   Chapter 2.8, Problem 12E

Chapter
Section
Textbook Problem

Suppose the alphabet consists of a through z , in natural order, followed by a blank, a comma, and a period, in that order. Associate these "letters" with the numbers 0 ,   1 ,   2 ,   ...   ,   28 , respectively, thus forming a 29 -letter alphabet, E . Use the affine cipher to decipher the message B   Z   Z   K   ,   A   U   Z   N   Z   G,   R   S   K   Z   ,   A   U   W   A   O if you know that the plaintext message begins with " b " and ends with ".". Write out the affine mapping f and its inverse.

To determine

Plain text in the 29 letter alphabet E, by using the affine cipher to decipher the message “ BZZK, AUZNZG, RSKZ, AUWAO” and the plain text message begins with “b” followed by “.” Write down the affine mapping f and f1.

Explanation

Given information:

(i) The 29-letter alphabet D -

If alphabet consist of a to z, in natural order followed by blank, a comma and a period, then have 29 “letters” that associates with the integer 0,1,2,3,...,28 as follows:

Alphabet: abcdef...vwxyz"blank"commaperiod

D:012345...2122232425262728

ii) Euclidean Algorithm:

Let a and b are integers with b>0, then the procedure

a=bq0+r10r1<bb=r1q1+r20r2<r1r1=r2q2+r30r3<r2rk=rk+1qk+1+rk+20rk+2<rk+1

is known as Euclidean Algorithm.

Since the integers r1,r2,,rk+2 are decreasing and are all non-negative, there is a smallest integer n, such that rn+1=0:

rn1=rnqn+rn+10=rn+1

If r0=b, then the last non-zero remainder rn is always greatest common divisor of a and b.

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:

As plaintext begins with “b” and ends with “.”

The ciphertext B and O corresponds to the plain text “b” and ”.”.

Translating these into the set D,

BtranslatetoE1btranslatetoE1

OtranslatetoE14.translatetoE28

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

1=a(1)+bmod2914=a(28)+bmod29

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

a(1)+b=1mod29…… (1)

a(28)+b=14mod29…… (2)

Subtract (1) from (2), gives

a(28)a(1)=141mod29

a(27)=13mod29

27a=13mod29

Now, to find the multiplicative inverse of 27 using Euclidean algorithm.

29=27+2

27=13(2)+1

Thus,

1=2713(2)

1=2713(2927)

1=2713(29)+13(27)

1=27(14)13(29)

Thus, 27(14)=1(mod29)

The inverse 271=14(mod29).

Consider, 27a=13(mod29)

a=27113(mod29)

a=(14)13(mod29)

a=182(mod29)

a=8(mod29)

Now, from (1), a(1)+b=1mod29

(8(mod29))(1)+b=1mod29

b=8(mod29)+1(mod29)

b=8+1(mod29)

b=7(mod29)

b=22(mod29) as 22=7(mod29)

Therefore, a=8 and b=22 is the solution of the system

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