Bounds for minimum feedback vertex sets in distance
graphs and circulant graphs
Hamamache Kheddouci, Olivier Togni
To cite this version:
Hamamache Kheddouci, Olivier Togni.
Bounds for minimum feedback vertex sets in
dis-tance graphs and circulant graphs. Discrete Mathematics and Theoretical Computer Science,
DMTCS, 2008, 10 (1), pp.57–70.
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DMTCS vol. , 2008, 57–70
Bounds for Minimum Feedback Vertex Sets in
Distance Graphs and Circulant Graphs
Hamamache Kheddouci
1and Olivier Togni
21
LIESP, Universit´e Claude Bernard Lyon1, 843, Bd. du 11 novembre 1918, 69622 Villeurbanne Cedex France, hkheddou@bat710.univ-lyon1.fr
2
LE2I, UMR CNRS 5158, Universit´e de Bourgogne, BP 47870, 21078 Dijon Cedex, France olivier.togni@u-bourgogne.fr
received april 10, 2006,revised july 25, 2007,accepted february 11, 2008.
For a setD⊂Zn, the distance graphPn(D)hasZnas its vertex set and the edges are between verticesiandjwith
|i−j| ∈D. The circulant graphCn(D)is defined analogously by considering operations modulon. The minimum
feedback vertex set problem consists in finding the smallest number of vertices to be removed in order to cut all cycles in the graph. This paper studies the minimum feedback vertex set problem for some families of distance graphs and circulant graphs depending on the value ofD.
Keywords:Feedback Vertex Set, Distance graph, Circulant graph.
1
Introduction
For a finite undirected graphG = (V, E), theminimum feedback vertex setproblem is to find a set of vertices inGof minimum cardinality whose removal induces a forest. We shall denote byV a Feedback Vertex Set (FVS for short) and by F(G)the minimum of the cardinalities over all the FVS:F(G) =
min{|V|, V is a FVS}. Aminimum feedback vertex set(MFVS) is a FVS of cardinalityF(G).
This parameter has received much attention since finding a MFVS of a graph has a lot of applications in various areas such as combinatorial circuit design, deadlock prevention in computer systems, VLSI testing, artificial intelligence [BG94] and converter placement in optical networks [Tog00].
The MFVS problem is known to be NP-complete for general graphs [GJ79]. For undirected graphs, the best known algorithm found in [BG94] and [BBF99] is a2-approximation. A lot of work has also been done on finding exact values for specific graphs such as hypercubes [FLP00, Pik03], meshes and butter-flies [Luc98, CKK02, PK05], pyramids and mesh of trees [Luc03]. Recently, an interesting connection was made in [FGR02] between the MFVS problem and the acyclic coloring problem, which allows the authors to derive bounds for a number of families of graphs such as graphs of maximum degree3and4, planar graphs andk-trees. The reader is referred to [FPR99] for a rather complete survey on feedback sets.
In this paper, we study the MFVS problem on finite distance graphs and circulant graphs.
The (infinite)distance graphP(D)with distance setD ={d1, d2, . . . , dk}, wheredi being positive
integers, has the setN of integers as vertex set, with two distinct verticesi, j ∈ Nbeing adjacent if
and only if|i−j| ∈D. Thefinite distance graphPn(D)is the subgraph ofP(D)induced by vertices 0,1, . . . , n−1. Remark that there aren−dichords of lengthdi≤ninPn(D). Thus the number of edges
ofPn(D)is|E|=kn−Pki=1di. The study of distance graphs was initiated by Eggleton et al. [EES85].
A large amount of work has focused on colorings of distance graphs [EES90, Voi99, EK03].
Thecirculant graphcan be defined as follows: letnbe a natural number and letD={d1, d2, . . . , dk},
with1≤d1< d2< . . . < dk≤n/2. Then the vertex set of the circulant graphCn(D)is{0,1, . . . , n− 1} and the set of neighbors of the vertex i is {(i ±dj) modn, j = 1,2. . . , k}. Circulant graphs
(also called loop networks) have been widely studied for their interesting properties in terms of diameter, symmetry and connectivity. A survey paper about loop networks is given in [BCH95].
To simplify, we will sometimes omit the brackets thus writingPn(d1, d2, . . . , dk)(Cn(d1, d2, . . . , dk),
respectively) instead ofPn({d1, d2, . . . , dk})(Cn({d1, d2, . . . , dk}), respectively).
Notice that the distance graphPn(D)is a subgraph of the circulant graphCn(D). Using this fact, we
will first find feedback vertex sets for finite distance graphsPn(D), for different values ofDin Section 2
and then transform these results to find feedback vertex sets for circulant graphsCn(D)with the same
setsDin Section 3.
Our main results are summarized in Table 1.
F(Pn(D)) F(Cn(D))
D Lower bound Upper bound Lower bound Upper bound
{1,2} ⌈n−2 3 ⌉ ⌊ n+4 3 ⌋ {1, t} ⌈n−t 3 ⌉ ⌈ n−2 3 ⌉ ∗ ⌈n+1 3 ⌉ ⌈ n+t 3 ⌉+ 1 {1,2,4} 4⌊n 8⌋ ⌈ n 2⌉ 4⌊ n 8⌋ ⌈ n+4 2 ⌉ {1, s, t} ⌈2n−s−t 5 ⌉ ⌈ n 2⌉ ⌈ 2n+1 5 ⌉ ⌈ n+t 2 ⌉ {1,2,3, . . . , t} ⌊ n t+1⌋(t−1), ifn≡0,1 mod (t+ 1) n−2⌊ n t+1⌋ −2, otherwise n−2⌊ n t+1⌋ −1∗∗ n−2⌊ n t+1⌋∗∗
∗Proposition 3 gives a better result in some cases.∗∗n−2⌊ n
t+1⌋+ 1ifn≡0 mod (t+ 1).
Tab. 1:Bounds for the MFVS of distance graphs and circulant graphs
Before presenting our results, let us remember a lower bound found in [CKK02] that will be used all along this paper:
Lemma 1 ([CKK02]) LetGbe a graph of ordernand sizem, with maximum degree∆. Then
F(G)≥ m−n+ 1 ∆−1 .
Proof: LetG′be a graph obtained fromGby removingf vertices and all edges incident to them. The
graphG′cannot be acyclic ifm′ ≥n′i.e. iff <lm−n+1 ∆−1
m
. Thus any FVS must have at leastlm∆−1−n+1m
vertices. ✷
2
Feedback vertex sets of distance graphs
2.1
D
=
{
1
, t
}
Proposition 1 For any integert≥2,
F(Pn(1, t)) = 0 ifn≤t, n−t 2 if n2 ≤t < n, n+2 4 if n3 ≤t < n 2.
Proof:LetG=Pn(1, t). Ift≥nthenGis a path.
If n2 ≤ t < nthen the maximum degree ofGis 2 ifn = t+ 1and3 otherwise. Pt+1(1, t)is a
cycle, thus F(Pt+1(1, t)) = 1 =
1 2
. For n ≥ t+ 2, applying Lemma 1 with m = 2n−t−1
givesF(Pn(1, t)) ≥
n−t
2
. On the other hand, we construct a FVSV in this way: we skip thetfirst vertices and we pick alternately one on two of then−t last vertices (see Figure 1). More precisely, V ={t+ 2i, 0≤i <⌈n−t
2 ⌉}.
0 1 2 3 t−1 t t+1
0 removed vertices ⌈n−t
2 ⌉removed vertices
Fig. 1:Structure of the FVSV (non filled circles represent vertices in the FVS).
Note that thetfirst vertices ofGdo not induce any cycle and then−tremaining vertices have degree at most1inG\V. SoV is a FVS.
If n3 ≤t < n
2, letnibe the number of vertices of degreeiinG. It is clear that the first and the last
vertices ofGare of degree two, son2= 2. Moreover, vertices1,2, . . . t−1andn−t, n−t+ 1, . . . , n−2
are of degree3, thusn3= 2t−2. Therefore,n4=n−n3−n2=n−2t.
LetV be a MFVS ofG. Assume that the vertices ofV are removed fromGone by one in a certain order. Suppose that|V| = α+β, whereαis the number of vertices that were of degree4just before they were removed (i.e. each removed vertex contributes to decrease the number of edges by4) andβ is the number of vertices that were of degree less than4just before they were removed (i.e. each removed vertex contributes to decrease the number of edges by3or less). Notice that theseαvertices are of degree
4inGand form an independent set. There aren4=n−2tvertices of degree4inGthat are consecutive
on the graph, thusα≤ ⌈n−2t
2 ⌉.
Letn′ (m′, respectively) be the number of vertices (edges, respectively) ofG′ = G\V. We have
m′≤n′−1and thus2n−t−1−4α−3β≤n−α−β−1or equivalently,
3α+ 2β ≥n−t.
AsV is a MFVS, we wantα+β to be minimized subject to the above constraints. In order to do so we must maximizeα. Therefore the solution is given by settingα=⌈n−2t
2 ⌉andβ =⌈ n−t−3α 2 ⌉. Thus F(G)≥ ⌈n−2t 2 ⌉+⌈ n−t−3⌈n−2t 2 ⌉ 2 ⌉=⌊ n+2 4 ⌋.
Now, we have to construct a FVS of cardinality⌊n+2
4 ⌋: we begin, as for the caset < n ≤ 2t, by
skipping the tfirst vertices and picking alternately one on two of the next t vertices. For then−2t remaining vertices, ift is odd then we skip three vertices and pick one and so on for the lastn−2t vertices (see Parta)of Figure 2) and iftis even, we skip one vertex then pick one, then skip three and pick one repetitively until the end (see Partb)of Figure 2).
More precisely, V = {2i+t,0 ≤ i ≤ t−1
2 } ∪ {2t+ 3 + 4i,0 ≤ i ≤ ⌊
n−2t−4
4 ⌋} for todd and
V ={2i+t,0≤i≤ t−2 2 } ∪ {2t+ 1 + 4i,0≤i≤ ⌊ n−2t−2 4 ⌋}forteven. Thus|V|=t−1 2 + 1 +⌊ n−2t−4 4 ⌋+ 1 =⌊ n+2
4 ⌋for oddtand|V|=
t−2 2 + 1 +⌊ n−2t−2 4 ⌋+ 1 =⌊ n+2 4 ⌋ for event. t−1 t t+1 2t−1 2t 0 1 2 t−1 t t+1 2t−1 2t 0 1 2
Part a): t is odd
Part b): t is even 0 0 ⌊n−2t 4 ⌋ t 2 t+1 2 ⌊n−2t−2 4 ⌋+ 1
Fig. 2:Structure of the FVSV (non filled circles represent vertices in the FVS).
The setV is a FVS because the first block oftvertices does not induce any cycle inG′ = G\V,
vertices of the second block oft vertices are of degree 2 inG′ and the third block ofn−2tvertices
induces disjoint subgraphsP3. Moreover, eachP3consists in two vertices of even indices and one vertex
of odd index and each vertex of the second block has the same parity as(t+ 1). Hence, only the vertex of odd index of eachP3is linked by a chord with a vertex of the second block. ✷
Proposition 2 For anynand anyt, witht≥2andn≥t,
n−t 3 ≤F(Pn(1, t))≤ n−2 3 .
Proof:LetG=Pn(1, t). The lower bound comes from Lemma 1 asGis of ordern, size2n−t−1and
maximum degree at most4.
For the upper bound, we cut the proof in two parts:
Case 1:t6≡0 mod 3. LetV ={3j−1, 1≤j≤ ⌊n
3⌋}be the vertex set which we remove fromG. By
contradiction, suppose thatG′=G\V contains a cycleC.
Letibe the vertex ofCof minimum index. Then verteximust be adjacent to verticesi+ 1andi+t inC. We havei6≡2 mod 3andi+ 16≡2 mod 3thusi≡0 mod 3andi+ 2cannot be a vertex ofC. Therefore, vertexi+ 1must be adjacent to vertexi+t+ 1inCand thusC is a cycle of order4 (see Figure 3).
i i+1 i+t i+t+1
Fig. 3:The cycleC
Ast6≡0 mod 3, it turns out that eitheri+t≡2 mod 3ori+t+1≡2 mod 3, which is a contradiction. Thus in that case,V is a FVS ofGof cardinality|V|=⌊n
3⌋=⌈
n−2 3 ⌉.
Case 2: t≡0 mod 3. Letqbe the integer such thatt= 3q. We are going to construct a FVS with a periodic pattern of period2t:
LetV ={x∈W , x≤n−1}, withW =S⌈ n
2t⌉−1
j=0 (Aj∪Bj), where
Aj ={2tj+ 3i, 1≤i≤q}andBj ={2tj+ 3(q+i)−1, 1≤i≤q}.
Figure 4 gives an example of such a set.
Fig. 4:FVS inP24(1,6)(non filled circles represent vertices in the FVS)
First observe thatAj ∩Bj =∅ for anyj. Next, to prove thatV is a FVS, as for the previous case,
suppose thatG\V contains a cycleC. Letαbe the vertex ofCof minimum index. The neighbors of αon the cycleCareα+ 1andα+t; then the only cases to study are whenα 6∈ V,α+ 1 6∈ V and α+t6∈V. If, for somej,0≤j≤ ⌈n
2t⌉ −1, the vertexαis of the form:
1. α= 2tjthenα+t= 2tj+ 3q∈V.
2. α= 2tj+ 3ifor somei,1≤i≤qthenα∈V.
3. α= 2tj+ 3i+ 1for somei,0≤i≤q−1, thenα+ 2 = 2tj+ 3(i+ 1)∈V; the cycleCmust be C= (α, α+ 1, α+t+ 1, α+t, α). Butα+t+ 1 = 2tj+ 3(q+i) + 2 = 2tj+ 3(q+i+ 1)−1∈V, a contradiction.
4. α= 2tj+ 3i+ 2for somei,0≤i≤q−1thenα+ 1 = 2tj+ 3(i+ 1)∈V. 5. α= (2j+ 1)t+ 3ifor somei,1≤i≤q−1thenα+t= 2(j+ 1)t+ 3i∈V.
6. α= (2j+ 1)t+ 3i+ 1for somei,0≤i≤q−1thenα+ 1 = 2tj+ 3(q+i+ 1)−1∈V. 7. α= (2j+ 1)t+ 3i+ 2for somei,0≤i≤q−1thenα∈V.
Now we have to show that|V| ≤n
3
. By definition, we have|Aj| = |Bj| = q. Thus, on a period
of2t = 6qvertices, the feedback vertex set has cardinality2q = 2t
3. It remains to see that for the last
kvertices ofG(k < 2t), at mostk−23 of them are inV. Ifk ≤tthen⌈k−3
3 ⌉of thesekvertices are
inV since in each setAj, we skip the three first vertices and then we pick one on three repetitively. If
t < k < 2tthen the number of removed vertices is q+⌈k−t−2 3 ⌉ = ⌈
k−2
3 ⌉(since for thet+ 1first
vertices we removeqof them and for the lastk−t−1vertices, we skip one and then we pick one on three repetitively, thus removing⌈k−t−2
3 ⌉of them).
In both cases, we proved thatV is a feedback set of cardinality at mostn
3 =⌈n−2 3 ⌉, which proves the proposition. ✷ For the particular caset= 2, the previous proposition gives a sharp result:
Corollary 1 For anyn≥3,
F(Pn(1,2)) = n−2 3 .
We shall now give an algorithm to find another feedback set forPn(1, t). This new set has cardinality
closer to the optimal for specific values ofncompared tot.
Proposition 3 For any t ≥ 2, and any n > 3t, lett′ = 2⌈t
2⌉and let rbe the integer such thatn ≡
rmod 3t′,0≤r <3t′. Then F(Pn(1, t))≤ n−r 3 if0≤r≤t, ⌊n 3− t 6⌋+ 3 ift < r <3t ′.
Proof:Letn= 3t′q+r, letG=P
n(1, t), letG′=P3t′q(1, t)and letG′′=Pr(1, t).
The idea is to construct a periodic feedback setV′of period3t′ for the3t′qfirst vertices ofG(i.e. for
G′) and to use Proposition 1 to complete the FVS for the remainingrvertices (i.e. forG′′). We now
distinguish two cases depending on the parity oft.
Iftis even thent′=t. The FVS is constructed as follows: cut the set of3tconsecutive vertices starting
from vertex 0 in three blocks of sizet, skip the first block, pick alternately one on two vertices of the second block and then again pick alternately one on two vertices of the last block, but starting from the second vertex of the block. Repeat the process for the next block of3tvertices, and so on. An illustration of this method is given in Parta)of Figure 5.
Precisely, the FVS is defined by: V′ = q−1 [ j=0 {3tj+t+ 2i,0≤i≤ t−2 2 } ∪ {3tj+ 2t+ 2i+ 1,0≤i≤ t−2 2 } .
0 1 t−1 t t+1 2t−1 0 1 t−1 t t+1 2t−1 2t 3t 4t−1 2t 2t+1 2t+2 2t+3 3t+2 3t+3 3t−1 t/2 b) a) 0 0 0 t/2 (t+1)/2 (t+1)/2 0
Fig. 5:Structure of the FVSV′for eventon Part a) and for oddton Part b) Hence, the cardinality ofV′satisfies|V′| = 2(t−2
2 + 1)q=tq =t
′q = n−r
3 . To see thatV′is a FVS,
observe that in each block of3tvertices, the first blockB1oftvertices does not induce any cycle, and
there is only one edge between the next blockB2 oftvertices and the last blockB3oftvertices since
the removed vertices are of different parity inB2andB3. Thus, no cycle inG\V′can cross a block of
3tvertices. Furthermore, as the vertices of the blockB3are of degree one except the first vertex who has
degree2, then they cannot form a cycle with thetfirst vertices of the next block of3tvertices.
Iftis odd thent′=t+ 1. The FVS is constructed as follows: take a block of3t′ = 3t+ 3consecutive
vertices starting at vertex 0, skip thetfirst ones, pick alternately one on two of the nexttvertices, skip three vertices and then again pick alternately one on two of the lasttvertices. Repeat the process for the next block of3t′vertices, and so on. An illustration of this method is given in Partb)of Figure 5.
Precisely, the FVS is defined by: V′= q−1 [ j=0 {3t′j+t+ 2i,0≤i≤ t−1 2 } ∪ {3t ′j+ 2t+ 2i+ 3,0≤i≤t−1 2 } .
Hence, the cardinality ofV′satisfies|V′|= 2(t−1
2 +1)q=t
′q=n−r
3 . Among the three vertices between
the first2tvertices (blockB1andB2) and the lasttvertices (blockB3) of any block of3t′vertices, only
one has degree 4 (but with two pendant edges incident to him) and the two other have degree 1. Moreover, removed vertices from the blockB2andB3have the same parity. Thus, as the length of the chord is odd,
there is no chordal edge between blockB2and blockB3. Consequently, no cycle can cross a block of3t′
vertices.
For0≤r≤t,G′′is a path and thusV′′=∅is a FVS. Fort < r≤3t, by Proposition 1, there exists a
FVSV′′ofG′′. Iftis odd and3t+ 1≤r≤3t′−1 = 3t+ 2then we add the one or two last vertices of
G′′in the FVSV′′.
To see that the setV =V′∪V′′is a FVS ofG, it remains to show that no cycle is formed when making
the union ofG′\V′andG′′\V′′. In fact, for both even and oddt, among the lastt−1vertices ofG′,
the remaining vertices inG′\V′are of degree 0. Therefore, only one chordal edge will link a vertex of
If0≤r≤tthenV satisfies|V|= n−3r. Ift < r≤2tthen|V|=n−r 3 +⌈ r−t 2 ⌉ ≤ 2n−2r+3r−3t+6 6 ≤ n 3 − t 6+ 1≤ ⌊ n 3 − t 6⌋+ 3. If2t < r <3t′then|V| ≤ n−r 3 +⌊ r+2 4 ⌋+ 2≤ 4n−4r+3r+6 12 + 2≤ n 3 − t 6+ 3. As|V|is an integer, then|V| ≤ ⌊n 3 − t 6⌋+ 3. ✷
Remark 1 The result of Proposition 3 is sharp whenn ≡ tmod 3t′, witht′ = 2⌈t
2⌉. We have then
F(Pn(1, t)) = n−3t.
2.2
D
=
{
1
, s, t
}
Proposition 4 For any integersn, sandt, witht≤n−1and2≤s < t, 2n−s−t 5 ≤F(Pn(1, s, t))≤ ln 2 m .
Proof: LetG =Pn(1, s, t). Notice that, for anyd ∈ D, there are exactlyn−dchords of lengthdin
Pn(D). Thus the lower bound comes from Lemma 1 asGis of ordern, size3n−s−t−1and maximum
degree6.
To prove the upper bound we will discuss two cases. In each one, we remove a vertex set fromGwhich generates a feedback.
Case 1: sortis odd. LetV = {2j+ 1, 0 ≤j ≤ ⌊n−2
2 ⌋}be the vertex set which we remove from
G. As a chord of odd length always joins two vertices of different parity and as we remove all vertices of odd index, then the subgraph induced byV \V is either an independent set (if bothsandtare odd) or a disjoint union of paths. SoV is a FVS and|V|=⌊n
2⌋.
Case 2:sandtare even. For eachk,1≤k≤n t
, ifkis even we remove fromGthe setVk={u= 2j+ 1 + (k−1)t, u≤n−1, 0≤j ≤ t−2
2 }, otherwise we removeVk ={u= 2j+ (k−1)t, u≤
n−1, 0≤j ≤t−2
2 }. See Figure 6 for an illustration whens= 4andt= 6.
Fig. 6:A FVS ofPn(1,4,6).
One can see that, assandtare even, the chords of lengthsortofGalways join vertices of the same parity. Moreover, as we remove alternately odd and even vertices by blocks oftvertices, all chords of lengthtare removed and all chords of lengthsbetween two consecutive blocks oftvertices are removed too. Thus in each block we obtain a forest and since chords of length one are all removed except those between some consecutive blocks, the setV =S
kVkis a FVS.
Notice that we have|Vk|= 2t and|V| ≤ ⌈n2⌉since in each block oftvertices, we alternately remove
Proposition 5 For any integern≥4 4jn 8 k ≤F(Pn(1,2,4))≤ ln 2 m .
Proof: To prove the lower bound, letV be any FVS ofG=Pn(1,2,4). Consider the first6vertices of
G. Note that three consecutive vertices inGform a cycle. Thus we have two cases to study: if only two vertices over the six are inV, then the only possibility is to remove vertices1and4(see Figure 7). But in this case, the two following vertices (6 and 7) must be inV, otherwise a cycle is produced. Then, among the 8 first vertices ofG, 4 of them are inV. If three or more vertices among the first 6 one are inV then half or more than half of the vertices are removed on this part ofG. Therefore, on the first six or eight vertices ofG, at least half of them are inV. Next, iterating this process on the six next vertices ofGand so on until the end of the graph, we obtain the desired lower bound.
Fig. 7: The graphP6(1,2,4)with two vertices removed. The three configurations on the top of the figure produce a
cycle.
The upper bound follows from Proposition 4. ✷
2.3
D
=
{
1
,
2
,
3
, . . . , t
}
Proposition 6 For anyn≥5andt≤ n−1 2 , F(Pn({1,2,3, . . . , t})) = j n t+1 k (t−1) ifn≡0,1 mod (t+ 1), n−2j n t+1 k −2 otherwise.
Proof:LetG=Pn({1,2, . . . , t})and letn=p(t+ 1) +q, with0≤q < t+ 1andp≥1.
The subgraphsGi,0≤i≤p−1ofGinduced by vertices{(t+ 1)i+j, 0≤j≤t}andGpinduced
by vertices{p(t+ 1) +j, 0 ≤j ≤ q−1} are cliques. So any FVS ofGmust contain at leastt−1
vertices of eachGi,0≤i≤p−1and at leastq−2vertices ofGpifq≥2. Thus, ifq≤1thenF(G)≥
p(t−1) =⌊ n
t+1⌋(t−1), otherwise,F(G)≥p(t−1) +q−2 =p(t+ 1) +q−2p−2 =n−2⌊
n t+1⌋ −2.
To show the equality, we give a FVS with the desired number of vertices (see Figure 8): V = {i+
k(t+ 1),2≤i≤t,0≤k≤p−1} ∪ {p(t+ 1) +i,2≤i≤q−1}.
By construction, there is only one chord between two consecutive blocks of(t+ 1)vertices (a chord of lengthtbetween the second vertex of the block and the first vertex of the next block). Thus the graph G\V is a path. Moreover, we have|V|=p(t−1)ifq≤1and|V|=p(t−1) +q−2ifq≥2. ✷
1 2 3 t t+1 2t+2 0
Fig. 8:A FVS ofPn(1,2,3, . . . , t)
Remark 2 For the general distance graph F(Pn(d1, d2, . . . , dt))and for any n ≥ 5 and 1 ≤ d1 <
d2 < . . . < dt, withdt ≤n−1and at most onedieven, we have thatF(Pn(d1, d2, . . . , dt)) ≤n2. Effectively, if we remember that a chord of odd length always joins vertices of different parity, we can see that the setV ={2j+ 1,0≤j≤ n−2
2 }is a feedback vertex set.
3
Feedback vertex sets of circulant graphs
In this section, we shall always assume thatt≤n
2.
Proposition 7 Letnbe an integer and letD⊂Znandm≥maxd∈D{d}. Then
F(Pn(D))≤F(Cn(D))≤F(Pn−m(D)) +m.
Proof: The lower bound is trivial asPn(D)is a partial subgraph ofCn(D). For the upper bound, if we
cutmconsecutive vertices ofCn(D), we obtain a graph isomorphic toPn−m(D). Asm≥maxd∈D{d},
no chord can cross this block ofmvertices. Thus it is sufficient to add thesemvertices to a MFVS of Pn−m(D)to obtain a FVS of the circulant graphCn(D). ✷
3.1
D
=
{
1
, t
}
Proposition 8 For anyn≥5andt < n
2, n+ 1 3 ≤F(Cn(1, t))≤ ⌈n+t 3 ⌉+ 1 ift≡0 mod 3, ⌈n+t 3 ⌉ ift6≡0 mod 3.
Proof: LetG = Cn(1, t). The lower bound comes from Lemma 1 asG is of ordern, size 2n and
maximum degree 4. In order to prove the upper bound, we consider the circulant graph as a distance graph plus some vertices and edges: LetBbe the subgraph ofGinduced by verticest, t+ 1, . . . , n−1. B is the distance graphPn−t(1, t)and sincen−t > t, by Proposition 2, there exists a FVSVBof B
of size⌈n−t−2
3 ⌉. A FVSVG forGcan be constructed by adding toVB a subsetN of the remainingt
vertices0,1, . . . , t−1:N={j, 0≤j≤t−1,(j+t)6∈VB}. ThenVG=VB∪N.
B\VB induces a forest inGand by the choice ofN, there is no edge between a vertex of the first
block oftvertices and a vertex of the second block oftvertices. Moreover, as it can be seen in the proof of Proposition 2, no two consecutive vertices ofPn−t(1, t)are inVB. Thus, among thetfirst vertices of
G, at least one of any two consecutive vertices is inN. Consequently, there is no possibility to form a cycle since thetfirst vertices ofGbecome of degree at most one.
Moreover, by the construction of VB (see proof of Proposition 2), if t 6≡ 0 mod 3 then we have |N| = t−t−2 3 = ⌈2t 3⌉. Thus |VG| ≤ ⌈n−3t−2⌉+ 2t 3 ≤ ⌈n+t 3 ⌉. If t ≡ 0 mod 3 then|N| = t−t−3 3 =⌈2t 3⌉+ 1. Thus|VG| ≤ ⌈ n+t 3 ⌉+ 1. ✷ For the special caset = 2, the previous proposition gives a quasi optimal result. The exact value is determined in the following proposition:
Proposition 9 For anyn≥4, we have
F(Cn(1,2)) =
n+ 2 3
.
Proof: By Lemma 1,F(Cn(1,2))≥ ⌈n+13 ⌉. Observe that in casen= 3k+ 1or3k, we have⌈
n+1 3 ⌉=
⌈n+2
3 ⌉, thus Proposition 8 gives the result. So let us study the case wheren= 3k+ 2. The proof is done
by contradiction. Assume thatF(Cn(1,2))≤ ⌈n+23 ⌉ −1, i.e.F(Cn(1,2))≤k+ 1. Then there exists a
subset of vertices, sayS, such that the induced subgraphG\S is a forest. Give a clockwise orientation to the cycleC=Cn(1). LetS={s0, s1, . . . , sq−1}, with1≤q≤k+ 1such that these vertices hold in
this order onC. Lets+
i (s
−
i , respectively) be the successor (predecessor, respectively) of vertexsiinCn
with respect to the orientation (indices are taken moduloq). Denote byCi the part of cycle given by the segment(s+i Cs
−
i+1). Suppose first that there exists0 ≤
j ≤ q−1, such that the edgesjsj+1 ∈ E(Cn), i.e. such that|Cj| = 0. LetPn−2(1,2)be given by
Cn(1,2)\ {sj, sj+1}. By Proposition 2,F(Pn−2(1,2)) = ⌈n−43 ⌉ = k. SoF(Cn(1,2)) ≥ k+ 2, a
contradiction.
Now suppose that for all0 ≤ i ≤ q−1, |Ci| > 0. Observe that there does not exist Cj, with 0 ≤ j ≤ q−1, such that |Cj| ≥ 3, otherwise we obtain a cycle of size 3. We deduce that for each
i,0≤i≤q−1, we have1≤ |Ci| ≤2. It means that it is possible, in this case, to construct a cycle which
passes by eachCiby taking all vertices of each segment: letxiandyibe the vertices ofCi, wherexi =yi
if|Ci| = 1andyi = x+i if|Ci| = 2. Then the cycle is given by(x0, y0, x1, y1, . . . , xq−1, yq−1, x0).
ThereforeG\Sis not a forest, a contradiction. Then any MFVS has cardinality at leastk+ 2 =⌈n+2 3 ⌉.
Thus, Proposition 8 givesF(Cn(1,2)) =⌈n+23 ⌉. ✷
Ifn= 2pis even andt= n
2 =p, then the circulant graphC2p(1, p)is cubic. In that case, it is easy to
compute the cardinality of a MFVS.
Proposition 10 for anyp≥2,
F(C2p(1, p)) = p+ 1 2 . Proof:By Lemma 1,F(C2p(1, p))≥ p+1 2 .
Ifpis even, then the setV ={0} ∪ {2j+ 1,0≤j≤ p2−1}is a FVS of cardinality p2+ 1 =⌈p+12 ⌉. Ifpis odd, then the setV ={2j,0≤j ≤p−12 }is a FVS of cardinalityp−12 + 1 =⌈p+12 ⌉. ✷
3.2
D
=
{
1
, s, t
}
Proposition 11 For any integersn, sandt, witht≤ n−1
2 and2≤s < t, 2n+ 1 5 ≤F(Cn(1, s, t))≤ n+t 2 .
Proof:This proposition immediately follows from Lemma 1, Proposition 4 and Proposition 7. ✷
Corollary 2 For any integern≥9,
4jn 8 k ≤F(Cn(1,2,4))≤ n+ 4 2 .
Proof:The lower bound follows from Proposition 5 asPn(1,2,4)is a subgraph ofCn(1,2,4). The upper
bound is given by Proposition 11 withm=t= 4. ✷
3.3
D
=
{
1
,
2
, . . . , t
}
Proposition 12 For anyn ≥5and3 ≤t ≤ n−1
2 , letpandqbe integers such thatn =p(t+ 1) +q,
q < t+ 1. Then
F(Cn(1,2, . . . , t)) =n−2p+ 1, ifq= 0;
n−2p−1≤F(Cn(1,2, . . . , t))≤n−2p, ifq≥1.
Proof: LetG=Cn(1,2, . . . , t)and letn= (t+ 1)p+q, withq < t+ 1. We cut up the vertices ofG
successively intop+ 1sets of vertices, in other words we have
Bi={(i(t+ 1), i(t+ 1) + 1, . . . ,(i+ 1)(t+ 1)−1}, for0≤i≤p−1, and
Bp={p(t+ 1), p(t+ 1) + 1, . . . , p(t+ 1) +q−1}.
As|Bi| ≤t+ 1, any subgraph ofGinduced byBi, for0 ≤ i ≤ p, is a clique. Thus any triplet of
vertices of this subgraph forms a cycle. So a FVS in this case must contain at leastt−1vertices of each setBi, for0≤i≤p−1andq−2vertices ofBpifq≥2.
However, there is still a ”big” cycle since if it remains two vertices in each blockBi(one vertex inBp
ifq= 1), then there is at least an edge between any two successive blocks. Thus any FVS must contain an extra vertex.
Consequently, ifq = 0then we haveF(Cn(1,2, . . . , t))≥p(t−1) + 1 =n−2p+ 1and ifq≥1,
thenF(Cn(1,2, . . . , t))≥p(t−1) + max(q−2,1) + 1≥p(t+ 1) +q−2p−1 =n−2p−1.
We construct a FVSV by taking thet−1first vertices of eachBi, with0 ≤ i ≤p−1, and theq
vertices ofBp. By construction, there is only one chord between two consecutive blocks (the chord of
lengthtjoining the last vertex of the first block with the last but one vertex of the next block). Ifq6= 0
thenG\V is the pathP= (t−1, t,2t,2t+ 1,3t+ 1,3t+ 2, . . . , p(t+ 1)−2, p(t+ 1)−1). Otherwise, ifq= 0thenG\V consists in the cycleC=P∪ {p(t+ 1)−1, t−1}. In this case, we add any vertex of the cycleCtoV.
Observe that ifq= 0then|V| ≤p(t−1)+1 =n−2p+1and ifq≥1then|V| ≤p(t−1)+q=n−2p.
Remark 3 From Proposition 7 and Remark 2, the following can be easily obtained: for anyn≥5and 1≤d1< d2< . . . < dt, withdt≤n−12 ,
i) ifnis even and if alldiare odd, thenF(Cn(d1, d2, . . . , dt))≤n2,
ii) ifnis odd and if alldiare odd, thenF(Cn(d1, d2, . . . , dt))≤n+2dt,
iii) ifnis even and if there exists only one evendi, with1≤i≤t, thenF(Cn(d1, d2, . . . , dt))≤
n+di
2
.
4
Conclusion
We have derived several bounds for the cardinality of a minimum feedback vertex set of distance and circulant graphs. The approximation ratios of our algorithms are in many cases equal or asymptotically close to one for circulant graphs and for distance graphs with chords of length at most half the number of vertices.
For isomorphism reasons, our results concerning circulant graphs defined by the setD = {1, t}can also be applied to many other circulants of degree4. For instance, the following isomorphism is easy to observe: ifgcd(d1, n) = 1orgcd(d2, n) = 1then there existstsuch thatCn(d1, d2)≡Cn(1, t).
Acknowledgements
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