Large Feedback Arc Sets, High Minimum Degree Subgraphs, and Long Cycles in Eulerian Digraphs

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1 Combiatorics, Probability ad Computig (013, c Cambridge Uiversity Press 013 doi: /s Large Feedback Arc Sets, High Miimum Degree Subgraphs, ad Log Cycles i Euleria Digraphs HAO HUANG 1,JIEMA, ASAF SHAPIRA 3, BENNY SUDAKOV 4 ad RAPHAEL YUSTER 5 1 Departmet of Mathematics, UCLA, Los Ageles, CA 90095, USA ( huaghao@math.ucla.edu Departmet of Mathematics, UCLA, Los Ageles, CA 90095, USA ( jiema@math.ucla.edu 3 School of Mathematics, Tel Aviv Uiversity, Tel Aviv 69978, Israel ( asafico@tau.ac.il 4 Departmet of Mathematics, ETH, 809 Zurich, Switzerlad ad Departmet of Mathematics, UCLA, Los Ageles, CA 90095, USA ( bsudakov@math.ucla.edu 5 Departmet of Mathematics, Uiversity of Haifa, Haifa 31905, Israel ( raphy@math.haifa.ac.il Received 1 February 01; revised 7 August 013; first published olie 1 September 013 A miimum feedback arc set of a directed graph G is a smallest set of arcs whose removal makes G acyclic. Its cardiality is deoted by β(g. We show that a simple Euleria digraph with vertices ad m arcs has β(g m / + m/, ad this boud is optimal for ifiitely may m,. Usig this result we prove that a simple Euleria digraph cotais a cycle of legth at most 6 /m, ad has a Euleria subgraph with miimum degree at least m /4 3. Both estimates are tight up to a costat factor. Fially, motivated by a cojecture of Bollobás ad Scott, we also show how to fid log cycles i Euleria digraphs. 010 Mathematics subject classificatio: Primary 05C0 Secodary 05C38 1. Itroductio Extremal problems related to the existece of various types of cycles i graphs are some of the most basic ad well-studied problems i graph theory. Somewhat surprisigly, Research supported by a UC Dissertatio Year Fellowship. Supported i part by ISF grat 4/11 ad a Marie Curie CIG grat Research supported i part by NSF grat DMS , by AFOSR MURI grat FA ad by a USA Israel BSF grat.

2 860 H. Huag, J. Ma, A. Shapira, B. Sudakov ad R. Yuster i may cases it turs out that problems that are very easy to solve i the settig of udirected graphs become much more challegig i the settig of digraphs. A prime example is the well-kow Caccetta Häggkvist cojecture [4] (see below for more details. I some other cases, a result that holds for udirected graphs might fail completely for geeral digraphs, ad so it is atural to fid families of digraphs for which the result still holds. Motivated by a cojecture of Bollobás ad Scott [3], we cosider i this paper extremal problems of the above two types. It is well kow that a udirected graph G with vertices ad m edges has a subgraph with miimum degree at least m/, ad so if m such a G also cotais a cycle of legth at least m/ + 1. It is atural to ask whether results of this type ca be exteded to digraphs. However, it turs out that these statemets are ofte trivially false eve for very dese geeral digraphs. For istace, a trasitive touramet does ot cotai ay cycle, ad its subgraphs always have zero miimum i-degree ad outdegree. Therefore, i order to obtai meaigful results as i the udirected case, it is ecessary to restrict to a smaller family of digraphs. A atural cadidate oe may cosider is the family of Euleria digraphs, i which the i-degree equals the out-degree at each vertex. I this paper we ivestigate several atural parameters of Euleria digraphs, ad study the coectios betwee them. I particular, the parameters we cosider are miimum feedback arc set, shortest cycle, logest cycle, ad largest miimum degree of ay subgraph. Throughout this paper, we always assume the Euleria digraph is simple, i.e., it has o multiple arcs or loops, but arcs i differet directios such as (u, v ad (v, u are allowed. For other stadard graph-theoretic termiology ivolved, the reader is referred to []. A feedback arc set of a digraph is a set of arcs whose removal makes the digraph acyclic. Give a digraph G, deote by β(g the miimum size of a feedback arc set. Computig β(g ad fidig a correspodig miimum feedback arc set is a fudametal problem i combiatorial optimizatio. It has applicatios i may other fields such as testig of electroic circuits ad efficiet deadlock resolutio (see, e.g., [8, 10]. However, computig β(g turs out to be difficult, ad it is NP-hard eve for touramets [1, 5]. Oe basic questio i this area is to boud β(g as a fuctio of other parameters of G, ad there are several papers (see, e.g., [6, 7, 11] studyig upper bouds for β(g of this form. However, much less is kow about lower bouds for β(g, perhaps because a geeral digraph could be very dese ad still have a small miimum feedback arc set. For example, a trasitive touramet has β(g = 0. Nevertheless, it is easy to see that ay Euleria digraph G with vertices ad m arcs has β(g m/, sice the arcs ca be decomposed ito a disjoit uio of cycles, each of legth at most, ad ay feedback arc set cotais at least oe arc from each cycle. I this paper we actually prove the followig much stroger lower boud for β(g, ad show that it is tight for a ifiite family of Euleria digraphs. Theorem 1.1. Every Euleria digraph G with vertices ad m arcs has β(g m / + m/. Furthermore,if m the there exists a Euleria digraph G with vertices ad m arcs with β(g =m / + m/.

3 Large Feedback Arc Sets ad Log Cycles i Euleria digraphs 861 As metioed earlier, may problems related to cycles i udirected graphs are much harder to solve i the settig of digraphs. Oe of the most famous problems of this type is the celebrated Caccetta Häggkvist cojecture [4]: every directed -vertex digraph with miimum out-degree at least r cotais a cycle with legth at most /r, which is ot completely solved eve whe restricted to Euleria digraphs (for more discussio, we direct the iterested reader to the surveys [9, 1]. I this paper we study the existece of short cycles i Euleria digraphs with a give order ad size. The girth g(g of a digraph G is defied as the legth of the shortest cycle i G. Combiig Theorem 1.1 ad a result of Fox, Keevash ad Sudakov [7] which coects β(g ad g(g for a geeral digraph G, we are able to obtai the followig corollary. Corollary 1.. Every Euleria digraph G with vertices ad m arcs has g(g 6 /m. We also poit out that the upper boud i Corollary 1. is tight up to a costat, sice the costructio of Theorem 1.1 also provides a example of Euleria digraphs with girth at least /m. A repeated applicatio of Corollary 1. gives a Euleria subgraph of the origial digraph G, whose arc set is a disjoit uio of Ω(m / cycles. Usig this fact we ca fid a Euleria subgraph of G with large miimum degree. Theorem 1.3. Every Euleria digraph G with vertices ad m arcs has a Euleria subgraph with miimum degree at least m /4 3. This boud is tight up to a costat for ifiitely may m,. I 1996, Bollobás ad Scott ([3], Cojecture 6 asked whether every Euleria digraph G with o-egative arc-weightig w cotais a cycle of weight at least cw(g/, where w(g is the total weight ad c is some absolute costat. For the uweighted case, i.e., w =1, this cojecture becomes: Is it true that every Euleria digraph with vertices ad m arcs cotais a cycle of legth at least cm/? Eve this special case is still wide ope after 15 years. A obvious cosequece of Theorem 1.3 is that every Euleria digraph cotais a cycle of legth at least 1 + m /4 3. This ca be slightly improved to 1 + m / 3 usig Theorem 1.1 ad the simple fact that ay digraph has a cycle of legth at least β(g/ (see Sectio 4. Whe the digraph is dese, i.e., m = c, our theorem provides a cycle of legth liear i, which partially verifies the Bollobás Scott cojecture i this rage. However, observe that whe m is small, i particular whe m = o( 3/, Theorem 1.3 becomes meaigless. Nevertheless, we ca always fid a log cycle of legth at least m/ + 1, as show by the followig propositio. 1 Propositio 1.4. Every Euleria digraph G with vertices ad m arcs has a cycle of legth at least 1+ m/. Together with Theorem 1.1 ad the fact that ay digraph 1 This propositio was also obtaied idepedetly by Jacques Verstraete.

4 86 H. Huag, J. Ma, A. Shapira, B. Sudakov ad R. Yuster has a cycle of legth at least β(g/, this implies that G has a cycle of legth at least 1 + max{m / 3, m/ }. The rest of this paper is orgaized as follows. I Sectio, we obtai our bouds for feedback arc sets by provig Theorem 1.1. Sectio 3 cotais the proofs of our results for the existece of short cycles, log cycles, ad subgraph with large miimum degree. The fial sectio cotais some cocludig remarks ad ope problems.. Feedback arc sets This sectio cotais the proofs of Theorem 1.1. Cosider some liear order of the vertex set of a Euleria digraph G =(V,E with vertices ad m arcs. Let v i be the ith vertex i this order. We say that v i is before v j if i<j.aarc(v i,v j isaforward arc if i<j,ad is a backward arc if i>j. Observe that every cycle cotais at least oe backward arc. Hece, β(g is precisely the miimum umber of backward arcs over all liear orderigs. We prove Theorem 1.1 by showig that ay liear order of V has at least as may backward arcs as the amout stated i the theorem. We first require the followig simple lemma. Here a cut is defied as a partitio of the vertices of a digraph ito two disjoit subsets. Lemma.1. I ay cut (A, V \ A of a Euleria digraph, the umber of arcs from A to V \ A equals the umber of arcs from V \ A to A. Proof. The sum of the out-degrees of the vertices of A equals the sum of the i-degrees of the vertices of A. Each arc with both edpoits i A cotributes oe uit to each of these sums. Hece, the umber of arcs with oly oe edpoit i A splits equally betwee arcs that go from A to V \ A ad arcs that go from V \ A to A. Proof of Theorem 1.1. First we costruct a ifiite family of Euleria digraphs that achieves the boud i Theorem 1.1. For ay positive itegers, m such that t := m/ is a iteger, we defie the Cayley digraph G(, m to have vertex set {0, 1,..., 1} ad arc set {(i, i + j :1 i, 1 j t}, where all additios are modulo. From the defiitio, it is easy to verify that G(, m is a Euleria digraph. Cosider a order of the vertex set such that vertex i is the ith vertex i this order. We observe that for t +1 i, vertex i has backward arcs (i, j, where 1 j t ( i ad there is o backward arc from vertex i for i t. Therefore, t ( t +1 β(g(, m t ( i = j = = m + m. i= t+1 Next we prove the boud for arbitrary Euleria digraph. Fix a Euleria digraph G with V = ad E = m. We claim that it suffices to oly cosider Euleria digraphs which are -cycle-free, i.e., betwee ay pair of vertices {i, j} there do ot exist arcs i two differet directios. Suppose there are k differet -cycles i G. By removig all of them, we delete exactly k arcs. Note that the resultig -cycle-free digraph G is still Euleria j=1

5 Large Feedback Arc Sets ad Log Cycles i Euleria digraphs 863 ad cotais m k arcs. Therefore if Theorem 1.1 is true for all -cycle-free Euleria digraphs, the β(g (m k + m k. Obviously, i ay liear order of V (G, exactly half of the k arcs deleted must be backward arcs. Therefore, β(g β(g (m k +k + m k ( m + k = + m k(m k + k k ( m + m k( + k k = m + m. The last iequality follows from the fact that m k (, sice m k couts the umber of pairs of vertices with oe arc betwee them. From ow o, we always assume that G is a -cycle-free Euleria digraph. I order to prove a lower boud o β(g, we fix a liear orderig v 1 <v < <v with the miimum umber, β(g, of backward arcs. It will be importat for the aalysis to cosider the legth of a arc (v i,v j, which is i j. Observe that the legth of ay arc is a iteger i {1,..., 1}. Moreover, we call a arc short if its legth is at most /. Otherwise, it is log. Partitio the arc set E ito two parts, S ad L, wheres cotais the short arcs ad L cotais the log arcs. For a vertex v i,lets i deote the umber of short arcs coectig v i with some v j where j>i. It is importat to ote that at this poit we claim othig regardig the directios of these arcs. Sice G is -cycle-free, s i i. As each short arc (v i,v j cotributes exactly oe to either s i or s j, we have that s i = S. We ow estimate the sum of the legths of the short arcs. Cosider some vertex v i. Sice G is -cycle-free, the s i short arcs coectig v i to vertices appearig after v i must have distict legths. Hece, the sum of their legths is at least s i = ( s i +1. Thus, deotig by w(s the sum of the legths of the short arcs, we have that ( si +1 w(s. (.1 Next we calculate the sum of the legths of the log arcs, which is deoted by w(l. There is at most oe log arc of legth 1. There are at most two arcs of legth, ad, more geerally, there are at most i arcs of legth i. Thus, if we deote by t i the umber of log arcs of legth i for i / +1 ad set t i =0 for i /, we have that t i i, ad w(l = i t i. (.

6 864 H. Huag, J. Ma, A. Shapira, B. Sudakov ad R. Yuster Obviously, t i + s i = L + S = m. Let A i = {v 1,...,v i } ad cosider the cuts C i =(A i,v \ A i fori =1,...,.Letc i deote the umber of arcs crossig C i (ad otice that c = 0. Sice a arc of legth x crosses precisely x of these cuts, we have that c i = w(s+w(l. (.3 Cosider a pair of cuts C i,c i+ / for i =1,..., /. If a arc crosses both C i ad C i+ / the its legth is at least / + 1. Hece, a short arc caot cross both of these cuts. Let y i deote the umber of log arcs that cross both of these cuts. By Lemma.1, c i / backward arcs cross C i ad c i+ / / backward arcs cross C i+ /, ad we have couted at most y i such arcs twice. It follows that the umber of backward arcs is at least 1 (c i + c i+ / y i. Averagig over all / such pairs of cuts, it follows that the umber of backward arcs is at least 1 / / ( 1 (c i + c i+ / y i. (.4 As each log arc of legth j crosses precisely j / pairs of cuts C i ad C i+ /,we have / y i = j / t j (j / =w(l L /. This, together with (.3 ad (.4, gives β(g 1 ( 1 (w(s+w(l (w(l L / / w(s w(l + L. (.5 / Note that whe =k is eve, the above iequality becomes w(s w(l β(g + L. Next we show that whe =k + 1 is odd, the same iequality still holds. To see this, first assume that w(s w(l. The, applyig iequality (.5, we have that for =k +1, β(g w(s w(l k + L w(s w(l + L. Next suppose that w(s <w(l. Istead of cosiderig the cuts C i ad C i+k,welookat the pair C i ad C i+k+1 for i =1,...,k. Moreover, deote by z i the umber of log arcs

7 Large Feedback Arc Sets ad Log Cycles i Euleria digraphs 865 that cross both of these cuts. By a argumet similar to that used earlier, the umber of backward arcs is at least 1 (c i + c i+k+1 z i for 1 i k, adc i /fori = k + 1. This provides k + 1 lower bouds for β(g, ad we will average over all of them. Sice each log arc of legth j crosses precisely j (k +1 pairs of cuts C i ad C i+k+1, we agai have that k z i = t j (j (k +1=w(L (k +1 L, j k+1 ad we have β(g 1 ( k k +1 1 ( 1 k +1 w(s w(l k + ( 1 (c i + c i+k+1 z i + c k+1 (w(s+w(l (w(l (k +1 L + L w(s w(l + L, whereweusethefactthatw(l >w(s. Usig our lower boud estimate (.1 for w(s ad the expressio (. for w(l, we obtai w(s w(l β(g + L 1 ( ( si +1 i t i + = 1 ( ( si +1 +( it i. t i (.6 Defie F(s 1,...,s ; t 1,...,t := ( si +1 +( it i. I order to fid a lower boud of β(g, we eed to solve the followig iteger optimizatio problem: F(m, := mi F(s 1,...,s ; t 1,...,t subject to s i i, t i i, s i + t i = m. Lemma. below provides a precise solutio to this optimizatio problem, which gives that F(m, =tm (t t/, where t = m/. Hece, if we assume that m = t k

8 866 H. Huag, J. Ma, A. Shapira, B. Sudakov ad R. Yuster with 0 k 1, the β(g 1 F(m, =tm t t = t + t = (t k tk t + t + t k = t(t k tk + ( k k = m + m. t t The last iequality is because 0 k 1, so 0 k/ < 1adk / k/. Note that equality is possible oly whe m is a multiple of. Lemma.. F(m, =tm (t t/, wheret = m/. Proof. The proof of this lemma cosists of several claims. We set a i = s i + t i. The 0 a i ( i, s i i, ad i a i = m, ad the objective fuctio becomes ( si +1 +( it i = 1 s i ( i 1/s i +( ia i. Sice s i is a iteger, this fuctio of s i is miimized whe s i = i if a i i, ad whe s i = a i if a i < i. Therefore, subject to i a i = m ad a i ( i, we wat to miimize F = ( ai +1 + (( i +1 +( i(a i ( i a i < i a i i = ( ai +1 + ( ( i ( ia i. (.7 a i < i a i i For coveiece, defie A = {i : a i < i}, adb = {i : a i i}. Claim 1. For ay i A, if we icrease a i by 1 the F icreases by a i + 1, ad if we decrease a i by 1 the F decreases by a i.forayj B, if we icrease (decrease a j by 1 the F icreases (decreases by j. ( Proof. Note that whe a i = i or a i = i 1, ai +1 =( iai ( i, therefore if ( we icrease a i by1forayi A, the cotributio of a i to F always icreases by ai ( + ai +1 = ai + 1. Whe we decrease a i by 1, F decreases by ( a i ( +1 ai = ai.itis also easy to see that for ay j B, if we icrease or decrease a j by 1, the cotributio of a j to F always icreases or decreases by j. Claim. F is miimized whe A = {1,...,l 1} ad B = {l,...,} for some iteger l. Proof. We prove Claim by cotradictio. Suppose this statemet is false. The F is miimized by some {a i } such that there exists i<j, i B ad j A. Now we decrease a i by 1 ad icrease a j by 1, which ca be doe sice a j < ( j. The by Claim 1,

9 Large Feedback Arc Sets ad Log Cycles i Euleria digraphs 867 F decreases by ( i (a j +1 (j 1 (a j +1=( j a j > 0 sice j A, which cotradicts the miimality of F. We have a i = m, which is fixed. The ext claim shows that i order to miimize F, we eed to take the variables whose idex is i B to be as large as possible, with at most oe exceptio. Claim 3. F is miimized whe A = {1,...,l 1} ad B = {l,...,} for some iteger l. Moreover, a i =( i foralli l +1. Proof. First ote that for i B, its cotributio to F is ( ia i ( i. The secod term is fixed, ad a i has coefficiet i which decreases i i. Therefore, whe F is miimized, if i is the largest idex i B such that a i < ( i, the all j<i i B must satisfy a j = j; otherwise we might decrease a j ad icrease a i to make F smaller. Therefore, if i>l, we have a i 1 = i + 1. Note that if we icrease a i by 1 ad decrease a i 1 by 1, by Claim 1 the target fuctio F decreases by a i 1 ( i = 1. Therefore the oly possibility is that i = l, which proves Claim 3. Claim 4. There is a extremal cofiguratio for which a i = l or a i = l +1 for i l 1, a l is betwee l ad ( l, ad a i =( i fori l +1. Proof. From Claim 3, we kow that i a extremal cofiguratio, a i < i for 1 i l 1, l a l ( l, ad a i =( i fori l + 1. Amog all extremal cofiguratios, we take oe with the largest l, ad for all such cofiguratios, we take oe for which a l is the smallest. For such a cofiguratio, if we icrease a j by1for some j A ad decrease a l by 1, the by Claim 1, F icreases by (a j +1 ( l, which must be o-egative. Suppose a j +1= l. Ifj is chaged to be i B, it cotradicts Claim 3 o matter whether l remais i B or is chaged to be i A; ifj remais i A, it cotradicts the maximality of l if l is chaged to be i A or cotradicts the miimality of a l if l remais i B. Therefore a j l for every 1 j l 1. We ext cosider two cases: either a l is equal to ( l, or strictly less tha ( l. Case 1: a l =( l. From the discussios above, we already kow that a j l for every 1 j l 1. I particular a l 1 = l, sice it is strictly less tha (l 1. If, for some j l 1, a j l +, the we ca decrease a j by 1 ad icrease a l 1 by 1, sice a j is strictly greater tha 0 ad a l 1 is strictly less tha ( l +1.ByClaim1,F decreases by a j ( l +1 1, which cotradicts the miimality of F. Hece we have that l a j l +1foreveryj l 1. Case : a l < ( l. If we decrease a j by 1 ad icrease a l by 1, F decreases by a j ( l by Claim 1. Therefore a j l by the miimality of F, ad hece a j = l for all 1 j l 1. I both cases, the extremal cofiguratio cosists of l or l + 1 for the first l 1 variables, a l is betwee l ad ( l, ad a i =( i fori l +1.

10 868 H. Huag, J. Ma, A. Shapira, B. Sudakov ad R. Yuster By Claim 4, we ca boud the umber of arcs m from both sides: l 1 m = a i + a i (l 1( l+( l+ ( i =( l( 1, l 1 m = a i + i=l a i < (l 1( l +1+ i=l i=l+1 ( i =( l +1( 1. Solvig these two iequalities, we get m l<+1 m 1 1. Let m = t k, wheret = m/ ad 0 k 1. It is ot difficult to check that if t k, l = t ad if t<k, l = t +1. Now let x be the umber of variables a 1,...,a l 1 which are equal to l + 1. Sice a i =( i fori l + 1, we have that x + a l = m (l 1( l a i = m ( ( l. (.8 Whe t k, the l = t ad i l+1 x + a l = m ( t =t k<t =( l, ad hece a l < ( l. By the aalysis of the secod case i Claim 4, a j = l = t for all j l 1, ad therefore x =0 ad a l =t k. Sice l = t, the usig the summatio formula k=1 k = k(k + 1(k +1/6, we have from (.7 that (with details of the calculatio omitted ( t +1 F = ( t 1 + t(t k = tm (t t/. i=l ( t + ( ( i i l+1 Now we assume t<k,adsol = t + 1. The, usig (.8 agai, ( i x + a l = m ( (t 1 = k +(t 1 > (t 1 = ( l. The oly possibility without cotradictig the secod case i Claim 4 is that a l =( l ad x = k. Thus there are k of a 1,...,a l 1 which are equal to l +1=t, ad the rest of k t are equal to t 1. Agai by (.7, ( ( t +1 t F = ( k+ (k t+ ( ( i ( i = tm (t t/. i l As we have covered both cases, we have completed the proof of Lemma.. 3. Short cycles, log cycles, ad Euleria subgraphs with high miimum degree I this sectio, we prove the existece of short cycles, log cycles, ad subgraphs with large miimum degree i Euleria digraphs. A importat compoet i our proofs is the

11 Large Feedback Arc Sets ad Log Cycles i Euleria digraphs 869 followig result by Fox, Keevash ad Sudakov [7] o geeral digraphs. We poit out that the origial Theorem 1. i [7] was proved with a costat 5, which ca be improved to 18 usig exactly the same proof if we further assume r 11. Theorem 3.1. If a digraph G o vertices has β(g > 18 /r, with r 11, theg cotais a cycle of legth at most r, i.e., g(g r. Applyig this theorem ad Theorem 1.1, we ca ow prove Corollary 1., which says that every Euleria digraph G with vertices ad m arcs cotais a cycle of legth at most 6 /m. Proof of Corollary 1.. Let r =6 /m. Give a Euleria digraph G with vertices ad m arcs, if G cotais a -cycle, the g(g 6 /m. So we may assume that G is -cycle-free ad thus m (. By Theorem 1.1, β(g m + m > m = 18 (6 /m. Sice r =6 /m > 6 / ( > 11, we ca use Theorem 3.1 to coclude that g(g r = 6 m. To see that this boud is tight up to a costat factor, we cosider the costructio of the Cayley digraphs i Theorem 1.1. It is ot hard to see that if k = m/, theshortest directed cycle i G(, m has legth at least /k /m. Next we show that every Euleria digraph with vertices ad m arcs has a Euleria subgraph with miimum degree Ω(m / 3. Proof of Theorem 1.3. We start with a Euleria digraph G with vertices ad m arcs. Note that Corollary 1. implies that every Euleria digraph with vertices ad at least m/ arcs cotais a cycle of legth at most 1 /m. I every step, we pick oe such cycle ad delete all of its arcs from G. Obviously the resultig digraph is still Euleria, ad this process will cotiue util there are less tha m/ arcs left i the digraph. Therefore through this process we obtai a collectio C of t arc-disjoit cycles C 1,...,C t,where t (m m//(1 /m m /4. Deote by H the uio of all these cycles, where obviously H is a Euleria subgraph of G. If H has miimum degree at least t/ m /4 3, the we are already doe. Otherwise, we repeatedly delete from H ay vertex v with degree d(v t/ 1, together with all the d(v cycles i C passig through v. This process stops after a fiite umber of steps. I the ed we delete at most ( t/ 1 t 1 cycles i C, so the resultig digraph H is o-empty. Moreover, every vertex i H has degree at least t/ m /4 3. Sice H is the disjoit uio of the remaiig cycles, it is also a Euleria subgraph of G, adwe coclude the proof of Theorem 1.3.

12 870 H. Huag, J. Ma, A. Shapira, B. Sudakov ad R. Yuster Figure 1. The Euleria digraph H(s, t with s = 3. Remark. The proof of Theorem 1.3 also shows that G cotais a Euleria subgraph with miimum degree Ω(m / 3 ad at least Ω(m arcs. To see that the boud i Theorem 1.3 is tight up to a costat, for ay itegers s, t > 0, we costruct a Euleria digraph H := H(s, t such that: V (H =(U 1 U s (V 1 V t, U i = V j = s for 1 i s, 1 j t, for ay 1 i t 1 ad vertices u V i, v V i+1,thearc(u, v E(H, for ay 1 i s ad every vertex u U i, there is a arc from u to the ith vertex i V 1, ad aother arc from the ith vertex i V t to u. It ca be verified that H(s, t is a Euleria digraph with (s + ts vertices ad s (t +1 arcs. Moreover, every cycle i H(s, t must pass through a vertex i U 1 U s, whose degree is exactly 1. Therefore ay Euleria subgraph of H(s, t has miimum degree at most 1. Next we defie the δ-blowup H(s, t, δ: for ay iteger δ>0, we replace every vertex i V (H(s, t with a idepedet set W i = δ, adeacharc(i, j E(H(s, t by a complete bipartite digraph with arcs directed from W i to W j. The blowup digraph H(s, t, δ is still Euleria, ad has = s(s + tδ vertices ad m = s (t +1δ arcs. Takig t =s, we have that for H(s, s, δ, m 3 = (s (s +1δ (s(s +sδ 3 = 1 ( + 1 δ 4 7 s 7 δ. Note that, similarly to the previous discussio o H(s, t, every cycle i the blowup H(s, s, δ cotais at least oe vertex with degree δ. Therefore, the miimum degree of ay Euleria subgraph of H(s, s, δ is at most δ 7 m 4. This implies that the boud i 3 Theorem 1.3 is tight up to a costat factor for ifiitely may m,. Before provig Propositio 1.4, let us recall the followig easy fact.

13 Large Feedback Arc Sets ad Log Cycles i Euleria digraphs 871 Propositio 3.. If a digraph G has miimum out-degree δ + (G, theg cotais a directed cycle of legth at least δ + (G+1. Proof. Let P = v 1 v v t be the logest directed path i G. The all the out-eighbours of v t must lie o this path, otherwise P will become loger. If i<t is miimal with (v t,v i E(G, the v i v t v i gives a cycle of legth at least d + (v t +1 δ + (G+1. This propositio, together with Theorem 1.3, shows that a Euleria digraph G with vertices ad m arcs cotais a cycle of legth at least 1 + m /4 3. As discussed i the Itroductio, this ca be slightly improved to 1 + m / 3, but these bouds become meaigless whe the umber of arcs m is small. However, we may use a differet approach to obtai a cycle of legth at least m/ +1. Proof of Propositio 1.4. To prove that ay Euleria digraph G with vertices ad m arcs has a cycle of legth at least m/ + 1, we use iductio o the umber of vertices. Note that the base case whe = is obvious, sice the oly Euleria digraph is the -cycle with m/ + 1 =. Suppose the statemet is true for 1. Cosider a Euleria digraph G with vertices ad m arcs. If its miimum degree δ + (G isatleast m/, by Propositio 3., G already cotais a cycle of legth at least 1 + m/. Therefore we ca assume that there exists a vertex v with m/ >d + (v :=t. AsG is Euleria, there exist t arc-disjoit cycles C 1,C,...,C t passig through v. If oe of these cycles has legth at least m/ + 1, the agai we are doe. Otherwise, C i m/ for all 1 i t. Now we delete from G the vertex v together with the arcs of the cycles C 1,...,C t. The resultig Euleria digraph has 1 vertices ad m arcs, where t m = m C i m t ( m/ m 1 1. By the iductive hypothesis, the ew digraph (therefore G has a cycle of legth at least 1+ ( m /( 1 1+ m 1 1 /( 1 1+ m/. 4. Cocludig remarks We ed with some remarks o the Bollobás Scott cojecture whose uweighted versio states that a Euleria digraph with vertices ad m arcs has a cycle of legth Ω(m/. The caoical proof for showig that a udirected graph with this may vertices ad edges has a cycle of legth m/ proceeds by first passig to a subgraph G with miimum degree at least m/ ad the applyig Propositio 3. to G. We ca the iterpret the secod statemet of Theorem 1.3 as statig that whe applied to Euleria digraphs, this approach ca oly produce cycles of legth O(m / 3. There is, however, aother way to show that a udirected graph has a cycle of legth m/ usig depth-first search (DFS. Recall that the DFS is a graph algorithm that visits

14 87 H. Huag, J. Ma, A. Shapira, B. Sudakov ad R. Yuster all the vertices of a (directed or udirected graph G as follows. It maitais three sets of vertices, lettig S be the set of vertices which we have completed explorig them, T be the set of uvisited vertices, ad U = V (G \ (S T, where the vertices of U are kept i a stack (a last-i first-out data structure. The DFS starts with S = U = ad T = V (G. While there is a vertex i V (G \ S, ifu is o-empty, let v be the last vertex that was added to U. Ifv has a eighbour u T, the algorithm iserts u ito U ad repeats this step. If v does ot have a eighbour i T the v is popped out from U ad is iserted ito S. IfU is empty, the algorithm chooses a arbitrary vertex from T ad pushes it to U. Observe crucially that all the vertices i U form a directed path, ad that there are o edges from S to T. Cosider ay DFS tree T of a udirected graph G rooted at some vertex v. Recall that ay edge of G is either a arc of T or a back arc, that is, a edge coectig a vertex v to oe of its acestors i T. Hece, if G has o cycle of legth at least t, the ay vertex of T seds at most t 1 arcs to his acestors i T. This meas that m t or that t m/. Note that this argumet shows that ay DFS tree of a udirected graph has depth at least m/. For directed graphs, however, ot all arcs are tree arcs or back arcs. Nevertheless, the set of back arcs form a feedback arc set, ad hece, if the logest cycle of a digraph G has legth t, the t β(g. It is atural to try ad adapt the DFS approach to the case of Euleria digraphs. Ufortuately, as the followig propositio shows, this approach fails i Euleria digraphs. Propositio 4.1. There is a Euleria digraph G with average degree at least /0 such that some DFS tree of G has depth 4. Proof. We first defie a graph G as follows. Let t be a positive iteger ad let G be a graph cosistig of t vertex sets V 1,...,V t, each of size t. We also have a special vertex r, sog has t + 1 vertices. We ow defie the arcs of G usig the followig iterative process. We have t iteratios, where i iteratio 1 j t we add the followig arcs; we have t arcs poitig from r to the t vertices of V j, the a matchig from the t vertices of V j to the vertices of V j+1, ad i geeral a matchig from V k to V k+1 for every j k t j. We fially have t arcs from V t j+1 to r. We ote that we ca ideed add a ew (disjoit from previous oes matchig betwee ay pair of sets (V k,v k+1 ieach of the t iteratios by relyig o the fact that the edges of the complete bipartite graph K t,t ca be split ito t perfect matchigs. Observe that i iteratio j we add t(t j +3 arcs to G. Hece G has t t(t j +3 t 3 j=1 arcs. Moreover it is easy to see from costructio that G is Euleria. To get the graph G we modify G as follows. For every vertex v t V i we add two ew vertices v i,v out ad add a 4-cycle (r, v i,v,v out,r. We get that G has 6t + 1 vertices ad more tha t 3 arcs, so settig =6t + 1 we see that G has average degree at least /0.

15 Large Feedback Arc Sets ad Log Cycles i Euleria digraphs 873 Now cosider a DFS tree of G which proceeds as follows. We start at r, ad the for every v V t go to v i, the to v, adthetov out.next,foreveryv V t 1 we go to v i, the to v, adthetov out. We cotiue i this way util we cover all the vertices of G. The DFS tree we thus get has r as its root, ad t paths of legth 3 (of type r, v i,v,v out attached to it. Observe that the above propositio does ot rule out the possibility that some DFS tree has depth Ω(m/. We ote that provig such a claim will imply that a Euleria digraph has a path of legth Ω(m/. It appears that eve this special case of the Bollobás Scott cojecture is still ope, so it might be iterestig to ivestigate this problem further. I fact, we suspect that if G is a coected Euleria digraph the for ay vertex v G there is a path of legth Ω(m/ startig at v. This statemet for udirected graphs follows from the DFS argumet at the begiig of this sectio. Ackowledgemet We would like to thak Jacques Verstraete for helpful iitial discussios. We also would like to thak the referee for carefully readig the mauscript ad providig may helpful suggestios. Refereces [1] Alo, N. (006 Rakig touramets. SIAM J. Discrete Math [] Bollobás, B. (1998 Moder Graph Theory, Vol. 184 of Graduate Texts i Mathematics, Spriger. [3] Bollobás, B. ad Scott, A. (1996 A proof of a cojecture of Body cocerig paths i weighted digraphs. J. Combi. Theory Ser. B [4] Caccetta, L. ad Häggkvist, R. (1978 O miimal digraphs with give girth. I Proc. 9th Southeaster Coferece o Combiatorics, Graph Theory, ad Computig: Boca Rato Cogress. Numer. XXI [5] Charbit, P., Thomassé, S. ad Yeo, A. (007 The miimum feedback arc set problem is NP-hard for touramets. Combi. Probab. Comput [6] Chudovsky, M., Seymour, P. ad Sulliva, B. (008 Cycles i dese digraphs. Combiatorica [7] Fox, J., Keevash, P. ad Sudakov, B. (010 Directed graphs without short cycles. Combi. Probab. Comput [8] Leiserso, C. ad Saxe, J. (1991 Retimig sychroous circuitry. Algorithmica [9] Nathaso, M. The Caccetta Häggkvist cojecture ad additive umber theory. [10] Shaw, A. (1974 The Logical Desig of Operatig Systems, Pretice Hall. [11] Sulliva, B. (008 Extremal problems i digraphs. PhD thesis, Priceto Uiversity. [1] Sulliva, B. A summary of results ad problems related to the Caccetta Häggkvist cojecture.