Floor Scale Modulo Lifting for QC-LDPC codes Nikita Polyanskii, Vasiliy Usatyuk, and Ilya Vorobyev Huawei Technologies Co., Moscow, Russia Email: nikitapolyansky@gmail.com, l@lcrypto.com, vorobyev.i.v@yandex.ru Abstract—In the given paper we present a novel approach for with respect to the number of small cycles. The performance constructingaQC-LDPCcodeofsmallerlengthbyliftingagiven of QC-LDPC codes obtained by the floor scale modulo lifting QC-LDPC code. The proposed method can be considered as a method is investigated by simulations in Section V. generalizationoffloorlifting.Alsoweproveseveralprobabilistic statements concerning a theoretical improvement of the method II. QC-LDPCCODES with respect to the number of small cycles. Making some A QC-LDPC code is described by a parity-check matrix H offline calculation of scale parameter it is possible to construct which consists of square blocks which could be either zero a sequence of QC-LDPC codes with different circulant sizes generated from a single exponent matrix using only floor and matrix or circulant permutation matrices. Let P = (Pij) be scale operations. The only parameter we store in memory is a the L×L circulant permutation matrix defined by constant needed for scaling. (cid:40) 7 Keywords: QC-LDPC code, floor lifting, modulo lifting, 1, if i+1≡j mod L P = 1 ij block cycle, girth. 0, otherwise. 0 2 I. INTRODUCTION Then Pk is the circulant permutation matrix (CPM) which b Low-density parity-check (LDPC) codes were first discov- shifts the identity matrix I to the right by i times for any k, e 0 ≤ k ≤ L−1. For simplicity of notation denote the zero ered by Gallager [1], generalized by Tanner [2], Wibberg [3] F matrix by P−1. Denote the set {−1,0,1,...,L−1} by A . andrediscoveredbyMacKayetal.[4],[5]andSipseretal.[6]. L 4 Quasi-cyclic low-density parity-check (QC-LDPC) codes LetthematrixHofsizemL×nLbedefinedinthefollowing manner are of great interest to researchers [7]–[11] since they can be T] encoded and decoded with low complexity and allow to reach  Pa11 Pa12 ··· Pa1n  I high throughput using linear-feedback shift register [12]–[14].  Pa21 Pa22 ··· Pa2n  cs. One advantage of QC-LDPC codes based on circulant H= ... ... ... ... , (1) permutationmatrices(CPM)isthatitiseasiertoanalyzetheir [ Pam1 Pam2 ··· Pamn code and graph properties than in the case of random LDPC 2 codes. The performance of LDPC codes is strongly affected where ai,j ∈ AL. Further we call L the circulant size of H. v In what follows a code C with parity-check matrix H will be by their graph properties such as the length of the shortest 1 referredtoasaQC-LDPCcode.LetE(H)=(E (H))bethe cycle, i.e., girth [15], [16], and trapping sets [17], [18] and ij 2 exponent matrix of H given by: 5 code properties, e.g., the distance of the code [19], [20] and   7 the ensemble weight enumerator [21]. a a ··· a 11 12 1n 0 The main contribution of the paper is a novel approach  a21 a22 ··· a2n  1. for constructing a quasi-cyclic LDPC code of smaller length E(H)= ... ... ... ... , (2) 0 by lifting a given QC-LDPC code. The proposed method 7 can be considered as a generalization of floor lifting method am1 am2 ··· amn 1 introduced in [22], [23]. Making some offline calculation it i.e., the entry Eij(H) = aij. The mother matrix M(H) is : v is possible to construct a sequence of QC-LDPC codes with a m × n binary matrix obtained from replacing −1’s and Xi different circulant sizes generated from a single exponent other integers by 0 and 1, respectively, in E(H). If there matrix of QC-LDPC code having the largest length. The only is a cycle of length 2l in the Tanner graph of M(H), it r a parameterwestoreinmemoryisaconstantneededforscaling is called a block-cycle of length 2l. Any block-cycle in in the lifting procedure. M(H) of length 2l corresponds both to the sequence of 2l The outline of the paper is as follows. In Section II, we CPM’s{Pa1,Pa2...,Pa2l}inHandsequenceof2l integers introduce some basic definitions and notations for our presen- {a1,a2...a2l} in E(H) which will be called exponent chain. tation.InSectionIII,wereviewstate-of-artliftingmethodsfor The following well known result gives the easy way to find QC-LDPC codes. Also assuming some natural assumption we cycles in the Tanner graph of parity-check matrix H. prove some probabilistic statements with respect to cycles of Proposition 1. [15]. An exponent chain forms a cycle in length4andprovideacomparisonbetweenliftingprocedures. the Tanner graph of H iff the following condition holds In Section IV, we present our floor scale modulo lifting 2l (cid:88) method for QC-LDPC codes and prove several probabilistic (−1)iai ≡0 mod L. statements concerning theoretical improvement of the method i=1 III. LIFTINGOFQC-LDPCCODES that a −b +d −c ≡0 mod 2. Therefore, the conditional 2 2 2 2 probability A. State-of-art Lifting Methods Consider a QC-LDPC code with mL0×nL0 parity-check Pr(C1 |C0)=Pr(a2−b2−c2+d2 =0|C0)= matrix H0 with circulant size L0, m × n exponent matrix Pr(a2−b2−c2+d2 =0|a2−b2−c2+d2 ≡0 mod 2)=3/4. E(H ) = (E (H )) and mother matrix M(H ). Given a 0 ij 0 0 set of circulant sizes {L }, L < L , lifting is a method of Indeed we have exactly 8 = 23 equiprobable choices for constructing QC-LDPC ckodeskwith m0Lk ×nLk parity-check a2,b2,c2,d2 depicted in Table I, 6 = (cid:0)42(cid:1) of which give the matrices H from H , which have the same mother matrix cycle. k 0 M(H ) = M(H ) and entries of exponent matrices E(H ) k 0 k TABLEI satisfy −1 ≤ E (H ) ≤ L − 1. Therefore, it suffices to ij k k POSSIBLECHOICESFORa2,b2,c2,d2 specify a formula using which we recalculate each value of E(H ) from E(H ). In paper [22] two lifting approaches are k 0 a2 b2 c2 d2 a2−b2−c2+d2 given. 0 0 0 0 0 Floor lifting is defined as follows: 0 0 1 1 0 0 1 0 1 0 Eij(Hk)=(cid:40)(cid:106)LLk0 ×Eij(H0)(cid:107), if Eij(H0)(cid:54)=−1, (3) 01 10 10 01 -22 −1, otherwise. 1 0 1 0 0 1 1 0 0 0 Modulo lifting is determined by the following equation: 1 1 1 1 0 (cid:40) E (H ) mod L , if E (H )(cid:54)=−1, Eij(Hk)= ij 0 k ij 0 (4) Now let us find the probability Pr(C1 |C0). Since −1, otherwise. Pr(C C ) Pr(C |C )= 1 0 Now we prove several probabilistic statements. 1 0 Pr(C ) 0 Consider an exponent chain of length 4 with exponent and Pr(C ) = 2q−1, it suffices to obtain Pr(C C ). Find the values a, b, c, d 0 2q 1 0 numberofall4-tuples(a,b,c,d),suchthata −b −c +d ≡0 (cid:20) (cid:21) 1 1 1 1 A= a b , mod q and a2−b2−c2+d2 (cid:54)≡0 mod 2q. We have q3 ways c d to choose a ,b ,c ,d and 10 ways to choose a ,b ,c ,d 1 1 1 1 2 2 2 2 whereeachelementischosenindependentlyandequiprobable for q >2. Therefore, fromtheset{0,1,...,2q−1},L0 =2qisacirculantsize,q > 10q3 5 5 2. Notice that the probability of the event C0: “the exponent Pr(C1C0)= (2q)4 = 8q and Pr(C1 |C0)= 4(2q−1). chain with exponent values a, b, c and d forms a cycle”, i.e., a−b−c+d ≡ 0 mod 2q, is equal to 1/(2q). Assume that Let us sum up the results in we use some lifting method to obtain exponent values a(cid:48), b(cid:48), Proposition 2. An exponent chain in E(H) of length 4, c(cid:48), d(cid:48) whichformsacycleintheparity-checkmatrixHwithcirculant size 2q, turns into a cycle in the parity-check matrix H(cid:48) with (cid:20) a(cid:48) b(cid:48) (cid:21) B = , circulantsizeqobtainedafterfloorliftingwithprobability3/4, c(cid:48) d(cid:48) while an exponent chain of length 4, which does not form a forcirculantsizeL =q.Weareinterestedintheprobabilities cycle, turns into a cycle with probability p (cid:44)5/(4(2q−1)). 1 fl of an event C : “the exponent chain with exponent values a(cid:48), 1 C. Modulo Lifting b(cid:48), c(cid:48) and d(cid:48) forms a cycle” given the event C and given 0 the event C0. In Sections III-B and III-C we obtain these Leta=a1q+a2,b=b1q+b2,c=c1q+c2,d=d1q+d2, probabilities for floor lifting and modulo lifting, respectively. wherea2,b2,c2,d2 ∈{0,1,...,q−1}.Itiseasytocheckthat Finally,wesummarizeresultsandcomparethesetwomethods a(cid:48) =a2, b(cid:48) =b2, c(cid:48) =c2, d(cid:48) =d2. Given the event C0 occurs, in Section III-D. we have q(a −b −c +d )+(a −b −c +d )≡0 mod 2q. B. Floor Lifting 1 1 1 1 2 2 2 2 It follows that Leta=2a +a ,b=2b +b ,c=2c +c andd=2d + 1 2 1 2 1 2 1 d2, where a2,b2,c2,d2 ∈ {0,1}. One can see that a(cid:48) = a1, a(cid:48)−b(cid:48)−c(cid:48)+d(cid:48) =a −b −c +d ≡0 mod q, 2 2 2 2 b(cid:48) =b , c(cid:48) =c , d(cid:48) =d . Given the event C occurs, i.e. 1 1 1 0 thustheconditionalprobabilityPr(C |C )=1.Letusobtain 1 0 2(a1−b1−c1+d1)+(a2−b2−c2+d2)≡0 mod 2q. probability Pr(C1 |C0). Since the event C1, i.e., a1−b1+d1−c1 ≡0 mod q, is equivalent Pr(C |C )= Pr(C1C0) to the condition a2 −b2 −c2 +d2 = 0. From C0 it follows 1 0 Pr(C0) and Pr(C )= 2q−1, we need to find Pr(C C ). Calculate the Let a = 2a + a , b = 2b + b , c = 2c + c and d = 0 2q 1 0 1 2 1 2 1 2 numberofall4-tuples(a,b,c,d),suchthata −b −c +d ≡0 2d +d , where a ,b ,c ,d ∈{0,1}. Suppose we use floor 2 2 2 2 1 2 2 2 2 2 mod q and a−b−c+d(cid:54)≡0 mod 2q. We have q3 ways to scale modulo lifting for L = q with scale value r = 2t+1, 1 choose a ,b ,c ,d and 8 ways to choose a ,b ,c ,d for 02. Therefore, B(r): 8q3 1 1 (cid:20) a(cid:48)(r) b(cid:48)(r) (cid:21) Pr(C C )= = and Pr(C |C )= . B(r)= , 1 0 (2q)4 2q 1 0 2q−1 c(cid:48)(r) d(cid:48)(r) As a result we have obtained the following where Proposition 3. An exponent chain in E(H) of length 4, (cid:22)2a r+a r(cid:23) a(cid:48)(r)= 1 2 ≡a r+a t mod q. whichformsacycleintheparity-checkmatrixHwithcirculant 2 1 2 size 2q, turns into a cycle in the parity-check matrix H(cid:48) with Othervaluesb(cid:48)(r),c(cid:48)(r)andd(cid:48)(r)arerepresentedinthesame circulant size q obtained after modulo lifting with probability way. By C (r) denote the event: “the exponent chain with 1,whileanexponentchainoflength4,whichdoesnotforma 1 cycle, turns into a cycle with probability p (cid:44)1/(2q−1). exponent values a(cid:48)(r), b(cid:48)(r), c(cid:48)(r) and d(cid:48)(r) forms a cycle”. mod One can see that D. Comparison 3 Pr(C (r)|C )=Pr(C (1)|C )= . Now summarize the results from Sections III-B and III-C 1 0 1 0 4 in the following Moreover Theorem 1. Suppose that in exponent matrix E(H) with C (r)∩C =C (1)∩C . 1 0 1 0 circulantsize2q wehavey exponentchainsoflength4,which donotformacycle,andxexponentchainsoflength4,which Proposition4. Letr1,r2betwodistinctintegers,suchthat formacycle.ThenmathematicalexpectationsECfl (ECmod) 00. Therefore, 2 2n √ Emin(X ,X ,...,X )=EminX+o( y)= 1√ 2 √Nr √ √ and minEX− yc +o( y)=p y− yc +o( y). (9) Nr fl Nr (cid:18) (cid:19) y Pr(X +X =n)=(2p )n(1−2p )y−n , 1 2 fl fl n C. Proof of Theorem 3 it follows the statement of the proposition. Proof:WeprovetheformulaforP only.Firstly,find fsml theprobabilityP thatthefirstk rowsofthematrixDcontain k B. Proof of Theorem 2 only zeros Proof: Consider a random vector Pk =(1−kpfl)y. Secondly, using the inclusion-exclusion principle we get X=(X ,X ,...,X ,X ), 1 2 Nr Nr+1 (cid:80)Nr P =(cid:88)Nr(−1)k−1(cid:18)Nr(cid:19)(1−kp )y. (10) where XNr+1 =y− Xi). fsml k fl i=1 k=1 One can see that X has a multinomial distribution with Now we prove an auxiliary statement. y! N(cid:80)rxi Lemma 1. The binomial identity Pr(x1,...,xNr+1)= x1!·...·xNr+1!pif=l0 (1−Nrpfl)xNr+1. (cid:88)n (−1)k(cid:18)n(cid:19)g(k)=0 k BytheCentrallimittheoremthedistributionofrandomvector k=0 X−√EX tendstonormaldistributionN(0,Σ)asy →∞,where holds for every polynomial g(k) with degree less than n. y Σ is the covariance matrix of X. Proof: Every polynomial g(k) of degree t can be repre- Let us prove that Emin(X ,X ,...,X )−EminX = sented in the following form √ 1 2 Nr o( y) as y →∞. t (cid:88) g(k)= c (k) , l l |Emin(X ,X ,...,X )−EminX| 1 2 Nr l=0 ≤yP(minX=XNr+1)≤yP(XNr+1 ≤y/(Nr+1)) where (cid:18) (cid:19) 1 (k) =k(k−1)...(k−l+1), ≤yP X ≤EX (6) l Nr+1 Nr+1(1−p N )(N +1) fl r r ans c are some coefficients. l From N ≤ϕ(2q)≤q it follows that 1−p N >p for For every l2. The following chain of inequalities takes place n (cid:18) (cid:19) n (cid:18) (cid:19) (cid:88) n (cid:88) n (−1)k (k) = (−1)k (k) k l k l 1−p N >p ⇒ fl r fl k=0 k=l pfl(Nr+1)Nr ≤Nr 0 and use the (n−l)! (n−k)!(k−l)! (1−pflNr)(Nr+1) k=l Chernoff inequality n (cid:18) (cid:19) n! (cid:88) n−l = (−1)k =0, (11) yP(XNr+1 ≤EXNr+1(1−δ))≤ye−δ2EX2Nr+1 =o(√y). (n−l)! k=l n−k therefore, n (cid:18) (cid:19) (cid:88) n (−1)k g(k)=0. k k=0 Finally, using the evident asymptotic (cid:18) (cid:19) 5 1 p = =O , as q →∞, fl 4(2q−1) q along with Lemma 1 and the equality (10), we obtain the statement of Theorem 3.

Floor Scale Modulo Lifting for QC-LDPC codes

Similar Floor Scale Modulo Lifting for QC-LDPC codes