Fourier Analysis of MAC Polarization

0.26 MB·English

by  Rajai Nasser

Fourier Analysis of MAC Polarization
Rajai Nasser, Emre Telatar
Ecole Polytechnique Fe´de´rale de Lausanne,
Lausanne, Switzerland
Email: {rajai.nasser, emre.telatar}@epfl.ch
Abstract
A problem of MAC polar codes which are based on MAC polarization is that they do not always achieve the
entire capacity region. The reason behind this problem is that MAC polarization sometimes induces a loss in the
capacity region. This paper provides a single letter necessary and sufficient condition which characterizes all the
MACs that do not lose any part of their capacity region by polarization.
I. INTRODUCTION
5
1 Polar coding is a low complexity coding technique invented by Arıkan which achieves the capacity of
0 symmetric binary input channels [1]. The probability of error of polar codes was shown to be o(2−N21−ǫ)
2
where N is the block length [2]. The polar coding construction of Arıkan transforms a set of identical
l
u and independent channels to a set of “almost perfect” or “almost useless channels”. This phenomenon is
J
called polarization.
0
Polarizing transformations can also be constructed for non-binary input channels. S¸as¸og˘lu et al. [3]
3
generalized Arıkan’s results to channels where the input alphabet size is prime. Park and Barg [4] showed

T] that if the size of the input alphabet is of the form 2r with r > 1, then using the algebraic structure
I Z in the polarizing transformation leads to a multilevel polarization phenomenon: while we do not
. 2r
s always have polarization to “almost perfect” or “almost useless” channels, we always have polarization
c
[ to channels which are easy to use for communication. Multilevel polarization can be used to construct

capacity achieving polar codes.
2
v Sahebi and Pradhan [5] showed that multilevelpolarization also happens if any Abelian group operation
6
on the input alphabet is used. This allows the construction of polar codes for arbitrary discrete memoryless
7
0 channels (DMC) since any alphabet can be endowed with an Abelian group structure. Polar codes for
6 arbitrary DMCs were also constructed by S¸as¸og˘lu [6] by using a special quasigroup operation that ensures
0
two-level polarization. The authors showed in [7] that all quasigroup operations are polarizing (in the
.
1
general multilevel sense) and can be used to construct capacity-achieving polar codes for arbitrary DMCs
0
5 [8].
1 In the context of multiple access channels (MAC), S¸as¸og˘lu et al. showed that if W is a 2-user MAC
:
v where the two users have F as input alphabet, then using the addition modulo q for the two users leads to
q
i
X a MAC polarization phenomenon [9]. Abbe and Telatar showed that for binary input MACs with m ≥ 2
r users, using the XOR operation for each user is MAC-polarizing [10]. A problem of the MAC polar code
a
construction in [9] and [10] is that they do not always achieve the entire capacity region. The reason
behind this problem is that MAC polarization sometimes induces a loss in the capacity region.
A characterization of all the polarizing transformations that are based on binary operations — in both
the single-user and the multipleaccess settings — can be found in [14] and [15]. Abelian group operations
are a special case of the characterization in [15]. Therefore, using Abelian group operations for all users
is MAC-polarizing.
This paper provides a necessary and sufficient condition which characterizes all the MACs that do not
lose any part of their capacity region by polarization. The characterization that we provide works in the
general setting where we have an arbitrary number of users and each user uses an arbitrary Abelian group
operation on his input alphabet.
We will show that the reason why a given MAC W loses parts of its capacity region by polarization is
becauseitstransitionprobabilitiesarenot“aligned”,whichmakesW “incompatible”withpolarization.The
1
“alignment” conditionwill be expressed in terms of the Fourier transforms of the transition probabilitiesof
W. The use of Fourier analysis in our study should not come as a surprise since the transition probabilities
of W− can be expressed as a convolution of the transition probabilities of W. This is what makes Fourier
analysis useful for our study because it turns convolutions into multiplications, which are much easier to
handle.
Note that there are alternate polar coding solutions which can achieve the entire capacity region without
any loss. These techniques, which are not based on MAC polarization, are hybrid schemes combining
single user channel polarization with other techniques. In [9], S¸as¸og˘lu et al. used the “rate splitting/onion
peeling” scheme of [12] and [13] to transform any point on the dominant face of an m-user MAC into a
corner point of a (2m−1)-user MAC and then applied single user channel polarization to achieve this
corner point. In [11], Arıkan used monotone chain rules to construct polar codes for the Slepian-Wolf
problem, but the same technique can be used to achieve the entire capacity region of a MAC.
Although the alternate solutions of [9] and [11] can achieve the entire capacity region, they are more
complicated than MAC polar codes (those that are based on MAC polarization). The alternate solution
in [9] requires more encoding and decoding complexity because it adds m − 1 virtual users. Arıkan’s
solution [11] does not add significant encoding and decoding complexity, but the code design is much
more complicated than that of MAC polar codes. So if we are given a MAC W whose capacity region is
preserved by polarization (i.e., MAC polar codes can achieve the entire capacity region of this MAC), then
using MAC polar codes for this MAC is preferable to the alternate solutions. One practical implication
of this study is that it allows a code designer to determine whether he can use the preferable MAC polar
codes to achieve the capacity region.
In section II, we introduce the preliminaries of this paper: we describe the MAC polarization process
and explain the discrete Fourier transforms on Abelian groups. In section III, we characterize the two-user
MACs whose capacity regions are preserved by polarization. Section IV generalizes the results of section
IV to MACs with arbitrary number of users.
II. PRELIMINARIES
Throughout this paper, G ,...,G are finite Abelian groups. We will use the addition symbol + to
1 m
denote the group operations of G ,...,G .
1 m
A. Polarization
W
Notation 1. Let W : G ×...×G −→ Z be an m-user MAC. We write (X ,...,X ) −→ Z to denote
1 m 1 m
the following:
• X1,...,Xm are independent random variables uniformly distributed in G1,...,Gm respectively.
• Z is the output of the MAC W when X1,...,Xm are the inputs.
Notation 2. Fix S ⊂ {1,...,m} and let S = {i ,...,i }. Define G as
1 |S| S
G := G = G ×...×G .
S i i1 i|S|
i∈S
Y
For every (x ,...,x ) ∈ G ×...×G , we write x to denote (x ,...,x ).
1 m 1 m S i1 i|S|
W
Notation 3. Let W : G × ... × G −→ Z and (X ,...,X ) −→ Z. For every S ⊂ {1,...,m}, we
1 m 1 m
write I (W) to denote I(X ;ZX ). If S = {i}, we denote I (W) by I (W).
S S Sc {i} i
I(W) := I (W) = I(X ,...,X ;Z) is called the symmetric sum-capacity of W.
{1,...,m} 1 m
Definition 1. The symmetric capacity region of an m-user MAC W : G ×...×G −→ Z is given by:
1 m
J(W) = (R ,...,R ) ∈ Rm : ∀S ⊂ {1,...,m}, R ≤ I (W) .
1 m i S
n Xi∈S o
2
Notation 4. {−,+}∗ := {−,+}n, where {−,+}0 = {ø}.
n≥0
[
Definition 2. Let W : G ×...×G −→ Z. We define the m-user MACs W− : G ×...×G −→ Z2
1 m 1 m
and W+ : G ×...×G −→ Z2 ×G ×...×G as follows:
1 m 1 m
1
W−(z ,z |u ,...,u ) = W(z |u +u ,...,u +u )
1 2 11 1m 1 11 21 1m 2m
|G |···|G |
1 m
u2X1∈.G1
.
.
u2m∈Gm
×W(z |u ,...,u ),
2 21 2m
and
1
W+(z ,z ,u ,...,u |u ,...,u ) = W(z |u +u ,...,u +u )
1 2 11 1m 21 2m 1 11 21 1m 2m
|G |···|G |
1 m
×W(z |u ,...,u ).
2 21 2m
For every s ∈ {−,+}∗, we define the MAC Ws as follows:
W if s = ø,
Ws :=
((...((Ws1)s2)...)sn if s = (s1,...,sn).
Remark 1. Let Um and U′m be two independent random variables uniformlydistributed in G ×...×G .
1 1 1 m
Let Xm = Um +U′m and X′m = U′m. Let (X ,...,X ) −W→ Z and (X′,...,X′ ) −W→ Z′. We have:
1 1 1 1 1 1 m 1 m
• I(W) = I(X1m;Z) = I(X1′m;Z′).
• I(W−) = I(U1m;ZZ′) and I(W+) = I(U1′m;ZZ′U1m).
Hence,
2I(W) = I(Xm;Z)+I(X′m;Z′) = I(XmX′m;ZZ′) = I(UmU′m;ZZ′)
1 1 1 1 1 1
= I(Um;ZZ′)+I(U′m;ZZ′Um) = I(W−)+I(W+).
1 1 1
Therefore, the symmetric sum-capacity is preserved by polarization. On the other hand, I might not be
S
preserved if S ( {1,...,m}.
For example, consider the two-user MAC case. Let W : G ×G −→ Z. Let (U ,V ) and (U ,V ) be
1 2 1 1 2 2
two independent random pairs uniformly distributed in G ×G . Let X = U +U , X = U , Y = V +V
1 2 1 1 2 2 2 1 1 2
W W
and Y = V . Let (X ,Y ) −→ Z and (X ,Y ) −→ Z . We have:
2 2 1 1 1 2 2 2
• I1(W−) = I(U1;Z1Z2V1) and I1(W+) = I(U2;Z1Z2U1V1V2).
• I2(W−) = I(V1;Z1Z2U1) and I2(W+) = I(V2;Z1Z2U1V1U2).
On the other hand, we have:
• I1(W) = I(X1;Z1Y1) = I(X2;Z2Y2).
• I2(W) = I(Y1;Z1X1) = I(Y2;Z2X2).
Therefore,
2I (W) = I(X ;Z Y )+I(X ;Z Y ) = I(X X ;Z Z Y Y ) = I(U U ;Z Z V V )
1 1 1 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2
= I(U ;Z Z V V )+I(U ;Z Z V V U ) ≥ I(U ;Z Z V )+I(U ;Z Z V V U ) (1)
1 1 2 1 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 2 1
= I (W−)+I (W+),
1 1
2I (W) = I(Y ;Z X )+I(Y ;Z X ) = I(Y Y ;Z Z X X ) = I(V V ;Z Z U U )
2 1 1 1 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2
= I(V ;Z Z U U )+I(V ;Z Z U U V ) ≥ I(V ;Z Z U )+I(V ;Z Z U U V )
1 1 2 1 2 2 1 2 1 2 1 1 1 2 1 2 1 2 1 2 1
= I (W−)+I (W+).
2 2
3
By induction on n ≥ 0, we can show that:
1
I (Ws) ≤ I (W), (2)
2n 1 1
s∈{−,+}n
X
1
I (Ws) ≤ I (W), (3)
2n 2 2
s∈{−,+}n
X
1
I(Ws) = I(W). (4)
2n
s∈{−,+}n
X
While(4) shows that polarizationpreserves the symmetricsum-capacity,(2) and (3) showthat polarization
may result into a loss in the capacity region.
Similarly, for the m-user case, we have
1
I (Ws) ≤ I (W), ∀S ( {1,...,m}.
2n S S
s∈{−,+}n
X
Definition 3. Let S ⊂ {1,...,m}. We say that polarization ∗-preserves I for W if for all n ≥ 0 we
S
have:
1
I (Ws) = I (W).
2n S S
s∈{−,+}n
X
If polarization ∗-preserves I for every S ⊂ {1,...,m}, we say that polarization ∗-preserves the
S
symmetric capacity region for W.
Remark 2. If polarization ∗-preserves the symmetric capacity region for W, then the entire symmetric
capacity region can be achieved by polar codes.
Section III provides a characterization of two-user MACs whose I are ∗-preserved by polarization.
1
Section IV generalizes the results of section III and provides a characterization of m-user MACs whose
I are ∗-preserved by polarization, where S ( {1,...,m}. This yields a complete characterization of the
S
MACs with ∗-preservable symmetric capacity regions.
B. Discrete Fourier Transform on finite Abelian Groups
A tool that we are going to need for the analysis of the polarization process is the discrete Fourier
transform (DFT) on finite Abelian groups. The DFT on finite Abelian groups can be defined based on
the usual multidimensional DFT.
Definition 4. (Multidimensional DFT) The m-dimensional discrete Fourier transform of a mapping f :
Z ×...×Z → C is the mapping fˆ: Z ×...×Z → C defined as:
N1 Nm N1 Nm
ˆ −j2πxˆ1x1...−j2πxˆmxm
f(xˆ1,...,xˆm) = f(x1,...,xm)e N1 Nm .
x1∈ZN1,X...,xm∈ZNm
Notation 5. For x = (x ,...,x ) ∈ Z ×...×Z and xˆ = (xˆ ,...,xˆ ) ∈ Z ×...×Z , define
1 m N1 Nm 1 m N1 Nm
hxˆ,xi ∈ R as:
xˆ x xˆ x
1 1 m m
hxˆ,xi := +...+ ∈ R.
N N
1 m
Using this notation, the DFT has a compact formula:
fˆ(xˆ) = f(x)e−j2πhxˆ,xi.
x∈ZN1X×...×ZNm
4
It is known that every finite Abelian group is isomorphic to the direct product of cyclic groups, i.e., if
(G,+) is a finite Abelian group then there exist m integers N ,...,N > 0 such that G is isomorphic to
1 m
Z ×...×Z . This allowsus to define aDFT on G usingthe multidimensionalDFT on Z ×...×Z :
N1 Nm N1 Nm
Definition 5. Let (G,+) be a finite Abelian group which is isomorphic to Z × ... × Z . Fix an
N1 Nm
isomorphism between G and Z ×...×Z . The discrete Fourier transform of a mapping f : G → C
N1 Nm
is the mapping fˆ: G → C defined as:
fˆ(xˆ) = f(x)e−j2πhxˆ,xi,
x∈G
X
where hxˆ,xi is computed by identifying xˆ and x with their respective images in Z ×...×Z by the
N1 Nm
fixed isomorphism.
In the rest of this section, we recall well known properties of DFT.
Proposition 1. The inverse DFT is given by the following formula:
1
f(x) = fˆ(xˆ)ej2πhxˆ,xi,
|G|
xˆ∈G
X
where hxˆ,xi is computed by identifying xˆ and x with their respective images in Z ×...×Z by the
N1 Nm
fixed isomorphism.
Remark 3. The DFT on G as defined in this paper depends on the fixed isomorphism between G and
Z ×...×Z . If the DFT is computed using a fixed isomorphism, the inverse DFT must be computed
N1 Nm
using the same isomorphism in order to have consistent computations.
Note that it is possible to define the DFT on finite Abelian groups in a canonical way without the need
to fix any isomorphism, but this requires the character theory of finite Abelian groups.
Definition 6. The convolution of two mappings f : G → C and g : G → C is the mapping f ∗g : G → C
defined as:
(f ∗g)(x) = f(x′)g(x−x′).
x′∈G
X
We will sometimes write f(x)∗g(x) to denote (f ∗g)(x).
Proposition 2. Let f : G → C and g : G → C be two mappings. We have:
• (\f ∗g)(xˆ) = fˆ(xˆ)gˆ(xˆ).
1
• (\f ·g)(xˆ) = (fˆ∗gˆ)(xˆ).
|G|
• If fa : G → C is defined as fa(x) = f(x−a), then fˆa(xˆ) = fˆ(xˆ)e−j2πhxˆ,ai.
• If f˜: G → C is defined as f˜(x) = f(−x), then fˆ˜(xˆ) = fˆ(xˆ)∗.
C. Useful notation
This subsection introduces useful notation that will be used throughout this paper. The usefulness of
this notation will be clear later. We added this subsection so that the reader can refer to it anytime.
W
Let W : G ×G −→ Z and let (X,Y) −→ Z. Define the following:
1 2
• YZ(W) := {(y,z) ∈ G2 ×Z : PY,Z(y,z) > 0}.
• For every (y,z) ∈ YZ(W), define py,z,W : G1 → [0,1] as py,z,W(x) = PX|Y,Z(x|y,z).
For every z ∈ Z, define:
• Yz(W) = {y ∈ G2 : PY,Z(y,z) > 0}.
• ∆Yz(W) := y1 −y2 : y1,y2 ∈ Yz(W) .
(cid:8) (cid:9)
5
• Xˆz(W) := xˆ ∈ G1 : ∃y ∈ Yz(W),pˆy,z,W(xˆ) 6= 0 .
• Dz(W) := Xˆz(W)×∆Yz(W) = (xˆ,y) : xˆ ∈ Xˆz(W), y ∈ ∆Yz(W) .
(cid:8) (cid:9)
Now define:
(cid:8) (cid:9)
• XˆZ(W) := (xˆ,z) : z ∈ Z, xˆ ∈ Xˆz(W) ,
• D(W) := (cid:8) Dz(W). (cid:9)
z∈Z
[
D. Pseudo quadratic functions
Definition 7. Let D ⊂ G ×G . Define the following sets:
1 2
• H1(D) = {x : ∃y, (x,y) ∈ D}.
• For every x ∈ H1(D), let H2x(D) = {y : (x,y) ∈ D}.
• H2(D) = {y : ∃x, (x,y) ∈ D}.
• For every y ∈ H2(D), let H1y(D) = {x : (x,y) ∈ D}.
We say that D is a pseudo quadratic domain if:
• H1y(D) is a subgroup of G1 for every y ∈ H2(D).
• H2x(D) is a subgroup of G2 for every x ∈ H1(D).
Definition 8. Let D ⊂ G ×G and let F : D → T be a mapping from D to T = {ω ∈ C : |ω| = 1}.
1 2
We say that F is a pseudo quadratic function if:
• D is a pseudo quadratic domain.
• For every y ∈ H2(D), the mapping x → F(x,y) is a group homomorphism from H1y(D),+ to
(T,·).
(cid:0) (cid:1)
• For every x ∈ H1(D), the mapping y → F(x,y) is a group homomorphism from H2x(D),+ to
(T,·).
(cid:0) (cid:1)
Definition 9. We say that W : G ×G −→ Z is polarization compatible with respect to the first user if
1 2
there exists a pseudo quadratic function F : D → T such that:
• D(W) ⊂ D ⊂ G1 ×G2.
• For every (xˆ,z) ∈ XˆZ(W) and every y1,y2 ∈ Yz(W), we have pˆy1,z(xˆ) = F(xˆ,y1 −y2)·pˆy2,z(xˆ).
III. TWO-USER MACS WITH ∗-PRESERVED I
1
In this section, we only consider two-user MACs W : G × G −→ Z, where G and G are finite
1 2 1 2
Abelian groups.
The following theorem is the main result of this paper:
Theorem 1. polarization ∗-preserves I for W if and only if W is polarization compatible with respect
1
to the first user.
Theorem 1 has the following implications:
W
• (Proposition 7) If G1 = G2 = Fq for a prime q and (X,Y) −→ Z, then polarization ∗-preserves I1
for W if and only if there exists a ∈ F such that I(X +aY;Y|Z) = 0.
q
• (Corollary 3) Polarization ∗-preserves the symmetric capacity region for the binary adder channel.
W
• (Proposition 8) If |G1| and |G2| are co-prime and (X,Y) −→ Z, then polarization ∗-preserves I1 for
W if and only if I(X;Y|Z) = 0 (i.e., if and only if the dominant face of J(W) is a single point).
The rest of this section is dedicated to prove Theorem 1.
6
A. Preserved and ∗− Preserved
Definition 10. Let W : G × G −→ Z. We say that I is preserved for W if and only if I (W−) +
1 2 1 1
I (W+) = 2I (W). We say that I is ∗− preserved for W if and only if I is preserved for W[n]− for every
1 1 1 1
n ≥ 0, where [n]− ∈ {−,+}n is the sequence containing n minus signs (e.g., [0]− = ø, [2]− = (−,−)).
Lemma 1. Polarization ∗-preserves I for W if and only if I is preserved for Ws for every s ∈ {−,+}∗.
1 1
Similarly,polarization∗-preserves I forW ifandonlyifI is∗− preserved for Ws foreverys ∈ {−,+}∗.
1 1
Proof: Polarization ∗-preserves I for W if and only if
1
1 1 1
∀n ≥ 0, I (W) = I (Ws) ⇔ ∀n ≥ 0, I (Ws) = I (Ws′)
1 2n 1 2n 1 2n+1 1
s∈{−,+}n s∈{−,+}n s′∈{−,+}n+1
X X X
⇔ ∀n ≥ 0, 2I (Ws) = (I (W(s,−))+I (W(s,+)))
1 1 1
s∈{−,+}n s∈{−,+}n
X X
⇔ ∀n ≥ 0, 2I (Ws)−I (W(s,−))−I (W(s,+)) = 0.
1 1 1
s∈{−,+}n
X (cid:0) (cid:1)
But since 2I (Ws) − I (W(s,−)) − I (W(s,+)) ≥ 0 (see (1)), we conclude that polarization ∗-preserves
1 1 1
I for W if and only if ∀n ≥ 0,∀s ∈ {−,+}n, I (W(s,−)) + I (W(s,+)) = 2I (Ws). In other words,
1 1 1 1
polarization ∗-preserves I for W if and only if I is preserved for Ws for every s ∈ {−,+}∗. Moreover,
1 1
we have
∀s ∈ {−,+}∗, I is preserved for Ws ⇔ ∀s ∈ {−,+}∗,∀n ≥ 0, I is preserved for W(s,[n]−)
1 1
⇔ ∀s ∈ {−,+}∗,I is ∗− preserved for Ws.
1
B. Necessary condition
In the rest of this section, we consider a fixed two-user MAC W : G × G −→ Z. For the sake of
1 2
simplicity, we write p (x) to denote p (x).
y,z y,z,W
According to (1), I is preserved for W if and only if I(U ;V |Z Z V ) = 0, which means that for every
1 1 2 1 2 1
z ,z ∈ Z and every v ,v ∈ G , if P (v ,z ,z ,v ) > 0 then P (u |v ,z ,z ,v ) does
1 2 1 2 2 V2,Z1,Z2,V1 2 1 2 1 U1|V2,Z1,Z2,V1 1 2 1 2 1
not depend on v .
2
In order to study this condition, we should keep track of the values of z ,z ∈ Z and v ,v ∈ G for
1 2 1 2 2
which P (v ,z ,z ,v ) > 0. But P (v ,z ,z ,v ) = P (v + v ,z )P (v ,z ), so
V2,Z1,Z2,V1 2 1 2 1 V2,Z1,Z2,V1 2 1 2 1 Y1,Z1 1 2 1 Y2,Z2 2 2
it is sufficient to keep track of the pairs (y,z) ∈ G ×Z satisfying P (y,z) > 0. This is where YZ(W)
2 Y,Z
and {Yz(W) : z ∈ Z} become useful.
The following lemma gives a characterization of two user MACs with preserved I in terms of the
1
Fourier transform of the distributions p .
y,z
Lemma 2. I is preserved for W if and only if for every y ,y ,y′,y′ ∈ G and every z ,z ∈ Z satisfying
1 1 2 1 2 2 1 2
• y1 −y2 = y1′ −y2′,
• y1,y1′ ∈ Yz1(W) and y2,y2′ ∈ Yz2(W),
we have
pˆ (xˆ)·pˆ (xˆ)∗ = pˆ (xˆ)·pˆ (xˆ)∗, ∀xˆ ∈ G .
y1,z1 y2,z2 y1′,z1 y2′,z2 1
Proof: Let U ,U ,V ,V ,X ,X ,Y ,Y ,Z ,Z be as in Remark 1. We know that I is preserved
1 2 1 2 1 2 1 2 1 2 1
for W if and only if I(U ;V |Z Z V ) = 0, which is equivalent to say that U is independent of V
1 2 1 2 1 1 2
conditionally on (Z ,Z ,V ).
1 2 1
7
In other words, for any fixed (z ,z ,v ) ∈ Z×Z×G satisfying P (z ,z ,v ) > 0, if v ,v′ ∈ G
1 2 1 2 Z1,Z2,V1 1 2 1 2 2 2
satisfy P (v |z ,z ,v ) > 0 and P (v′|z ,z ,v ) > 0, then we have
V2|Z1,Z2,V1 2 1 2 1 V2|Z1,Z2,V1 2 1 2 1
∀u ∈ G , P (u |v ,z ,z ,v ) = P (u |v′,z ,z ,v ),
1 1 U1|V2,Z1,Z2,V1 1 2 1 2 1 U1|V2,Z1,Z2,V1 1 2 1 2 1
This condition is equivalent to say that for every z ,z ∈ Z and every v ,v ,v′ ∈ G satisfying
1 2 1 2 2 2
P (z ,z ,v +v ,v ) > 0 and P (z ,z ,v +v′,v′) > 0 we have
Z1,Z2,Y1,Y2 1 2 1 2 2 Z1,Z2,Y1,Y2 1 2 1 2 2
∀u ∈ G , P (u |z ,z ,v +v ,v ) = P (u |z ,z ,v +v′,v′).
1 1 X1−X2|Z1,Z2,Y1,Y2 1 1 2 1 2 2 X1−X2|Z1,Z2,Y1,Y2 1 1 2 1 2 2
Bydenotingv +v ,v ,v +v′ andv′ asy ,y ,y′ andy′ respectively(sothaty −y = y′−y′ = v ),wecan
1 2 2 1 2 2 1 2 1 2 1 2 1 2 1
deduce that I is preserved for W if and only if for every y ,y ,y′,y′ ∈ G and every z ,z ∈ Z satisfying
1 1 2 1 2 2 1 2
y −y = y′ −y′, P (z ,z ,y ,y ) > 0 and P (z ,z ,y′,y′) > 0 (i.e., y ,y′ ∈ Yz1(W)
1 2 1 2 Z1,Z2,Y1,Y2 1 2 1 2 Z1,Z2,Y1,Y2 1 2 1 2 1 1
and y ,y′ ∈ Yz2(W)), we have
2 2
∀u ∈ G , P (u |z ,z ,y ,y ) = P (u |z ,z ,y′,y′).
1 1 X1−X2|Z1,Z2,Y1,Y2 1 1 2 1 2 X1−X2|Z1,Z2,Y1,Y2 1 1 2 1 2
On the other hand, we have:
P (u |z ,z ,y ,y ) = P (u +u |z ,y )P (u |z ,y )
X1−X2|Z1,Z2,Y1,Y2 1 1 2 1 2 X1|Z1,Y1 1 2 1 1 X2|Z2,Y2 2 2 2
uX2∈G1
= p (u +u )p (u ) = (p ∗p˜ )(u ),
y1,z1 1 2 y2,z2 2 y1,z1 y2,z2 1
uX2∈G1
wherep˜ (x) = p (−x).SimilarlyP (u |z ,z ,y′,y′) = (p ∗p˜ )(u ).Therefore,
y2,z2 y2,z2 X1−X2|Z1,Z2,Y1,Y2 1 1 2 1 2 y1′,z1 y2′,z2 1
for every u ∈ G , we have
1 1
(p ∗p˜ )(u ) = (p ∗p˜ )(u ),
y1,z1 y2,z2 1 y1′,z1 y2′,z2 1
which is equivalent to pˆ (uˆ )·pˆ (uˆ )∗ = pˆ (uˆ )·pˆ (uˆ )∗ for every uˆ ∈ G .
y1,z1 1 y2,z2 1 y1′,z1 1 y2′,z2 1 1 1
Lemma 2 characterizes the MACs W for which I is preserved. In the next few lemmas we characterize
1
the MACs for which I is ∗− preserved.
1
Lemma 3. Suppose that I is ∗− preserved for W. Fix n > 0 and let (U ,V ) be a sequence of
1 i i 0≤i<2n
random pairs which are independent and uniformly distributed in G ×G . Let
1 2
1 1
F = .
0 1
(cid:20) (cid:21)
Define X2n−1 = F⊗n·U2n−1 and Y2n−1 = F⊗n·V2n−1, and for each 0 ≤ i < 2n let (X ,Y ) −W→ Z . We
0 0 0 0 i i i
have the following:
• The MAC (U0,V0) −→ Z02n−1 is equivalent to W[n]−.
• I(U0;V12n−1|Z02n−1V0) = 0.
Proof: We will show the lemma by induction on n > 0. For n = 1, the claim follows from Remark
1 and from the fact that I is preserved for W if and only if I(U ;V |Z Z V ) = 0 (see (1)).
1 0 1 0 1 0
Now let n > 1 and supposethat theclaim is truefor n−1. Let N = 2n−1. We haveX2n−1 = F⊗n·U2n−1
0 0
and Y2n−1 = F⊗n ·V2n−1, i.e., X2N−1 = F⊗n ·U2N−1 and Y2N−1 = F⊗n ·V2N−1. Therefore, we have:
0 0 0 0 0 0
• X0N−1 = F⊗(n−1) ·(U0N−1 +UN2N−1) and XN2N−1 = F⊗(n−1) ·UN2N−1.
• Y0N−1 = F⊗(n−1) ·(V0N−1 +VN2N−1) and YN2N−1 = F⊗(n−1) ·VN2N−1.
This means that (UN−1+U2N−1,VN−1+V2N−1,ZN−1) and (U2N−1,V2N−1,Z2N−1) satisfy the conditions
0 N 0 N 0 N N N
of the induction hypothesis. Therefore,
• I(U0 +UN;V1N−1 +VN2N+1−1|Z0N−1,V0 +VN) = 0.
8
• I(UN;VN2N+1−1|ZN2N−1,VN) = 0.
Moreover, since (UN−1 +U2N−1,VN−1 +V2N−1,ZN−1) is independent of (U2N−1,V2N−1,Z2N−1), we
0 N 0 N 0 N N N
can combine the above two equations to get:
I(U +U ,U ;VN−1 +V2N−1,V2N−1|Z2N−1,V +V ,V ) = 0,
0 N N 1 N+1 N+1 0 0 N N
which can be rewritten as
I(U U ;VN−1V2N−1|Z2N−1V V ) = 0. (5)
0 N 1 N+1 0 0 N
On the other hand, it also follows from the induction hypothesis that:
• The MAC (U0 +UN,V0 +VN) −→ Z0N−1 is equivalent to W[n−1]−.
• The MAC (UN,VN) −→ ZN2N−1 is equivalent to W[n−1]−.
This implies that the MAC (U ,V ) −→ Z2N−1 is equivalent to W[n]−. Now since I is ∗− preserved for
0 0 0 1
W, I must be preserved for W[n−1]−. Therefore,
1
I(U ;V |Z2N−1V ) = I(U ;V |ZN−1Z2N−1V ) (=a) 0, (6)
0 N 0 0 0 N 0 N 0
where (a) follows from (1). We conclude that:
I(U ;V2N−1|Z2N−1V ) = I(U ;V |Z2N−1V )+I(U ;VN−1V2N−1|Z2N−1V V )
0 1 0 0 0 N 0 0 0 1 N+1 0 0 N
≤ I(U ;V |Z2N−1V )+I(U U ;VN−1V2N−1|Z2N−1V V ) (=b) 0,
0 N 0 0 0 N 1 N+1 0 0 N
where (b) follows from (5) and (6).
2n−1
Lemma 4. For every n > 0, if X2n−1 = F⊗nU2n−1, then U = (−1)|i|bX , where |i| is the number
0 0 0 i b
i=0
of ones in the binary expansion of i. X
Proof:Wewillshowthelemmabyinductiononn > 0.Forn = 1,thefactthatX1 = F⊗1·U1 = F·U1
0 0 0
1
implies that X = U +U and X = U . Therefore U = X −X = (−1)|i|bX .
0 0 1 1 1 0 0 1 i
i=0
Now let n > 1 and suppose that the claim is true for n −1. Let NX= 2n−1. The fact that X2N−1 =
0
F⊗n ·U2N−1 implies that:
0
• X0N−1 = F⊗(n−1) ·(U0N−1 +UN2N−1).
• XN2N−1 = F⊗(n−1) ·UN2N−1.
We can apply the induction hypothesis to get:
N−1
• U0 +UN = (−1)|i|bXi.
i=0
X
N−1
• UN = (−1)|i|bXi+N.
i=0
X
Therefore,
N−1 N−1 N−1 N−1
U = (−1)|i|bX − (−1)|i|bX = (−1)|i|bX + (−1)1+|i|bX
0 i i+N i i+N
i=0 i=0 i=0 i=0
X X X X
N−1 2N−1 N−1 2N−1
= (−1)|i|bX + (−1)1+|i−N|bX (=a) (−1)|i|bX + (−1)|i|bX
i i i i
i=0 i=N i=0 i=N
X X X X
2N−1
= (−1)|i|bX ,
i
i=0
X
9
where (a) follows from the fact that for 2n = N ≤ i < 2N = 2n+1, we have |i−N| = |i−2n| = |i| −1.
b b b
Lemma 5. If I is ∗− preserved for W, then for every n > 0, every y ,...,y ,y′,...,y′ ∈ G and
1 1 2n 1 2n 2
every z ,...,z ∈ Z satisfying
1 2n
2n 2n
• yi = yi′,
i=1 i=1
• yX1 ∈ Yz1(XW),...,y2n ∈ Yz2n(W), and
• y1′ ∈ Yz1(W),...y2′n ∈ Yz2n(W),
we have
2n 2n
pˆ (xˆ) = pˆ (xˆ), ∀xˆ ∈ G .
yi,zi yi′,zi 1
i=1 i=1
Y Y
Proof: Fix xˆ ∈ G . If pˆ (xˆ) = 0 for every (y,z) ∈ YZ(W), then we clearly have
1 y,z
2n 2n
pˆ (xˆ) = pˆ (xˆ).
yi,zi yi′,zi
i=1 i=1
Y Y
Therefore, we can assume without loss of generality that there exists (y,z) ∈ YZ(W) which satisfies
pˆ (xˆ) 6= 0.
y,z
Let U2n+1−1, V2n+1−1, X2n+1−1, Y2n+1−1 and Z2n+1−1 be as in Lemma 3 and let N = 2n+1 so that we
0 0 0 0 0
have
I(U ;VN−1|ZN−1V ) = 0. (7)
0 1 0 0
Since XN−1 = F⊗(n+1) ·UN−1 and YN−1 = F⊗(n+1) ·VN−1, Lemma 4 implies that
0 0 0 0
N−1 N−1
U = (−1)|i|bX and V = (−1)|i|bY . (8)
0 i 0 i
i=0 i=0
X X
Notice that 0 ≤ i < N = 2n+1 : |i| ≡ 0 mod 2 = 0 ≤ i < N = 2n+1 : |i| ≡ 1 mod 2 = 2n.
b b
Let k ,...,k be the elements of 0 ≤ i < N : |i| ≡ 0 mod 2 and let l ,...,l be the elements of
1 2n b 1 2n
(cid:12)(cid:8) (cid:9)(cid:12) (cid:12)(cid:8) (cid:9)(cid:12)
0 ≤ i < N : (cid:12)|i| ≡ 1 mod 2 . (cid:12) (cid:12) (cid:12)
b
(cid:8) (cid:9)
Define (y˜,y˜′,z˜) as follows:
i i i 0≤i<N
(cid:8) (cid:9)
• For every 1 ≤ i ≤ 2n, let y˜ki = yi, y˜k′i = yi′ and z˜ki = zi (where yi,yi′,zi are given in the hypothesis
of the lemma).
• For every 1 ≤ i ≤ 2n, let y˜li = y˜l′i = y and z˜li = z (where (y,z) is any fixed pair in YZ(W)
satisfying pˆ (xˆ) 6= 0).
y,z
Now let v˜N−1 = (F⊗(n+1))−1 ·y˜N−1 and v˜′N−1 = (F⊗(n+1))−1 ·y˜′N−1. We have
0 0 0 0
N−1 2n 2n
v˜ (=a) (−1)|i|by˜ = (y˜ −y˜ ) = y −2ny
0 i ki li i
!
i=0 i=1 i=1
X X X
2n 2n N−1
(=b) y′ −2ny = (y˜′ −y˜′ ) = (−1)|i|by˜′ (=c) v˜′,
i ki li i 0
!
i=1 i=1 i=0
X X X
2n 2n
where (a) and (c) follow from Lemma 4. (b) follows from the fact that y = y′. Therefore,
i i
i=1 i=1
X X
(v˜ ,z˜N−1) = (v˜′,z˜N−1). (9)
0 0 0 0