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Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
ISBN: 9781285463230

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BuyFindarrow_forward

Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
ISBN: 9781285463230
Textbook Problem

Suppose that in a long ciphertext message the letter x occurred most frequently, followed in frequency by c . Using the fact that in the 26 -letter alphabet A , described in Example 4 , e occurs most frequently, followed in frequency by t , read the portion of the message

R N C Y X R N C H F T enciphered using an affine mapping on A . Write out the affine mapping   f and its inverse.

To determine

Plain text in the 26 letter alphabet A, e occurs most frequently, followed in frequency by t, enciphered the message ‘RNCYXRNCHFT’ using an affine mapping on A and write affine mapping and its inverse.

Explanation

Given information:

The 26-letter alphabet A -

If alphabet consist of a to z, in natural order, then have 26 “letters” that associates with the integer 0,1,2,3,...,25 as follows:

Alphabet: abcdef...vwxyz

A:012345...2122232425

Formula used:

(i) Affine cipher:

Define the mapping f:AA by f(x)=ax+bmodn, where a,b are fixed integers and order pair a,b forms a key to affine cipher.

(ii) Cancellation law:

If ax=ay(modn) and (a,b)=1, then x=y(modn)

Calculation:

As in long cipher text message the letter x occurred most frequently, followed by frequency by c and in the 26 letter alphabet A, e occurs most frequently, followed in frequency by t

It seems reasonable that the cipher text letters x and c corresponds to the plaintext e and t.

Translating these into the set A,

xtranslatetoA23etranslatetoA4

ctranslatetoA2ttranslatetoA19

Now, to determine the key from the following system of equations for a and b.

23=a(4)+bmod262=a(19)+bmod26

Use a=b(modn)b=a(modn)

a(4)+b=23mod26…… (1)

a(19)+b=2mod26 …… (2)

Subtract (1) from (2), gives

a(19)a(4)=223mod26

a(15)=21mod26

a(15)=5mod26

15a=5mod26

By using cancellation law,

3a=1mod26

a=9(mod26)

Now, from (1),

a(4)+b=23mod26

(9(mod26))(4)+b=23mod26

b=23(mod26)36(mod26)

b=2336(mod26)

b=13(mod26)

b=13(mod26)

Therefore, a=9 and b=13

Now, to define affine mapping as, f:AA such that, f(x)=9x+13mod26, where A={0,1,2,3,

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