Visual Cryptography and Contrast Bounds - Visual Cryptography and Secret Image Sharing - page 244

or equivalently

2 3

X X X

2 jT 0 j1 2 jT 0 j1jTj

4 5 =

T 0

ST ( T 0
jSjjTj mod 2

;6=T 0 f1;:::;ng STf1;:::;ng

jSjjTj mod 2

2 3

5 + T 0 (1) jT 0 j+jSj 1
2

X X X

4

2 jT 0 j1 2 jT 0 j1jTj T 0 :

;6=T 0 f1;:::;ng ST ( T 0

jSj6jTj mod 2

STf1;:::;ng
jSj6jTj mod 2

(Note that the last summand is equal to T 0 for jT 0 j6jSj mod 2 and equal
to 0 otherwise.)
Comparing coecients for each T 0 we obtain

X

2 njSj1 2 jT 0 j1 2 jT 0 j1jTj =

ST ( T 0
jSjjTj mod 2

0 1

X

(1) jT 0 j+jSj 1
2

2 njSj1 2 jT 0 j1

@

2 jT 0 j1jTj

A +

; (8.33)

ST ( T 0
jSj6jTj mod 2

but this is true since

(1) jT 0 jjSj = (1 2) jT 0 jjSj

jT 0 jjSj
i

jT 0 jjSj

X

(2) i =

i=0

jT 0 jjSj
i

jT 0 jjSj
i

jT 0 jjSj jT 0 jjSj

X X

2 i 2 i

=

i=0
i even

i=0
i odd

X X

2 jT 0 jjTj 2 jT 0 jjTj =

STT 0
jT 0 jjTj mod 2

STT 0
jT 0 j6jTj mod 2

jT 0 jjSj
i

X

, since = 1:

STT 0
jTjjSj=i

Thus,

0 1

X X

2 jT 0 jjTj =

@

2 jT 0 jjTj

A + (1) jT 0 jjSj

STT 0
jT 0 jjTj mod 2

STT 0
jT 0 j6jTj mod 2