Python | Rotate Rectangle and get new boundariy dimensions - python-3.x

So we got the rectangle A and we rotate it around its corner by x degree. Now I want to know how to calculate the boundaries of the new rectangle.
What I mean with boundaries (blue rect):
known values are inner rectangle width/height/center/corners
Thanks in advance!

Bounding rectange has dimensions:
New_Height = Old_Width * Abs(Sin(Fi)) + Old_Height * Abs(Cos(Fi))
New_Width = Old_Width * Abs(Cos(Fi)) + Old_Height * Abs(Sin(Fi))

Related

PyQt creating color circle

I want to add a color circle to the widget placeholder there:
I already tried this library:
https://gist.github.com/tobi08151405/7b0a8151c9df1a41a87c1559dac1243a
But if the window wasnt a quadrat, the color circle didnt worked.
I have already tried the other solution, but there is no method to get the color value.
How to create a "Color Circle" in PyQt?
Could you recommend/ show me a way of creating my own, so I can add them there?
Thanks!
The widget assumes that its shape is always a square; the code provides a custom AspectLayout for that, but it's not necessary.
The problem comes from the fact that when the shape is not a square the computation of the color is wrong, as coordinates are not properly mapped when a dimension is much bigger than the other. For instance, if the widget is much wider than tall, the x coordinate is "shifted" since the circle (which is now an actual ellipse) is shown centered, but the color function uses the minimum size.
The solution is to create an internal QRect that is always displayed at the center and use that for both painting and computation:
class ColorCircle(QWidget):
# ...
def resizeEvent(self, ev: QResizeEvent) -> None:
size = min(self.width(), self.height())
self.radius = size / 2
self.square = QRect(0, 0, size, size)
self.square.moveCenter(self.rect().center())
def paintEvent(self, ev: QPaintEvent) -> None:
# ...
p.setPen(Qt.transparent)
p.setBrush(hsv_grad)
p.drawEllipse(self.square)
p.setBrush(val_grad)
p.drawEllipse(self.square)
# ...
def map_color(self, x: int, y: int) -> QColor:
x -= self.square.x()
y -= self.square.y()
# ...
Note that the code uses numpy, but it's used for functions that do not really require such a huge library for an application that clearly doesn't need numpy's performance.
For instance, the line_circle_inter uses a complex way to compute the position for the "cursor" (the small circle), but that's absolutely unnecessary, as the Hue and Saturation values already provide "usable" coordinates: the Hue indicates the angle in the circle (starting from 12 hour position, counter-clockwise), while the Saturation is the distance from the center.
QLineF provides a convenience function, fromPolar(), which returns a line with a given length and angle: the length will be the radius multiplied by the Saturation, the angle is the Hue multiplied by 360 (plus 90°, as angles start always at 3 o'clock); then we can translate that line at the center of the circle and the cursor will be positioned at the second point of the segment:
def paintEvent(self, event):
# ...
p.setPen(Qt.black)
p.setBrush(self.selected_color)
line = QLineF.fromPolar(self.radius * self.s, 360 * self.h + 90)
line.translate(self.square.center())
p.drawEllipse(line.p2(), 10, 10)
The map color function can use the same logic, but inverted: we construct a line starting from the center to the mouse cursor position, then the Saturation is the length divided by the radius (sanitizing the value to 1.0, as that's the maximum possible value), while the Hue is the line angle (minus 90° as above) divided by 360 and sanitized for a positive 0.0-1.0 range.
def map_color(self, x: int, y: int) -> QColor:
line = QLineF(QPointF(self.rect().center()), QPointF(x, y))
s = min(1.0, line.length() / self.radius)
h = (line.angle() - 90) / 360 % 1.
return h, s, self.v

How to translate points on image after cropping it and resizing it?

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

Find coordinates of isosceles triangle with maximum area bounded by ellipse

"Redirected" here from math overflow:
https://mathoverflow.net/questions/372704/find-coordinates-of-isosceles-triangle-with-maximum-area-bounded-by-ellipse
I have a window with an ellipse inscribed inside it. The ellipses radii are screen_width / 2 and screen_height / 2. I want to find the coordinates of the maximum isosceles triangle that will fit in the ellipse without overflowing.
The direction of the triangle's tip is a enum parameter (i.e., N, E, S, W). From what I've read, there is not a unique solution, but the maximum area is a simple formula and there is a way to find a triangle that solves the problem. That way, however, is merely hinted at, and probably involves using linear algebra to normalize the eclipse and isosceles triangle to a unit circle and an equilateral triangle, but no such formula seems to exist online.
An equilateral triangle inscribed in a circle is the triangle that covers the max area of the circle (some theorem that you should look up).
An ellipse is a "squished" circle, therefore, if we squish a circle with an inscribed equilateral triangle, providing we do that along a line of symmetry, we end up with a max area isosceles triangle (two sides get resized by a common factor, the 3rd side gets stretched by another factor).
The angles follow the inscribed angle theorem and complementary angle theorem
Considering your screen is wider than it is high, the coordinates of the 3 apex of the triangle are as follows (in screen coordinates, with the origin at top left)
top: (w/2, 0) # this one does not change
bot_left = (w/2 - w*cos(pi/6)/2, h/2 + h*sin(pi/6)/2)
bot_right = (w/2 + w*cos(pi/6)/2, h/2 + h*sin(pi/6)/2)
Adding to #Reblochon's answer, Here is a complete example. I attempted it, so why not share it :)
import pygame
from math import sin, cos, pi
pygame.init()
SW = 600
SH = 600
WIN = pygame.display
D = WIN.set_mode((SW, SH))
radiiX = SW/2
radiiY = SH/2
def ellipse(center, rx, ry):
global gotPositions
angle = 0
while angle < 6.28:
angle += 0.0005
pygame.draw.circle(D, (255, 255, 0), (int(center[0]), int(center[1])), 2)
x = center[0] + sin(angle)* radiiX
y = center[1] + cos(angle)* radiiY
D.set_at((int(x), int(y)), (255, 255, 0))
top= (SW/2, 0) # this one does not change
bot_left = (SW/2 - SW*cos(pi/6)/2, SH/2 + SH*sin(pi/6)/2)
bot_right = (SW/2 + SW*cos(pi/6)/2, SH/2 + SH*sin(pi/6)/2)
points = [top, bot_left, bot_right]
while True:
D.fill((0, 0, 0))
events = pygame.event.get()
for event in events:
if event.type == pygame.QUIT:
pygame.quit()
ellipse([radiiX, radiiY], radiiX, radiiY)
pygame.draw.lines(D, (255, 255, 0), True, points)
pygame.display.flip()
based on notes from Reblochon Masque
def inner_rect (self):
rect = self.outer_rect () # bounding box of ellipse
x, y, w, h = rect
r = self.child.orientation.radians () # direction of triangle
pts = inscribe_polygon (3, r)
pts = graphics_affines (pts) # from cartesian
pts = scale_points (pts, rect) # scale points to ellipse dims
o, r = bounding_rect (pts)
xmin, ymin = o
dx, dy = r
return (xmin, ymin, dx, dy)

Finding the characteristics of a hand written Arrows with opencv

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:

Drawing circles in python pixel by pixel

Hey guys so I have a question. I would like to draw a circle in python pixel by pixel but I can't seem to get it.
It was a question we talked about in my programming class today and my teacher said it was possible by using the distance formula to find the distance between the center of the window and the points on the outside of the circle and then just telling the program to fill it in if it fits that case.
So this is not homework, purely for my own curiosity. I have been working on it for hours and I feel like I am close but definitely missing a key element.
My goal is to put a red circle in the middle of the window, which has a width and height between 300 and 400.
I said that the center point is = to (w/2, h/2) and I want to fill in every point that is a fixed distance away from that.
I chose (100, 50) as a random point that would be on the circle and then set the variable distance = to the distance formula using those previous two points.
The two for-loops are there to iterate over every pixel in the window. I was able to create a box right where I want my circle... But I can't seem to get anything close to a circle to appear
here is my code, I named my function Japanese flag as it seemed appropriate:
import cImage
import math
def japanese_flag(w, h):
newimg = EmptyImage(w, h)
distance = int(math.sqrt((50 - h / 2) ** 2 + (100 - w / 2) ** 2))
for col in range(w):
for row in range(h):
if h/2 - distance < row < (h/2 + distance) and w / 2 - distance < col < (w / 2 + distance):
newpixel = Pixel(255, 0, 0) # this denotes the color red
newimg.setPixel(col, row, newpixel)
return newimg
def show_one_flag(func_name):
import random
width = random.randint(300, 400)
height = random.randint(200, 300)
win = ImageWin(str(func_name),width,height)
newimg = func_name(width, height)
newimg.draw(win)
show_one_flag(japanese_flag)
Use squared radius for simpler comparison
SqDist = (50 - h / 2) ** 2 + (100 - w / 2) ** 2
To draw only circumference, compare with some tolerance level
Eps = 1
...
if (Math.Abs((h/2 - row) ** 2 + (w/2 - col) ** 2 - SqDist) < Eps)
draw pixel
To draw filled circle:
if ((h/2 - row) ** 2 + (w/2 - col) ** 2 < SqDist)
draw pixel

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