Answers and Solutions to (Even Numbered) Suggested Exercises in Sections of Grimaldi s Discrete and Combinatorial Mathematics

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Leonard Grant
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1 Answers and Solutions to (Even Numbered) Suggested Exercises in Sections of Grimaldi s Discrete and Combinatorial Mathematics Section a. r = = + = c + e. So the error pattern is e =. The probability of an error occurring at the fourth position while all other bits are transmitted correctly is p( p) 8 = (.5)(.95) b. r = = + = c + e. So this time the error pattern is e =. The probability of this happening is p 2 ( p) 7 = (.5) 2 (.95) 7.7. c. The probability that a single error occurs is 9p( p) 8 = 9(.5)(.95) d. The probability that a double error occurs is ( 9 2) p 2 ( p) 7 = 36(.5) 2 (.95) e. The probability that a double error occurs is ( 9 3) p 3 ( p) 6 = 47(.5) 3 (.95) f. The probability that a double error occurs is ( 7 3) p 3 ( p) 6 = 56(.5) 3 (.95) The encoding is by five-times repetition code, and the decoding is by majority rule. a. c = E() = is sent. This will be decoded correctly if the received word r has more s than s. The probability of this is ( p) 5 + ( ) 5 4 p( p) 4 + ( 5 3) p 2 ( p) 3 = (.95) 5 + 5(.5)(.95) 4 + (.5) 2 (.95) b. If c = E() = is sent, then we can treat each digit of the message word as we have done in (a). Therefore probability of correct decoding of this signal is ( ( p) 5 + ( ) 5 4 p( p) 4 + ( 5 3) p 2 ( p) 3) c. If r = is received in the (, 2) five-times repetition code, then since the st, 3rd, 5th, 7th and 9th bits are,,,,, respectively, it decodes as by the majority rule; and since the 2nd, 4th, 6th, 8th and th bits are,,,,, respectively, it decodes as by the majority rule. Therefore D(r) =. d. If r =, r = or r =, then D(r) =. (There are = 253 more possible values of r which decode as.) e. To have D(r) = w, more than half of the bits in the odd numbered positions of r must be the same as the first bit of w, and more than half of the bits in the even numbered positions of r must be the same as the second bit of w. Therefore, D (w) = (2 5 /2) 2 = The code in Example 6.24 of the textbook is the (6,2) triple repetition code. We will decode by majority rule. a. r = = c = = w =. b. r = = c = = w =. c. r = = c = = w =. d. r = = c = = w =.

2 If the minimum distance is 9, then we can detect errors of weight 8, and correct errors of weight a. The parity-check matrix is. Since the minimum distance is d = 3, we can detect detect errors of weught 2, and correct single errors. i. r =. Hr t = = Since Hr t is not the zero vector, there are errors. Since Hr t is equal to the third column of H, we will assume that there is exactly one error and it is in the third bit. Therefore e = and c = r + e = + =, and w =. ii. r =. Hr t = = Since Hr t is the zero vector, there cannot be two or less errors. So we assume that there are no errors. Therefore c = r =, and w =. iii. r =. Hr t = = Since Hr t is not the zero vector, there are errors. If we assume that there is exactly one error, then since Hr t is the same as the fifth column of H, this error must be at the fifth bit. So e =, c = r + e = + =, and w =. iv. r =. Hr t = = 2

3 Since Hr t is not the zero vector, there are errors. Since Hr t is equal to the fifth column of H, we will assume that there is exactly one error and it is in the fifth bit. Therefore e = and c = r + e = + =, and w =. v. r =. Hr t = = Since Hr t is not the zero vector, there are errors. Since Hr t is equal to the fourth column of H, we will assume that there is exactly one error and it is in the fourth bit. Therefore e = and c = r + e = + =, and w =. vi. r =. Hr t = = Since Hr t is not the zero vector, there are errors. Since Hr t is not equal to any column of H, there are at least two errors and we cannot correct them. c and w can be and, and, or and, respectively. vii. r =. Hr t = = Since Hr t is not the zero vector, there are errors. Since Hr t is equal to the fourth column of H, we will assume that there is exactly one error and it is in the fourth bit. Therefore e = and c = r + e = + =, and w =. vi. r =. Hr t = = Since Hr t is not the zero vector, there are errors. Since Hr t is not equal to any column of H, there are at least two errors and we cannot correct them. c and w can be and, and or and, respectively. 3

4 a. Since H = if we write H = [B I 3 ] then the generating matrix G is [I 3 B t ]. G = The set of code words C is {,,,,,,, }. b. This code does not correct all single errors. For instance, if r = received, then the sent code word can be c = with error e =, or c = with error e =. This can also be seen from the parity-check matrix H. Since the second and the fifth columns of H are the same, a single error in the second bit and a single error in the fifth bit cannot be told apart a. Let G = [ ]. Then, since, E() = and [ ] G = [ ], and E() = and [ ] G = [ ], G is the generating matrix for the (9,) nine-times repetition code. b. G = [I 3 A] where A = [ ]. Therefore, the parity-check matrix is H = [A t I 8 ]. That is, H = Let M(n,k) Z n 2 contain the maximum number of code words among all codes of length n with minimum distance 2k +. For any x Z n 2, consider S(x,k), the sphere of radius k and center x. This set consists of all y Z n 2 that differ from x at at most k bits. Therefore it contains k i= i) elements. Since the distance between any two elements of M(n,k) is at least 2k +, if c,c 2 are distinct elements of M(n,k) then the spheres M(c,k) and S(c 2,k) are disjoint. Therefore, M(n,k) k i= i) 2 n, or in other words, M(n,k) 2 n k i= i). This is called the Hamming bound. 4

5 On the other hand, if there is an element x Z n 2 whose distance to every c M(n,k) is at least 2k +, then we can add it to the code to obtain a code whose minimum distance is still 2k +, but which has more elements than M(n,k). This contradicts the maximality of the number of elements of M(n,k). Therefore, every x is contained in S(c, 2k) for some c M(n,k). Since S(c, 2k) = 2k i= i other words, This is called the Gilbert bound. ), this time we have 2 n M(n,k) 2k i= i), or in 2 n 2k ) M(n,k). i= i The parity-check matrix is. We use the decoding table in Table 6.9 of the textbook. a. The syndrome of r = is Hr t = =, in other words. The coset leader for this syndrome is e =. So the code word is c = r + e =, and the message word is w =. The syndrome of r = is Hr t = =, in other words. The coset leader for this syndrome is e =. So the code word is c = r + e =, and the message word is w =. The syndrome of r = is Hr t = =, in other words. The coset leader for this syndrome is e =. So the code word is c = r + e =, and the message word is w =. 5

6 The syndrome of r = is Hr t = =, in other words. The coset leader for this syndrome is e =. So the code word is c = r+e =, and the message word is w =. (However, note that now we are using the eight row of the table. For this row, the choice of coset leader was not unique, because the code cannot correct double errors. In a table which uses a different coset leader for this row, this received message would be corrected differently.) The syndrome of r = is Hr t = =, in other words. The coset leader for this syndrome is e =. So the code word is c = r + e =, and the message word is w =. The syndrome of r = is Hr t = =, in other words. The coset leader for this syndrome is e =. This means there is no error, and the code word is c = r + e =, and the message word is w =. The syndrome of r = is Hr t = =, in other words. The coset leader for this syndrome is e =. So the code word is c = r + e =, and the message word is w =. 6

7 The syndrome of r = is Hr t = =, in other words. The coset leader for this syndrome is e =. So the code word is c = r + e =, and the message word is w =. b. The coset leader for the coset {,,,,,,, } is not unique. e = and e = are the two possibilities. Table 6.9 chooses the first one, and this causes r = to be corrected as c =, and decoded as w =. If we make a different table which chooses e = as the coset leader, then the error pattern for r = will be e =, and therefore c = r + e = and w = a. The parity-check matrix for the (7,4) Hamming code is H = If we let H = [B I 3 ], then the generating matrix for the (7,4) Hamming code is G = [I 4 B t ]. G = Therefore, the messages,,,,, are encoded as,,,,,, respectively. b. If r =, then Hr t = [] t, and this is the fifth column of H. Therefore, e =, c = r + e =, and w =. If r =, then Hr t = [] t, and this is the fourth column of H. Therefore, e =, c = r + e = and w =. If r =, then Hr t = [] t, and this is the fifth column of H. Therefore, e =, c = r + e = and w =. If r =, then Hr t = [] t, and this is the first column of H. Therefore, e =, c = r + e = and w =. c. The decoding table is as follows: 7

8 Hr t e d. The answers are the same. For instance, if r =, the syndrome Hr t = gives the coset leader is e =. Then the code word is c = r+e =, and the message word is w = The rate of the (3,) triple repetition code is /3.33. The rate of the (7,4) Hamming code is 4/

Linear Codes and Syndrome Decoding

Linear Codes and Syndrome Decoding These notes are intended to be used as supplementary reading to Sections 6.7 9 of Grimaldi s Discrete and Combinatorial Mathematics. The proofs of the theorems are left