Single-Source/Sink Network Error Correction Is as Hard as Multiple-Unicast

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1 Snge-Source/Snk Network Error Correcton Is as Hard as Mutpe-Uncast Wentao Huang and Tracey Ho Department of Eectrca Engneerng Caforna Insttute of Technoogy Pasadena, CA Mchae Langberg Department of Eectrca Engneerng Unversty at Buffao, SUNY Buffao, NY Joerg Kewer Department of ECE New Jersey Insttute of Technoogy Newark, NJ Abstract We study the probem of communcatng over a snge-source snge-termna network n the presence of an adversary that may jam a snge nk of the network. If any one of the edges can be jammed, the capacty of such networks s we understood and foows drecty from the connecton between the mnmum cut and maxmum fow n snge-source sngetermna networks. In ths work we consder networks n whch some edges cannot be jammed, and show that determnng the network communcaton capacty s at east as hard as sovng the mutpe-uncast network codng probem for the error-free case. The atter probem s a ong standng open probem. I. INTRODUCTION The probem of network error correcton concerns reabe transmsson of nformaton n a network wth pont-to-pont noseess channes, n the presence of an adversary. The adversary contros a set A of channes n the network and may corrupt the nformaton transmtted on these channes n an arbtrary way. A network error correcton code, frst ntroduced by Ca and Yeung [1], [2], s a network code that can correct adversara errors njected nto the network from a set of channes A, for a A A, where A s a prescrbed coecton of subsets of channes that characterzes the strength of the adversary. For snge-source mutcast, under the smpfyng assumpton that a channes have unt capacty and A s the coecton of a subsets contanng Z channes, [1], [2] show that the cut-set bound s tght and characterzes the network error correcton capacty, whch can be acheved by a near code. Under smar settngs a varety of works, e.g., [3], [4], [5], [6], [7], [8] have proposed dfferent effcent capactyachevng codes and strateges. However, under sghty more genera settngs such that channe capactes are non-unform or A has a more genera structure, much ess s known about the network error correcton capacty and achevabe strateges. Km et a. [9] study a mode n whch channe capactes are arbtrary and show that capacty upper bounds based on cut-set approaches are generay not tght. [9] aso constructs exampes where near codes are nsuffcent to acheve capacty. Kosut et a. [10] study a mode n whch the adversary contros network nodes nstead of channes, whch s a speca case of network error correcton for non-unform A n the sense that A may ncude subsets of dfferent szes. In ths case [10] constructs an exampe that near codes are nadequate to acheve capacty. Achevabe strateges under ths node adversary mode are aso studed n [11], [12], [13], whereas determnng the capacty regon remans an open probem. As opposed to the we studed and we understood settng of [1], [2], the subtety of fndng and achevng the network error correcton capacty n the more genera settngs above motvates us to examne the fundamenta compexty of the genera network error correcton probem. In ths paper, we show that sovng the snge-uncast network error correcton probem wth genera A s as hard as sovng the mutpe-uncast network codng probem wth no error). Specfcay, we convert any unt rate k-uncast network codng probem nto a correspondng network error correcton probem wth a snge source, a snge snk, and a snge adversara channe chosen from a subset of channes, such that the unt rate k-uncast s feasbe wth zero error f and ony f the zero-error network error correcton capacty s k. Under the vanshng error mode, we show a smar but sghty weaker resut. Specfcay, n ths case f the unt rate k-uncast s feasbe, then a network error correcton rate of k s feasbe. Conversey, f a network error correcton rate of k s feasbe, then the unt rate k-uncast s asymptotcay feasbe. Our resuts add to the portfoos of probems that are connected to mutpe-uncast network codng, whch s a ong standng open probem not presenty known to be n P, NP or undecdabe [14], [15], [16], [17]. Prevousy, equvaence resuts have been estabshed, e.g., between mutpe-uncast network codng and mutpe-mutcast network codng [18], [19], ndex codng [20], [21], secure network codng [22], [23] and two-uncast network codng [24]. The remander of the paper s structured as foows. In Secton II, we present the modes and defntons of mutpeuncast network codng and snge-source snge-snk network error correcton. In Secton III and IV, we prove the reducton from mutpe-uncast to network error correcton for the zero error mode and vanshng error mode, respectvey. Fnay, we concude the paper n Secton V. II. MODELS A. Mutpe-uncast Network Codng We mode the network to be a drected graph G = V, E), where the set of vertces V represents network nodes and the

2 set of edges E represents network channes. Each edge e E has a capacty c e, whch s the maxmum number of bts 1 that can be transmtted on e n one transmsson. An nstance I = G, S, T, B) of the mutpe-uncast network codng probem, ncudes a network G, a set of source nodes S V, a set of termna nodes T V and an S by T requrement matrx B. The, j)-th entry of B equas 1 f termna j requres the nformaton from source and equas 0 otherwse. We assume that B s a permutaton matrx and so each source s pared wth a snge termna. Let st) be the source that s requred by termna t. Denote [n] {1,.., n }, then each source s S s assocated wth a message, whch s a rate R s random varabe M s unformy dstrbuted over [2 nrs ]. The messages for dfferent sources are ndependent. A network code of ength n s defned as a set of encodng functons φ e for every e E and a set of decodng functons φ t for each t T. For each e = u, v), the encodng functon φ e s a functon takng as nput the random varabes assocated wth ncomng edges of node u and the random varabe M u f u S, and maps to vaues n [2 nce ]. For each t T, the decodng functon φ t maps a random varabes assocated wth the ncomng edges of t, to a message ˆM st) wth vaues n [2 nr st) ]. A network code {φ e, φ t } e E,t T s sad to satsfy a termna t under transmsson m s, s S) f ˆMst) = m st) when M s, s S) = m s, s S). A network code s sad to satsfy the mutpe-uncast network codng probem I wth error probabty ɛ f the probabty that a t T are smutaneousy satsfed s at east 1 ɛ. The probabty s taken over the jont dstrbuton on random varabes M s, s S). Namey, the network code satsfes I wth error probabty ɛ f Pr M s,s S) { } t s satsfed under M s, s S) 1 ɛ t T For an nstance I of the mutpe-uncast network codng probem, rate R s sad to be feasbe f R s = R, s S, and for any ɛ > 0, there exsts a network code wth suffcenty arge ength that satsfes I wth error probabty at most ɛ. Rate R s sad to be feasbe wth zero error f R s = R, s S and there exsts a network code that satsfes I wth zero error probabty. Rate R s sad to be asymptotcay feasbe f for any δ > 0, rate 1 δ)r s feasbe. The capacty of I refers to the supremum over a rates R that are asymptotcay feasbe and the zero-error capacty of I refers to the supremum over a rates R that are feasbe wth zero error. The gven mode assumes a sources transmt nformaton at equa rate. There s no oss of generaty n ths assumpton as a varyng rate source s can be modeed by severa equa rate sources coocated at s. 1 For convenence we assume that the network channes transmt bnary symbos. Our resuts can be naturay extended to the genera q-ary case. 1) B. Snge-Source Snge-Snk Network Error Correcton An nstance I c = G, s, t, A) of the snge-source sngetermna network error correcton probem ncudes a network G, a source node s V, a termna node t V and a coecton of subsets of channes A 2 E susceptbe to errors. In ths probem the channes are not aways reabe and an error s sad to occur n a channe f the output of the channe s dfferent from the nput. More precsey, the output of a channe e s the nput sgna superposed by an error sgna r e, and we say there s an error on channe e f r e 0. For a subset A A of channes, an A-error s sad to occur f an error occurs n every channe n A. For an nstance I c of the sngesource snge-termna network error correcton probem, a network code {φ e, φ t, } e E,t T s sad to satsfy a termna t under transmsson m s f ˆMs = m s when M s = m s, gven the occurrence of any error pattern r = r e, e E) that resuts n an A-error, for a A A. A network code s sad to satsfy probem I c wth error probabty ɛ f the probabty that t s satsfed s at east 1 ɛ. The probabty s taken over the dstrbuton on the source message M s. For an nstance I c of the snge-source snge-termna network error correcton probem, rate R s sad to be feasbe f R s = R and for any ɛ > 0, there exsts a network code wth suffcenty arge ength that satsfes I c wth error probabty at most ɛ. Rate R s sad to be feasbe wth zero error f R s = R and there exsts a network code that satsfes I c wth zero error probabty. Rate R s sad to be asymptotcay feasbe f for any δ > 0, rate 1 δ)r s feasbe. The capacty of I c refers to the supremum over a rates R that are asymptotcay feasbe and the zero-error capacty of I c refers to the supremum over a rates R that are feasbe wth zero error. Throughout the paper we denote by R A the set of a possbe error patterns r that resut n A-errors, where A A. III. REDUCTION FROM MULTIPLE-UNICAST TO NETWORK ERROR CORRECTION: ZERO ERROR CASE In ths secton we reduce the mutpe-uncast network codng probem wth no error) to the snge-source sngetermna network error correcton probem wth at most a snge adversara channe. We start wth the zero-error case. Theorem 1. Gven any mutpe-uncast network codng probem I wth source-destnaton pars {s, t ), = 1,..., k}, a correspondng snge-source snge-snk network error correcton probem I c = G, s, t, A) n whch A ncudes sets wth at most one edge can be constructed as specfed n Fgure 1, such that the zero-error capacty of I c s k f and ony f unt rate s feasbe wth zero error n I. Proof: The zero-error capacty of I c s upper bounded by k, because t s the mn-cut from s to t. We show that the feasbty of a zero-error rate k n I c mpes the feasbty of unt zero-error rate n I. Suppose a zero-error rate k s acheved n I c by a network code wth ength n, and denote the source message by M, then M s unformy dstrbuted over [2 nk ]. For any edge e E,

3 Fg. 1: In the snge-source snge-termna network error correcton probem I c, the source s wants to communcate wth the termna t. N s a genera network wth pont-topont noseess channes. A edges outsde N.e., edges for whch at east one of ts end-pont does not beong to N ) have unt capacty. There s at most one error n ths network, and ths error can occur at any edge except {a, b, 1 k}. Namey, A ncudes a sngeton sets of a snge edge n the network except {a } and {b }, = 1,..., k. Note that there are k branches n tota but ony the frst and the k-th branches are drawn expcty. The mutpe-uncast network codng probem I s defned on the network N, where the k source-destnaton pars are s, t ), = 1,..., k, and a channes are error-free. we denote by em, r) : [2 nk ] R A [2 n ] the sgna receved on edge e when the source message equas m and the error pattern r occurs n the network. When the context s cear, we may denote em, r) smpy by e. Let bm, r) = b 1 m, r),..., b k m, r)), then because the edges b 1,..., b k form a cut-set from s to t, bm, r) must be njectve wth respect to m due to the zero error decodabty constrant. Formay, for two dfferent messages m 1 m 2, t foows from the zero error decodabty constrant that bm 1, r 1 ) bm 2, r 2 ), r 1, r 2 R A. Note that the codoman of b s [2 n ] k, whch has the same sze as the set of messages [2 nk ]. Therefore denote by bm) bm, 0), then bm) s a bjectve functon and bm, r) = bm), r R A. Smary am, r) = a 1 m, r),..., a k m, r)) s aso a bjectve functon of the message, regardess of the error patterns. For any e E, denote em) em, 0). For = 1,..., k, we cam that for any two messages m 1, m 2 [2 nk ] such that a m 1 ) a m 2 ), t foows that x m 1 ) x m 2 ), y m 1 ) y m 2 ) and z m 1 ) z m 2 ). Suppose for contradcton that there exst m 1, m 2 such that a m 1 ) a m 2 ) and such that the cam s not true,.e., x m 1 ) = x m 2 ) or y m 1 ) = y m 2 ) or z m 1 ) = z m 2 ). Frst consder the case that x m 1 ) = x m 2 ). Because of the one-to-one correspondence between m and a, there exsts a message m 3 m 1, m 2 and such that am 3 ) = a 1 m 1 ),..., a 1 m 1 ), a m 2 ), a +1 m 1 ),..., a k m 1 )). Then x m 1 ) = x m 3 ) because by hypothess x m 1 ) = x m 2 ). Consder the foowng two scenaros. In the frst scenaro, m 1 s transmtted, and an error turns y m 1 ) nto y m 3 ); n the second scenaro, m 3 s transmtted, and an error turns z m 3 ) nto z m 1 ). Then the cut-set sgnas a 1,..., a 1, x, y, z, a +1,..., a k are exacty the same n both scenaros, and so t s mpossbe for t to dstngush m 1 from m 3, a contradcton to the zero error decodabty constrant. Therefore x m 1 ) x m 2 ). Wth a smar argument t foows that y m 1 ) y m 2 ) and z m 1 ) z m 2 ), and the cam s proved. The cam above suggests that x, y and z, as functons of a, are njectve. They are aso surjectve functons because the doman and codoman are both [2 n ]. Hence there are oneto-one correspondences between a, x, y and z. Next we show that for any two messages m 1, m 2, f b m 1 ) b m 2 ), then z m 1) z m 2). Suppose for contradcton that there exsts m 1 m 2 such that b m 1 ) b m 2 ) and z m 1) = z m 2). Then f m 1 s transmtted and an error r 1 turns x m 1 ) nto x m 2 ), the node B w receve the same sgnas as n the case that m 2 s transmtted and an error r 2 turns y m 2 ) nto y m 1 ). Therefore b m 1, r 1 ) = b m 2, r 2 ). But, as shown above, because b m 1, r 1 ) = b m 1 ) and b m 2, r 2 ) = b m 2 ), t foows that b m 1 ) = b m 2 ), a contradcton. Ths suggests that f z m 1) = z m 2) then b m 1 ) = b m 2 ) and therefore b s a functon of z. The functon s surjectve because b takes a 2 n possbe vaues. Then snce the doman and the codoman are both [2 n ], t foows that b must be a bjectve functon of z. Wth the same argument t foows that b s aso a bjectve functon of x. Hence z s a bjecton of a, a s a bjecton of x, x s a bjecton of b, and b s a bjecton of z. Therefore for a 1 k, z s a bjecton of z, and therefore unt rate s feasbe wth zero error n I. Conversey, we show that the feasbty of the unt zero-error rate n I mpes the achevabty of a zero-error rate of k n I c. A constructve scheme s shown n Fgure 2. In I c, the source ets M = M 1,..., M k ), where the M s are..d. unformy dstrbuted over [2 n ]. Let the network code be a M) = x M) = y M) = z M) = z M) = M, = 1,..., k, and et node B, = 1,..., k, perform majorty decodng. It s straghtforward to see that the scheme ensures that b M) = M under a error patterns n R A. Therefore rate k s feasbe wth zero error n I c. Ths rate acheves capacty snce t s equa to the mn-cut from s to t. IV. REDUCTION FROM MULTIPLE-UNICAST TO NETWORK ERROR CORRECTION: VANISHING ERROR CASE In ths secton we show that a smar but sghty weaker resut hods under the vanshng error mode.

4 between z and z and fnay we show that ths mpes the asymptotc feasbty of unt rate n I. Fg. 2: A scheme to acheve zero-error rate k n I c gven that unt rate s feasbe wth zero error n I. M = M 1,..., M k ) and node B performs majorty decodng. Theorem 2. Gven any mutpe-uncast network codng probem I wth source-destnaton pars {s, t ), = 1,..., k}, a correspondng snge-source snge-snk network error correcton probem I c = G, s, t, A) n whch A ncudes sets wth at most a snge edge can be constructed as specfed n Fgure 1, such that f unt rate s feasbe n I then rate k s feasbe n I c. Conversey, f rate k s feasbe n I c then unt rate s asymptotcay feasbe n I. We frst gve a smpe emma. Lemma 1. Let X, Y, Z be three arbtrary random varabes. Then Proof: IX; Z) IX; Y ) + IY ; Z) HY ). IX; Z) = HZ) HZ X) HZ) HZ, Y X) = HZ) HY X) HZ Y, X) HZ) HY X) HZ Y ) = HY ) HY X) + HZ) HZ Y ) HY ) = IX; Y ) + IY ; Z) HY ) In the foowng we prove Theorem 2. We frst show the feasbty of rate k n I c mpes the asymptotc feasbty of unt rate n I. We w take the foowng path. In I c, we appy the network code that acheves rate k wth message unformy dstrbuted over a seected subset of [2 nk ]. Then we show that for = 1,..., k, ths nduces a arge mutua nformaton between random varabes a and b, between b and z, and between z and a. Hence t mpes a arge mutua nformaton Step 1: Seect a subset of messages. Suppose n I c a rate of k s acheved by a network code wth ength n and wth a probabty of error ɛ. Let M be the source message unformy dstrbuted over M = [2 kn ] and et ˆM be the output of the decoder at the termna. Partton M nto good and bad messages M g + M b n the way that m M g f the network code satsfes t under transmsson m,.e., the termna decodes successfuy ˆM = m when M = m for a r R A. Therefore f m M b then there exsts r R A such that ˆM m f M = m and r occurs,.e., r resuts n a decodng error. By the hypothess on the probabty of error t foows that M b 2 kn ɛ. For = 1,..., k, et x m, r) : M R A [2 n ] be the sgna receved from channe x when m s transmtted by the source and the error pattern r happens. Let x m) = x m, 0), xm, r) = x 1 m, r),..., x k m, r)) and xm) = x 1 m),..., x k m)). We defne functons a, b, y, z, z, a, b, y, z, z n a smar way. Lemma 2. There exsts M M g such that for any m 1, m 2 M, m 1 m 2, t foows that am 1 ) am 2 ), bm 1 ) bm 2 ) and z m 1 ) z m 2 ), and such that M 2 kn 1 ɛ ), where ɛ = 4ɛ. Proof: As a and b are cut-set sgnas, for any m 1, m 2 M g, m 1 m 2, t foows that am 1 ) am 2 ) and bm 1 ) bm 2 ). Settng B g = {bm) : m M g }, t hods that B g 1 ɛ)2 kn, and settng B b = [2 n ] k \B g, t hods that B b 2 kn ɛ. For any m M g, t s sad to be a poor message f there exsts another m M g such that z m) = z m ). Consder an arbtrary poor message m 1. By defnton there exsts m 2 M g, m 2 m 1, such that z m 1 ) = z m 2 ). Snce bm 1 ) bm 2 ), there exsts j such that b j m 1 ) b j m 2 ). Let r 1 be the error pattern that changes the sgna on x j to x j m 2 ), and et r 2 be the error pattern that changes the sgna on y j to y j m 1 ). Then f m 1 s sent and r 1 happens, node B j w receve the same nputs as n the stuaton that m 2 s sent and r 2 happens. Therefore b j m 1, r 1 ) = b j m 2, r 2 ) and t foows that ether b j m 1, r 1 ) b j m 1 ) or b j m 2, r 2 ) b j m 2 ). In the former case, the tupe of sgnas b 1 m 1, r 1 ),..., b j m 1, r 1 ),..., b k m 1, r 1 )) = b 1 m 1 ),..., b j m 1, r 1 ),..., b k m 1 )) w be decoded by the termna to message m 1 because by hypothess m 1 M g, whch s decodabe under any error r R A ). It s therefore an eement of B b as t does not equa bm 1 ). Smary, n the atter case, b 1 m 2, r 2 ),..., b j m 2, r 2 ),..., b k m 2, r 2 )) = b 1 m 2 ),..., b j m 2, r 2 ),..., b k m 2 )) w be decoded by the termna to message m 2 and s an eement of B b. For ẑ [2 n ] k, et Mẑ ) be the set of messages {m M g : z m) = ẑ }. Then f Mẑ ) > 1, by the argument above, there are at east Mẑ ) 2 Mẑ ) 3 eements of B b, such that each of them w be decoded by the termna to some message

5 m Mẑ ). Let M poor be the set of a poor messages, then M poor = Mẑ ) 3 B b 3ɛ 2 kn. ẑ : Mẑ ) >1 Let M = M g \M poor, then M = M g M poor 1 ɛ )2 kn, where ɛ = 4ɛ. Ths proves the asserton. Let A = {am) m M } and A = [2 n ] k \A, then A = M 1 ɛ )2 kn snce by Lemma 2, am 1 ) am 2 ) for m 1, m 2 M, m 1 m 2. Therefore A 2 kn ɛ. Smary et B = {bm) m M } and B = [2 n ] k \B, then B 2 kn ɛ. For = 1,..., k, et A = {a m) m M }, then A 1 ɛ )2 kn mpes that A 1 ɛ )2 n. For â A, et Mâ ) = {m M : a m) = â } and defne Nâ ) = Mâ ). Furthermore defne A, = {â A Nâ ) 1 ɛ )2 k 1)n }. We show that the sze of A, s arge. Consder any â A \A,, then by defnton {a 1,..., a k ) A : a = â } < 1 ɛ )2 k 1)n. And because {a 1,..., a k ) [2 n ] k : a = â } = 2 k 1)n, there are at east ɛ 2 k 1)n eements of A such that ther -th entry equas to â. Therefore A \A, ɛ 2 k 1)n A 2 kn ɛ, and A \A, 2n /. So A, A 2n / 1 ɛ 1/)2 n. Defne B, B,, Z and Z, smary, then t foows from the same argument that B,, Z, 1 ɛ 1/)2 n. Step 2: Connect a and b. Let M be the random varabe that s unformy dstrbuted over M. In the foowng we show that f M s the source message then Ia ; b )/n 1 as ɛ 0, = 1,..., k. We start by ower boundng the entropy Hb ). Consder any ˆb B,, then Pr{b = ˆb } = Nˆb ) M 1 ɛ )2 k 1)n 2 kn = 1 ɛ 2 n 2) Pr{b = ˆb } = Nˆb ) M 2k 1)n 1 ɛ )2 kn = 1 2 n 1 ɛ ) Therefore, Hb ) = ˆb B ˆb B, Pr{ˆb } ogpr{ˆb }) Pr{ˆb } ogpr{ˆb }) 3) a) B, 1 ) ɛ 1 2 n og 2 n 1 ɛ ) 1 ɛ 1 ) 2 n 1 ɛ 2 n og2 n 1 ɛ )) = 1 ɛ 1 ) 1 ɛ )n + og1 ɛ )) 4) where a) s due to 2) and 3). Smary t foows that, Ha ) 1 ɛ 1 ) 1 ɛ )n + og1 ɛ )) 5) Hz ) 1 ɛ 1 ) 1 ɛ )n + og1 ɛ )), 6) In the next step we upper bound Hb a ). Reca that Mâ ) = {m M : a m) = â }, we frst prove a usefu emma. Lemma 3. Suppose {b m) : m Mâ )} = {ˆb 1) L),..., ˆb }, then there exst L 1)Nâ ) dstnct eements of B such that each of them w be decoded by the termna to some message m Mâ ). Proof: Consder arbtrary â A, by hypothess there exst L messages m 1,..., m L Mâ ) such that b m j ) = ˆbj), j = 1,..., L. For j = 1,..., L, et r j be the error pattern that changes the sgna on z to be z m j). Then f an arbtrary message m 0 Mâ ) s transmtted by the source and r j happens, the node B w receve the same nputs as n the stuaton that m j s sent and no error happens. Therefore j) b m 0, r j ) = ˆb, and so bm 0, r j ) takes L dstnct vaues for j = 1,.., L. Snce m 0 M s decodabe under any error pattern r R A, t foows that bm 0, r j ) w be decoded by the termna to m 0 for a j = 1,..., L. Among these L vaues,.e., {bm 0, r j ), j = 1,..., L}, ony one s equa to bm 0 ), and the remanng L 1 of them are eements of B. Sum over a m 0 Mâ ) and the asserton s proved. Partton A, nto A,L=1 +A,L>1, such that every eement of A,L=1 has a correspondng L = 1 as defned n Lemma 3. Then t foows from Lemma 3 that: and so A,L>1 1 ɛ ) 2 k 1)n B 2 kn ɛ A,L>1 ɛ 2 n 1 ɛ. 7) We are ready to upper bound Hb a ). Hb a ) = Pr{â } Pr{ˆb â } og Pr{ˆb â } where I 1 = I 2 = I 3 = â A ˆb B I 1 + I 2 + I 3, 8) â A \A, â A,L>1 â A,L=1 Pr{â } ˆb B Pr{â } ˆb B Pr{â } ˆb B We now bound I 1, I 2 and I 3 respectvey. I 1 Pr{â } og 2 n = n â A \A, â A \A, Pr{â } n M â A, Nâ ) M Pr{ˆb â } og Pr{ˆb â } Pr{ˆb â } og Pr{ˆb â } Pr{ˆb â } og Pr{ˆb â }

6 n 1 A, 1 ) ɛ )2 k 1)n M n 1 1 1/ ɛ )2 n 1 ɛ )2 k 1)n ) 2 kn < 1/ + ɛ )n. 9) I 2 = n â A,L>1 â A,L>1 Pr{â } og 2 n Pr{â } n A,L>1 2 k 1)n 2 kn 1 ɛ ) b) ɛ 1 ɛ )1 ɛ n, 10) ) where b) foows from 7). Fnay, by defnton f â A,L=1, then there s a unque ˆb B such that for a messages m Mâ), t foows that b m) = ˆb. Therefore I 3 = 0. Substtutng 10) and 9) nto 8) we have ) 1 Hb a ) < + ɛ ɛ n + 1 ɛ )1 ɛ ) n. Together wth 4), t foows Ia ; b ) > 1 ɛ 1 ) 1 ɛ )n + og1 ɛ )) ) 1 + ɛ ɛ n 1 ɛ )1 ɛ n. 11) ) Step 3: Connect z and b. Next we show that Ib ; z )/n 1, as ɛ 0, n, for = 1,..., k, by upper boundng Hz b ). We frst make some usefu observatons. Lemma 4. Let Mẑ ) = {m M : z m) = ẑ }, then for any m 1, m 2 Mẑ ) such that b m 1 ) b m 2 ), there exsts an eement of B that w be decoded by the termna to ether m 1 or m 2. Proof: Consder any m 1, m 2 Mẑ ) such that b m 1 ) b m 2 ). Let r 1 be the error pattern that changes the sgna on x to be x m 2 ), and et r 2 be the error pattern that changes the sgna on y to be y m 1 ). Then f m 1 s transmtted by the source and r 1 happens, the node B w receve the same nputs as n the stuaton that m 2 s transmtted and r 2 happens. Therefore b m 1, r 1 ) = b m 2, r 2 ), and so ether b m 1, r 1 ) b m 1 ) or b m 2, r 2 ) b m 2 ) because by hypothess b m 1 ) b m 2 ). Consder the frst case that b m 1, r 1 ) b m 1 ), then the tupe of sgnas b 1 m 1, r 1 ),..., b k m 1, r 1 )) = b 1 m 1 ),..., b m 1, r 1 ),..., b k m 1 )) w be decoded by the termna to message m 1 because by hypothess m 1 M whch s decodabe under any error pattern r R A. Therefore t s an eement of B snce t does not equas bm 1 ). Smary n the atter case, b 1 m 2, r 2 ) b 1 m 2 ), then b 2 m 2, r 2 ),..., b k m 2, r 2 )) = b 2 m 2 ),..., b m 2, r 2 ),..., b k m 2 )) s an eement of B and w be decoded by the termna to m 2. Therefore n ether case we fnd an eement of B that w be decoded to ether m 1 or m 2. Lemma 5. Let Mẑ, ˆb ) = {m M : z m) = ẑ, b m) = ˆb }, ˆb,ẑ = arg maxˆb B Mẑ, ˆb ) and Nẑ ) = Mẑ ), 1 then there are at east 2 Nẑ ) Mẑ, ˆb,ẑ ) ) dstnct eements of B that w be decoded by the termna to some messages n Mẑ ). Proof: Let W := Mẑ ), and we descrbe an teratve procedure as foow. Pck arbtrary m 1, m 2 W such that b m 1 ) b m 2 ), and then deete them from W and repeat unt there does not exst such m 1, m 2. By Lemma 4, each par of eements deeted from W w generate a dstnct eement of B, whch w be decoded by the termna to ether m 1 or m 2. After the teratve procedure termnates, t foows that W Mẑ, ˆb,ẑ ), because otherwse there must exst m 1, m 2 W such that b m 1 ) b m 2 ). Therefore at east Nẑ ) Mẑ, ˆb,ẑ ) eements are deeted and the emma s proved. Z We are now ready to upper bound Hz b ). Reca that = {z m) : m M }. We have, Hb z ) = where and ẑ Z Pr{ẑ } ˆb B Pr{ˆb ẑ } og Pr{ˆb ẑ } = I 4 + I 5, 12) I 4 = < ẑ Z ẑ Z Pr{ẑ } Pr{ˆb,ẑ ẑ } og Pr{ˆb,ẑ ẑ } Pr{ẑ } 1 13) I 5 = Pr{ẑ } Pr{ˆb ẑ } og Pr{ˆb ẑ } ẑ Z ˆb ˆb,ẑ Pr{ẑ } Pr{ˆb ẑ } og 2 kn ˆb ˆb,ẑ ẑ Z = kn ẑ Z = kn ẑ Z kn 2 kn 1 ɛ ) c) kn 2 B 2 kn 1 ɛ ) = kn 2 ɛ 2 kn 2 kn 1 ɛ ) ˆb ˆb,ẑ Pr{ẑ, ˆb } Nẑ ) Mẑ, ˆb,ẑ ) ẑ Z M Nẑ ) Mẑ, ˆb,ẑ ) ) = 2kɛ n, 14) 1 ɛ

7 where c) foows from Lemma 5. Substtutng 13) and 14) to 12) we have Hb z ) 1 + 2kɛ n. 15) 1 ɛ Step 4: Connect a and z. Fnay we dscuss the connecton between a and z. We may assume that z = a wthout oss of generaty n the foowng sense. For every network code that acheves rate k wth error probabty ɛ n I c, we can modfy t sghty to obtan a new code such that z = a, and such that the code on a other edges and at the termna are the same as the orgna code. Ths modfcaton s feasbe because the encodng functon z of the orgna code s a functon of a, and so f we et z = a, then the node s, = 1,.., k, aways has enough nformaton to reproduce the orgna network code. Snce from the perspectve of the termna, the modfed code s the same as the orgna code, t aso acheves rate k wth error probabty ɛ n I c. Hence, Ia ; z ) = Ha ). 16) Step 5: Connect z and z. By 11), 15), 16) and Lemma 1, we have Iz ; z ) Ia ; z ) + Ia ; b ) + Ib ; z ) Ha ) Hb ) = Ia ; b ) + Ib ; z ) Hb ) = Ia ; b ) Hb z ) > 1 ɛ 1 ) 1 ɛ )n + og1 ɛ )) ) 1 + ɛ ɛ n 1 ɛ )1 ɛ ) n 1 2kɛ n. 17) 1 ɛ Reca that 17) s obtaned under the assumpton that the source message s unformy dstrbuted over M, and hence a s unformy dstrbuted over A. Therefore the random varabes {a 1 M ),..., a k M )} are not ndependent. Now we consder the case that a s unformy dstrbuted over [2 n ] k and the network code s the same as before. Specfcay, the decodng functon and the encodng functons at a edges except a 1,..., a k are the same as the network code that acheves rate k wth error probabty ɛ n I c. We are nterested n the mutua nformaton between z and z under ths settng where the random varabes {a 1,..., a k } are ndependent. Let a [2 n ] k denote that the dstrbuton of a s unform over [2 n ] k, I a [2n ] kz ; z ) = Note that Pr a [2 n ] k {a A }I a [2n ] kz ; z a A ) + Pr a [2 n ] k {a A }I a [2 n ] kz ; z a A ). 18) I a [2n ] kz ; z a A ) = I a A z ; z ) = Iz ; z ), 19) whch s exacty the resut we computed n 17). And Pr {a A } = A /2 kn 1 ɛ ). 20) a [2 n ] k Therefore by 19) and 20), I a [2 n ] kz ; z ) 1 ɛ )Iz ; z ). Then by 17), for any ɛ 1 > 0, by frst choosng a suffcenty arge and then a suffcenty sma ɛ and a suffcenty arge n, t foows that I a [2n ] kz ; z )/n > 1 ɛ 1, = 1,..., k. Hence unt rate s asymptotcay feasbe n I by the channe codng theorem [25]. Ths competes the proof of the frst part of the theorem.. Conversey, we show that f unt rate s feasbe n I, then rate k s feasbe n I c. Agan we use the constructve scheme n Fgure 2. In I c, the source ets M = M 1,..., M k ), where the M s are..d. unformy dstrbuted over [2 n ]. Let the network code be a M) = x M) = y M) = z M) = z M) = M, = 1,..., k, and et node B, = 1,..., k performs majorty decodng. The termna t w not decode an error as ong as the mutpe-uncast nstance I does not commt an error. Ths happens wth probabty at east 1 ɛ, whch mpes the feasbty of rate k n I c. V. CONCLUSION We summarze the resuts of ths paper n Fg. 3 whch expresses the possbe connectons between the feasbty of a genera mutpe-uncast network codng nstance I and ts reduced snge-uncast network error correcton nstance I c. Our resuts present an equvaence between mutpe-uncast and network error correcton under zero-error communcaton. Namey, determnng the feasbty of zero-error unt rate n I s equvaent to determnng the feasbty of zero-error rate k n I c. Ths s expressed n Fg. 3 by the fact that the two states n the frst row are connected by a snge sod edge and there are no other edges connected to these two states. For the vanshng error mode, however, the mpcaton of our resuts are more nvoved. In our reducton there s a sght sackness, whch gves rse to the two dashed nes n Fg. 3. For exampe, consder the case that unt rate s feasbe but not wth zero error) n I. Then t foows that rate k s feasbe but not wth zero error) n I c. Ths fact s expressed by the soe sod ne eavng the state that unt rate s feasbe but not wth zero error) n I. However, f rate k s feasbe but not wth zero error) n I c, then there are potentay two possbtes: a) that unt rate s feasbe but not wth zero error) n I; and b) that unt rate s asymptotcay feasbe but not exacty feasbe) n I. Opton a) s represented by a sod ne, as ndeed we have observed nstance I wth a correspondng I c that fts ths settng. Opton b) s represented by a dashed ne, as on one the one hand t has not been rued out by our anayss, but on the other hand we are not aware of how to construct nstances I wth a correspondng I c that ft ths settng. A n a, under the vanshng error mode, the two dashed nes n Fg. 3 do not aow us to drecty determne the feasbty of I based on the feasbty of I c. Nevertheess, we may consder the foowng probem on I whch can be

8 Fg. 3: Equvaence between mutpe-uncast and network error correcton under the constructon of Fg. 1. In our reducton, gven an nstance I of the mutpe-uncast network codng probem, we construct an nstance I c of the snge-uncast network error correcton probem. Ths fgure expresses our current understandng of the reaton between achevabty n I and I c. A sod ne between two states means that there exst nstances of I and I c wth the correspondng achevabty. If there s no ne between two states, there do not exst nstances of I and I c wth the correspondng achevabty. A dashed ne between two states means that whether there s a sod ne between the two states or not s st an open probem. soved by the study of I c usng our resuts: Gven an nstance I, partay determne between the three possbe settngs n the foowng manner: f unt rate s feasbe but not wth zero error) n I, answer yes; f unt rate s not asymptotcay feasbe n I, answer no; f unt rate s asymptotcay feasbe but not exacty feasbe n I, then any answer s consdered correct. By our resuts, answerng yes f and ony f rate k s feasbe but not wth zero error) n I c soves the probem above. Whether the parta dstncton probem above on I s a dffcut one compared to the standard feasbty of mutpeuncast network codng) s yet to be estabshed. Fnay we note that our reducton and anayss n the paper consder the case that the connectons between source destnaton pars n I have unt rate. Our reducton can be adapted to connectons wth dfferent rates. 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