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I am creating a program which allows a user to annotate images with points.
This program allows user to zoom in an image so user can annotate more precisely.
Program zooms in an image doing the following:
Find the center of image
Find minimum and maximum coordinates of new cropped image relative to center
Crop image
Resize the image to original size
For this I have written the following Python code:
import cv2
def zoom_image(original_image, cut_off_percentage, list_of_points):
height, width = original_image.shape[:2]
center_x, center_y = int(width/2), int(height/2)
half_new_width = center_x - int(center_x * cut_off_percentage)
half_new_height = center_y - int(center_y * cut_off_percentage)
min_x, max_x = center_x - half_new_width, center_x + half_new_width
min_y, max_y = center_y - half_new_height, center_y + half_new_height
#I want to include max coordinates in new image, hence +1
cropped = original_image[min_y:max_y+1, min_x:max_x+1]
new_height, new_width = cropped.shape[:2]
resized = cv2.resize(cropped, (width, height))
translate_points(list_of_points, height, width, new_height, new_width, min_x, min_y)
I want to resize the image to original width and height so user always works on same "surface"
regardless of how zoomed image is.
The problem I encounter is how to correctly scale points (annotations) when doing this. My algorithm to do so was following:
Translate points on original image by subtracting min_x from x coordinate and min_y from y coordinate
Calculate constants for scaling x and y coordinates of points
Multiply coordinates by constants
For this I use the following Python code:
import cv2
def translate_points(list_of_points, height, width, new_height, new_width, min_x, min_y):
#Calculate constants for scaling points
scale_x, scale_y = width / new_width, height / new_height
#Translate and scale points
for point in list_of_points:
point.x = (point.x - min_x) * scale_x
point.y = (point.y - min_y) * scale_y
This code doesn't work. If I zoom in once, it is hard to detect the offset of pixels but it happens. If I keep zooming in, it will be much easier to detect the "drift" of points. Here are images to provide examples. On original image (1440x850) I places a point in the middle of blue crosshair. The more I zoom in the image it is easier to see that algorithm doesn't work with bigger cut-ofs.
Original image. Blue crosshair is middle point of an image. Red angles indicate what will be borders after image is zoomed once
Image after zooming in once.
Image after zooming in 5 times. Clearly, green point is no longer in the middle of image
The cut_off_percentage I used is 15% (meaning that I keep 85% of width and height of original image, calculated from the center).
I have also tried the following library: Augmentit python library
Library has functions for cropping images and resizing them together with points. Library also causes the points to drift. This is expected since the code I implemented and library's functions use the same algorithm.
Additionally, I have checked whether this is a rounding problem. It is not. Library rounds the points after multiplying coordinates with scales. Regardless on how they are rounded, points are still off by 4-5 px. This increases the more I zoom in the picture.
EDIT: A more detailed explanation is given here since I didn't understand a given answer.
The following is an image of right human hand.
Image of a hand in my program
Original dimension of this image is 1440 pixels in width and 850 pixels in height. As you can see in this image, I have annotated right wrist at location (756.0, 685.0). To check whether my program works correctly, I have opened this exact image in GIMP and placed a white point at location (756.0, 685.0). The result is following:
Image of a hand in GIMP
Coordinates in program work correctly. Now, if I were to calculate parameters given in first answer according to code given in first answer I get following:
vec = [756, 685]
hh = 425
hw = 720
cov = [720, 425]
These parameters make sense to me. Now I want to zoom the image to scale of 1.15. I crop the image by choosing center point and calculating low and high values which indicate what rectangle of image to keep and what to cut. On the following image you can see what is kept after cutting (everything inside red rectangle).
What is kept when cutting
Lows and highs when cutting are:
xb = [95,1349]
yb = [56,794]
Size of cropped image: 1254 x 738
This cropped image will be resized back to original image. However, when I do that my annotation gets completely wrong coordinates when using parameters described above.
After zoom
This is the code I used to crop, resize and rescale points, based on the first answer:
width, height = image.shape[:2]
center_x, center_y = int(width / 2), int(height / 2)
scale = 1.15
scaled_width = int(center_x / scale)
scaled_height = int(center_y / scale)
xlow = center_x - scaled_width
xhigh = center_x + scaled_width
ylow = center_y - scaled_height
yhigh = center_y + scaled_height
xb = [xlow, xhigh]
yb = [ylow, yhigh]
cropped = image[yb[0]:yb[1], xb[0]:xb[1]]
resized = cv2.resize(cropped, (width, height), cv2.INTER_CUBIC)
#Rescaling poitns
cov = (width / 2, height / 2)
width, height = resized.shape[:2]
hw = width / 2
hh = height / 2
for point in points:
x, y = point.scx, point.scy
x -= xlow
y -= ylow
x -= cov[0] - (hw / scale)
y -= cov[1] - (hh / scale)
x *= scale
y *= scale
x = int(x)
y = int(y)
point.set_coordinates(x, y)
So this really is an integer rounding issue. It's magnified at high zoom levels because being off by 1 pixel at 20x zoom throws you off much further. I tried out two versions of my crop-n-zoom gui. One with int rounding, another without.
You can see that the one with int rounding keeps approaching the correct position as the zoom grows, but as soon as the zoom takes another step, it rebounds back to being wrong. The non-rounded version sticks right up against the mid-lines (denoting the proper position) the whole time.
Note that the resized rectangle (the one drawn on the non-zoomed image) blurs past the midlines. This is because of the resize interpolation from OpenCV. The yellow rectangle that I'm using to check that my points are correctly scaling is redrawn on the zoomed frame so it stays crisp.
With Int Rounding
Without Int Rounding
I have the center-of-view locked to the bottom right corner of the rectangle for this demo.
import cv2
import numpy as np
# clamp value
def clamp(val, low, high):
if val < low:
return low;
if val > high:
return high;
return val;
# bound the center-of-view
def boundCenter(cov, scale, hh, hw):
# scale half res
scaled_hw = int(hw / scale);
scaled_hh = int(hh / scale);
# bound
xlow = scaled_hw;
xhigh = (2*hw) - scaled_hw;
ylow = scaled_hh;
yhigh = (2*hh) - scaled_hh;
cov[0] = clamp(cov[0], xlow, xhigh);
cov[1] = clamp(cov[1], ylow, yhigh);
# do a zoomed view
def zoomView(orig, cov, scale, hh, hw):
# calculate crop
scaled_hh = int(hh / scale);
scaled_hw = int(hw / scale);
xlow = cov[0] - scaled_hw;
xhigh = cov[0] + scaled_hw;
ylow = cov[1] - scaled_hh;
yhigh = cov[1] + scaled_hh;
xb = [xlow, xhigh];
yb = [ylow, yhigh];
# crop and resize
copy = np.copy(orig);
crop = copy[yb[0]:yb[1], xb[0]:xb[1]];
display = cv2.resize(crop, (width, height), cv2.INTER_CUBIC);
return display;
# draw vector shape
def drawVec(img, vec, pos, cov, hh, hw, scale):
con = [];
for point in vec:
# unpack point
x,y = point;
x += pos[0];
y += pos[1];
# here's the int version
# Note: this is the same as xlow and ylow from the above function
# x -= cov[0] - int(hw / scale);
# y -= cov[1] - int(hh / scale);
# rescale point
x -= cov[0] - (hw / scale);
y -= cov[1] - (hh / scale);
x *= scale;
y *= scale;
x = int(x);
y = int(y);
# add
con.append([x,y]);
con = np.array(con);
cv2.drawContours(img, [con], -1, (0,200,200), -1);
# font stuff
font = cv2.FONT_HERSHEY_SIMPLEX;
fontScale = 1;
fontColor = (255, 100, 0);
thickness = 2;
# draw blank
res = (800,1200,3);
blank = np.zeros(res, np.uint8);
print(blank.shape);
# draw a rectangle on the original
cv2.rectangle(blank, (100,100), (400,200), (200,150,0), -1);
# vectored shape
# comparison shape
bshape = [[100,100], [400,100], [400,200], [100,200]];
bpos = [0,0]; # offset
# random shape
vshape = [[148, 89], [245, 179], [299, 67], [326, 171], [385, 222], [291, 235], [291, 340], [229, 267], [89, 358], [151, 251], [57, 167], [167, 164]];
vpos = [100,100]; # offset
# get original image res
height, width = blank.shape[:2];
hh = int(height / 2);
hw = int(width / 2);
# center of view
cov = [600, 400];
camera_spd = 5;
# scale
scale = 1;
scale_step = 0.2;
# loop
done = False;
while not done:
# crop and show image
display = zoomView(blank, cov, scale, hh, hw);
# drawVec(display, vshape, vpos, cov, hh, hw, scale);
drawVec(display, bshape, bpos, cov, hh, hw, scale);
# draw a dot in the middle
cv2.circle(display, (hw, hh), 4, (0,0,255), -1);
# draw center lines
cv2.line(display, (hw,0), (hw,height), (0,0,255), 1);
cv2.line(display, (0,hh), (width,hh), (0,0,255), 1);
# draw zoom text
cv2.putText(display, "Zoom: " + str(scale), (15,40), font,
fontScale, fontColor, thickness, cv2.LINE_AA);
# show
cv2.imshow("Display", display);
key = cv2.waitKey(1);
# check keys
done = key == ord('q');
# Note: if you're actually gonna make a GUI
# use the keyboard module or something else for this
# wasd to move center-of-view
if key == ord('d'):
cov[0] += camera_spd;
if key == ord('a'):
cov[0] -= camera_spd;
if key == ord('w'):
cov[1] -= camera_spd;
if key == ord('s'):
cov[1] += camera_spd;
# z,x to decrease/increase zoom (lower bound is 1.0)
if key == ord('x'):
scale += scale_step;
if key == ord('z'):
scale -= scale_step;
scale = round(scale, 2);
# bound cov
boundCenter(cov, scale, hh, hw);
Edit: Explanation of the drawVec parameters
img: The OpenCV image to be drawn on
vec: A list of [x,y] points
pos: The offset to draw those points at
cov: Center-Of-View, where the middle of our zoomed display is pointed at
hh: Half-Height, the height of "img" divided by 2
hw: Half-Width, the width of "img" divided by 2
I have looked through my code and realized where I was making a mistake which caused points to be offset.
In my program, I have a canvas of specific size. The size of canvas is a constant and is always larger than images being drawn on canvas. When program draws an image on canvas it first resizes that image so it could fit on canvas. The size of resized image is somewhat smaller than size of canvas. Image is usually drawn starting from top left corner of canvas. Since I wanted to always draw image in the center of canvas, I shifted the location from top left corner of canvas to another point. This is what I didn't account when doing image zooming.
def zoom(image, ratio, points, canvas_off_x, canvas_off_y):
width, height = image.shape[:2]
new_width, new_height = int(ratio * width), int(ratio * height)
center_x, center_y = int(new_width / 2), int(new_height / 2)
radius_x, radius_y = int(width / 2), int(height / 2)
min_x, max_x = center_x - radius_x, center_x + radius_x
min_y, max_y = center_y - radius_y, center_y + radius_y
img_resized = cv2.resize(image, (new_width,new_height), interpolation=cv2.INTER_LINEAR)
img_cropped = img_resized[min_y:max_y+1, min_x:max_x+1]
for point in points:
x, y = point.get_original_coordinates()
x -= canvas_off_x
y -= canvas_off_y
x = int((x * ratio) - min_x + canvas_off_x)
y = int((y * ratio) - min_y + canvas_off_y)
point.set_scaled_coordinates(x, y)
In the code below canvas_off_x and canvas_off_y is the location of offset from top left corner of canvas
I'm trying to retrieve the orientation of a hand written arrows:
after removing shadows and applying binarization and dilating the lines, here are the images:
Now I'd like to get the orientation of the arrow so I have tried using HoughLines,
lines = cv2.HoughLines(edges, rho=1, theta=np.pi / 180, threshold=20)
But is seems it generates too many lines (around 54 lines), I'd like it to generate only 3 lines so I would be able to find the intersection of those lines. I can group the lines into groups of similar angle (+/-20 degrees) and then average the angle. but I'm not sure what should be rho of an average line, can somebody please give a simple example?
Is there any other approach which may be more accurate?
I'll be glad to hear, thank you all
I suggest a different approach. In summary, the approach goes as follows (made it in a hurry, might need some tuning):
Find the center of the minimum area rectangle (rotated rectangle) that encloses the whole arrow. (The circle drawn in the third image)
Find the center of gravity for all white points. It will be shifted a bit towards the actual head of the arrow. (Drawn in 4th pic as the origin of the eigenvector)
Find eigenvectors for all white points.
Find the displacement vector (the center of gravity - the center of the rotated rectangle)
Now:
Arrow angle(unoriented): is the angle of the first eigenvector
Arrow direction: is the sign of the dot product of (the first eigenvector and the centers' displacement vector)
Code:
Parts related to PCA are inspired by and mostly copied from this. I only made a minor change to the "getOrientation" method, added the following lines before it returns
angle = (angle - math.pi) * 180 / math.pi
return angle, (mean[0,0]), (mean[0,1]), p1
Code implementing the logic above:
#threshold
_, img = cv2.threshold(img, 128, 255, cv2.THRESH_OTSU)
imshow(img)
#close the image to make sure the contour is connected)
st_el = cv2.getStructuringElement(cv2.MORPH_RECT, (5, 5))
img = cv2.morphologyEx(img, cv2.MORPH_CLOSE, st_el)
imshow(img)
#get white points
pnts = cv2.findNonZero(img)
#min area rect
rect_center = cv2.minAreaRect(pnts)[0]
#draw rect center
cv2.circle(img, (int(rect_center[0]), int(rect_center[1])), 3, 128, -1)
imshow(img)
angle, pca_center, eigen_vec = getOrientation(pnts, img)
cc_vec = (rect_center[0] - pca_center[0], rect_center[1] - pca_center[1])
dot_product = cc_vec[0] * eigen_vec[0] + cc_vec[1] * eigen_vec[1]
if dot_product > 0:
angle *= -1
print ("Angle = ", angle)
imshow(img)
Edit
I suggest a simpler method. This new method does not depend on PCA for finding the unoriented angle [0 - 180]. Instead, uses the min area rectangle angle immediately. And uses the contour momentum for finding the center of gravity.
Simpler Method Code:
#get white points
pnts = cv2.findNonZero(img)
#min area rect
rect_center, size, angle = cv2.minAreaRect(pnts)
#simple fix for angle to make it in [0, 180]
angle = abs(angle)
if size[0] < size[1]:
angle += 90
#find center of gravity
M = cv2.moments(img)
gravity_center = (M["m10"] / M["m00"], M["m01"] / M["m00"])
#rot rect vec based on angle
angle_unit_vec = (math.cos(angle * 180 / math.pi), math.sin(angle * 180 / math.pi))
#cc_vec = gravity center - rect center
cc_vec = (gravity_center[0] - rect_center[0], gravity_center[1] - rect_center[1])
#if dot product is negative add 180 -> angle between [0, 360]
dot_product = cc_vec[0] * angle_unit_vec[0] + cc_vec[1] * angle_unit_vec[1]
angle += (dot_product < 0) * 180
#draw rect center
cv2.circle(img, (int(rect_center[0]), int(rect_center[1])), 3, 128, -1)
cv2.circle(img, (int(gravity_center[0]), int(gravity_center[1])), 3, 20, -1)
imshow(img)
print ("Angle = ", angle)
Edit2:
This edit includes these changes:
Use cv2.fitLine() and use the fitted line angle for orientation.
Replace angle_unit_vec with a vector that has the gravity center as the origin and goes parallel to the fitted line.
Code
#get white points
pnts = cv2.findNonZero(img)
#min area rect
rect_center, size, angle = cv2.minAreaRect(pnts)
#fit line to get angle
[vx, vy, x, y] =cv2.fitLine(pnts, cv2.DIST_L12, 0, 0.01, 0.01)
angle = (math.atan2(vy, -vx)) * 180 / math.pi
M = cv2.moments(img)
gravity_center = (M["m10"] / M["m00"], M["m01"] / M["m00"])
angle_vec = (int(gravity_center[0] + 100 * vx), int(gravity_center[1] + 100 * vy))
#cc_vec = gravity center - rect center
cc_vec = (gravity_center[0] - rect_center[0], gravity_center[1] - rect_center[1])
#if dot product is positive add 180 -> angle between [0, 360]
dot_product = cc_vec[0] * angle_vec[0] + cc_vec[1] * angle_vec[1]
angle += (dot_product > 0) * 180
angle += (angle < 0) * 360
#draw rect center
cv2.circle(img, (int(rect_center[0]), int(rect_center[1])), 3, 128, -1)
cv2.circle(img, (int(gravity_center[0]), int(gravity_center[1])), 3, 20, -1)
imshow(img)
print ("Angle = ", angle)
Output:
Using code from edit2:
First image:
Second image:
Third image:
I have two spheres that are intersecting, and I'm trying to find the intersection point nearest in the direction of the point (0,0,1)
My first sphere's (c1) center is at (c1x = 0, c1y = 0, c1z = 0) and has a radius of r1 = 2.0
My second sphere's (c2) center is at (c2x = 2, c2y = 0, c2z = 0) and has a radius of r2 = 2.0
I've been following the logic on this identical question for the 'Typical intersections' part, but was having some trouble understanding it and was hoping someone could help me.
First I'm finding the center of intersection c_i and radius of the intersecting circle r_i:
Here the first sphere has center c_1 and radius r_1, the second c_2 and r_2, and their intersection has center c_i and radius r_i. Let d = ||c_2 - c_1||, the distance between the spheres.
So sphere1 has center c_1 = (0,0,0) with r_1 = 2. Sphere2 has c_2 = (2,0,0) with r_2 = 2.0.
d = ||c_2 - c_1|| = 2
h = 1/2 + (r_1^2 - r_2^2)/(2* d^2)
So now I solve the function of h like so and get 0.5:
h = .5 + (2^2 - 2^2)/(2*2^2)
h = .5 + (0)/(8)
h = 0.5
We can sub this into our formula for c_i above to find the center of the circle of intersections.
c_i = c_1 + h * (c_2 - c_1)
(this equation was my original question, but a comment on this post helped me understand to solve it for each x,y,z)
c_i_x = c_1_x + h * (c_2_x - c_1_x)
c_i_x = 0 + 0.5 * (2 - 0) = 0.5 * 2
1 = c_i_x
c_i_y = c_1_y + h * (c_2_y - c_1_y)
c_i_y = 0 + 0.5 * (0- 0)
0 = c_i_y
c_i_z = c_1_z + h * (c_2_z - c_1_z)
c_i_z = 0 + 0.5 * (0 - 0)
0 = c_i_z
c_i = (c_i_x, c_i_z, c_i_z) = (1, 0, 0)
Then, reversing one of our earlier Pythagorean relations to find r_i:
r_i = sqrt(r_1*r_1 - hhd*d)
r_i = sqrt(4 - .5*.5*2*2)
r_i = sqrt(4 - 1)
r_i = sqrt(3)
r_i = 1.73205081
So if my calculations are correct, I know the circle where my two spheres intersect is centered at (1, 0, 0) and has a radius of 1.73205081
I feel somewhat confident about all the calculations above, the steps make sense as long as I didn't make any math mistakes. I know I'm getting closer but my understanding begins to weaken starting at this point. My end goal is to find an intersection point nearest to (0,0,1), and I have the circle of intersection, so I think what I need to do is find a point on that circle which is nearest to (0,0,1) right?
The next step from this solutionsays:
So, now we have the center and radius of our intersection. Now we can revolve this around the separating axis to get our full circle of solutions. The circle lies in a plane perpendicular to the separating axis, so we can take n_i = (c_2 - c_1)/d as the normal of this plane.
So finding the normal of the plane involves n_i = (c_2 - c_1)/d, do I need to do something similar for finding n_i for x, y, and z again?
n_i_x = (c_2_x - c_1_x)/d = (2-0)/2 = 2/2 = 1
n_i_y = (c_2_y - c_1_y)/d = (0-0)/2 = 0/2 = 0
n_i_z = (c_2_z - c_1_z)/d = (0-0)/2 = 0/2 = 0
After choosing a tangent and bitangent t_i and b_i perpendicular to this normal and each other, you can write any point on this circle as: p_i(theta) = c_i + r_i * (t_i * cos(theta) + b_i sin(theta));
Could I choose t_i and b_i from the point I want to be nearest to? (0,0,1)
Because of the Hairy Ball Theorem, there's no one universal way to choose the tangent/bitangent to use. My recommendation would be to pick one of the coordinate axes not parallel to n_i, and set t_i = normalize(cross(axis, n_i)), and b_i = cross(t_i, n_i) or somesuch.
c_i = c_1 + h * (c_2 - c_1)
This is vector expression, you have to write similar one for every component like this:
c_i.x = c_1.x + h * (c_2.x - c_1.x)
and similar for y and z
As a result, you'll get circle center coordinates:
c_i = (1, 0, 0)
As your citate says, choose axis not parallel to n vect0r- for example, y-axis, get it's direction vector Y_dir=(0,1,0) and multiply by n
t = Y_dir x n = (0, 0, 1)
b = n x t = (0, 1, 0)
Now you have two vectors t,b in circle plane to build circumference points.
I want to make a program which plots a Sierpinsky triangle (of any modulo). In order to do it I've used TkInter. The program generates the fractal by moving a point randomly, always keeping it in the sides. After repeating the process many times, the fractal appears.
However, there's a problem. I don't know how to plot points on a canvas in TkInter. The rest of the program is OK, but I had to "cheat" in order to plot the points by drawing small lines instead of points. It works more or less, but it doesn't have as much resolution as it could have.
Is there a function to plot points on a canvas, or another tool to do it (using Python)? Ideas for improving the rest of the program are also welcome.
Thanks. Here's what I have:
from tkinter import *
import random
import math
def plotpoint(x, y):
global canvas
point = canvas.create_line(x-1, y-1, x+1, y+1, fill = "#000000")
x = 0 #Initial coordinates
y = 0
#x and y will always be in the interval [0, 1]
mod = int(input("What is the modulo of the Sierpinsky triangle that you want to generate? "))
points = int(input("How many points do you want the triangle to have? "))
tkengine = Tk() #Window in which the triangle will be generated
window = Frame(tkengine)
window.pack()
canvas = Canvas(window, height = 700, width = 808, bg = "#FFFFFF") #The dimensions of the canvas make the triangle look equilateral
canvas.pack()
for t in range(points):
#Procedure for placing the points
while True:
#First, randomly choose one of the mod(mod+1)/2 triangles of the first step. a and b are two vectors which point to the chosen triangle. a goes one triangle to the right and b one up-right. The algorithm gives the same probability to every triangle, although it's not efficient.
a = random.randint(0,mod-1)
b = random.randint(0,mod-1)
if a + b < mod:
break
#The previous point is dilated towards the origin of coordinates so that the big triangle of step 0 becomes the small one at the bottom-left of step one (divide by modulus). Then the vectors are added in order to move the point to the same place in another triangle.
x = x / mod + a / mod + b / 2 / mod
y = y / mod + b / mod
#Coordinates [0,1] converted to pixels, for plotting in the canvas.
X = math.floor(x * 808)
Y = math.floor((1-y) * 700)
plotpoint(X, Y)
tkengine.mainloop()
If you are wanting to plot pixels, a canvas is probably the wrong choice. You can create a PhotoImage and modify individual pixels. It's a little slow if you plot each individual pixel, but you can get dramatic speedups if you only call the put method once for each row of the image.
Here's a complete example:
from tkinter import *
import random
import math
def plotpoint(x, y):
global the_image
the_image.put(('#000000',), to=(x,y))
x = 0
y = 0
mod = 3
points = 100000
tkengine = Tk() #Window in which the triangle will be generated
window = Frame(tkengine)
window.pack()
the_image = PhotoImage(width=809, height=700)
label = Label(window, image=the_image, borderwidth=2, relief="raised")
label.pack(fill="both", expand=True)
for t in range(points):
while True:
a = random.randint(0,mod-1)
b = random.randint(0,mod-1)
if a + b < mod:
break
x = x / mod + a / mod + b / 2 / mod
y = y / mod + b / mod
X = math.floor(x * 808)
Y = math.floor((1-y) * 700)
plotpoint(X, Y)
tkengine.mainloop()
You can use canvas.create_oval with the same coordinates for the two corners of the bounding box:
from tkinter import *
import random
import math
def plotpoint(x, y):
global canvas
# point = canvas.create_line(x-1, y-1, x+1, y+1, fill = "#000000")
point = canvas.create_oval(x, y, x, y, fill="#000000", outline="#000000")
x = 0 #Initial coordinates
y = 0
#x and y will always be in the interval [0, 1]
mod = int(input("What is the modulo of the Sierpinsky triangle that you want to generate? "))
points = int(input("How many points do you want the triangle to have? "))
tkengine = Tk() #Window in which the triangle will be generated
window = Frame(tkengine)
window.pack()
canvas = Canvas(window, height = 700, width = 808, bg = "#FFFFFF") #The dimensions of the canvas make the triangle look equilateral
canvas.pack()
for t in range(points):
#Procedure for placing the points
while True:
#First, randomly choose one of the mod(mod+1)/2 triangles of the first step. a and b are two vectors which point to the chosen triangle. a goes one triangle to the right and b one up-right. The algorithm gives the same probability to every triangle, although it's not efficient.
a = random.randint(0,mod-1)
b = random.randint(0,mod-1)
if a + b < mod:
break
#The previous point is dilated towards the origin of coordinates so that the big triangle of step 0 becomes the small one at the bottom-left of step one (divide by modulus). Then the vectors are added in order to move the point to the same place in another triangle.
x = x / mod + a / mod + b / 2 / mod
y = y / mod + b / mod
#Coordinates [0,1] converted to pixels, for plotting in the canvas.
X = math.floor(x * 808)
Y = math.floor((1-y) * 700)
plotpoint(X, Y)
tkengine.mainloop()
with a depth of 3 and 100,000 points, this gives:
Finally found a solution: if a 1x1 point is to be placed in pixel (x,y), a command which does it exactly is:
point = canvas.create_line(x, y, x+1, y+1, fill = "colour")
The oval is a good idea for 2x2 points.
Something remarkable about the original program is that it uses a lot of RAM if every point is treated as a separate object.
I have gotten as far as making a set of rays, but I need to connect them. Any help? My code is as follows
from math import *
from graphics import *
i = 1
segments = 15
lastPoint = Point(100,0)
print("Begin")
win = GraphWin("Trigonometry", 1500, 1500)
while i<=segments:
angle =i*pi/segments
y = int(sin(angle)*100)
x = int(cos(angle)*100)
i = i+1
p = Point(x,y)
l = Line(p, lastPoint)
l.draw(win)
print(p.x, p.y)
print("End")
OP code draws only "rays" because, while point p lays on the circle, lastPoint doesn't change between iterations.
We have to update the value of lastPoint to literally the last point calculated in order to draw the arc as a series of consecutive segments.
Here is a modified code, with further explanations as asked by OP in his comment:
from math import *
from graphics import *
# Function to calculate the integer coordinates of a Point on a circle
# given the center (c, a Point), the radius (r) and the angle (a, radians)
def point_on_circle( c, r, a ) :
return Point( int(round(c.x + r*cos(a))), int(round(c.y + r*sin(a))) )
# Define the graphical output window, I'll set the coordinates system so
# that the origin is the bottom left corner of the windows, y axis is going
# upwards and 1 unit corresponds to 1 pixel
win = GraphWin("Trigonometry", 800, 600)
win.setCoords(0,0,800,600)
# Arc data. Angles are in degrees (more user friendly, but later will be
# transformed in radians for calculations), 0 is East, positive values
# are counterclockwise. A value of 360 for angle_range_deg gives a complete
# circle (polygon).
angle_start_deg = 0
angle_range_deg = 90
center = Point(10,10)
radius = 200
segments = 16
angle_start = radians(angle_start_deg)
angle_step = radians(angle_range_deg) / segments
# Initialize lastPoint with the position corresponding to angle_start
# (or i = 0). Try different values of all the previous variables
lastPoint = point_on_circle(center, radius, angle_start)
print("Begin")
i = 1
while i <= segments :
# update the angle to calculate a new point on the circle
angle = angle_start + i * angle_step
p = point_on_circle(center, radius, angle)
# draw a line between the last two points
l = Line(p, lastPoint)
l.draw(win)
print(p.x, p.y)
# update the variables to move on to the next segment which share an edge
# (the last point) with the previous segment
i = i + 1
lastPoint = p
print("End")