On the Deterministic Sum-Capacity of the Multiple Access Wiretap Channel

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by  Rick Fritschek

On the Deterministic Sum-Capacity of the Multiple
Access Wiretap Channel
Rick Fritschek Gerhard Wunder
Heisenberg Communications and Information Theory Group Heisenberg Communications and Information Theory Group
Freie Universität Berlin, Freie Universität Berlin,
Takustr. 9, D–14195 Berlin, Germany Takustr. 9, D–14195 Berlin, Germany
Email: [email protected] Email: [email protected]
7 Abstract—We study a deterministic approximation of the are scaled such that alignment can be achieved. The intended
1 two-user multiple access wiretap channel. This approximation messages are recovered by minimum-distance decoding and
0 enables results beyond the recently shown 2 secure degrees of
2 freedom(s.d.o.f.)fortheGaussianmultipleac3cesschannel.While the error probability is bounded by usage of the Khintchine-
n the s.d.o.f. were obtained by real interference alignment, our Groshev theorem of Diophantine approximation theory. The
a approach uses signal-scale alignment. We show an achievable disadvantage of the method is that these results only hold
J scheme which is independent of the rationality of the channel for almost all channel gains. This is unsatisfying for se-
5 gains. Moreover, our result can differentiate between channel crecy purposes, since it leaves an infinite amount of cases
strengths, in particular between both users, and establishes a
2 where the schemes do not work, e.g. rational channel gains.
secrecy rate dependent on this difference. We can show that the
Moreover, secrecy should not depend on the accuracy of
resultingachievablesecrecyratetendstothes.d.o.f.forvanishing
]
T channel gain differences. Moreover, we extend the s.d.o.f. bound channelmeasurements.Realinterferencealignmentispartofa
towardsageneralboundforvaryingchannelstrengthsandshow broaderclassofinterferencealignmentstrategies.Interference
I
. thatourachievableschemereachestheboundforcertainchannel alignment (IA) was introduced in [5] and [6], among others,
s
c gain parameters. We believe that our analysis is the first step and its main idea is to design signals such that the caused
[ towards a constant-gap analysis of the Gaussian multiple access
wiretap channel. interference overlaps(aligns) and therefore uses less signal

1 dimensions. The resulting interference-free signal dimensions
v I. INTRODUCTION can be used for communication. IA methods can be di-
0 In this paper we study the secure capacity of an approx- vided into two categories, namely the vector-space alignment
8
imation of the Gaussian multiple access wiretap channel. approach and the signal-scale alignment approach [7]. The
3
The wiretap channel was first proposed by Wyner in [1], formerusestheclassicalsignallingdimensionstime,frequency
7
0 and solved in its degraded version. The result was later and multiple-antennas for the alignment, while the latter uses
. extendedtothegeneralwiretapchannelbyCsiszarandKörner the signal strength for alignment. Real interference alignment
1
0 in [2]. Moreover, the Gaussian equivalent was studied by and signal-strength deterministic models are examples for
7 Leung-Yan-Cheon and Hellman in [3]. The wiretap channel signal-scale alignment. Signal-strength deterministic models
1 and its modified version served as an archetypical channel are based on an approximation of the Gaussian channel. An
:
v for physical-layer security investigations. However, in recent example for such an approximation is the linear deterministic
i years, the network nature of communication, i.e. support of model (LDM), introduced by Avestimehr et al. in[8]. It is
X
multiple users, becomes increasingly important. A straight- based on a binary expansion of the transmit signal, and an
r
a forward extension of the wiretap channel to multiple users approximation of the channel gain to powers of two. The
was done in [4], where the Gaussian multiple access wiretap resulting binary expansion gets truncated at the noise level
channel (GMAC-WT) was introduced. A general solution for which yields a noise-free binary signal vector and makes the
the secure capacity of this multi-user wiretap set-up was out model deterministic. It has been shown that various Gaussian
of reach and investigations focused on the secure degrees of channels(i.e.[9],[10],[11],[12])canbeapproximatedbythe
freedom (s.d.of.) of these networks. Degrees of freedom are LDM such that the deterministic capacity is within a constant
used to gain insights into the scaling behaviour of multi- bit-gap of the Gaussian channel. Moreover, layered lattice
user channels. They measure the capacity of the network, codingschemescanbeusedtotransfertheachievablescheme
normalized by the single-link capacity, as power goes to to the Gaussian model.
infinity. This also means that the d.o.f. provide an asymptotic Previous work and Contributions:Previousresultsonthe
view on the problem at hand. This simplifies the analysis multiple access wiretap channel include the sum secure d.o.f.
and enables asymptotic solutions of channel models where of K(K−1) for the K-user case in [13]. There, the authors
K(K−1)+1
no finite power capacity results could be found. An example used a combination of real interference alignment together
for a technique which yields d.o.f. results is real interference with cooperative jamming [14]. The idea is that the users
alignment. It uses integer lattice transmit constellations which can allocate a small part of the signalling dimensions with
Z1 the probability of error P(n) = P[(Wˆ ,Wˆ ) (cid:54)= (W ,W )]
e 1 2 1 2
UsWer11 X1 b h11 b Y1 gmoeesssatgoezWeroi,sassaind tgooebsetionfoinrfimnaittiyonl-itmheno→re∞ticPael(lny) s=ecu0re. Aif
h21 the eavesdropper cannot reconstruct the message W from the
channel observation Yn. This means that the uncertainty of
Z 2
h12 2 the message is almost equal to its entropy, given the channel
User2 Eavesdropper
W2 X2 b h22 b Y2 observation.ConsideringbothmessagesW1,W2,wehavethat
1H(W ,W |Yn)≥ 1H(W ,W )−(cid:15), (2)
n 1 2 2 n 1 2
Figure1. TheGaussianmultipleaccessWiretapchannel. which leads to I(W ,W ;Yn)≤(cid:15)n for any (cid:15)>0. A secrecy
1 2 2
rater issaidtobeachievableifitisachievablewhileobeying
the secrecy constraint (2).
uniform distributed random bits. Those random bits are send
such that they occupy a small space at the legitimate receiver, A. The Linear Deterministic Model
while overlapping with the signals at the eavesdropper. The We investigate the linear deterministic model (LDM) of
crypto-lemmashows,thattheeavesdroppercannotrecoverthe the multiple access wiretap channel as simplification of the
signal without knowledge of the random bits, hence securing corresponding Gaussian model. This approximation models a
communication. The next step is to transition from the d.o.f. signal in the channel as bit-vector X, which is achieved by
result,toaconstant-gapcapacityresult.In[15]thepreviously a binary expansion of the input signal X. The elements of
known 12 d.o.f.resultofthewiretapchannelwithahelper[13] the resulting vector are referred to as bit-levels. The addition
wasextendedtowardsaconstant-gapresultforcertainchannel of signals is modelled by binary addition. Carry-overs are
gains. This was achieved by approximating the problem with neglected and the addition is limited to the specific bit-level.
the LDM and then transferring the results to the Gaussian The resulting bit-vectors get truncating at the noise bit-level,
model. We will use the same strategy for the multiple access whichyieldsadeterministicmodel.Moreover,specificchannel
wiretap channel, i.e. restrict the problem to the two user gains shift the bit-vectors for a certain number of bit-levels
case and approximate it by the LDM. We develop a novel with a shift-matrix S, which is defined as
communication scheme which achieves the 2 d.o.f. result as
3 0 0 ··· 0 0
baseline and extends to a generalized s.d.o.f. characterization.
1 0 ··· 0 0
Moreover, we transfer bounding techniques of [13] to the  
dwehteicrhmianlsisotiicnsceluttdinegsaansydmexmteentrdictahlecmhatonwnealrdgsaiangs.eneralbound, S=0... 1... ·.·..· 0... 0.... (3)
0 0 ··· 1 0
II. SYSTEMMODEL
The model therefore shifts an incoming bit vector for q−n
The Gaussian multiple access wiretap channel is defined
positions with Y = Sq−nX, where q := max{n}. Channel
as a system consisting of two transmitters and two receivers.
gains are represented by n -bit levels which corresponds to
Each transmitter i has a channel input X which it sends ij
i
(cid:100)logSNR(cid:101) of the original channel. With this definitions, the
over the channel to the legitimate receiver Y , whereas the
1
system model can be approximated by
eavesdropper Y tries to intercept the messages. The channel
2
itself is modelled with additive white Gaussian noise, Z1,Z2. Y1 =Sq−n11X(cid:48)1⊕Sq−n21X(cid:48)2 (4a)
The system equations can therefore be written as Y =Sq−n22X(cid:48) ⊕Sq−n12X(cid:48), (4b)
2 2 1
Y =h X +h X +Z (1a)
1 11 1 21 2 1 where q :=max{n ,n ,n ,n }.For ease ofnotation,we
11 12 21 22
Y2 =h22X2+h12X1+Z2, (1b) denote X1 = Sq−n11X(cid:48)1 and X2 = Sq−n21X(cid:48)2. Furthermore,
where the channel inputs satisfy an input power constraint we denote Sq−n22X(cid:48)2 and Sq−n12X(cid:48)1 by Xˆ2 and Xˆ1, respec-
tively. We also assume that n = n =: n , and denote
E{|X |2} ≤ P for each i. Moreover, the Gaussian noise 22 12 E
i n − n =: n with n =: n and n =: n . We may
terms are assumed to be independent and zero mean with 1 2 ∆ 11 1 21 2
assume w.l.o.g. that n > n , where we leave out the case
unit variance, Zi ∼ CN(0,1). A (2nR1,2nR2,n) code for that n = n , see rem1ark (32). To specify a particular range
the multiple access wiretap channel will consist of a message 1 2
of elements in a bit-level vector we use the notation a to
pair (W ,W ) uniformly distributed over the message set [i:j]
1 2 indicate that a is restricted to the bit-levels i to j. Bit levels
[1 : 2nR1]×[1 : 2nR2] with a decoding and two randomized
are counted from top, most significant bit in the expansion, to
encodingfunctions.Encoder1assignsacodewordXn(w )to
1 1 bottom. If i = 1, it will be omitted a , the same for j=n
each message w1 ∈ [1 : 2nR1], while the encoder 2 assigns a . [:j]
a codeword X2n(w2) to each message w2 ∈ [1 : 2nR2]. The [i:]
decoderassignsanestimate(wˆ1,wˆ2)∈[1:2nR1]×[1:2nR2] Remark 1. The assumption that n22 = n12 = nE, i.e.
to each observation of Yn. A rate is said to be achievable the eavesdropper receives the signals with equal strength,
1
if there exist a sequence of (2nR1,2nR2,n) codes, for which does not influence the achievable secrecy sum-rate. Consider
X
a channel with n (cid:54)= n , for example n > n . The 1
22 12 22 12 Message
part of X(cid:48) which is received above n at the eavesdropper, n
2 12 X ∆
Xˆ , cannot be utilized since it cannot be jammed. 2 Jamming
On2e,[:nc2a2n−nth1e2]refore achieve the same rate by ignoring the top 3n∆ Y
1,c
n22−n12 bitsofX(cid:48)2.Thesameargumentholdsforn12 >n22. Xˆ1 Xˆ2
III. MAINRESULT
A. Achievable Secrecy Rate n
c
Theorem 1. The achievable secrecy sum-rate R of the
ach
linear deterministic multiple access wiretap channel with
symmetric channel gains at the eavesdropper is
(cid:40)
Rach = 2332(((cid:98)(cid:98)33nnnncc∆∆(cid:99)(cid:99)33nn∆∆))++nQp.+Q. ffoorr nn2E≥>nnE2, (5) nQ
where n =n +n , n =n −n and n
c E ∆ p 1 c p

q for n <n
 Q ∆ Y Y
1 2
Q= n for 2n >n ≥n (6)
∆ ∆ Q ∆
n∆+q for nQ ≥2n∆, Figure2. Illustrationoftheachievablescheme.TheprivatepartY1,pcanbe
usedfreelyandis,inthiscase,allocatedbyUser1.ThecommonpartY1,c
with nQ =nc−(cid:98)3nnc∆(cid:99)3n∆ and q =nQ−(cid:98)nn∆Q(cid:99)n∆. ubseetwseoeunrbaloitghnmsigenntalsst,ratotegmyi.nTimheizsetrtahteegeyffeexcptloofitjsatmhemcinhganantetlhgeariencdeiivffeerreYnc1e,
1) Case 1 (n ≥ n ): First of all, we introduce the whilejammingallsignalpartsattheeavesdropperY2.
2 E
commonandprivatepartsofthereceivedsignalY .Wecount
1
the bits from the top (most-significant bit) downwards. The
seen that every 3n -part of Y allocates 2n bits for the
common part will be denoted by Y and consist of the top ∆ 1,c ∆
1,c
messages. This results in the common secrecy rate
n =n +n bitsofY .Theremainder,theprivatepart,will
c E ∆ 1
sbiegndaelnoptaerdtsaswYhi1c,hpaarnednhoatsrne1c−eivnecd:=byntphebietsa.vIetsodnrolyppceorn.taOinusr rc =((cid:98)3nnc∆(cid:99)3n∆)32 +Q, (8)
strategy is to deploy a cooperative jamming scheme such that where Q specifies the rate part of the remainder term. In the
minimal jamming is done to Y , while maximal jamming is remainder part we allocate all remaining bits as message bits,
1,c
received at Y2. We denote the part of X1 in Y1,c by X1,c. as long as nQ < n∆. For 2n∆ > nQ ≥ n∆, we allocate the
Moreover, we denote the part of X2 which gets received at first n∆ bits of nQ for the message. And for nQ ≥2n∆, we
Y2 by X2,c. We partition these common signals into 3n∆- allocate the first n∆ bits as well as the last q bits, where q is
bit parts and partition these parts again into n -bit parts. For defined as
∆
X1,c, in every 3n∆-bit part we use the first n∆ bits for the q =nQ−(cid:98)nn∆Q(cid:99)n∆. (9)
message and the next n bits for jamming, while the last n
∆ ∆ This results in
bits willnot beused. For X , inevery 3n -bit part,the first
2,c ∆ 
n∆ bits will be used for jamming. The next n∆ bits will be q for nQ <n∆
used for the message and the last n∆ bits left free. There will Q= n∆ for 2n∆ >nQ ≥n∆ (10)
be a reminder part with n +q for n ≥2n .
∆ Q ∆
nQ =nc−(cid:98)3nnc∆(cid:99)3n∆ bits. (7) Together with the private rate term, we achieve
Tpahretsr,eumntiinldner pbairtst aforelloawllosctahteeds.aTmheedscehsiegmneruisledsesaisgntheeds3unc∆h R= 32((cid:98)3nnc∆(cid:99)3n∆)+np+Q. (11)
Q
thatthejammingpartsofX1,c andX2,c overlapatY1,c,while 2) Case 2 (n2 ≥nE): The achievable scheme is the same
the message parts of one signal overlap with the non-used as in the previous case, except that we do not have a private
part of the other signal. However, due to the signal strength part. We therefore have an achievable rate of
datifYfer2e,nsceeeFni∆g,.2th.eSejacmurmeicnogmpmaurtnsicoavteiorlnapiswthiethrefthoerempreosvsaidgeeds R= 32((cid:98)3nnc∆(cid:99)3n∆)+Q. (12)
bythecrypto-lemma,aslongasweuseaBern(1)distribution Remark 2. The bit level shift between Y and Y of n
2 1 2 ∆
for the jamming bits. The whole private part can be used bits makes it impossible to divide Y in exclusively private
1
for messaging and its sum-rate is therefore r = n . The and common parts. In our division, the bottom n bits of
p p ∆
achievable secure rate for the common part consists of the x areonlyreceivedatY andthereforeprivate.Hence,the
1,c 1
rate for the 3n partitions and the reminder part. It can be common rate r is not purely made of common signal parts.
∆ c
Nevertheless, our choice of division reaches the upper bound and it therefore holds that
and fits into the scheme.
H(Xn )≤H(Yn )+I(Xn;Yn |Yn )−n(R +(cid:15) ). (15)
2,c 1,c 1 1,p 1,c 1 3
Remark3. Ourschemereliesonthesignalstrengthdifference
The same can be shown for H(Xn ), where it holds that
between both users. Our scheme would not work, if n1 =n2, 1,c
while having equal channel gains at the eavesdropper. In that H(Xn )≤H(Yn )+I(Xn;Yn |Yn )−n(R +(cid:15) ). (16)
1,c 1,c 2 1,p 1,c 2 4
casewewouldnothaveanysignalstrengthdiversitytoexploit
which results in a singularity point where the secrecy rate is Moreover, we have that
zero.
I(Xn;Yn |Yn )+I(Xn;Yn |Yn )
1 1,p 1,c 2 1,p 1,c
B. Converse = 2H(Yn |Yn )−H(Yn |Yn ,Xn)−H(Yn |Yn ,Xn)
1,p 1,c 1,p 1,c 1 1,p 1,c 2
=2H(Yn |Yn )− H(Xn |Yn )−H(Xn |Yn )
Theorem2. Thesecrecysum-rateRach ofthelineardetermin- 1,p 1,c 2,p 1,c 1,p 1,c
istic multiple access wiretap channel with symmetric channel = H(Yn |Yn ). (17)
1,p 1,c
gains at the eavesdropper is bounded from above by
The key idea for the various cases is now to bound the term
rUB =(cid:40)232nnc++n1pn+ 13n∆ ffoorr nn2 ≥>nnE. (13) Hap(pYro1np,rci|aYte2nw),aoyr,etoqubiveaalebnletlytoHu(sYe (1n1|5Y)2na)ndfo(r1n6E) o>nn(124,)i.nWane
3 c 3 ∆ E 2 start with the first case:
Proof: We start with some general observations and 1) Case 1 (n2 ≥ nE): Here we have a none vanishing
derivations before handling the different cases explicitly. We private part, due to the definition of Y1n,c and therefore need
begin with the following derivations toboundthetermH(Y1n,c|Y2n).Notethatduetothedefinition
of Yn we have that H(Xn ) = H(Xˆn). We look into the
1,c 2,c 2
n(R1+R2) = H(W1,W2) first term of equation (14) and show that
= H(W ,W |Yn)+I(W ,W ;Yn)
1 2 1 1 2 1 H(Yn |Yn)=H(Yn ,Yn)−H(Yn)
≤ I(W ,W ;Yn)+n(cid:15) 1,c 2 1,c 2 2
1 2 1 ≤H(Yn ,Xˆn,Xˆn)−H(Yn)
(a) 1,c 1 2 2
≤ I(W1,W2;Y1n)−I(W1,W2;Y2n)+n(cid:15)2 =H(Xˆn,Xˆn)+H(Yn |Xˆn,Xˆn)−H(Yn)
1 2 1,c 1 2 2
≤ I(W1,W2;Y1n,Y2n)−I(W1,W2;Y2n)+n(cid:15)2 ≤H(Xˆn)+H(Xˆn)−H(Yn|Xˆn)
1 2 2 2
≤ I(W1,W2;Y1n|Y2n)+n(cid:15)2 +H(Yn |Xˆn,Xˆn)
1,c 1 2
≤ I(Xn,Xn;Yn|Yn)+n(cid:15)
1 2 1 2 2 =H(Xˆn)+H(Yn |Xˆn,Xˆn). (18)
= H(Yn|Yn)−H(Yn|Yn,Xn,Xn)+n(cid:15) 2 1,c 1 2
1 2 1 2 1 2 2
Observe that the second term of equation (18) is depended on
(=b) H(Yn|Yn)+n(cid:15)
1 2 2 the specific regime. We can bound this term by
≤ H(Yn |Yn)+H(Yn |Yn,Yn )+n(cid:15) (14)
1,c 2 1,p 2 1,c 2 H(Yn |Xˆn,Xˆn)≤n(n −n )=nn . (19)
1,c 1 2 c E ∆
where we used basic techniques such as Fano’s inequality
Note that the choice of Xˆn in (18) as remaining signal part
and the chain rule. Step (a) introduces the secrecy constraint 2
was arbitrary due to our assumption that both signals Xˆn and
(2), while we used the chain rule, non-negativity of mutual 1
Xˆn havethesamesignalstrength.Moreover,itfollowsonthe
informationandthedataprocessinginequalityinthefollowing 2
lines. Step (b) follows from the fact that Yn is a function of same lines that
1
(thXen1p,rXivan2t)e.pNaortt1eothfaYt dnu,eittfooltlhoewdsetfihnatitHion(Yofnth|eYcno,mYmnon)=an0d H(Y1n,c|Y2n)≤H(Xˆn1)+nn∆. (20)
1 1,p 2 1,c
for n ≥n . Looking at this result, its intuitive that one can also show the
E 2
We now extend the strategy of [13], of bounding a single stronger result
signal part, to asymmetrical channel gains
H(Yn |Yn)≤H(Xn ) (21)
1,c 2 1,c
nR =H(W )
1 1 for the case that n ≥n . This can be shown by considering
2 E
≤I(W1;Y1n)−n(cid:15)3 a similar strategy as in (18)
≤I(Xn;Yn)−n(cid:15)
1 1 3 H(Yn |Yn)=H(Yn ,Yn)−H(Yn)
1,c 2 1,c 2 2
=I(Xn;Yn )+I(Xn;Yn |Yn )−n(cid:15)
1 1,c 1 1,p 1,c 3 ≤H(Yn,Xn ,Xn )−H(Yn)
2 1,c 2,c 2
=H(Yn )−H(Yn |Xn)+I(Xn;Yn |Yn )−n(cid:15)
1,c 1,c 1 1 1,p 1,c 3 =H(Xn ,Xn )+H(Yn|Xn ,Xn )−H(Yn)
1,c 2,c 2 1,c 2,c 2
=H(Yn )−H(Xn )+I(Xn;Yn |Yn )−n(cid:15)
1,c 2,c 1 1,p 1,c 3 ≤H(Xn )+H(Xn )−H(Yn|Xn )
1,c 2,c 2 1,c
+H(Yn|Xn ,Xn )−H(Yn)
1ThecommonpartisdefinedasYn =Yn ,andtheprivatepartas 2 1,c 2,c 2
Yn =Yn . 1,c 1,[:nc] =H(Xn )+H(Yn|Xn ,Xn ), (22)
1,p 1,[nc+1:] 1,c 2 1,c 2,c
where Bounding H(Yn|Yn) by H(Xn) requires more work. We
1 2 1
H(Yn|Xn ,Xn )≤(n −n )+ =0. (23) have a redundancy in the negative entropy terms, with which
2 1,c 2,c E 2
we can cancel the H(Xˆn|Xn) term in the following way
2 2
We combine one sum-rate inequality (14) with (18) and
H(Yn|Yn)≤H(Xn,Xn)+H(Xˆn|Xn)−H(Yn )
one with (22). Moreover, we plug (15) and (16) into the 1 2 1 2 2 2 2,c
corresponding bound, which yields ≤H(Xn)+H(Xn)−H(Yn |Xn)
1 2 2,c,[:n2] 1
+H(Xˆn|Xn)−H(Yn |Xn,Yn )
2 2 2,c,[n2+1:] 1 2,c,[:n2]
n(2R +R −(cid:15) )≤H(Yn )+I(Xn;Yn |Yn ) =H(Xn)−H(Yn |Xn,Yn )
1 2 6 1,c 2 1,p 1,c 1 2,c,[n2+1:] 1 2,c,[:n2]
+H(Yn |Yn,Yn ) +H(Xˆn|Xn)
1,p 2 1,c 2 2
≤H(Xn)−H(Yn |Xn,Yn ,Xn)
and 1 2,c,[n2+1:] 1 2,c,[:n2] 2
+H(Xˆn|Xn)
n(R +2R −(cid:15) )≤H(Yn )+I(Xn;Yn |Yn ) 2 2
1 2 7 1,c 1 1,p 1,c =H(Xn)−H(Yn |Xn,Xn)
+H(Y1n,p|Y2n,Y1n,c)+nn∆. +H(1Xˆn|Xn) 2,c,[n2+1:] 1 2
2 2
A summation of these results gives =H(Xn)−H(Xˆn |Xn)+H(Xˆn|Xn)
1 2,c,[n2+1:] 2 2 2
3n(R +R )−n(cid:15) ≤2H(Yn )+I(Xn;Yn |Yn ) =H(Xn). (26)
1 2 8 1,c 1 1,p 1,c 1
+I(Xn;Yn |Yn )+nn
2 1,p 1,c ∆ Now we can bound one (24) with (25) and one with (26).
+2H(Yn |Yn,Yn ). Moreover, we use (15) and (16) on the result. Note that due
1,p 2 1,c
to our regime, (15) becomes
Using (17), and the fact that H(Yn |Yn,Yn ) ≤ nn and
1,p 2 1,c p
H(Y1n,p|Y1n,c)≤nnp results in H(Xn2)≤H(Y1n)−n(R1+(cid:15)3), (27)
3n(R +R )−n(cid:15) ≤2H(Yn )+3nn +nn while (16) becomes
1 2 8 1,c p ∆
≤2nnc+3nnp+nn∆. H(Xn1)≤H(Y1n)−n(R2+(cid:15)4). (28)
Dividing by 3n and letting n→∞ shows the result. Putting everything together results in
2) Case 2 (n >n ): First, we assume that n ≥n , and
E 2 E 1 3n(R +R )−n(cid:15) ≤2H(Yn)+nn
include a short proof for n > n ≥ n at the end of this 1 2 8 1 ∆
1 E 2
≤2nn +nn .
subsection. For Case 2, the private part Y is zero, due to c ∆
1,p
the definition of the private part and nE >n2. It follows that Dividing by 3n and letting n→∞ shows the result.
(14) is We need to modify a bound on H(Yn|Yn), if the signal
1 2
n(R1+R2)≤H(Y1n|Y2n). (24) strength nE lies in between n1 and n2. In (25), we see that
Moreover, H(Xn) = H(Xn ) ≤ H(Xˆn), which is why we H(Xn)−H(Yn |Xn)≤n(n −n )+. (29)
2 2,c 2 1 2,c 2 1 E
need to bound (24) by H(Xn) and H(Xn). We therefore
2 1 Moreover, we have that H(Xˆn|Xn) ≤ n(n −n )+. Both
modify (22) to fit our purpose in the following way 2 2 E 2
changes cancel and we get the same result as (25). The result
H(Yn|Yn)=H(Yn,Yn)−H(Yn) follows on the same lines as in the previous derivation.
1 2 1 2 2
≤H(Xn,Xn)+H(Yn|Xn,Xn)−H(Yn)
1 2 2 1 2 2 IV. CONCLUSION
=H(Xn,Xn)+H(Yn |Xn,Xn)
1 2 2,c 1 2 A. Discussion of the Results
+H(Yn |Xn,Xn,Yn )
2,p 1 2 2,c WehaveapproximatedtheGaussianmultipleaccesswiretap
−H(Yn )−H(Yn |Yn ) channel with the linear deterministic model. This enables a
2,c 2,p 2,c
≤H(Xn,Xn)+H(Yn |Xn,Xn)−H(Yn ), simplified look at the model with an emphasis on the role
1 2 2,c 1 2 2,c
of channels gains in the secrecy sum-rate analysis. We used
where Y2n,c = Y2n,[:n1] and Y2n,p = Y2n,[n1+1:]. Now, we can a signal-scale alignment approach to provide information-
show that theoreticsecurity.Ourapproachdistinguishesbetweenchannel
gain differences which is reflected in the achievable sum-
H(Yn|Yn)≤H(Xn,Xn)+H(Yn |Xn,Xn)−H(Yn )
1 2 1 2 2,c 1 2 2,c rate. Our results agree with previous s.d.o.f. results, as our
=H(Xn,Xn)+H(Xˆn|Xn)−H(Yn ) secrecy sum-rate approaches the s.d.o.f. asymptotically for
1 2 2 2 2,c
≤H(Xn1)+H(Xn2)−H(Y2n,c|Xn2) n∆ → 0 and nE ≥ min{n2,n1}, i.e. without channel gain
difference between users and without private part, see Fig. 3.
+H(Xˆn|Xn)
2 2 Moreover, we have shown upper bounds which are tight for
=H(Xn2)+H(Xˆn2|Xn2) certainn∆ ranges.Wenotethattheachievablesum-ratevaries
≤H(Xn)+nn . (25) betweenbeingaboveandbelowthe 2 threshold.Aninteresting
2 ∆ 3
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n 2
/ 3 Many-to-OneandOne-to-ManyGaussianInterferenceChannels,”IEEE
h
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0.5
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1 3 1 2
2 4 ∞ TheoryWorkshop(ITW),2014IEEE,Nov2014,pp.107–111.
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networks,” IEEE Transactions on Information Theory, vol. 60, no. 6,
Figure3. Thegraphicshowstheachievablesecrecyrate,normalizedbythe pp.3359–3378,June2014.
strongestsignalnmax=max{n1,n2},inblue.Moreover,itshowstheupper [14] E. Tekin and A. Yener, “The general gaussian multiple-access and
bound rUB in green. The regime is nE > min{n1,n2}, which results in two-waywiretapchannels:Achievableratesandcooperativejamming,”
the private rate being zero. The x-axis visualizes the fraction α = n2. At Information Theory, IEEE Transactions on, vol. 54, no. 6, pp. 2735–
n1 2751,June2008.
α=0, there is no penalty in utilizing X2 for jamming, and nmax bits can
[15] R. Fritschek and G. Wunder, “Towards a constant-gap sum-capacity
bsatnerdoanctghheeirevrthaetdaen.aFXpop(cid:48)rr.oαIanc→theercs1ht,ahntehgeisn.gdd.iofthf.efereornofclee23snbomeftawbxo.eteFhnourXse11rs<ainndαthXe≤s2c∞hgee,mtsXesr(cid:48)2mesagulellettssr rIensteurltnaftoirontahleSgyamupssoisainumwiornetaInpfocrhmanantieolnwTihtheoaryh(eIlSpIeTr,)”,Jinuly20210616I,EpEpE.
inamirror-symm1etricbehaviour.Wealsoshowaredcurvefor 2n1+n2. 2978–2982.
3 2
question is, to which channel gain the 2 d.o.f. refer to. The
3
best d.o.f. representation in our scheme is 2 of n1+n2, which
3 2
is represented in the red curve in Fig. 3.
B. Future Research Directions
We think that the results of this paper are important for
a constant-gap sum-capacity analysis of the Gaussian wiretap
channel.TheachievableschemecanbetranslatedtotheGaus-
sian equivalent by using layered lattices codes. The signals
have to be partitioned into n intervals, where every partition
∆
represents a single-link using one lattice code. Moreover, our
upperboundsguidethedesignofsmartgeniesfortheGaussian
channel. We expect that the techniques of [13] together with
smartgenies,whichpartitionthetermsaccordingly,willresult
inboundswhichleadtoaconstant-gapresultfortheGaussian
multiple access wiretap channel.
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