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APPENDIX
Proof of Proposition 1
Proof. The derivation is started from formula (4).
First, the means (moments of order one), that are
present under powers, should be multiplied by suit-
able unity terms: µ
1,0
·
·
x
2
−x
1
x
2
−x
1
, µ
0,1
·
·
y
2
−y
1
y
2
−y
1
. This allows
to extract the denominators and form the normaliz-
ing constant 1/(D(x
2
− x
1
)
p
(y
2
− y
1
)
q
) in front of the
summation. Then, the powers are expanded by means
of the binomial theorem, grouping the terms into the
ones dependent on the current pixel index (x,y) and
the ones independent of it. This yields:
1
D(x
2
− x
1
)
p
(y
2
− y
1
)
q
∑
x
1
6x6x
2
∑
y
1
6y6y
2
p
∑
l=0
p
l
x
l
−x
1
− µ
1,0
x
1
,y
1
x
2
,y
2
(x
2
− x
1
)
p−l
·
q
∑
m=0
q
m
y
m
−y
1
− µ
0,1
x
1
,y
1
x
2
,y
2
(y
2
− y
1
)
q−m
· i(x,y)
. (17)
Finally, by changing the order of summations one ar-
rives at:
1
D(x
2
− x
1
)
p
(y
2
− y
1
)
q
·
p
∑
l=0
p
l
−x
1
− µ
1,0
x
1
,y
1
x
2
,y
2
(x
2
− x
1
)
p−l
·
q
∑
m=0
q
m
−y
1
− µ
0,1
x
1
,y
1
x
2
,y
2
(y
2
− y
1
)
q−m
·
∑
x
1
6x6x
2
∑
y
1
6y6y
2
x
l
y
m
i(x,y)
|
{z }
∆
x
1
,y
1
x
2
,y
2
(ii
l,m
)
. (18)
The underbrace indicates how the expensive summa-
tion over all pixels in the rectangle gets replaced by
the cheap constant-time computation of the growth of
a suitable integral image. Note also that the required
normalizer D is calculated by the growth of the zeroth
order integral image D = ∆
x
1
,y
1
x
2
,y
2
(ii
0,0
).