Stopping Set Elimination for LDPC Codes

Transcription

1 Sopping Se Eliminaion for LDPC Codes Anxiao (Andrew) Jiang, Pulakesh Upadhyaya, Ying Wang, Krishna R. Narayanan Hongchao Zhou, Jin Sima, and Jehoshua Bruck Compuer Science and Engineering Deparmen, Texas A&M Universiy Elecrical and Compuer Engineering Deparmen, Texas A&M Universiy School of Informaion Science and Engineering, Shandong Universiy Elecrical Engineering Deparmen, California Insiue of Technology Absrac This work sudies he Sopping-Se Eliminaion Problem, namely, given a sopping se, how o remove he fewes erasures so ha he remaining erasures can be decoded by belief propagaion in k ieraions (including k = ). The NPhardness of he problem is proven. An approximaion algorihm is presened for k =. And efficien exac algorihms are presened for general k when he sopping ses form rees. I. INTRODUCTION In his paper, we sudy a basic heoreical problem for LDPC codes: when he erasures in a noisy LDPC codeword canno be correced by he decoder, how o remove he fewes erasures so ha he remaining erasures become decodable? The problem has several applicaions: Disribued Sorage. Disribued file sysems like HDFS have been widely used in big daa applicaions. Typically, hey sore daa in blocks, and ECCs are applied over he blocks (where each block is seen as a codeword symbol of he ECC). Binary LDPC codes are naurally an aracive candidae for disribued sorage, as hey have excellen code raes, good localiy (e.g., a missing block can be recovered by a local disk from a few neighboring blocks), and excellen compuaional simpliciy (only XOR is used for decoding, since when each block has bis, he decoding can be seen as binary LDPC codes being decoded in parallel). Meanwhile, almos all big IT companies sore muliple copies of heir daa a differen locaions. So when one sie loses some blocks in an LDPC code and canno recover hem by iself, i needs o rerieve some los blocks from oher remoe sies. Since communicaion wih remoe sies is much more cosly han accessing local disks, i is desirable o minimize he number of blocks rerieved from remoe sies as long as he remaining erasures become decodable. Saellie-o-Ground Communicaion wih Feedback. Consider saellie-o-ground communicaion, where daa (e.g., big sensing images) are pariioned ino packes (i.e., blocks), and LDPC codes are applied over he packes (similar o he case for disribued sorage). As he channel is noisy, some packes received by he ground may be un-decodable, and he ground will reques he saellie o reransmi some of hose los packes. Since he saellieo-ground communicaion can be cosly, i is desirable o minimize he number of reransmied packes. Le us define he problem more specifically. Le he LDPC code s decoder be he following widely-used ieraive beliefpropagaion (BP) algorihm: in each ieraion, use every pariycheck equaion involving exacly one erasure o decode ha erasure; and repea unil every equaion involves zero or a leas wo erasures. If he decoding fails, hen we are lef wih a sopping se, which is a se of erasures such ha every pariy-check equaion involving any of hem involves a leas wo of hem. If we represen he LDPC code by a biparie Tanner graph, hen a sopping se is a subse of variable nodes (represening erasures) such ha a check node adjacen o any of hem is adjacen o a leas wo of hem. We illusrae he average sizes of Sopping Ses for differen raw bi-erasure raes (RBERs) in Fig.. I is for an (89,756) LDPC code of rae.9 and regular degrees (d v =, d c = 9). (For RBERs near he code s decoding hreshold, he uncorrecable bi-erasure raes (UBER) afer BP decoding is shown in Fig. (a).) For RBERs in he full range from o, he average sopping-se sizes (namely, average number of un-decodable erasures afer BP-decoding) are shown in Fig. (b). I can be seen ha he average sopping-se size increases approximaely linearly (from o 89) as RBER increases from o. (a) Uncorrecable Bi Erasure Rae (UBER) Raw Bi Erasure Rae (RBER) (b) Average sopping se size Raw Bi Erasure Rae (RBER) Fig.. Saisics of an (89,756) LDPC code. (a) UBER for differen RBERs near he code s decoding hreshold. (b) Average sopping-se size for differen RBERs. The problem o sudy can now be defined formally as follows. Le G = (V C, E) be a biparie graph, where V (represening erasures) is a subse of he variable nodes in an LDPC code s Tanner graph, C is a subse of he check nodes in he same Tanner graph such ha every node in C is adjacen o a leas one node in V, and E is he se of edges in he Tanner graph wih one endpoin in V and anoher

2 endpoin in C. If every node in C has degree wo or more, hen G is called a Sopping Graph and V is called a Sopping Se. Now le k be an ineger parameer. If an ieraive BP algorihm (as inroduced earlier) ha runs on G can decode all he variable nodes in V (where every variable node in V is an erasure) wihin k ieraions, hen V is called a Decodable Se (or simply decodable); oherwise, i is a Non-Decodable Se (or simply non-decodable). (Here we inroduce he parameer k o make he problem more general, and o conrol no only he decodabiliy of erasures bu also he ime for decoding.) Noe ha a Sopping Se mus be a Non-Decodable Se, bu no vice versa. The problem we sudy, called Sopping-Se Eliminaion (SSE k ) Problem, is as follows. Definiion. Given a Sopping Graph G = (V C, E), how o remove he minimum number of variable nodes from V such ha he remaining variable nodes can be decoded by BP decoding wihin k ieraions? If he consrain on k ieraions does no exis, we can see k as and call i he SSE Problem. The res of he paper is organized as follows. In Secion II and III, we prove he NR-hardness of he SSE Problem and he SSE k Problem for finie k, respecively. In Secion IV, we presen an approximaion algorihm for he laer problem. In Secion V, we presen efficien algorihms ha reurn opimal soluions for he SSE Problem when he Sopping Ses form ree srucures. In Secion VI, we presen conclusions. Due o space limiaion, we skip some deails in proofs and analysis. Ineresed readers can refer o he full paper [] for he deails. II. NP-HARDNESS OF SSE PROBLEM In his secion, we prove ha he SSE Problem is NPhard. The proof has wo seps: firs, using he well-known Se Cover Problem, we prove ha a relaed covering problem where nearly all elemens are covered which we call he Pseudo Se Cover Problem is NP-complee; hen, we reduce he laer problem o he SSE Problem. A. NP-compleeness of Pseudo Se Cover Problem Consider he well-known Se Cover Problem. Le T = {,,, n } be a universe of n elemens. Le S, S,, S m be m subses of T such ha T = m i= S i. (Each S i is said o cover is elemens.) Le k m be a posiive ineger. The Se Cover Problem asks if here exis k subses S i, S i,, S ik such ha T = k j= S i j. We now define a Pseudo Se Cover Problem ha differs only in is quesion: i asks if here exis k subses S i, S i,, S ik such ha k j= S i j T. Theorem. The Pseudo Se Cover Problem is NP-complee. Proof: We consruc a polynomial-ime reducion from he NP-complee Se Cover Problem o he Pseudo Se Cover Problem. Le an insance of he Se Cover Problem have inpu parameers T = {,,, n }, S, S,, S m and k m as inroduced above. For he corresponding insance of he Pseudo Se Cover Problem, le is universe of elemens be T = {,,, n, n+ }, where n+ is a new elemen, and le is subses be S, S,, S m, S m+, where S m+ = { n+ }. I is simple o see ha he mapping beween he wo insances akes polynomial (in fac, linear) ime. Le us now prove he following claim: he Se Cover Problem has k subses covering all he n elemens in T if and only if he Pseudo Se Cover Problem has k subses covering a leas T = n elemens in T. Consider one direcion of he proof: suppose ha he Se Cover Problem has k subses covering all elemens of T. Then he same k subses cover exacly n elemens of T. (The only uncovered elemen is n+.) Now consider he oher direcion of he proof: suppose ha he Pseudo Se Cover Problem has k subses S i, S i,, S ik covering a leas n elemens in T. Wihou loss of generaliy (WLOG), assume ha i < i < < i k. There are hree possible cases: Case : The k subses cover all he n + elemens of T. Then i k = m +, and he remaining k subses cover all he elemens in T. By adding o he k remaining subses any oher subse in {S, S,, S m }, we ge k subses covering all elemens of T for he Se Cover Problem. Case : The k subses cover n elemens of T, including n+. Then i k = m +, and here mus be exacly one uncovered elemen in T. Say ha uncovered elemen is i, and le S j (where j m) be any subse ha conains j. (Such a subse S j mus exis.) By replacing S ik = S m+ by S j, we ge k subses ha cover all he elemens of T. Case : The k subses cover n elemens of T, bu no covering n+. Then he same k subses cover all he elemens of T. So he reducion exiss, and he conclusion holds. B. NP-hardness of SSE Problem We now prove he NP-hardness of he SSE Problem by using a reducion from he Pseudo Se Cover Problem. Le us begin wih some consrucions. Consider he biparie graph shown in Fig. (a). I consiss of four variable nodes (s i, j, u i,j and w i,j ) and hree check nodes (c i,j, c i,j and c i,j ). We denoe i by D i,j o indicae ha i connecs node s i and node j. We prove some basic propery i has on ieraive BP decoding. Lemma. In he graph D i,j ha conains he variable nodes s i, j, u i,j, w i,j as a Sopping Se, if he value of he variable node s i becomes known, he BP decoding algorihm will recover he values of all he hree remaining variable nodes. On he oher hand, if he value of he variable node j becomes known, he BP decoding algorihm will no recover he value of any of he oher hree variable nodes. Proof: If he value of s i becomes known, by using he check nodes c i,j and c i,j, he BP decoding algorihm will

3 (a) c i,j c i,j u i,j (e) (c) S S S 4 5 s s s 4 5 (d) s s s 4 5 c, c, c, c, c, c, c,4 c,4 c, c, c, c, c,4 c,4 c,5 c,5 u, w, u, w, u, w, u,4 w,4 u, w u,, w, u,4 w,4 u,5 w,5 c, c, c, c,4 c, c, c,4 c,5 (f) c j s i w i,j c i,j (b) s i j g i,j s s s 4 5 c c c Fig.. (a) A biparie graph D i,j ha connecs variable nodes s i and j. (b) A symbol for he graph D i,j. (c) An insance of he Pseudo Se Cover Problem, where T = {,,, 4, 5 } and S = {,, 4 }, S = {, }, S = {, 4, 5 }. (d) The corresponding graph G I. (e) The corresponding graph G I wih full deails. (f) The corresponding graph G II. recover he values of u i,j and w i,j, respecively. Then via he check node c i,j, i will recover he value of j. If he value of j becomes known, since c i,j has degree, he BP algorihm will no recover any more values. The graph D i,j will be viewed as a gadge ha connecs node s i wih node j. To simplify he presenaion, in he following, we ofen represen i by he symbol shown in Fig. (b), where he gae g i,j represens he five nodes (c i,j, c i,j, c i,j, u i,j, w i,j ) and heir inciden edges. The direcion of he gae g i,j indicaes he direced propery shown in he above lemma: decoding s i leads o decoding j, bu no vice versa. Consider he Pseudo Se Cover Problem wih inpu parameers T = {,,, n }, S, S,, S m and k m as inroduced earlier. To reduce i o he SSE Problem, we will map every insance of he Pseudo Se Cover Problem o some insance of he SSE Problem. Le us sar by building a biparie graph G I. We sar by assigning m+n nodes: for every subse S i (for i m) or elemen j (for j n) in he Pseudo Se Cover Problem, here is a corresponding variable node s i or j in G I. Then, whenever he Pseudo Se Cover Problem has j S i, we connec nodes s i and j by he biparie graph D i,j. The graph obained his way is G I. An example is shown below. Example 4. Le he Pseudo Se Cover Problem be T = {,,, 4, 5 } and S = {,, 4 }, S = {, }, S = {, 4, 5 }. (As k is irrelevan o he mapping, we do no specify i.) I is illusraed in Fig. (c), where here is an edge beween S i and j if and only if j S i. The corresponding graph G I is shown in boh Fig. (d) and (e), where he symbol for each D i,j is used in Fig. (d), and he full deails of G I are shown in Fig. (e). I is easy o see he correspondence beween G I and he Pseudo Se Cover Problem. We now creae a biparie graph G II as follows. Given graph G I, we add m + addiional check nodes c, c, c,, c m. For i m and j n, add an edge beween he check node c i and he variable node j. For i m, add an edge beween he check node c i and he variable node s i. The graph obained his way is G II. (For example, following Example 4, G II is as shown in Fig. (f).) In he following, we consider only cases where n >. (The case n = is rivial.) I is hen simple o see ha in G II, he degree of every check node is a leas wo. So i is a Sopping Graph, namely, an insance of he SSE Problem. Lemma 5. If for he Pseudo Se Cover Problem, here exis k subses ha cover a leas n elemens of T, hen for he corresponding graph G II, k variable nodes can be removed so ha he remaining variable nodes form a Decodable Se. Proof: Suppose ha S i, S i,, S ik are k chosen subses ha cover a leas n elemens of T. Le us remove he corresponding k variable nodes s i, s i,, s ik from he graph G II. Since removing a variable node is equivalen o urning he node from an erasure o a known value, by he direced propery of D i,j proved earlier, we know ha he BP decoding algorihm will recover he values of a leas n variable nodes among,,, n. Tha is because if an elemen j is covered by some chosen subse S ir (where r k), since he value of he variable node s ir is now known, via he gadge D ir,j, he BP decoding algorihm can recover he value of j. We now show ha he BP decoding algorihm can recover he values of all n variable nodes,,, n. From he above discussion, we know ha a mos one of hem say x is no decoded ye. So he BP algorihm can use he check node c (which has degree n) o recover he value of x as x = i n,i x i. Since he values of,,, n are all known now, for i =,,, m, he BP decoding algorihm can use he check node c i o recover he value of s i (if is value is no already known). So all he variable nodes can recover heir values. Therefore, he remaining variable nodes form a Decodable Se. When a se of variable nodes S V is removed from a Sopping Graph G = (V C, E), if he remaining nodes of V become decodable, we call S an Eliminaion Se of size S. Lemma 6. If G II has an Eliminaion Se of size k m, hen G II has an Eliminaion Se of size k ha is also a subse of {s, s,, s m }. Proof: Le X = {x, x,, x k } be an Eliminaion Se of G II, where each x i is a variable node. Le us creae a se Y = {y, y,, y k } {s,, s m } as follows. For i =,,, k, do:

4 If x i {s, s,, s m }, le y i = x i. If x i is eiher u i,j or w i,j namely, i is a variable node in he gadge D i,j (more specifically, g i,j ) ha connecs s i and j le y i = s i if s i is no in Y ye, and le y i be any node in {s, s,, s m } ha is no ye in Y oherwise. If x i = j for some j n, le s i be a node such ha here is a gadge D i,j connecing s i and j. (Such a node s i mus exis because in he Pseudo Se Cover Problem, j is covered by a leas one subse.) If s i is no in Y ye, le y i = s i ; oherwise, le y i be any node in {s, s,, s m } ha is no ye in Y. Wih he above consrucion, for any node x i in X, here exiss a node s i in Y such ha eiher s i = x i, or s i and x i exis in he same gadge D i,j for some j. By he direced propery of gadges D i,j, we see ha when he values of variable nodes in Y are known, he BP algorihm can decode all he variable nodes in X; and since s an Eliminaion Se, he BP algorihm can consequenly decode all he variable nodes in G II. So Y is an Eliminaion Se of size k ha is a subse of {s, s,, s m }. Lemma 7. If G II has an Eliminaion Se of size k {s i, s i,, s ik } {s, s,, s m }, hen for he corresponding Pseudo Se Cover Problem, he k subses S i, S i,, S ik cover a leas n elemens of T. Proof: The proof is by conradicion. Suppose ha S i, S i,, S ik cover a mos n elemens of T. Then in G II, when he values of {s i, s i,, s ik } are known, he BP algorihm can use he gadges D i,j o decode a mos n variable nodes among,,, n. Then he BP algorihm ges suck because i canno use any check node o decode any more variable node: For any check node c i (where i m), a leas wo adjacen nodes in {,,, n } are no decoded ye. So he BP algorihm canno use c i o decode more variable nodes. For any gadge D i,j ha connecs s i and j, if s i / {s i, s i,, s ik }, by he direced propery of he gadge, he BP algorihm canno use i o decode s i wheher he node j has been decoded or no. Tha means {s i, s i,, s ik } is no an Eliminaion Se, which is a conradicion. Tha leads o he conclusion. By combining he above wo lemmas, we ge: Lemma 8. If G II has an Eliminaion Se of size k m, hen for he corresponding Pseudo Se Cover Problem, here exis k subses ha cover a leas n elemens of T. We now prove our main resul here. Theorem 9. The SSE Problem is NP-hard. Proof: The SSE Problem is an opimizaion problem. Le us consider is decision problem: given a Sopping Graph G = (V C, E) and a posiive ineger k, does i have an Eliminaion Se of size k? Le us call his decision problem P sse. I is clear ha P sse NP. We have shown a mapping ha maps every insance of he Pseudo Se Cover Problem o an insance of P sse. The mapping akes polynomial ime. By combining Lemma 5 and Lemma 8, we see ha he answer o he Pseudo Se Cover Problem is yes (namely, here exis k subses ha cover a leas n elemens of T ) if and only if he answer o P sse is yes (namely, G II has an Eliminaion Se of size k). So he mapping is a polynomial-ime reducion. By Theorem, he Pseudo Se Cover Problem is NP-complee. So P sse is NP-complee, which leads o he conclusion. III. NP-HARDNESS OF SSE k PROBLEM FOR FINITE k We now consider a new quesion: if k is finie (or even a consan), does he SSE k problem become polynomialime solvable? A posiive answer seems possible a firs sigh, because having a small k pus more localized consrains on soluions. For example, if k =, o correc all remaining erasures in jus one ieraion, in he subgraph induced by he remaining variable nodes and heir adjacen check nodes, every variable node needs o be adjacen o a leas one check node of degree one. Tha is a very local propery for he biparie graph and can possibly make he problem simpler. However, our sudy below will give a negaive answer. We will prove ha even he SSE Problem is NP-hard. There have been a number of works on he node-deleion problem [], [], [4], [6], which can be generally saed as follows: find he minimum number of verices o delee from a given graph so ha he remaining subgraph saisfies a propery π. They focus on properies ha are herediary on induced subgraphs, namely, whenever a graph G saisfies π, by deleing nodes from G, he remaining subgraphs also saisfies π. However, he SSE k Problem is no herediary, because removing a check node can urn a decodable graph (he desired propery) ino an un-decodable one. We now prove he NP-hardness of he SSE Problem. We use a reducion from he NP-complee No-all-equal SAT Problem [5], defined as follows: le x, x,, x n be n Boolean variables. A lieral is eiher x i or x i (namely, he NOT of x i ) for some i {,,, n}. Le a clause be a se of hree lierals. Le S = {C, C,, C k } be a se of k clauses. The quesion is: Is here a ruh assignmen o he n Boolean variables such ha for every clause in S, he hree lierals in he clause are neiher all rue nor all false (namely, every clause has a leas one rue lieral and also a leas one false lieral)? (If he answer is yes, he problem is called saisfiable.) By convenion, rue is also represened by, and false is also represened by. We give an example of he No-allequal SAT Problem. Example. Consider he following insance of he No-allequal SAT Problem. Le n = 4 and k = 5. Le he Boolean variables be x, x, x, x 4, and le he se of clauses

5 be C = (x, x, x ), C = ( x, x, x 4 ), C = (x, x, x 4 ), C 4 = (x, x, x 4 ), C 5 = ( x, x, x ). The above insance is saisfiable because we can le he ruh assignmen be x =, x =, x =, x 4 =, x 5 =. Correspondingly, he clauses become C = (,, ), C = (,, ), C = (,, ), C 4 = (,, ), C 5 = (,, ). None of he clauses is (,, ) (namely, all rue) or (,, ) (namely, all false). A. Reducing No-all-equal SAT Problem o SSE Problem In his subsecion, we consruc a reducion ha maps every insance of he No-all-equal SAT Problem o an insance of he SSE Problem. For every Boolean variable x i of he No-all-equal SAT Problem (for i n), we creae a graph as shown in Fig. (a), which will be called he gadge V i. I is a biparie graph of hree variable nodes and hree check nodes. (Here nodes Xi and Xi represen he rue and false values of x i, respecively.) For every clause C j of he No-all-equal Problem (for j k), we creae wo graphs as shown in Fig. (b), which will be called gadges Uj and U j, respecively. (Here for =,,, nodes A j and B j represen he rue and false values of he -h lieral in clause C j, respecively.) We hen connec hem ino one larger gadge W j as shown in Fig. (c), where for =,,, wo pahs are used o connec he nodes A j and Bj. (For example, he wo pahs beween and have nodes d j, and he four check nodes by hem.) In he final graph corresponding o he insance, he gadge V i will be conneced o he res of he graph only hrough nodes Xi and Xi. So o simplify he presenaion, we someimes represen V i by he symbol in Fig. (d), where he wo inerface nodes Xi, are shown and he remaining deails are hidden. Also in he final graph, he gadge W j will be conneced o he res of he graph only hrough nodes A j, A j,,, B j, ; so we someimes represen i by he symbol in Fig. (e). We now connec he gadges for clauses o he gadges for Boolean variables. Consider a clause C j, and assume is lierals are C j = (l, l, l ). For =,,, if l = x i for some i n, we connec A j o and connec Bj o Xi (hrough some inermediae nodes) as shown in Fig. (f). Oherwise l = x i for some i n, and we connec A j o Xi and connec Bj o as shown in Fig. (g). Example. Assume ha a clause is C j = (l, l, l ) = (x, x, x 4 ). Is gadge W j is conneced o he gadges V, V, V 4 as in Fig. (h). To simplify he presenaion of he graph, we represen he connecion beween a node A j (or B j ) and a node x i (or x i ) by a recangle ha is generally denoed by he H bar. Then he graph in Fig. (h) is simplified as he presenaion in Fig. (i), which shows he connecions more clearly. However, i should be noed ha each A j, B j, x i or x i is conneced o an H bar via wo edges, no one. By now, we have consruced he whole graph ha corresponds o an insance of he No-all-equal Problem. The graph will be denoed by G sse. Le us see an example. (a) (c) (f) (h) (i) (j) X gadge V i a j X X A H y i V i p j W j (b) gadge Wj V X X V X X q j W j gadge U j gadge U j a j b j (g) V X X 4 W j V X X 4 V 4 V 4 X 4 H H H H H W A A B B B V X X X X V X 4 (d) (e) V i p j b j gadge V i V i gadge W j W j an "H" bar V X X 4 W j B j B j H H H H H H H H H H H H (k) v 5 v 9 v 5 v 4 v v v v v v 6 v v 7 v 8 v 6 v 7 v 4 v (l) v A v v v 4 v 5 v 6 v 7 v8 v 9 v v v v v 4 v 5 v 6 v 7 W A A B B B Fig.. (a) The gadge corresponding o a Boolean variable x i, for i =,,, n. (b) Two gadges corresponding o a clause C j, for j =,,, k. (c) The conneced gadge corresponding o a clause C j, for j =,,, k. (d) Symbol for V i. (e) Symbol for W j. (f) Connec clause gadge o Boolean variable gadge: case one. (g) Connec clause gadge o Boolean variable gadge: case wo. (h) An example of connecing a clause gadge o variable gadges. (i) Simplified represenaion of he graph in (h). (j) The graph G sse corresponding o he No-all-equal Problem where n = 4, k =, C = (x, x, x 4 ), C = (x, x, x ), where gadges are represened by symbols. (k) A Sopping Tree G = (V C, E). (l) Is BFS (Breadh-Firs Search) ree G BF S. V 4 X 4 q j

6 Example. For he No-all-equal Problem, le n = 4 and k =. Le he wo clauses be C = (x, x, x 4 ), C = (x, x, x ). Then he corresponding graph G sse is shown in Fig. (j), where is gadges are represened by symbols for clariy. I is easy o see ha G sse is a biparie graph, where every check node has degree more han one. (Specifically, every check node has degree wo.) So G sse is a Sopping Graph. The subsequen analysis will prove ha he No-all-equal SAT Problem is saisfiable if and only if G sse has an Eliminaion Se of size n+k such ha afer is nodes are removed, he BP algorihm can decode he remaining variable nodes in jus one ieraion. B. Properies of Reducion The mapping from any insance of he No-all-equal SAT Problem o a graph G sse has been shown. We now analyze is properies. Due o space limiaion, we only presen skeches of proofs. For deailed proofs, please see []. Le he biparie graph G sse be G sse = (V sse C sse, E sse ), where V sse is he se of variable nodes, C sse is he se of check nodes, and E sse is he se of edges. We now define he conceps of Inerface Nodes, One-Ieraion Eliminaion Se and Canonical Eliminaion Se. Definiion. Le I sse {X j i i n, j } { i i k, j } {Bj i i k, j } be a subse of variable nodes in G sse. Every node in I sse is called an Inerface Node. (As an example, he inerface nodes are shown as circles in Fig. (j).) Definiion 4. Le T V sse be a se of variable nodes in G sse. If afer removing T from G sse, he BP algorihm can decode he remaining variable nodes in one ieraion, hen T is called a One-Ieraion Eliminaion Se. If T is a one-ieraion eliminaion se and T I sse, hen T is called a Canonical Eliminaion Se. Lemma 5. If G sse has a One-Ieraion Eliminaion Se of α nodes, hen G sse also has a Canonical Eliminaion Se of a mos α nodes. Skeches of proof: Le F V sse be a One-Ieraion Eliminaion Se of α nodes. We will prove he exisence of a Canonical Eliminaion Se ˆF Isse wih ˆF α nodes. Noe ha he nodes in G sse are in hree kinds of gadges: gadge V i, gadge W j, or H bar. (See Fig. (j) for an illusraion.) The main idea of he proof is o ransform F ino ˆF by swiching nodes of F o inerface nodes. For example, consider a gadge V i (for i n). (See Fig. (a) for an illusraion.) Noe ha Xi and X i are is only wo nodes connecing o nodes ouside V i in G sse. If y i F, Xi F and X i F, we can delee y i from F and sill ge a one-ieraion eliminaion se, because y i s wo neighboring check nodes already have degree afer Xi and Xi are removed from G sse. If y i F, Xi / F and Xi / F, we can replace y i by Xi and sill ge a one-ieraion eliminaion se, because afer Xi is removed from G sse, boh Xi and y i will have a neighboring check of degree, and more edges will be removed in he par of he graph G sse ha is ouside V i. Similar analysis can be applied o oher cases of V i and o oher gadges (W j and H bar) o show ha non-inerface nodes in F can be eiher deleed or replaced by inerface nodes o ge ˆF. Tha leads o he conclusion. Some properies of Canonical Eliminaion Ses are shown in he nex lemma. We firs define endpoins of an H bar. Definiion 6. Le u be any node in {A j j k, } {Bj j k, }, and le v be any node in {Xi i n} {X i i n}. If u and v are conneced by an H bar, hen hey are called he wo endpoins of ha H bar. Example 7. In Fig. (f), he endpoins of H bars are (A j, ) and (B j, X i ). In In Fig. (g), such endpoin pairs are (A j, X i ) and (B j, ). Lemma 8. For he graph G sse, a Canonical Eliminaion Se F has he following properies: () Propery : For i =,, n, eiher Xi F or X i F ; () Propery : For j =,,, k and =,,, eiher A j F or Bj F ; () Propery : For j =,,, k, F {A j, A j, } and F {Bj, B j, } ; (4) Propery 4: If u and v are he wo endpoins of an H bar, hen eiher u F or v F. Skeches of proof: For he gadge V i (see Fig. (a)), if neiher X i nor X i is in F, hen he BP algorihm canno decode y i in one ieraion since boh of y i s neighboring check nodes will have degree. So Propery is rue. The oher hree properies can be proved similarly. Corollary 9. If F is a One-Ieraion Eliminaion Se of G sse, hen F n + k. Proof: If F is a Canonical Eliminaion Se, by Propery and Propery in Lemma 8, we ge F n+k. Then by Lemma 5, he same conclusion holds for any One-Ieraion Eliminaion Se. Definiion. Le F be a Canonical Eliminaion Se of G sse. If F = n + k, hen F is called an Ideal Eliminaion Se of G sse. (Here Ideal means of minimum possible size. Noe ha an Ideal Eliminae Se may or may no exis for G sse.) The nex lemma easily follows from Lemma 8. Lemma. An Ideal Eliminaion Se F of G sse has hese properies: () Propery : For i =,,, n, eiher Xi or X i is in F, bu no boh; () Propery : For j =,,, k and =,,, eiher A j or B j is in F, bu no boh; () Propery : For j =,,, k, in he se {A j, A j, }, a leas one node is in F, and a leas one node is no in F. The same is rue for he se {Bj, B j, }; (4) Propery 4: If u and v are he wo endpoins of an H bar, hen eiher u or v is in F, bu no boh.

7 Given an Ideal Eliminaion Se of G sse, we can consruc a soluion o he No-all-equal SAT Problem as follows. Definiion. Le F be an Ideal Eliminaion Se of G sse. A corresponding soluion Sol(F ) for he No-all-equal SAT Problem is consruced as follows: i n, he Boolean variable x i = (namely, x i is rue) if and only if X i F. Clearly, in he above soluion Sol(F ), a Boolean variable x i = (namely, x i is false) if and only if X i F. Lemma. Le F be an Ideal Eliminaion Se of G sse, and le Sol(F ) be is corresponding soluion o he No-all-equal SAT Problem. Then for j k and, he -h lieral in he clause C j is rue if and only if A j / F. Proof: Le lj denoe he -h lieral in he clause C j. Consider wo cases: Case : lj is x i for some i n. In his case, by he consrucion of G sse, A j is conneced o by an H bar. lj is rue if and only if F, which by Propery 4 of Lemma happens if and only if A j / F. Case : lj is x i for some i n. In his case, by he consrucion of G sse, A j is conneced o X i by an H bar. lj is rue if and only if x i is false, which happens if and only if Xi F, which by Propery 4 of Lemma happens if and only if A j / F. So in boh cases, he conclusion holds. Lemma 4. If F is an Ideal Eliminaion Se of G sse, hen Sol(F ) is a saisfying soluion o he No-all-equal SAT Problem. Proof: For j k, le A j F and A j / F. By Propery of Lemma, such wo inegers, {,, } exis. Consider he clause C j. By Lemma, he -h lieral of C j is false, and -h lieral of C j is rue. So for he No-allequal SAT Problem, every clause has a leas one rue lieral and a leas one false lieral. So Sol(F ) is a saisfying soluion o he No-all-equal SAT Problem. The above lemma is useful for he scenario where G sse has a One-Ieraion Eliminaion Se of n + k nodes. We now consider anoher possible scenario: he No-all-equal SAT Problem is saisfiable. Given a saisfying soluion o he No-all-equal SAT Problem, we can consruc an Ideal Eliminaion Se of G sse. We firs define he corresponding se. Definiion 5. Le π be a saisfying soluion o he No-allequal SAT Problem; ha is, wih he soluion π, every clause has a leas one rue lieral and a leas one false lieral. A corresponding se of nodes, F(π), in G sse is consruced as follows: For i =,,, n, if x i = in he soluion π, hen X i F(π) and X i / F(π); oherwise, X i / F(π) and X i F(π). For j =,,, k and =,,, if he -h lieral of clause C j is rue given he soluion π, hen A j / F(π) and Bj F(π); oherwise, A j F(π) and B j / F(π). Lemma 6. Le π be a saisfying soluion o he No-all-equal SAT Problem. Then F(π) is an Ideal Eliminaion Se of G sse. Skeches of proof: We firs prove he asserion: F(π) is an One-Ieraion Eliminaion Se of G sse. Consider he gadge V i, for i n. (See Fig. (a).) By he consrucion of F(π), eiher Xi or X i is in F(π). Eiher way, afer he node in F(π) is removed, he oher wo variable nodes in V i will have neighboring check nodes of degree. The oher gadges (W j and H bar) can be analyzed similarly. So he asserion holds. Then i can be seen ha all he nodes in F(π) are Inerface Nodes and F(π) = n + k. So F(π) is an Ideal Eliminaion Se of G sse. Theorem 7. The SSE Problem is NP-hard. Skeches of proof: By Lemmas 5, 4, 6 and Corollary 9, he No-all-equal SAT Problem is saisfiable if and only if he corresponding SSE Problem has a one-ieraion eliminaion se of size n + k. IV. APPROXIMATION ALGORITHM FOR SSE PROBLEM In his secion, we presen an approximaion algorihm for he SSE problem, for Sopping Graphs whose degrees of variable nodes and check nodes are upper bounded by d v and d c, respecively. Is approximaion raio is d v (d c ). (Clearly, he same resul also applies o regular (d v, d c ) LDPC codes and irregular codes wih he same consrain on maximum degrees.) Noe ha he opimizaion objecive is o minimize he size of he eliminaion se (namely, he number of removed variable nodes). So he approximaion raio means he maximum raio of he size of an eliminaion se produced by he approximaion algorihm o he size of an opimal (i.e., minimum) eliminaion se. Definiion 8. In he Sopping Graph G = (V C, E), v V, define is variable-node neighborhood as Λ(v) {u V {v} c C such ha (u, c) E and (v, c) E}. Tha is, every variable node in Λ(v) shares a common neighboring check node wih v. The algorihm will assign hree colors o variable nodes: Iniially, every variable node is of he color whie. I means ha his variable node canno be decoded by one ieraion of BP-decoding ye. As he algorihm proceeds, if a variable node s color urns black, i means he algorihm has included i in he Eliminaion Se (namely, he algorihm has removed i). As he algorihm proceeds, if a variable node s color urns gray, i means he variable node is no ye removed, bu i will be decodable afer one ieraion of BP decoding. The algorihm works as follows: () as iniializaion, make every variable node whie; () choose an arbirary whie

8 variable node v. Le U v denoe he se of variable nodes in Λ(v) ha are currenly whie or gray; urn he colors of he nodes in U v o black, and urn he color of v o gray. For every check node c ha is conneced o a leas one variable node in U v, check if exacly one of c s neighboring variable node is whie and all c s oher neighboring variable nodes are black; if so, urn ha neighboring variable node s color from whie o gray. () repea he previous sep unil all variable nodes are eiher black or gray. Then reurns he se of black variable nodes as he Eliminaion Se. The algorihm can be shown o have ime complexiy O(d vd c V ). We now analyze he approximaion raio of he algorihm. Say ha he algorihm uses oally ieraions o idenify a sequence of whie variable nodes ˆv, ˆv,, ˆv in he Sopping Graph G = (V C, E), and urns he variable nodes in Uˆv, Uˆv,, Uˆv black. Le us define a sequence of subgraphs G, G,, G accordingly. Definiion 9. Le G = G. For i =,,,, le G i be obained from G i by removing he nodes in Uˆvi {ˆv i } {check nodes adjacen o ˆv i } and heir inciden edges. Noe ha for i =,,,, in he i-h ieraion, he algorihm removes only he variable nodes in Uˆvi (namely, urning hem black) from he subgraph G i, no ˆv i or is adjacen check nodes. (I urns ˆv i o gray.) However, once Uˆvi is removed, all he nodes in Λ(ˆv i ) are removed, so ˆv i and is adjacen check nodes become disconneced from he res of he graph (which is G i ). Therefore i becomes sufficien o consider he SSE Problem for G i in he nex ieraion, and i can be seen ha ˆv i+, Uˆvi+, {check nodes adjacen o ˆv i+ }, ˆv i+, Uˆvi+, {check nodes adjacen o ˆv i+ },, ˆv, Uˆv, {check nodes adjacen o ˆv } are all nodes in G i. Lemma. For i =,,,, every one-ieraion eliminaion se for G i conains a leas one variable node in Uˆvi+ {ˆv i+ }. Proof: Consider a one-ieraion eliminaion se S for G i. Eiher ˆv i+ S, or ˆv i+ needs o be decodable in one ieraion afer S is removed. The laer requires ˆv i+ o have a neighboring check node c such ha all c s oher neighboring variable nodes in G i are included in S, and hose nodes are all in Uˆvi+. Tha leads o he conclusion. Lemma. For i =,,,, le α i denoe he minimum size of a one-ieraion eliminaion se for G i. Then α i i. Proof: The proof is by inducion, bu in he reverse order for i (i.e, from i =, down o ). When i =, clearly α i i =, so he conclusion holds for he base case. Now assume ha he conclusion holds for α, α,, α i+, and consider he case for α i. Consider an opimal (i.e., minimum-sized) one-ieraion eliminaion se S for G i. Define Y S (Uˆvi+ {ˆv i+ }). By Lemma, S removes a leas one variable node in {ˆv i+ }, so Y. Le G be he biparie graph Uˆvi+ obained by removing he variable nodes in Y from G i (and heir inciden edges), and le α denoe he minimum size of a one-ieraion eliminaion se for G. Then α i = S = Y + α α +. G i+ is obained from G i by removing he variable nodes in Uˆvi+ {ˆv i+ }, which is a superse of Y. So G i+ can also be obained from G by removing he variable nodes in (Uˆvi+ {ˆv i+ }) Y. So α i+ α. By he inducion assumpion, we ge α i+ (i + ). By combining he above resuls, we ge α i α + α i+ + (i + ) + = i. Theorem. Le d v and d c denoe he maximum degrees of variable nodes and check nodes, respecively, in he Sopping Graph G = (V C, E). Then he above algorihm has an approximaion raio of d v (d c ). Proof: By seing i = in Lemma, we ge α, namely, any one-ieraion eliminaion se for G removes a leas variable nodes. The algorihm removes he nodes in Uˆv Uˆv Uˆv, whose size is i= Uˆv i = i= Uˆv i i= Λ(ˆv i) d v (d c ). So he approximaion raio is a mos d v (d c ). V. ALGORITHM FOR SSE k AND SSE PROBLEMS In his secion, we presen an algorihms for he SSE k Problem for general k, including k =, when he Sopping Graph is a ree (or a fores). The algorihm oupus an opimal soluion and has linear ime complexiy. The Sopping Graph G = (V C, E) can be a ree, especially when he RBER is low. In his case, we call G a Sopping Tree. Noe ha if G is a fores, he SSE k Problem can be solved for each of is ree componens independenly. Given a Sopping Tree G = (V C, E), we can pick an arbirary variable node v V as he roo, run Breadh-Firs Search (BFS) on G saring wih v, and label he nodes of G by v, v,, v V + C based on heir order of discovery in he BFS. (Noe ha he roo node v is labelled by v, and siblings nodes in he BFS ree always have consecuive labels.) We denoe he resuling BFS ree by G BF S. For any non-roo node v in G BF S, le π(v) denoe is paren. Le G sub denoe he subree of G BF S obained his way: if we remove he subree rooed a π(v V + C ) from G BF S, he remaining subgraph is G sub. Example. A Sopping Tree and is BFS ree are shown in Fig. (k) and (l), respecively. (Noe ha he node labels v, v,, v 7 in Fig. (k) are no known a priori; insead, hey are obained afer we run BFS on he graph wih v as is roo.) Here v V + C = v 7, π(v 7 ) = v, and G sub is he subree in he dashed circle in Fig. (l). The algorihm firs runs BFS on G o ge he ree G BF S ha labels nodes by v, v,, v V + C, where v is he roo. Then i processes he nodes in he reverse order of heir labels, and keeps reducing he SSE k Problem acually, a more general form of he SSE k Problem, which shall be called he

9 gsse k Problem o smaller and smaller subrees. Le us now define his gsse k Problem. Definiion 4. [gsse k Problem] Le G = (V C, E) be a Sopping Graph. and le k be a non-negaive ineger. Every variable node v V is associaed wih wo parameers δ(v) {,,, k, } and ω(v) {,,, k, } saisfying he condiion ha eiher δ(v) = or ω(v) =, bu no boh; and when he BP decoder runs on G, v s value can be recovered (namely, v can become a non-erasure) by he end of he δ(v)-h ieraion auomaically (namely, wihou any help from neighboring check nodes). Then, how o remove he minimum number of variable nodes from V such ha for every remaining variable node v wih ω(v) k, i can be correced by he BP decoder in no more han ω(v) ieraions? (By defaul, if ω(v) =, v has o be removed from V because he BP decoder sars wih he s ieraion.) A soluion o he gsse k Problem (namely, he se of removed nodes) is called a g-eliminaion Se. We see ha if δ(v) = and ω(v) = k for every v V, hen he gsse k Problem is idenical o he SSE k Problem. In G BF S, le τ {,,, V + C } denoe he minimum ineger such ha v τ eiher is a sibling of v V + C or is v V + C iself. (So v τ, v τ+,, v V + C are siblings.) Define P {i τ i V + C, ω(v i ) k} and Q {i τ i V + C, δ(v i ) k}. Since v V, eiher δ(v) or ω(v) is bu no boh, P and Q form a pariion of he se {τ, τ +,, V + C }. By convenion, for he empy se, we say max i δ(v i ) = max i ω(v i ) =. We firs make some observaions. Lemma 5. Suppose max i P ω(v i ) > max i Q δ(v i ). Le i be an ineger in P such ha ω(v i ) = max i P ω(v i ). Then here exiss a minimum-sized g-eliminaion Se for G BF S ha includes he nodes in {v i i P, i i } bu no v i. Proof: Any g-eliminaion Se for G BF S has o include a leas P nodes in {v i i P} because oherwise he un-included nodes in {v i i P} will never be correced. I is an opimal sraegy o include he P nodes in {v i i P, i i } in he g-eliminaion Se because heir ω(v i ) values impose more resricive requiremens han ω(v i ) does. Now le T be a g-eliminaion Se for G BF S ha includes all he nodes in {v i i P, i i }. If v i T, we can replace i by π(π(v V + C )) in T and ge anoher g-eliminaion Se T for G BF S, wih T T (since π(π(v V + C )) may already be in T ). (Noe ha wih T, since max i P ω(v i ) > max i Q δ(v i ), he check node π(v V + C ) can help correc v i by ieraion max i Q δ(v i ) + max i P ω(v i ) = ω(v i ); and since π(π(v V + C )) T, he BP decoding in G sub becomes independen of he subree rooed a π(v V + C ).) So here exiss a minimum-sized g-eliminaion Se for G BF S ha includes he nodes in {v i i P, i i } bu no v i. Lemma 6. Suppose max i P ω(v i ) max i Q δ(v i ). Then here exiss a minimum-sized g-eliminaion Se for G BF S ha conains all he nodes in {v i i P}. Proof: If P =, he conclusion auomaically holds. If P = and max i P ω(v i ) =, any g-eliminaion Se for G BF S has o include {v i i P}, so he conclusion also holds. Now consider he case where max i P ω(v i ) >. Le T be a minimum-sized g-eliminaion Se for G BF S. T has o include a leas P nodes in {v i i P} because oherwise he un-included nodes in {v i i P} canno be correced (using he check node π(v V + C )). If T = P, he le j P be an ineger such ha v j / T. If T {v i i Q} =, hen v j canno be correced by ieraion ω(v j ) max i P ω(v i ) max i Q δ(v i ) because no all nodes in {v i i Q} will be correced by ieraion ω(v j ), so his is an impossible case. So T {v i i Q}. Le m Q be an ineger such ha v m T ; hen we can replace v m by v j in T and ge anoher g- Eliminaion Se T for G BF S because v m helps decoding more han v j : v m can be correced auomaically. Since T = T and {v i i P} T, he conclusion holds. The nex wo lemmas show how o reduce he gsse Problem from G BF S o is subree G sub. In some cases, in he derived gsse Problem for G sub, he values of δ(π(π(v V + C ))) and ω(π(π(v V + C ))) in G sub may be differen from heir original values in G BF S ; and in such cases, o avoid confusion, we will denoe he ree G sub by Ĝ sub. Lemma 7. Suppose max i P ω(v i ) max i Q δ(v i ). Consider five cases: ) Case : If Q > and max i Q δ(v i ) = k, le S be a minimum-sized g-eliminaion Se for G sub. ) Case : If Q >, max i Q δ(v i ) < k and δ(π(π(v V + C ))) k, le S be a minimum-sized g- Eliminaion Se for Ĝ sub where δ(π(π(v V + C ))) is changed o min{δ(π(π(v V + C ))), max i Q δ(v i ) + }. ) Case : If Q > and ω(π(π(v V + C ))) max i Q δ(v i ) < k, le S be a minimum-sized g- Eliminaion Se for G sub. 4) Case 4: If Q > and max i Q δ(v i ) < ω(π(π(v V + C ))) k, le S be a minimum-sized g- Eliminaion Se for Ĝ sub where δ(π(π(v V + C ))) is changed o max i Q δ(v i ) + and ω(π(π(v V + C ))) is changed o. 5) Case 5: If Q =, here are wo sub-cases: () if ω(π(π(v V + C ))) =, le S be a minimumsized g-eliminaion Se for G sub ; () oherwise, le S be a minimum-sized g-eliminaion Se for Ĝ sub where δ(π(π(v V + C ))) is changed o and ω(π(π(v V + C ))) is changed o. Then S {v i i P} is a minimum-sized g-eliminaion Se for G BF S. Proof: By Lemma 6, here exiss a minimum-sized g- Eliminaion Se for G BF S ha conains all he nodes in {v i i P}. Now consider only minimum-sized g-eliminaion Ses for

10 G BF S ha conain all he nodes in {v i i P}. See he nodes in {v i i P} as removed (because nodes in an Eliminaion Se are removed before decoding begins); hen o prove he conclusion, we jus need o prove his asserion: when P =, S is a minimum-sized g-eliminaion Se for G BF S. For Case, since max i Q δ(v i ) = k, he subree rooed a π(v V + C ) canno help correc he node π(π(v V + C )) in he firs k ieraions. Every node v wih ω(v) is in G sub and has ω(v) k. So finding a minimum-sized g-eliminaion Se for G BF S is equivalen o finding such as se for G sub. So he asserion holds. For Case, if we compare G sub and Ĝsub, we see ha hey differ only in heir values of δ(π(π(v V + C ))). (For Ĝ sub, ha value is min{δ(π(π(v V + C ))), max i Q δ(v i ) + }.) Now observe he check node π(v V + C ) and is neighboring variable nodes: when BP decoder runs on G BF S, all he nodes in {v i i Q} can be correced auomaically by ieraion max i Q δ(v i ) < k; so by using he check node π(v V + C ), he node π(π(v V + C )) can be correced by ieraion max i Q δ(v i ) + k. Tha is equivalen o urning δ(π(π(v V + C ))) ino min{δ(π(π(v V + C ))), max i Q δ(v i ) + } and urning G sub ino Ĝsub when i comes o BP decoding. Tha leads o he asserion. The remaining hree cases can be proved similarly. (For deails, please see [].) Lemma 8. Suppose max i P ω(v i ) > max i Q δ(v i ). Le i be an ineger in P such ha ω(v i ) = max i P ω(v i ). Consider wo cases: ) Case : If max i P ω(v i ) > δ(π(π(v V + C ))), le S be any minimum-sized g-eliminaion Se for G sub. ) Case : If max i P ω(v i ) δ(π(π(v V + C ))), le S be any minimum-sized g-eliminaion Se for Ĝ sub where δ(π(π(v V + C ))) is changed o and ω(π(π(v V + C ))) is changed o min{ω(π(π(v V + C ))), max i P ω(v i ) }. Then S {v i i P, i i } is a minimum-sized g-eliminaion Se for G BF S. Proof: By Lemma 5, here exiss a minimum-sized g- Eliminaion Se for G BF S ha includes he nodes in {v i i P, i i } bu no v i. Le T be such a minimum-sized g- Eliminaion Se for G BF S. For Case, when he g-eliminaion Se for G BF S is T, he subree rooed a he check node π(v V + C ) canno help correc he node π(π(v V + C )). Insead, hose nodes of T ha are in G sub will be a g-eliminaion Se for G sub, and he BP decoder will correc he un-removed nodes in G sub (wihin each of heir required number of ieraions ω(v)). If π(π(v V + C )) T, he check node π(v V + C ) will correc v i in he s ieraion; oherwise, he BP decoder will correc π(π(v V + C )) in a mos δ(π(π(v V + C ))) ieraions, so π(v V + C ) will correc v i in a mos δ(π(π(v V + C )))+ ω(v i ) ieraions. Since T s size is minimized, he number of nodes of T ha are in G sub is also minimized. Tha leads o he conclusion. For Case, when he g-eliminaion Se for G BF S is T, he BP decoder needs o correc he node π(π(v V + C )) by ieraion max i P ω(v i ) < δ(π(π(v V + C ))) because only hen will he check node π(v V + C ) help correc he node v i by ieraion max i P ω(v i ) = ω(v i ). Tha is equivalen o urning ω(π(π(v V + C ))) ino min{ω(π(π(v V + C ))), max i P ω(v i ) }, urning δ(π(π(v V + C ))) ino and urning G sub ino Ĝsub when i comes o BP decoding. Tha leads o he conclusion. We can design an algorihm for SSE k as follows: () run BFS on G o ge G BF S and as iniializaion, le ω(v) = k and δ(v) = for every v V ; () use Lemma 7 and 8 repeaedly o reduce he graph in he gsse Problem from G BF S o is subree G sub (or Ĝsub), and hen o smaller and smaller subrees in he same way, unil he subree conains only he roo node v ; during his reducion process, more and more nodes (namely eiher {v i i P} in Lemma 7 or {v i i P, i i } in Lemma 8) are included in he Eliminaion Se; () in he las sep, when he subree conains only v, include v in he Eliminaion Se if and only if ω(v ) k a ha momen. The above algorihm can be implemened by processing he nodes in he reverse order of heir labels from v V + C back o v and has ime complexiy O( V + C ). Due o space consrains, we omi is pseudo code here. We can see ha i reurns an opimal (i.e., minimum-sized) k-ieraion Eliminaion Se of G = (V C, E). For he special case of k =, he algorihm can be simplified: for every check node, include all bu one of is children in he Eliminaion Se S; also include v in S. For is deails, please see []. VI. CONCLUSIONS This paper sudies he Sopping-Se Eliminaion Problem moivaed by several applicaions. The NP-hardness of boh SSE and SSE Problems is proven. An approximaion algorihm is presened for he SSE Problem. And linearime algorihms ha reurn opimal soluions are presened for he SSE and SSE k Problems when he Sopping Ses have ree srucures. REFERENCES [] B. Addis, M. D. Summa and A. Grosso, Removing Criical Nodes from a Graph: Complexiy Resuls and Polynomial Algorihms for he Case of Bounded Treewidh, available a hp : // online.org/db F ILE//7/.pdf. [] T. Fujio, Approximaing Node-Deleion Problems for Marodial Properies, in Journal of Algorihms, vol., pp. 7, 999. [] A. Jiang, P. Upadhyaya, Y. Wang, K. R. Narayanan, H. Zhou, J. Sima and J. Bruck, Sopping Se Eliminaion for LDPC Codes, available a hp : //faculy.cse.amu.edu/ajiang/sse.pdf. [4] M. Kumar, S. Mishra, N. S. Devi and S. Saurabh, Approximaion Algorihms for Node Deleion Problems on Biparie Graphs wih Finie Forbidden Subgraph Characerizaion, in Theoreical Compuer Science, vol. 56, pp. 9 96, 4. [5] T. J. Schaefer, The Complexiy of Saisfiabiliy Problems, in Proc. h Annual ACM Symposium on Theory of Compuing, pp. 6 6, 978. [6] M. Yannakakis, Node-Deleion Problems on Biparie Graphs, in SIAM Journal on Compuing, vol., no., pp. 7, May 98.