On Approximation Properties of the Independent Set Problem for Degree 3 Graphs Piotr Berman1 ? and Toshihiro Fujito2 ?? 1 Dept. of Computer Science and Engineering, The Pennsylvania State University, University Park, PA 16802 USA
[email protected]
2 Dept. of Electrical Engineering, Hiroshima University, 1{4{1 Kagamiyama, Higashi{Hiroshima 739 JAPAN
[email protected]
The main problem we consider in this paper is the Independent Set problem for bounded degree graphs. It is shown that the problem remains MAX SNP {complete when the maximum degree is bounded by 3. Some related problems are also shown to be MAX SNP {complete at the lowest possible degree bounds. Next we study better poly{time approximation of the problem for degree 3 graphs, and improve the previously best ratio, 45 , to arbitrarily close to 65 . This result also provides improved poly{time approximation ratios, B5+3 + , for odd degree B . Abstract.
1
Introduction
The area of ecient approximation algorithms for NP {hard optimization problems has recently seen dramatic progress with a sequence of breakthrough achievements. Even when restricted only to the area of constant bound approximation the following remarkable results have been obtained in the last few years. The subclass of NP optimization problems, called MAX SNP , consisting solely of constant ratio approximable problems was introduced and shown to have many natural complete problems by Papadimitriou and Yannakakis [14]. It turns out that actually all the NP optimization problems approximable within some constant factors belong to MAX SNP (via approximation preserving reduction) [11]. In 1992 the notion of MAX SNP {hardness became to bear much more significance when Arora et al. established that there do not exist polynomial time approximation schemes (PTAS) for MAX SNP {hard optimization problems unless P = NP [1]. It has thus become possible to classify qualitatively many of NP {hard optimization problems according to their approximation properties, and as the area becomes more mature the research focus has to be shifted also to more quantitative classi cation of approximability. The main problem we treat in this paper is a well{studied one: bounded degree Independent Set problem (called MAX IS{B when the maximum vertex degree ? ??
This work was partially supported by NSF Grant CCR-9114545 Part of this work was done while the author was at Dept. of CSE, Penn State.
of input graphs is bounded by B ). This is one of the original and prototypical MAX SNP {complete problems [14], and its approximation ratio has been greatly improved over the years by a number of new techniques and analysis. The rst nontrivial performance ratio B appeared implicitly in the Lovasz's algorithmic proof [12] of Brooks' coloring theorem [5]. Hochbaum [9] developed a heuristic with a ratio B2 using this coloring technique coupled with a method of Nemhauser and Trotter [13]. Halldorsson and Radhakrishnan [7] recently showed that the greedy heuristic actually delivers a better ratio, B3+2 . Very recently a progress in getting even better bounds became swift, probably in the wake of emergence of non{approximability results. Berman and Furer [4] discovered new heuristics whose performance ratios are arbitrarily close to B5+3 for even B and B +3:25 for odd B . Soon afterwards, Halldorsson and Radhakrishnan [8] obtained 5 B ). asymptotically better ratios, B6 + o(1) and O( log log B In this paper we pay a special attention to MAX IS{3. MAX IS{B is NP { complete even when instance graphs are restricted to be cubic and planar [6]. It is also known that MAX IS (unbounded degree) admits a PTAS when graphs are planar [2]. In section 2 we will show, however, that MAX IS{3 is MAX SNP {complete. As by-products a few other problems (such as MAX 3{SET PACKING{2 and MAX TRIANGLE PACKING{4) are shown to remain MAX SNP {complete at the lowest possible degree bounds. Next we study better approximation of MAX IS{3 in sections 3 through 5. The previously best ratio was 45 , and we improve it to arbitrarily close to 65 . Currently the best performance guarantee for MAX IS{B is given by Berman{ Furer's algorithm for B up to at least 613 [8]. The new ratio for MAX IS{3 provides further improvements on their ratios for odd B , matching their performance guarantee formula for even B . It should be also noted that with a method of Nemhauser{Trotter this result gives approximation of the Minimum Vertex Cover problem for graphs of degree B within a factor 2 0 B5+3 + , which is an improvement over the previous best for small B .
2 MAX Bounds
SNP {Completeness with
the Lowest Degree
Many of MAX SNP {complete problems (especially graph problems) involve some parameters in their de nitions, and even if they don't we may introduce such parameters to consider some restricted versions of original problems. It would be then important and interesting to know the exact boundaries of MAX SNP { completeness of those problems in terms of their parameters. Maximum Independent Set problem is known to remain NP {complete even when input graphs are restricted to be cubic and planar. However, for planar graphs (of unbounded degree) MAX IS admits a PTAS [2]. On the other hand we will show that it becomes MAX SNP {complete as soon as degree 3 nodes are allowed. MAX 3SAT{B is a restriction of MAX 3SAT s.t. in any instance the number of occurrences of any variable is bounded by B . We reduce MAX 3SAT{ B to MAX IS{3 using the \ring of trees" construction of Kann [10].
Theorem 1.
MAX IS{3 is
MAX SNP {complete.
Suppose a variable u occurs d times in a given 3SAT{B instance. 3 Let K be a large enough power of two (it suces to take K = 2blog2 ( 2 B +1)c ). Prepare K identical cycles of length 2d (called \rings") and sequentially index the nodes in each cycle from 1 to 2d. Also prepare 2d complete binary trees with K leaves each. Join these rings and trees by overlapping, in the identical fashion, leaves of each tree with nodes of the same index from each ring. Label the roots of these trees as ui and ui; i = 1; 1 1 1 ; d, alternatively in the order of their indices. Construct a \ring of trees" this way for every variable. Each clause c will be represented by a clique of size jcj. Suppose the ith occurrence of a variable u is in a clause c. Connect ui or ui , depending on whether u appeaars in positive or not, to the corresponding node in c{clique. This way clause cliques and rings of trees are connected together. Note that every node has its degree bounded by 3. Let A be a (disjoint) union of rings of trees corresponding to all variables. An independent set is said to be consistent if it includes all ui 's and none of ui 's, or vice versa, for every variable u. For a MAX IS{3 instance thus constructed it can be shown that 1) An independent set maximum in A is cosistent, and 2) An independent set maximum in A is larger than an independent set not maximum in A. We deduce that an optimal solution for MAX IS{3 consists of a maximum independent set in A plus a collection of nodes, one each from a clique corresponding to a satis ed clause in an optimal solution for MAX 3SAT{B. It follows that an optimal value is scaled up only by some constant factor because the size of A can be bounded by K and d 3 1 (number of clauses) and at least half of clauses can be always satis ed. Secondly, from any MAX IS{3 solution we can nd a consistent solution of no smaller size, and hence, a solution for MAX 3SAT{B. The value of a solution thus obtained is no further away from the optimum than the original one. 2 There are some other MAX SNP {complete problems whose structures are closely related to MAX IS{3 and MAX IS{B . Consider for instance the following problems:
Proof(Sketch).
P
MAX 3{DIMENSIONAL MATCHING{B Given three sets W; X; Y and a set M W 2 X 2 Y s.t. the number of occurrences of any element of W; X; or Y in M is bounded by B , nd the largest matching, i.e., a subset M 0 M s.t. no two elements of M 0 agree in any coordinate. MAX 3{SET PACKING{B Given a collection C of subsets of a set S where every c 2 C contains at most three elements and every element s 2 S is contained in at most B of the subsets in C , nd a largest collection of mutually disjoint sets in C . MAX TRIANGLE PACKING{B Given a graph of maximum vertex degree bounded by B nd a largest collection of mutually (vertex) disjoint 3{cliques.
Using MAX SNP {completeness of 3{DIMENSIONAL MATCHING{3 Kann showed that MAX IS{5, MAX 3{SET PACKING{3 and MAX TRIANGLE
PACKING{6 are MAX SNP {complete as well [10]. Reducing MAX IS{3 to these problems instead, we can improve the degree bounds in these problems to the best possible ones:
Corollary 2. are
3
MAX 3{SET PACKING{2 and MAX TRIANGLE PACKING{4
MAX SNP {complete.
Better Approximation of MAX IS{3
The rest of the paper is devoted to proving the next theorem:
Theorem 3. 1 O (n2+ ).
MAX IS{3 can be approximated within a factor
6 5
+
in time
Our algorithm, which is the subject of section 4, extends the idea of Berman and Furer's heuristic given in [4]. However, to improve the existing approximation ratio we apply some general reductions, old and new, and construct solutions with care. Consequently much more careful and involved analysis is called for, and it will be given in section 5. When combined with our algorithm for MAX IS{3, Berman{Furer's algorithm for MAX IS{B does better for every odd B , giving the same performance guarantee formula as the one for even B . Therefore,
Corollary 4. MAX IS{B can be approximated in poly{time within a factor ar2. bitrarily close to B5+3 for all B
De nitions and Notation. For a graph G = (V; E ) and a node set U V , G(U ) denotes a subgraph of G induced by U . The neighborhood set of U , denoted N (U ), is the set of nodes adjacent to U . The degree of a node v 2 V is d(v) = jN (v)j. A neighborhood set and degree can be given conditional to an arbitrary node set (instead of V ). NW (U ) denotes N (U ) \ W , and dW (v) is jNW (v )j. An
acronym MIS will be used for a maximum independent set. The independence of a graph G is the cardinality of a MIS in G, and denoted (G). Our local search technique is based on augmentation of an independent set U by a node set I called an improvement. I is an improvement for U if G(I ) is connected and the symmetric di erence U 8 I is a larger independent set. An s{improvement is the one that adds s and removes s 0 1 from U , and a solution is s{optimal if it has no s{improvement. number
4
Approximation Algorithm
We adopt two main ingredients of Berman{Furer's algorithm for MAX IS{B : The rst method is to nd and apply every s{improvement to a current solution A, where s = O(k log n) for jV j = n (Note: because the size of improvements is O(log n) the algorithm runs in polynomial time, as one can show that the
number of \candidates" for s{improvements of any U V is smaller than n3s ). The second method is to nd a MIS in V n A, that is larger than A (Note: after failing with the rst method, G(V nA) is a degree 2 graph). The resulting solution thus satis es the two properties: (1) A is s{optimal, and (2) (G(V n A)) jAj. The properties (1) and (2) assure that ( 54 + k1 )jAj (G); moreover, this inequality is tight. To improve the performance, we will assure that the output satis es several additional properties that will enable us to prove a stronger inequality. Our algorithm (see Figure 1) has the following new features; 1. We add the rst Repeat{loop as the preprocessing. As a result, we reduce the problem, without loss of approximation quality, to the case where (a) all nodes have degree 2 or 3, and nodes of degree 2 form an independent set (Branchy reduction). (b) (G) 21 jV j (Nemhauser{Trotter reduction). (c) for every node subset U V of size at most k, U is not a MIS in G(U [ N (U )) (Small Commitment reduction). 2. Initial solutions are constructed in a special manner. First, all of the degree 2 nodes are collected in set A1 (after Branchy reduction, A1 is independent), and then, an independent set A2 is found, recursively, from G(V n(A[N (A))). The union A1 [ A2 is used as an initial solution. This feature allows us to analyze the performance of the algorithm only for the case when at most 54 of degree 2 nodes belong to some MIS of G. 3. Before starting the local improvement procedure (in the second Repeat{ loop), a current solution is replaced by an independent set A of no smaller size s.t. G(V n A) is acyclic.
Branchy Reduction. A graph G is called branchy if every node has degree 2 or more, and no two degree 2 nodes are adjacent (the notion of branchy graphs and reduction was used before in the context of weighted feedback vertex set problem, see [3]). The branchy reduction of an instance graph G for MAX IS is described as follows: 1. While there is a node of degree 0 or 1, remove it (along with its neighbor, if any) from G and store in set S . 2. While there is a path (v1 ; v2 ; v3 ; v4 ) where both v2 and v3 have degree 2, remove v2 and v3 from G, insert edge fv1 ; v4 g and store the path in set P . The next lemma states that branchy reduction is indeed a lossless reduction.
Lemma 5. Let G0 = (V 0; E 0 ) be a branchy reduction of G = (V; E ), and let S and P be the sets formed in that reduction. Then, 1. For any independent set A0 for G0 , an independent set A for G can be constructed in linear time s.t. jAj = jA0 j + jS j + jP j. 2. There is an optimal independent set A for G s.t. A \ V 0 is an independent set for G0 with jAj 0 jS j 0 jP j elements.
Input:
A degree 3 graph G = (V; E ). An approximation of a maximum independent set.
Output:
Repeat
oldsize jV j Do Branchy reduction Do Nemhauser{Trotter reduction Do Small Commitment reduction until jV j = oldsize /* until no more reduction is applicable */ A1 fv 2 V jd(v ) = 2g A2 a recursively computed independent set in G(V n (A1 [ N (A1 )))
A
A1
[ A2
Repeat
oldsize jAj Do Acyclic Complement procedure Do all possible improvements of size 3k log n to Find an optimal solution A3 in G(V n A) If A3 larger than A then A A3 until jAj = oldsize
Fig. 1.
A
A 56 {Approximation Algorithm for MAX IS{3
Nemhauser{Trotter reduction. Nemhauser and Trotter [13] studied the so-
lution to linear program relaxation of MAX IS problem, and showed that any graph G(V ) can be reduced to G(V 0 ), without loss of approximation quality, s.t. (G(V 0 )) 21 jV 0 j.
Small Commitment reduction. This reduction repetitively looks for the following kind of situation: for some A V; jAj k, and A is a MIS for G(A [ N (A)). If found we proceed exactly like in Nemhauser{Trotter reduction. Acyclic Complement procedure. Given an independent set I in a connected degree 3 graph G = (V; E ), which is not K4 , this procedure nds an independent set I 0 in polynomial time s.t. 1) jI 0j jI j and 2) G(V n I 0) is acyclic (details omitted in this short version).
5
Algorithm Analysis and Approximation Ratio
5.1 Node Type Classi cation and Normalization To analyze the structural properties of approximate solutions produced by our algorithm, it will be convenient to compare the relative sizes of node subsets de ned by a solution and one xed optimal solution. Given a graph G = (V; E ), which is obtained from an input graph by preprocessing, V is partitioned into four sets A; B; C; and D, with the following interpretations:
{ A[C is an independent set constructed by our algorithm (A for \approximate
" solution). [ C is a maximum independent set (B for \best" solution and C for \common" portion). { D is V n (A [ B [ C ), i.e., the set of remaining nodes.
{B
Assuming that an approximate solution, A [ C , is 1{optimal, the following observations are easy to verify; 1. 2. 3. 4.
an A{node cannot be adjacent to A [ C and must be adjacent to B . a B {node cannot be adjacent to B [ C and must be adjacent to A. a C {node must be nonadjacent to A [ B [ C . a D{node must be adjacent to both A [ C and B [ C .
Based on the partition of V thus de ned, every node in D will be further classi ed according to its \connectivity pattern"; if a node is of degree 3 (2) and its three neighbors belong to node sets X; Y; and Z (X and Y ), respectively, we say the node has a type [XY Z ] ([XY ]). The use of preprocessing and the acyclic complement procedures as described in section 4 is to \normalize" an input graph and the nal solution:
Lemma 6. The node sets A; B; C; and D satisfy the following properties: Norm 1. C is maximal (given our solution, we consider a MIS with a maximum overlap).
2. B [ D, the complement of a solution A [ C , contains no cycles. 3. Degree 2 nodes form an independent set in graph G. 4. Every node in A has at least 2 neighbors in B . 5. There is no node of type [AB ]. Now because of Norm 5 each node in D is classi ed into one of the following
Norm Norm Norm Norm types:
degree 2. [AC ][BC ][CC ][CD] degree 3. [AAB ] [AAC ] [ABB ] [ABC ] [ABD] [ACC ] [ACD] [BBC ] [BCC ] [BCD] [CCC ] [CCD] [CDD] 5.2 Approximation Ratio of 65
Denote the cardinalities of the sets, A; B; C; and D by respective lower case letters, and set i = b 0 a. Also denote the number of degree 2 nodes in a set X by x2 . The number of D{nodes of type [XY Z ] ([XY ]) is denoted [xyz ] ([xy ]). The (in)equalities given in Figure 2 will be proven in the subsections below. Assume for now that all these (in)equalities hold. Summing them up with the 6 12 16 ; 13 ; 1; 133 ; 134 ; 0 133 ; and 13 , yields the respective multiplicative factors of 137 ; 13 inequality k+1 1 1 (a + c) 5 i + (a2 + c2 + 27[aab] + 6[abb] + 10[abd] + 33[aac] + 6[acc] k 13 13 +10[acd] + 13[cdd] + 6[bbc] + 20[ac] + [bc] + 14[cd])
k +1 c 3[bc] + 2[bbc] + [bcd] + [cdd] + [cd] + [bcc] + [cc] 0 (2[acc] + [aac] + k [abc] + [acd] + [ac]) k +1 c [bbc] + [bc] + [bcc] + [cc] + 1 ([bcd] + [cdd] + [cd] + [ccd]) Ineq 2. k 2 1 Ineq 3. c i + [ccc] + [acc] + [aac] + [ac] + [cc] + ([cdd] + [ccd] + [acd] + [cd]) 2 k +1 Ineq 4. a 3 i + a + 2[ aab ] + 2[ aac ] + [ abb ] + [ abd] + [acc] + [acd] + [abc] + [ac] 2 k Ineq 5. d c + i Ineq 6. b2 = 3i + a2 + ([aab] + 2[aac] + [acc] + [acd] + [ac]) 0 ([abb] + 2[bbc] + [bcc] + [bcd] + [bc]) Ineq 7. 3c 0 c2 = 3[ccc] + 2([acc] + [bcc] + [ccd] + [cc]) + [aac] + [abc] + [bbc] + [acd] + [bcd] + [cdd] + [ac] + [bc] + [cd] 1 Ineq 8. a2 + [bc] + [cc] 4 (b2 + c2 + [ac] + [cd]) Ineq 1.
Fig. 2.
List of inequalities relating node set sizes
Since all the variables are nonnegative, it follows that
k+1 1 (a + c) 5 i: k 13 Therefore,
jB [ C j = b + c = a + c + i 1 + 1 < 6 + 1 k 1 jA [ C j a + c a+c 5 5k k+1 5 13
and the desired ratio is obtained.
Proof of Ineq 1. The method of \local accounting" used here is also applied to proving Ineq 2 and 4. In a nutshell, we assign potential to every node in C [ D so that the inequality in question states that the total potential of C [ D cannot be positive. We will partition C [ D into \connected components" to study each of them one at a time. A component with positive potential will either contain an s{improvement, or a k{commitment, or we will be able to \factor it out" and reduce the problem to a smaller one. There are 14 di erent D{node types that are adjacent to C ; for the sake of the proof we name them as follows:
{
enforcers solitaires: [BC ]; [BBC ] half-solitaires: [BCD]; [CDD]; [CD] pairs: [BCC ]; [CC ] absorbers: [ACC ]; [AAC ]; [ABC ]; [ACD ]; [AC ]
{ { others: [CCC ]; [CCD]
We will view pairs as \edges" connecting their neighbors in C ; in this manner we divide C into connected components. Each enforcer will have a weight, which is initially 1; larger weights result from reductions that \factor out" a connected components. The potential is assigned to nodes of C [ D as follows:
{ enforcer of type [ ]: c 0 wk01 where c is the coecient of [ ] in Ineq 1 and w is the weight. { absorber of type [ ]: c (note that it is negative) { C {node: 0 k+1 k { other nodes: 0 Now Ineq 1 is equivalent to p(C [ D) 0.
De nition 7. 1. For a node set Y , p(Y ) denotes the sum of potentials of nodes in Y . 2. If U D then IU = U [ NC (U ); if set U is independent, IU is the improvement candidate de ned by U . 3. If U D then w(U ) is the maximum weight of an independent subsest of U that consists of enforces only; given X C , such a subset would provide \the best candidate" to de ne an improvement containing X . 4. For a subset X of C de ne Ne (X ) to be the set of enforcers adjacent to X and ab(X ) to be the number of edges that join X with absorbers. We decompose p(C [ D) into a sum of expressions that refer to some connected fragments of C [ D. Let X be a connected component of C de ned by pairs. Let Ne (X ) be the set of enforcers adjacent to X and ab(X ) be the number of edges from X to adjacent absorbers. It is easy to see that p(C [ D) is the sum, over the components X of C , of
pX = p(X [ Ne (X )) 0 ab(X ): If no pX is positive we are done. Otherwise, we apply a case analysis to positive pX 's; to classify the cases, we de ne wX = w(Ne (X )). First we consider the cases when wX k. Let x = jX j. Obviously, Ne (X ) contains at least x 0 1 pairs. If Ne (X ) contains x + 1 (or more) pairs, these pairs, together with X , form a k{improvement, a contradiction. We get a similar contradiction if Ne (X ) contains x pairs and a (half-)solitaire, or x 0 1 pairs and either of: 3 half-solitaires, one solitaire and one half-solitaire, two solitaires, or two nonadjacent half-solitaires. If Ne (X ) consists of pairs only, and jNe (X )j x then pX < 0. As a conclusion we consider only the cases when Ne (X ) consists of x 0 1 pairs and either one of: a solitaire, or two half-solitaires adjacent to each other. Consider rst the case when Ne (X ) consists of x 0 1 pairs and a solitaire of type [BC ] (i.e., with coecient 2 0 (w 0 1)=k). We will remove X , Ne (X ) and the adjacent edges from consideration and assign the potential pX (if positive) to nodes of type [CCC ] or [CCD] that are adjacent to X . Below we analyze why it works. As a technical preliminary, observe that at least two edges go from X to nonenforcers: if X = fug for some u, then u has three neighbors as a neighbor of a degree 2 node (Norm 3), hence two non-enforcer neighbors; otherwise (X ,pairs) is a tree with at least two leaves and each leaf has a non-enforcer neighbor, in
particular, a leaf adjacent to [BC ] has three neighbors|one pair, one [BC ] and one non{enforcer. Next, one can easily compute that pX = 20wX =k 0ab(X ). Thus if ab(X ) = 2, pX < 0 and we are satis ed. If ab(X ) = 1, then an edge goes from X to a [CCC ] or [CCD] node, say u. If u is of type [CCC ], then after removing X [ Ne (X ) from considration u appears to be [CC ], a pair. As we observed above, pairs and other enforcers lead to small improvements if adjacent to the same component of C in excessive numbers. Assume that after removing X [Ne (X ), a set I C [Ne (C ) [ fug becomes an improvement. If u 62 I , then I was an improvement even before the removal, otherwise I [ X [ Ne (X ) is an improvement. Thus an s-improvement containing u after the removal translates into (s + wX )-improvement that is \genuine". Therefore, after the removal we will measure the improvement size correctly if we give u the weight of 1 + wX . We also rede ne p(u) from 0 to 1 0 wX =k. This way the total potential is preserved, and the potential of u agrees with the formula for pairs. Observe that we can handle identically the case when u is of type [CCD]; now u appears to be a half-solitaire after the removal, and the potential of the half-solitaires has the same formula. In such cases we will say that u is promoted (from \other" to a pair or a half-solitaire). Finally, when ab(X ) = 0, X is adjacent to two \other" nodes and they both can be promoted. In the short version of this paper we skip the technicalities involved in the assignment of potential and weight in this case. The case when Ne (X ) consists of x 0 1 pairs and a solitaire of type [BBC ] is similar, only simpler. Now, pX = 1 0 wX =k 0 ab(X ) and at least one edge goes from X to non-enforcers. If ab(X ) > 0, then pX < 0, otherwise the edge mentioned above goes to an \other" node and we can perform a promotion. The case when Ne (X ) consists of x01 pairs and two half-solitaires adjacent to each other is very similar to the latter, but for one additional subcase|when no edge joins X with non-enforcers, i.e., when Ne (X ) = N (X ). Let S be a maximum independent set in X [ N (X ). If jS j > x, then S 8 X is a k -improvement|and a contradiction. If jS j = x, then X is a MIS for X [ N (X ). Since jX j = x k , such situation would be eliminated by the Small Commitment reduction, so again, a contradiction. Observe that it is not possible that this situation occurred only after we performed some removals|removing a component does not change the set of neighbors of another (only the classi cation of the neighbors may change). Now to prove Ineq 1 it remains to handle the case when a component X has weight wX k and positive potential pX . It is easy to see that in this case X [Ne (X ) contains an improvement. Using a similar reasoning to that of Berman and Furer [4] we can show that X [ Ne (X ) contains a (4k log n)-improvement. In a nutshell, we very easily nd such improvements if even 1=(k log n) proportion of Ne (X ) consists of (half-)solitaires: such an improvement is a path in X [ Ne(X ) that joins two solitaires via C -nodes and pairs. In the remaining case Ne(X ) contains roughly (1+1=k )x pairs and we can use Lemma 3.1 of [4]. [The detailed reasoning will be in the full version which contains an accounting necessitated
by our \weight" system. ]
Proof of Ineq 2. The proof of Ineq 2 is quite similar to that of Ineq 1, but simpler, as we do not use the promotions. It is omitted in this short version. Proof of Ineq 4. This inequality is proven in two stages. First, nodes in B with only one neighbor in A are called solitaires and those with exactly two neighbors are called pairs. Let b(1) and b(2) be their respective numbers. First we prove Lemma 8.
If there is no improvement of size
2b(1) + b(2)
maxf3k log n; 4k + 2k log ng,
k+1 a k
The proof is similar to that of Ineq 1, but simpler because we do not have half-solitaires, solitaires with potential 3 and absorbers, and we have only one type of \`other" node: [AAA]. The mechanism of promotions works as before, with one exception. Given a component X of A de ned by pairs that is adjacent to x 0 1 pairs and one solitaire (x = jX j), we must argue that this component is adjacent to at least one \other" node. When we dealt with components of C , this followed from the fact that the algorithm uses Small Commitment reduction, but now this argument is not applicable as X can have neighbors in D, and these cannot be promoted. Nevertheless, we still can obtain a contradiction. Assuming that X is not adjacent to any \other" node, we have jNB (X )j = x, consequently, C [ X [ B n N (X ) is an independent set of the same size as C [ B , but with a larger overlap with \our" independent set C [ A, and that contradicts the way we have chosen C [ B among maximum independent sets of G (Norm 1). Given Lemma 8, it suces to show that 2b(1) + b(2) = 3i + a2 + 2[aab] + 2[aac] + [abb] + [abd] + [acc] + [acd] + [abc] + [ac]: This fact follows simply by expanding the equality
Pu2A dB (u) = Pu2B dA(u).
Proofs of the remaining inequalities. Ineq 3 follows from the fact that our solution, A[C , is at least as large as the largest independent set in B [D. Because B [ D contains no cycles the latter can be estimated as B , plus all the nodes in D without neighbors in B [ D (types [CCC ]; [ACC ]; [AAC ]; [AC ] and[CC ]) plus half of the nodes in D without neighbors in B but with some in D (types [CCD]; [CDD]; [ACD] and[CD]). Note that D contains no odd{length cycles because of Norm 2. Ineq 5 follows from the fact that we use Nemhauser{Trotter reduction, consequently, the size of a MIS, a + i + c, does not exceed one half of the total number of nodes, i.e., 2a + i + c + d.
P
P
Ineq 6 is an equality obtained from u2A dB (u) = u2B dA (u) and b = a + i. Similarly, Ineq 7 is an equality obtained from u2C d(u) = u2C dD (u) = u2D dC (u). Lastly, Ineq 8 uses the fact that we reduced the problem to the case when no MIS contains more than 4=5 of degree 2 nodes (and Norm 5).
P
P
P
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