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Dieter van Melkebeek, the conference chair of the Computational Complexity Conference (CCC), has started a discussion on whether to end the IEEE “sponsorship” of CCC. This is, I realize, a supremely boring topic, which seemingly affects only the handful of people who are tasked with organizing the conference. It does, however, affect to some extent all who ever paid a registration fee or published a paper in CCC, and I would like to discuss why I am, in the strongest possible way, in favor of leaving IEEE.

First of all, the term “sponsoring” here is a bit misleading. IEEE sponsors CCC not in the way Coca Cola sponsors the Olympics, but more in the way pirates sponsor navigation, or Elsevier sponsors scholarly publications. That is, it’s not that IEEE subsidizes the conference in exchange for exposure and good will; IEEE takes a huge chunk of our registration money, in exchange for making the lives of the local organizers harder. (I don’t mean to single out IEEE, when a professional society sponsors a conference it always works this way, but IEEE is particularly skilled at making the lives of the organizers hard.)

Last year I was an organizer of the complexity conference at Stanford. We had 108 attendees, who paid a total of $37,425 of registrations, or $346.5 each on average. How did we spend this money?

IEEE charged a 20% overhead to all expenses; that’s about $60 per person and about $6,500 for the whole conference. This, I should emphasize, is for doing literally nothing, except creating problems. For example, our conference could not be “approved” until all the paperwork of the previous conference was closed; it wasn’t closed because they were expecting some banking documents from them and I don’t remember if the issue was that the document does not exist in Portugal, or that they had already received the document and they had lost it. We were not allowed to make the conference web site go live until this was resolved.

One of the perks of organizing the conference with IEEE is that they provide free banking in the United States. (If the conference were organized by a created-for-this-purpose non-profit, it could have its own permanent bank account for little or no fee.) Three weeks after we sent all the paperwork to have our IEEE bank account set up, we get an email from the person we paid $6,500 to “assist” us, saying “URGENT, URGENT, you have to send us the paperwork for the bank account”. After we reminded them that they had had it for three weeks, they replied, “oh yes, we do,” no apology. (And it still took a while to get the checks.)

The free banking does not include free registration handling, however. Here I accept responsibility for being foolish enough, after all this, to trust IEEE with their registration service. At a cost of more than $26 per person (that’s almost $3,000) they produced a registration website that seemed put together by a high school intern, and which required filling up five pages of… well, if you attended the conference you remember it.

Finally, IEEE press charged almost $2,000 (almost $18 per person) for the proceedings. Which proceedings, you may ask, since the conference had no proceedings? That’s a very good question. The charge of $1,925 was to take our papers, take the copyright, and then put the papers were it is impossible to find them, even if you subscribe to the IEEE digital library. (Seriously, try to google a paper appeared in an ACM conference and one appeared in an IEEE conference and see if you are able to get the IEEE paper. Say what you want about the ACM, at least they know how to build a web site.)

In summary, IEEE took $6,500 for doing nothing but causing delays, $3,000 for a terrible registration site, and $2,000 to hide our papers. That’s more than $100 per attendee, and about 30% of how we spent the registration fee. The rest went on food and room rent.

Why are we still doing this? One reason is that the only people that are really affected by this system are the local organizers, and once one is done with the job, one doesn’t want to hear of or talk about IEEE any more. The other reason is that there is a big initial effort that is needed to make the conference independent. One needs to start some kind of entity to own the conference (the IACR, for example, was founded pretty much for the purpose of running CRYPTO and the other crypto conferences), which needs to have a statute, officers, a bank account and all kind of paperwork needs to be done. After that, however, it’s smooth sailing and considerable savings.

Here is one thing that IEEE does: if the conference runs a deficit, they cover it. We did, in fact run a deficit last year, of about $1,000; so IEEE “covered” it, but that just means that instead of $6,500 for the “sponsorship” they got $5,500. If, going forward, we budget with the intention of putting away $5,000 to $7,000 per year, the registration costs will go down slightly and over a few years we can put away, say $20,000 that can be a cushion for any kind of unforeseen event, and from that point forward budget to a balanced budget, with notably lower registration fees, and with some years running a surplus and some years running a deficit. Plus, we own our papers, and people can find them!

Ok, this is as boring as could be expected. I thank all those that have read so far, and, for the sanity of future local organizers and the principle of keeping our hard-earned taxpayer money and our papers, please support making CCC an independent conference.

Link to CCC forum

I would like to pass along the following piece of information that I received from Tal Rabin: the next bi-annual Women in Theory Workshop will be this May 28-30, 2014, in New York City, immediately preceding STOC 2014, which will also take place there.

Applications are due by January 20, 2014.

The relevant information is at womenintheory.wordpress.com

The Simons Institute for the Theory of Computing at Berkeley has started operations a couple of months ago, it has been off to a great start. This semester there is a program on applications of real analysis to computer science, in which I am involved, and one on “big data,” whose workshops have been having such high attendance that they had to be organized offsite.

The institute itself is housed in a beautiful circular three-story building, half of whose second floor is an open space with sofas, whiteboards, tall ceilings, big windows, and exposed pipes on the ceiling, for an added loft-like look. If it had a ball pit, a foosball table, and free sushi it would look like the offices of a startup.

Next year, there will be programs on spectral graph theory, on applications of algebraic geometry, and on information theory.

Junior people (senior graduate students, postdocs, and junior assistant professors who have received their PhD no longer than six years ago) can participate in the program of the institute as “fellows.” Information on the fellowships is at simons.berkeley.edu/fellows2014; the deadline to apply is December 15.

The theoretical computer science traveling circus returns to the San Francisco Bay Area: FOCS 2013 will be in Berkeley in about 3 weeks. The early registration deadline is **this Friday**, October 4.

The week before FOCS, the Simons Institute will have a workshop on parallel and distributed algorithms for learning and optimization, as part of its program on big data.

The conference will be held in the same hotel in which FOCS 2006 was held, in the Berkeley marina. It is not a bad idea to rent a car, which will make it easier to go to downtown Berkeley and to San Francisco for dinner. You will get to drive on the new and improved Eastern span of the Bay Bridge, which is totally safe, even if the bolts are defective and will need to be replaced, or so we are being told.

There is plenty to do around San Francisco, and this list of links maintained by the Simons institute is very helpful. A couple of suggestions specific to the weekend of the 26th: open studios will run in the Mission/Castro/Noe Valley area, and there are good coffee, food and drink options in the area after seeing the art. On Friday evening, the De Young museum is open until 9pm, and it has a full bar and live music in the lobby; downstairs there is a temporary exhibition of Bulgari jewels. The De Young was designed by Herzog and de Meuron, who designed the Tate Modern in London and the National Stadium in Beijing; it is one of the few beautiful modern buildings in San Francisco. On the 27th, there is a pre-Halloween costume 5k run in Golden Gate park. Also on the 27th, James Franco will be at the Castro theater to talk about his new book and about how awesome he is.

Having (non-rigorously) defined the Laplacian operator in manifolds in the previous post, we turn to the proof of the Cheeger inequality in manifolds, which we restate below.

Theorem 1 (Cheeger’s inequality)Let be an -dimensional smooth, compact, Riemann manifold without boundary with metric , let be the Laplace-Beltrami operator on , let be the eigenvalues of , and define theCheeger constantof to bewhere the is the boundary of , is the -dimensional measure, and is -th dimensional measure defined using . Then

We begin by recalling the proof of the analogous result in graphs, and then we will repeat the same steps in the context of manifolds.

Theorem 2 (Cheeger’s inequality in graphs)Let be a -regular graph, be its adjacency matrix, be its normalized Laplacian matrix, be the eigenvalues of , and define for every subset of vertices . Define the conductance of aswhere is the number of edges with one endpoint in and one endpoint in . Then

**1. Proof the Cheeger inequality in graphs **

We will use the *variational characterization* of the eigenvalues of the Laplacian of a graph .

and if is a minimizer in the above expression then

Following the definition of we see that

and so the minimum in (3) is 0, and it is achieved for . This means that

The expression in the right-hand-side of (4) is an important one, and it is called the Rayleigh quotient of , which we will denote by :

It is also useful to consider the variant of the Rayleigh quotient where there are no squares; this does not have a standard name, so let us call it the Rayleigh quotient and denote it by :

The proof of the graph Cheeger inequality now continues with the proof of the following three facts.

Lemma 3 (Rounding of embeddings)For every non-negative vector there is a value such that

Lemma 4 (Embedding of into )For every non-negative vector , we have

Lemma 5 (From an eigenvector to a non-negative vector)For every such that there is a non-negative such that and such that

Now let us start from a function that optimizes (4), so that and , then apply Lemma 5 to find a function such that the volume of the vertices having positive coordinates in is at most and such that . Then consider the vector such that ; by Lemma 4, we have , and by Lemma 3 there is a threshold such that the set has conductance . Since is a subset of the vertices having positive coordinates in , we have , and so

which is the Cheeger inequality for graphs. It remains to prove the three lemmas.

*Proof:* of Lemma 3. For each threshold , define the set

The idea of the proof is that if we pick at random then the probability that an edge belongs to is proportional to and the probability that is proportional to , so that the expected number of edges in is proportional to the numerator of and the expected number of vertices in is proportional to the denominator of ; if is small, it is not possible for to always be large for every .

To avoid having to normalize the range of to be between and , instead of taking averages over a random choice of , we will consider the integral over all values of . We have

because we can write , where if and otherwise, and we see that only the values of between and make , so .

We also have

and if we denote by the threshold such that is smallest among all the , then

so that

*Proof:* of Lemma 3. Let us consider the numerator of ; it is:

(we used Cauchy-Swarz)

(we used the definition of and Cauchy-Swarz again)

And so

*Proof:* of Lemma 5. Let be the median of , and consider defined as . We have

because the numerators of and are the same (the additive term cancels). The denominators are such that

because and the vector are orthogonal, and so by Pythagoras’s theorem the length-squared of equals the length-squared of plus the length-squared of .

Let us define and so that . We use the following fact:

Fact 6Let be disjointly supported non-negative vectors (“disjointly supported” means that they are non-zero on disjoint subsets of coordinates), then

*Proof:* The numerator of is

and, using orthogonality and Pythagoras’s theorem, the denominator of is

The fact now follows from the inequality

The lemma now follows by observing that and are non-negative and disjointly supported, so

and that both and have at most non-zero coordinate.

**2. Proof of the Cheeger inequality in manifolds **

We will now translate the proof of the graph Cheeger inequality to the setting of manifolds.

As you may remember, we started off by saying that is symmetric and so all its eigenvalues are real and they are given by the variational characterization. Now we are already in trouble because the operator on manifolds cannot be thought of as a matrix, so what does it mean for it to be symmetric? The consequence of symmetry that is exploited in the analysis of the spectrum of symmetric matrices is the fact that if is symmetric, then for every we have

and the property makes no references to coordinates, and it is well defined even for linear operators over infinite-dimensional spaces, provided that there is a notion of inner product. If we the define the inner product

on functions , and more generally

for functions , where is a vector space with inner product , then we can say that an operator is *self-adjoint* if

for all (appropriately restricted) functions . If is compact, this property is true for the Laplacian, and, in particular, and are adjoints of each others, that is,

(The discrete analog would be that is the transpose of .)

Self-adjointness (and appropriate conditions on ) imply a version of the spectral theorem and of the variational characterization. In particular, all eigenvalues of are real, and if there is a minimum one then it is

and if is a minimizer of the above, then

(The minimization is quantified over all functions that are square-integrable, and the minimum is achieved because if is compact then the space of such functions is also compact and the cost function that we are minimizing is continuous. In this post, whenever we talk about “all functions,” it should be understood that we are restricting to whatever space of functions makes sense in the context.)

From the property that and are adjoint, we have

so

where the Rayleigh quotient

is always non-negative, and it is zero for constant , so we see that and

By analogy with the graph case, we define the “ Rayleigh quotient”

And we can prove the analogs of the lemmas that we proved for graphs.

Lemma 7 (Rounding of embeddings)For every non-negative function there is a value such that

where the Cheeger constant of a subset of the manifold is

Lemma 8 (Embedding of into )For every non-negative function , we have

Lemma 9 (From an eigenfunction to a non-negative function)For every function such that there is a non-negative such that and such that

Let us see the proof of these lemmas.

*Proof:* of Lemma 7. For each threshold , define the set

Let be a threshold for which is minimized

We will integrate the numerator and denominator of over all . The *coarea formula* for nonnegative functions is

and we also have

which combine to

so that

*Proof:* of Lemma 8. Let us consider the numerator of ; it is:

We can apply the chain rule, and see that

which implies

and, after applying Caucy-Swarz,

And so

*Proof:* of Lemma 9. Let be a median of , and consider defined as . We have

because the numerators of and are the same (the derivatives of functions that differ by a constant are identical) and the denominators are such that

where we used the fact the integral of is zero.

Let us define and so that . We use the following fact:

Fact 10Let be disjointly supported non-negative functions (“disjointly supported” means that they are non-zero on disjoint subsets of inputs), then

*Proof:* We begin with the following observation: if is a non-negative function, and , then , because has to be a local minimum.

Consider the expression occurring in the numerator of . We have

But

because for every at least one of or is zero, and so at least one of or is zero.

Using this fact, we have that the numerator of is equal to the sum of the numerators of and :

and the denominator of is also the sum of the denominators of and :

because for every . The fact now follows from the inequality

The lemma now follows by observing that and are non-negative and disjointly supported, so

and that both and have a support of volume at most .

If anybody is still reading, it is worth observing a couple of differences between the discrete proof and the continuous proof.

The Rayleigh quotient is defined slightly differently in the continuous case. It would correspond to defining it as

in the discrete case.

If are disjointly supported and nonnegative, the sum of the numerators of the Rayleigh quotients and can be strictly smaller than the numerator of , while we always have equality in the continuous case. In the discrete case, the sum of the numerators of and can be up to twice the numerator of (this fact is useful, but it did not come up in this proof), while again we have exact equality in the continuous case.

The chain rule calculation

corresponds to the step

In the continuous case, and are “infinitesimally close”, so we can approximate by .

The foundations of cryptography were laid down in 1982, the annus mirabilis that saw the publications of the work of Blum and Micali on pseudorandom generators, of Goldwasser and Micali on rigorous definitions of encryption, and of Yao, who gave a more general definitional approach. The paper of Shafi Goldwasser and Silvio Micali, in particular, introduced the incredibly influential concept of *indistinguishability* of distributions, and the idea of defining security in terms of *simulation of an ideal model* in which the security requirements are self-evident. (For example, because in the ideal model an adversary is not able to access the channel that we use to send encrypted data.) Almost every definition of security in cryptography follows the simulation approach, which also guides proofs of security. Shafi and Silvio both went on to do foundational work in cryptography, complexity theory, and algorithms, including their work on zero knowledge, secure multiparty computation, and property testing.

So it was with much joy, this early morning in Japan, that I heard the news that Shafi Goldwasser and Silvio Micali have been named as recipients of this year’s Turing award.

Omer Reingold has more information about their work. With no offense to colleagues around my age and younger, Shafi and Silvio are also representative of a time when leading theoretical computer scientists were more *interesting* people. They both have incredible charisma.

My favorite memory of Shafi and Silvio is from the time I interviewed for a faculty job at MIT. Shafi was in the last weeks of her pregnancy and did not make an appointment to see me, but then the day of my interview she changed her mind and showed up in Silvio’s office halfway through my meeting with him.

Silvio had been looking at my schedule and was giving me advice on how to talk to various people. Shafi asked what we were talking about, and then proceeded to give the opposite advice that Silvio had been giving me. The two of them spent the rest of the meeting arguing with each other.

*Oh man, not another election! Why do we have to choose our leaders? Isn’t that what we have the Supreme Court for?*

– Homer Simpson

Nate Silver is now putting Barak Obama’s chance of reelection at around 85%, and he has been on the receiving end of considerable criticism from supporters of Mitt Romney. Some have criticized his statistical analysis by pointing out that he has a soft voice and he is not fat (wait, what? read for yourself – presumably the point is that Silver is gay and that gay people cannot be trusted with such manly pursuits as statistics), but the main point seems to be: if Romney wins the election then Silver and his models are completely discredited. (E.g. here.) This is like someone saying that a die has approximately a 83% probability of not turning a 2, and others saying, if I roll a die and it turns a 2, this whole “probability” thing that you speak of is discredited.

But still, when someone offers predictions in terms of probability, rather than simply stating that a certain outcome is more likely, how can we evaluate the quality of such predictions?

In the following let us assume that we have a sequence of binary events, and that each event has a probability of occurring as a and of occurring as . A predictor gives out predicted probabilities , and then events happen. Now what? How would we score the predictions? Equivalently, how would we fairly compensate the predictor?

A simple way to “score” the prediction is to say that for each event we have a “penalty” that is , or a score that is . For example, the prediction that the correct event happens with 100% probability gets a score of 1, but the prediction that the correct event happens with 85% probability gets a score of .85.

Unfortunately this scoring system is not “truthful,” that is, it does not encourage the predictor to tell us the true probabilities. For example suppose that a predictor has computed the probability of an event as 85% and is very confident in the accuracy of the model. Then, if he publishes the accurate prediction he is going to get a score of .85 with probability .85 and a score .15 with probability .15. So he is worse off than if he had published the prediction of the event happening with probability 100%, in which case the expected score is .85. In general, the scheme makes it always advantageous to round the probability to 0% or 100%.

Is there a truthful scoring system? I am not sure what the answer is.

If one is scoring multiple predictions of independent events, one can look at all the cases in which the prediction was, say, in the range of 80% to 90%, and see if indeed the event happened, say, a fraction between 75% and 95% of the times, and so on.

One disadvantage of this approach is that it seems to require a discretization of the probabilities, which seems like an arbitrary choice and one that could affect the final score quite substantially. Is there a more elegant way to score multiple independent events without resorting to discretization? Can it be proved to be truthful?

Another observation is that such an approach is still not entirely truthful if it is applied to events that happen sequentially. Indeed, suppose that we have a series of, say, 10 events for which we predicted a 60% probability of a 1, and the event 1 happened 7 out of 10 times. Now we have to make a prediction of a new event, for which our model predicts a 10% probability. We may then want to publish a 60% prediction, because this will help even out the “bucket” of 60% predictions.

I don’t think that there is any way around the previous problem, though it seems clear that it would affect only a small fraction of the predictions. (The complexity theorists among the readers may remember similar ideas being used in a paper of Feigenbaum and Fortnow.)

Surely the task of scoring predictions must have been studied in countless papers, and the answers to the above questions must be well known, although I am not sure what are the right keywords to use to search for such work. In computer science, there are a lot of interesting results about using expert advice, but they are all concerned with how you score *your own way of picking which expert to trust* rather than the experts themselves. (This means that the predictions of the experts are not affected by the scoring system, unlike the setting discussed in this post.)

Please contribute ideas and references in the comments.

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