)
Angle1 = mod2pi(startAng)
Angle2 = mod2pi(endAng)
# make the angle from angle1 to angle2 to be <= 180 degrees
rAngle = mod2pi( mod2pi(Angle2 - Angle1) + 2π)
if rAngle >= π
Angle1, Angle2 = Angle2, Angle1 # Swaps the values
end # if
ys = Int[]
xs = Int[]
break_point = 0
if ellipse.fill == false
break_point = ((ellipse.ρy - ellipse.thickness) / ellipse.ρy) ^ 2 + ((ellipse.ρx - ellipse.thickness) / ellipse.ρx) ^ 2
end
for i in ellipse.center.y - ellipse.ρy : ellipse.center.y + ellipse.ρy
for j in ellipse.center.x - ellipse.ρx: ellipse.center.x + ellipse.ρx
y = i - ellipse.center.y
x = j - ellipse.center.x
val = (x / ellipse.ρy) ^ 2 + (y / ellipse.ρx) ^ 2
# atan(y, x) returns the angle in the correct quadrant [−π,π], not like atan(y/x)
# But make it to be in the range [0, 2π)by means of mod2pi()
ang = mod2pi( atan(y, x) )
# Test if the angle lies betwen the startAngle and the endAngle
if (Angle1 <= Angle2)
AngleIsBetween = ang >= Angle1 && ang <= Angle2
else
AngleIsBetween = ang >= Angle1 || ang <= Angle2
end # if
if val < 1 && val >= break_point && AngleIsBetween
push!(ys, i)
push!(xs, j)
end
end
end
for (yi, xi) in zip(ys, xs)
drawifinbounds!(img, yi, xi, color)
end
img
end
I'm trying to align multiple line objects along a human body circumference depending on the orientation of the triangles from the mesh. I would like to put the lines parallel to the mesh. I correctly assign the position for the lines along the circumference, but I also need to add the rotation of the lines such that to be parallel with the body.
The body is a mesh formed by multiple triangles and every line is "linked" with a triangle.
All I have is:
3 points for the closest triangle from the mesh for every line
The normal of the triangle
The positions for the instantiated lines (2 points, start and end)
I need to calculate the angle for every X, Y, Z axes for the line such that the normal of the triangle is perpendicular with the line mesh. I don't know how to get the desired angle. I really appreciate if someone would like to help me.
input:
FVector TrianglePoints[3];
FVector Triangle_Normal; //Calculated as (B-A)^(C-A), where A,B,C are the points of the triangle
FVector linePosition; //I also have the start line and the endLine position if that helps
ouput:
//FRotator rotation(x,y,z), such that the triangle normal and the line object to be perpendicular.
An overview of the circumference line construction. Now the rotation is calculated using the Start position and End position for each line. When we cross some irregular parts of the mesh we want to rotate the lines correctly. Now the rotation is fixed, depending just on the line start and end position.
If I have understood correctly your goal, here is some related vector geometry:
A,B,C are the vertices of the triangle:
A = [xA, yA, zA],
B = [xB, yB, zB]
C = [xC, yC, zC]
K,L are the endpoints of the line-segment:
K = [xK, yK, zK]
L = [xL, yL, zL]
vectors are interpreted as row-vectors
by . I denote matrix multiplication
by x I denote cross product of 3D vectors
by t() I denote the transpose of a matrix
by | | I denote the norm (magnitude) of a vector
Goal: find the rotation matrix and rotation transformation of segment KL
around its midpoint, so that after rotation KL is parallel to the plane ABC
also, the rotation is the "minimal" angle rotation by witch we need to
rotate KL in order to make it parallel to ABC
AB = B - A
AC = C - A
KL = L - K
n = AB x AC
n = n / |n|
u = KL x n
u = u / |u|
v = n x u
cos = ( KL . t(v) ) / |KL|
sin = ( KL . t(n) ) / |KL|
U = [[ u[0], u[1], u[2] ],
[ v[0], v[1], v[2] ],
[ n[0], n[1], n[2] ],
R = [[1, 0, 0],
[0, cos, sin],
[0, -sin, cos]]
ROT = t(U).R.U
then, one can rotate the segment KL around its midpoint
M = (K + L)/2
Y = M + ROT (X - M)
Here is a python script version
A = np.array([0,0,0])
B = np.array([3,0,0])
C = np.array([2,3,0])
K = np.array([ -1,0,1])
L = np.array([ 2,2,2])
KL = L-K
U = np.empty((3,3), dtype=float)
U[2,:] = np.cross(B-A, C-A)
U[2,:] = U[2,:] / np.linalg.norm(U[2,:])
U[0,:] = np.cross(KL, U[2,:])
U[0,:] = U[0,:] / np.linalg.norm(U[0,:])
U[1,:] = np.cross(U[2,:], U[0,:])
norm_KL = np.linalg.norm(KL)
cos_ = KL.dot(U[1,:]) / norm_KL
sin_ = KL.dot(U[2,:]) / norm_KL
R = np.array([[1, 0, 0],
[0, cos_, sin_],
[0,-sin_, cos_]])
ROT = (U.T).dot(R.dot(U))
M = (K+L) / 2
K_rot = M + ROT.dot( K - M )
L_rot = M + ROT.dot( L - M )
print(L_rot)
print(K_rot)
print(L_rot-K_rot)
print((L_rot-K_rot).dot(U[2,:]))
A more inspired solution was to use a procedural mesh, generated at runtime, that have all the requirements that I need:
Continuously along multiple vertices
Easy to apply a UV map for texture tiling
Can be updated at runtime
Isn't hard to compute/work with it
In a game, an area is defined by triangles that never overlap, and characters are defined by circles.
How can I know whether the full character's collision circle is contained within these triangles?
Example image:
Here, the red parts are outside triangles, so the circle isn't contained within them. Is there an algorithm that can detect this?
I've only came up with "non-perfect" solutions, like sampling points at the border of the circle, then testing if each is inside a triangle.
So basically, the triangles form a domain with polygonal boundary and you want to check if a disk, defined by a center point and a radius is contained inside the domain. So if you start with the triangles, you have to find a way to extract the polygonal boundary of your domain and represent it as a 2D array (matrix) of shape n rows and two columns so that every row is the two coordinates of a vertex point of the polygonal boundary line and the points are ordered so that they are consecutive order along the boundary in a counterclockwise position, i.e. when you walk in a direction from point of index i to the next point i+1 the domain stays on your left. For example, here is the representation of a polygonal boundary of a domain like yours:
a = 4/math.sqrt(3)
Pgon = np.array([[0,0],
[a,0],
[2*a,-1],
[2*a+4,0],
[2*a+4,4],
[2*a,4],
[2*a,2],
[a,1],
[a,4],
[0,0]])
Observe that the first and the last points are the same.
In such a scenario, maybe you can try the following algorithm:
import numpy as np
import math
def angle_and_dist(p1, p2, o):
p12 = p2 - p1
op1 = p1 - o
op2 = p2 - o
norm_p12 = math.sqrt(p12[0]**2 + p12[1]**2)
norm_op1 = math.sqrt(op1[0]**2 + op1[1]**2)
norm_op2 = math.sqrt(op2[0]**2 + op2[1]**2)
p12_perp = np.array([ - p12[1], p12[0] ])
h = - op1.dot(p12_perp)
theta12 = op1.dot(op2) / (norm_op1*norm_op2)
theta12 = math.acos( theta12 )
if h < 0:
theta12 = - theta12
if op1.dot(p12) > 0:
return theta12, norm_op1
elif op2.dot(p12) < 0:
return theta12, norm_op2
else:
return theta12, h/norm_p12
def is_in_polygon(p, disk):
o, r = disk
n_p = len(p)-1
index_o = 0
h_min = 400
for i in range(n_p):
theta, h = angle_and_dist(p[i,:], p[i+1,:], o)
index_o = index_o + theta
if 0 <= h and h < h_min:
h_min = h
if theta <= math.pi/100:
return 'center of disc is not inside polygon'
elif theta > math.pi/100:
if h_min > r:
return 'disc is inside polygon'
else:
return 'center of disc is inside polygon but disc is not'
a = 4/math.sqrt(3)
Pgon = np.array([[0,0],
[a,0],
[2*a,-1],
[2*a+4,0],
[2*a+4,4],
[2*a,4],
[2*a,2],
[a,1],
[a,4],
[0,0]])
# A test example:
#disc = (np.array([3*a/4, 2]), a/4-0.001)
disc = (np.array([3*a/4, 2]), math.sqrt(3)*a/8 - 0.0001)
print(is_in_polygon(Pgon, disc))
I'm currently working on a game in LBP2 that has modify the way a controller gives input. This question:
How can I convert coordinates on a square to coordinates on a circle?
Has helped me quite a lot with what I am doing, but I do have one problem. I need the inverse function of the one they give. They go from square -> circle, and I've tried searching all over for how to map a circle to a square.
The function given in the previous question is:
xCircle = xSquare * sqrt(1 - 0.5*ySquare^2)
yCircle = ySquare * sqrt(1 - 0.5*xSquare^2)
From Mapping a Square to a Circle
My question is given xCircle and yCircle... how do I find xSquare and ySquare?
I've tried all of the algebra I know, filled up two pages of notes, tried to get wolfram alpha to get the inverse functions, but this problem is beyond my abilities.
Thank you for taking a look.
x = ½ √( 2 + u² - v² + 2u√2 ) - ½ √( 2 + u² - v² - 2u√2 )
y = ½ √( 2 - u² + v² + 2v√2 ) - ½ √( 2 - u² + v² - 2v√2 )
Note on notation: I'm using x = xSquare , y = ySquare, u = xCircle and v = yCircle;
i.e. (u,v) are circular disc coordinates and (x,y) are square coordinates.
For a C++ implementation of the equations, go to
http://squircular.blogspot.com/2015/09/mapping-circle-to-square.html
See http://squircular.blogspot.com
for more example images.
Also, see http://arxiv.org/abs/1509.06344 for the proof/derivation
This mapping is the inverse of
u = x √( 1 - ½ y² )
v = y √( 1 - ½ x² )
P.S. The mapping is not unique. There are other mappings out there. The picture below illustrates the non-uniqueness of the mapping.
if you have xCircle and yCircle that means that you're on a circle with radius R = sqrt(xCircle^2 + yCircle^2). Now you need to extend that circle to a square with half-side = R,
if (xCircle < yCircle)
ySquare = R, xSquare = xCircle * R/yCircle
else
xSquare = R, ySquare = yCircle * R/xCircle
this is for the first quadrant, for others you need some trivial tweaking with the signs
There are many ways you could do this; here's one simple way.
Imagine a circle of radius R centred on the origin, and a square of side 2R centred on the origin, we want to map all of the points within and on the boundary of the circle (with coordinates (x,y)) to points within and on the boundary of the square. Note that we can also describe points within the circle using polar coordinates (r, ø) (that's supposed to be a phi), where
x = r cos ø,
y = r sin ø
(ie r^2 = x^2 + y^2 and r <= 1). Then imagine other coordinates x' = a(ø) x = a(ø) r cos ø, and y' = a(ø) y (ie, we decide that a won't depend on r).
In order to map the boundary of the circle (r = 1) to the boundary of the square (x' = R), we must have, for ø < 45deg, x' = a(ø) R cos ø = R, so we must have a(ø) = 1/cos ø. Similarly, for 45 < ø < 90 we must have the boundary of the circle map to y' = R, giving a(ø) = 1/sin ø in that region. Continuing round the circle, we see that a(ø) must always be positive, so the final mapping from the circle to the square is
x' = a(ø) x,
y' = a(ø) y
where
ø = |arctan y/x| = arctan |y/x|
and
a(ø) = 1/cos ø, when ø <= 45 deg (ie, when x < y), and
a(ø) = 1/sin ø, when ø > 45 deg.
That immediately gives you the mapping in the other direction. If you have coordinates (x', y') on the square (where x' <= R and y' <= R), then
x = x'/a(ø)
y = y'/a(ø)
with a(ø) as above.
A much simpler mapping, though, is to calculate the (r, ø) for the desired position on the circle, and map that to x' = r and y' = ø. That also maps every point in the circle into a rectangle, and vice versa, and might have better properties, depending on what you want to do.
So that's the real question: what is it you're actually aiming to do here?
I was implementing the solution above but the results are not satisfiying.
The square coordinates are not exact.
Here is a simple counter-example:
Consider the point (x,y)=(0.75, 1) on the square.
We map it to the circle with (u,v)=(0.53, 0.85) on the circle.
Applying the expression above we get the new square coordinates
(x',y')=(u/v,r)=(0.625543242, 1) with r=(u^2+v^2)^(1/2).
This point is close but not the expected precise solution.
I solved a root finding problem in order to get the inverse expression of the mapping from square to circle like above.
you need to solve the system equations like above:
I) u = x*(1-y^2/2)^(1/2)
II) v = y*(1-x^2/2)^(1/2)
One ends up with 8 root points as solution. One of the roots I implemented into Excel-VBA which I present here below and it works very fine.
' given the circle coordinates (u,v) caluclates the x coordinate on the square
Function circ2sqrX(u As Double, v As Double) As Double
Dim r As Double, signX As Double, u2 As Double, v2 As Double, uuvv As Double, temp1 As Double
u2 = u * u
v2 = v * v
r = Sqr(u2 + v2)
signX = 1
If v = 0 Or u = 0 Then
circ2sqrX = u
Exit Function
End If
If u < 0 Then
signX = -1
End If
If Abs(u) = Abs(v) And r = 1 Then
circ2sqrX = signX
Exit Function
End If
uuvv = (u2 - v2) * (u2 - v2) / 4
temp1 = 2 * Sqr(uuvv - u2 - v2 + 1)
circ2sqrX = -((temp1 - u2 + v2 - 2) * Sqr(temp1 + u2 - v2 + 2)) / (4 * u)
End Function
' given the circle coordinates (u,v) caluclates the y coordinate on the square
' make use of symetrie property
Function circ2sqrY(u As Double, v As Double) As Double
circ2sqrY=circ2sqrX(v,u)
End Function
I'm using OpenCV to fit a line from a set of points using cvFitLine()
cvFitLine() returns a normalized vector that is co-linear to the line and a point on the line.
See details here
Using this information how can I get the equation of a line so that I can draw the line?
If cvFitLine() returns normalized vector (vx,vy) and point (x0,y0), then the equation of the line is
(x,y) = (x0,y0) + t*(vx,vy)
where t runs from −∞ to +∞.
This is what you asked for, but probably isn't immediately helpful in drawing the line. You would want to clip it either to the screen boundaries, or perhaps the bounding box of the the original set of points. To clip a line to a rectangle, just solve for values of t where the line crosses the boundary of the rectangle.
Just draw a big line instead of solving for the boundaries. eg:
cv.Line(img, (x0-m*vx[0], y0-m*vy[0]), (x0+m*vx[0], y0+m*vy[0]), (0,0,0))
will do it for example.. for m large enough :)
This just spells out #brainjam's answer in python for any passers by.
The formula for a line using a unit vector (vx, vy) and some point on the line (x0, y0) is:
(x, y) = (x0, y0) + t*(vx, vy)
The return from cv2.fitLine() is:
np.array([vx, vy, x0, y0])
In the example case, I have a line spanning the height of my image, so I want to find the t0 and t1 that intersect with y=0 and y=img.shape[0] (the top/bottom boundaries).
# get the fitLine for your set of points in the array, `line`
fit_line = cv2.fitLine(line, cv2.DIST_L2, 0, 0.01, 0.01)
# compute t0 for y=0 and t1 for y=img.shape[0]: (y-y0)/vy
t0 = (0-fit_line[3])/fit_line[1]
t1 = (img.shape[0]-fit_line[3])/fit_line[1]
# plug into the line formula to find the two endpoints, p0 and p1
# to plot, we need pixel locations so convert to int
p0 = (fit_line[2:4] + (t0 * fit_line[0:2])).astype(np.uint32)
p1 = (fit_line[2:4] + (t1 * fit_line[0:2])).astype(np.uint32)
# draw the line. For my version of opencv, it wants tuples so we
# flatten the arrays and convert
# args: cv2.line(image, p0, p1, color, thickness)
cv2.line(img, tuple(p0.ravel()), tuple(p1.ravel()), (0, 255, 0), 10)
I used a strategy similar to Karpathy up there but used an extra function. As you can see, I'm using cvClipLine to trim the line to the size of the image, which is unnecessary but does add a little niceness.
Also the multiplier here is defined as theMult = max(img->height,img->width) so we dont get numbers that might one day overflow or something.
void drawLine(IplImage * img, float line[4], int thickness,CvScalar color)
{
double theMult = max(img->height,img->width);
// calculate start point
CvPoint startPoint;
startPoint.x = line[2]- theMult*line[0];// x0
startPoint.y = line[3] - theMult*line[1];// y0
// calculate end point
CvPoint endPoint;
endPoint.x = line[2]+ theMult*line[0];//x[1]
endPoint.y = line[3] + theMult*line[1];//y[1]
// draw overlay of bottom lines on image
cvClipLine(cvGetSize(img), &startPoint, &endPoint);
cvLine(img, startPoint, endPoint, color, thickness, 8, 0);
}
Adding to #brainjam answer:
To clip to the bounding box of original set of points:
// std::vector<Point2i> points = ...
//lineParams: [vx,vy, x0,y0]: (normalized vector, point on our contour)
Vec4f lineParams; fitLine(points, lineParams, CV_DIST_L2, 0, 0.01, 0.01);
// derive the bounding xs of points
decltype(points)::iterator minXP, maxXP;
std::tie(minXP, maxXP) = std::minmax_element(points.begin(), points.end(), [](const Point2i& p1, const Point2i& p2){ return p1.x < p2.x; });
// derive y coords of fitted line
float m = lineParams[1] / lineParams[0];
int y1 = ((minXP->x - lineParams[2]) * m) + lineParams[3];
int y2 = ((maxXP->x - lineParams[2]) * m) + lineParams[3];
line(clearTarget, Point(minXP->x, y1), Point(maxXP->x, y2), Scalar(255, 255, 255), 2);
To clip to the entire image boundaries substitute minXP->x to 0 and maxXP->x to image.cols - 1, which was originally answered in https://stackoverflow.com/a/14192660/2380455
we use a " Vec4f fitedLine;" for fitted Line
in fitLine we have 4 parameters
if we consider Line relation az bellow:
Y - Y0 = M (X - X0)
we have
Y0 = FitedLine[3];
X0 = FitedLine[2];
m = FitedLine[1]/FitedLine[0];
so we have a Line equation we can find other points on it.