Coding Theory Pythagorean Codes Massoud Malek. By looking at the legs of a right-angle triangle (a, b, c), Pythagoras gave us the first theorem:

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1 By looking at the legs of a right-angle triangle (a, b, c), Pythagoras gave us the first theorem: a 2 + b 2 = c 2 The leg c, is called the hypotenuse. If no other number except one divides both a and b, then the triangle is called primitive Pythagorean triangle or just PPT. The smallest and best-known PPT is (3, 4, 5). Ancient clay tablets from Babylonia indicate that more than 1000 years before Pythagoras, the Babylonians demonstrated an awareness of at least 15 triangles. In ancient Egypt, the (3, 4, 5) triangle was used in constructions of pyramids and temples. Plato (380 B. C.) is attributed with the formula: <1, n> = (n 2 1, 2 n, n 2 + 1), for n > 1. A more general formula for obtaining all triples was given by Euclid in c. 300 B.C. Lemma 1. For positive integers m and n, where 1 m < n, the triple <m, n> = (n 2 m 2, 2 m n, n 2 + m 2 ) is Pythagorean. Moreover, if m and n are relatively prime of opposite parity, then they generate a primitive triple. Basic Properties. In a primitive Pythagorean triangle T = (a, b, c), the following conditions hold: The hypotenuse and one of the legs are always odd and the other leg is divisible by 4. Exactly one of the legs is divisible by 3 and also, exactly one of the three sides is divisible by 5. One leg may be divisible by both 3 and 5, as in (15, 8, 17) or (11, 60, 61). The largest number that always divides a b c is 60. All prime factors of c are of the form 4 n + 1. The hypotenuse of every PPT exceeds the odd leg by twice the square of an integer r and the even leg by the square of an odd integer s, such that c = a + 2 r 2 = b + s 2. Generating Matrices. A matrix is a generating matrix, if its product by some vector, generates a Pythagorean triple. In 1934, Berggren defined the following matrices: A = 2 1 2, B = 2 1 2, and C = ; and proved that T = (a, b, c) is primitive, if and only if T = (a, b, c) = (3, 4, 5) M, California State University, East Bay 1

2 where the generating matrix M, is a finite product of the above matrices Consider the matrix G = and the vector u [m, n] = ( m 2, m n, n 2), where m and n are positive integers. Then u [m, n] G = ( m 2, m n, n 2) = ( m 2 +2 m n, 2 m n + 2 n 2, m 2 +2 m n+2 n 2) = ( (m + n) 2 n 2, 2 (m + n) n, (m + n) 2 + n 2) = <n, m + n>. If n and m + n are relatively prime of opposite parity, then by Lemma 1, u [m, n] G is a primitive Pythagorean triple. Lemma 2. Let T = <m, n> be a PTT. Then T = u [n m, m] G Proof. We have u [n m, m] G = <m, n m + m> = <m, n>= T. Theorem 1. For fixed positive integers m 0 and n 0, where m 0 is odd, define the following sets of PPTs: R m0 = {T n = (a n, b n, c n ) = u[m 0, n] G : n = 1, 2, 3,..., gcd (m 0, n) = 1}, and S n0 = {T m = (a m, b m, c m ) = u[m, n 0 ] G : m = 1, 3, 5,..., gcd (m, n 0 ) = 1}. Then (i) P = R 2 i+1 = S j ; i=0 j=1 (ii) R m0 S n0 = u[m 0, n 0 ] G, whenever gcd(m 0, n 0 ) = 1; (iii) for all T n R m0, c n b n = m 2 0 ; (iv) for all T m S n0, c m a m = 2 n 2 0. (v) for i 1 i 2 and j 1 j 2, R 2 i1 +1 R 2 i2 +1 = and S j1 S j2 =. Proof. (i) follows from Lemma 2. (ii) If gcd (m 0, n 0 ) = 1, then R m0 S n0 contains only one PPT, which is u[m 0, n 0 ] G. From the definitions of u [m, n] G, we obtain T = (a, b, c) = u [m, n] G = ( m m n, 2 m n + 2 n 2, m m n + 2 n 2). For the remainder of the proof, notice that c b = m 2 and c a = 2 n 2. Thus for the same a, b, and c ; m and n must be unique. Hence for i 1 i 2 and j 1 j 2, R 2 i1 +1 R 2 i2 +1 = and S 2 j1 S 2 j2 =. California State University, East Bay 2

3 From the fact that u [m, n] G = <n, m + n>, we conclude that R 1 = { T n = (a n, b n, c n ) = u [1, n] G = ( n, 2 n + 2 n 2, n + 2 n 2) : n = 1, 2, 3,... } is the set of all PPTs generated by <n, n + 1> (two consecutive positive integers) and S 1 = { T m = (a m, b m, c m ) = u [m, 1] G = ( m 2 +2 m, 2 m + 2, m 2 +2 m + 2 ) : m = 1, 3, 5,... } is the set of all PPTs generated by <1, m + 1> (Plato s formula). Theorem 2. Let T = (a, b, c) be a PPT. [ (i) If T R m0 ; then T = u m 0, a m2 0 2 m 0 [ (ii) If T S n0 ; then T = u n 0 + ] ( G = a, a2 m m 2 0 n a, n 0 ] G = Proof. (i) If T R m0 ; then for some n with gcd (m 0, n) = 1, )., a2 + m m 2 0 ( a, 2 n 0 n a, a + 2 n2 0 T = (a, b, c) = u[m 0, n] G = ( m m 0 n, 2 n (m 0 + n), m m 0 n + 2 n 2 ) ). ( ) a m From a = m m 0 n, we obtain n = 0. Hence 2 m 0 [ T = u m 0, a ] ( m2 0 G = a, a2 m m 0 2 m 2, a2 + m 4 ) m 2. 0 (ii) If T S n0 ; then for some odd m with gcd (m, n 0 ) = 1, T = (a, b, c) = u[m, n 0 ] G = ( m m n 0, 2 n 0 (m + n 0 ), m n 0 (m + n 0 ) ). By solving the quadratic equation a = m m n 0 for a positive m, we obtain ( ) m = n 0 + n 20 + a. Hence [ ] ( ) T = u n 0 + n a, n 0 G = a, 2 n 0 n a, a + 2 n2 0. ). Generating Pythagorean triples with Sequence of Numbers. The Fibonacci numbers are the numbers in the following integer sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,... The sequence F n of Fibonacci numbers is defined by the recurrence relation: F n = F n 1 + F n 2, with seed values F 0 = 0 and F 1 = 1. California State University, East Bay 3

4 The Lucas numbers are the numbers in the following integer sequence: 2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 121, 197, 318,... The sequence of Lucas numbers L n, uses exactly the same recurrence relation as the sequence of Fibonacci numbers F n, but with the seed values L 0 = 2 and L 1 = 1. Clearly any increasing sequence of positive integers may produce Pythagorean triples. Note that F 3 n and L 3 n are even, so the Pythagorean triples < F 3 n+1, F 3 n+2 > and < L 3 n+1, L 3 n+2 > are not primitive. A generalization of the Fibonacci numbers defined by the linear recurrence equation: H n [r, s] = r H n 2 [r, s] + s H n 1 [r, s] with the seed values H 0 [r, s] and H 1 [r, s] is called Horadam sequence. In all that follows, r, s, and the seed values are chosen in a way that for n 1, they generate increasing sequences of positive integers. Note that only H 0 [r, s] may be zero. For r = s = 1, and the seed values H 0 [1, 1] = 0, and H 1 [1, 1] = 1 ; the Horadam sequence reduces to the Fibonacci numbers. By choosing H 0 [1, 1] = 2, instead of H 0 [1, 1] = 0, we obtain the Lucas numbers. Note that consecutive numbers in both Haradam sequences: H n [1, 2] : H n [2, 1] : are relatively prime odd numbers and none of the Pythagorean triples, generated by <H n [1, 2], H n+1 [1, 2]> and <H n [2, 1], H n+1 [2, 1]> are primitive, but every single Pythagorean triple, generated by u [ H n [1, 2], H n+1 [1, 2] ] G and u [ H n [2, 1], H n+1 [2, 1] ] G is primitive. So as long as we chose a sequence of odd numbers, not necessary increasing, where two consecutive numbers are relatively prime, then by using the matrix G, we may generate PPTs with any two consecutive members of that sequence. Let T n = (a n, b n, c n ) be a Pythagorean triangle: If the two non-hypotenuse legs of T n differ by 1, then T n = (3, 4, 5) A n. Note that if n is odd then b n = a n 1 and if n is even, then b n = a n + 1. If the hypotenuse and the odd leg of T n differ by 2, then T n = (3, 4, 5) B n. If the hypotenuse and the even leg of T n differ by 1, then T n = (3, 4, 5) C n. California State University, East Bay 4

5 Application to Coding Theory and Cryptography. Primitive Pythagorean triple can lend itself to Coding Theory and Cryptography. There are infinitely many ways to encode letters of alphabet with PPTs. Here is a simple way to create a Pythagorean alphabet in Coding Theory, by the use of a Horadam sequence and PPTs: First, choose a Horadam sequence such as H n [1, 2] : Then assign a number in that sequence to each letter. Suppose H 6 [1, 2] = 99, is chosen for the letter F, which is the sixth letter of the alphabet. Then assign to F, a codeword, as follows: u[ 99, 239 ] G = ( ), so F. Another way of defining a class of codewords, is by assigning a positive integer to each letter of the alphabet; then multiply the PPT (3, 4, 5) by the matrix A, B, or C. If we assign 6 to the sixth letter of the alphabet, F ; Then using the matrices A, B, or C, we obtain: (3, 4, 5) A 6 = ( ), so F ; (3, 4, 5) B 6 = ( ), so F ; (3, 4, 5) C 6 = ( ), so F. Finally, we may define a class of codewords by selecting either R m0 or S n0. Suppose, we choose R m0, where m 0 is a prime odd number; then we assign a positive integer k with gcd (m 0, k) = 1 to each letter of the alphabet. Then, we create a Pythagorean alphabet using the generating matrix G. Note that PPTs in corollaries 4.5 and 4.6 may be used if we choose R 1. From the fact that the last digit of any integer, could not be different from 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 ; we conclude that any set of PPTs may be partitioned into a finite number of disjoint classes, based on the last digits of a, b, and c, of T = (a, b, c). Here are the first 12 PPTs T k = (a k, b k, c k ), generated by (3, 4, 5) A k, where k = 0, 1, 2, 3,..., 11 : Since there are spaces between words, we may assign the PTT (3, 4, 5) to a blank space. According to the above table, the set of T n is partitioned into six disjoint classes based on the last digits of a k, b k, and c k, with coset leaders: (3, 4, 5), (1, 0, 9), (9, 0, 9), (7, 6, 5), (9, 0, 1), and (1, 0, 1). California State University, East Bay 5

6 Here are the first 27 PPTs T k = (a k, b k, c k ), generated by (3, 4, 5) B k, where k = 0, 1, 2, 3,..., 26 : (3, 4, 5) (5, 8, 7) (5, 2, 7) (3, 6, 5) (9, 0, 1) A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Notice that the odd legs and the hypothenuses differ by two.these triples are also in S 1. Here are the first 26 PPTs T k = (a k, b k, c k ), generated by (3, 4, 5) C k, where k = 0, 1, 2, 3,..., 26 : (3, 4, 5) (5, 2, 3) (7, 4, 5) (9, 0, 1) (1, 0, 1) A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Notice that the even legs and the hypothenuses differ by one. These triples are also in R 1. Error Detection The fact that the matrix A (resp. B and C) partitions the set of PPTs generated by (3, 4, 5) and A n (resp. B n and C n ) into six (resp. 5) disjoint classes; the received codeword must be in one of those six (resp. 5) classes. If the last digits of a 6, b 6, and c 6 do not match the digits of any of the coset leaders, errors in the codeword is detected. Once the error is detected, it could eventually be corrected either directly or by a retransmission. Although coset leader decoding detects some error in the case of transmitting a codeword assigned to F, when A 6 is used; one could improve the error detection by adding 345 as parity-check digits at the beginning or the end of the codeword in the following ways: (3, 4, 5) A or F. As long as the parity-check digits are not corrupted, the message could be decoded by those digits. Similarly, we could use coset leaders as parity-check digits for codewords obtained from B 6 and C 6 : (3, 4, 5) B or F ; (3, 4, 5) C or F. In order to increase the probability of correcting errors and decreasing the number of retransmission; California State University, East Bay 6

7 one could choose k s with large gaps between them. For example, if 1 is chosen to create a primitive triple for the letter A, then instead of choosing 2 for the letter B and 3 for the letter C, one could choose 12 for B and 13 for C; and so on. In this case, each class contains triples with large distance between them. This way, it is most likely that the closest codeword to the received word is the word sent. Parity-check Digits. Let v = ( a, b, c ) be a PTT. Then define v = ( â, b, ĉ ) ( a, b, c ) mod 10. If v is a message, then v = ( â, b, ĉ ) must be a cost leader. One way to define a Pythagorean code is to define the codeword as follows: w = ( a, b, c, â, b, ĉ ). Clearly the last three digits are the parity-check digits. Data Compression. Sometimes, it is necessary to shorten the length of the codewords without compromising the integrity of the message. For example, to shorten the PTT assigned to F, using (3, 4, 5) C 6 ; one could remove the hypothenuse or the even leg; since the even leg and the hypothenuse differ only by one. So the shorter message will be F or F. Now the codeword including the parity-check digits become or Since in all the triples generated by (3, 4, 5) and a power of B, the odd leg and the hypothenuse differ by two; the codeword could be shorten to F. In the case of the matrix A, although the odd leg and even leg differ by one, but for an even n the even leg is larger and for the odd n the odd leg is larger. To avoid confusion, one could assign PTTs to the letters of the alphabet, by using only odd powers or even powers of A. If we select R m0 which contains PPTs, where the hypotenuses exceed the even legs by m 2 0 or S n 0 which contains PPTs, where the hypotenuses exceeds the odd legs by 2 n 2 0. Then removing the hypotenuse of the encoded word in these sets, would not alter the message and its integrity. According to Theorem 2, from the odd leg a, of the PTT, T = ( a, b, c ) R m0, we may obtain the even leg b = a2 m m 0. Thus we could also remove the even leg b. Hence, we only need the odd leg. Suppose R 7 = {T n = (a n, b n, c n ) = u[7, n] G : n = 1, 2,..., 6, 8, 9,..., 13, 15, 15,..., 20, 22,..., 27, 29} is used to generate the alphabet. Notice that n must be different from 7, 14, 21, and 28 which are multiples of 7. The fact that c b = 49 = 7 2, for all c s and b s implies that each message could be shorten, by removing the hypotenuse. According to Theorem 2, from the odd leg a, we may obtain the value of California State University, East Bay 7

8 the even leg b = a Thus we could also remove the even leg b. 14 In order to be able to detect some transmission errors, we should add to the codeword, the last two digits of the coset leader of the class which contains the word. Here is the short form of the alphabet ( n = 1, 2,..., 26) in R 7 using only he odd leg and the paritycheck digits: R 7 (3, 6, 5) (7, 6, 5) (1, 0, 9) (5, 8, 7) (9, 0, 9) A B C D E F K G Q I J O L H M N S P Q R W X T U V Y Z We assign the PTT (3, 4, 5) to a blank space. Here are the short form of alphabets in S 1 and R 1, respectively, without using the hypothenuse: S 1 (3, 4, 5) (5, 8, 7) (5, 2, 7) (3, 6, 5) (9, 0, 1) 3 4 A 15 8 B C D E F G H I J K L M N O P Q R S T U V W X Y Z R 1 (3, 4, 5) (5, 2, 3) (7, 4, 5) (9, 0, 1) (1, 0, 1) 3 4 A 5 12 B 7 24 C 9 40 D E F G H I J K L M N O P Q R S T U V W X Y Z Error Correction and Decoding. We assume that R 7 with the coset leaders (3, 4, 5), (5, 2, 3), (7, 4, 5), (9, 0, 1), and (1, 0, 1) and two-digit parity check was used. Suppose the message w 1 = and w 2 = were received. Since (3, 6, 7) and (9, 7, 5) are not coset leaders; but is in our table, generated by u[7, 6] G, representing the letter F ; and is very close to generated by u[7, 22] G, representing the letter S. We therefore conclude that it is most likely that v 1 = and v 2 = are the transmitted codewords. California State University, East Bay 8

9 In order to increase the probability of correcting errors or to increase the security, one could choose three different classes of PTT, such as a combination of : R m1, R m2, R m3, S n1, S n2, and S n3 ; then assign for example, three PPTs to each letter in the following way: Choose the odd leg of the PPT from the first class, the even leg of the second PPT from the second class, and the hypotenuse from the third PPT from the third class. This way, the security is increased and if one of the legs is error-free, then the codeword may be identified. Exercises. 1. Suppose the following words using odd legs and parity-check digits were received: Decode the message, assuming that at most one error occurred in each received word. 2. The word is known to be a codeword. The messages were encoded using the odd legs and parity-check digits. Identify and correct (if possible) the following compressed received words: The word [ ] is known to be a codeword. The messages were encoded using the even legs and parity-check digits. Identify and correct (if possible) the following compressed received words: MATLAB. G = [ ; ; ] ; inputm = What is the value of m? ; m = input (inputm) for n = 1 : 27, vm = [ m 2, m n, n 2 ] ; wm = vm G, end ; inputn= What is the value of n? ; n = input (inputn) for m = 1 : 2 : 27, vn = [ m 2, m n, n 2 ] wn = vn G, end ; A = [ ; ; ] ; B = [ ; 2 1 2; ] ; C = [ ; ; ] ; u = [ ] ; w1 = u ; w2 = u ; w3 = u ; inputpa = Choose an positive integer pa less than 11, to generate PPTs with the matrix A ; pa = input (inputpa) for k = 1 : pa, w1 (k, :) = u A k ; end ; [ w1 ] inputpb = Choose an positive integer pb, to generate PPTs with the matrix B ; pb = input (inputpb) for k = 1 : pb, w2 (k, :) = u B k ; end ; [ w2 ], inputpc = Choose an positive integer pc, to generate PPTs with the matrix C ; pc = input (inputpc) for k = 1 : pc, w3 (k, :) = u C k ; end ; [ w3 ], California State University, East Bay 9