What did I learn What did I learn about about cut, cut,

expansion expansion and and densitydensity problems problems

Guy KortsarzGuy Kortsarz

The character flaw of Michael The character flaw of Michael Myers, Freddy Kruger and Jason Myers, Freddy Kruger and Jason Voorhees: Cut problems.Voorhees: Cut problems.

Input:Input: an edge weighted graph an edge weighted graph G(V,E) G(V,E) and a and a

collection of pairs collection of pairs {{SSii,,tti i }} Output:Output: a minimum cost collection of edges whose a minimum cost collection of edges whose

removal disconnects all pairsremoval disconnects all pairs

We look at it as a fractional relaxation that We look at it as a fractional relaxation that sayssays: Put lengths on edges : Put lengths on edges so that the distance so that the distance

between between SSi i andand ttii is at least is at least 1 1 for every for every i.i.

The The “good cut point” “good cut point” lemmalemma

s t

0.2

0.1

0.4

0.01

0.050.07

0.1

0.1

0.2

0.2

0.1

0.5

0.20.3

0.3

0.1

There is a radios r0.5 so that if you cut at r the cut cots value is only O(log n)factor larger than the LP value inside the sphere.

0.1

RemarksRemarks

The fact that The fact that r<0.5r<0.5 is important. is important. Means that no Means that no s,t s,t pair can pair can bothboth be in the be in the

sphere.sphere. The total cost is roughly The total cost is roughly O(log n)O(log n) times times

the LP value.the LP value. Needs to give some non zero value for Needs to give some non zero value for

radios radios 00 for this to work. for this to work.

Vertex separators in Vertex separators in undirected graphsundirected graphs

Say that we have a collection of Say that we have a collection of {s,t} {s,t} pairs so that the distance between every pairs so that the distance between every pair is at least pair is at least d.d.

Say that you want to delete Say that you want to delete verticesvertices and and cut cut all all such pairs.such pairs.

A fractional solution of A fractional solution of n/d n/d is giving every is giving every vertex value vertex value 1/d1/d..

Small vertex separatorSmall vertex separator

There is a collection of at most There is a collection of at most

O(log n)n/d O(log n)n/d vertices who separate all the vertices who separate all the pairs of distance at least pairs of distance at least dd..

Not that far from the fractional solution.Not that far from the fractional solution. This is not written anywhere (but not hard This is not written anywhere (but not hard

to prove).to prove). Credit: the upper bound due to Credit: the upper bound due to GargGarg

VaziraniVazirani and and YanakakisYanakakis

Separating pairs of distance Separating pairs of distance al least al least dd: directed graphs: directed graphs

How many edges do you need to remove How many edges do you need to remove if we have a collection if we have a collection {s,t} {s,t} of pairs so of pairs so that that dist(s,t)≥d?dist(s,t)≥d?

We worked on this problem without We worked on this problem without knowing that its closely related to the knowing that its closely related to the Maximum Number of Disjoint Paths in Maximum Number of Disjoint Paths in Directed graphsDirected graphs..

Thus some results were knownThus some results were known Chekuri and Khanna Chekuri and Khanna O(nO(n44/d/d44) ) edges.edges. Hajiaghayi, Leighton Hajiaghayi, Leighton O(nO(n33/d/d33)) edges edges

What we proved What we proved

We proved that there exists a separator with We proved that there exists a separator with O(nO(n22/d/d22) ) edges (edges (K, NutovK, Nutov))

Good achievement .Good achievement . Then I found out: The above easily implies an Then I found out: The above easily implies an

O(nO(n2/32/3)) approximation to the approximation to the “Maximum “Maximum number of disjoint paths problem on directed number of disjoint paths problem on directed graphs”graphs”

Thus I started to Thus I started to suspectsuspect this lemma was this lemma was proved before.proved before.

It was done 3 months It was done 3 months before in a before in a SODASODA paper paper

The authors The authors Varadarajan Varadarajan andand VenkataramanVenkataraman. . Hence the Hence the O(nO(n2/32/3) ) was known.was known.

The algorithm: among the non joined pairs The algorithm: among the non joined pairs that are reachable take the that are reachable take the shortest pathshortest path..

Still we proved a ratio Still we proved a ratio O(nO(n2/32/3/opt/opt1/31/3)) ratio for ratio for unit capacities. unit capacities.

Also Also O(opt)O(opt) is known (see later is known (see later Anupam Anupam GuptaGupta))

Clearly Clearly sqrt{n}sqrt{n} ratio is only ratio is only if if opt=sqrt{n}opt=sqrt{n}

Best ratio known by Best ratio known by Amit AgarwalAmit Agarwal, , Noga Alon Noga Alon and and Moses CharikarMoses Charikar. . O(nO(n1111/n/n2323) ) ratio.ratio.

The paper seems very complex to meThe paper seems very complex to me For the unit capacity case since we have For the unit capacity case since we have

sqrt{n} sqrt{n} only for only for opt=sqrt{n}opt=sqrt{n} maybe there is a maybe there is a simpler algorithm that returns simpler algorithm that returns o(n) o(n) edges?edges?

Hardness of approximation: Hardness of approximation: LabelcoverLabelcover hardhard. . ChuzhoyChuzhoy and and Khanna.Khanna.

Both the breaking of the Both the breaking of the sqrt{n}sqrt{n} and the and the lower bound are huge achievments.lower bound are huge achievments.

The elegant and simple The elegant and simple paper by Anupam Guptapaper by Anupam Gupta The ratio is only The ratio is only O(sqrt{n}) O(sqrt{n}) but very elegant.but very elegant. The paper of The paper of Agarwal et al Agarwal et al used ideas from used ideas from

the paper of Gupta.the paper of Gupta. Also from Also from my paper with Nutovmy paper with Nutov. Reducing . Reducing

reachability among pairs.reachability among pairs. And now to another less known but very And now to another less known but very

simple simple cut lemmacut lemma.. In this case we may charge In this case we may charge allall edges. edges.

The ideas of The ideas of GuptaGupta

Solve the LP. Add all edges with Solve the LP. Add all edges with

xxee ≥1/sqrt{n} ≥1/sqrt{n} to the solution. to the solution. Consider a non separated pair Consider a non separated pair ssjj t tjj And now we show that there is a cut with And now we show that there is a cut with 1/3 ≤ r≤ 1/3 ≤ r≤

2/3 2/3 distance from distance from ssj j of value at most the of value at most the fractional fractional optopt. .

For this, break the distances of the LP to For this, break the distances of the LP to multiplication of a small multiplication of a small ..

We do get We do get x x eeccee if we go from if we go from 1/3 1/3 toto 2/3 2/3 and and multiply the costs bymultiply the costs by

The The I’thI’th farctional value farctional value adds the cost crossing adds the cost crossing ii

to to (i+1)(i+1) times times 1/3≤i1/3≤i≤2/3≤2/3 Cut(iCut(i)≤opt)≤optff

It can not be that for every i:It can not be that for every i:

Cut(iCut(i)>3 opt)>3 optff

This means that there is a cut whose This means that there is a cut whose value is at most value is at most 33optoptff

Note that here Note that here everyevery (relevant) edge for (relevant) edge for s s jj t tjj is charged.is charged.

Loss of reachabilityLoss of reachabilityLet Let (u,v) (u,v) be an edge on a path from be an edge on a path from ssjj to to ttjj. .

Note that there is a path of length Note that there is a path of length 1/31/3 in the in the LPLP distances that distances that uu was was able to reach and now it does not.able to reach and now it does not.

To make length To make length 1/3 1/3 you need you need (sqrt{n}) (sqrt{n}) vertices. Thus vertices. Thus (u,v) (u,v) is is charged at most charged at most O(sqrt{n}) O(sqrt{n}) times times hence hence sqrt{n} sqrt{n} ratio. In addition as ratio. In addition as each time charged each time charged 1/31/3 and disjoint and disjoint 1/3, so O(opt 1/3, so O(opt ff)) ratio follows. ratio follows.

Loss of reachabilityLoss of reachabilityLet Let (u,v) (u,v) be an edge on a path from be an edge on a path from ssjj to to ttjj. .

Note that there is a path of length Note that there is a path of length 1/31/3 in the in the LPLP distances that distances that uu was was able to reach and now it does not.able to reach and now it does not.

To make length To make length 1/3 1/3 you need you need (sqrt{n}) (sqrt{n}) vertices. Thus vertices. Thus (u,v) (u,v) is is charged at most charged at most O(sqrt{n}) O(sqrt{n}) times times hence hence sqrt{n} sqrt{n} ratio. In addition as ratio. In addition as each time charged each time charged 1/31/3 and disjoint and disjoint 1/3, so O(opt 1/3, so O(opt ff)) ratio follows. ratio follows.

The Max Cut problemThe Max Cut problem

Given Given G(V,E) G(V,E) separate separate V V intointo S S andand V-S V-S and and maximizes maximizes e(S,V-S)e(S,V-S)

One of the best papers ever:One of the best papers ever: 0.878 0.878 ratioratio

by by M. X. Goemans M. X. Goemans and and D. P. WilliamsonD. P. Williamson. . Introduced Introduced PSDPSD to approximation. to approximation.

This ratio is tight assuming the This ratio is tight assuming the Unique Game Unique Game ConjectureConjecture of of KhotKhot

In my opinion the hard instances for In my opinion the hard instances for UGCUGC are are expanders with hidden dense subgraph inside. expanders with hidden dense subgraph inside.

ConductanceConductance

Conductance: Conductance: e(S,V-S)/deg(S) e(S,V-S)/deg(S) (a.k.a sparsest (a.k.a sparsest cut) cut)

Approximation AlgorithmsApproximation AlgorithmsO(log n) O(log n)

[Bourgain], [Leighton, Rao],[Bourgain], [Leighton, Rao], [Linial,London,Rabinovich][Linial,London,Rabinovich] Uses (quite elementary) embedding of a Uses (quite elementary) embedding of a metric into metric into LL1 1 O(log n)O(log n) stretch factor.stretch factor.

The best known ratio The best known ratio

By By Arora, Rao and Vazirani.Arora, Rao and Vazirani.

Approximation ratio Approximation ratio

Uses the so called negative type Uses the so called negative type metric.metric.

Uses what the authors call Uses what the authors call expander expander flowsflows

We now focus on the We now focus on the small set small set expansion conjectureexpansion conjecture..

The small set expansion The small set expansion conjectureconjecture

Let Let ≤0.5 ≤0.5 be a constant and let be a constant and let be an be an arbitrary small constant.arbitrary small constant.

Let Let (S)=e(S,V-S)/deg(S) (S)=e(S,V-S)/deg(S) be the be the expansion factor of expansion factor of SS..

Let Let X X be the collection of all subsets of be the collection of all subsets of VV with size with size nn

Consider Consider (()=)=MinMinSSXX(S)(S)

The conjectureThe conjecture

It is hard to distinguish It is hard to distinguish between the between the following two cases:following two cases:

(a)(a) (()≤)≤ andand (b) (b) (()≥1-)≥1- This is a This is a weaker weaker assumption than the assumption than the

Unique Game conjecture by Unique Game conjecture by KhotKhot.. If we prove the If we prove the SSECSSEC we prove the we prove the UGCUGC

A closely related problemA closely related problem

There is an alternative way to disprove There is an alternative way to disprove the conjecture (due to the conjecture (due to Gandhi Gandhi and and KK).).

By describing a very simple problem that By describing a very simple problem that

has has ratio 2ratio 2 even in the weighted case. even in the weighted case. And we will show that under the And we will show that under the SSECSSEC

this is the best ratio.this is the best ratio. Or a Or a 2-2- ratio for the ratio for the problem disproves theproblem disproves the SSEC SSEC..

The problem The problem

Given: a graph Given: a graph G(V,E)G(V,E) and a number and a number kk.. Required: Find a set Required: Find a set UU of of k k vertices with vertices with

minimum number minimum number of touching edgesof touching edges A touching edge: at least one endpoint in A touching edge: at least one endpoint in UU Remark: our main results were for the Remark: our main results were for the

weighted case. We improved a result by weighted case. We improved a result by ShmoysShmoys et al and a different one by et al and a different one by HochbaumHochbaum et al. et al.

A trivial ratio of A trivial ratio of 2 2

•Let OPT, |OPT|=k, be the best solution

•Let U be the k least degrees vertices , thus deg(OPT)≥ deg(U)

•Clearly t(OPT)≥deg(OPT)/2•: t(U)≤deg(U)≤deg(OPT) ≤2t(OPT)

The weighted caseThe weighted case

Vertices have Vertices have weightsweights. . Minimum edges under cost at least Minimum edges under cost at least kk: :

Find a set Find a set U U of cost at least of cost at least k k and and minimize the number of edges touching minimize the number of edges touching UU..

We explain the ratio We explain the ratio 22 for a related for a related question: maximum touching edges question: maximum touching edges m’m’ and and maximize maximize the vertex cost.the vertex cost.

Some ideas of how to give Some ideas of how to give ratio ratio 22 for for Maximum cost at Maximum cost at

mostmost M M edgesedges We use We use Dynamic ProgrammingDynamic Programming.. We guess the number We guess the number PP of edges of edges

between the optimum set between the optimum set OPTOPT and and V-V-OPTOPT..

We guess the sum of degree of We guess the sum of degree of OPTOPT whom may be whom may be 2M2M. A serious technical . A serious technical problem: we are only able to compute problem: we are only able to compute A[n, P, M]A[n, P, M]..

The reason for thatThe reason for that

This is the only way, it seems, to assure This is the only way, it seems, to assure feasibility.feasibility.

Indeed if Indeed if deg(U)≤M deg(U)≤M then then t(U)≤Mt(U)≤M.. The question is do we loose a lot by bounding The question is do we loose a lot by bounding

the sum of degrees by the sum of degrees by M M while the sum may while the sum may bebe 2M 2M??

One more detail: we need to guess the One more detail: we need to guess the highest cost vertex in highest cost vertex in OPT OPT and add it the our and add it the our solution.solution.

IfIf deg(U)≤M deg(U)≤M, , how much how much cost we loose?cost we loose?

Let Let OPT= AOPT= A {x} {x} B B so that so that deg(A+x) deg(A+x) isis the first to be above the first to be above MM

Thus Thus A A isis a feasible solution for a feasible solution for MM.. Clearly Clearly BB too is a feasible solution for too is a feasible solution for MM

because because deg(A+x)>Mdeg(A+x)>M and the total at most and the total at most 2M2M

One of One of AA or or BB has has ½½ the weight. The fact the weight. The fact that we guess the highest cost vertex in that we guess the highest cost vertex in OPT OPT compensate for compensate for xx..

Thus ratio Thus ratio 22..

What is the properties of a What is the properties of a good solution?good solution?

Let us check the case of a d regular graphs. The question is if the edges are internal or

external

S

What is the properties of a What is the properties of a good solutiongood solution

In this example most of the edges in S stay inside. Which means that t(S) is close to kd/2

U

What is the properties of a What is the properties of a good solutiongood solution

But S can behave badly. Namely most edges go to

V-UIn this case t(S) close to dk.

U

Is the Is the Small SET Expansion Small SET Expansion ConjectureConjecture reliable?reliable?

Opinions vary. I think: Opinions vary. I think: VERY RELIABLEVERY RELIABLE.. We tried to disprove the We tried to disprove the SSGE SSGE and and

failed.failed. Namely we tried to give a ratio Namely we tried to give a ratio 2-2- for for

the problem descrbed before.the problem descrbed before. It seems that the It seems that the SSECSSEC is related to the is related to the

a a Dense k-subgraphDense k-subgraph

Partition via Sparsest cutPartition via Sparsest cut Motivation:Motivation:

1.1. Natural Social Natural Social Communities[MSST08,ABL10,…] Communities[MSST08,ABL10,…]

2.2. Better clusters (AGM)Better clusters (AGM)3.3. Easier to compute (GLMY)Easier to compute (GLMY)4.4. Useful for DistributedUseful for Distributed Computation (AGM)Computation (AGM)

Good Clusters Good Clusters Low Conductance? Low Conductance? Inside: Well-connected, Inside: Well-connected, Toward outside: Not so well-connected.Toward outside: Not so well-connected.

Overlapping ClusteringOverlapping Clustering

Find a set of (at most Find a set of (at most KK) overlapping clusters: ) overlapping clusters:

each cluster degree sumeach cluster degree sum≤ ≤ BB, ,

coveringcovering all nodesall nodes, ,

and minimize:and minimize: Maximum conductance of clusters (Maximum conductance of clusters (Min-MaxMin-Max)) Sum of the conductance of clusters (Sum of the conductance of clusters (Min-SumMin-Sum))

Overlapping vs. non-overlapping variants?Overlapping vs. non-overlapping variants?

Summary of ResultsSummary of Results[Khandekar, K, Mirrokni.][Khandekar, K, Mirrokni.]Overlap vs. no-overlap:Overlap vs. no-overlap:

Min-Sum: Within a Min-Sum: Within a factor 2 using Uncrossingfactor 2 using Uncrossing.. Min-Max: Might be arbitrarily different.Min-Max: Might be arbitrarily different.

A lot of follow up work.A lot of follow up work.

Arora et al: Finding overlapping communities in social networks: Toward a rigorous approach.

M. Balcan et al: Again, rigorous study of overlapping clusters

And much more. Now last topic: Density

The densest subgraph The densest subgraph problemproblem

Let Let e(S)e(S), be the number of edges in the graph , be the number of edges in the graph induced by induced by SS..

This problem requires finding a subset This problem requires finding a subset SS of of VV that maximizes that maximizes e(S)/|S|e(S)/|S|..

A faster algorithm, approximates the best A faster algorithm, approximates the best density by density by 2 2 but get but get O(n)O(n) time which is much time which is much faster than flow.faster than flow.

Was done by Was done by K,PelegK,Peleg in in 19921992. Also . Also Charikar Charikar 19981998..

Very extensively cited for social networks. Very extensively cited for social networks. Almost always attribute the result to Almost 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 optimum We show that all vertices in the optimum

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

less than less than increases the optimum.increases the optimum. Therefore we can iteratively remove Therefore we can iteratively remove

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

The The 22 approximation approximation continuedcontinued

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

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

The The dense k-subgraph dense k-subgraph problemproblem

Input: A graph Input: A graph G(V,E)G(V,E) and a number and a number kk

Required: a set Required: a set UU of size of size kk with with maximum maximum e(U)e(U)

I started working on this circa I started working on this circa 19931993..

What did I learn? What did I learn?

Unfortunately, not muchUnfortunately, not much

It may be fair to say that this is a It may be fair to say that this is a central central problem. Amazing number of applications (too problem. Amazing number of applications (too long to list).long to list).

A big disappointment: under A big disappointment: under PPNP NP there is there is still no hardnessstill no hardness

No No PTASPTAS under under NPNP is not sub exponential. is not sub exponential. KhotKhot

Harness under assumptions on Harness under assumptions on random 3-random 3-SAT. SAT. FeigeFeige

A fact about walks (and A fact about walks (and a one line proof) a one line proof)If the minimum degree is If the minimum degree is then then clearly the number of walks of length clearly the number of walks of length d d is at least is at least nndd

What about the average degree?What about the average degree?

It turns out that this is also true for It turns out that this is also true for

the average degree.the average degree.

The number of walks with The number of walks with respect the the average respect the the average degree degree

It turns out that the number of walks It turns out that the number of walks is also at least is also at least nndd

I do not know the proof to this.I do not know the proof to this.

But I discovered years later that is But I discovered years later that is was known from was known from 1920’s!1920’s!

A problem is hard: bypass the proof.A problem is hard: bypass the proof.

Do not solve problems: Do not solve problems: bypass thembypass themBy the claim above the number of By the claim above the number of walks of length walks of length dd is at least is at least nndd

We bypass this hard proof. We give We bypass this hard proof. We give a one line or so proof that there is a one line or so proof that there is alwaysalways u,v u,v so that so that Walks(u,v)≥ Walks(u,v)≥ dd/n/n

Uses a known fact: the largest Uses a known fact: the largest eigenvalue of the matrix of eigenvalue of the matrix of G G is at is at least the average degree inleast the average degree in G G..

[Feige, K, Peleg]: [Feige, K, Peleg]: nn approximation for approximation for <1/3<1/3• Consider the matrix of the graph.Consider the matrix of the graph.

•Let Let 11≥ ≥ 22≥ ≥ 33 ≥… ≥… be the (real) be the (real)

eigenvalues of eigenvalues of A.A.

•Well known: Well known: trace(A)=trace(A)= i i ii

•Well known: the eigenvalues of Well known: the eigenvalues of AAkk

areare IIk k

Well known that Well known that AAkk[i,j] [i,j] are the are the number of walks fromnumber of walks from i i toto j j of length of length kk..

The proofThe proof

•AAkk[i,j][i,j] A Akk[i,j]=[i,j]= ii II2k 2k ≥ ≥ 11

2k 2k ≥≥2k2k

• This implies that there is an This implies that there is an i i and and

aa j j so that so that AA2l2l[i,j] ≥[i,j] ≥2k2k/n/n22

The claim follows by taking square The claim follows by taking square root of every side.root of every side.

Why did we count walks?Why did we count walks?

Walks are example of trees that Walks are example of trees that exist if the graph is dense enough exist if the graph is dense enough but not if it is random.but not if it is random.

The additional insightThe additional insightBhaskara, Charikar, Chlamtac, Bhaskara, Charikar, Chlamtac, Feige Vijaraghavan. Feige Vijaraghavan. nn1/41/4 ratioratio

Almost Almost 2020 years later! years later!

•The intuition comes from comparing The intuition comes from comparing random graph to a random graph random graph to a random graph with a dense subgraph with a dense subgraph implanted implanted in in the random graph.the random graph.

•It also turns out that walks are It also turns out that walks are not not the best thingthe best thing to count. to count.

Counting local treesCounting local treesIt turns out that in a dense graph It turns out that in a dense graph some trees appear some trees appear moremore than in than in G(n,p)G(n,p). For example walks.. For example walks.

Ad hoc-Ad hoc- systematic. systematic.

The state of the art: we can not tell The state of the art: we can not tell between a graph chosen from between a graph chosen from G(n,1/sqrt{n}) G(n,1/sqrt{n}) and the same graph and the same graph with with sqrt{n} sqrt{n} vertices changed to vertices changed to G(sqrt{n},1/nG(sqrt{n},1/n1/41/4))

Log densities Log densities • Take two graphs. The original Take two graphs. The original

graph that has, say graph that has, say n n vertices and vertices and

average degree average degree d.d.

•The log density of a graph is The log density of a graph is

log log n n dd

Say that the Dense subgraph hasSay that the Dense subgraph has k k vertices and average degree vertices and average degree pp..

The log density is The log density is log log kk p p

The relation of the log The relation of the log densitiesdensities

A non trivial graph is returned only if the A non trivial graph is returned only if the log density of the log density of the Dense k-subgraphDense k-subgraph is is larger than the log density of the graph.larger than the log density of the graph.

Consider the two graphs discussed:Consider the two graphs discussed: G(n,1/sqrt{n}) G(n,1/sqrt{n}) and the same graph with and the same graph with sqrt{n} sqrt{n} vertices changed to vertices changed to G(sqrt{n},1/nG(sqrt{n},1/n1/41/4))..

The log densities The log densities

The initial graph has The initial graph has nn vertices and vertices and sqrt{n} sqrt{n} average degree.average degree.

The Dense The Dense kk-subgraph has -subgraph has sqrt{n} sqrt{n} vertices vertices and has average degree and has average degree nn1/41/4

This is problematic because they have the This is problematic because they have the same log densities: same log densities:

log log nn sqrt{n}= log sqrt{n}= log sqrt{n}sqrt{n} n n1/41/4 =1/2 =1/2

Techniques Techniques

This same result was achieved by This same result was achieved by

by two groups.by two groups.

One using LP One using LP lift and project.lift and project.

The other group showed that The other group showed that

combinatorial techniques suffice.combinatorial techniques suffice.

The technical difference is that The technical difference is that FKP FKP

counted counted walkswalks while the new paper while the new paper

counts counts CaterpillarsCaterpillars..

Something is missing in Something is missing in our lower bound our lower bound understandingunderstanding• We keep adding assumptions. The We keep adding assumptions. The SSECSSEC, the , the 2 to 1 unique game 2 to 1 unique game conjectureconjecture. . Projection game Projection game conjectureconjecture..

•Countless paper show hardness Countless paper show hardness just by saying the paper is just by saying the paper is Dense k-Dense k-sub graph hard sub graph hard to approximate.to approximate.

Technically, there is no Technically, there is no value to such hardnessvalue to such hardness• Still, I would not try to give a Still, I would not try to give a polylogpolylog ratio to problems that are ratio to problems that are dense k-subgraph dense k-subgraph hardhard..

•An example of a simple such An example of a simple such problem. problem. k-Steiner Forestk-Steiner Forest: connect : connect kk of the pairs. At best we can expect of the pairs. At best we can expect poly ratios. poly ratios.

•It seem OK assume the exponential It seem OK assume the exponential time hypothesis namely that 3-SAT time hypothesis namely that 3-SAT can not be solved in time can not be solved in time 22o(n)o(n)

The exponential time hypotesis The exponential time hypotesis is totally central to fixed is totally central to fixed

parameter tractabilityparameter tractability.. Assuming that we can show that Assuming that we can show that PP Quasi(P). Quasi(P). A favorite open problem. Say that in Clique A favorite open problem. Say that in Clique

you need to return you need to return (1) (1) vertices.vertices. Also say that Also say that opt=logloglog nopt=logloglog n.. Any inapproximability for Any inapproximability for CliqueClique or or Set CoverSet Cover?? The conjecture is that no The conjecture is that no g(opt) g(opt) ratio is ratio is

possible in time possible in time f(opt)f(opt)Poly(n) Poly(n) for any for any g,f.g,f.

Essential Essential open problemsopen problems

1) Is the randomality gap 1) Is the randomality gap nn1/41/4 the right inapproximability the right inapproximability for the for the Dense k-subgraphDense k-subgraph problem? If not give some problem? If not give some strong inapproximability.strong inapproximability.

2) Inpproximability of2) Inpproximability of

(log n) (log n) for for Undirected Undirected

Multicut Multicut (?)(?)