enter image description here


A Pixel World is a PNG image where gravitational, electromagnetic, and nuclear forces no longer exist. Special forces known as "pixel forces" are all that remain. We define this force as

Fp->q = a * (p * q) / (r * r) * r̂

  • F is the pixel force that p exerts on q
  • a is Adams' constant, defined as a = 4.2 * 10 ^ 1
  • p and q are charges on two pixels
  • r is the distance from p to q
  • is the direction of r in radians, measured counterclockwise* from the positive x-axis

*There are infinitely many acceptable values for any given direction. For example, 6.28, 0, -6.28, and -12.57 radians are all equivalent, and would all be accepted.

Three types of pixels exist:

  • A red pixel holds a charge of positive one
  • A black pixel holds a charge of negative one
  • A white pixel holds a charge of zero

The bottom left pixel of pixel world is located at (0, 0). The positive y axis falls up the page, and the positive x axis falls to the right. The total force on a pixel is simply the vector sum of all forces exerted upon that pixel. Like charges repel, and opposite charges attract.

Given a file path to a Pixel World, as well as two integers x and y, output the total force exerted on the pixel at location (x, y) in the format <magnitude>\n<direction>. Output must be accurate to at least 2 decimal places, but you may output more if you want. Direction must be output in radians. The integers x and y are guaranteed to be within the world boundary.


You must include a program, and a command that can be used to run your program. Two examples:

python AbsolutelyPositive.py "C:\Pixtona.png" 50 40
java UltimateAttraction "C:\Jupix.png" 30 30

Example Image

In the image below, there is a black pixel at (100, 104).


enter image description here

Example Output

This output does not correspond to the above input.


Need help getting started?

I wrote some example code here. I haven't tested it, so use it at your own risk.

  • \$\begingroup\$ Can it be PPM instead of PNG? and typically, images put 0,0 in top left corner, do we have to rotate it? \$\endgroup\$
    – user137
    Commented Aug 29, 2014 at 15:27
  • \$\begingroup\$ @user137 Everyone has to handle the PNG image format. I'm considering posting some Java code to get people started. I don't want to start adding optional requirements. The bottom left was selected as (0,0) because this allows us to use a right hand coordinate system and measure radians counterclockwise from the positive x-axis. If we place (0,0) in the top left, and radians would be measured clockwise from the positive x-axis. No matter which way you do it, some standard will unfortunately be broken. Couldn't you just flip the image anyway? \$\endgroup\$
    – Rainbolt
    Commented Aug 29, 2014 at 15:33
  • 2
    \$\begingroup\$ I love the intro picture... \$\endgroup\$ Commented Aug 29, 2014 at 17:28
  • \$\begingroup\$ I think the choice for PNG will really hold people back, except those with ready knowledge of Java — which, as far as I know, is the only language with something like ImageIO. C++, for one, doesn't have it. \$\endgroup\$
    – tomsmeding
    Commented Aug 30, 2014 at 9:40
  • \$\begingroup\$ Nice challenge. PNG is fine IMO. Btw, here are some questions I have while attempting it. 1. I'm not sure if is it me, but I can't seem to find the black pixel in (99,101) in the image. 2. Is the example output the answer for the test image? 3. For the direction, if the angle is < pi degrees clockwise from the +ve x-axis, is negative values (i.e. 0 < Angle < pi) allowed, or must it be (pi < Angle < 2*pi) ? \$\endgroup\$
    – Vectorized
    Commented Aug 30, 2014 at 15:47

1 Answer 1


Python - 355

import sys,Image,math as m
c=lambda i,j:2*(a[i,j][0]>200)-(a[i,j][1]>200)-1
for j in range(h):
 for i in range(w):q=c(i,h-1-j);y=Y-j;x=X-i;r=x*x+y*y;f=r and 42.*p*q/r;V+=m.cos(t(x,y))*f;H+=m.sin(t(x,y))*f


import sys
import Image
import math as m

X,Y=map(int,sys.argv[2:]) # The X and Y coordinates of the test pixel

i=Image.open(sys.argv[1]) # Open the image
w,h=i.size # Get the width and height of the image
a=i.load() # Get a pixel access object of the image
V=0 # V = vector sum of Vertical Forces
H=0 # H = vector sum of Horizontal Forces

# Function to calculate the charge of the a pixel at x=i, y=j
def c(i,j):
    global a
    if a[i,j][0]>200: # If Red > 200
        if a[i,j][2]>200: # If Green > 200
            return 0 # We assume that pixel is White
            return 1 # We assume that pixel is Red
    return -1 # Else, we asusme that pixel is Black

p=c(X,h-1-Y) # Assign the charge of the test pixel to p

for j in range(h): # For every y value...
    for i in range(w): # For every x value...
        q=c(i,h-1-j) # Assign the charge of the current pixel to q
        y=Y-j # The y distance of the test pixel from the current pixel
        x=X-i # The x distance of the test pixel from the current pixel
        rSquared=x*x+y*y # The r-squared distance between the 2 pixels
        f=rSquared and 42.*p*q/rSquared # If rSquared is > 0, calculate the force. 
                                        # Otherwise, the force is zero
        V+=m.cos(m.atan2(x,y))*f # Add the Y component of the force to V
        H+=m.sin(m.atan2(x,y))*f # Add the X component of the force to H


Some little program used to create tests:

import Image
width  = 100 # Define your own image width
height = 100 # Define your own image height
image = Image.new("RGB", (width, height), "white")
pixels = image.load()
blacks = [(0,0), (1,1)] # Define your own black pixels
reds   = [(0,1), (1,0)] # Define your own red pixels
for x,y in blacks:
    pixels[x,height-1-y] = (0,0,0)
for x,y in reds:
    pixels[x,height-1-y] = (255,0,0)
image.save("y.png") # Save image

Sample tests:

enter image description here

Users-MacBook-Air:pngforces User$ python z3.py y.png 0 0
59.3969696197 0.785398163397

enter image description here

Users-MacBook-Air:pngforces User$ python z3.py y.png 50 50
0.0084 3.92699081699

enter image description here

Users-MacBook-Air:pngforces User$ python z3.py y.png 50 50
0.0084 0.785398163397

Just posting first... If there are any mistakes or issues, drop a comment down below, but will probably take quite some time to respond cuz I am really busy with other stuff.

  • 2
    \$\begingroup\$ I love the test program! It's a lot better than zooming in 400% on a little image, making dots by hand, then trying to record the locations of all the little dots. You must be a programmer! \$\endgroup\$
    – Rainbolt
    Commented Aug 30, 2014 at 18:24

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