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Affine Cipher - Decryption (Known Plaintext Attack)

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In this video I talk about ways to decrypt the Affine Cipher when the key is NOT known. Specifically, I go over an example of the known plaintext attack. 3^(-1) = 9 (mod 26) math worked out (Euclidean Algorithm): 1. Forwards: 26 = 3(8) + 2 3 = 2(1) + 1 2. Backwords: 1 = 3 - 2(1) 1 = 3(1) - (26 - 3(8))(1) 1 = 3(1) - 26 + 3(8) 1 + 26 = 3(9) 1 (mod 26) = 3(9) Hence 9 is the inverse. Links: -Affine Encryption: https://www.youtube.com/watch?v=_E8rSP0uAIY -Affine Decryption (Known Key): https://www.youtube.com/watch?v=XFxFPBKFVe8 -Euclidean Algorithm: https://www.youtube.com/watch?v=K5nbGbN5Trs
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Text Comments (22)
Koding Kyle (1 month ago)
My alpha was negative?
Divya S (1 month ago)
thank you it's was useful
Rhys Jennings (2 months ago)
I know this video is pretty old, but thank you very much. Helped heaps :)
Danielle Hibbs-Heiser (2 months ago)
SO what happens if you were to try and find the inverse and you end with a 0 and not with a one. So finding 6^-(1)
PRABHU S Teaching (3 months ago)
Muaaz Hammad (7 months ago)
how can someone get first two letters of plane text???? its illogical .. you should do this with frequency analysis
Peter Walker (2 months ago)
Ha Ha! But Question Paper has Theoretically.
Bruna (8 months ago)
totally saved me on my midterm!! thank you so much <3
Emre Can (11 months ago)
thx, it was helpful to understand :)
Alex lou (11 months ago)
so i got a 2 letters fr encrypted as eu in mod 28 and the number i am supposed to get the multiplicate inverse for dosnt have a multiplicate inverse what do i do wrong?
ellieneedsalife (1 year ago)
thank you, this has just saved my grade during a uni exam
Aaron Titiner (1 year ago)
How does 25*9 equal 17(mod26)? I can't figure it out
Sebastian St Johnston (6 months ago)
Aaron Titiner (1 year ago)
Figured it out :) hahaha
Marius L. (1 year ago)
Very helpful, thanks!
Mahmoud Shoair (1 year ago)
can u make a video for hill Cipher - Decryption (Known Plaintext Attack)
Der Lukas (2 years ago)
u usually don't need "known plaintext", since the affine cipher is linear and can be represented as a substitution cipher, which allows statistical attacks.
Theoretically (2 years ago)
+mouad marmouchi https://www.youtube.com/watch?v=LaWp_Kq0cKs&feature=gp-n-y&google_comment_id=z12nxpijksndw5nov04ch1ow5zymsxaq1as
mouad (2 years ago)
+Der Lukas can you explain how we can find key without "known plaintext" pls ?
Sawyer Tealey (2 years ago)
email me at [email protected] I need help with this cypher
Trisnet Corelone (3 years ago)
thank you very much for this. I spend a day watching this video in order to conquer a similar question. The question was the final boulder and I completed. Thank you again. Gained yourself another subscriber
Ahmad Alhonainy (3 years ago)
Thanks alot

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