A Relation Between Weight Enumerating Function and Number of Full Ran Sub-matrices Mahesh Babu Vaddi and B Sundar Rajan Department of Electrical Communication Engineering, Indian Institute of Science, Bengaluru 56001, KA, India E-mail: {vaddi, bsrajan}@iiscacin arxiv:19010571v1 [csit] 17 Jan 019 Abstract In most of the networ coding problems with messages, the existence of binary networ coding solution over F depends on the existence of adequate sets of -dimensional binary vectors such that each set comprises of linearly independent ( vectors In a given n (n binary matrix, there exist n binary sub-matrices of size Every possible submatrix may be of full ran or singular depending on the columns present in the matrix In this wor, for full ran binary matrix G of size n satisfying certain condition on minimum Hamming weight, we establish a relation between the number of full ran sub-matrices of size and the weight enumerating function of the error correcting code with G as the generator matrix We give an algorithm to compute the number of full ran submatrices I INTRODUCTION Networ coding is a technique to enhance the rate of information transmission in a networ by taing advantage of redundancy in demands of various receivers Networ coding was introduced in [1] An acyclic networ can be represented as an acyclic directed graph D = (V,E, where V is the set of all nodes and E is the set of all edges in the networ Every edge E in D can carry one message symbol from the given finite field F q There exists a unique node S in D, called the source node, the source node has message symbols x i F q for i [1 : ] There exist some receiver nodes in the networ and each receiver wants some subset of message symbols {x 1,x,,x } The objective in networ coding is to satisfy the demands of all receivers by minimizing the number of transmissions in the edges of the networ A solution to the networ coding problem is the design of -dimensional precoding vectors (these precoding vectors are called global encoding ernels in networ coding to each edge in the networ such that these precoding vectors satisfy some linear independence constraints An -dimensional linear networ code on an acyclic networ over a field F q consists of a global encoding mapping f ni,n j : F q F q for each edge (n i,n j E in the networ The existence of linear networ coding solutions was studied in [] and [3] A construction of linear networ coding for multicast networ coding problems was studied in [4] Depending on the networ topology and requirement of the receivers, a networ may or may not have a solution in F In most of the networ coding problems, the existence of binary networ coding solution over F depends on the existence of adequate sets of -dimensional binary vectors such that each set comprises of linearly independent vectors The problem of index coding with side-information was introduced by Bir and Kol [5] In index coding problems, we often require the condition that a specific number of submatrices in a given n matrix needs to have full ran In index coding, the default choice for this n matrix is an n Vandermonde matrix The Vandermonde matrix exists over higher fields and the field size required depends on n Hence, it is useful to design binary n matrices satisfying the given linear independence conditions A Motivating Example Consider the networ coding problem as shown in Fig 1 In this networ coding problem, the source comprises of two messages and there exist six receivers, each of the receiver wants both the messages For this networ, linear solution over F is not possible [6] This is nown as the ( 4 combinational networ However, for the networ coding problem shown in Fig 1, if we remove any one receiver out of six receivers, then linear solution over F is possible as shown in Table I This follows from the fact that for any binary matrix of size 4, there exist atmost 5 full ran submatrices of size and this also ( means that binary coding solution does not exist for the 4 combinational networ Hence, the number of full ran submatrices in a binary matrix of size n is useful in analysing some multicast networ coding problems u 1 u 3 u x 1 x S R 1 R R 3 R 4 R 6 {x 1,x } {x 1,x } {x 1,x } {x 1,x } {x 1,x } {x 1,x } Fig 1: An instance of networ coding problem with no linear binary solution A linear (n, error correcting code over the field F q is a -dimensional subspace of F n q Let C denote the (n, linear u 4 R 5
Receivers Present R 1,R,R 3,R 4,R 5 R 1,R,R 3,R 4,R 6 R 1,R,R 3,R 5,R 6 R 1,R,R 4,R 5,R 6 R 1,R 3,R 4,R 5,R 6 R,R 3,R 4,R 5,R 1 Global encoding ernels for Out(S f s,u1 = [1 0] T,f s,u = [0 1] T f s,u3 = [1 1] T,f s,u4 = [1 1] T f s,u1 = [1 0] T,f s,u = [1 1] T f s,u3 = [0 1] T,f s,u4 = [1 1] T f s,u1 = [1 0] T,f s,u = [1 1] T f s,u3 = [1 1] T,f s,u4 = [0 1] T f s,u1 = [1 1] T,f s,u = [1 0] T f s,u3 = [0 1] T,f s,u4 = [1 1] T f s,u1 = [1 1] T,f s,u = [1 0] T f s,u3 = [1 1] T,f s,u4 = [0 1] T f s,u1 = [1 1] T,f s,u = [1 1] T f s,u3 = [1 0] T,f s,u4 = [0 1] T TABLE I: Global encoding ernels at Out(S for the six different networ coding problems obtained after removing one receiver R i for i [1 : 6] in Fig 1 Global encoding ernels for the edges in Out(u i is equal tof s,ui fori [1 : 4] error correcting code Let G and H be the generator and parity chec matrix of C The weight enumerating function of C is given by the polynomial n W C (x,y = A d x n d y d, d=0 where A d is the number of codewords with Hamming weight d and x,y are indeterminates We refer A d s for d [0 : n] as weight enumerator coefficients (WECs The linear bloc code generated by H as generator matrix is called dual of C and is denoted byc T The relation between weight enumerator W C (x,y of C and weight enumerator W C T(x,y of C T was derived by MacWilliams in [7] W C (x,y and W C T(x,y are related as W C T(x,y = 1 C W C(x+y,x y (1 For more details on error correcting codes and details on weight enumerators, the readers are referred to [8] Given a subspace V of F n, the space of all vectors orthogonal to V in F n is called orthogonal complement of V For any given arbitrary full ran matrix G of size n, from the fundamental theorem of linear algebra, the null space is the orthogonal complement of the row space The dimension of the row space of G is, the dimension of null space of G is and the sum of the dimension of the null space and row space is equal to n For any given arbitrary full ran matrix A of size n, we refer B of size ( n as orthogonal complement of A if the ran of B is and AB T is all zero matrix Let G be a full ran n binary matrix Let H be an orthogonal complement of G If n, let d be the minimum Hamming weight of the error correcting code generated byh Else, letd be the minimum Hamming weight of the error correcting code generated by G B Contributions The contributions of this paper are summarized below: For a given binary full ran n matrix G, we show that the number of full ran submatrices in G are equal to the number of full ran ( ( submatrices in the orthogonal complement of G In any binary full ran matrix G of size n, if 3d > max(,, we establish a relation between the weight enumerating function W C (x,y of C generated by G and the number of full ran submatrices of size ing We explicitly quantify the number of full ran submatrices of size in G For a binary full ran matrix G of size n satisfying the condition 3d > max(,, we give an algorithm to compute the number of full ran submatrices of size in G II ON WEIGHT ENUMERATOR FUNCTION AND THE NUMBER OF FULL-RANK MATRICES Lemma 1 and Lemma given below are useful to derive the main result in this section Lemma 1 Let G be a binary n matrix Let G (s be the n symmetric matrix obtained from G by using elementary row operations The matrix G (s is of the form [I : P ( ] If the columns corresponding to the indices {i 1,i,,i } [1 : n] are linearly dependent in G, then the columns corresponding to the indices {i 1,i,,i } are linearly dependent in G (s also Proof Let r i and r i be the ith row of G andg (s respectively for every i [1 : ] Let c j and c j be the jth column of G and G (s respectively for every j [1 : n] Let g i,j and g i,j be the element in the ith row and jth column of G and G (s respectively for every i [1 : ] and j [1 : n] In the matrix G (s, let the ith row be a linear combination of rows in the matrix G We can write r i = f i (r 1,r,,r for i [1 : ], ( where f i is some linear function over F From (, we can write g i,j as g i,j = f i (g 1,j,g,j,,g,j for i [1 : ] and j [1 : n] (3 Given that the columns corresponding to the indices {i 1,i,,i } [1 : n] are linearly dependent in G There exists a 1,a,,a F such that From (4, we have a 1 c i 1 +a c i ++a c i = 0 (4 a 1 g 1,i 1 +a g 1,i ++a g 1,i = 0 a 1 g,i 1 +a g,i ++a g,i = 0 a 1 g,i 1 +a g,i ++a g,i = 0 (5
From (5, for every i [1 : ], we have From (6, we have a 1 f i (g 1,i 1,g,i 1,,g,i 1 + a f i (g 1,i,g,i,,g,i ++ a f i (g 1,i,g,i,,g,i = 0 (6 a 1 g i,i1 +a g i,i ++a g i,i = 0 for every i [1 : ] (7 From (7, we have a 1 c i1 +a c i ++a c i = 0 That is, the columns c i1,c i,,c i of G (s are linearly dependent Lemma Let G be a binary n matrix of the form [I : P ( ] Define the ( n matrix H as given below H = [P T ( : I ( (] If {i 1,i,,i } [1 : n] be indices of any linearly dependent columns in G, then the ( columns corresponding to the indices[1 : n]/{i 1,i,,i } inhare linearly dependent Proof We have GH T = [I : P ( ][P T ( : I ( (] T = P ( +P ( = 0 ( The ran of the matrices G and H are and n respectively and every row in G is orthogonal (inner product is zero to every other row in H over F Hence, the row space spanned by G and H are orthogonal complements Let {i 1,i,,i } [1 : n] be indices of any linearly dependent columns in G Consider a binary vector v F n such that v has 1s in i 1,i,,i positions and 0 in other positions The vectorv must be in the null space ofgand row space of H Consider a binary vectorṽ F n such that it has 0 in i 1,i,,i positions and 1 in other positions The vector ṽ must be in the row space of G (row space of G and H are orthogonal complements Hence, we have Hṽ = 0 and the columns corresponding to the indices [1 : n]/{i 1,i,,i } in H are linearly dependent Corollary 1 Let G be a binary n matrix of the form [I : P ( ] Define the ( n matrix H as given below H = [P T ( : I ( (] If {i 1,i,,i } [1 : n] be indices of any linearly independent columns in G, then the (n columns corresponding to the indices [1 : n]/{i 1,i,,i } in H are linearly independent Remar 1 The number of linearly independent submatrices of size in G is equal to the number of linearly independent submatrices of size ( ( in H Example 1 Consider the matrices G and H given below 1 0 0 0 1 1 1 0 1 0 0 1 1 0 G = 0 0 1 0 1 0 1,H = 0 0 0 1 0 1 1 [ ] 1 1 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 1 0 0 1 The columns with indices {1,, 3, 5} are linearly dependent in G Hence, the columns [1 : 7]\{1,,3,5} = {4,6,7} are linearly dependent in H The columns with indices {1,4,6,7} are linearly independent in G Hence, the columns [1 : 7]\{1,4,6,7}= {,3,5} are linearly independent in H Definition 1 Let G = [I : P ( ] be a binary full ran matrix of size n We can select any set of t columns for 1 t from the columns of P ( in 1 ways Let {c i1,c i,,c it } be the t selected columns in P ( The effective distance d (e (i 1,i,,i t of these t columns is defined as d (e (i 1,i,,i t = d H (c i1 +c i ++c it +t, where d H (c i1 + c i + + c it is the Hamming weight of ( dimensional binary vector c i1 +c i ++c it Example Consider the matrix G given below [ ] 1 0 0 1 1 1 G = 0 1 0 1 1 0 0 0 1 1 0 1 The effective distances of the matrix G are as follows: d (e (4 = d H (111+1 = 4, d (e (5 = d H (110+1 = 3, d (e (6 = d H (101+1 = 3, d (e (4,5 = d H (111+110+ = 3, d (e (4,6 = d H (111+101+ = 3, d (e (5,6 = d H (101+110+ = 4, d (e (4,5,6 = d H (111+110+101+3 = 4 Theorem 1 establishes the relation between weight enumerating function and the number of singular matrices of a full ran binary matrix of size n Theorem 1 Consider an arbitrary binary full ran matrix G of size n (n Let C be the linear bloc code generated by G as generator matrix and C T be the dual code of C Let D be the number of singular submatrices of size in G If 3d > max(,, then D =, (8 where d=d A d A d s are the weight enumerating coefficients of W C T(x,y if A d s are the weight enumerating coefficients of W C (x,y if <
Proof Let G (s = [I P ( ] be the systematic generator matrix for C From Lemma 1, the number of linearly dependent -column sets in G is equal to the number of linearly dependent -column sets in G (s Let H = [P T I ( (] be the parity chec matrix for C Let c j be the jth column of G (s for j [1 : n] Case (i :- For every {i 1,i,,i t } [ + 1 : n], let j 1,j,, j dh(c i1 +c i ++c it be the positions of 1s present in c i1 + c i + +c it Note that c j1,c j,,c jdh (c i1 +c i ++c it are the columns of identity matrix present in G (s with 1s in j 1,j,, j dh(c i1 +c i ++c it positions respectively Hence, in the matrix G, the d (e (i 1,i,,i t number of columns {c i1,c i,,c it } {c j1,c j,,c }{{} } jdh (c i1 +c i +,c it }{{} t columns d H(c i1 +c i ++c it are linearly dependent If d (e (i 1,i,,i t, then, any combination of d (e (i 1,i,,i t columns in the remaining n d (e (i 1,i,,i t along with the d (e (i 1,i,,i t columns {c i1,c i,,c it } {c j1,c j,,c jdh (c i1 +c i +,c it }of G are linearly dependent Hence, there exists ( n d (e (i 1,i,,i d (e (i 1,i,,i t sets of size, which are linearly dependent and which comprise {c i1,c i,,c it } The total number of linearly dependent sets of -columns are ( n d (e (i 1,i,,i t D = d (e (i 1,i,,i t {i 1,i,,i t} [+1:n] ( n d (e (i 1,i,,i t = (9 We have {i 1,i,,i t} [+1:n] d (e (i 1,i,,i t = d H (c i1 +c i ++c it +t = d H (c i1 +c i ++c it +d H (sum of any t columns of I = Hamming weight of codeword generated by i 1,i,,i t }{{} [1:] By using (10, we can write (9 as ( n d (e (i 1,i,,i t D = = {i 1,i,,i t} [+1:n] n d=d A d = d=d A d rows of H (10, (11 where A d s are the weight enumerating coefficients of C T The last equality in (11 follows from the fact that the codewords with weight from +1 to n in H do not yield any set of columns which are linearly dependent Now, if 3d > max(, =, we would show that there is no double counting in (9 From (10, the minimum value 1 3 4 5 6 7 8 1 i 1 i i 3 i 4 i d1 Fig 1 3 4 d 1 3 4 5 6 7 8 1 Fig 3 of d (e (i 1,i,,i t for any {i 1,i,,i t } [ + 1 : n] is d Consider a codeword of H with Hamming weight d 1 d and having 1s in i 1,i,,i d1 positions Hence, the i 1,i,,i d1 th columns with any combination of ( d 1 columns in the remaining (n d 1 columns of G are linearly dependent There exists ( n d 1 d 1 sets of size, which are linearly dependent and which comprisei 1,i,,i d1 columns Let these ( n d 1 d 1 sets of size columns be S1 Consider any other codeword of H with Hamming weight d d and having 1s in If 1,,, d positions Hence, the 1,,, d th columns with any combination of ( d columns in the remaining (n d columns of G are linearly dependent There exists ( n d d sets of size, which are linearly dependent and which comprisei 1,i,,i d columns Let these ( n d d sets of size columns be S We need to show all the column sets in S 1 are distinct to the columns sets in S Let there exists a column set which is common in both S 1 and S With out loss of generality, we assume that this common set is as shown in Figure and Figure 3 Because the indices in Figure and Figure 3 are equal, the d 1 vacant positions in Figure should have indices from the set { 1,,, d } which are not present in the set {i 1,i,,i d1 } Similarly, The d vacant positions in Figure 3 should have indices from the set {i 1,i,,i d1 } which are not present in the set { 1,,, d } But, we have {i 1,i,,i d1 } and { 1,,, d } differ in atleast d indices (d is the minimum Hamming distance of H Hence, we have This implies ( d 1 +( d d d +d 1 +d 3d This is a contradiction because we assumed 3d > Hence, there exists no column set which are common in both S 1 and S Case (ii < :- In Lemma, we proved that the number of linearly dependent n column sets in H are same as that of number of linearly dependent column sets in G (s Let H (s = [I ( ( P T ] That is, H(s is obtained by rearranging the columns of H The number of linearly dependent column sets in H and H (s are same The matrix
H (s satisfies the condition given in case (i Hence, for H (s, we can write (11 as D = A d, (1 wherea d s are the weight enumerating coefficients of(c T T = C The number of column sets that are linearly dependent is equal to the number of singular matrices in G This completes the proof Theorem Consider an arbitrary binary full ran matrix G of size n (n Let C be the linear bloc code generated by G as generator matrix and C T be the dual code of C Let I be the number of full ran submatrices of size in G If 3d > max(,, then where I = ( n A d, A d s are the WECs of W C T(x,y if A d s are the WECs of W C (x,y if < Proof In a binary matrix of size n, there exist ( n binary sub-matrices of size Each of this sub-matrix may be of full ran or singular Thus, we have ( n D +I = where D is the number of singular sub-matrices in G The remaining proof follows from Theorem 1 Remar To calculate D and I, we use the weight enumerating coefficients of either C or C T that satisfy the condition length of the code dimension < Remar 3 For a given binary full ran matrix of size n, to calculate I by using brute-force technique requires the ran computation of ( n number of matrices Whereas, by using Theorem, one only needs to now the weight enumerating coefficients of the given n matrix or its orthogonal complement The weight enumerating coefficients required in Theorem of any arbitrary n matrix needs the computation of Hamming weight of min(, binary vectors of size n For a binary full ran matrix G of size n satisfying the condition 3d > max(,, Algorithm 1 finds the number of full ran submatrices of size in G Example 3 Consider matrix G given below 1 1 1 0 1 0 0 1 0 1 1 0 0 1 G = 1 1 1 1 1 1 1 0 1 1 0 0 1 1 Algorithm 1 Algorithm to find the number of full ran submatrices of size in a given full ran matrix G of size n Step 1 11: Convert the matrix G in symmetric form by using elementary row operations Let this symmetric matrix be G (s = [I : P ( ] (Lemma 1 1: If < n, Go to Step 13: Else G (s = [P T ( : I ( (] (Lemma Step 1: Find the weight enumerating coefficients of G (s Let A d for d [1 : n] be the weight enumerating coefficients of G (s : I = ( n A d Step 3 Exit ( n d (Theorem For the matrix G, we have = 4,n = 7 By using the elementary row operations, the matrix G can be converted into symmetric matrix G (s the matrix G (s is given below 1 0 0 0 1 1 1 G (s 0 1 0 0 1 1 0 = [I 4 4 : P 4 3 ] = 0 0 1 0 1 0 1 0 0 0 1 0 1 1 For the matrix G (s, we have Hence, we need to calculate D and I from weight enumerating coefficients of C T The generator matrix of C T is given below [ ] 1 1 1 0 1 0 0 [P T 3 4 : I 3 3 ] = 1 1 0 1 0 1 0 1 0 1 1 0 0 1 The weight enumerating function of C T can be calculated by enumerating all the 3 = 8 codewords generated by above 3 7 matrix The weight enumerating function is given by W C T(x,y = x 7 +7x 4 y 4 (13 From (13, we have d = 4,A 4 = 7 and A d = 0 for d 4 The and d in this example satisfy the condition 3d > max(, From Theorem 1, we have ( ( ( n d n 4 4 D = A d = A 4 = 7 = 7 4 From Theorem, we have I = ( n D = 35 7 = 8 The 4-column sets corresponding to the 7 singular 4 4 matrices are given below D 1 = {1,,3,5}, D = {1,,4,6}, D 3 = {1,3,4,7} D 4 = {3,4,5,6}, D 5 = {,3,6,7}, D 6 = {,4,5,7} D 7 = {1,5,6,7} The 4-column sets corresponding to the 8 full-ran 4 4 matrices are given below
I 1 = {1,,3,4}, I = {1,,3,5}, I 3 = {1,,3,7} I 4 = {1,,4,6}, I 5 = {1,,4,7}, I 6 = {1,,5,6} I 7 = {1,,5,7}, I 8 = {1,,6,7}, I 9 = {1,3,4,5}, I 10 = {1,3,4,6}, I 11 = {1,3,5,6}, I 1 = {1,3,5,7}, I 13 = {1,3,6,7}, I 14 = {1,4,5,6}, I 15 = {1,4,5,7}, I 16 = {1,4,6,7}, I 17 = {,3,4,5}, I 18 = {,3,4,6}, I 19 = {,3,4,7}, I 0 = {,3,5,6}, I 1 = {,3,6,7}, I = {,4,5,6}, I 3 = {,4,5,7}, I 4 = {,5,6,7}, I 5 = {3,4,5,7}, I 6 = {3,4,6,7}, I 7 = {3,5,6,7} I 8 = {4,5,6,7} Note 1 The matrix G given in Example 3 is a generator matrix of(7, 4 Hamming code Hence, in the generator matrix of (7,4 Hamming code, there exist 8 full ran submatrices of size 4 4 From Lemma, in the parity chec matrix of (7,4 Hamming code, there exist 8 full ran submatrices of size 3 3 Example 4 Consider matrix G given below 1 1 1 1 1 1 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 G = 0 0 1 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 1 For the matrix G, we have = 7,n = 10 and the matrix G (s is given below 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 1 1 G (s = [I 7 7 : P 7 3 ] = 0 0 0 1 0 0 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 0 1 1 0 1 For the matrix G (s, we have Hence, we need to calculate D and I from weight enumerating coefficients of C T The generator matrix of C T is given below [P T 3 7 : I 3 3] = [ ] 1 0 1 1 0 1 1 1 0 0 1 1 1 0 1 1 0 0 1 0 0 1 1 1 0 1 1 0 0 1 The weight enumerating function of C T can be calculated by enumerating all the 3 = 8 codewords generated by above 3 10 matrix The weight enumerating function is given by W C T(x,y = x 10 +7x 4 y 6 (14 From (14, we have d = 6,A 6 = 7 and A d = 0 for d 6 The and d in this example satisfy the condition 3d > max(, From Theorem 1, we have D = A d = A 6 ( n 6 ( 4 = 6 = 8 3 From Theorem, we have ( n I = D = 10 8 = 9 Example 5 Consider a generator matrix G of (n = 15, = 11 Hamming code The weight enumerating function of the dual of the (15,11 Hamming code is W C T(x,y = x 15 +15x 7 y 8 In this example, we have n and hence d is the minimum Hamming weight of the error correcting code generated by H and d = 8 The and d in this example satisfy the condition 3d > max(, From Theorem 1, we have D = A d = A 8 ( n 8 ( 7 = 15 = 55 4 From Theorem, we have ( 15 I = D = 1,365 55 = 840 11 Note In a generator matrix of (15, 11 Hamming code, there exist 840 full ran submatrices of size 11 11 From Lemma, in the parity chec matrix of (15,11 Hamming code, there exist 840 full ran submatrices of size 4 4 III SCOPE FOR FUTURE WORK In this paper, for a n binary full ran matrix satisfying a condition of minimum Hamming weight, we quantified the number of binary full ran submatrices of size in terms of weight enumerating function Some of the interesting problems are given below Establishing the relation between the number of full ran submatrices of size for any arbitrary full ran binary matrix of size n is an interesting problem For given positive integers and n (n >, finding the maximum value of I and the corresponding matrix G is an interesting problem Extending the result given in this paper to other fields is an interesting problem ACKNOWLEDGEMENT This wor was supported partly by the Science and Engineering Research Board (SERB of Department of Science and Technology (DST, Government of India, through JC Bose National Fellowship to B Sundar Rajan
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