August 2011
Intermediate to advanced
332 pages
9h 10m
English
Content preview from Lattice Basis Reduction
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The LLL Algorithm 65
Step (3) sets k = 2.
Step (4)(a) calls reduce(2,1); since |µ
21
| = 27/127 < 1/2 the procedure
performs no action. Step (4)(b) tests the exchange condition; we have
γ
∗
2
=
5621
127
<
15400
127
= (1 − µ
2
21
)γ
∗
1
.
Hence Step (4)(b)(iii) calls exchange(2); the current state (iteration 1) is
3 −2 6 −1
−2 7 7 −5
2 −8 −9 −7
8 −9 6 −4
,
1 0 0 0
27
50
1 0 0
−
1
2
−
3725
5621
1 0
41
25
−
3064
5621
350695
765183
1
,
50
5621
50
765183
5621
399424
765183
.
Since k = 2 we do not decrement k in Step (4)(b)(iv); we return to the top of
the loop with k = 2.
Step (4)(a) calls reduce(2,1); since |µ
21
| = 27 /50 > 1/2 and ⌈µ
21
⌋ = 1,
the pro c e dure reduces y
2
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