The Extended Euclidean Algorithm-Computer Security-Lecture Handouts, Lecture notes of Computer Security

Rocky Hiranandani gave this handout to help with Computer Security course at Baddi University of Emerging Sciences and Technologies. It includes: Extended, Euclidean, Algorithm, Method, Back, Substitution, Gretest, Common, Divisor

Typology: Lecture notes

2011/2012

Uploaded on 07/07/2012

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The Extended Euclidean Algorithm
Example 1: m= 65, n = 40
Step 1: The (usual) Euclidean algorithm:
(1) 65 = 1 ·40 +25
(2) 40 = 1 ·25 +15
(3) 25 = 1 ·15 +10
(4) 15 = 1 ·10 +5
10 = 2 ·5
Therefore: gcd(65,40) = 5.
Step 2: Using the method of back-substitution:
5(4)
=15 10
(3)
=15 (25 15) =2·15 25
(2)
=2(40 25) 25 =2·40 3·25
(1)
=2·40 3(65 40) =5·40 3·65
Conclusion: 65(3) +40(5) =5.
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The Extended Euclidean Algorithm

Example 1: m = 65, n = 40

Step 1: The (usual) Euclidean algorithm:

(1) 65 = 1 · 40 + 25 (2) 40 = 1 · 25 + 15 (3) 25 = 1 · 15 + 10 (4) 15 = 1 · 10 + 5 10 = 2 · 5

Therefore: gcd(65, 40) = 5.

Step 2: Using the method of back-substitution:

5 (4) = 15 − 10 (3) = 15 − (25 − 15) = 2 · 15 − 25 (2) = 2(40 − 25) − 25 = 2 · 40 − 3 · 25 (1) = 2 · 40 − 3(65 − 40) = 5 · 40 − 3 · 65

Conclusion: 65(−3) + 40(5) = 5.

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The Extended Euclidean Algorithm

Example 2: m = 1239, n = 735

Step 1: The (usual) Euclidean algorithm:

(1) 1239 = 1 · 735 + 504 (2) 735 = 1 · 504 + 231 (3) 504 = 2 · 231 + 42 (4) 231 = 5 · 42 + 21 42 = 2 · 21

Therefore: gcd(1239, 735) = 21.

Step 2: Using the method of back-substitution:

21 (4) = 231 − 5 · 42 (3) = 231 − 5(504 − 2 · 231) = 11 · 231 − 5 · 504 (2) = 11(735 − 504) − 5 · 504 = 11 · 735 − 16 · 504 (1) = 11 · 735 − 16(1239 − 735) = 27 · 735 − 16 · 1239

Conclusion: 1239(−16) + 735(27) = 21.

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