Cyclotomic FFT of Length 2047 Based on a Novel 11-point Cyclic Convolution

1 Cyclotomic FFT of Length 2047 Based on a No v el 11-point Cycli c Con v olution Meghanad D. W agh, Ni ng Chen, and Zhi yuan Y an Abstract In this manuscrip t, we prop ose a novel 11- point cyclic conv olution algor ithm based on alterna te Fourier transform . W ith the p roposed bilinear form, we c onstruct a length-204 7 cyclotomic FFT . I . I N T RO D U C T I O N Discrete Fourier trans forms (DFTs) over finite fields have wides pread applications in error c orrection coding [1]. For Reed-Solomon (RS) codes , all syndrome -based bounded d istance de coding methods in v olve DFTs ov er finite fields [1]: sy ndrome computation and the Chie n search are both evaluations of polynomials and hence can be viewed as DFTs ; in v erse DFTs a re used to recov er transmitted cod ew ords in transform-domain decod ers. Thus ef ficient DFT algorithms can be used to reduc e the complexity of RS decoders. F or example, u sing the prime-f actor fast Fourier transform (FFT) in [2], T ruong et al. propos ed [3] an in verse-free transform-domain RS de coder with s ubstantially lower complexity than time-domain de coders; FFT tec hniques are used to compute syndromes for time-domain deco ders in [4]. Cyclotomic FFT w as prop osed recently in [5] and two variations were su bseque ntly considered in [6], [7]. Compared with other FFT techniques [2], [8], CFF Ts in [5]–[7] a chieve signific antly lo wer mu l- tiplicati ve complexities, w hich makes them very attracti ve. B ut their additiv e complexities (numbers of additions requ ired) are very high if impleme nted directly . A common su bexpression elimination (CSE) algorithm was propo sed to significantly reduce the a dditi ve co mplexities of CFFTs in [9]. Along with those full CF FTs, reduced-complexity partial and dual partial CFFTs were us ed to de sign low complexity RS decod ers in [10]. The lengths of CF FTs in [9] are only up to 1 023 while longer CFFTs are required to decode long RS code s. T o pursuit a length-204 7 CF FT , 11-point cyclic con v olution over cha racteristic-2 fields is n ecess ary , which is not readily av ailable in the literature. In this manus cript, we first propose a novel 11-point cyclic co n v olution for cha racteristic-2 fields in Section II . Base d on this cyclic con voluti on, a length-204 7 CFFT is pres ented in Section III . Using the same app roach, CFFTs of any lengths tha t divide 2047 can also be cons tructed. June 1 1, 201 8 DRAFT 2 I I . 1 1 - P O I N T C Y C L I C C O N V O L U T I O N OV E R C H A R AC T E R I S T I C - 2 F I E L D S W e first de ri ve a fast cyclic c on v olution of 1 1 p oints over the rea l field. Denote the cyclic co n v olution of 11 point sequ ences x and y by the sequen ce z . 1 The sequen ce z may be computed by Fourier transforming x and y , multiplying the transforms point-by-point and finally , in verse Fourier transforming the produc t seq uence . Let X , Y and Z denote the Fourier transforms of x , y a nd z respec ti vely . As defined by the Fourier transform, X 0 = 10 X i =0 x i Y 0 = 10 X i =0 y i . (1) W e express the rest comp onents, X ′ and Y ′ (over reals) u sing the b asis h 1 , W , W 2 , . . . , W 9 i where W denotes the 11 th primiti v e root of unity . This b asis is sufficient beca use W 11 − 1 = 0 yields W 10 = − 1 − W − W 2 − · · · − W 9 . Thus , X ′ = P 9 i =0 X ′ i W i and Y ′ = P 9 i =0 Y ′ i W i , in wh ich X ′ i = ( x i − x 10 ) Y ′ i = ( y i − y 10 ) . (2) W e will ca ll the vector ( X 0 , X ′ 0 , X ′ 1 , . . . , X ′ 9 ) as the Alternate F ourier transfor m (AFT) of seq uence x . No te that AF T is simply the DFT compone nts X 0 and X ′ in their special bases . F rom (1) and (2) it is obvious that the AFT computation may be described as a multiplication with a 11 × 11 ma trix B with structure B =            1 1 . . . 1 1 − 1 I 10 − 1 . . . − 1            where I 10 is a 10 × 10 iden tity matrix. Alternately , given the AFT of x , on e can determine x by using matrix B − 1 giv en by B − 1 = 1 11   1 A 1 A 2 A 3   (3) where length-10 ro w A 1 = (10 , − 1 , − 1 , . . . , − 1) , length-10 column A 2 = (1 , 1 , . . . , 1) T and 10 × 10 submatrix A 3 has 10 o n the first upper diago nal and -1 everywhere e lse. Now consider the product of B − 1 and an AF T vector: B − 1   U 0 U ′   = 1 11   1 A 1 A 2 A 3     U 0 U ′   =   V 0 V ′   1 In this manuscript, vectors and matrices are represented by boldface letters, and scalars by normal letters. June 11, 2018 DRAFT 3 where U 0 , U ′ , an d V 0 , V ′ are a ppropriate partitions of the AFT a nd the signal vectors. V alues o f V 0 and V ′ can be c omputed as V 0 = (1 / 11) U 0 + (1 / 11) A 1 U ′ and V ′ = (1 / 11) A 2 U 0 + (1 / 11) A 3 U ′ . Note that A 1 and A 3 are related as A 1 = − (1 , 1 , . . . , 1) A 3 . This implies tha t the sum of the c omponents of (1 / 11 ) A 3 U ′ giv es − (1 / 11) A 1 U ′ . Furthermore, A 2 contains only 1’ s. Th us the computation of V 0 and V ′ reduces to V 0 = (1 / 11) U 0 − (1 / 11) X ( A 3 U ′ ) V ′ = (1 / 11) [ U 0 , U 0 , . . . , U 0 ] T + (1 / 11) A 3 U ′ . (4) Relation (4) s hows tha t the in verse of an AFT only need s an ev aluation of (1 / 11) A 3 U ′ . T o c ompute cyc lic con volution o f x a nd y , one s hould multiply the Fourier transforms of x an d y and then take the in verse Fourier transform of the prod uct. W e us e AFT ins tead of c lassical Fourier transform. Multiplying X 0 and Y 0 is simple, but since X ′ and Y ′ are expressed in a ba sis with 10 e lements, their product ma y be dif ficult. Simi larly in verse AFT requires multiplication by matrix A 3 which may be complicated. Howev er , we now show that bo th these two difficult computation stag es are eq ui valent to only a T oe plitz product (i.e., p roduct of a T oe plitz matrix and a vector) [11]. The pointwise multiplication results are Z 0 = X 0 Y 0 and Z ′ defined as  9 X i =0 X ′ i W i  9 X i =0 Y ′ i W i  = 9 X i =0 Z ′ i W i . (5) V ector Z ′ can be computed through the matrix prod uct ( Z ′ 0 , Z ′ 1 , . . . , Z ′ 9 ) T = M ( X ′ 0 , X ′ 1 , . . . , X ′ 9 ) T where the eleme nts of matrix M are M k ,j = Y ′ k − j + Y ′ k − j +11 − Y ′ 10 − j . (6) Note tha t in (6), Y ′ i are con sidered as zero outside its valid range, i.e., Y ′ i = 0 if i < 0 or i > 9 . The terms in (6) are eas y to deduc e from (5). Matrix element M k ,j sums up tho se terms in Y ′ that after multiplication with X ′ j W j result in W k terms. For example, since product of X ′ j W j and Y ′ k − j W k − j results in X ′ j Y ′ k − j W k , we get the first term in (6) as given. Secon d term of (6) can be similarly ar gued. The third term is du e to the prod uct ( X ′ j W j )( Y ′ 10 − j W 10 − j ) = X ′ j Y ′ 10 − j W 10 = − X ′ j Y ′ 10 − j P 9 i =0 W i . Computing in v erse DFT o f Z req uires o ne to multi ply A 3 and vector ( Z ′ 0 , Z ′ 1 , . . . , Z ′ 9 ) T where A 3 is the matrix defined in (3). Thu s one has to compu te R ( X 1 (0) , X 1 (1) , . . . , X ′ 9 ) T where the 10 × 10 matrix R = (1 / 11) A 3 M . June 11, 2018 DRAFT 4 W e now show by direc t computation tha t R is a T oeplitz matrix. F rom the structure of A 3 , we have R i,j = 1 11 9 X k =0 ,k 6 = i +1 − M k ,j + 10 11 M i +1 ,j = − 1 11 9 X k =0 M k ,j + M i +1 ,j . (7) From (6), us ing the a ppropriate ranges for the three terms we now get 9 X k =0 M k ,j = − 9 X k =0 Y ′ 10 − j + j − 2 X k =0 Y ′ k − j + p + 9 X k = j Y ′ k − j = − 10 Y ′ 10 − j + 9 X s =11 − j Y ′ s + 9 − j X s =0 Y ′ s = 9 X s =0 Y ′ s − 11 Y ′ 10 − j (8) Finally , c ombining (6), (7) a nd (8) gives R i,j = Y ′ i − j +1 + Y ′ i − j +12 − 1 11 9 X s =0 Y ′ s . (9) Since R i,j is a func tion of on ly i − j , R is a T oep litz matrix. Thus Z ′ = RX ′ is comp uted as         Z ′ 0 Z ′ 1 . . . Z ′ 9         =         Y ′ 1 Y ′ 0 0 Y ′ 9 . . . Y ′ 3 Y ′ 2 Y ′ 1 Y ′ 0 0 . . . Y ′ 4 . . . . . . . . . . . . . . . . . . 0 Y ′ 9 Y ′ 8 . . . . . . Y ′ 1                 X ′ 0 X ′ 1 . . . X ′ 9         +         P 9 i =0 X ′ i P 9 i =0 Y ′ i P 9 i =0 X ′ i P 10 i =1 Y i . . . P 9 i =0 X ′ i P 9 i =0 Y ′ i         . Recall that Y ′ i is assumed z ero if it s index is outside the vali d range fr om 0 to 9. Th us in (9), exac tly one of the first two terms is valid for any combination of i a nd j . Fig. 1 illustrates the b ilinear cyclic con vol ution algorithm of length 11 b ased on this discuss ion. The multiplication of the 10 × 10 T oeplitz matrix R with a vector c an be obtained using the T oeplitz product algo rithms of lengths 2 a nd 5. The matrix R can be split into four 5 × 5 s ubmatrices and the vector X ′ can be split into two leng th-5 vectors. By the definition of T oe plitz matrices, the T oeplitz product RX ′ can be c omputed as RX ′ =   R 0 R 1 R 2 R 0     X ′ 0 X ′ 1   =   R 0 ( X ′ 0 + X ′ 1 ) + ( R 1 − R 0 ) X ′ 1 R 0 ( X ′ 0 + X ′ 1 ) + ( R 2 − R 0 ) X ′ 0   . Although the cyclic c on v olution is deri ved over the r eal field, it can be easily con v erted to c haracteristic- 2 fields. Based on the me thod in [12], we multiply both sides of all equations above by 11 modulus 2. In June 11, 2018 DRAFT 5 x 0 x 1 y 1 y 0 x 2 z z z 0 1 2 y 10 ( + + + ) x 9 z z 9 10 R x 10 Matrix Toeplitz 10 point Fig. 1. 11-point cyclic con volu tion based on AFT the con verted form, X = T x , Y = T y , and z = S Z . Thus we ob tain 11-point cyc lic conv olution over characteristic-2 fields. T o find its b ilinear form, we nee d the bilinear form of T oeplitz product of len gth 10. The bilinear form of length-5 T oeplitz p roduct over characteristic-2 fields v = Q ( T 5) ( R ( T 5) r · P ( T 5) u ) is given in Appendix I, whe re · s tands for pointwise multiplication. Based on the leng th-10 T oeplitz product, the bilinear form of 11-point cyclic con vol ution over GF(2 m ) is given by z = Q (11) ( R (11) y · P (11) x = S      1 0 0 0 0 Q ( T 5) Q ( T 5) 0 0 Q ( T 5) 0 Q ( T 5)                      1 0 0 R ( T 5) Π 0 0 R ( T 5) Π 1 0 R ( T 5) Π 2         T y ·         1 0 0 P ( T 5) Π 3 0 P ( T 5) Π 4 0 P ( T 5) Π 5         T x         . Details of matrices S , T , Π 0 , . . . , Π 5 are giv en in Appen dix II. The propose d leng th-11 cyclic con volution needs only 43 multiplications. W e compa re it with cyclic con vol utions of other lengths from [1], [13], [14] in T able I. I I I . C Y C L OT O M I C F F T OV E R GF(2 11 ) Based on the derived 11-point cyclic con volution over GF(2 m ) , we c an co nstruct a l ength-2047 cyclotomic FFT over GF(2 11 ) . In this man uscript, we focus on d irect CFFT as in [5] since it was June 11, 2018 DRAFT 6 T ABLE I M U LT I P L I C A T I V E C O M P L E X I T Y O F C Y C L I C C O N VO L U T I O N n 2 3 4 5 6 7 8 9 10 11 Mult. 3 4 9 10 12 13 27 19 30 43 shown in [9] all variants of CFFTs have the same multiplicati ve complexity and they ha ve the same additiv e complexity under direct implementation. Gi ven a primiti ve e lement α ∈ GF(2 m ) , the DFT of a vec tor f = ( f 0 , f 1 , . . . , f n − 1 ) T is define d as F ,  f ( α 0 ) , f ( α 1 ) , . . . , f ( α n − 1 )  T , where f ( x ) , P n − 1 i =0 f i x i ∈ GF(2 m )[ x ] . W e choose the field generated by the polynomial x 11 + x 2 + 1 . In this field, there are on e size-1 coset and 186 size -11 cosets. W e permute the input f to f ′ such that f ′ = ( f 0 , f 1 , f 2 , . . . , f 186 ) and each size-11 vector f i contains the compon ents of f whose indices are in the same coset ( k i , k i 2 , . . . , k i 2 m i − 1 ) mo d 2047 where m i | 11 is the coset size. T hus the p olynomial f ( x ) is divided into parts, each one is L i ( x k i ) = P m i − 1 j =0 f k i 2 j mod 2047 ( x k i ) 2 j . Henc e L i ( x ) ’ s are linea rized polyn omials. Each e lement α k i can be decom- posed with resp ect to a basis β i = ( β i, 0 , β i, 1 , . . . , β i,m i − 1 ) such tha t α j k i = P 10 s =0 a i,j,s β i,s , a i,j,s ∈ GF(2) . So e ach compon ent of DFT is factored into f ( α j ) = 186 X i =0 L i ( α j k i ) = 186 X i =0 m i X s =0 a i,j,s L i ( β i,s ) = 186 X i =0 10 X s =0 a i,j,s  10 X p =0 β 2 p i,s f k i 2 p  . In matrix form, it is F = aLf , in which L is a b lock diagona l matrix with eac h diagonal block being L i =         β i, 0 β 2 i, 0 . . . β 2 m i − 1 i, 0 β i, 1 β 2 i, 1 . . . β 2 m i − 1 i, 1 . . . . . . . . . . . . β i,m i − 1 β 2 i,m i − 1 . . . β 2 m i − 1 i,m i − 1         . Using a normal basis as β i , the matrix L i becomes a cyclic matrix and L i f i becomes a s ize- m i cyclic con vol ution. For length-204 7 CFF T , m i is 1 or 11. Th us we obtain a length-2047 CFFT using the bilinear form of 11-point cyclic con voluti on as F = a         1 Q (11) . . . Q (11)                         1 R (11) . . . R (11)                 1 β 1 . . . β 186         ·         1 P (11) . . . P (11)                 f 0 f 1 . . . f 186                 . June 11, 2018 DRAFT 7 It requires 78 12 multiplications to c ompute the constructed le ngth-2047 CFFT . Unde r direct impleme n- tation, it requires 215 4428 additions. W ith i ncomplete optimization using the CSE algorithm [9], it s additiv e c omplexity c an be reduced to 529720 . W e comp are its complexity with those of sho rter CFF Ts in T able II . In T able II, our numbers of additions for CFFTs of leng ths 7 , 15 , · · · , 1023 a re rep roduced from [9]. T ABLE II C O M P L E X I T Y O F F U L L C Y C L OT O M I C F F T n Mult. Additions Ours [5] Direct 7 6 24 25 34 15 16 74 77 154 31 54 299 315 570 63 97 759 805 2527 127 216 2576 2780 9684 255 586 6736 7919 37279 511 1014 23130 2664 3 141710 1023 2 827 75360 - 536093 2047 7 812 52 9720 - 2154428 A P P E N D I X I T O E P L I T Z P RO D U C T O F L E N G T H 5 T oeplitz product of len gth 5 as            v 0 v 1 v 2 v 3 v 4            =            r 4 r 5 r 6 r 7 r 8 r 3 r 4 r 5 r 6 r 7 r 2 r 3 r 4 r 5 r 6 r 1 r 2 r 3 r 4 r 5 r 0 r 1 r 2 r 3 r 4                       u 0 u 1 u 2 u 3 u 4            can be d one in bilinear form as v = Q ( T 5) ( R ( T 5) r · P ( T 5) u ) . June 11, 2018 DRAFT 8 R ( T 5) =                                        1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0                                        P ( T 5) =                                        1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 0 0 1 0 0 1 0 0 1 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 1 0 1 0 0 0 1 1 1 1 0 1 1                                        Q ( T 5) =            0 0 0 0 1 0 0 0 0 1 0 1 1 1 0 0 0 1 0 0 0 1 0 0 1 0 1 1 0 0 1 0 0 0 1 0 1 0 1 1 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1            . A P P E N D I X I I B I L I N E A R F O R M O F 1 1 - P O I N T C Y C L I C C O N VO L U T I O N OV E R C H A R A C T E R I S T I C - 2 F I E L D S z = S      1 Q ( T 5) Q ( T 5) 0 Q ( T 5) 0 Q ( T 5)                      1 R ( T 5) Π 0 R ( T 5) Π 1 R ( T 5) Π 2         T y ·         1 P ( T 5) Π 3 P ( T 5) Π 4 P ( T 5) Π 5         T x         . June 11, 2018 DRAFT 9 T =                           1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 1                           S =                              1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1                              Π 0 =                        1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1                        Π 3 =            1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1            Π 1 =                        1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0                        Π 4 =            0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1            June 11, 2018 DRAFT 10 Π 2 =                        0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0                        Π 5 =            1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0            R E F E R E N C E S [1] R. 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