Main question: What is the efficiency of the code of Example 2 of the notes? (Ef

Question

Main question:

What is the efficiency of the code of Example 2 of the notes?

(Effi- ciency is defined in Chapter 7.) Suppose you use the code of Exercise 1 with p = 127 (a prime). Then each element of F_127 is uniquely representable by a number m with 0 m < 127, which in turn corresponds to a seven-tuple of bits, the digits in the base 2 representation of the number m. (For example, 113 corresponds to 1110001: 113 = 64+32+16+1.) A coded message C is a seven-tuple of elements of F_127, each element in turn representable by a seven-tuple of bits. Hence C corresponds to a bit message with 49 bits. Then R also has 49 bits

**** reference ****

1. Similar to Example 2, consider the two error correcting Reed-Solomon code with (m, e, n) = (2,2,6), with the field F-E, with p a large prime. Use (ao, ai, a2, a3, a4,A5, a6) = (-3,-2,-1, 0, 1, 2, 3). You want to send the plaintext message w = (15,1,-2) to Bob, and want Bob to be able to correct two errors. You and Bob agree to use the Reed-Solomon code as just described. Find the encoded 7-tuple C for the plaintext message w that you to Bob.

Explanation / Answer

Here is the solution as per the given criteria, please go through it:-

As per the given data:-

m=2

e=2

n=6

i.e, (m,e,n) = (2,2,6)

(a0,a1,a2,a3,a4,a5,a6) = (-3,-2,-1,0,1,2,3)

Here,

w = (15,1,-2)

and, as we have seen in the example,

we find a polynomial of degree 2

w(x) = 15+x-2x2

so, we want to find the code vector C and that is given by

C = (w(-3),w(-2),w(-1),w(0),w(1),w(2),w(3))

so,

w(-3) = 15-3-2x(9)

= 15-3-18

= -6

w(-2) = 15-2-2x(4)

= 15-2-8

= 5

w(-1) = 15-1-2x(1)

= 15-1-2

= 12

w(0) = 15-0-2x(0)

= 15-0-0

= 15

w(1) = 15+1-2x(1)

= 15+1-2

= 14

w(2) = 15+2-2x(4)

= 15+2-8

= 9

w(3) = 15+3-2x(9)

= 15+3-18

= 0

Therefor,

C = (-6,5,12,15,14,9,0)

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Thankyou

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