August 2011
Intermediate to advanced
332 pages
9h 10m
English
Content preview from Lattice Basis Reduction
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Deep Insertions 97
Using equation (5.6) we obtain
z
∗
k−1
=
1
z
∗
k−2
·z
∗
k−2
×
|y
∗
k
|
2
+ µ
2
k,k−1
|y
∗
k−1
|
2
y
∗
k−2
− µ
k,k−2
|y
∗
k−2
|
2
y
∗
k
+ µ
k,k−1
y
∗
k−1
!
.
Another application of equation (5.6) gives the formula we want e xpressing
z
∗
k−1
in terms of the y
∗
i
:
z
∗
k−1
=
1
|y
∗
k
|
2
+ µ
2
k,k−1
|y
∗
k−1
|
2
+ µ
2
k,k−2
|y
∗
k−2
|
2
×
|y
∗
k
|
2
+ µ
2
k,k−1
|y
∗
k−1
|
2
y
∗
k−2
−µ
k,k−2
|y
∗
k−2
|
2
y
∗
k
+ µ
k,k−1
y
∗
k−1
.
(5.9)
From this we can easily write down the formula for |z
∗
k−1
|
2
. Equation (5.8)
also shows that for i ≤ k−2 we have
ν
k−1,i
=
z
k−1
·z
∗
i
z
∗
i
· z
∗
i
=
y
k−2
· z
∗
i
z
∗
i
·z
∗
i
=
y
k−2
·y
∗
i
y
∗
i
· y
∗
i
(i < k−2)
y
k−2
·z
∗
k−2
z
∗
k−2
·z
∗
k−2
(i = k−2)
Using equations (5.6) and (5.7) we obtain
ν
k−1,i
=
µ
k−2,i
(i < k−
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