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Elements Of Modern Algebra

8th Edition
Gilbert + 2 others
ISBN: 9781285463230

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Section
BuyFindarrow_forward

Elements Of Modern Algebra

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

In the 27 -letter alphabet A described in Example 2 , use the affine cipher with key a   =   7  and  b   =   5 to encipher the following message. all systems go

What is the inverse mapping that will decipher the ciphertext?

Example 2 Translation Cipher Associate the n letters of the "alphabet" with the integers 0 ,   1 ,   2 ,   3   .....   n 1 . Let A   =   { 0 ,   1 ,   2 ,   3   .....   n  -  1 } and define the mapping

f : A   A by

f ( x )   = x + k m o d n where k is the key, the number of positions from the plaintext to the ciphertext. If our alphabet consists of a through z , in natural order, followed by a blank, then we have 27 "letters" that we associate with the integers 0 ,   1 ,   2 ,   ...   ,   26 as follows:

Alphabet: a b c d e f ... v w x y z "blank" A: 0 1 2 3 4 5 21 22 23 24 25 26

To determine

Cipher text in 27 letter alphabet A, by using the affine cipher with key a=7 and b=5 to encipher the message, “all systems go” and find the inverse mapping that will decipher the cipher text.

Explanation

Given information:

(i) a=7 and b=5

(ii) The 27 letter alphabet A -

If alphabet consist of a to z, in natural order followed by a blank, then have 27 “letters” that associates with the integer 0,1,2,3,...,26 as follows:

Alphabe: abcdef...vwxyz"blank"

A:012345...212223242526

Formula used:

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.

Calculation:

To use affine mapping with, a=7 and b=5 as the key in 27 letter alphabet

Define the map f:AA where A={0,1,2,...,26} given by,

f(x)=7x+5mod27

If key, a=7 and b=5, then the plain text message translates into the cipher text message as follows:

Translate the statement “all systems go” to the 27 alphabet A:

0111126182418194121826614

Now to affine cipher f(x)=7x+5mod27

f(0)=7(0)+5mod27=5mod27=5

f(11)=7(11)+5mod27=82mod27=1

f(26)=7(26)+5mod27=187mod27=

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