Flow Decomposition for Multi-User Channels - Part I
aa r X i v : . [ c s . I T ] N ov Flow Decomposition for Multi-User Channels- Part I
Jonathan Ponniah
Department of Electrical EngineeringSan Jose State University
Liang-Liang Xie
Department of Electrical and Computer EngineeringUniversity of Waterloo
Abstract —A framework based on the idea of flow decomposition is proposed to characterize the decode-forward region for generalmulti-source, multi-relay, all-cast channels with independentinput distributions. The region is difficult to characterize directlywhen deadlocks occur between two relay nodes, in which bothnodes benefit by decoding after each other. Rate-vectors in thedecode-forward region depend ambiguously on the outcomes ofall deadlocks in the channel. The region is characterized indi-rectly in two phases. The first phase assumes relays can operatenon-causally. It is shown that every rate-vector in the decode-forward region corresponds to a set of flow decompositions, whichdescribe the messages decoded at each node with respect to themessages forwarded by all the other nodes. The second phaseimposes causal restrictions on the relays. Given an arbitraryset of (possibly non-causal) flow decompositions, necessary andsufficient conditions are derived for the existence of an equivalentset of causal flow decompositions that achieves the same rate-vector region.
I. I
NTRODUCTION
Multi-node channels in network information theory arenot fully understood. Despite decades of research, only thecapacity of the one-way point-to-point channel is known. Thetwo-way channel is still unsolved as are all other channels withthree or more nodes. We examine multi-node all-cast channelsin which the source nodes send common messages to all ofthe destination nodes with the help of the relay nodes. Thegeneral discrete-memoryless channel models the input-outputdynamics. Our analysis is confined to decode-forward schemes.Each relay decodes source messages and forwards some ofthese messages to destinations and other relays. Our objectiveis to characterize the region of rate vectors achieveable bydecode-forward schemes, otherwise known as the decode-forward region.The original decode-forward scheme, first proposed for theone-source one-relay channel, also extends to other channelswith particular numbers of source and relay nodes. In moregeneral multi-node channels, deadlocks between pairs of relaynodes occur when both relays in a pair have an incentive towait for the other to forward messages before decoding theirown. Each deadlock yields two different outcomes dependingon which of the two relays decode first, and each outcomegenerates different rate-vector regions.To send a message, each source transmits a sequence ofsymbols through the channel. The mapping from messages tosequences is defined by a source codebook. Each transmittedsymbol requires one use of the channel by the sources. The symbols received at any given node are correlated with thesymbols simultaneously transmitted by the other nodes. Relaysand destinations decode source messages if there are codebooksequences correlated (or “typical”) with the sequences ofsymbols they receive. A message “hop” occurs when a relaydecodes and forwards a message. Source messages hop fromnode to node until they arrive at each destination.Every message hop between two nodes has an encodingdelay, defined as the number of channel uses from the start ofthe first node sending the message until the start of the second.Without loss of generality, the encoding delays can be positiveinteger-valued multiples of some fixed block of channel uses.In each block, the sources send unique messages and the relaysforward unique combinations of past messages to other nodes.To decode any set of messages, a node first identifies allof the prior blocks in which other nodes transmit some ofthe messages in the set, then identifies which codebook se-quences are jointly typical with the sequences received duringthese blocks. This general procedure is called joint decoding.Since the knowledge of one message in any simultaneouslytransmitted set helps remove some uncertainty about the othermessages (even if the messages are independent), it is better todecode messages together (or “jointly”) rather than separately.The messages decoded by each node in a given block areunique and determined by the desired rate vector and themessages forwarded by other nodes in previous blocks. Theencoding delays, which determine the messages forwarded byeach relay, are constrained by the assumption of causality;relays can only forward messages they have already decoded.Any relay that jointly decodes a set of source messages toachieve a desired rate-vector must be the last node to jointlydecode this set out of all of the preceding nodes. Encodingdelays at this relay must be larger than the encoding delaysof all the preceding hops. A deadlock occurs if the encodingdelays at one relay can only satisfy these causality constraintsat the expense of another relay.An outer-bound on the decode-forward region can be de-rived from the capacity of the point-to-point channel. Thisouter-bound depends on the sequences of nodes traversed bythe message hops from each source. We call each sequencea flow . For some arbitrarily assigned flows and some relay inthe channel, the corresponding “super-source” consists of allthe nodes preceding the relay. The “super-channel” is the thechannel seen by the relay free of any interference from nodesot in the super-source. The sum rate of the source nodes in thesuper-source cannot exceed the capacity of the point-to-pointsuper-channel. Invoking this argument for any set of flowscreates an outer-bound on the decode-forward region. It turnsout that this outer-bound is tight when deadlocks are absentfrom the channel. When deadlocks are present, the decode-forward region is too complicated to express explicitly; theoutcomes of each deadlock introduce unique constraints thathave no simple interpretation.The decode-forward region is difficult to characterize di-rectly, so we propose an indirect approach in two phases. Thefirst phase allows relays to operate non-causally. For arbitrarilyassigned flows and encoding delays, we show each node canachieve any rate-vector in the outer-bound by decoding a corre-sponding set of source messages. We introduce the concept offlow decompositions to describe the source messages decodedat a particular node relative to the messages encoded by all theother nodes. Every rate-vector in the decode-forward regioncorresponds to a set of flow decompositions, where each flowdecomposition in the set belongs to a unique node in thechannel.The second phase reimposes causal restrictions on the relays.Flow decompositions are causal by definition if the correspond-ing relays only forward messages they have already decoded.The previous result and the assumption of causality imply thatevery rate-vector in the decode-forward region corresponds toa set of causal flow decompositions. An arbitrarily chosenset (which may include non-causal flow decompositions) is“feasible” by definition if there exists another set of causalflow decompositions that achieves the same rate-vector region.We derive necessary and sufficient conditions that determinewhether or not an arbitrary set of flow decompositions is feasi-ble. These conditions emerge naturally from the two-way two-relay channel, the simplest channel in which deadlocks occur,and restrict the flow decompositions that can be simultaneouslyassigned to the affected nodes.No single decode-forward scheme is universally better thanthe others. Each scheme achieves a different region of rate-vectors. Collectively, these overlapping and interlocking re-gions recover the outer-bound defined by the point-to-pointsuper-channel. Our approach exploits this underlying structure.However, we do not express the decode-forward region in theconventional way for channels with deadlocks. We also do notaddress which rate-vectors outside the decode-forward regionare achievable. This fundamental open problem is not fullysolved, even for the one-relay channel.Concerning the organization of the paper, Section II definesthe concept of flow, Section III defines the concept of flowdecomposition and states the theorem that gives this conceptsignificance, Section IV works through an important examplethat demonstrates some of the key ideas in the proof, SectionV provides this proof, and Section VI concludes the paper andsets the stage for the one to follow. II. F
LOWS
Let N denote the set of all nodes, I ⊆ N the setof nodes with inputs into the channel,
S ⊆ I the set ofsource nodes,
D ⊆ N the set of destination nodes, and
Z ⊆ I the set of relay nodes, where
Z ⊆ D and Z and S are not necessarily disjoint. Every destination decodes allof the source messages. Let y D := {h d, y d i : d ∈ D} and x I := {h i, x i i : i ∈ I} . The input-output dynamics aremodeled by the discrete memoryless channel: ( Y i ∈I X i , p ( y D | x I ) , Y d ∈D Y d ) . (1)Decode-forward schemes describe both the order in whichsource messages “hop” from one node to another until theyreach the destination nodes as well as the encoding delaysinduced by each hop. In each “block” of n channel uses,the source nodes generate new messages and the relay nodesforward messages from the past. Encoding delays refer to thedifference between the blocks in which the first and secondnodes of a hop transmit the same message. Hops also occurbetween disjoint sets of nodes if the nodes in each set transmitthe same message simultaneously, but both sets transmit thesame message in different blocks. The encoding delays arenon-negative integer multiples of n and the entire transmissionperiod occurs over B blocks.A flow f ( s, d ) is a sequence of hops that starts at source s ∈ S and finishes at destination d ∈ D . Formally, f ( s, d ) = Z k −→ Z k −→ · · · → Z q k q −→ d where Z := { s } , Z l ⊆Z \ { s, d } for each l = 2 , . . . , q , Z l ∩ Z l ′ = {} for all l, l ′ =1 , . . . , q when l = l ′ , and k l ∈ N is the one-hop encoding delayof a message from s leaving any node in Z l . By convention || f ( s, d ) || := q . For any Z ⊆ Z , Z ∈ f ( s, d ) if Z ⊆ Z l for some ≤ l ≤ q . Note that d / ∈ f ( s, d ) . For any i ∈ Z l , k s,i = P le =1 k e is the encoding delay between source s andnode i . This definition extends to nodes that do not forwardmessages from s . If i / ∈ f ( s, d ) for any d ∈ D then k s,i := ∞ .A flow set F := { f ( s, d ) : s ∈ S , d ∈ D} specifies a flowfor every source-destination pair ( s, d ) . The space of flow setsis denoted by F . Each flow set induces a multi-edge directedgraph on N . This induced graph may have cycles. Source s ∈S generates the message m s ( b ) ∈ { , . . . , nR s } in block b and node i ∈ Z sends the index w i ( b ) ∈ { , . . . , n ( P s ∈ S R s ) } assigned to the message vector ¯ w i ( b ) := {h s, m ( b − k s,i ) i : s ∈ S } in the same block, where k s,i is finite for each s ∈ S ,and R s is the rate of source s . Example 1.
Let f (1 ,
3) = 1 −→ ∞ −→ as depicted in Figure1(i). Then k , = 1 , k , = ∞ , and k , = k , + k , = ∞ .In block b , node 1 sends m ( b ) ∈ { , . . . , nR } and node 2forwards m ( b − . Node 3 does not forward messages. Each node i ∈ I has a codebook consisting of n P s ∈ S R s n -length codewords generated by an i.i.d distribution on X .Every index w ∈ { , . . . , n P s ∈ S R s } corresponds to a uniquecodeword ¯ x i ( w ) . In block b , node i transmits the codeword ¯ x i ( w i ( b )) . ∞ ∞ ∞ ∞ ∞ (i) (ii) (iii) (iv) (v)Fig. 1: The relay channel (i) f (1 ,
3) = 1 −→ ∞ −→ ; (ii) ¯ L = ( { } , { } ) ; (iii) ¯ L = ( { } , {} , { } ) ; (iv) ¯ L = ( { } , { } ) ; (v) f (1 ,
3) = 1 −→ ∞ −→ and ¯ L = ( { } , {} , { } ) . Original flows are depicted with dottedlines and virtual flows with solid lines in (ii)-(iv).Let ¯ R := {h s, R s i : s ∈ S} denote the vector of ratesallocated to the source nodes. For any S ⊆ S \ { d } , let F d ( S ) := { i ∈ f ( s, d ) : s ∈ S } denote the set of nodes cov-ered by the flows terminating at node d , let ˜ F d ( S ) := I\ F d ( S ) ,and let R S := P s ∈ S R s . Since R S is the rate of the “super-source” F d ( S ) , it follows that R S < I ( X F d ( S ) ; Y d | X ˜ F d ( S ) ) , (2)by invoking the capacity of the point-to-point channel on the“super-channel” between F d ( S ) and node d . Let R d ( F ) denotethe region of rate vectors ¯ R := {h s, R s i : s ∈ S} that satisfy(2) for every subset S ⊆ S \ { d } . We have the followingouter-bound on the decode-forward region: [ F ∈F \ d ∈D R d ( F ) . (3) Example 2.
Let f (1 ,
3) = 1 −→ ∞ −→ and F = { f (1 , } as shown in Figure 1(i). Then R ( F ) = { R : R
Decoding schemes specify the message vectors decoded ineach block at each node. For any d ∈ D , a layered partition ¯ L d := { L d, , . . . , L d, | ¯ L d |− } of F d ( S ) , is a vector of sets,some possibly empty, that satisfies the following conditions: L d,l ⊆ F d ( S ) for every l = 0 , . . . , | ¯ L d | ; L d,l ∩ L d,m = {} , for l, m = 0 , . . . , | ¯ L d | and l = m ; L d, | ¯ L d |− = {} , and F d ( S ) = ∪ | ¯ L d |− l =0 L d,l . If i ∈ L d,l , then LAYER ( i ) = l . Each “layer”corresponds to some past block of channel uses. Deeper layersgo deeper into the past. The layered partition ¯ L d decomposes F so that each node in F d ( S ) helps node d decode somesubset of a message vector in some previous block. The pair ( F, ¯ L d ) is a flow decomposition and describes the messagevectors encoded and decoded at node d in each block. Example 3.
Let f (1 ,
3) = 1 −→ ∞ −→ . Figure 1(ii) depicts ¯ L = ( { } , { } ) , Figure 1(iii) depicts ¯ L = ( { } , {} , { } ) ,Figure 1(iv) depicts ¯ L = ( { } , { } ) . Example 4.
Let f (1 ,
3) = 1 −→ ∞ −→ . Figure 1(v) depicts ¯ L = ( { } , {} , { } ) . Let ¯ m d ( b ) := {h s, m s ( b s ) i : s ∈ S} denote the messagevector node d decodes in block b where ≤ b s ≤ b for each s ∈ S . This message vector determines the side information atnode d , namely the set of message vectors ∪ b − q =1 ¯ m d ( q ) node d decodes prior to block b . Node i carries useful informationto node d in block b − l if ¯ m d ( b ) splits ¯ w i ( b − l ) into aset of messages that node d decodes in block b and a set ofmessages that node d knows from its side information. Let A l := { i : ¯ w i ( b − l ) ∩ ¯ m d ( b ) = {} , ¯ w i ( b − l ) ⊆ ∪ bq =1 ¯ m d ( q ) } and ˜ A l := { i : ¯ w i ( b − l ) ⊆ ∪ b − q =1 ¯ m d ( q ) } . Equivalently, ˜ A l := { i : ¯ w i ( b − l ) ⊆ ∪ bq =1 ¯ m d ( q ) } \ A l . If L d,l ⊆ A l ∪ ˜ A l for every ≤ l ≤ | ¯ L d | − , then ¯ m d ( b ) is a splitting vector .Define the mapping u : S × F d ( S ) → N as follows: for s ∈S and i ∈ f ( s, d ) , let u ( s, i ) := LAYER ( i )+ k s,i . Let M ( s ) := { j : j = arg min i ∈ f ( s,d ) u ( s, i ) } . For each source s ∈ S , thereis a virtual source v ( s ) := { j : j = arg min i ∈ M ( s ) || f ( s, i ) ||} .The virtual source v ( s ) is a set of nodes Z ∈ f ( s, d ) thatappear to node d as the original source s . The encoding delay k s,v ( s ) is well defined, since the nodes in v ( s ) experience thesame encoding delay from source s by construction. Set: ¯ m d ( b ) := {h s, m s ( b − k s,v ( s ) − LAYER ( v ( s ))) i : s ∈ S} (4) Lemma 1. ¯ m d ( b ) is a splitting vector for ( F, ¯ L d ) roof. Given f ( s, d ) := Z k −→ Z k −→ · · · → Z q k q −→ d , let f ( v ( s ) , d ) := Z ′ k ′ −→ Z ′ k ′ −→ · · · Z ′ p k ′ p −→ d be the flow thatsatisfies Z ′ = v ( s ) , Z ′ ⊆ Z q − ( p − , and Z ′ m = Z q − ( p − m ) forall ≤ m ≤ p ≤ q . For i ∈ f ( v ( s ) , d ) let k v ( s ) ,i := k s,i − k s,v ( s ) . Let f ( s, v ( s )) := Z ′ k ′ −→ Z k ′ −→ · · · k ′ p − −−−→ Z ′ p be theflow that satisfies Z ′ p = v ( s ) , Z ′ p ⊆ Z p , and Z ′ m = Z m for allof ≤ m ≤ p − ≤ q − . For i ∈ f ( s, v ( s )) let k i,v ( s ) := k s,v ( s ) − k s,i . We first derive the following inequalities: LAYER ( v ( s )) − LAYER ( i ) ≤ k v ( s ) ,i ∀ i ∈ f ( v ( s ) , d ) (5) LAYER ( i ) − LAYER ( v ( s )) > k i,v ( s ) ∀ i ∈ f ( s, v ( s )) (6)By construction, LAYER ( v ( s )) + k s,v ( s ) ≤ LAYER ( i ) + k s,i for all i ∈ f ( v ( s ) , d ) which implies (5) and LAYER ( v ( s )) + k s,v ( s ) < LAYER ( i ) + k s,i for all i ∈ f ( s, v ( s )) which implies(6). Fix l ∈ { , . . . , | ¯ L d | − } and i ∈ L d,l . For every s ∈ S there are two cases to consider. If i ∈ f ( v ( s ) , d ) then (5)implies h s, m ( b − k s,i − l ) i ∈ ∪ bq =1 ¯ m d ( q ) . If i ∈ f ( s, v ( s )) then (6) implies h s, m ( b − k s,i − l ) i ∈ ∪ b − q =1 ¯ m d ( q ) whichconcludes the proof.For f ( s, d ) = Z k −→ Z k −→ · · · → Z q k q −→ d , the virtualflow g ( s, d ) := Z ′ k ′ −→ Z ′ k ′ −→ · · · Z ′ p k ′ p −→ d seen by node d satisfies four conditions: Z ′ l ⊆ Z j l for each ≤ l ≤ p ≤ q ,where j , j , . . . , j p is a subsequence of , , . . . , q ; i ∈ Z ′ l ifand only if i ∈ f ( s, d ) and LAYER ( v ( s )) − LAYER ( i ) = k v ( s ) ,i ; k ′ l = P j l +1 − e = j l k e ; and Z ′ := { v ( s ) } . Only the nodes in thevirtual flow help the destination decode the source, so thevirtual flow g ( s, d ) appears to node d as the original flow f ( s, d ) . Since every subsequent node in a virtual flow movesfrom a lower layer to a higher layer, { g ( s, d ) : s ∈ S} generates a multi-edge directed acyclic graph. Example 5.
Let f (1 ,
3) = 1 −→ ∞ −→ and ¯ L = ( { } , { } ) ,as depicted in Figure 1(ii). Since LAYER (2) − LAYER (1) =1 ≤ k , , v (1) = 1 satisfies (5) and g (1 ,
3) = 1 −→ ∞ −→ .Then ¯ m ( b ) = {h , m ( b − i} . Example 6.
Let f (1 ,
3) = 1 −→ ∞ −→ and ¯ L =( { } , {} , { } ) as depicted in Figure 1(iii). Since LAYER (1) − LAYER (2) = 2 > k , , v (1) = 2 satisfies (6) and g (1 ,
3) =2 ∞ −→ . Then ¯ m ( b ) = {h , m ( b − i} . Example 7.
Let f (1 ,
3) = 1 −→ ∞ −→ and ¯ L = ( { } , { } ) as depicted in Figure 1(iv). Since LAYER (1) − LAYER (2) = − < k , , v (1) = 1 satisfies (5) and g (1 ,
3) = 1 ∞ −→ . Then ¯ m ( b ) = {h , m ( b ) i} . Example 8.
Let f (1 ,
3) = 1 −→ ∞ −→ and ¯ L =( { } , {} , { } ) as depicted in Figure 1(v). Since LAYER (1) − LAYER (2) = 2 = k , , v (1) = 1 satisfies (5) and g (1 ,
3) =1 −→ ∞ −→ . Then ¯ m ( b ) = {h , m ( b − i} . For any subset S ⊆ S and ≤ l ≤ | ¯ L d | − , let A l ( S ) := { i ∈ g ( s, d ) : s ∈ S } ∩ L d,l (7) ˜ A l ( S ) := ( ∪ lq =0 L d,q ) \ A l ( S ) (8) It follows from Lemma 1 that A l ( S ) = A l and ˜ A l ( S ) ⊆ ˜ A l .We rely on the context to convey that A l ( S ) and ˜ A l ( S ) dependon a particular ( F, ¯ L d ) . Example 9.
Let f (1 ,
3) = 1 −→ ∞ −→ and ¯ L = ( { } , { } ) ,as depicted in Figure 1(ii). Then A ( { } ) = { } and A ( { } ) = { } . Example 10.
Let f (1 ,
3) = 1 −→ ∞ −→ and ¯ L =( { } , {} , { } ) as depicted in Figure 1(iii). Then A ( { } ) = { } , A ( { } ) = {} , and A ( { } ) = {} . Example 11.
Let f (1 ,
3) = 1 −→ ∞ −→ and ¯ L = ( { } , { } ) as depicted in Figure 1(iv). Then A ( { } ) = { } , A ( { } ) = {} , and A ( { } ) = {} . Example 12.
Let f (1 ,
3) = 1 −→ ∞ −→ and ¯ L =( { } , {} , { } ) as depicted in Figure 1(v). Then A ( { } ) = { } , A ( { } ) = {} , and A ( { } ) = { } . In block b , node d decodes ¯ m d ( b ) as defined in (4) byfinding the message vector {h s, m i : s ∈ S} that satisfiesthe following typicality checks for ≤ l ≤ | ¯ L d | − : ( { ¯ x i ( w i ( b − l )) : i ∈ A l ( S ) } , { ¯ X i ( b − l ) : i ∈ ˜ A l ( S ) } , ¯ Y d ( b − l )) ∈ T ( n ) ǫ ( X ∪ lq =0 L d,q , Y d ) . An error event occurs if some subset of source messages S ⊆ S is decoded incorrectly. The probability of such anevent goes to zero if the following constraint is satisfied: R S < | ¯ L d |− X l =0 I ( X A l ( S ) ; Y d | X ˜ A l ( S ) ) . (9)Let R ( F, ¯ L d ) denote the region of rate vectors {h s, R s i : s ∈ S} that satisfy (9) for every S ⊆ S \ { d } . Example 13.
Let f (1 ,
3) = 1 −→ ∞ −→ , F = { f (1 , } , and ¯ L = ( { } , { } ) as depicted in Figure 1(ii). Then ¯ m ( b ) = {h , m ( b − i} is the unique m ∈ { , . . . , nR } that satisfies (¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y ) and (¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , Y ) . It follows that R ( F, ¯ L ) = { R : R < I ( X ; Y ) + I ( X ; Y | X ) = I ( X X ; Y ) } . Remark 1.
The rate region R ( F, ¯ L ) achieved in Example13 coincides with the decode-forward outerbound R ( F ) inExample 2. Example 14.
Let f (1 ,
3) = 1 −→ ∞ −→ , F = { f ( , } ,and ¯ L = ( { } , {} , { } ) as depicted in Figure 1(iii). Then ¯ m ( b ) = {h , m ( b − i} is the unique m ∈ { , . . . , nR } that satisfies (¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y ) . It follows that R ( F, ¯ L ) = { R : R < I ( X ; Y ) } . Remark 2.
In Example 14, node 1 is excluded from the virtualflow and the typicality check because the message it sends inblock b − is already decoded by node 3 in block b . Thecorresponding achievable region R ( F, ¯ L ) is less than thedecode-forward outerbound in Example 2. xample 15. Let f (1 ,
3) = 1 −→ ∞ −→ , F = { f (1 , } , and ¯ L = ( { } , { } ) as depicted in Figure 1(iv). Then ¯ m ( b ) = {h , m ( b ) i} is the unique m ∈ { , . . . , nR } that satisfies (¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y ) . It follows that R ( F, ¯ L ) = { R : R < I ( X ; Y ) } . Remark 3.
Node 2 is excluded from the virtual flow and thetypicality check because the message it sends in block b − is already decoded by node 3 in block b . The less restrictiveconstraint R < I ( X ; Y | X ) is also achievable since node3 has already decoded the message that node 2 sends in block b . However both regions are less than the decode-forwardouterbound in Example 2. Example 16.
Let f (1 ,
3) = 1 −→ ∞ −→ , F = { f (1 , } and ¯ L = ( { } , {} , { } ) as depicted in Figure 1(v). Then ¯ m ( b ) = {h , m ( b − i} is the unique m ∈ { , . . . , nR } thatsatisfies (¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y ) and (¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , Y ) . It follows that R ( F, ¯ L ) = { R : R < I ( X ; Y ) + I ( X ; Y | X ) = I ( X X ; Y ) } . Remark 4.
The rate regions R ( F, ¯ L ) in Examples 13 and16 are the same even though the corresponding flow decom-positions are different. Flow decompositions that recover thesame region of rate-vectors are equivalent by definition. Although different flow decompositions recover differentrate-vector regions, these regions share a fundamental relation-ship.
Theorem 1. If ¯ R ∈ R d ( F ) then ¯ R ∈ R ( F, ¯ L d ) for some ¯ L d .Proof. See Section V.Examples 6 and 7 achieve strictly lower rates (seen inExamples 13 and 15) than Example 5, which matches theouter-bound. The diamond relay channel is a more suitableexample of the concepts introduced in this section.IV. E
XAMPLE : T HE D IAMOND R ELAY C HANNEL
The diamond relay channel is defined by the flows f (1 ,
5) =1 −→ −→ ∞ −→ and f (2 ,
5) = 2 −→ ∞ −→ , bothdepicted in Figure 2(i), where S = { , } , Z = { , , } and D = { } . In block b , node 1 encodes ¯ w ( b ) = {h , m ( b ) i} ,node 2 encodes ¯ w ( b ) = {h , m ( b − i , h , m ( b ) i} , node3 encodes ¯ w ( b ) = {h , m ( b − i} , and node 4 encodes ¯ w ( b ) = {h , m ( b − i} . Let F = { f (1 , , f (2 , } . Bydefinition, R ( F ) is set of rate vectors ¯ R = ( R , R ) thatsatisfies: R < I ( X X X ; Y | X ) (10) R < I ( X X ; Y | X X ) (11) R + R < I ( X X X X ; Y ) (12)Four different decoding schemes collectively achieve all therate vectors in R ( F ) . A. The first decoding scheme
Set ¯ L = ( { , } , {} , { } , { } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (2) = 1 ≤ k , and LAYER (1) − LAYER (3) = 3 ≤ k , which satisfy (5). Similarly, v (2) = 4 since LAYER (2) − LAYER (4) = 2 > which satisfies (6). Then g (1 ,
5) = 1 −→ −→ ∞ −→ and g (2 ,
5) = 4 ∞ −→ . The virtualflows are depicted in Figure 2(ii).In block b , node 5 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b − i} by finding the unique pair m ∈{ , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfythe following typicality checks: (¯ x ( m ) , ¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , X , Y )(¯ x ( m ) , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , Y ) (13) (¯ x ( m ) , ¯ X ( b − , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , Y ) In each block, node 2 sends its own source message and asource message from node 1. The source message generatedby node 2 in block b − has already been decoded by node 5in block b , so ¯ x in (13) effectively depends on m alone. Theprobability of error goes to zero if ¯ R satisfies the followingconditions: R < I ( X ; Y | X ) + I ( X ; Y | X X )+ I ( X ; Y | X X X )= I ( X X X ; Y | X ) (14) R < I ( X ; Y | X ) (15) R + R < I ( X X ; Y ) + I ( X ; Y | X X )+ I ( X ; Y | X X X X )= I ( X X X X ; Y ) (16)Each of the three inequalities above addresses an error event.The probability that m is decoded correctly and m is not,goes to zero if (14) is satisfied. Similarly, the probability that m is decoded correctly and m is not, goes to zero if (15) issatisfied. Finally, the probability that both m and m are notdecoded correctly goes to zero if (16) is satisfied. B. The second decoding scheme
Set ¯ L = ( { } , { } , { } , { } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (2) = 1 ≤ k , and LAYER (1) − LAYER (3) = 3 ≤ k , which satisfy (5). Similarly, v (2) = 2 since LAYER (2) − LAYER (4) ≤ k , = 1 which satisfies (5).Then g (1 ,
5) = 1 −→ −→ ∞ −→ and g (2 ,
5) = 2 −→ ∞ −→ .The virtual flows are depicted in Figure 2(iii).In block b , node 5 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b − i} by finding the unique pair m ∈{ , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfythe following typicality checks: (¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y )(¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , Y )(¯ x ( m , m ) , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , Y )(¯ x ( m ) , ¯ X ( b − , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ (i) (ii) (iii) (iv) (v)Fig. 2: The diamond relay channel (i) f (1 ,
5) = 1 −→ −→ ∞ −→ and f (2 ,
5) = 2 −→ ∞ −→ (ii) ¯ L = ( { , } , {} , { } , { } ) (iii) ¯ L = ( { } , { } , { } , { } ) (iv) ¯ L = ( { } , {} , { } , { , } ) (v) ¯ L = ( { } , {} , {} , { , } , { } ) . Original flows are depictedwith dotted lines and virtual flows with solid lines in (ii)-(v). ∈ T ( n ) ǫ ( X , X , X , Y ) The probability of error goes to zero if ¯ R satisfies the followingconditions: R < I ( X ; Y ) + I ( X ; Y | X X )+ I ( X ; Y | X X X )= I ( X ; Y ) + I ( X X ; Y | X X ) (17) R < I ( X ; Y | X ) + I ( X ; Y | X X ) (18) R + R < I ( X ; Y ) + I ( X ; Y | X )+ I ( X ; Y | X X ) + I ( X ; Y | X X X X )= I ( X X X X ; Y ) (19) C. The third decoding scheme
Set ¯ L = ( { } , {} , { } , { , } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (2) = 0 ≤ k , and LAYER (1) − LAYER (3) = 3 = k , which satisfy (5). Similarly, v (2) = 2 since LAYER (2) − LAYER (4) = 1 = k , which satisfies (5).Then g (1 ,
5) = 1 −→ ∞ −→ and g (2 ,
5) = 2 −→ ∞ −→ . Thevirtual flows are depicted in Figure 2(iv).In block b , node 5 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b − i} by finding the unique pair m ∈{ , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfythe following typicality checks: (¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y )(¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , Y )(¯ x ( m ) , ¯ x ( m ) , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , X , Y ) (20)The message from node 1 forwarded by node 2 in block b − has already been decoded by node 5 in block b , so ¯ x in (20)effectively depends on m alone. The probability of error goesto zero if ¯ R satisfies the following conditions: R < I ( X ; Y ) + I ( X ; Y | X X X ) (21) R < I ( X ; Y | X ) + I ( X ; Y | X X X ) (22) R + R < I ( X ; Y ) + I ( X ; Y | X )+ I ( X X ; Y | X X X )= I ( X X X X ; Y ) (23) D. The fourth decoding scheme
Set ¯ L = ( { } , {} , {} , { , } , { } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (2) = 0 ≤ k , and LAYER (1) − LAYER (3) = 3 = k , which satisfy (5). Similarly, v (2) = 2 since LAYER (2) − LAYER (4) = 1 = k , which satisfies (5).Then g (1 ,
5) = 1 −→ ∞ −→ and g (2 ,
5) = 2 −→ ∞ −→ . Thevirtual flows are depicted in Figure 2(v).In block b , node 5 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b − i} by finding the unique pair m ∈{ , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfythe following typicality checks: (¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y )(¯ x ( m ) , ¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , Y )(¯ x ( m ) , ¯ X ( b − , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , X , Y ) The message from node 1 forwarded by node 2 in block b − has already been decoded by node 5 in block b , so ¯ x in (20)effectively depends on m alone. The probability of error goeso zero if ¯ R satisfies the following conditions: R < I ( X ; Y ) + I ( X ; Y | X X ) (24) R < I ( X ; Y | X X ) + I ( X ; Y | X X X )= I ( X X ; Y | X X ) (25) R + R < I ( X ; Y ) + I ( X X ; Y | X )+ I ( X ; Y | X X X X )= I ( X X X X ; Y ) (26) E. The achievability of R ( F ) To prove that R ( F ) is achievable, we show that any ratevector in the region defined by (10)-(12) is in the regiondefined by (14)-(16) or (17)-(19) or (21)-(23) or (24)-(26).Suppose ¯ R is not in (14)-(16). Since (14) and (16) define theboundaries of R ( F ) , ¯ R must violate (15). Then (16) impliesthat ¯ R satisfies (17). If ¯ R satisfies (18) the proof is finishedsince (19) is a boundary of R ( F ) . If ¯ R violates (18) then(19) implies that ¯ R satisfies (21). If ¯ R satisfies (22) the proofis finished since (23) is a boundary of R ( F ) . Otherwise (23)implies that ¯ R satisfies (24). Here the proof is finished since(25) and (26) are boundaries of R ( F ) . Remark 5.
A fundamentally distinctive feature of the flowdecomposition framework is that the encoding scheme isfixed and determines the boundaries of the decode-forwardregion, but the decoding scheme is variable and dependson the desired rate-vector in the region. The proof aboveexplicitly depends on the particular flows and encoding delaysof the channel, and does not extend to general channels witharbitrary flows. Appendix A and B use the same methodologyto prove the achievability of R d ( F ) for the two-source coop-erative multi-access channel and the two-source multi-accessrelay channel respectively. V. P
ROOF OF T HEOREM R ∈ R d ( F ) and construct a flow decomposition with thespecial property of being “complete”. If R / ∈ R ( F, ¯ L d ) , then ( F, ¯ L d ) is successively modified (or shifted) until an ( F, ¯ L ′ d ) is created for which ¯ R ∈ R ( F, ¯ L ′ d ) .Given any ( F, ¯ L d ) and S ⊆ S \ { i } , let G d ( S ) := { i ∈ g ( s, d ) : s ∈ S } denote the set of nodes covered by the virtualflows. By definition, ( F, ¯ L d ) is complete if F d ( S ) = G d ( S ) .More generally, ( F, ¯ L d ) is complete on S ⊆ S , by definition,if F d ( S ) = G d ( S ) . Example 17.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) =2 −→ ∞ −→ , and F = { f (1 , , f (2 , } . If ¯ L =( { , } , {} , { } , { } ) , then ( F, ¯ L ) is complete on S = { } and S = { , } as depicted in Figure 2(ii). Example 18.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) =2 −→ ∞ −→ , and F = { f (1 , , f (2 , } . If ¯ L =( { } , { } , { } , { } ) , then ( F, ¯ L ) is complete on S = { } , S = { , } , and S = { } as depicted in Figure 2(iii). Example 19.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) =2 −→ ∞ −→ , and F = { f (1 , , f (2 , } . If ¯ L =( { } , { } , { , } ) , then ( F, ¯ L ) is complete on S = { , } and S = { } as depicted in Figure 2(iv). Example 20.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) =2 −→ ∞ −→ , and F = { f (1 , , f (2 , } . If ¯ L =( { } , { , } , { } ) , then ( F, ¯ L ) is complete on S = { , } and S = { } as depicted in Figure 2(v). Lemma 2.
The constraints (2) and (9) coincide at S = S if ( F, ¯ L d ) is complete.Proof. F d ( S ) = G d ( S ) . Hence, F d ( S ) = ∪ | ¯ L d |− l =0 A l ( S ) . Lemma 3.
A complete ( F, ¯ L d ) exists.Proof. In Section III, the set of virtual flows { g ( s, d ) : s ∈ S} derives from a particular flow decomposition ( F, ¯ L d ) . To proveLemma 3, it is useful to derive some corresponding ¯ L d froma particular set of virtual flows { g ( s, d ) : s ∈ S} and “initialconditions” { LAYER ( v ( s )) : s ∈ S} . Let g ( s, d ) = Z k −→ Z k −→ · · · → Z q k q −→ d where Z := v ( s ) . The corresponding ¯ L d by definition must for each l = 1 , . . . , q satisfy: LAYER ( Z l ) = LAYER ( v ( s )) − l − X e =1 k e . (27)For any { g ( s, d ) : s ∈ S} such that G d ( S ) = F d ( S ) , itfollows from (27) that every corresponding ¯ L d is completelydetermined by the choice of { LAYER ( v ( s )) : s ∈ S} . Givenany directed graph defined by the flows { f ( s, d ) : s ∈ S} there exists a directed acyclic spanning subgraph defined bydisjoint virtual flows { g ( s, d ) : s ∈ S} . Since G d ( S ) = F d ( S ) ,setting LAYER ( v ( s )) = P q − e =1 k e for each s ∈ S suffices todefine a corresponding ¯ L d . Example 21.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) =2 −→ ∞ −→ and F = { f (1 , , f (2 , } . Let g (1 ,
5) =1 −→ −→ ∞ −→ and g (2 ,
5) = 4 ∞ −→ where v (1) = 1 and v (2) = 4 . Then { g (1 , , g (2 , } is a directed acyclicspanning subgraph of F where g (1 , and g (2 , are disjoint.Setting LAYER ( v (1)) = 3 and LAYER ( v (2)) = 0 gives ¯ L = ( { , } , {} , { } , { } ) as depicted in Figure 2(ii). The flow decomposition ( F, ¯ L d ) bifurcates F d ( S ) into I ⊆I by definition, if there exists an l ′ such that I ⊆ ∪ | ¯ L d |− l ≥ l ′ L d,l and { F d ( S ) \ I } ⊆ ∪ l ′ l =0 L d,l . For any S ⊆ S , define theboolean variable E ( S ) := TRUE if ( F, ¯ L d ) bifurcates F d ( S ) into F d ( S ) , and E ( S ) := FALSE if otherwise.
Example 22.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ , F = { f (1 , , f (2 , } and ¯ L = ( { , } , {} , { } , { } ) as depicted in Figure 2(ii). Then ( F, ¯ L ) bifurcates F ( { , } ) into F ( { } ) since F ( { } ) ⊆ L , ∪ L , ∪ L , ∪ L , and F ( { , } ) \ F ( { } ) ⊆ L , . emark 6. In Example 22, the constraint (9) evaluated at S = { } is given by (14) and coincides with the boundarycondition (2) at S = { } given by (10). Example 23.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ , F = { f (1 , , f (2 , } , and ¯ L =( { } , {} , {} , { , } , { } ) as depicted in Figure 2(v). Then ( F, ¯ L ) bifurcates F ( { , } ) into F ( { } ) since F ( { } ) ⊆ L , ∪ L , and F ( { , } ) \ F ( { } ) ⊆ L , ∪ L , ∪ L , ∪ L , . Remark 7.
In Example 23, the constraint (9) evaluated at S = { } is given by (25) and coincides with the boundarycondition (2) at S = { } given by (11). Given ( F, ¯ L d ) , let δ ( S ) := 1 − E ( S ) and let ( F, ¯ L ′ d ) := SHIFT (( F, ¯ L d ) , S ) for some S ⊆ S , if for every i ∈ G d ( S ) : LAYER ′ ( i ) = ( l i ∈ A l ( S ) \ A l ( S ) l + δ ( S ) i ∈ A l ( S ) (28) Example 24.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ and F = { f (1 , , f (2 , } . Let ¯ L = ( { , } , {} , { } , { } ) as depicted in Figure 2(ii) and ¯ L = ( { } , { } , { } , { } ) as depicted in Figure 2(iii). Then ( F, ¯ L ′ ) = SHIFT (( F, ¯ L ) , { } ) . Example 25.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ and F = { f (1 , , f (2 , } . Let ¯ L = ( { } , { } , { } , { } ) as depicted in Figure 2(iii) and ¯ L = ( { } , {} , { } , { , } ) as depicted in Figure 2(iv). Then ( F, ¯ L ′ ) = SHIFT (( F, ¯ L ) , { } ) . Example 26.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ and F = { f (1 , , f (2 , } . Let ¯ L = ( { } , {} , { } , { , } ) as depicted in Figure 2(iv) and ¯ L = ( { } , {} , {} , { , } , { } ) as depicted in Figure 2(v).Then ( F, ¯ L ′ ) = SHIFT (( F, ¯ L ) , { } ) . Example 27.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ and F = { f (1 , , f (2 , } .Let ¯ L = ( { } , {} , { } , { , } ) as depicted in Figure 2(iv).Then ( F, ¯ L ) = SHIFT (( F, ¯ L ) , { } ) since ( F, ¯ L ) bifurcates F ( { , } ) into F ( { } ) . Lemma 4. ( F, ¯ L ′ d ) is complete if ( F, ¯ L d ) is complete.Proof. From the definition of
SHIFT in (28).Given some ( F, ¯ L d ) and ¯ R ∈ R d ( F ) , let V ⊆ S denotethe set of source nodes that satisfy (9) for all S ⊆ V , let U ⊂ S denote the largest subset of S that violates (9), andlet A l ( · ) and ˜ A l ( · ) be defined in (7) and (8). Let ( F, ¯ L ′ d ) := SHIFT (( F, ¯ L d ) , U ) . The corresponding A ′ l ( · ) , ˜ A ′ l ( · ) , V ′ and U ′ are defined with respect to ( F, ¯ L ′ d ) . Lemma 5. V ′ = ( S \ U ) ∪ V .Proof. The proof is by contradiction. Suppose there is some S ⊆ ( S \ U ) ∪ V that violates (9) for ( F, ¯ L ′ d ) . By assumption, R S > | ¯ L ′ d | X l =0 I ( X A ′ l ( S ) ; Y d | X ˜ A ′ l ( S ) ) (29) Case 1: S ∩ U = {} . Then, R U ∪ S > | ¯ L d | X l =0 I ( X A l ( U ) ; Y d | X ˜ A l ( U ) )+ | ¯ L ′ d | X k =0 I ( X A ′ k ( S ) ; Y d | X ˜ A ′ k ( S ) ) (30) ≥ | ¯ L d | X l =0 ( I ( X A l ( U ) ; Y d | X ˜ A l ( U ) )+ I ( X A ′ l ( S ) ; Y d | X ˜ A ′ l ( S ) )) (31) ≥ | ¯ L d | X l =0 ( I ( X A l ( U ) ; Y d | X ˜ A l ( U ) )+ I ( X A l ( S ) \ A l ( U ) ; Y d | X ˜ A l ( S ) \ A l ( U ) )) (32) ≥ | ¯ L d | X l =0 I ( X A l ( U ) ∪ A l ( S ) ; Y d | X ˜ A l ( U ) \ A l ( S ) ) (33) ≥ | ¯ L d | X l =0 I ( X A l ( U ∪ S ) ; Y d | X ˜ A l ( U ∪ S ) ) (34)which contradicts the assumption that U is the largest subsetthat violates (9) for ( F, ¯ L d ) . (34) follows from (29) and thedefinition of U . To justify (31)-(33), fix any l ∈ { , . . . , ¯ L d } .Since A l ( S ) \ A l ( U ) ⊆ A ′ l ( S ) and ˜ A l ( S ) \ A l ( U ) ⊆ ˜ A ′ l ( S ) ,(32) follows from (31). Since { A l ( S ) \ A l ( U ) } ∪ { ˜ A l ( S ) \ A l ( U ) } ⊆ ˜ A l ( U ) , (33) follows from (32). Case 2: S ∩ U = {} . Since { S \ U } ∩ U = {} , the firstcase implies: R S \ U < | ¯ L ′ d | X l =0 I ( X A ′ l ( S \ U ) ; Y d | X ˜ A ′ l ( S \ U ) ) (35)It follows that, R S ∩ U > | ¯ L ′ d | X l =0 I ( X A ′ l ( S ∩ U ) ; Y d | X ˜ A ′ l ( S ∩ U ) \ A ′ l ( S \ U ) ) (36) ≥ | ¯ L d | X l =0 I ( X A l ( S ∩ U ) ; Y d | X ˜ A l ( S ∩ U ) ) (37)which contradicts the assumption that all subsets of V satisfy(9) for ( F, ¯ L d ) . Note that { S ∩ U } ⊆ V since S ⊆ ( S \ U ) ∪ V .To justify (36)-(37), fix any l ∈ { , . . . , ¯ L ′ d } . Then, I ( X A ′ l ( S \ U ) ; Y d | X ˜ A ′ l ( S \ U ) )+ I ( X A ′ l ( S ∩ U ) ; Y d | X ˜ A ′ l ( S ∩ U ) \ A ′ l ( S \ U ) ) (38) ≥ I ( X A ′ l ( S \ U ) ∪ A ′ l ( S ∩ U ) ; Y d | X ˜ A ′ ( S \ U ) \ A ′ l ( S ∩ U ) ) (39) ≥ I ( X A ′ l ( S ) ; Y d | X ˜ A ′ ( S ) ) . (40)ince ˜ A ′ l ( S ∩ U ) \ A ′ l ( S \ U ) ⊆ ˜ A ′ l ( S \ U ) and A ′ l ( S ∩ U ) ⊆ ˜ A ′ l ( S \ U ) , (39) follows from (38). By inspection, (36) followsfrom (29), (35), and (40). Since ˜ A l − ( S ∩ U ) ⊆ ˜ A ′ l ( S ∩ U ) \ A ′ l ( S \ U ) and A l − ( S ∩ U ) = A ′ l ( S ∩ U ) , (37) follows from(36) which finishes the proof of Lemma 5. Lemma 6. U U ′ Proof.
The proof is by contradiction. Suppose U ⊂ U ′ . Then, R U ′ > | ¯ L ′ d | X l =0 I ( X A ′ l ( U ′ ) ; Y d | X ˜ A ′ l ( U ′ ) ) (41) ≥ | ¯ L d | X l =0 I ( X A l ( U ′ ) ; Y d | X ˜ A l ( U ′ ) ) (42)which contradicts the assumption that U is the largest subsetthat violates (9) for ( F, ¯ L d ) . Since ˜ A ′ l ( U ′ ) = ˜ A l ( U ′ ) and A l ( U ) ⊆ A l ( U ′ ) , (42) follows from (41) which finishes theproof of Lemma 6.Lemmas 5 and 6 are difficult to demonstrate in channelswith fewer than three sources. Examples of (29)-(42) areprovided in Appendix C for the three-source multiple-accessrelay channel.For any ¯ R ∈ R d ( F ) , define the sequence of flow decom-positions { ( F, ¯ L d ) k } and the sequence of sets { U k } where ( F, ¯ L d ) k +1 = SHIFT (( F, ¯ L d ) k , U k ) , ( F, ¯ L d ) is complete, and U k is the largest subset of S that violates (9) with respectto ( F, ¯ L d ) k . The sequence { U k } converges, by definition, if U k = U all but finitely often (a.b.fo), for some possibly empty U ⊆ S . In addition, let { V k } be the sequence of sets such thatall subsets of V k satisfy (9) with respect to ( F, ¯ L d ) k . Thesequence { V k } converges , by definition, if V k = S a.b.f.o.Note that { V k } converges iff { U k } converges to {} . Lemma 7. { V k } converges.Proof. Since ( F, ¯ L d ) is complete (Lemma 3), each flow de-composition in the sequence { ( F, ¯ L d ) k } is complete (Lemma4). Then V k ⊂ V k +1 for any k such that U k = U k − , wherethe inclusion is strict (Lemma 2, Lemma 5 and Lemma 6).Therefore { U k } must converge to some U ⊆ S . It remains toshow that U is empty.Suppose U is non-empty. First we show that E ( U ) = TRUE with respect to { ( F, ¯ L d ) k } a.b.f.o. By inspection of ( ?? ), { ( F, ¯ L d ) k } bifurcates F d ( S ) into some G d ( U ) a.b.f.o. If G d ( U ) = F d ( U ) then E ( U ) = TRUE . Suppose G d ( U ) = F d ( U ) . For some s ∈ U and ( F, ¯ L d ) ∈ { ( F, ¯ L d ) k } , considerthe case i ∈ f ( v ( s ) , d ) and i / ∈ G d ( U ) . The property of v ( s ) expressed in (5) implies that LAYER ( v ( s )) − LAYER ( i )
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ , and F = { f (1 , , f (2 , } .Let ¯ L = ( { } , {} , { } , { , } ) as depicted in Figure 3(i) ∞ ∞ ∞ ∞ (i) (ii)Fig. 3: The Shift Operation. f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ , and F = { f (1 , , f (2 , } . (i) ¯ L ′ = ( { } , {} , { } , { , } ) . (ii) ¯ L ′ = ( {} , { } , { } , { } , { } ) . ( F, ¯ L ′ ) = SHIFT (( F, ¯ L ) , { } ) . (and Figure 2(iv)). It follows that v (1) = 1 and g (1 ,
5) =1 −→ . Now let ( F, ¯ L ′ ) = SHIFT (( F, ¯ L ) , { } ) . Then ¯ L ′ = ( {} , { } , { } , { } , { } ) as depicted in Figure 3(ii). Itfollows that v ′ (1) = 1 and g ′ (1 ,
5) = 1 −→ −→ ∞ −→ . Remark 8.
In Example 28 prior to the shift, ∈ f (1 , but / ∈ g (1 , since LAYER (1) − LAYER (2) = 0 < k , = 1 .Hence, G ( { } ) = { , } 6 = F ( { } ) = { , , } . After theshift ∈ g (1 , since LAYER (1) − LAYER (2) = 1 = k , .Hence G ′ ( { } ) = { , , } = F ( { } ) = { , , } . Now assume i ∈ f ( s, v ( s )) and i / ∈ G d ( U ) . The property of v ( s ) expressed in (6) implies that LAYER ( i ) − LAYER ( v ( s )) >k i,v ( s ) . The same arguments from the previous case apply.Therefore { ( F, ¯ L d ) k } bifurcates F d ( S ) into F d ( U ) a.b.f.o. Example 29.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) =2 −→ ∞ −→ , and F = { f (1 , , f (2 , } . Let ¯ L =( { , } , {} , { } , { } ) as depicted in Figure 2(ii). It followsthat v (2) = 4 and g (2 ,
5) = 4 ∞ −→ . Now let ( F, ¯ L ′ ) = SHIFT (( F, ¯ L ) , { } ) . Then ¯ L ′ = ( { } , { } , { } , { } ) as de-picted in Figure 2(ii). It follows that v ′ (2) = 2 and g ′ (2 ,
5) =2 −→ ∞ −→ . Remark 9.
In Example 29 prior to the shift, ∈ f (2 , but / ∈ g (2 , since LAYER (2) − LAYER (4) = 2 > k , = 1 .Hence G ( { } ) = { } 6 = F ( { } ) = { , } . After the shift, ∈ g ′ (2 , since LAYER (2) − LAYER (4) = 1 = k , . Hence G ′ ( { } ) = { , } = F ( { } ) = { , } . Now we show that U must be empty. Suppose U isnot empty. Since each flow decomposition in { ( F, ¯ L d ) k } isomplete and { ( F, ¯ L d ) k } bifurcates F d ( S ) into F d ( U ) a.b.f.o.,it follows that { ( F, ¯ L d ) k } is complete on U and that (2) and (9)agree on S = U a.b.f.o, which is a contradiction.This finishesthe proof of Lemma 7. Example 30.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ , F = { f (1 , , f (2 , } and ¯ L = ( { , } , {} , { } , { } ) as depicted in Figure 2(ii). Remark 10.
In Example 30, ( F, ¯ L ) is complete on { , } andbifurcates F ( { , } ) into F ( { } ) . Hence, ( F, ¯ L ) is completeon { } . The constraint (9) evaluated at S = { } is given by(14) and coincides with the boundary condition (2) evaluatedat S = { } given by (10). Example 31.
Let f (1 ,
5) = 1 −→ −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ , F = { f (1 , , f (2 , } and ¯ L = ( { } , {} , {} , { , } , { } ) as depicted in Figure 2(v). Remark 11.
In Example 31, ( F, ¯ L ) is complete on { , } andbifurcates F ( { , } ) into F ( { } ) . Hence, ( F, ¯ L ) is completeon { } . The constraint (9) evaluated at S = { } is given by(25) and coincides with the boundary condition (2) evaluatedat S = { } , given by (11). Lemma 7 finishes the proof of Theorem 1. The flow decom-positions in Figure 2(iii) and Figure 3(ii) are “equivalent” inthe sense that they share the same virtual flows, the samenon-empty layers, and the same achievable region of ratevectors. Another example of equivalent flow decompositionsis discussed in Remark 4. Any flow decomposition in Figure2 can be “shifted” into a flow decomposition equivalentwith any other in Figure 2. The concept of equivalent flowdecompositions will play a fundamental role in characterizingthe decode-forward region for multi-source multi-relay all-castchannels. VI. C
ONCLUSION
It is conventional in network information theory to expressrate-vector regions in terms of mutual informations. Thisapproach is inadequate for general multi-node channels. Theinterdependencies between the messages decoded by eachnode and the messages encoded by all the other nodesmake a direct characterization of the decode-forward regionintractable.We provide a way of circumventing this problem by describ-ing instead, any encoding and decoding scheme at a given nodethat recovers a particular rate-vector in the decode-forwardregion. Flow decompositions are mathematical abstractionsof these schemes. Theorem 1 implies that every rate-vectorin the decode-forward region corresponds to a set of causalflow decompositions, each assigned to a unique node in thechannel. The companion paper in Part II characterizes thedecode-forward region indirectly, by identifying necessary andsufficient conditions that determine whether an arbitrary setof (possibly non-causal) flow decompositions maps to anequivalent set of causal flow decompositions. A
PPENDIX AT HE C OOPERATIVE M ULTIPLE -A CCESS C HANNEL
The cooperative multiple-access channel is defined by theflows f (1 ,
3) = 1 −→ ∞ −→ and f (2 ,
3) = 2 −→ ∞ −→ , both depicted in Figure 4(i), where S = { , } , Z = { , } and D = { } . In block b , node 1 en-codes ¯ w ( b ) {h , m ( b ) i , h , m ( b − i} and node 2 encodes ¯ w ( b ) = {h , m ( b ) i , h , m ( b − i} .Let F = { f (1 , , f (2 , } . By definition, R ( F ) is set ofrate vectors ¯ R = ( R , R ) that satisfies: R < I ( X X ; Y ) (43) R < I ( X X ; Y ) (44) R + R < I ( X X ; Y ) (45)Three different decoding schemes collectively achieve allthe rate vectors in R ( F ) . A. The first decoding scheme
Set ¯ L = ( { } , { } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (2) = − ≤ k , which satisfies (5). Simi-larly, v (2) = 2 since LAYER (2) − LAYER (1) = 1 = k , whichsatisfies (5). Then g (1 ,
3) = 1 ∞ −→ and g (2 ,
5) = 4 ∞ −→ asdepicted in Figure 4(ii).In block b , node 3 decodes ¯ m ( b ) = {h , m ( b ) i , h , m ( b − i} by finding the unique pair m ∈ { , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfy the following typi-cality checks: (¯ x ( m , m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y )(¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , Y ) (46)In each block, node 2 sends its own source message and asource message from node 1. The message m ( b − , thoughencoded by node 2 in block b − has already been decodedby node 3 in block b , so ¯ x in (46) effectively depends on m alone. The probability of error goes to zero if ¯ R satisfies thefollowing conditions: R < I ( X ; Y ) (47) R < I ( X ; Y ) + I ( X ; Y | X )= I ( X X ; Y ) (48) R + R < I ( X ; Y ) + I ( X ; Y | X )= I ( X X ; Y ) (49)Each of the three inequalities above addresses an error event.The probability that m is decoded correctly and m is not,goes to zero if (47) is satisfied. Similarly, the probability that m is decoded correctly and m is not, goes to zero if (48) issatisfied. Finally, the probability that both m and m are notdecoded correctly goes to zero if (49) is satisfied. B. The second decoding scheme
Set ¯ L = ( { , } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (2) = 0 ≤ k , which satisfies (5). Simi-larly, v (2) = 2 since LAYER (2) − LAYER (1) = 0 ≤ k , which ∞ ∞ ∞ ∞
21 3 ∞ ∞ ∞∞ (i) (ii) (iii) (iv)Fig. 4: The cooperative multiple-access channel (i) f (1 ,
3) = 1 −→ ∞ −→ and f (2 ,
3) = 2 −→ ∞ −→ (ii) ¯ L = ( { } , { } ) (iii) ¯ L = ( { , } ) (iv) ¯ L = ( { } , { } ) . Original flows are shown in dotted lines and virtual flows in solid lines in (ii)-(iv)satisfies (5). Then g (1 ,
3) = 1 ∞ −→ and g (2 ,
3) = 2 ∞ −→ asdepicted in Figure 4(iii).In block b , node 3 decodes ¯ m ( b ) = {h , m ( b ) i , h , m ( b ) i} by finding the unique pair m ∈ { , . . . , nR } and m ∈ { , . . . , nR } that jointlysatisfies the following typicality check: (¯ x ( m ) , ¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , X , Y ) The probability of error goes to zero if ¯ R satisfies the followingconditions: R < I ( X ; Y | X ) (50) R < I ( X ; Y | X ) (51) R + R < I ( X X ; Y ) (52) C. The third decoding scheme
Set ¯ L = ( { } , { } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (2) = 1 ≤ k , which satisfies (5).Similarly, v (2) = 2 since LAYER (2) − LAYER (1) = − < k , which satisfies (5). Then g (1 ,
3) = 1 −→ ∞ −→ and g (2 ,
3) = 2 ∞ −→ as depicted in Figure 4(iv).In block b , node 3 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b ) i} by finding the unique pair m ∈{ , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfiesthe following typicality checks: (¯ x ( m , m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y )(¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , Y ) (53)The message m ( b − forwarded by node 1 in block b − has already been decoded by node 3 in block b , so ¯ x in (53)effectively depends on m alone. The probability of error goesto zero if ¯ R satisfies the following conditions: R < I ( X ; Y ) + I ( X ; Y | X )= I ( X X ; Y ) (54) R < I ( X ; Y ) (55) R + R < I ( X ; Y ) + I ( X ; Y | X ) (56) = I ( X X ; Y ) D. The achievability of R ( F ) To show that R ( F ) is achievable, we show that any ratevector in the region defined by (43)-(45) is in the regiondefined by (47)-(49) or (50)-(52) or (54)-(56). Suppose ¯ R isnot in (47)-(49). Since (48) and (49) define the boundaries of R ( F ) , ¯ R must violate (47). Then (49) implies that ¯ R satisfies(51). If ¯ R satisfies (50) the proof is finished since (52) is aboundary of R ( F ) . If ¯ R violates (50) then (52) implies that ¯ R satisfies (55). Now the proof is finished since (54) and (56)are boundaries of R ( F ) .A PPENDIX BT HE T WO -S OURCE M ULTIPLE -A CCESS R ELAY C HANNEL
The multiple-access relay channel is defined by the flows f (1 ,
3) = 1 −→ ∞ −→ and f (2 ,
3) = 2 −→ ∞ −→ , bothdepicted in Figure 5(i), where S = { , } , Z = { } and D = { } . In block b , node 1 encodes ¯ w ( b ) {h , m ( b ) i} , node2 encodes ¯ w ( b ) = {h , m ( b ) i} , and node 3 encodes ¯ w ( b ) = {h , m ( b − i , h , m ( b − i} .Let F = { f (1 , , f (2 , } . By definition, R ( F ) is set ofrate vectors ¯ R = ( R , R ) that satisfies: R < I ( X X ; Y | X ) (57) R < I ( X X ; Y | X ) (58) R + R < I ( X X X ; Y ) (59)Three different decoding schemes collectively achieve allthe rate vectors in R ( F ) . A. The first decoding scheme
Set ¯ L = ( { , } , { } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (3) = 1 = k , which satisfies (5).Similarly, v (2) = 2 since LAYER (2) − LAYER (3) = 0 ≤ k , which satisfies (5). Then g (1 ,
3) = 1 −→ ∞ −→ and g (2 ,
4) = 2 −→ ∞ −→ as depicted in Figure 5(ii).In block b , node 3 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b ) i} by finding the unique pair m ∈
21 3 ∞ ∞ ∞ ∞ ∞ ∞ ∞∞ (i) (ii) (iii) (iv)Fig. 5: The two-source multiple-access relay channel (i) f (1 ,
4) = 1 −→ ∞ −→ and f (2 ,
4) = 2 −→ ∞ −→ (ii) ¯ L = ( { , } , { } ) (iii) ¯ L = ( { } , { , } ) (iv) ¯ L = ( { , } , { } ) . Original flows are depicted in dotted lines and virtualflows in solid lines in (ii)-(iv). { , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfiesthe following typicality checks: (¯ x ( m ) , ¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , X , Y )(¯ x ( m ) , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , Y ) (60)In each block, node 3 sends source messages from node 1 andnode 2. The message m ( b − , though encoded by node 3in block b has already been decoded by node 4 in block b , so ¯ x in (60) effectively depends on m alone. The probabilityof error goes to zero if ¯ R satisfies the following conditions: R < I ( X ; Y | X X ) + I ( X ; Y | X )= I ( X X ; Y | X ) (61) R < I ( X ; Y | X ) (62) R + R < I ( X X ; Y ) + I ( X ; Y | X X )= I ( X X X ; Y ) (63)Each of the three inequalities above addresses an error event.The probability that m is decoded correctly and m is not,goes to zero if (61) is satisfied. Similarly, the probability that m is decoded correctly and m is not, goes to zero if (62) issatisfied. Finally, the probability that both m and m are notdecoded correctly goes to zero if (63) is satisfied. B. The second decoding scheme
Set ¯ L = ( { } , { , } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (3) = 1 ≤ k , which satisfies (5).Similarly, v (2) = 2 since LAYER (2) − LAYER (3) = 1 ≤ k , which satisfies (5). Then g (1 ,
4) = 1 −→ ∞ −→ and g (2 ,
4) = 2 −→ ∞ −→ as depicted in Figure 5(iii).In block b , node 3 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b − i} by finding the unique pair m ∈{ , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfiesthe following typicality check: (¯ x ( m , m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y )(¯ x ( m ) , ¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X X X Y ) The probability of error goes to zero if ¯ R satisfies the followingconditions: R < I ( X ; Y ) + I ( X ; Y | X X )= I ( X X ; Y | X ) (64) R < I ( X ; Y ) + I ( X ; Y | X X )= I ( X X ; Y | X ) (65) R + R < I ( X ; Y ) + I ( X ; Y | X X )= I ( X X X ; Y ) (66) C. The third decoding scheme
Set ¯ L = ( { , } , { } ) . It follows that v (1) = 1 since LAYER (1) − LAYER (2) = − ≤ k , which satisfies (5).Similarly, v (2) = 2 since LAYER (2) − LAYER (3) = 1 = k , which satisfies (5). Then g (1 ,
3) = 1 ∞ −→ and g (2 ,
4) = 2 −→ ∞ −→ as depicted in Figure 5(iv).In block b , node 5 decodes ¯ m ( b ) = {h , m ( b ) i , h , m ( b − i} by finding the unique pair m ∈ { , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfies the following typ-icality checks: (¯ x ( m ) , ¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , X , Y ) (67) (¯ x ( m ) , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , Y ) The message m ( b − forwarded by node 3 in block b hasalready been decoded by node 4 in block b , so ¯ x in (67)effectively depends on m alone. The probability of error goesto zero if ¯ R satisfies the following conditions: R < I ( X ; Y | X ) (68) R < I ( X ; Y | X ) + I ( X ; Y | X X )= I ( X X ; Y | X ) (69) R + R < I ( X X ; Y ) + I ( X ; Y | X X )= I ( X X X ; Y ) (70) . The achievability of R ( F ) To show that R ( F ) is achievable, we show that any ratevector in the region defined by (57)-(59) is in the regiondefined by (61)-(63) or (64)-(66) or (68)-(56). Suppose ¯ R isnot in (61)-(63). Since (61) and (63) define the boundaries of R ( F ) , ¯ R must violate (62). Then (63) implies that ¯ R satisfies(64). If ¯ R satisfies (65) the proof is finished since (66) is aboundary of R ( F ) . If ¯ R violates (65) then (66) implies that ¯ R satisfies (68). Now the proof is finished since (69) and (70)are boundaries of R ( F ) .A PPENDIX CT HE T HREE -S OURCE M ULTIPLE -A CCESS R ELAY C HANNEL
The three-source multiple-access relay channel is defined bythe flows f (1 ,
5) = 1 −→ ∞ −→ , f (2 ,
3) = 2 −→ ∞ −→ and f (3 ,
5) = 3 −→ ∞ −→ depicted in Figure 6(i), where S = { , , } , Z = { } and D = { } . In block b , node 1 encodes ¯ w ( b ) = {h , m ( b ) i} , node 2 encodes ¯ w ( b ) = {h , m ( b ) i} ,node 3 encodes ¯ w ( b ) = {h , m ( b ) i} and node 4 encodes ¯ w ( b ) = {h , m ( b − i , h , m ( b − i , h , m ( b − i} .Let F = { f (1 , , f (2 , , f (3 , } . By definition, R ( F ) isset of rate vectors ¯ R = ( R , R , R ) that satisfies: R < I ( X X ; Y | X X ) (71) R < I ( X X ; Y | X X ) (72) R < I ( X X ; Y | X X ) (73) R + R < I ( X X X ; Y | X ) (74) R + R < I ( X X X ; Y | X ) (75) R + R < I ( X X X ; Y | X ) (76) R + R + R < I ( X X X X ; Y ) (77)The proofs in Appendices A and B rely on the two-dimensional structure of the rate regions in channels with twosources, and do not extend to higher-dimensional regions suchas R ( F ) . In general, the number of constraints needed toexpress the decode-forward region increases exponentially inthe number of sources, which is evident when comparing (71)-(77) with (57)-(59). The number of decode-forward schemesrequired in the proof of achievability is also subject to expo-nential scaling.Instead of explicitly characterizing these schemes and piec-ing their rate regions together as before, we will focus onfour particular decode-forward schemes (out of the manyother possibilities) in order to illustrate the arguments inLemmas 5 and 6. These lemmas are the basis for the proofof Theorem 1. Using the notation of layered partitions, thedecode-forward schemes of interest are: ¯ L = ( { , } , { , } ) , ¯ L = ( { } , { , , } , ¯ L = ( { , , } , { } ) , and ¯ L =( { } , { , } , { } ) , depicted in Figures 6(ii), (iii), (iv), and (v)respectively. A. The first decoding scheme
Set ¯ L = ( { , } , { , } ) . It follows that g (1 ,
5) = 1 −→ ∞ −→ , g (2 ,
5) = 2 ∞ −→ , and g (3 ,
5) = 3 −→ ∞ −→ asdepicted in Figure 6(ii). In block b , node 5 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b ) i , h , m ( b − i} by finding theunique triple m ∈ { , . . . , nR } , m ∈ { , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfies the followingtypicality checks: (¯ x ( m , m ) , ¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , X , Y ) (78) (¯ x ( m ) , ¯ x ( m ) , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , X , Y ) The message m ( b − forwarded by node 4 in block b hasalready been decoded by node 5, so ¯ x in (78) effectivelydepends on m and m alone. The probability of error goesto zero if ¯ R satisfies the following conditions: R < I ( X ; Y | X X X ) + I ( X ; Y | X ) (79) R < I ( X ; Y | X ) R < I ( X ; Y | X X X ) + I ( X ; Y | X ) R + R < I ( X ; Y | X X X ) + I ( X X ; Y ) (80) R + R < I ( X X ; Y | X X ) + I ( X ; Y | X )= I ( X X X ; Y | X ) R + R < I ( X ; Y | X X X ) + I ( X X ; Y ) R + R + R < I ( X X ; Y | X X ) + I ( X X ; Y )= I ( X X X X ; Y ) B. The second decoding scheme
Set ¯ L = ( { } , { , , } ) . It follows that g (1 ,
5) = 1 −→ ∞ −→ , g (2 ,
5) = 2 −→ ∞ −→ , and g (3 ,
5) = 3 −→ ∞ −→ asdepicted in Figure 6(iii). In block b , node 5 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b − i , h , m ( b − i} by findingthe unique triple m ∈ { , . . . , nR } , m ∈ { , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfies the followingtypicality checks: (¯ x ( m , m , m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y )(¯ x ( m ) , ¯ x ( m ) , ¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , X , Y ) The probability of error goes to zero if ¯ R satisfies the followingconditions: R < I ( X ; Y | X X X ) + I ( X ; Y ) (81) R < I ( X ; Y | X X X ) + I ( X ; Y ) R < I ( X ; Y | X X X ) + I ( X ; Y ) R + R < I ( X X ; Y | X X X ) + I ( X ; Y ) (82) R + R < I ( X X ; Y | X X ) + I ( X ; Y ) R + R < I ( X X ; Y | X X ) + I ( X ; Y ) R + R + R < I ( X X X ; Y | X ) + I ( X ; Y )= I ( X X X X ; Y ) C. The third decoding scheme
Set ¯ L = ( { , , } , { } ) . It follows that g (1 ,
5) = 1 −→ ∞ −→ , g (2 ,
5) = 2 ∞ −→ , and g (3 ,
5) = 3 ∞ −→ as depicted in Figure 6(iv). In block b , node 5 decodes
21 4 3 ∞ ∞ ∞ ∞ ∞ ∞
01 1 1 ∞ ∞ ∞ ∞∞ ∞ ∞∞ ∞ (i) (ii) (iii) (iv) (v)Fig. 6: The three-source multiple-access relay channel (i) f (1 ,
5) = 1 −→ ∞ −→ , f (2 ,
5) = 2 −→ ∞ −→ and f (3 ,
5) = 3 −→ ∞ −→ (ii) ¯ L = ( { , } , { , } ) (iii) ¯ L = ( { } , { , , } ) (iv) ¯ L = ( { , , } , { } ) , ¯ L = ( { } , { , } , { } ) .Original flows are depicted in dotted lines and virtual flows in solid lines in (ii)-(iv). ¯ m ( b ) = {h , m ( b − i , h , m ( b ) i , h , m ( b ) i} by findingthe unique triple m ∈ { , . . . , nR } , m ∈ { , . . . , nR } and m ∈ { , . . . , nR } that jointly satisfies the followingtypicality checks: (¯ x ( m ) , ¯ x ( m ) , ¯ x ( m )) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , X , X , Y ) (83) (¯ x ( m ) , ¯ X ( b − , ¯ X ( b − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , X , Y ) The messages m ( b − and m ( b − forwarded by node 4in block b have already been decoded by node 5 in block b , so ¯ x in (83) effectively depends on m alone. The probabilityof error goes to zero if ¯ R satisfies the following conditions: R < I ( X ; Y | X X X ) + I ( X ; Y | X X )= I ( X X ; Y | X X ) R < I ( X ; Y | X X ) (84) R < I ( X ; Y | X X ) R + R < I ( X ; Y | X X X ) + I ( X X ; Y | X )= I ( X X X ; Y | X ) R + R < I ( X ; Y | X X X ) + I ( X X ; Y | X )= I ( X X X ; Y | X ) R + R < I ( X X ; Y | X ) (85) R + R + R < I ( X ; Y | X X X ) + I ( X X X ; Y )= I ( X X X X ; Y ) (86) D. The fourth decoding scheme
In block b , node 5 decodes ¯ m ( b ) = {h , m ( b − i , h , m ( b ) i , h , m ( b − i} by finding the unique triple m ∈ { , . . . , nR } , m ∈ { , . . . , nR } and m ∈{ , . . . , nR } that jointly satisfies the following typicality checks: (¯ x ( m ) , ¯ Y ( b )) ∈ T ( n ) ǫ ( X , Y )(¯ x ( m ) , ¯ x ( m ) , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , Y ) (87) (¯ x ( m ) , ¯ X ( b − , ¯ X ( − , ¯ X ( b − , ¯ Y ( b − ∈ T ( n ) ǫ ( X , X , X , X , Y ) The messages m ( b − and m ( b − forwarded by node 4in block b − have already been decoded by node 5 in block b ,so ¯ x in (87) effectively depends on m alone. The probabilityof error goes to zero if ¯ R satisfies the following conditions: R < I ( X ; Y | X X X ) + I ( X ; Y | X X )= I ( X X ; Y | X X ) R < I ( X ; Y ) R < I ( X ; Y | X X ) R + R < I ( X ; Y | X X X ) + I ( X ; Y | X X )+ I ( X ; Y )= I ( X X ; Y | X X ) + I ( X ; Y ) (88) R + R < I ( X ; Y | X X X ) + I ( X X ; Y | X )= I ( X X X ; Y | X ) R + R < I ( X ; Y | X X ) + I ( X ; Y ) R + R + R < I ( X ; Y | X X X ) + I ( X X ; Y | X )+ I ( X ; Y )= I ( X X X X ; Y ) Scenarios one and two described below will demonstrate thearguments in Lemma 5 and 6 respectively. . Scenario One
Let ¯ L = ( { , , } , { } ) as depicted in Figure 6(iv). Pickany ¯ R ∈ R ( F ) and let V = { } and U = { , } . Recallthat V is the largest subset of S such that all subsets of V satisfy (9) and U is the largest subset of S that violates (9). Bydefinition of U and V , ¯ R violates (84) and (85). Set ( F, ¯ L ′ ) = SHIFT (( F, ¯ L ) , U ) . Then ¯ L ′ = ( { } , { , , } ) as depicted inFigure 6(iii). We will check that V ′ = { , } . Suppose, as inthe hypothesis of Lemma 5, that there is some S ⊆ V ′ = { , } that violates (9). There are two cases to consider. Case 1: S ∩ U = {} . Since U = { , } and S ⊆ V ′ = { , } , this condition implies S = { } . From (81) and thedefinition of S , R > I ( X ; Y | X X X ) + I ( X ; Y ) . From(85) and the definition of U , R + R > I ( X X ; Y | X ) .Therefore, R + R + R > I ( X ; Y | X X X ) + I ( X ; Y )+ I ( X X ; Y | X ) (89) = I ( X X X X ; Y ) (90)which violates (86) and contradicts the assumption that U = { , } is the largest subset of S that violates (9) for ( F, ¯ L ) . Inthis example U = { , , } . Note that (89) and (90) correspondwith (31) and (34) respectively. Case 2: S ∩ U = {} . Suppose S = { , } . From (82) andthe definition of S , R + R > I ( X X ; Y | X X ) + I ( X ; Y ) . (91)Note that (91) corresponds with (29). From the chain rule: I ( X ; Y | X X X ) + I ( X ; Y ) + I ( X ; Y | X X ) ≥ I ( X X ; Y | X X ) + I ( X ; Y ) . (92)Note that (92) corresponds with (38)-(40). From Case 1: R < I ( X ; Y | X X X ) + I ( X ; Y ) (93)which corresponds with (35) since S \ U = { } . It followsfrom (91)-(93) that R > I ( X ; Y | X X ) (94)which contradicts the implication that ¯ R satisfies (84) bydefinition of V = { } . Note that (94) corresponds with (36)-(37) since S ∩ U = { } . F. Scenario Two
Let ¯ L = ( { , } , { , } ) as depicted in Figure 6(ii). Pickany ¯ R ∈ R ( F ) and let V = {} and U = { } . Set ( F, ¯ L ′ ) = SHIFT (( F, ¯ L ) , U ) . Then ¯ L ′ = ( { } , { , } , { } ) as depictedin Figure 6(v). Let U ′ be the largest subset of S that violates(9) for ( F, ¯ L ′ ) . We will show that U ′ = { , } is impossible.Note that U ⊂ U ′ . From (88) and the definition of U ′ , R + R > I ( X X ; Y | X X ) + I ( X ; Y ) (95) > I ( X ; Y | X X X ) + I ( X X ; Y ) , (96)which violates (80) and contradicts the assumption that U = { } is the largest subset of S that violates (9) for ( F, ¯ L ) . In this example U = { , } . Note that (95) and (96) correspondwith (41) and (42) respectively. This argument could bereplayed for any U ′ ⊆ S such that U ⊂ U ′ .R EFERENCES[1] T. Cover and A. El-Gamal, “Capacity theorems for the relay channel,”
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