While developing GEM (GEstures with Mouse) the biggest problem i faced is, "the user drawn gesture is most of the times smaller / bigger than already defined gesture" but they are almost same in shape. So, i used a logic to prove SIMILAR Objects (Different in size but with same shapes ) are EQUAL.
LOGIC:
Let us consider two similar circles A, B as shown in picture.
(x1, y1) = (6, 4)
(x2, y2) = (8, 6)
(x3, y3) = (6, 8)
(x4, y4) = (4, 6)
The co-ordinates of Circle B are
(X1, Y1) = (6, 3)
(X2, Y2) = (9, 6)
(X3, Y3) = (6, 9)
(X4, Y4) = (3, 6)
now we re-calculate the above co-ordinates using following formulas
X = [X - [Xmax+Xmin]/2] / [Xmax-Xmin]+ 0.5 ---------(1)
where Xmin = min(x1,x2,x3,x4) for circle A
Xmax = max(x1,x2,x3,x4) for circle A
X = x1, x2, x3, x4 for circle A
Xmin = min(X1,X2,X3,X4) for circle B
Xmax = max(X1,X2,X3,X4) for circle B
X = X1, X2, X3, X4 for circle B
Y = [Y - [Ymax+Ymin]/2] / [Ymax-Ymin]+ 0.5 ------------(2)
where Ymin = min(y1,y2,y3,y4) for circle A
Ymax = max(y1,y2,y3,y4) for circle A
Y = y1, y2, y3, y4 for circle A
Ymin = min(Y1,Y2,Y3,Y4) for circle B
Ymax = max(Y1,Y2,Y3,Y4) for circle B
Y = Y1, Y2, Y3, Y4 for circle B
Re-Calculating Circle A points:
Xmin = min(6, 8, 4, 6) = 4
Xmax = max(6, 8, 4, 6) = 8
Ymin = min(4, 6, 6, 8) = 4
Ymax = max(4, 6, 6, 8) = 8
new x1 = (6 - (8+4) / 2)/4 + 0.5 = 0.5
new y1 = (4 - (8+4) / 2)/4 + 0.5 = 0.0
in the same manner
new x2 = 1.0
new y2 = 0.5
new x3 = 0.5
new y3 = 1.0
new x4 = 0.0
new y4 = 0.5
Re-Calculating Circle B points:
Xmin = min(6, 9, 6, 3) =3
Xmax = max(6, 9, 6, 3) =9
Ymin = min(3, 6, 9, 6) =3
Ymax = max(3, 6, 9, 6) =9
new X1 = (6 - (9+3)/2)/(9-3) + 0.5 = 0.5
new Y1 = (3 - (9+3)/2)/(9-3) + 0.5 = 0.0
in the same manner
new X2 = 1.0
new Y2 = 0.5
new X3 = 0.5
new Y3 = 1.0
new X4 = 0.0
new Y4 = 0.5
After Re-calculating Circle A & B points using Formula (1) and (2)
x1 = X1; x2 = X2; x3 = X3; x4 = X4; and
y1 = Y1; y2 = Y2; y3 = Y3; y4 = Y4;
Hence, SIMILAR Circles A and B are EQUAL
Note: Not only circles using this formulas any two similar drawing shapes can be proved to equal.
