Floor Scale Modulo Lifting for QC-LDPC codes

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by  Nikita Polyanskii

Floor Scale Modulo Lifting for QC-LDPC codes
Nikita Polyanskii, Vasiliy Usatyuk, and Ilya Vorobyev
Huawei Technologies Co., Moscow, Russia
Email: [email protected], [email protected], [email protected]
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 0<r <2q,whichiscoprimewith2q.Thenweobtainmatrix
2 2 2 2 2 2 2 2
q >2. 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) 0<r1,r2 <2q,(r1,2q)=1,(r2,2q)=1andr1 (cid:54)≡r2(q+1)
of the number of cycles after floor lifting (modulo lifting) for mod 2q. Then
circulant size q are as follows:
Pr(C (r )∩C (r )∩C )=0.
1 1 1 2 0
3 5 1
ECfl = 4x+ 4(2q−1)y, ECmod =x+ (2q−1)y. In other words, for any scale values r1 and r2 fulfilled the
condition of Proposition 4 if the start exponent chain in the
Note that ECfl ≥ ECmod when y ≥ (2q − 1)x. Since matrix A does not form a cycle then at least one exponent
usually we try to eliminate short cycles in matrix E(H), the chain in the matrices B(r ) and B(r ) does not form a cycle
1 2
number y is likely to be much greater than (2q −1)x. So, too.
we can conclude that modulo lifting is better than floor lifting Proof: Let u be such integer that u r ≡ 1 mod 2q.
1 1 1
with respect to the number of short cycles. Note that C (u )=C . Therefore,
0 1 0
IV. FLOORSCALEMODULOLIFTINGOFQC-LDPC Pr(C (r )∩C (r )∩C )=Pr(C (r u )∩C (r u )∩C )=0.
1 1 1 2 0 1 1 1 1 2 1 0
CODES
Assume events C (r u ) and C (r u ) occur. Thus, a −b −
1 1 1 1 2 1 1 1
Now we introduce the proposed lifting method which we
c +d ≡0 mod q and
1 1
call floor scale modulo lifting:
(cid:40)−1, E (H )=−1, (a1−b1−c1+d1)r2(cid:48) +(a2−b2−c2+d2)t(cid:48)2 ≡0 mod q,
ij 0
Eij(Hk)= (cid:106)LLk0((r×Eij(H0)) mod L0)(cid:107), otherwise, where
(5) 1+2t(cid:48) =r(cid:48) ≡r u mod q, 0<r(cid:48) <2q.
2 2 2 1 2
where special parameter r is called a scale value.
From
Define A(r):
(a −b −c +d )t(cid:48) ≡0 mod q,
(cid:20) (cid:21) 2 2 2 2 2
a(r) b(r)
A(r)= c(r) d(r) , (a2−b2−c2+d2)∈[−2,2], t(cid:48)2 ∈[1,q−1]
where and 2t(cid:48) +1 (cid:54)= q+1 it follows that a −b −c +d = 0.
2 2 2 2 2
Hence (a −b −c +d )≡0 mod q and
a(r)≡ra mod 2q, b(r)≡rb mod 2q, 1 1 1 1
2(a −b −c +d )+(a −b −c +d )
c(r)≡rc mod 2q, d(r)≡rd mod 2q. 1 1 1 1 2 2 2 2
=a−b−c+d≡0 mod 2q,
ByC (r)denotetheevent:“theexponentchainwithexponent
0
valuesa(r),b(r),c(r)andd(r)formsacycle”.Noticethatfor i.e., we prove that C (r )∩C (r )⇒C .
1 1 1 2 0
r coprime with 2q, i.e. (r,2q) = 1, elements of matrix A(r) Remark 1. Note that if r ≡ r (q + 1) mod 2q, then
1 2
have the same distribution as matrix A. Moreover, exponent r ≡ r (q+1) mod 2q. Therefore, we can choose a set R
2 1
chainsfrommatricesAandA(r)formacyclesimultaneously. of scale values of cardinality ϕ(2q)/2 (ϕ(n) is Euler’s totient
function) for even q and ϕ(2q) for odd q, such that for every we are interested in the probability that after lifting for the
r ,r ∈R the conditions of Proposition 4 are fulfilled. circulant size L = q we will not obtain any cycle of length
1 2 1
Consider a floor scale modulo lifting with a family R = 4. We again assume that all events C are independent for
1
{r ,r ,...,r } of N scale values, such that for any two all exponent chains, i.e., all columns of matrix D are chosen
1 2 Nr r
scale values r , r ∈ R the conditions of Proposition 4 are independently.
i j
satisfied. Let D = (D ) be an N ×y matrix, where the i- Theorem3.[28]. Theprobabilityoftheabsenceofcycles
ij r
th row corresponds to scale values r ∈ R, and each column of length 4 in the parity-check matrix with circulant size
i
corresponds to one exponent chain of length 4 in E(H). We q obtained after modulo lifting, floor lifting and floor scale
setD to1ifthej-thexponentchainformsacycleafterfloor modulo lifting is as follows
ij
scale modulo lifting with scale value r , and to 0 otherwise.
i P =(1−p )y =1−yp +O(q−2), q →∞
The first x columns, which corresponds to cycles in exponent mod mod mod
matrixwithcirculantsize2q,equaltothecolumnofoneswith P =(1−p )y =1−yp +O(q−2), q →∞
fl fl fl
probability 3/4 and to the column of zeros with probability
1/4. The rest y columns equal to the column of zeros with
probability 5N P (N )=(cid:88)Nr(−1)k−1(cid:18)Nr(cid:19)(1−kp )y
1−N p =1− r fsml r k fl
r fl 4(2q−1) k=1
(cid:40)
and to the column of weight 1 with one at position i with = 1−O(q−Nr), if y ≥Nr, q →∞,
probabilitypfl =5/(4(2q−1))foreachi∈[1,Nr].LetXi be 1, if y <Nr, q →∞.
equal to the number of ones in the i-th row. We are interested
In this case we see that floor scale modulo lifting is much
intheminimumnumberofcyclesmin(X ,X ,...,X ).For
1 2 Nr better than modulo and floor lifting.
further calculations we assume that all columns of matrix
Table II shows one of possible advantages of the proposed
D are chosen independently. Under this assumption exact
lifting approach. We compare the floor lifting length adaption
formulas for the mathematical expectation EC (N ) =
fsml r of QC-LDPC codes used in IEEE 802.16 for rate 1/2 with
3x/4+Emin(X ,X ,...,X ) could be easily written out,
1 2 Nr the proposed floor scale modulo lifting. We apply the lifting
but they rather messy. We provide only formula for the case
methodstothe12×24mothermatrix.Wehavefoundoptimal
N =2 in the form of
r r scalevalueforourliftingapproachwithrespecttogirthand
Proposition 5.[28]. Supposewehaveanexponentmatrix
number of exponent chains which form cycles of the minimal
E(H) with circulant size 2q having x exponent chains of
length. In Table II for each circulant size the optimal r scale
length 4, which form a cycle in H, and y exponent chains of
value, girth and the number of cycles are depicted. Note that
length 4, which do not form a cycle. Then the mathematical
the QC-LDPC code of IEEE 802.16 standard was optimized
expectation EC (2) of the number of cycles of length 4 in
fsml with considering floor lifting method. If the QC-LDPC code
the parity-check matrix of circulant size q obtained after floor
withthemaximallengthsizeisnotoptimizedwithconsidering
scale modulo lifting with N =2 scale values, which satisfies
r floor or modulo lifting method, then the superiority of the
the conditions of Proposition 4, is described by the following
proposedfloorscalemoduloliftingwillbemoreconspicuous.
expression
y (cid:32) (cid:0) n (cid:1)(cid:33)
3 (cid:88) n (cid:98)n(cid:99) V. SIMULATIONRESULTS
EC (2)= x+ 1− 2
fsml 4 2 2n
n=0 QC-LDPC codes of smaller lengths can be obtained by
(cid:18) (cid:19)
×(2p )n(1−2p )y−n y . liftingexponentmatrixofQC-LDPCcodesofmaximallength.
fl fl n TheirperformanceoveranAWGNchannelwithBPSKmodu-
lation was analyzed by computer simulations. Figure 1 shows
The proof of Proposition 5 is provided in the full version
theframeerrorrate(FER)performanceofrate4over5AR4JA
of the given paper [28].
code defined by protograph of size 3×11 from [24]. We use
Ify →∞theasymptoticbehaviorofEC (N )isgiven
fsml r
native lifting for fixed circulant sizes {16,32,64,128} and
by
floor modulo scale lifting beginning from parity-check matrix
Theorem 2. [28]. The mathematical expectation of the
H of circulant size 128 which goes down to circulant sizes
number of cycles of length 4 after floor scale modulo lifting
{16,32,64}. BP decoder with 100 iterations is used.
has the following asymptotic form
Figure 2 shows the SNR required to achieve 10−2 FER
3 √ √
EC = x+p y−c y+o( y), if y →∞, performance over an AWGN channel with QPSK modulation
fsml 4 fl Nr for3familiesofQC-LDPCcodeswithrate8over9.Families
where c does not depend on y. A [26] and B [25] are the industrial state-of-art QC-LDPC
Nr
Let us consider another scenario. Suppose that the number codes with their own lifting. For family C we applied floor
ofcyclesoflength4inmatrixHwithcirculantsizeL =2qis modular scale lifting. Layered normalized offset min-sum
0
equalto0,andthenumbery ofexponentchainsisfixed.Now decoderwith15iterationswasusedinsimulations.Normalize
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APPENDIX Let denote as minn(X) the function which equals
max(min(min(X),n),−n). This function is continuous and
A. Proof of Proposition 5
bounded, hence
Proof: Consider a random variable min(X ,X ). Then
1 2 (cid:18) (cid:19)
X−EX
lim Emin √ =
Emin(X ,X )=EE(min(X ,X )|X +X ) y→∞ y
1 2 1 2 1 2
X−EX
y
= (cid:88)Emin(Y,n−Y)Pr(X1+X2 =n), nl→im∞yl→im∞Eminn √y
n=0 = lim Emin N(0,Σ)=
n
n→∞
where Y ∼B(n,1). From EminN(0,Σ)=−c , (8)
2 Nr
Emin(Y,n−Y)= n(cid:32)1− (cid:0)(cid:98)nn/2(cid:99)(cid:1)(cid:33) where cNr =EmaxN(0,Σ)>0. 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 l<n
r fl r fl
q >2. 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 <Nr+1⇒ =(cid:88)n (−1)k n!k! =(cid:88)n (−1)k n!k!
1 k!(n−k)!(k−l)! k!(n−k)!(k−l)!
<1. (7) k=l k=l
(1−p N )(N +1)
fl r r n! (cid:88)n (n−l)!
= (−1)k
Denote 1 as 1−δ for some δ >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.