A survey on approximating graph spanners Guy Kortsarz, Rutgers Camden

A survey on A survey on approximating graph approximating graph

spannersspanners

Guy Kortsarz, Rutgers Guy Kortsarz, Rutgers CamdenCamden

Spanners: definition for Spanners: definition for unweighted undirected unweighted undirected

graphs. graphs. Peleg and UpfalPeleg and Upfal..•Input: Input: An undirected graph An undirected graph

G(V,E)G(V,E) and and an integer an integer kk

• Required: Required: a subgrapha subgraph G’ G’ so that so that for every for every uu and and v v V V::

DistDistG’G’

(u,v)/Dist(u,v)/DistGG

(u,v)≤ k(u,v)≤ k

It is enough to take care It is enough to take care of edgesof edges

• Say we are talking about the lengths and Say we are talking about the lengths and weights that are weights that are 11. A.K.A Basic Spanners.. A.K.A Basic Spanners.

• Say that you have a Say that you have a kk-spanner -spanner G(V,E’) G(V,E’) of of a graph a graph G(V,E)G(V,E)

• Let Let E-E’ E-E’ be thebe the omitted omitted edges edges• What happens when you add an What happens when you add an omittedomitted

edge is added to a edge is added to a kk-spanner -spanner G(V,E’)G(V,E’)??• Then a cycle of size at most Then a cycle of size at most k+1k+1 must be must be

closed.closed.

A look on an omitted edge A look on an omitted edge ee

For example a For example a 44-spanner-spanner

A look on an omitted edge A look on an omitted edge ee

For example a For example a 44-spanner: the omitted -spanner: the omitted edge creates a cycle of size edge creates a cycle of size 4+1=54+1=5

It enough that all edges are “spanned”, to get distance between any two vertices is at most k the original one.

An example of a An example of a 22--spannersspanners

• The original graph:The original graph:

A A 22-spanner-spanner

• A few of the edge that were part of A few of the edge that were part of triangles triangles were removedwere removed..

ApplicationsApplications

• In geometry.In geometry.• Small routing tables: Small routing tables: spanners have less edges. spanners have less edges.

Thus smaller tables. But not much larger Thus smaller tables. But not much larger distancedistance

• Synchronizers: make non synchronized distributed computation, synchronized.

• Parallel distributed and streaming algorithms.• Distance oracles. Handle queries about distance

between two vertices quick by preprocessing.• Property testing• Minimum time broadcast.

Directed spanners, Directed spanners, applications:applications:

•Transitive Closure SpannersTransitive Closure Spanners See See Grigorescu, Jung, Raskhodnikova, Woodruff, SODA’09

•Property testing. See a survey by Raskhodnikova.

•Access Control hierarchies. See a paper by De Santis et al.

TC spannersTC spanners

• Transitive closure spanners.Transitive closure spanners.• Given G. Find Given G. Find H H with same with same TC TC of of GG, ,

and has diameter and has diameter kk. See a paper by . See a paper by Bhattacharyya Bhattacharyya et al.et al.

• Studied in access control (but were not Studied in access control (but were not given a name). Very long time ago.given a name). Very long time ago.

• In the last 20 years many In the last 20 years many knownknown properties of transitive closure were properties of transitive closure were rediscoveredrediscovered..

TC spannersTC spanners: remarkable : remarkable many applicationsmany applications

• Circuit construction. See Circuit construction. See Chandra Chandra et alet al et al et al STOC 1983STOC 1983..

• Data structures. See Data structures. See Alon et alAlon et al 19981998..• Property testing: Property testing: RaskhodnikovaRaskhodnikova, ,

FOCS 2012FOCS 2012..• Lower bounds for stretch (a.k.a Lower bounds for stretch (a.k.a

Lipshitzness). See Lipshitzness). See JR JR 20112011. And . And much more.much more.

2-spanners2-spanners

• There is a difficulty. Unlike There is a difficulty. Unlike k≥3k≥3 thre are thre are not necessarilynot necessarily 2 2 spanners with few spanners with few edges.edges.

• The only The only 22-spanners of a complete -spanners of a complete bipartite graph is the graph itself.bipartite graph is the graph itself.

• Also true for Also true for TC spannersTC spanners if all the if all the directed edges go from left to right.directed edges go from left to right.

• Like in Like in 2-SAT2-SAT and and 2-Coloring 2-Coloring and other and other problems, problems, 22-spanners is different than the -spanners is different than the rest.rest.

For For kk at least at least 3 3 there are there are spanners with few edgesspanners with few edges

• As we shall see: As we shall see: 3-spanners3-spanners with with O(n*sqrt{n})O(n*sqrt{n}) edges always exist, and the edges always exist, and the same goes for same goes for 4-spanners. 4-spanners. And this is tight. And this is tight.

• The larger The larger kk is, the smaller is the upper is, the smaller is the upper bound bound

on the number of edges in the best on the number of edges in the best spanner.spanner.

• Remarkable fact: maximum number of Remarkable fact: maximum number of edges in a graph with girthedges in a graph with girth g g not known.not known.

• Maybe for Maybe for 40 40 years the upper and lower years the upper and lower bound are quite far!bound are quite far!

General lengthsGeneral lengths

• The definition is the same, just that the The definition is the same, just that the length is weighted length. Also length is weighted length. Also weights=lengthsweights=lengths..

• Definition: The weighted distance of Definition: The weighted distance of every pair of vertices in every pair of vertices in G’G’ must be no must be no larger thanlarger than k k times their weighted times their weighted distance in distance in GG..

• It is not enough that an It is not enough that an omittedomitted edge edge closes a cycle of length at most closes a cycle of length at most k+1k+1..

• However, However, it works if the the omitted edge it works if the the omitted edge is the largest in the closed cycle.is the largest in the closed cycle.

An edge that can be An edge that can be removedremoved


9

7

6 8

4

Removing the heaviest edge Removing the heaviest edge in a cycle of length at most in a cycle of length at most

k+1k+1 is OK. is OK.


7

6 8

4

A generalization of the A generalization of the Kruskal algorithm:Kruskal algorithm:

• Sort the edges of the graph in increasing Sort the edges of the graph in increasing weights.weights.

c(ec(e11) ≤ c(e) ≤ c(e22) ≤ c(e) ≤ c(e33) ≤……….. ≤ c(e) ≤……….. ≤ c(emm))• Go over all edges from small cost to large.Go over all edges from small cost to large.• For the next edge For the next edge eeii, , if the edge does not if the edge does not

close a cycle of length at mostclose a cycle of length at most k+1 k+1 with with previously added edges, add previously added edges, add eei i to to G’G’ or or else else i=i+1i=i+1

• This algorithm is due to This algorithm is due to I. Althofer, G. Das, D. Dobkin, D. Joseph, and J. Soares.

The resulting graph is a The resulting graph is a kk--spannerspanner

• If an edge If an edge ee is missing, then by is missing, then by construction, this edge is the most heavy construction, this edge is the most heavy edge in a cycle of length at most edge in a cycle of length at most k+1k+1..

• This is because we go over edges in non This is because we go over edges in non decreasing costs.decreasing costs.

• If we reach a cycle of size If we reach a cycle of size k+1k+1, then it , then it means that previous edges were not means that previous edges were not removed.removed.

• This implies that This implies that e e is the largest edge in is the largest edge in a cycle of length at most a cycle of length at most k+1k+1 and it is and it is safe to remove it.safe to remove it.

Girth Girth k+2k+2

• We observe that the resulting graph has We observe that the resulting graph has girthgirth at least at least k+2k+2

• The The girthgirth is the size of the minimum simple is the size of the minimum simple cycle.cycle.

• Observe that when we reach the largest Observe that when we reach the largest edge edge ee of a cycles with at most of a cycles with at most k+1k+1 edges edges it it will be removed.will be removed.

• Therefore, there are no Therefore, there are no k+1k+1 size or smaller size or smaller cycles.cycles.

• Graphs with large girth have “few’’ edges. Graphs with large girth have “few’’ edges.

Example: graphs with girth Example: graphs with girth 55 and and 66

• We show that graphs with girth We show that graphs with girth 5 5 and and 6 6 have have O(n*sqrt{n})O(n*sqrt{n}) edges. edges.

• First remove all vertices of degree First remove all vertices of degree strictly smaller than strictly smaller than m/n m/n..

• Here Here mm is the numbers of edges and is the numbers of edges and nn is the number of vertices.is the number of vertices.

• Since we have removed at most Since we have removed at most nn vertices and each vertex removes less vertices and each vertex removes less than than m/nm/n edges it is clear that the edges it is clear that the resulting graph is not empty.resulting graph is not empty.

Two layers BFS graphTwo layers BFS graph

• All the vertices seen below areAll the vertices seen below are distinctdistinct as otherwise there is a cycle as otherwise there is a cycle of length at most of length at most 44..

m/n

(m/n)-1 (m/n)-1

Number of edges Number of edges

• This implies that This implies that m/n(m/n-1)≤ nm/n(m/n-1)≤ n, or , or mm22/n/n22-m/n -m/n ≤n≤n

• AsAs m/n<n m/n<n we get that we get that mm22/n/n22<2n <2n oror m m22<2n<2n33

• Thus Thus m=O(n sqrt{n})m=O(n sqrt{n})• A matching lower bound. A graph of girth A matching lower bound. A graph of girth 66

that has that has ΩΩ(n*sqrt{n})(n*sqrt{n}) edges. edges.• A A projective plane projective plane for our needs is a bipartite for our needs is a bipartite

graph with graph with nn vetices on each side and degree vetices on each side and degree ӨӨ(sqrt{n}) (sqrt{n}) thusthus containscontains ӨӨ(n*sqrt{n})(n*sqrt{n}) edges. edges.

• The main property: every pair of vertices in the same The main property: every pair of vertices in the same side share at most side share at most 11 neighbor. neighbor.

Girth Girth 66

• There could not be a cycle of size There could not be a cycle of size 44::

A cycle of length 4 implies that two vertices on the same side share two neighbors. Contradiction

Girth Girth 66

• There could not be a cycle of size There could not be a cycle of size 44::

Therefore girth 6

General bounds on the General bounds on the minimum number of edges minimum number of edges

for a given girthfor a given girth• It is known that there is always a It is known that there is always a 2k-12k-1

spanner with spanner with O(nO(n1+1/k1+1/k)) edges. edges.• By the above we know that there is always By the above we know that there is always

a a 3,4 3,4 spanners with spanners with O(n*sqrt{n})O(n*sqrt{n}) edges. edges. But better than But better than sqrt{n}sqrt{n} approximation is approximation is known (later).known (later).

• For if we want a For if we want a 33-spanner put -spanner put k=2k=2. This . This indeed gives indeed gives O(n* sqrt{n})O(n* sqrt{n}) edges. The edges. The number minimum number of edges for number minimum number of edges for k=3k=3 and and k=4k=4 is the same as the bad is the same as the bad example is a bipartite graph.example is a bipartite graph.

There are spanners with There are spanners with few edges but what about few edges but what about

the cost?the cost?• An interesting result by An interesting result by Chandra et alChandra et al..• The same algorithm described before The same algorithm described before

returns a graph with small returns a graph with small weightweight..• It is roughly of cost It is roughly of cost nn1/t 1/t times the cost times the cost

of the minimum spanning tree (on top of the minimum spanning tree (on top of the fact that there are just few of the fact that there are just few edges),edges),

• For For log nlog n it gives about it gives about O(1)O(1) times the times the MST MST weight, and weight, and O(n)O(n) edges. edges.

The story of the minimum The story of the minimum cost cost

22-spanner problem.-spanner problem.• The edges have length The edges have length 1. 1. We show an We show an

approximation for costsapproximation for costs 1 1,, even thougheven though works works for arbitrary costs.for arbitrary costs.

• In 1992, In 1992, K, PelegK, Peleg: the : the 22-spanner problem admits -spanner problem admits an an O(log n)O(log n) ratio approximation in polynomial ratio approximation in polynomial timetime

• In 1998: In 1998: KortsarzKortsarz: unless : unless P=NPP=NP there exists a there exists a constant so that constant so that c*log n/kc*log n/k approximation is not approximation is not possible for min cost possible for min cost kk-spanners. -spanners. ΩΩ(log n) (log n) for for k=2k=2. . TIGHTTIGHT..

• I know of oneI know of one k k for which the approximation and for which the approximation and the lower bound are equal . the lower bound are equal . UnnaturalUnnatural k. k.

• See a paper by See a paper by Dinitz Dinitz and and KK from from ICALP 2012ICALP 2012..

The idea for the 2-spanner The idea for the 2-spanner O(logn)O(logn) ratio approximation ratio approximation

algorithmalgorithm• A covering problem. A universe A covering problem. A universe UU of of

covering items.covering items.• A non decreasing function A non decreasing function f: 2f: 2SS---> R+---> R+

given indirectly via an oracle.given indirectly via an oracle.• The cost is sub-additive. Namely,The cost is sub-additive. Namely,• c(Ac(A B)≤ c(A)+c(B) B)≤ c(A)+c(B). . • The cost of a set The cost of a set AA is given by an oracle. is given by an oracle. • We look for a minimum cost We look for a minimum cost SS so that so that

f(S)=f(U)f(S)=f(U)..

The density theorem The density theorem

• SS is initiated to is initiated to ..

• At iteration i we add a set Si to S.

• Let OPT be the optimum solution and opt its cost.• The density condition: (f(S Si)-f(S))/c(Si)≤ (f(OPT S)-f(S))/opt

• Intuition: f(OPT S)-f(S)=f(OPT)-f(S)=f(U)-f(S).

• Therefore the density of adding Si is no worse than the density of adding OPT .

The The density theorem density theorem

• Say that Say that SS is initiated to is initiated to . . • Now say that at any iteration we meet the Now say that at any iteration we meet the

density condition density condition Si so that: so that: (f(Si+S)-f(S))/c(Si)≥(f(U)-f(S))/opt

• Then the final set S has cost bounded by

(ln(U)+1) opt

ProofProof

• LetLet S Si i be thebe the LAST LAST set added in iteration set added in iteration ii..• We get We get (f(S+ SjSj)-f(S))/c(SjSj)≥(f(U)-

f(S))/opt• And therefore:• f(U)-f(S1)≤ (1-c(S1)/opt) f(U)• Similarly:• f(U)-f(S1+S2)≤

(1-c(S1)/opt) (1-c(S2)/opt ) f(U)

Proof continuedProof continued

• f(U) -f(U) - j≤ i-1 f(Sj)≥ 1.• We may assume that the cost of every We may assume that the cost of every

set added is at most set added is at most optopt, therefore , therefore c(Sc(Sii) ) ≤ opt• Therefore it remains to bound: Therefore it remains to bound: j≤ i-1 c(Si) Let us concentrate on what happens Let us concentrate on what happens

before before Si is added.

By the previous claimsBy the previous claims

• 1 ≤ f(U)-f(S1+S2 +……Si-1)≤

Πj≤ i-1(1-c(Si)/opt)· f(U)• 1/f(U) ≤ Πj≤ i-1(1-c(Sj)/opt)·

• Take ln and use ln(1+x) ≤ x: -ln( f(U))≤ i≤ j -c(Sj) )/opt

j≤ i-1 c(Sj) ≤ opt ln( f(U)) and so the ratio of ln( f(U))+1

follows.

One of many ways to get a One of many ways to get a good density solution for good density solution for

subadditive costssubadditive costs• Decompose Decompose OPT=OPT=ii OPT OPTii . . OPTOPTii OPT OPT ..

• The frequency property: each covering item The frequency property: each covering item participates in at most participates in at most c c different different OPTOPTi i ..

• By the frequency property and sub-additivity By the frequency property and sub-additivity c(c(ii OPT OPTii )≤ )≤ c(OPT c(OPTii) ≤ c*opt) ≤ c*opt..

• We need that We need that f(OPT)=f(f(OPT)=f(ii OPT OPTii) ≤ ) ≤ f(OPT f(OPTii))

• Density: Density: f(OPT f(OPTii)/ )/ c(OPT c(OPTii) ≥f(OPT)/c*opt) ≥f(OPT)/c*opt

By AveragingBy Averaging

• There exists an i so that There exists an i so that

f(OPTf(OPTii)/c(OPT)/c(OPTii)≥f(OPT)/c*opt)≥f(OPT)/c*opt..

• The question is if we can find or approximate The question is if we can find or approximate the type of structure that the type of structure that OPTOPTi i is.is.

• The decomposition in the The decomposition in the 22-spanner -spanner problem: a collection of stars. For each problem: a collection of stars. For each vertex the vertex and all its optimal edges vertex the vertex and all its optimal edges are some are some OPTOPTii

Applying the decomposition Applying the decomposition schemescheme

• We get We get c=2 c=2 as every edge participated in as every edge participated in 22 stars.stars.

• For a vertexFor a vertex v v that is connected in the that is connected in the optimum to optimum to QQN(v) N(v) let let f(v,Q)f(v,Q) be the number be the number of edges in the graph induced by of edges in the graph induced by QQ..

• We clearly get that We clearly get that f(OPT)=f(f(OPT)=f(vv f(v,Q f(v,Qvv) )≤ ) )≤

f(v,Qf(v,Qvv) ) ..

• Can we find the best star? Namely, the one Can we find the best star? Namely, the one who minimizes who minimizes f(v, Qf(v, Qvv)/|Q)/|Qvv||??

The set The set QQvv N(v) N(v) soso

• Algorithm: check every vertex Algorithm: check every vertex v v • For a vertex For a vertex v v look at the graph look at the graph

induced by induced by N(v)N(v)• Find a desnsest subgraph Find a desnsest subgraph S(v)S(v) in in

N(v)N(v)• Return the edges fromReturn the edges from v v to to S(v)S(v) that that

is the most dense set over all is the most dense set over all v v and and iterateiterate

S

The problem we need to The problem we need to solve is the densest solve is the densest

subgraphsubgraph• Let Let e(S)e(S), , SSN(v)N(v) be the number of edges in the be the number of edges in the graph induced by graph induced by SS..

• This problem requires finding a subset of the This problem requires finding a subset of the vertices with maximum density vertices with maximum density e(S)/|S|e(S)/|S| and can be and can be solved exactly via flow. This implies (by the density solved exactly via flow. This implies (by the density claim) an claim) an O(log d)O(log d) ratio follows for ratio follows for dd the average the average degree.degree.

• A faster algorithm, approximates the best density A faster algorithm, approximates the best density by by 2 2 but get but get O(n)O(n) time and not flow time. Adds time and not flow time. Adds 22 to to the ratio (so negligible).the ratio (so negligible).

• Was done by Was done by K,PelegK,Peleg in in 19921992. Also . Also Charikar Charikar 19981998..• Very extensively cited for social networks. Almost Very extensively cited for social networks. Almost

always attribute the result to always attribute the result to CharikarCharikar..

A quick approximation for A quick approximation for densest subgraphdensest subgraph

• Let Let bebe e(OPT)/|OPT| e(OPT)/|OPT| • We show that all vertices in the We show that all vertices in the

optimum have degree at least optimum have degree at least ..• Otherwise, removing a vertex with Otherwise, removing a vertex with

degree less than degree less than increases the increases the optimum.optimum.

• Therefore we can iteratively remove Therefore we can iteratively remove vertices of degree strictly less than vertices of degree strictly less than . . We never remove a vertex of We never remove a vertex of OPTOPT and and so the remaining graph is not empty.so the remaining graph is not empty.

The The 22 approximation approximation continuedcontinued

• The degree of all vertices in the final graph The degree of all vertices in the final graph is at least is at least ..

• Say that there are Say that there are ii vertices. The density is vertices. The density is the sum of the degrees, over the sum of the degrees, over 22,, divided by divided by ii..

• In other words the density is at least In other words the density is at least

i*i* /(2i)= /(2i)= /2 /2

Thus ratioThus ratio 2 2. With a bit of data structure, . With a bit of data structure,

O(n)O(n) running time. running time.

A reduction from Set-Cover A reduction from Set-Cover to the minimum size to the minimum size 22--

spannerspanner

A

C

S

SC

h E

C

A reduction from Set-A reduction from Set-CoverCover

A

C

S

SC

hE

C

The edges between A and S are spanned via the edges of h


A

C

S

SC

hE

CFew edges in the set cover instance, so add them. Will be negligible in size.


A

C

S

SC

hE

CIf there is an edge in the optimum between a vertex of A and a vertex of E, replace it by a path of length 2 via S


A

C

S

SC

hE

CThis means that every vertex in A is joined to a set-cover because the edges of A to E can not span themselves


A

C

S

SC

hE

CIf the size of the set cover is opt the size of the 2-spanner is about n*opt and the inapproximability is implied.

How hard is it to How hard is it to approximate approximate

spanners for spanners for k≥3k≥3??• Strong hardness is Strong hardness is exp(log exp(log 1-1- n) n)• Weak hardness is Weak hardness is (log n)/k(log n)/k• K. 98K. 98. First hardness. Weak hardness for . First hardness. Weak hardness for

w(e)=l(e)=1w(e)=l(e)=1. Later similar methods . Later similar methods employed for hardness for employed for hardness for Buy at Bulk.Buy at Bulk.

• Elkin PelegElkin Peleg: Strong hardness for: : Strong hardness for: • 1) 1) l(e)=w(e)l(e)=w(e). . • 2) 2) w=1w=1 arbitrary length arbitrary length • 3)Unit length, arbitrary weights 3)Unit length, arbitrary weights • 4) Basic but directed spanners.4) Basic but directed spanners.

A question posed in A question posed in 19921992

• Does the case Does the case w=l=1w=l=1, , k=3k=3 admit strong admit strong hardness?hardness?

• In In 20002000 Elkin and PelegElkin and Peleg showed conditional showed conditional hardness. If hardness. If Min-RepMin-Rep (see (see K. 98K. 98) with large ) with large super girth is strongly hard, then super girth is strongly hard, then approximating basic approximating basic kk-spanners for -spanners for k≥ 3k≥ 3 is is strongly hard.strongly hard.

• Dinic, Kortsarz and RazDinic, Kortsarz and Raz ICALP 2012ICALP 2012 . . Min-Min-RepRep with large supergirth is indeed strongly with large supergirth is indeed strongly hard, thus basic spanners as well. Solved hard, thus basic spanners as well. Solved after 20 years.after 20 years.

A technique employed for A technique employed for directed Steiner Forestdirected Steiner Forest

• Feldman, K. and NutovFeldman, K. and Nutov..

The following picture:The following picture:

s t

At most n2/5 vertices in every layer

LP flow at least ¼ between every pair s,t

An edge with large An edge with large xxee

• Between every two layers there is at most Between every two layers there is at most nn4/54/5 edges.edges.

• Let Let xxe e be the largest capacity. Thus via every be the largest capacity. Thus via every edge at most edge at most xxe e flow unit pass from flow unit pass from ss to to tt..

• The total flow between The total flow between ss and and tt is at least is at least ¼ ¼..

• Therefore Therefore nn4/54/5** x xee≥ ¼≥ ¼

• Therefore there is an edge of value about Therefore there is an edge of value about 1/n1/n4/54/5

• Iterative rounding gives ratioIterative rounding gives ratio nn4/54/5

Approximating directed Approximating directed spannersspanners

• KrauthgamerKrauthgamer and and DinicDinic employed (part of employed (part of our) techniques to get an our) techniques to get an nn2/32/3 approximation for unit length directed approximation for unit length directed kk--spanners. The ratio does not depend on spanners. The ratio does not depend on kk. . As happens many times, the techniques As happens many times, the techniques were invented independently.were invented independently.

• Improvement: non iterative but randomized Improvement: non iterative but randomized rounding gets about rounding gets about nn1/21/2 ratio. Very clever ratio. Very clever trick!trick!

• Due to Due to Berman, Bhattacharyya, Berman, Bhattacharyya, Makarychev, Raskhodnikova, YaroslavtsevMakarychev, Raskhodnikova, Yaroslavtsev. .

Other resultsOther results

• For For k=3k=3 they get ratio they get ratio nn1/3 1/3 ratio for the ratio for the directed case. Note that even for directed case. Note that even for undirected graphs undirected graphs nn1/21/2 is trivial but is trivial but nn1/31/3 not.not.

• They also improve the result for They also improve the result for Directed Directed Steiner forestSteiner forest about 3 years after it was about 3 years after it was established. The new best ratio is established. The new best ratio is nn2/32/3..

• So far we dealt with weighted spanner.So far we dealt with weighted spanner.• Consider spanners with Consider spanners with w=l=1w=l=1 but we but we

want minimum want minimum maximum degreemaximum degree..

The 20 years patternThe 20 years pattern

• In In 19931993 K., PelegK., Peleg approximation of approximation of 1/4 1/4

for minimum maximum degree for minimum maximum degree 22--spannerspanner

• A related remark: in A related remark: in 19961996 Feige, K, Feige, K, Peleg: a Peleg: a nn1/3-1/60 1/3-1/60 for thefor the dense k-dense k-subgraphsubgraph problem. problem.

• Improved by Improved by BhsaskaraBhsaskara, , Chlamtac, Charikar,Charikar, FeigeFeige and and Vijayaraghavan Vijayaraghavan almost almost 2020 years later years later to aboutto about nn1/41/4 ratio.ratio.

ImprovementsImprovements

• The minimization version of the denseThe minimization version of the dense k k--subgraph problem.subgraph problem.

• Given a bound Given a bound BB on the edges, find a on the edges, find a minimum size collection minimum size collection V’V’ of vertices so that of vertices so that e(V’)≥Be(V’)≥B. Its ratio is never worse than the . Its ratio is never worse than the dense dense kk-subgraph ratio.-subgraph ratio.

• Improvement 1: Improvement 1: Chlamtac, Dinic and Krauthgamer. Ratio 0.172 for minimum for minimum maximum degree maximum degree 22-spanner-spanner

• Improvement 2: Improvement 2: nn0.1720.172 ratio for the ratio for the minimization version of dense minimization version of dense kk-subgraph. In -subgraph. In my opinion great result.my opinion great result.

Remarks on this Remarks on this algorithmalgorithm

• The improvement for Dense The improvement for Dense kk-subgraph can -subgraph can be thought of: instead of counting walks, be thought of: instead of counting walks, count Caterpillars. Still, improves count Caterpillars. Still, improves 2020 years years result.result.

• The new results of The new results of CDKCDK uses the uses the Sherali-Adams hierarchy. Also a notion of . Also a notion of faithful faithful roundingrounding which is new and very interesting. which is new and very interesting.

• In 1993 In 1993 K, PelegK, Peleg said that Bounded degree said that Bounded degree 2-spanner related to 2-spanner related to Dense k-SubgraphDense k-Subgraph..

• Showed to be indeed the case by Showed to be indeed the case by CDKCDK..

Open problems (in Open problems (in approximation as otherwise approximation as otherwise

too many)too many)• The caseThe case k=2 k=2 for unit length and arbitrary for unit length and arbitrary

cost and even on directed graphs is tight. cost and even on directed graphs is tight. Almost all other problems large gap between Almost all other problems large gap between ratio and inapproximability.ratio and inapproximability.

• New variants: New variants: Augmenting spannersAugmenting spanners, , Client-Client-server spannersserver spanners, , All server Client-server All server Client-server spannersspanners. All these problems invented by . All these problems invented by Elkin and PelegElkin and Peleg. Also . Also Transitive closure Transitive closure spannersspanners. . Tree spannersTree spanners. . Additive spannersAdditive spanners..

• In addition a slightly new concept is In addition a slightly new concept is Fault Fault tolerant spannerstolerant spanners..

Documents

A survey on approximating graph spanners Guy Kortsarz, Rutgers Camden