On the Sum Capacity of A Class of Cyclically Symmetric Deterministic Interference Channels

0.14 MB·English

by  Bernd Bandemer

On the Sum Capacity of A Class of Cyclically
Symmetric Deterministic Interference Channels
Bernd Bandemer, Gonzalo Vazquez-Vilar, and Abbas El Gamal
Stanford University, Information Systems Laboratory
350 Serra Mall, Stanford, CA 94305, USA
email: {bandemer, gvazquez}@stanford.edu,[email protected]
9 Abstract—Certain deterministic interference channels have α V1 W3
been shown to accurately model Gaussian interference channels
0 in the asymptotic low-noise regime. Motivated by this corre- M1→X1 Y1→Mˆ1
0 spondence, we investigate a K user-pair, cyclically symmetric, β
2 deterministicinterferencechannelinwhicheachreceiverexperi- V2
n encesinterferenceonlyfromitsneighboringtransmitters(Wyner
a model). We establish the sumcapacity for a large set of channel α W1
J parameters,thusgeneralizingpreviousresultsforthe2-paircase. M2→X2 Y2→Mˆ2
6
β
2 V3
I. INTRODUCTION

] The Gaussian interference channel (G-IC) is one of the α W2
T
most important and practically relevant models in multiple
I M3→X3 Y3→Mˆ3
. user information theory. Although the capacity region of W3
s β
c this channel is not known in general, significant progress V1
[ has been made recently toward finding capacity under weak

interference [1]–[3] and bounds that are provably close to Fig. 1. Cyclically symmetricdeterministic interference channel, K=3.
1
v capacity [4]–[6]. In [4], the capacity region for the two user-
8 pair G-IC is established to within one bit using new outer
6 boundsanda simplifiedHan-Kobayashiachievabilityscheme. defined as Z = [0N×N,IN]T. Further, let U,D ∈ F22N×2N
0 The same asymptotic result is derived in [5] by making a be the upshift and downshift matrix, respectively, such that
.4 correspondence between the G-IC in low-noise regime and U[x1,x2,...,x2N−1,x2N]T = [x2,x3,...,x2N,0]T and
1 a class of deterministic, finite-field interference channels [7]. D[x1,x2,...,x2N−1,x2N]T = [0,x1,...,x2N−2,x2N−1]T.
0 Some progress toward generalizing this result to more than We also use the standard notation An =(A1,A2,...,An).
9
two user-pairs has been made in [6], where the solution is We referto a K user-pairinterferencechannelas cyclically
0
: found for the fully symmetric case. symmetricifthechannelisinvarianttocyclicrelabelingofthe
v Motivatedbythese recentresults, we considera class of K pairs, i.e., renaming i as i+1 for i<K, and K as 1.
i
X user-pair (K ≥3), cyclically symmetric, deterministic, finite- We investigate the class of cyclically symmetric, determin-
r field interferencechannelsin which each receiver experiences istic, finite-field interference channels with K = 3 user-pairs
a interference only from its two nearest neighbors as in the depicted in Fig. 1. The channel is stationary and memory-
Wyner model [8]. We determine the sum capacity of this less across multiple channel uses. The channel inputs are
channel for a wide range of interference parameters. Because X1,X2,X3 ∈FN2 anditsoutputsareY1,Y2,Y3 ∈F22N,where
of symmetry in the channel and in the data rates, it suffices N isthenumberofinputbitpipesateachsender.Theoutputs
to consider the K =3 case depicted in Fig. 1. We focus our of the channel are given by
discussion on this case while keeping in mind that all results Y =ZX +V +W ,
generalize immediately to the K user-pair Wyner case. 1 1 2 3
Y =ZX +V +W ,
2 2 3 1
II. CHANNELDEFINITION Y =ZX +V +W ,
3 3 1 2
Let F denote the binary finite field and let I be the
2
identity matrix. The zeropadding operator Z ∈ F2N×N is where + is the modulo-2 addition operator, and Vk =
2 U(α−1)NZX , W =D(1−β)NZX for every k.
k k k
Thechannelisparameterizedbythetriple(N,α,β), which
Wegratefully acknowledge thecontribution ofDr.AydinSezgin, whoorigi- we constrain to α ∈ [1,2], β ∈ [0,1], and αN,βN ∈ Z.
nallysuggestedthechannelthatweconsiderinthiswork.BerndBandemeris The parametersα and β characterizethe amountof up/down-
supportedbyanEricandIlleanaBenhamouStanfordGraduateFellowshipand
shiftonthecrosslinksandthuslooselycorrespondtochannel
US Army grant W911NF-07-2-0027-1. Gonzalo Vazquez-Vilar is supported
byFundacionPedro-Barrie delaMazaGraduateScholarship. gains. Since by the definition of our channel, for each user-
β
pair there is always exactly one interferer being up-shifted
and one being down-shifted, our channel is a special case of 1 Legend
the class of cyclically symmetric, deterministic, finite field, /
Wynerconnectedchannels.Notethattheup-shiftedVk retains X Df Ee Blinoeuanrdparieycebsetween
the complete information of X , while the down-shifted W
k k 2/3 2/3
incurs clipping at the low end of the vector. Da Ed Localminimum
Transmitter k ∈ {1,2,3} wishes to convey an independent Dc Db Ec Eb Ea (dsym=1/2)
message Mk atdata rate Rk to its correspondingreceiver.We 0.5 Be Bf Bg Localmaximum
define a (2nR1,2nR2,2nR3,n) code, probability of error, and 3/5 Ba (withdsym value)
achievabilityofagivenratetriple(R ,R ,R )inthestandard Bb Ab
1 2 3
way[9].ThecapacityregionC ofthechannelistheclosureof Bd Bc Boundarybetween
Aa achieving schemes
the set of all achievable rate triples. Define the sum capacity
as R := sup{R +R +R | (R ,R ,R ) ∈ C} and the
Σ 1 2 3 1 2 3 1
symmetric capacity as R := sup{R | (R,R,R) ∈ C}. By 0
sym α
symmetryofthechannelandconvexityofthecapacityregion, 1 1.5 2
R = 3R . Furthermore, define the symmetric generalized Fig.2. IllustrationofTheorem1inthe(α,β)parameterplane.Theresult
Σ sym applies everywhere exceptinregion“X”.
degrees of freedom d := R /N, i.e., the symmetric
sym sym
capacitynormalizedwithrespectto theinterference-freecase.
Before we state our main result, define the function N (1−(α−β))N
(α−β)N
V(x):= 1+|x−1| = x2 if x≥1 ZX1 ZX1
2 (1− x if x<1.
2 V2 V2
Remark 1: This definition is useful in the context of de- W3 W2 W2 W3 W2 W2
terministic finite-field interference channels. For example, the
αN T3 T3
symmetricgeneralizeddegreesoffreedomofthetwouser-pair βN
symmetric deterministic interference channel with parameters (a) Interferers donotoverlap. (b) Interferers overlap.
(N,α), with α∈[0,∞), are known [4], [5] to be
Fig. 3. Components of received signal Y1 for the converse proof. The
d =min 1,V(α),V(2α) . componentsareshownsideways,withthebottompipeontheleft.Thedotted
sym vertical line symbolizes the “noise level”, i.e., the lower end of the vector
III. M(cid:8)AIN RESULT (cid:9) wherefurther down-shifts cause lossofinformation. Thereceived signalY1
isthemodulo-2sumofthethreecomponents.
Our main result establishes the symmetric generalized de-
greesoffreedomfortheclassofinterferencechannelsdefined
above for a large set of (α,β) parameters. information to the receivers, thus effectively degenerating the
Theorem 1: The symmetric generalized degrees of free- three user-pair case to the two-pair case.
dom of the class of three user-pair, cyclically symmetric, Hence we focus on proving the bound V(α−β) by gen-
deterministic,finite field interferencechannelwith parameters eralizing the methods introduced in [10] to the case at hand.
(α,β)∈[1,2]×[0,1], where α≥2β or α≥ β +1, is First note that Fano’s inequality implies (with some abuse of
2
notation for brevity) for every k
d =min 1,V(α),V(β),V(2β),V(α−β) .
sym
Fig. 2 depicts o(cid:8)ur result. The claimed dsym is p(cid:9)iecewise nRk ≤I(Xkn;Ykn).
linear in (α,β), and the figure shows the linear regionsin the
parameterplanewiththeirrespectiveminimumandmaximum A. Without overlap between interferers
values of d . Some of the linear pieces are subdivided (for
sym Firstconsiderα−β ≥1,whichcorrespondstothefirstline
example,“Ea”and“Eb”)toindicatethatdifferentachievability
in the definition of V(α−β). In this case, the two interfering
schemes are needed even within a single linear piece (see
signals do not overlapwithin the receivedsignal, as shown in
Section V).
Fig. 3(a). For example, at receiver 1, the sparsity patterns of
Remark 2: The theorem implies that d is independent
sym V and W are disjoint. We can write
2 3
of N. For fixed α and β, all valid values of N (satisfying
αN,βN ∈Z) yield the same dsym. I(Xn;Yn)(=a)I(Xn;YnWn)
1 1 1 1 2
IV. CONVERSE PROOF =I(X1n;W2n)+I(X1n;Y1n |W2n)
The upper bounds 1, V(α), V(β), and V(2β) follow in a (=b)H(Yn |Wn)−H(Yn |Xn,Wn),
1 2 1 1 2
straightforwardway fromthe knowndegreeof freedomresult
ofthetwouser-paircase[4],[10].Thiscanbeshownbygiving where (a) is with equality since W is not interfered with in
2
the complete signal Xn of one of the interferers as genie Y , and (b) uses the independencebetween X and W . Now
k 1 1 2
consider the last term. Again, repeating the same for all three rates, we arrive at
H(Yn |Xn,Wn)=H(ZXn+Vn+Wn |Xn,Wn) n R ≤H(Yn |Wn,Tn)−H(Tn |Tn)−H(Wn |Wn)
1 1 2 1 2 3 1 2 k k 1 2 3 1 1 1 1
=H(Vn+Wn |Wn) +H(Yn |Wn,Tn)−H(Tn |Tn)−H(Wn |Wn)
2 3 2 P 2 3 1 2 2 2 2
(=a)H(W3n)+H(V2n |W2n) +H(Y3n |W1n,T2n)−H(Tn3 |T3n)−H(Wn3 |W3n).
=H(Wn)+H(Wn |Wn), Since T and T together form W , which together with W
3 2 2 k k k k
forms X , we can write
where(a)followsfromthefactthatV andW donotoverlap k
2 3
and differenttransmitters’signals are independent,and W2 is nR1 ≤H(X1n)=H(T1n)+H(Tn1|T1n)+H(W1|T1n,Tn1)
the partofX thatis notcontainedin W (see Fig.3(a)).We
conclude that2 2 =W1n
I(Xn;Yn)=H(Yn |Wn)−H(Wn)−H(Wn |Wn). Using this expression and its equivalent for R2 and|R{3zw}ith
1 1 1 2 3 2 2 the previous inequality, we obtain
WritingananalogousequationforI(Xn;Yn)andI(Xn;Yn),
2 2 3 3 2n R ≤H(Yn |Wn,Tn)+H(Tn)+H(Yn |Wn,Tn)
and adding all three of them, we arrive at k k 1 2 3 1 2 3 1
+H(Tn)+H(Yn |Wn,Tn)+H(Tn)
n R ≤H(Yn |Wn)+H(Yn |Wn)+H(Yn |Wn) P 2 3 1 2 3
k k 1 2 2 3 3 1 ≤n H(Y |W ,T )+H(T )+H(Y |W ,T )
−H(Wn)−H(Wn |Wn)−H(Wn) 1 2 3 1 2 3 1
P 1 1 1 2 +H(T )+H(Y |W ,T )+H(T ) .
−H(Wn |Wn)−H(Wn)−H(Wn |Wn) (cid:0) 2 3 1 2 3
2 2 3 3 3 Again, the right hand side is maximized by choos(cid:1)ing all X
=H(Yn |Wn)+H(Yn |Wn)+H(Yn |Wn) k
1 2 2 3 3 1 components independently according to Bern(1/2), yielding
−H(Xn)−H(Xn)−H(Xn)
1 2 3
2 R ≤3N +3N(1−(α−β)),
k k
Considering that nR ≤H(Xn), we conclude that
k k d ≤1− α−β,
P sym 2
2n R ≤H(Yn |Wn)+H(Yn |Wn)+H(Yn |Wn)
k k 1 2 2 3 3 1 which matches the definition of V(α−β) for α−β <1.
≤nH(Y |W )+nH(Y |W )+nH(Y |W ),
P 1 2 2 3 3 1
V. ACHIEVABILITY PROOF
wheresingle-letterizationisperformedbyusingthechainrule
The set of interest {(α,β)} is divided into regions “Aa” to
and omitting part of the conditioning. The right hand side of
“Df”asshownin Fig.2. Ineachregion,we use thefollowing
the last equation is maximized by letting each input bit pipe
coding scheme. For every sender k, we set
be independent Bern(1/2). Thus
X =GD ,
2 R ≤3N(α−β), and finally, k k
k k
dsym P= RNsym ≤ α−2β. where G ∈ F2N×dsymN is the assignment matrix, and Dk ∈
FdsymN is a vector of i.i.d. Bern(1/2) message bits.
B. With overlap between interferers 2
We constrain the coding scheme in several ways, namely,
Now consider the case where α − β < 1, i.e., the two
(a) there is no coding across multiple channel uses, (b) all
interferingsignalsateachreceiveroverlapinsignalspace,see transmittersusethesameG, and(c)theproposedGmatrices
Fig. 3(b). Define the top (1−(α−β))N part of X as T .
k k will have at most one non-zero element per row, i.e., each
We will augment the genie information Wn of the previous
2 pipe in X is assigned either an information bit or a zero.
subsection by Tn. This is exactly the part of the X -based k
3 3 Whiletheseassumptionsmayseemoverlyrestrictive,theyare
interference that overlaps with the X -based interference.
2 sufficient for our purposes. Indeed, it is surprising that such
Similar to the previous section, we conclude
a constrained set of codes is able to meet the upper bound of
I(Xn;Yn)=I(Xn;Yn,Wn,Tn) Section IV.
1 1 1 1 2 3
=I(Xn;Wn,Tn)+I(Xn;Yn |Wn,Tn) Remark 3: If the number of input pipes N is small, it can
1 2 3 1 1 2 3 severely limit our options in terms of assignment matrices.
=H(Yn |Wn,Tn)−H(Yn |Xn,Wn,Tn),
1 2 3 1 1 2 3 The following argument can circumvent this problem by
The last term becomes expanding a given channel to one with more input pipes.
To this end, consider L ≥ 2 subsequent channel uses, with
H(Yn |Xn,Wn,Tn)=H(ZX +Vn+Wn |Xn,Wn,Tn)
1 1 2 3 1 2 3 1 2 3 channel inputs Xk,1,...,Xk,L. By interleaving these vectors
=H(Vn+Wn |Wn,Tn) into a supersymbol X = L (I ⊗e )X , and likewise
2 3 2 3 k l=1 N l k,l
(=a)H(Tn |Tn)+H(Wn |Wn), for the outputs Yk, it can bPe shown that the resulting channel
3 3 2 2 {X ,X ,X }→{Ye,Y ,Y }isinfact(LN,α,β)asdefined
1 2 3 1 2 3
where T denotes the part of W that is not included in T . in Section II. (Heere, ⊗ denotes the Kronecker product, and
3 3 3
Its size is N(α−1). We are allowed to separate the terms in eleis tehe leth columneofeILe.) Through this method, a channel
(a) because the overlapping part is resolved by T . with a given N can be expanded to one with LN input
3
B − − T
otto 31 ε− 31 ε+ ε− 31 ε+ op 1. Direct readout:
m − − − Consider the situation in Fig. 6(b). If a data block (i) does
21 2 21 2
δ δ δ δ δ δ δ
notoverlapwithanyotherdatablockand(ii)islocatedabove
X :
k the noise level, then its data content can be read out directly
from the received signal.1 A block that has been read out is
Fig.4. ProposedGassignmentforregion“Df”,expressedintermsofthe
transmitvector. Theblocklengths aregivenasfractions ofN,suchthatthe thenremovedfromthereceivedsignal.Iftheblockhasatwin,
sumofallblocklengths is1. it is removed as well.
ZX1 6 4 3 4 5 3 2. Overlapping twins scenario (A):
ConsiderFig.6(c).Iftwotwinpairsexistsuchthat(i)they
V2 2 1 2 1 have the same block length, b1 = b2, (ii) they have the same
W3 4 3 4 3 separation,s1 =s2, (iii) the relativeshift betweenthe pairs is
less than the separation, c < s , and (iv) the dashed sections
1
Fig.5. ReceivedsignalY1in“Df”,atα=1.6,β=0.9,withdsym =0.55. of (A) in Fig. 6(c) do not overlap with any other data block
Blocks indifferent rowscarrydifferent data. and are above the noise floor, then both twin pairs can be
decoded and canceled from the received signal.2 To see this,
consider the following successive decoding argument [6]. Let
pipes. Note that d is unaffected by this transformation
sym thetwocopieswithinatwin beinreverseorderofeachother.
since it is normalized by the number of pipes. In light of
First, the leftmost part of the left blue twin is read out. Its
thistransformation,we assumefromnowonthatN is(orhas
data reappears on the right side of the right blue twin, thus
been made) large enough such that any fraction of N that we
revealinga chunkof data on the rightside of the rightyellow
incur corresponds to an integer number of pipes.
twin. This data in turn is replicated on the left side of the left
Optimal assignment matrices G for all regions in Fig. 2 yellow twin, which exposes a new part of the left blue twin.
are listed in Table 1. An interactive online animation is also The process repeats until both twins are completely decoded.
available at [11].Each row in the table containsthe definition
3. Overlapping twins scenario (B):
of a region in terms of affine constraints in (α,β) and a
Thisruleisa variantofthepreviousone,wherepattern(B)
representationof G by means of the resulting transmit vector
replaces pattern (A) in Fig. 6(c).
X . In the following we discuss the details for one particular
k
example, which is representative for all other cases. In our example, the sequence of steps that completely
Example (Region “Df”): This region is parameterized by decodes X1 is annotated in Fig. 5: First, block 1 is decoded
(α,β) = (4/3+ε, 2/3+δ) with ε ≤ 2δ, ε ≥ 1δ, δ ≤ 1. via directreadout(rule 1). The now-knowndata block and its
2 3
Fig. 4, copiedfromTable 1, representsanoptimalassignment twin are removed from the received signal Y1. The same rule
G by means of the resulting transmit vector. The vector X allows block 2 to be decoded, which is then removed from
k
is subdivided into data blocks (hatched) that correspond to Y1. Each removal step makes more room for subsequent rule
non-zerorowsofG,andzero blocks(gray)thatcorrespondto applications. Next, rule 2 is applied to the two pairs of twins
all-zerorowsof G. Some datablocksoccurtwice. We denote 3. Continuingin the same fashion, the removalof blocks1, 2
suchblockpairsastwins.Twinscarrythesamedatabits,albeit
1It is crucial that both (i) and (ii) hold for all (α,β) in the region, since
in reverse order as discussed later. The length of each block
thelengthandlocation oftheblocks inFig.5change whenαandβ vary.
as a fraction of N is annotated in the figure. 2Again,conditions (i)–(iv)mustholdforall(α,β)intheregion.
ToproveachievabilityofTheorem1,werequirethetransmit
vector to be both valid and decodable. By valid we mean
(a) all block lengths are non-negative for the range of (ε,δ) Legend
that constitute the region, (b) the sum of the block lengths is nootherdatablocksallowed
1, and (c) adding the sizes of all data blocks, counting twins otherdatablocksallowed
onlyonce,resultsinthedesiredd asclaimedinTheorem1 (a) Legend. (b) Directreadout.
sym
(2/3−δ/2inourexample).Bydecodable,wemeanthatusing
this transmit vector assignment, the receiver can recover all b1 b1
desired data blocks from the received signal. c s1
Toverifydecodability,considerFig.5,whichusesthesame b2 b2
s2
conventionsasFig.3.Thereceiverseesthesumofdatablocks
from different transmitters, each characterized by its length
(A)
and shift location. Different blocks may or may not overlap.
(B)
Decoding is performed sequentially, block by block. In each
(c) Overlapping twins.
step,oneofthreerulesisappliedinordertodecodeadditional
datablocks,whicharethenremovedfromthereceivedsignal. Fig. 6. Rules for verifying decodability. Legend (a) applies to “direct
The three decoding rules are as follows. readout”,shownin(b),andtwovariantsof“overlappingtwins”,shownin(c).
and3enablesthetwotwinpairs4tobedecodedusingrule3. [5] G.BreslerandD.Tse,“Thetwo-userGaussianinterference channel: a
Finally,datablocks5and6canberecoveredbydirectreadout deterministicview,”Euro.Trans.Telecomm.,vol.19,no.4,pp.333–354,
Jun.2008,arXiv:0807.3222.
(rule1),whichcompletesthedecodingprocess.Bysymmetry,
[6] S.A.JafarandS.Vishwanath, “Generalized degrees offreedomofthe
thesignalsattheothertworeceiverscanbesimilarlydecoded.
symmetricK userGaussianinterference channel,” arXiv:0804.4489.
TheassignmentsforallotherregionsaslistedinTable1can
[7] A. S. Avestimehr, S. N. Diggavi, and D. N. C. Tse, “A deterministic
beshowntobevalidanddecodableusingthesameprocedure. approach towirelessrelaynetworks,” inProc.Allerton,Sep.2007.
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(α,β) Regionconstraints dsym Assignment(showngraphically andascorresponding listofblocklengths)
Aa (2+ε, δ≥0, δ≤1+ε, 1+1ε−1δ
2 2
1 1 1 1
δ) ε≤−δ (δ| − ε− δ|1+ε| − ε− δ)
2 2 2 2
Ab (2+ε, ε≤0, δ≤ 1, 1−δ
2
δ) ε≥−δ (δ|1−δ)
Ba (6/5+ε, ε≥3δ, ε≤ 1 +δ, 3−1ε+1δ
5 5 2 2
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2/5+δ) ε≥ − δ ( −ε+δ| − ε− δ| − ε− δ|− +2ε+δ| − ε− δ| − ε− δ| +ε)
10 2 5 5 2 2 5 2 2 5 5 2 2 5 2 2 5
Bb (6/5+ε, ε≥−1δ, ε≤ 1+δ, 3−1ε+1δ
3 5 5 2 2
1 1 1 1 1 1 3 1 1 3 1 1 1 1 1
2/5+δ) ε≤ − δ, ε≥3δ ( −ε+δ| − ε− δ| ε+ δ| −2ε−δ| ε+ δ| − ε− δ| +ε)
10 2 5 5 2 2 2 2 5 2 2 5 2 2 5
Bc (6/5+ε, ε≥ 1δ, ε≤ 1 +δ, 3−1ε+1δ
2 5 5 2 2
1 1 1 1 3 1 1 3 1 1 1 1
2/5+δ) ε≤− δ ( + δ| +ε| − ε− δ| +ε| − ε− δ| +ε| + δ)
3 5 2 5 2 2 5 2 2 5 5 2
Bd (6/5+ε, ε≤ 1δ, ε≤−2δ, 3 + 1ε
2 5 2
1 1 1 1 1 1 1 1 1
2/5+δ) ε≥− (−ε+ δ| +ε| −ε+ δ| +ε| − ε−δ| +ε| − ε−δ| +ε| −ε+
5 2 5 2 5 2 5 2 5
1 1 1
δ| +ε| −ε+ δ)
2 5 2
Be (6/5+ε, ε≤ 1δ, ε≥−2δ, 3 −δ
2 5
1 1 1 1 1 1 1 1 1 1
2/5+δ) δ≤ (−ε+ δ| +ε|−ε+ δ| +ε| −2δ| +ε|−ε+ δ| +ε| −ε+ δ)
10 2 5 2 5 5 5 2 5 2
Bf (6/5+ε, ε≥ 1δ, ε≤3δ, 3 −δ
2 5
1 1 1 1 1 1 1
2/5+δ) ε≤ − δ ( −ε+δ| −ε+δ|2ε−δ| −2ε−δ|2ε−δ| −ε+δ| +ε)
10 2 5 5 5 5 5
Bg (6/5+ε, ε≤3δ, δ≤ 1 , 3 −δ
10 5
1 1 1 1 1 1 1 1 1
2/5+δ) ε≥ − δ ( −ε+δ| −ε+δ| −2δ| − +2ε+δ| −2δ| −ε+δ| +ε)
10 2 5 5 5 5 5 5 5
Da (4/3+ε, ε≥2δ, ε≥−δ, 2−1ε+1δ
3 2 2
1 1 1 1 1 1 1 1
2/3+δ) ε≤ +δ ( −δ| ε+ δ| −δ| ε+ δ| −ε+δ)
3 3 2 2 3 2 2 3
Db (4/3+ε, ε≤−δ, ε≥ 1δ, 2 +δ
2 3
1 1 1 1 1 1
2/3+δ) δ≥− ( + δ| −δ| + δ)
6 3 2 3 3 2
Dc (4/3+ε, ε≤ 1δ, ε≥2δ, 2 +δ
2 3
1 1 1 1 1 1 1 1
2/3+δ) δ≥− (−ε+ δ| +ε| −ε+ δ| +2ε−2δ| −ε+ δ| +ε| −ε+ δ)
6 2 3 2 3 2 3 2
Df (4/3+ε, ε≤2δ, ε≥ 1δ, 2 − 1δ
2 3 2
1 1 1 1 1 1
2/3+δ) δ≤ ( −δ|ε− δ| −δ| −ε+2δ|ε− δ| −δ| −ε+2δ)
3 3 2 3 2 3
Ea (2+ε, ε≤0, δ≥− 1 , 2 +δ
15 3
1 1 1
2/3+δ) ε≥3δ (−3δ| +2δ| −3δ| +5δ| −3δ| +2δ)
3 3 3
Eb (2+ε, ε≤0, δ≤− 1 , 2 +δ
15 3
1 1 1 1 1 1
2/3+δ) δ≥− , ε≥3δ (−3δ| +2δ| +2δ| − −5δ| +2δ| +2δ)
6 3 3 3 3 3
Ec (2+ε, ε≤3δ, ε≥−1 +δ, 2+1ε−1δ
3 3 2 2
1 1 3 1 1 1 1 1 1 1 1 1 1 1 1
2/3+δ) ε≤− −2δ (− ε− δ| + ε+ δ| + ε+ δ| − −ε−2δ| + ε+ δ| + ε+
3 2 2 3 2 2 3 2 2 3 3 2 2 3 2
1 1 3
δ| − ε+ δ)
2 2 2
Ed (2+ε, ε≤3δ, ε≥−1−2δ, 2+1ε−1δ
3 3 2 2
1 1 3 1 1 1 1 3 1 1 3 1 1 1 1 3
2/3+δ) ε≥− +δ, ε≤−3δ (− ε− δ| + ε+ δ|− ε− δ| +ε+2δ|− ε− δ| + ε+ δ|− ε+ δ)
3 2 2 3 2 2 2 2 3 2 2 3 2 2 2 2
Ee (2+ε, ε≤0, δ≤ 1 +ε, 2+1ε−1δ
3 3 2 2
1 1 1 1 3 1 1 3
2/3+δ) δ≥− ε ( −δ| −δ| ε+ δ| −δ| ε+ δ| −ε)
3 3 3 2 2 3 2 2
Table 1. Assignments thatachieve dsym asstatedinTheorem1.