Announcements: Assignment 1 due tomorrow in class. Questions? Roll Call Today: Vigenere ciphers

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Announcements: Assignment 1 due tomorrow in class. Questions? Roll Call Today: Vigenere ciphers

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Announcements: Assignment 1 due tomorrow in class. Questions? Roll Call Today: Vigenere ciphers

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DTTF/NB479: DszquphsbqizDay 5

- Announcements:
- Assignment 1 due tomorrow in class.

- Questions?
- Roll Call
- Today: Vigenere ciphers
- Pronunciation?

Vigenere Ciphers

- Idea: the key is a vector of shifts
- Ex. Use a word like hidden (7 8 3 3 4 13).
- Example:
- The recent development of various methods of
7 8 3 3 413 7 8 3 3 413 7 8 3 3 413 7 8 3 3 413 7 8 3 3 4 13 7 8 3 3 413 7 8

015 7 20 815112122 6 8 811191718161720 1 17 8 25132416172322 2511 11017 7 5 2113

- aph uiplvw giiltrsqrub ri znyqrxw zlbkrhf vn

- The recent development of various methods of

- Encryption:
- Repeat the vector as many times as needed to get the same length as the plaintext
- Add this repeated vector to the plaintext.
- Demo

Source

The recent development of various methods of modulation such as PCM and PPM which exchange bandwidth for signal-to-noise ratio has intensified the interest in a general theory of communication. A basis for such a theory is contained in the important papers of Nyquist and Hartley on this subject. In the present paper we will extend the theory to include a number of new factors, in particular the effect of noise in the channel, and the savings possible due to the statistical structure of the original message and due to the nature of the final destination of the information.

(Claude Shannon’s A Mathematical Theory Of Communication)

Security

- The vector isn’t known
- It’s length isn’t even known!
- With shift cipher, most frequent cipher letter is probably e.
- But here, e maps to H, I, L, … (spread out!)

- Consider 4 attacks:
- Known plaintext?
- Chosen plaintext?
- Chosen ciphertext?
- Ciphertext only?

- What would you do to simplify the problem?
- You can even assume that you know how long the key is…
- 3-4 min with a partner

H 0.061

I 0.070

J 0.002

K 0.008

L 0.040

M 0.024

N 0.067

O 0.075

P 0.019

Q 0.001

R 0.060

S 0.063

T 0.091

U 0.028

V 0.010

W 0.023

X 0.001

Y 0.020

Z 0.001

A 0.082

B 0.015

C 0.028

D 0.043

E 0.127

F 0.022

G 0.020

Graph:

- Consider Gadsby by Ernest Vincent Wright, February 1939:
- http://www.spinelessbooks.com/gadsby/01.html

- What do you notice about it?

- Assume we know the key length, L, …
- We’ll see how to find it shortly

- Method 1:
- Parse out the characters at positions p = 1 (mod L)
- These have all been shifted the same amount
- Do a frequency analysis to find shift
- The most frequent letter should be e, given enough text. Can verify to see how shift affects other letters.

- This gives the first letter of the key
- Repeat for positions p = 2, p = 3, … p = L
- Problem: involves some trial and error.
- For brute force to work, would need to brute force all letters of key simultaneously: _____ possibilities

- Parse out the characters at positions p = 1 (mod L)

- Consider A = (0.0820 0.0150 0.0280 0.0430 0.1270 0.0220 0.0200 0.0610 0.0700 0.0020 0.0080 0.0400 0.0240 0.0670 0.0750 0.0190 0.0010 0.0600 0.0630 0.0910 0.0280 0.0100 0.0230 0.0010 0.0200 0.0010)
- Ai = A displaced i positions to the right
- A0 .* A0 = ?
- A0 .* A1 = ?
- Ai .* Aj depends on _____ only.
- Max occurs when _____. 3 reasons why:

- What if the frequency of letters in the plaintext approximates A?
- Then for each k, frequency of each group of letters in position p = k (mod L) in the ciphertext approximates A.
- Then loop, displacing the ciphertext by i, and counting the number of coincidences.
- Get max when displace by correct key length
- So just look for the max!
shift

APHUIPLVWGIILTRSQRUBRIZNYQRXWZLBKRHFVN (0)

NAPHUIPLVWGIILTRSQRUBRIZNYQRXWZLBKRHFV (1)

VNAPHUIPLVWGIILTRSQRUBRIZNYQRXWZLBKRHF (2)

…

KRHFVNAPHUIPLVWGIILTRSQRUBRIZNYQRXWZLB (6) 5matches

Take any random pair in the ciphertext:

The letter in the top row is shifted by i (say 0)

The letter in the bottom row is shifted by j (say 2)

Prob(both ‘A’) = P(‘a’)*P(‘y’) = 0.082 * 0.020

Prob they are the same (any letter) is ______

When i=j, get max # of coincidences

Occurs when we shift by the correct key length

Demo

- Method 1
- Parse out the characters at positions p = 1 (mod L)
- These have all been shifted the same amount
- Do a frequency analysis to find shift
- The most frequent letter should be e, given enough text. Can verify to see how shift affects other letters.

- This gives the first letter of the key
- Repeat for positions p = 2, p = 3, … p = L

- Parse out the characters at positions p = 1 (mod L)

- Method 2
- Parse out the characters at positions p = 1 (mod L)
- These have all been shifted the same amount
- Get the whole freq. distribution W = (0.05, 0.002, …)
- W approximates A. Calculate
- Max occurs when we got the shift correct.
- Note we use the distribution of all 26 letters…more robust than just using 1 (‘e’).

- This gives the first letter of the key
- Repeat for positions p = 2, p = 3, … p = L

- Parse out the characters at positions p = 1 (mod L)

- Method 1
- Pros?

- Method 2
- Pros?

- Of my code…

- http://www.usafa.af.mil/df/dfcs/acis/applets/Vigenere.html
- Play with now and for homework
- Thanks to Dr. Dino Schweitzer, USAFA, who I met last Friday, for pointing me to his demo!
- Aren’t you glad I was at SIGCSE?

- What if we modified the Vigenere cipher so that each individual letter was not simply shifted, but the result of an affine function?