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MAX-FLOW

Given some capacities for the edges and , find a flow from to that satisfies the constraints and maximizes the value ().

MIN-FLOW

Given , costs for the edges and , find a flow from to that has value () and minimizes the total cost ().

MIN-CIRCULATION

Given costs for the edges , find a circulation that minimizes the total cost ().

MAX-MULTICOMMODITY-FLOW

Given commodities , where is the tuple (ie source, destination and demand), and capacities for the edges , find flows from to that satisfies the capacity constraints () and maximizes the value ().

MIN-MULTICOMMODITY-FLOW

Given commodities , where is the tuple (ie source, destination and demand), and costs for the edges , find flows from to that satisfies the demand() and minimizes the costs ().

All these problems can be solved with linear programming, by having a variable for every vertex and putting the constraints, optimization as linear functions on them. But the problem is when one requires a solution for the variables that takes integer values(integral solution), these becomes hard. There is one exception of MAX-FLOW, because of the MAX-FLOW MIN-CUT theorem, if the capacities are integral, the maximum flow has integral value. Finding an integral solution for the MIN-MULTICOMMODITY-FLOW is NP-HARD, even if there are only two commodities.

The MIN-MULTICOMMODITY-FLOW is used to model congestion and routing in the internet. The costs on the edges are just the latencies(the amount of time required for a unit of data to be transfered between the ends of a link). One would like to find how the packets from s to s need to be routed so as to minimize the total latency. But in actual networks the latency of a link increases when more flow is routed through it due to queuing in the devices. So we need the latency on the edges to be increasing functions of the flow rather than just scalars. Also in the internet routing of packets are not controlled by a central authority. So what is assumed is each packet takes a route so as to minimize its latency. Such a flow is called the Nash flow. But such a process need not minimize the total latency incurred by all the packets. The ratio between the largest latency of Nash flows by the Optimal total latency is called the Price of Anarchy.

Though I didn’t understand most of the talks, I got a vague idea of the problems and how they are solved. Streaming algos are heavily used in applications like web search, databases, routers etc. These algos work on a looong stream of data(say of length ), over which they can do only a few passes(, a const, or ), keeping only small memory( or ). Though they have been around for some time, the big push in the area came after the ams paper(all the speakers referenced this paper and for this the authors won the godel prize). This paper gave approximate randomized one pass algorithm for frequency moments using small space. One another important feature is that they also proved lowerbounds using communication complexity. This was an awesome idea and has been successfully used to obtain lower bounds for other streaming problems.

Some of the problems were based on the specifiations that these applications require and some purely theoretical. The first problem is to define a good model of computation that abstracts actual appliccations. Since the applications are different you need to look at different models. People talked about models of distributed streaming computations, models in which on can annotate the stream or write to it, copies of streams can be maintained, multiple passes allowed(the ams paper just looked at simple one pass read only stream).

Many streaming algos work by making a small sketch of the streaming data in there small memory. Later it extracts the requred information about the stream from the sketch. Usually hash functions are used to make sketches.

People seems to have mastered the art of using communication complexity to prove lowerbounds. Even for multiple passes they consider the communication complexity for multiple rounds of communication. There was some talk on 2 other methods called information complexity and round elemibation.

Theorem[by Omer Reingold] : Given a undirected graph and , the question “Is there a path between and ?” can be answered in space

All graphs mentioned in the proof are undirected and may have multiple edges. denotes .

The problem is equivalent to outputting all vertices reachable from in space, or more generally seperating the connected components in . But one of the vertices may be reachable only by an length path from , and it might be a degree vertex. Since this length path can be an arbitrary one, the only way i can think of is by brute force checking of all length paths. But the number of such paths can be exponential in !. If degree is , it can be as large as . However if the graph is such that all nodes reachable from are at a distance , then brute force is a viable method. Such a property is true in special graphs called expanders which are graphs whose adjacency matrix have small second eigen values. The proof basically constructs an expander from an arbitrary undirected graph preserving the reachability between vertices in space.

My aim is just to show the logic that leads to the proof. So i just state the main lemmas without proofs and some definitions. Proofs of the lemmas could be found in Emanuele Viola’s Lecture notes.

First step is to reduce the reachability question in between vertices to reachability question between large connected subsets in some which is -regular graph and has a self loop on every node.

Reduction 1:There is an algorithm which on input and , that uses only spaces and outputs

• a -regular graph with a self-loop on all verices
• , such that
• the induced subgraphs of on and are connected

such that

there is a path between in there is path between to .

But this is actually not a simpler problem. The trouble I mentioned about still remains valid here. But this makes the problem simple for spectral analysis.

Note that if you have a graph without self loops and then add self loops to all nodes, length paths in the latter corresponds to length paths in the former(this is a many-one, surjective map). Another reason for doing this is to make the graph non-bipartite, which is required for the application of the next result.

Its a well known fact that the second eigen value of the adjacency matrix is related to how connected a graph is. Lower the eigen value, the graph is better connected. Following is how the second eigen value is defined.

Definition 1 :

where the normailized adjacency matrix of , and is the all 1′s vector.

For a complete graph and if and only if the graph is disconnected or it is bipartite. For a connected undirected graph .

Lemma 1 : For any connected, non-bipartite undirected graph on vertices

for some constant .

It is known that if is small, the number of edges between subsets of the vertex set is proportional to the product sizes of the subsets with a small additive term. This is known as the Expander mixing lemma. The following is a corollary of this lemma

Lemma 2 : For a -regular graph and , if then there is an edge between and .

So Lemma 2 says that, if is small, then we just need to check whether there is a direct edge to . This can be done provided the degree of nodes in are not high ie exponential in (can happen if there are multiple edges). The naive method to do here is to make a graph in which edges consists of length paths in orginal graph, called . Its easy to prove that this will make the second eigen value . Doing this times will make but the degree becomes . The above method actually corresponds to making a graph whose edges represent long paths in orginal graph as mentioned in the begining. Next is a method for reducing without blowing up the degree. This corresponds to considering only a small subset of the exponential number of paths possible and it will be proved that this is good enough.

Definition 2 :

• is said to be an -graph iff its vertex set is , it is -regular, and has .
• If is an -graph and is an -graph then is a -graph. For the th neighbour of denoted by is where is the th neigbour of in , and is the th neighbour of in (according to the induced ordering on the neighbours due to the ordering on the vertex set).

The operation is similar to graph squaring but the number of neighbours becomes instead of in . This is especially good if the . The next lemma states that the operation also decreases the eigen value.

Lemma 3 :

Instead of considering all vertices at distance , we are considering only some neigbours. Now we will apply the operation times to bring down to and the degree increases to . But after each operation, the degree increases, and so we need a larger for doing the next operation. Next lemma say there exists a sequence of ‘s with small ‘s.

Lemma 4 : There exists a sequence of -graphs where .

Corollary If and then . So after operations,

If , for ,

In the above corollary what is effectively done is that we have constructed a graph , in which edges represent long walks in . Clearly if all length paths were considered() as edges(this corresponds to squaring the graph many times), one cannot even find all the neighbours of a node in log space. So we consider only a carefully choosen sparse subset() by means of the operation. Here is an example of the above procedure done on the board(my mobile has a bad camera, but the image should be viewable at the full size)

Now for this to be done by a log space machine, the th neighbour (, and all other )of any vertex in should be computable in log space. Each th neighbour is obtained by doing a walk of length on . This walk is specified by a sequence of choices of neighbours . The following figure illustrates how to compute given and .

The neighbour (, and all other )of any vertex in is computable in space , for .

Lemma 5 : Reachability question in ( regular) between large connected subsets can be reduced in log space to the question whether there is an edge between large subsets in some graph in which neigbours of any vertex can be computed in log space.

Chernoff bound is by far the most used result in randomized algorithms. But there are many different versions of it, with very ugly proofs involving playing with inequalities relating to functions. Here is a simple proof for the additive version of the same.

‘s are iid 0-1 random variables with expectation . Then

for all

This is the Bernstein’s trick. That is multiplying by and then doing the full summation starting from 0. For terms in the full summation from 0 to , and we are multiplying by a small quantity. For terms to , we are multiplying by a large quantiy. But if , what we hope is these terms are extremely small and even multiplying with will not increase them by much. And both these facts hopefully give a good bound. Now we choose an minimizing above.

Above is minimized for , for , this is for .

Then

For , .

So the coefficient of in the exponent in the formula above is .

Hence .

If you want to see the longer ugly proof, here it is:

The usual proof for Chernoff bound(see this) gives

where is the binary relative entropy, .

If i take , then and (sum of 2 numbers in the denominator are 1).

So Taylor’s Theorem says that for nice functions.

Therefore which gives

Now plug complement in the original proof as the here. Then and . This will give you the expression

References

See this link if you are confused about the n different chernoff bounds. (http://www.cs.berkeley.edu/~jordan/courses/174-spring02/recitation/lec10.pdf)
There is a similar bound for hyper-geometric distributions called Chavatal’s Bound. Here you can find a nice writeup on it.

Go to http://www.wordle.net/create if you want to create one or your blog. It just extracts the rss feeds and make this images.

Being a beneficiary I offcourse support the protest. But instead of just demanding a stipend hike, a broader goal must be set. That is to remove the innefficency in the system. This would ensure more support to the cause(even from within the high ranks of the system) and will hopefully give a permanent solution. But some narrow minded people say why care about what happens after we leave the institute and all I want is my stipend to be hiked as soon as possible.

One strategy is to interview half the people, find the best score and accept the next guy with a higher score. So if the best guy is in the second half and second best guy is in the first half which happens with probability , she will be able to get the best candidate. One could generalize the above and decide to find the best score among the first candidates and choose the next guy who beats this score. With some asymptotic analysis, the probability of finding the best guy comes to where .(see http://en.wikipedia.org/wiki/Secretary_problem#Deriving_the_optimal_policy). This is maximized when giving a success probability of . It can be proved that this is the best that can be achieved.

The probabilists way of stating this is a sequence of random variables of scores. Since all candidates have different scores, is a random permutation of the scores(as we assumed that she calls candidates in a random order, and lets also assume that the scores are independent of what time they are taken). To make things slightly more complicated, let be the indicator random variable for .  After having seen (these form distinct numbers with gaps between them),   is equally likely to be in anyone of  the gaps. Therefore

Claim : is independent of

Proof : Following the logic which gave the probability above, whether is the maximum of whats seen till now doesnt depend on the ordering of .

We want to know what the best stratergy is.

Suppose we have already seen , we have no option but to choose the th candidate.

If , with probability 1 we have the best guy, else its 0. For later purposes lets define

Now again Whats the best strategy to follow once we have seen .

If ,

1. with probability we have the best guy
2. if we skip this guy with probability it will be found in the next step if the best strategy is followed from next step.

So we can compute these 2  values and choose the option with higher probability.  Let above 2 values

If ,

1. with probability we have the best guy
2. if we skip this guy with probability it will be found in the next step if the best strategy is followed from next step.

So the best thing is to skip. Let second value. And we can continue this inductively. Suppose be the probability of success if the best strategy is  followed from step given .

If then else  (ie if current is not the best among the candidates so far seen, he is definitely not best among all and so  we skip, hoping that best guy is yet to come).

As is increasing in , and it can be proved that is decreasing(i think after some threshold) at some point they cross. So from this point onwards we will choose the first guy that has best score so far(ie max is obtained for option 1). But we had already proved that best among such strategies is obtained by setting the threshold point to to obtain a probability of getting best guy as .[this is slightly rough write up, i hope to improve it later]

XOR(symmetric difference) of a n matching and a k matching is a set of even length cycles, even length alternating paths, odd length paths with 1 edge more from n matching(say a of them) and odd length augmentable paths(say b of them) for the k matching, such that a-b=n-k and b<=n-k.