1
CSE20 Lecture 6: Number Systems
5. Residual Numbers (cont) &
6. Cryptography
Docsity.com
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Guidelines and tips
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Cryptography - Discrete Mathematics - Lecture Slides, Slides of Discrete Mathematics
These lecture slides are very helpful for the student of discrete mathematics. The major points in these exam paper are: Cryptography, Residual Numbers, Number Systems, Inverse Conversion, Chinese Remainder Theorem, Corresponding Number, Number Theory, Public Key for Coding, Secret Key to Decode, Knowledge of Function
Typology: Slides
2012/2013
1 / 15
This page cannot be seen from the preview
Don't miss anything!
Related documents
Partial preview of the text
Download Cryptography - Discrete Mathematics - Lecture Slides and more Slides Discrete Mathematics in PDF only on Docsity!
1
CSE20 Lecture 6: Number Systems
5. Residual Numbers (cont) &
6. Cryptography
Residual Numbers
• Introduction
• Definition
• Operations
• Inverse Conversion
2
Chinese Remainder Theorem
Given a residual number (r 1 , r 2 , …, r k ) with moduli (m 1 , m 2 , …, m k ), where all m i are mutually prime, set M= m 1 ×m 2 × …×m k , and M i =M/m i
- Find S i that (M i
×S
i )%m i
= 1 (S
i an inverse of M i in mod m i
- The corresponding number x = (∑ i=1,k
(M
i
S
i r i
))%M.
4
Example
Given (m 1 ,m 2 ,m 3 )=(2,3,7), M=2× 3 ×7=42, we have M 1 =m 2 ×m 3
=3×7=21 (M
1
S
1 )%m 1
=(21S
1
M
2 =m 1 ×m 3
=2×7=14 (M
2
S
2 )%m 2
=(14S
2
M
3 =m 1 ×m 2
=2×3=6 (M
3
S
3 )%m 3
=(6S
3
Thus, (S 1
, S
2
, S
3
For a residual number (0,2,1): x=(M 1
S
1 r 1
+ M
2
S
2 r 2
+ M
3
S
3 r 3
)%M
=(21× 1 × 0 + 14× 2 × 2 + 6× 6 × 1 )%
5
Example: iClicker
Given (m 1 ,m 2 ,m 3 )=(2,3,5), M=2× 3 ×5=30, we have M 1 =m 2 ×m 3
=3×5=15 (M
1
S
1 )%m 1
=(15S
1
M
2 =m 1 ×m 3
=2×5=10 (M
2
S
2 )%m 2
=(10S
2
M
3 =m 1 ×m 2
=2×3=6 (M
3
S
3 )%m 3
=(6S
3
Thus, (S 1
, S
2
, S
3 ) is A. (1, 1, 1) B. (1, 2, 1) C. (2, 1, 2) D. None of the above 7
Example: iClicker
Given (m 1 ,m 2 ,m 3 )=(2,3,5), M=2× 3 ×5=30, we have M 1 =m 2 ×m 3
=3×5=15 (M
1
S
1 )%m 1
=(15S
1
M
2 =m 1 ×m 3
=2×5=10 (M
2
S
2 )%m 2
=(10S
2
M
3 =m 1 ×m 2
=2×3=6 (M
3
S
3 )%m 3
=(6S
3
For a residual number (x 1 ,x 2 ,x 3 )=(1,2,3), the corresponding number x is A. 5 B. 19 C. 23 D. None of the above 8
6. Cryptography
1. Introduction
2. RSA Protocol
3. Remarks
10
6.1 Cryptography: Introduction
- Application of residual number systems
- Number theory (skip)
- Show the basic concept and process
- Many variations 11
6.2 RSA Protocol
- N=pq where p & q are primes and kept secret.
- e is mutually prime to f(N)=(p-1)(q-1)
- d is the inverse of e mod f(N), i.e. (ed)%f(N)= Theorem: S(P(M))=P(S(M))=M for 0<=M<N Note that S(P(M))=M ed %N Theorem: M f(N) %N=1 for 0<=M<N Assumption: p & q are hard to find. Consequently, it is difficult to derive d. 13 P(X)=X e %N (^) S(X)=Xd%N M (^) M (e,N) (d,N)
6.2 RSA Protocol
- N=pq where p & q are primes and kept secret.
- e is mutually prime to f(N)=(p-1)(q-1)
- d is the inverse of e mod f(N), i.e. (ed)%f(N)= Example: N=pq=3x11=33, f(N)=(3-1)(11-1)= Let e=3, then d=7 (3x7%20=1). M=9 => P(9)= 3 %33=3 => S(3)= 7 %33=? 14 P(X)=X e %N (^) S(X)=Xd%N M (^) M (e,N) (d,N)