A Fast Detection Method for Bottle Caps Surface Defect Based on Sparse Representation - Intelligent Computing for Sustainable Energy and Environment - page 80

where R i denotes the value of the bin which is obtained by projecting, i

∈

[0 ,N

−

1] is the number of the bin, which represents the position of the bin on
the horizontal axis, N is the diagonal length of the template image, θ [0 , 359]

is the value of the projection angle, which denotes the number of the histogram.
Furthermore, the defect dictionary D d will be proposed. In most cases, the
defect whose width or height is more than 5 pixels cannot be tolerated. Therefore,
a circular region with the 5-pixel diameter is regarded as minimum basic unit,
whose projection can be got as p i =[3 , 5 , 5 , 5 , 3] T ,where i is the descriptor of the
central location. Motivated by the work in [9], we propose a defect dictionary
D d that can be obtained by successive change the position of p i ,where i =
0 , 1 ,

∈

1. D d is similar as the 'trivial templates' in literature [9], which is
constituted as follows

··· N −

⎛ ⎞

p 0

0

⎝ ⎠

. . .

D d = (8)

p N− 1

0

Due to five nonzero terms adjacent in the atoms of D d , the residual less than
the width of 5 pixels is ignored.

4 Fast Algorithm

When the template dictionary D t and the defect dictionary D d are obtained,
the further task is to detect for sample bottle caps. Assume the vertical pro-
jection histogram and the horizontal projection histogram of the sample ROI
are denoted by X 1 and X 2 respectively. Using expression (6), the defect sparse
factors α d and α d can be obtained with respect to X 1 and X 2 respectively. The
non-zero elements in α d and α d corresponds to defects, see Fig.3. As shown in
Fig.3, because two defect sparse factors are orthogonal, the location and size of
the defect can be determined. In order to get α d and α d quickly, a novel method
which includes two steps is proposed. The detail process will be addressed as the
following example of α d . The way of solving α d is similar as that of α d .
Firstly, according to (6), α t should be solved. α t is a template sparse factor
with respect to X 1 . The element in the sparse factor α t , which corresponds
to the best match column in the template dictionary D t , is set to one, and
the other elements of the sparse factor α t are set to zero. Thus, the following

equation is obtained, α t 0 = α t 1 = 1. Without loss of generality, we assume
that the best match column is j th. Obviously, the following resolutions is hold,
α t = s 0 ··· 0 T , s∈ R .

Secondly, α d is obtained. Substituting α t 1 = 1 into (6), we have
α d =argmin α d 1 ,

s 359 T = 0

s j− 1 s j s j +1 ······ ··· 010 ······

st. X 1 − D d α t 2 D t α t − ε. (9)