chinese remainder theorem crt chinese remainder theorem
play

Chinese Remainder Theorem (CRT) Chinese Remainder Theorem (CRT) n r - PowerPoint PPT Presentation

Aug 22, 2023 •458 likes •1.41k views

Chinese Remainder Theorem (CRT) Chinese Remainder Theorem (CRT) n r 1 (mod m 1 ) gcd(m i , m j ) = 1 r (mod m ) r 2 (mod m 2 ) r k (mod m k ) 1 Chinese Remainder

  1. Chinese Remainder Theorem (CRT) Chinese Remainder Theorem (CRT) n  r 1 (mod m 1 ) gcd(m i , m j ) = 1   r (mod m )  r 2 (mod m 2 )  • • •  r k (mod m k )          1

  2. Chinese Remainder Theorem (CRT) Chinese Remainder Theorem (CRT) n  r 1 (mod m 1 ) gcd(m i , m j ) = 1  solution:  r (mod m )  r 2 (mod m 2 ) m = m 1 m 2 ꞏ ꞏ ꞏ m k • • •  r k (mod m k ) z i = m / m i -1  1 (mod m i ) (since gcd(z i , m i ) = 1) *  z i -1  Z mi s.t. z i ꞏ z i k k n   z i ꞏ z i -1 ꞏ r i (mod m) i=1        1

  3. Chinese Remainder Theorem (CRT) Chinese Remainder Theorem (CRT) n  r 1 (mod m 1 ) gcd(m i , m j ) = 1  solution:  r (mod m )  r 2 (mod m 2 ) m = m 1 m 2 ꞏ ꞏ ꞏ m k • • •  r k (mod m k ) z i = m / m i -1  1 (mod m i ) (since gcd(z i , m i ) = 1) *  z i -1  Z mi s.t. z i ꞏ z i k k n   z i ꞏ z i -1 ꞏ r i (mod m) i=1  ex: r 1 =1, r 2 =2, r 3 =3 1 2 3  m 1 =3, m 2 =5, m 3 =7 m = 3 ꞏ 5 ꞏ 7     1

  4. Chinese Remainder Theorem (CRT) Chinese Remainder Theorem (CRT) n  r 1 (mod m 1 ) gcd(m i , m j ) = 1  solution:  r (mod m )  r 2 (mod m 2 ) m = m 1 m 2 ꞏ ꞏ ꞏ m k • • •  r k (mod m k ) z i = m / m i -1  1 (mod m i ) (since gcd(z i , m i ) = 1) *  z i -1  Z mi s.t. z i ꞏ z i k k n   z i ꞏ z i -1 ꞏ r i (mod m) i=1  ex: r 1 =1, r 2 =2, r 3 =3 1 2 3  m 1 =3, m 2 =5, m 3 =7 m = 3 ꞏ 5 ꞏ 7 z 1 =35, z 2 =21, z 3 =15    1

  5. Chinese Remainder Theorem (CRT) Chinese Remainder Theorem (CRT) n  r 1 (mod m 1 ) gcd(m i , m j ) = 1  solution:  r (mod m )  r 2 (mod m 2 ) m = m 1 m 2 ꞏ ꞏ ꞏ m k • • •  r k (mod m k ) z i = m / m i -1  1 (mod m i ) (since gcd(z i , m i ) = 1) *  z i -1  Z mi s.t. z i ꞏ z i k k n   z i ꞏ z i -1 ꞏ r i (mod m) i=1  ex: r 1 =1, r 2 =2, r 3 =3 1 2 3  m 1 =3, m 2 =5, m 3 =7 m = 3 ꞏ 5 ꞏ 7 z 1 =35, z 2 =21, z 3 =15 35 ꞏ 2 + 3 (-23) = 1 35 2 + 3 ( 23) 1 z -1 =2 z -1 =1 z -1 =1 z 1 2, z 2 1, z 3 1  1

  6. Chinese Remainder Theorem (CRT) Chinese Remainder Theorem (CRT) n  r 1 (mod m 1 ) gcd(m i , m j ) = 1  solution:  r (mod m )  r 2 (mod m 2 ) m = m 1 m 2 ꞏ ꞏ ꞏ m k • • •  r k (mod m k ) z i = m / m i -1  1 (mod m i ) (since gcd(z i , m i ) = 1) *  z i -1  Z mi s.t. z i ꞏ z i k k n   z i ꞏ z i -1 ꞏ r i (mod m) i=1  ex: r 1 =1, r 2 =2, r 3 =3 1 2 3  m 1 =3, m 2 =5, m 3 =7 m = 3 ꞏ 5 ꞏ 7 z 1 =35, z 2 =21, z 3 =15 z -1 =2 z -1 =1 z -1 =1 z 1 2, z 2 1, z 3 1 n  35ꞏ2ꞏ1 + 21ꞏ1ꞏ2 + 15ꞏ1ꞏ3  157  52 (mod 105) 1

  7. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 ) 2

  8. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 2

  9. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2 2

  10. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2 (mod m 2 ) 2

  11. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2 (mod m 2 ) (mod m 1 ) 2

  12. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2  n  r 1 (m 2 m 2 -1 ) + r 2 (m 1 m 1 -1 ) (mod m 1 m 2 ) 2

  13. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2  n  r 1 (m 2 m 2 -1 ) + r 2 (m 1 m 1 -1 ) (mod m 1 m 2 ) 2

  14. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2  n  r 1 (m 2 m 2 -1 ) + r 2 (m 1 m 1 -1 ) (mod m 1 m 2 ) Verification 2

  15. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2  n  r 1 (m 2 m 2 -1 ) + r 2 (m 1 m 1 -1 ) (mod m 1 m 2 ) n mod m 1 = 1 Verification 2

  16. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2  n  r 1 (m 2 m 2 -1 ) + r 2 (m 1 m 1 -1 ) (mod m 1 m 2 ) n mod m 1 = 1 Verification n mod m 2 = 2

  17. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2 mod m 1  n  r 1 (m 2 m 2 -1 ) + r 2 (m 1 m 1 -1 ) (mod m 1 m 2 ) n mod m 1 = r 1 1 1 Verification 2

  18. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2  n  r 1 (m 2 m 2 -1 ) + r 2 (m 1 m 1 -1 ) (mod m 1 m 2 ) n mod m 1 = r 1 0 1 1 Verification + 2

  19. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2 mod m 2  n  r 1 (m 2 m 2 -1 ) + r 2 (m 1 m 1 -1 ) (mod m 1 m 2 ) n mod m 1 = r 1 0 1 1 Verification + n mod m 2 = r 2 2

  20. CRT, gcd(m 1 , m 2 )=1 , g ( 1 , 2 )  n  r 1 (mod m 1 )  n r 1 (mod m 1 ) gcd(m 1 , m 2 ) 1 gcd(m 1 , m 2 ) = 1  r 2 (mod m 2 )   s, t such that m 1 s + m 2 t = 1 -1 + m 2 m 2 -1 = 1 i.e. m 1 m 1 1 1 2 2  n  r 1 (m 2 m 2 -1 ) + r 2 (m 1 m 1 -1 ) (mod m 1 m 2 ) n mod m 1 = r 1 0 1 1 Verification + n mod m 2 = 0 r 2 2

  21. Manually Incremental Calculation Manually Incremental Calculation n  1 (mod 3)  2 (mod 5)  3 (mod 7)  3 (mod 7) 21

  22. Manually Incremental Calculation Manually Incremental Calculation n  1 (mod 3)  2 (mod 5)  3 (mod 7)  3 (mod 7) satisfying the 1 st eq ^  n 1  1 (mod 3)  n 1  1 (mod 3) … satisfying the 1 eq. r 1 22

  23. Manually Incremental Calculation Manually Incremental Calculation n  1 (mod 3) n  1 (mod 3)  2 (mod 5)  2 (mod 5)  3 (mod 7)  3 (mod 7) satisfying the 1 st eq ^  n 1  1 (mod 3)  n 1  1 (mod 3) … satisfying the 1 eq. 23

  24. Manually Incremental Calculation Manually Incremental Calculation n  1 (mod 3) n  1 (mod 3)  2 (mod 5)  2 (mod 5)  3 (mod 7)  3 (mod 7) satisfying the 1 st eq ^  n 1  1 (mod 3)  n 1  1 (mod 3) … satisfying the 1 eq.  3 ꞏ (-3) + 5 ꞏ 2 = 1 24

  25. Manually Incremental Calculation Manually Incremental Calculation n  1 (mod 3) n  1 (mod 3)  2 (mod 5)  2 (mod 5)  3 (mod 7)  3 (mod 7) satisfying the 1 st eq ^  n 1  1 (mod 3)  n 1  1 (mod 3) … satisfying the 1 eq. inverse of 3 (mod 5)  3 ꞏ (-3) + 5 ꞏ 2 = 1 25

  26. Manually Incremental Calculation Manually Incremental Calculation n  1 (mod 3) n  1 (mod 3)  2 (mod 5)  2 (mod 5)  3 (mod 7)  3 (mod 7) satisfying the 1 st eq ^  n 1  1 (mod 3)  n 1  1 (mod 3) … satisfying the 1 eq. inverse of 3 (mod 5)  3 ꞏ (-3) + 5 ꞏ 2 = 1 inverse of 5 (mod 3) 26

  27. Manually Incremental Calculation Manually Incremental Calculation n  1 (mod 3) n  1 (mod 3)  2 (mod 5)  2 (mod 5)  3 (mod 7)  3 (mod 7) satisfying the 1 st eq ^  n 1  1 (mod 3)  n 1  1 (mod 3) … satisfying the 1 eq. inverse of 3 (mod 5)  3 ꞏ (-3) + 5 ꞏ 2 = 1 inverse of 5 (mod 3) ^  n 2  2 ꞏ 3 ꞏ (-3) + 1 ꞏ 5 ꞏ 2 ^ n 1 27

  28. Manually Incremental Calculation Manually Incremental Calculation n  1 (mod 3) n  1 (mod 3)  2 (mod 5)  2 (mod 5)  3 (mod 7)  3 (mod 7) satisfying the 1 st eq ^  n 1  1 (mod 3)  n 1  1 (mod 3) … satisfying the 1 eq. inverse of 3 (mod 5)  3 ꞏ (-3) + 5 ꞏ 2 = 1 inverse of 5 (mod 3) ^  n 2  2 ꞏ 3 ꞏ (-3) + 1 ꞏ 5 ꞏ 2 ^ r 2 n 1 28

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend

Notes 12 Spring 2005 Clancy/Wagner RSA and the Chinese remainder theorem The Chinese remainder

Notes 12 Spring 2005 Clancy/Wagner RSA and the Chinese remainder theorem The Chinese remainder

CS 70 Discrete Mathematics for CS Notes 12 Spring 2005 Clancy/Wagner RSA and the Chinese remainder theorem The Chinese remainder theorem Suppose we have a system of simultaneous equations, like maybe this one: ( mod 5 ) x 2 ( mod

185 views • 3 slides
Chinese Remainder Theorem explained with rotations Reference: Antonella Perucca, The Chinese

Chinese Remainder Theorem explained with rotations Reference: Antonella Perucca, The Chinese

Chinese Remainder Theorem explained with rotations Reference: Antonella Perucca, The Chinese Remainder Clock , The College Mathematics Journal, 2017, Vol. 48, No. 2, pp. 82-89. One step-wise rotation Consider a circle with one marked point on

230 views • 12 slides
RSA, the Chinese Remainder Theorem, and Remote Coin Flipping CS70 Summer 2016 - Lecture 7B David

RSA, the Chinese Remainder Theorem, and Remote Coin Flipping CS70 Summer 2016 - Lecture 7B David

RSA, the Chinese Remainder Theorem, and Remote Coin Flipping CS70 Summer 2016 - Lecture 7B David Dinh 02 August 2016 UC Berkeley Agenda RSA The Chinese remainder theorem Eulers Criterion Blums coin-flipping scheme Slides marked with

1.03k views • 102 slides
RSA, the Chinese Remainder Theorem, and Remote Coin Flipping CS70 Summer 2016 - Lecture 7B David

RSA, the Chinese Remainder Theorem, and Remote Coin Flipping CS70 Summer 2016 - Lecture 7B David

RSA, the Chinese Remainder Theorem, and Remote Coin Flipping CS70 Summer 2016 - Lecture 7B David Dinh 02 August 2016 UC Berkeley Agenda RSA The Chinese remainder theorem Eulers Criterion Blums coin-flipping scheme Slides marked with

487 views • 25 slides
A Mec hanical Pro of of the Chinese Remainder Theorem Da vid M. Russino Adv anced

A Mec hanical Pro of of the Chinese Remainder Theorem Da vid M. Russino Adv anced

A Mec hanical Pro of of the Chinese Remainder Theorem Da vid M. Russino Adv anced Micro Devices, Inc. david.russinoff@a md. co m http://www.onr.c om /us er /ru ss /da vid Informal Statemen t L et m ; : : :

375 views • 14 slides
Equidistribution from the Chinese Remainder Theorem E. Kowalski ETH Zrich April 23, 2020

Equidistribution from the Chinese Remainder Theorem E. Kowalski ETH Zrich April 23, 2020

Equidistribution from the Chinese Remainder Theorem E. Kowalski ETH Zrich April 23, 2020 [Joint work with K. Soundararajan, arXiv:2003.12965 ] Motivation Hooley (1964): the fractional parts of the roots modulo q x of a fixed

332 views • 21 slides
Lecture 7.7: The Chinese remainder theorem Matthew Macauley Department of Mathematical Sciences

Lecture 7.7: The Chinese remainder theorem Matthew Macauley Department of Mathematical Sciences

Lecture 7.7: The Chinese remainder theorem Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Lecture 7.7: The Chinese remainder

629 views • 10 slides
Numb3rs 11 2 10 3 The Chinese Remainder Theorem 9 4 8 5 7 6 Chiming Clocks Two clocks,

Numb3rs 11 2 10 3 The Chinese Remainder Theorem 9 4 8 5 7 6 Chiming Clocks Two clocks,

0 12 1 Numb3rs 11 2 10 3 The Chinese Remainder Theorem 9 4 8 5 7 6 Chiming Clocks Two clocks, with a hours and b hours on their dials 0 1 12 Say they both start at 0, and move one step every 2 11 10 3 minute 9 4 e.g.,

310 views • 13 slides
Constructing Ideal Secret Sharing Schemes based on Chinese Remainder Theorem Yu Ning, Fuyou

Constructing Ideal Secret Sharing Schemes based on Chinese Remainder Theorem Yu Ning, Fuyou

Constructing Ideal Secret Sharing Schemes based on Chinese Remainder Theorem Yu Ning, Fuyou Miao*, University of Science and Technology of China Contributions Generalization of existing CRT-based (t,n)-SS from Integer Ring to

548 views • 26 slides
Simplifying Pseudo-Boolean Constraints in Residual Number Systems Michael Codish Department of

Simplifying Pseudo-Boolean Constraints in Residual Number Systems Michael Codish Department of

Simplifying Pseudo-Boolean Constraints in Residual Number Systems Michael Codish Department of Computer Science Ben Gurion University Beer-Sheva , Israel Joint work: Yoav Fekete SAT 2014 Chinese Remainder Theorem Chinese Remainder Theorem

1.5k views • 46 slides
a (mod m ) b is congruent to modulo a m b mod mod a m b m

a (mod m ) b is congruent to modulo a m b mod mod a m b m

Number Theory and its Applications Modular Exponentiation Euclidean Algorithm for GCD Solving Linear Congruences Chinese Remainder Theorem and Application to Arithmetic with large numbers Covered in Sections 3.6 and 3.7 Based

489 views • 19 slides
Homomorphisms and Chinese Remainder Algorithms (cont.) L. Yohanes Stefanus L. Yohanes Stefanus

Homomorphisms and Chinese Remainder Algorithms (cont.) L. Yohanes Stefanus L. Yohanes Stefanus

Homomorphisms and Chinese Remainder Algorithms(cont.) Homomorphisms and Chinese Remainder Algorithms (cont.) L. Yohanes Stefanus L. Yohanes Stefanus Slide 07.1 Homomorphisms and Chinese Remainder Algorithms(cont.) The Chinese Remainder

354 views • 31 slides
CSCI 246 Class 5 RATIONAL NUMBERS, QUOTIENT REMAINDER THEOREM Quiz Questions Lecture

CSCI 246 Class 5 RATIONAL NUMBERS, QUOTIENT REMAINDER THEOREM Quiz Questions Lecture

CSCI 246 Class 5 RATIONAL NUMBERS, QUOTIENT REMAINDER THEOREM Quiz Questions Lecture 8: Give the divisors of n when: n = 10 n = 0 Lecture 9: Say: 10 = 3*3 +1 Whats the quotient q and the remainder r? Lecture

552 views • 26 slides
Section4.3 Polynomial Division; The Remainder Theorem and the Factor Theorem

Section4.3 Polynomial Division; The Remainder Theorem and the Factor Theorem

Section4.3 Polynomial Division; The Remainder Theorem and the Factor Theorem PolynomialLongDivision Example of Long Division of Numbers Lets compute 823 5 : Example of Long Division of Numbers Lets compute 823 5 : 164 5 )823 5 32

926 views • 64 slides
T-79.159 Cryptography and Data Security Lecture 6: Modular Arithmetic Kaufman et al: Ch 7

T-79.159 Cryptography and Data Security Lecture 6: Modular Arithmetic Kaufman et al: Ch 7

T-79.159 Cryptography and Data Security Lecture 6: Modular Arithmetic Kaufman et al: Ch 7 -Prime numbers Stallings: Ch 8 -Euclids algorithm -Chinese remainder theorem -Eulers totient function -Eulers theorem 1 Prime Numbers

730 views • 6 slides
DTTF/NB479: Dszquphsbqiz Day 8 Announcements: Please use pencil on quizzes if possible

DTTF/NB479: Dszquphsbqiz Day 8 Announcements: Please use pencil on quizzes if possible

DTTF/NB479: Dszquphsbqiz Day 8 Announcements: Please use pencil on quizzes if possible Questions? Today: Congruences Chinese Remainder Theorem Modular Exponents Hill Cipher implementation Encryption Easy to do in MATLAB.

466 views • 14 slides
RSA Example A small (and insecure) example [Stinson]: - Bob: - chooses p = 101, q = 113 -

RSA Example A small (and insecure) example [Stinson]: - Bob: - chooses p = 101, q = 113 -

RSA Example A small (and insecure) example [Stinson]: - Bob: - chooses p = 101, q = 113 - computes n = pq = 11413 and (n) = (p-1)(q-1) = 11200 - chooses e = 3533 (note: gcd(e, (n))=1) - computes d = e -1 mod (n) = 6597 - publishes n and

492 views • 11 slides
Fine scales of decay and an application to decay of waves in a viscoelastic boundary damping model

Fine scales of decay and an application to decay of waves in a viscoelastic boundary damping model

Fine scales of decay and an application to decay of waves in a viscoelastic boundary damping model (International Workshop on Operator Theory and its Applications) Reinhard Stahn (TU Dresden) (joint with Jan Rozendaal and David Seifert) August

936 views • 38 slides
MTH314: Discrete Mathematics for Engineers Lecture 9a: Public-Key Cryptography: Proofs Dr Ewa

MTH314: Discrete Mathematics for Engineers Lecture 9a: Public-Key Cryptography: Proofs Dr Ewa

MTH314: Discrete Mathematics for Engineers Lecture 9a: Public-Key Cryptography: Proofs Dr Ewa Infeld Ryerson University Dr Ewa Infeld Ryerson University MTH314: Discrete Mathematics for Engineers Chinese Remainder Theorem Theorem Suppose

853 views • 40 slides
Objectives Euclidean Algorithm to compute gcd (Greatest Common Divisor) to compute

Objectives Euclidean Algorithm to compute gcd (Greatest Common Divisor) to compute

More Number Theoretic Results Debdeep Mukhopadhyay Assistant Professor Department of Computer Science and Engineering Indian Institute of Technology Kharagpur INDIA -721302 Objectives Euclidean Algorithm to compute gcd (Greatest

151 views • 14 slides
Number Theory (II) Cunsheng Ding HKUST, Hong Kong November 10, 2015 Cunsheng Ding (HKUST, Hong

Number Theory (II) Cunsheng Ding HKUST, Hong Kong November 10, 2015 Cunsheng Ding (HKUST, Hong

Number Theory (II) Cunsheng Ding HKUST, Hong Kong November 10, 2015 Cunsheng Ding (HKUST, Hong Kong) Number Theory (II) November 10, 2015 1 / 21 Contents The Discrete Logarithm Problem 1 Diffie-Hellman Key Exchange Protocol 2 Linear

578 views • 21 slides
Discrete Mathematics & Mathematical Reasoning Multiplicative Inverses and Some Cryptography

Discrete Mathematics & Mathematical Reasoning Multiplicative Inverses and Some Cryptography

Discrete Mathematics & Mathematical Reasoning Multiplicative Inverses and Some Cryptography Colin Stirling Informatics Some slides based on ones by Myrto Arapinis Colin Stirling (Informatics) Discrete Mathematics (Chap 4) Today 1 / 13

629 views • 45 slides
Handout 9 Summary of this handout: RSA Generating Prime Numbers Arithmetic Modulo a

Handout 9 Summary of this handout: RSA Generating Prime Numbers Arithmetic Modulo a

06-20008 Cryptography The University of Birmingham Autumn Semester 2012 School of Computer Science Eike Ritter 22 November, 2012 Handout 9 Summary of this handout: RSA Generating Prime Numbers Arithmetic Modulo a Composite IV.4 RSA

498 views • 10 slides
Fields and Polynomials (a 3 ,b 3 ) (a 2 ,b 2 ) (a 5 ,b 5 ) b 1 (a 4 ,b 4 ) a 1 First, a little

Fields and Polynomials (a 3 ,b 3 ) (a 2 ,b 2 ) (a 5 ,b 5 ) b 1 (a 4 ,b 4 ) a 1 First, a little

15-251: Great Theoretical Ideas in Computer Science Fall 2016 Lecture 24 November 17, 2016 Fields and Polynomials (a 3 ,b 3 ) (a 2 ,b 2 ) (a 5 ,b 5 ) b 1 (a 4 ,b 4 ) a 1 First, a little more Number Theory Bezouts identity Let a,b be

1.07k views • 72 slides

More recommend