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Cryptanalysis of RSA Factorization - Library(ISI Kolkata) - Indian ...

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Chapter 7: Approximate Integer Common Divisor Problem 134<br />

Let X 2 = X 3 = ··· = X k = X be the common upper bound on each co-ordinate<br />

<strong>of</strong> the root (˜x 2 ,...,˜x k ). The shift polynomials from Equation (7.20) contribute<br />

P ′ 1 =<br />

m∏<br />

(X r a m−r<br />

r=0<br />

1 ) (k+r−2<br />

r ) = X<br />

η 4<br />

a η 5<br />

with η 4 = ∑ m<br />

r=0 r( )<br />

k+r−2<br />

r , η5 = ∑ m<br />

r=0 (m − r)( )<br />

k+r−2<br />

r , to the determinant <strong>of</strong> L ′ .<br />

(Note that this P 1 ′ is same as P 1 in Corollary 7.9). The shift polynomials from<br />

Equation (7.21) contribute<br />

P ′ 2 =<br />

t∏<br />

i 2 =1<br />

(X i 2<br />

X m ) (k+m−2 m ) = X<br />

η 6<br />

with η 6 = ∑ t<br />

i 2 =1 (i 2+m) ( )<br />

k+m−2<br />

m , to the determinant <strong>of</strong> L ′ . The dimension <strong>of</strong> L ′ is<br />

ω ′ =<br />

m∑<br />

( ) ( )<br />

k +r −2 m+k −2<br />

+t .<br />

r m<br />

r=0<br />

Now, we have ( )<br />

k+r−2<br />

r =<br />

r k−2<br />

(k−2)! +o(rk−2 ). Using Lemma 4.1 and neglecting lower<br />

order terms, we obtain<br />

P ′ 1 ≈ X ∑ m<br />

r=0 r rk−2<br />

∑ m rk−2<br />

r=0 (m−r)<br />

(k−2)! a<br />

(k−2)!<br />

1 ≈ X 1<br />

1<br />

m k 1 m k<br />

(k−2)! k a<br />

(k−2)! k−1 − 1 m k<br />

(k−2)! k<br />

1 ,<br />

P 2 ′ ≈ X ∑ t<br />

i 2 =1 (i 2+m) mk−2<br />

(k−2)!<br />

≈ X 1<br />

(k−2)! (t2 m k−2 +tm k−1) 2<br />

, and<br />

m∑<br />

ω ′ ≈<br />

r=0<br />

r k−2 mk−2<br />

+t<br />

(k −2)! (k −2)!<br />

Following Theorem 2.23, the required condition is<br />

≈<br />

det(L ′ ) = P ′ 1P ′ 2 < g mω′ ,<br />

m k−1 mk−2<br />

+t<br />

(k −1)(k −2)! (k −2)!<br />

where g is the common divisor. Let X = a α+β . Then putting the values <strong>of</strong> g,X<br />

in det(L ′ ) = P ′ 1P ′ 2 < g mω′ , we get,<br />

( m<br />

k<br />

k + mk−2 t 2<br />

2<br />

+m k−1 t<br />

)(α+β)+ mk<br />

k −1 − mk<br />

k<br />

< (1−α)<br />

(m k−1 t+ mk<br />

k −1<br />

)<br />

. (7.22)

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