Pattern recognition

Answers

• Chrisir

  November 2014 edited November 2014

  I think you got very far indeed already...

  scale

  you scale the thing: when you measure the distance from the border to the 1st white pixel (measure from bottom and from top or from all 4 sides)

  then on the left image distances are say 322 in sum, in the right 98 in sum.

  then you need to scale the smaller by a certain value (maybe 322/98? Not sure)

  for rotating

  for rotating: when you have only 90° (or mirroring on itself) you could produce those ca. 3 images automatically and compare the result to the given image...

  such things are done by neural networks, but this is hard too I guess, although there's a chapter in Shiffmanns book on that topic.

  ;-)
• tlecoz

  November 2014 edited November 2014

  hello !

  1) define the Xmin, Ymin, Xmax, Ymax of all your points

  2) normalizes each values

  point.x = (point.x - Xmin) / (Xmax - Xmin);
  point.y = (point.y - Ymin) / (Ymax - Ymin);
  

  3) set a new position for your points, located from the center instead of using the topleft corner (then we will be able to rotate it), and rescale the values (multiply them by 20 for exemple, to get a 20x20 picture (then a more simple picture than original one)

  float newScale = 20.0:
  float halfScale = newScale / 2;    
  
  float dx = point.x - 0.5;
  float dy = point.y - 0.5;
  float d = sqrt(dx*dx+dy*dy) * halfScale ;
  float angle = atan2(dy,dx);
  point.x = halfScale  + cos(angle) * d;
  point.y = halfScale  + sin(angle) * d;
  

  4)create an array of array of int, rotate your first 20x20 picture 360 times (for each degree) and register inside the arrays the différent point position. (actually, I think 360 time is too much, 50 times with rotation incremented by 360/50 should be enough)

  5) take you second picture, for the comparison, grab its pixel-array

  6) loop on you array of rotated-int-array and compute "the distance" (how much pixels doesn't match on the 2 pictures) , and register that distance in a object located in an ArrayList (one object by pixel, then 20x20 objects)

    int i,j,nbPixel = 20*20,nbRotation=50;
    int[] rotatedShape;
    int d;
    for(i=0;i<nbRotation;i++){
       rotatedShape = rotations[i];
       d = 0:
       for(j=0;j<nbPixel;j++){
           if(rotatedShape[j] != secondShape[j]) d++
       }
       distanceList.get(i).distance = d;
    }
  

  7) use Collections.sort to sort your arrayList by distance

  Collections.sort(distanceList);

  8) then, if the less distance found is less than a certain value, you can consider that the two pictures match.

  if(distanceList.get(0).distance < distanceMin){
     println("YEAH")
  }
  

  Good luck !

  • don't know about Shiffmanns book, he probably acts in a different way -