What is that angle?

Question

The goal of this challenge is to determine the angle of a line in a image.

Rules on the image:

• The image background will be white (#FFFFFF)
• The stroke of the line will be black (#000000)
• The line will NOT be anti-aliased
• The image will be 100x100 pixels
• The line will start at the center of the image
• The line will start pointing down (6-OClock)
• The line will be 50 pixels long
• The angle of the line will be measured going counterclockwise from the starting position
• The image codec will be either .jpg or .png

Input format will be a file name passed by the command line arg, script input, or function arg. Output format is simple - just output the number of degrees (e.g. 90).

Answers can be ±1 degree of the stated measure. Here are a few example images:

A reference image at 45 degrees with gray background

0 degrees

45 degrees

50 degrees

130 degrees

230 degrees

324 degrees

Here is the code used to create the images (this is coded with Processing):

int deg = 45;

int centX = width/2, centY = height/2;

background(255);
noSmooth();
line(centX,
     centY,
     centX + sin(radians(deg))*50,
     centY + cos(radians(deg))*50);

saveFrame("line-"+deg+".png");// image codec can be changed here. use '.png' or '.jpg'

Answer 1

JavaScript (ES6), 225 227 244 bytes

Let's get the ball rolling:

f=s=>{(i=new Image).src=s;c=document.createElement`canvas`.getContext`2d`;c.drawImage(i,0,0,100,100);for(a=360;a--,r=a/180*(m=Math).PI;)if(!c.getImageData(50+48*m.cos(r),50+48*m.sin(r),1,1).data[1]){alert((450-a)%360);break}}

Simply pass the URL of the image to the function:

f('90deg.png');

Alerts degrees within the ±1 range. Passed all test cases.

Ungolfed

f=s=>{
    // create new image and set source
    (i=new Image).src=s;
    // create canvas and get context
    c=document.createElement`canvas`.getContext`2d`;
    // set width/height to 100px and draw image on canvas
    c.drawImage(i,0,0,100,100);
    // check whether for any degree on the theoretical circle a black pixel is found
    for(a=360;a--,r=a/180*(m=Math).PI;)
        if(!c.getImageData(50+48*m.cos(r),50+48*m.sin(r),1,1).data[1]){
            // wait, it should be ccw and the board is rotated 90 degrees
            alert((450-a)%360);
            break
        }
}

Edits

• Saved 17 bytes – figured I don't need to set the width and height of the canvas element.
• Saved 2 bytes by negating the condition.

Answer 2

Pyth - 28 26 bytes

Uses same sort of brute force strategy as the js answer.

f!@F+]'zm+50s*48.t.tT7d_U2

Takes input as filename from stdin.

f                     Filters from 1 till predicate is matched
 !                    Boolean not so that only pixel with zero value matched
  @F+]                Folds by indexing to get pixel value  
   'z                 Reads image filename input
   m         _U2      Maps over both trig ratios
    +50               Adds 50 to pixel value
     *48              Multiplies pixel value by 48
      .t    d         Takes trig ratio with appropriate option
        .t 7          Degrees to radians
          T           Filter var

Answer 3

Matlab, 118 104 bytes

I generate a matrix of the same size as the image with complex numbers (0 in the center) and exctract from that matrix the values which are on the line. The argument of the mean of those is then displayed.

Thanks to @ThomasKwa for suggesting an improvement in accuracy which also resulted in shorter code!!!

I=imread(input('','s'));
[y,x]=ndgrid(-50:49);
c=y+i*x;
disp(mod(angle(mean(c(~I(:,:,1))))*180/pi+360,360))

Answer 4

Matlab, 86 77 bytes

Here's another way using Matlab:

[I,J]=find(~im2bw(imread(input('','s'))));mode(mod(round(atan2d(J-51,I-51)),360))

This reads the file (stolen from flawr), and finds the indices of the black pixels. Then, it works out the vector that points from the centre of the image to each black pixel, and uses atan2d to find the angle, rounding to get integer angles, and doing mod(...,360) to get results in the right range. To get the correct angle (there is a bit of error for the pixels close to the centre), take the most commonly calculated angle.

Thanks to slvrbld for the im2bw suggestion!

Answer 5

Labview, 10098 Bytes

Let's put another labview code out there.

Since there is no official way to count bytes in labview i use the size of the file when saved. Alternatively counting every wire and function as 1and the case as 2 it would come out to 71.

Load image, flatten to 1D, scan for 0s from both sides and take the first, calc back to point and use geometry to get angle.