Application of hyperplane arrangements to weight enumeration Relinde Jurrius (joint work with Ruud Pellikaan) Universit´ e de Neuchˆ atel May 25, 2015 1/21
� � Coding theory � channel message message noise 2/21
� Coding theory encoding decoding � � � codeword � channel � received word � message message noise 2/21
Coding theory Code Set of codewords ( ≈ vectors) of fixed length n . d ( x , y ) The number of places on which two vectors differ. d The minimal distance between codewords. 3/21
Coding theory Code Set of codewords ( ≈ vectors) of fixed length n . d ( x , y ) The number of places on which two vectors differ. d The minimal distance between codewords. Linear code Linear subspace C ⊆ F n q of dimension k . Generator matrix Some k × n matrix G whose rows span C . 3/21
Coding theory Example The [7 , 4] Hamming code over F 2 has generator matrix 1 0 0 0 1 1 0 0 1 0 0 1 0 1 G = . 0 0 1 0 0 1 1 0 0 0 1 1 1 1 The minimum distance is 3. 4/21
Coding theory Codes are equivalent if generator matrices are the same up to • left multiplication by nonsingular k × k matrix over F q (i.e., same rowspace); • permutation of columns; • multiplication of column by element of F ∗ q . 5/21
Coding theory Codes are equivalent if generator matrices are the same up to • left multiplication by nonsingular k × k matrix over F q (i.e., same rowspace); • permutation of columns; • multiplication of column by element of F ∗ q . We restrict to projective codes: they have generator matrix where • no column is zero; • no column is a multiple of another column. So, all columns coordinatize a different projective point. 5/21
Weight enumeration Weight The number of nonzero coordinates in a vector. For linear codes: minimum distance = minimum nonzero weight. 6/21
Weight enumeration Weight The number of nonzero coordinates in a vector. For linear codes: minimum distance = minimum nonzero weight. Weight enumerator n � A w X n − w Y w W C ( X , Y ) = w =0 where A w = number of words of weight w . 6/21
Weight enumeration Example The [7 , 4] Hamming code over F 2 has generator matrix 1 0 0 0 1 1 0 0 1 0 0 1 0 1 G = . 0 0 1 0 0 1 1 0 0 0 1 1 1 1 The weight enumerator is equal to W C ( X , Y ) = X 7 + 7 X 4 Y 3 + 7 X 3 Y 4 + Y 7 . 7/21
Weight enumeration Extension code [ n , k ] code C ⊗ F q m over some extension field F q m generated by the words of C . Generator matrix All extension codes of C have generator matrix G . 8/21
Weight enumeration Extension code [ n , k ] code C ⊗ F q m over some extension field F q m generated by the words of C . Generator matrix All extension codes of C have generator matrix G . Extended weight enumerator n � A w ( T ) X n − w Y w , W C ( X , Y , T ) = w =0 where A w ( q m ) = number of words of weight w in C ⊗ F q m . Fact: the A w ( T ) are polynomials of degree at most k . 8/21
Weight enumeration Example The [7 , 4] Hamming code has extended weight enumerator X 7 + W C ( X , Y , T ) = 7( T − 1) X 4 Y 3 + 7( T − 1) X 3 Y 4 + 21( T − 1)( T − 2) X 2 Y 5 + 7( T − 1)( T − 2)( T − 3) XY 6 + ( T − 1)( T 3 − 6 T 2 + 15 T − 13) Y 7 9/21
Why do we study this? The extended weight enumerator is interesting because: • Determines the probability of undetected error in error-detection. • Determines the probability of decoding error in bounded distance decoding. • Connection to Tutte polynomial in matroid theory. • Connection to zeta function of (algebraic geometric) codes. . . . and of course because it is an invariant of linear codes. 10/21
Weight enumeration ✞ ☎ 0 ✝ ✆ 1 × k k × n 1 × n message m generator matrix G codeword c 11/21
Weight enumeration ✞ ☎ 0 ✝ ✆ 1 × k k × n 1 × n message m generator matrix G codeword c Theorem c j = 0 ⇐ ⇒ m lies in hyperplane H j Weight enumeration = counting points in (intersections of) hyperplanes. 11/21
Codes and hyperplane arrangements Columns of a generator matrix G of a linear [ n , k ] code form a linear hyperplane arrangement in F k q . Notation: ( H 1 , . . . , H n ). • One-to-one correspondence between equivalence classes. • Independent of choice of G , so notation: A C . • Also valid over an extension field F m q . Theorem A w ( T ) = number of points from vectorspace over field of T elements that are on n − w hyperplanes. 12/21
Codes and hyperplane arrangements Example Let q > 2 and C generated by 1 1 0 0 0 0 G = 0 1 1 1 1 0 , H 2 0 1 1 0 1 a H 1 H 6 H 3 where a � = 0 , 1. H 4 H 5 The extended weights are given by A 0 ( T ) = 1 The zero word is on all hyperplanes. 13/21
Codes and hyperplane arrangements Example Let q > 2 and C generated by 1 1 0 0 0 0 G = 0 1 1 1 1 0 , H 2 0 1 1 0 1 a H 1 H 6 H 3 where a � = 0 , 1. H 4 H 5 The extended weights are given by A 1 ( T ) = 0 No points are on 5 hyperplanes. 13/21
Codes and hyperplane arrangements Example Let q > 2 and C generated by 1 1 0 0 0 0 G = 0 1 1 1 1 0 , H 2 0 1 1 0 1 a H 1 H 6 H 3 where a � = 0 , 1. H 4 H 5 The extended weights are given by A 2 ( T ) = T − 1 One projective point is on 4 hyperplanes. 13/21
Codes and hyperplane arrangements Example Let q > 2 and C generated by 1 1 0 0 0 0 G = 0 1 1 1 1 0 , H 2 0 1 1 0 1 a H 1 H 6 H 3 where a � = 0 , 1. H 4 H 5 The extended weights are given by A 3 ( T ) = T − 1 One projective point is on 3 hyperplanes. 13/21
Codes and hyperplane arrangements Example Let q > 2 and C generated by 1 1 0 0 0 0 G = 0 1 1 1 1 0 , H 2 0 1 1 0 1 a H 1 H 6 H 3 where a � = 0 , 1. H 4 H 5 The extended weights are given by A 4 ( T ) = 6( T − 1) Six projective points are on 2 hyperplanes. 13/21
Codes and hyperplane arrangements Example Let q > 2 and C generated by 1 1 0 0 0 0 G = 0 1 1 1 1 0 , H 2 0 1 1 0 1 a H 1 H 6 H 3 where a � = 0 , 1. H 4 H 5 The extended weights are given by A 5 ( T ) = (6( T +1) − 1 · 4 − 1 · 3 − 6 · 2)( T − 1) = (6 T − 13)( T − 1) Six lines with T + 1 points; minus the points counted before. 13/21
Codes and hyperplane arrangements Example Let q > 2 and C generated by 1 1 0 0 0 0 G = 0 1 1 1 1 0 , H 2 0 1 1 0 1 a H 1 H 6 H 3 where a � = 0 , 1. H 4 H 5 The extended weights are given by A 6 ( T ) = ( T − 1)( T − 2)( T − 3) The total number of projective points is T 2 + T + 1. 13/21
Geometric lattice To formalize this counting, we use the geometric lattice associated to the arrangement. Notation: L . Elements All intersections of hyperplanes Ordering x ≤ y if y ⊆ x Minimum Whole space F k q Maximum Zero vector 0 ∈ F k q Rank Codimension of x in F k q Atoms The hyperplanes of the arrangement 14/21
Geometric lattice M¨ obius function For all x ≤ y , we have µ L ( x , x ) = 0 and � � µ L ( x , z ) = µ L ( z , y ) = 0 . x ≤ z ≤ y x ≤ z ≤ y Characteristic polynomial � µ L (ˆ 0 , x ) T r ( L ) − r ( x ) χ L ( T ) = x ∈ L 15/21
Geometric lattice Example 123456 123 14 15 16 24 25 26 3456 1 2 3 4 5 6 ∅ 16/21
Coboundary polynomial Coboundary polynomial The coboundary of a geometric lattice is defined by � � µ L ( x , y ) S a ( x ) T r ( L ) − r ( y ) χ L ( S , T ) = x ∈ L x ≤ y ∈ L where a ( x ) is the number of atoms smaller then x . We write: n � � S i χ i ( T ) , χ L ( S , T ) = with χ i ( T ) = 1] ( T ) . χ [ x , ˆ i =0 x ∈ L a ( x )= i 17/21
Coboundary polynomial Theorem χ i ( T ) = A n − i ( T ) Proof: For every point in F k q m there is a unique biggest element of L that contains the point. A n − i ( q m ) number of points in F k = q m on exactly i hyperplanes � number of points in F k = q m in x but not in any y > x x ∈ L a ( x )= i 18/21
Coboundary polynomial Well-known fact: χ L ( q m ) number of points in F k = q m not in the arrangement q m in ˆ 0 but not in any y > ˆ number of points in F k = 0 This means that: A n − i ( q m ) � number of points in F k = q m in x but not in any y > x x ∈ L a ( x )= i � 1] ( q m ) = χ [ x , ˆ x ∈ L a ( x )= i χ i ( q m ) = So by interpolation, χ i ( T ) = A n − i ( T ). 19/21
Summary • Codes are linear subspaces of F n q . • Extending the underlying field gives extension codes C ⊗ F q m , and we define the extended weight enumerator W C ( X , Y , T ). • By viewing the columns of G as hyperplanes, we associate an arrangement to a code. • Finding the extended weight enumerator means counting points in intersections of hyperplanes. • This counting can be done using the geometric lattice associated with the arrangement. • The coboundary polynomial is equivalent to the extended weight enumerator. 20/21
Thank you for your attention. 21/21
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