>> Konstantin Makarychev: It's a great pleasure to introduce Yury Makarychev
from Toyota Technology Institute at Chicago. And visiting Microsoft Research
for three days, and today he will tell us about satisfiability of ordering
CSPs above average.
>> Yury Makarychev: Thank you for the introduction. So I will talk about
satisfiability of ordering CSPs above average. And I will show you that this
problem is fixed parameter tractable. This is joint work with Konstantin and
with Yuan Zhou from MIT.
So let me first define the problem, define what ordering CSPs are. So in an
ordering CSP we are given a set of N variables x1 et cetera xn, and m
constraints pi 1 et cetera pi m. And we want to find a linear ordering of
the variables that maximizes the number of satisfied constraints.
What are the constraints? So each constraint pi r depends on at most k
variables where k is a given constant. And each constraint prescribes what
relative ordering of the variable -- of the variables it depends on are valid
and which are not. Or, in other words, for each constraint we have a list of
possible relative orderings of the variables it depends on. And if the
ordering of the variables is in this list, then the constraint is satisfied;
otherwise, it's not.
And let me just show you two very well-known examples of ordering CSPs. The
first example is the maximum acyclic subgraph problem. In this problem we
are given a directed graph G on a set of vertices x1 et cetera xn. And our
goal is to find a linear ordering of the vertices that maximizes the number
of forward-going edges.
And it's easy to see that this problem is just an ordering CSP of arity 2.
So each edge xi, xj corresponds to a constraint xi is less than xj. And then
the number of forward-going edges is equal exactly to the number of satisfied
constraints for every ordering of xi. All right.
Another well-known example is the betweenness problem. And in this
problem -- so this problem is basically an ordering CSP of arity 3. And in
this problem all constraints of the form xi lies between xj and xk, or more
explicitly of the form xj is less than xi less than xk, or xk is less than xi
less than xj.
What is known about ordering CSPs? There is a very simple trivial
approximate algorithm for the problem. Let's just randomly permute all xi's
and out with a random ordering. How well does this algorithm perform? For
the maximum acyclic subgraph problem, this algorithm satisfies each
constraint with probability 1/2 because for every xi and xj two possibilities
are equally likely; that xi is less than xj and that xj is less than xi. And
there are four this algorithm satisfies m/2 constraints in expectation.
Unless you know this number, the expected number of constraints, this
algorithm satisfies by AVG, average.
For the betweenness problem, this algorithm satisfies each constraint with
probability 1/3 because each of the three vertices the constraint depends on
are equally likely to be between the other two. And so in expectation this
algorithm satisfies m/3 constraints.
Now, this algorithm is very basic, almost trivial algorithm. And it's
natural to ask if we can do better than this algorithm. And in general this
problem for different types of constraints satisfaction -- constraint
satisfaction problems has been studied a lot. So the basic question is to
get some advantage over a trivial random algorithm, get advantage over
random.
And the first result in the theory was due to Hastad who showed that the -and he started regular CSPs, not ordering CSPs -- that are 3-CNF and 3-XOR
CSPs are approximation resistant, meaning that no approximation algorithm
gets a better approximation [inaudible] than this trivial random assignment
algorithm. And if there is no trivial approximation for a constraint
satisfaction problem, we call this constraint satisfaction problem and the
corresponding predicate approximation resistant.
And it's an interesting question to classify all like regular predicates,
which of them are approximation resistant, which of them are not. We still
don't have the complete classification. For instance, Austrin and Mossel
proved there's a very wide class of predicates, predicates that could kind of
come from pairwise independence are approximation resistant. There are also
some positive results showing that some CSPs are not approximation resistant.
For instance, Hastad showed that any 2-CSP is not approximation resistant,
and also Hastad showed that bounded occurrence CSPs are not approximation
resistant. But there are many, many results in the theory and there are
still many open problems.
This problem, the advantage over random, was also studied for ordering CSPs.
And let me first tell you about positive results and then I will tell you
about negative results. So first Berger and Shor a long time ago in 1990
showed that we can get advantage over random for bounded occurrence instances
of the maximum acyclic subgraph ->>:
So that's just bounded DB?
>> Yury Makarychev: Yes. And then in a joint work with Charikar and
Konstantin we showed how to get some advantage over random for the maximum
acyclic subgraph in general graphs and the dependence is different.
Guruswami and Zhou showed that we can get advantage over random for any
ordering CSPs of arity 3 that have bounded occurrences of variables. And
this result was generalized to all bounded occurrence CSPs by Konstantin.
On the negative side, Guruswami, Hastad, Manokaran, Raghavendra and Charikar
showed there is no nontrivial multiplicative approximation algorithm for any
ordering CSP of any arity k. So unlike assignment CSPs, there is no better
approximation algorithm for any ordering CSPs, or formally this means that no
polynomial-time algorithm can find a solution satisfying at least 1 plus
epsilon times average constraints for instances with optimum greater even for
instances with optimum greater than 1 minus epsilon m for every constant
value of epsilon.
So, in other words, no algorithms gets -- so we can't get any constant factor
advantage over random. Right? And then it's natural to ask if we can get at
least some additive advantage over random. So we cannot improve the
approximation factor, maybe we can just do a little bit better than random.
And this question was posed by Gutin, van Iersel, Mnich and Yeo who
conjectured that there is a fixed-parameter algorithm that decides whether
the optimum is greater than average plus t or not. And here t is a fixed
parameter. So, in other words, this means that there isn't an algorithm
whose running time is some function of t, and this function may be, for
instance, exponential in t times a polynomial m plus n that answers this
question. So they decide whether the optimum is greater than average plus t
or not.
What is known about this conjecture? First of all, let me mention that a
similar conjecture was proved for assignment CSP. So Alon, Gutin, Kim,
Szeider and Yeo showed that for any assignment CSP, regular CSP, this -- like
the counterpart of this conjecture is true.
For ordering CPS, there has been a number of papers and by Gutin et al. who
showed that this conjecture is true for the maximum acyclic subgraph problem,
for the betweenness problem, and finally for all ordering CSPs of arity 3.
However, in general the problem seems to be much more difficult, and in
particular Gutin et al. know that it appeared -- noted that it appears
technically very difficult to extend results obtained for arities 2 and 3 to
arities greater than 3.
And in this paper we show that this conjecture is true for CSPs, for ordering
CSPs of all arities. And also we generalize this result and show that
actually a similar conjecture is true for a very wide class of constraint
satisfaction problems, constraint satisfaction problems that have some
structural properties, and in particular this class includes all ordering
CSPs. And I will say a few words about that later.
Also, to prove our result, we prove a new Bonami-type lemma for the
Efron-Stein decomposition, and we believe that this result is of independent
interest. And I will say a few words about this result later and I will also
define what the Efron-Stein decomposition is.
At the high level, we use the approach of Alon et al. that they used for
proving the counterpart of this conjecture for regular CSPs. And recall that
in this problem we are given an instance of an ordering CSP and the parameter
t, and we need to decide whether the optimum is greater than average plus t
or not.
And we prove that there are two possibilities. The first possibility is that
the instance depends on at most quadratic and t number of variables. So in
this case we can just try all possible orderings of these t squared
variables, find the best solution, and test whether the value of the solution
is greater than average plus t or not.
>>:
K has to be fixed also?
>> Yury Makarychev: K is fixed. [inaudible] k is fixed and t changes but
the running time may depend on -- for different values of k we can even
different algorithms.
>>: But when you said -- was that fixed parameter tractable for all arity?
You mean -- you meant for all ->> Yury Makarychev: For every k there exists -- it's true. But the
polynomials may be different. In particular here, yeah, this constant ck
depends on k. Any questions? Let me know if you have any questions about
the problem or the statements.
And also here, I mean, this set of variables is called a kernel. And so in
this case we have a kernel of quadratic and t size. And in the second case
the instance depends on more than ckt squared variables. And in this case we
show that actually the optimal value is greater than average plus t. So in
this case we know that the answer is positive.
Moreover, in this case we can actually find a solution whose value is greater
than average plus T, but of course not necessarily the optimal solution.
>>:
Question.
What's a canonical problem I should think for large K?
>> Yury Makarychev:
>>:
So for ordering CSP?
Yeah.
>> Yury Makarychev: Yeah, so we can always assume that all constraints of
the form xi1 is less than xi2 et cetera is less than xik.
So it'd be replaced because for every constraint we have a subset -- so each
constraint is just a conjunction of such, of such clauses. And then we can
just replace it with these clauses. Then the number of constraints will go
up but the optimum value will not change and the expected number of satisfied
constraints will not change.
>>:
So each one is [inaudible] is basically [inaudible]?
>> Yury Makarychev:
questions? Okay.
Yes.
Yes.
For such constraints, yes.
Any other
Now, how did we prove these, there are only these two possibilities? So
first let's see a random ordering of x1 et cetera xn. And let's denote by Z,
capital Z, the number of constraints satisfied by a random ordering.
And then by definition the expectation of Z is equal to this number AVG.
Right? Now, first we prove that if the instance depends on at least r
variables, then the variance of this random variable Z is large. It's at
least some positive number constant a times r. And so if we are not in case
1, if the instance depends on many variables, then we know that the variance
is high.
Now we prove theorem 2 that says that if we are not in the first case, or if
the variance is high, then the optimum is grater than the average plus some
positive constant times square root r. And here I want to know that this
result is not trivial in this sense that of course if the variance is high
then the standard deviation is high and we know that this random variable Z
deviates from the expected value by at least square root r. Right? But the
distribution of Z is not necessarily symmetric and, moreover, in most
cases -- in most cases it's very asymmetric.
So we don't know whether it deviates only in one direction or in both
directions. So it's possible that it takes some variables that are much
smaller than the expectation, but it never takes variables that are
considerably greater than the expectation.
And what this theorem shows, that actually in this case for this random
variable Z we have -- we still have a guarantee that Z attains a value that
are considerably larger than average.
And now as I corollary we get that if the variance of Z is at least at
squared then the optimum is greater than average plus t. So we are in case 2
and we know that the answer is positive. So this is our plan.
>>:
Sorry, what are the conditions?
>> Yury Makarychev:
constraints ->>:
So Z is our random variable, the number of satisfied
[inaudible] any ordering CSP at all --
>> Yury Makarychev:
For any ordering CSP.
>>: It doesn't -- you're not assuming the boundedness and -- there's no -yeah, okay.
>> Yury Makarychev: I mean, we assume that k is fixed, the number of
variables each constraint depends on. But other than that. And, yeah, I
mean, this constant be -- depends on k.
So now we need to prove two theorems with theorem 1 and theorem 2. And to do
so we use the Fourier analysis and more specifically the Efron-Stein
decomposition. Also in order to prove this theorem 2, we prove a Bonami-type
lemma for the Efron-Stein decomposition, and I will tell you about that in a
few moments.
And so we want to use Fourier analysis to solve this problem. One problem is
that Fourier analysis works especially well on product spaces. And in our
case the set of solutions is not a product space. Right? So the set of
solution is a set of permutations of variables x1 et cetera xn, and it's not
that convenient to use Fourier analysis on this set.
So let's instead assume that each variable xi belongs to the segment 0, 1.
And then for if we assign values to x1 et cetera xn from this segment 0, 1,
then we can consider the corresponding linear ordering of xi's. Right? Xi's
are just ordered according to their values.
Moreover, if we assign values to xi's randomly, then this ordering will be
also random and we'll be distributed uniformly. Right? So this is our plan.
And here I want to mention that it's very important that this domain, 0, 1,
is a continuous infinite domain. So some previous approaches to this problem
and some of the papers I mentioned on ordering CSPs uses so-called
[inaudible] technique. So what they did, they considered a fixed domain of
bounded size and assumed that xi -- xi's belong to this domain.
In this problem, we can't use this approach. So we cannot use a fixed domain
of constant size instead of this segment, 0, 1, because if we try to do that,
we will not preserve the optimal value. And so this approach will not work.
In order to preserve the optimal value, we can try -- we can take away large
domain of size n or polynomial mn. But that approach will not work because
then all the parameters will depend on n, and in particular the size of the
kernel will also depend on n. And so we will not -- the result we want to
get. So it's really crucial here that our domain is fixed, it doesn't depend
on n, and we can use it.
Now I want to define the Efron-Stein decomposition. And, roughly speaking,
the Efron-Stein decomposition is very similar to the Fourier analysis on the
Boolean cube. So if you have a function on the Boolean cube, then we can
write it a sum of -- sum over all subsets s of 1 of the set of numbers from
[inaudible] the Fourier coefficient corresponding to this set S times the
character for S. And it has the following very nice properties. First of
all, all these functions FS cut S depend on variables in this variables xi
with i in S. Then all these functions are mutually orthogonal. And in
particular, because of that, the variance of f is just equal to the sum of
variances of its Fourier terms. And also we have linearity. We have this
property for the standard Fourier analysis of Boolean functions.
Now, Efron-Stein
assume that f is
to real numbers,
in the form like
decomposition has very similar properties. So now let's
a function from the cube with segment 0, 1 to the power of n
and the Efron-Stein decomposition is a representation of f
this.
And actually we can -- we can write the Efron-Stein decomposition for
function to find many probability space. So here this set doesn't have -- we
want to be of a set of the set -- the segment 0, 1.
And it has very similar properties. So, first of all, each function fs
depends only on variables xi with i in S. Then all functions fs are mutually
orthogonal. And because of that, the variance of f is equal to the sum of
variances of this function fs. And the decomposition is linear.
Let me now define the Efron-Stein decomposition. And they will not be very
precise, but let me just tell a few words that explain what the Efron-Stein
decomposition is. And let's first start with the case n is equal to 2. And,
moreover, let's assume that f is a product of two functions -- is a product
of two functions g and h. And let's write f as the expectation of f. So g
empty set is the expectation of f. And g1 of x1 is just g of x1 minus the
expectation of g.
And similarly let h empty set be the expectation of h and h2 of x2 to be h
minus its expectation. Then clearly f is equal to this product. Now we
expand this product and we get this expression. And in this expression the
first term is this function f empty set. This function depends only on x1.
So let f1 be this function, this will be f2, and this will be f1, 2.
And now let's assume that n is greater than 2 and again f is the product of
terms like this, then we just decompose gi in a similar way, so we write that
gi is equal to its expectation plus gi minus its expectation. We substitute
each gi in this product by the corresponding expression and then expand that
and we get 2 to the power of n terms. And each term is a term in the
Efron-Stein decomposition of f.
And in general, of course, f is not necessarily a product of N functions but
we can extend this definition by linearity and we will get a valid definition
for all functions f.
So this is the idea behind the Efron-Stein decomposition.
questions? Should I say more about this?
>>:
Do you have any
[inaudible] corresponds [inaudible]?
>> Yury Makarychev: So in particular what we can do, we can consider the
standard Fourier decomposition of f, right? And let's say this function is
maybe sine and cosine functions, and then we can represent f as the sum of
say sine of x1 times cosine of x2, sine of x3 with some coefficient. So it
may be sine of 5x5 and so on. And then we apply these to each term.
>>: So is there a way to [inaudible] description of this decomposition using
just directly the Fourier decomposition, the usual ->> Yury Makarychev: Yes. So what we can do, so we have the Fourier
decomposition. So suppose that we have a Fourier basis of 50, 51, 52, et
cetera, infinite Fourier basis, and let's assume that 50 is just -- is the
constant function. So it's equal to 1. Right?
Then what we can write that f is equal to sum of f -- sorry -- phi -- let's
say i1 of -- okay, 5i1 of x1 phi, let's say, i11, i12 of x2, and so on, phi
in of xn. And now what we can do, we can define fs is the sum of these terms
where this index i -- so sum of those terms k size that ikj is equal to 0 if
j is ns.
So, in other words, okay, let me say that in other terms. So what we do, we
have this infinite number of terms here. Right? And now we kind of -- and
we have now -- now we want to find these 2 to the power of n terms. So for
every set we have a term, and now we want to assign each term to one of these
sets. And the terms for each set will be just the sum of the functions which
we assign to the set.
And so we look at one product like this and look at those terms, at those
functions phi that are now to go to 1. So basically, I mean, if, for
instance, this function is equal to 1, this is x2, this is -- see, I don't
know, sine of x3 of i and this is again 1, then this -- in this term -- this
term depends on nx2 and x3 nontrivial. So we putted it in this [inaudible]
to the set 2 and 3. And this is how we get this decomposition. This is
another explanation.
Is it -- also we can actually write very simple explicit formulas for
computing the Efron-Stein decomposition. And the Efron-Stein decomposition
is defined uniquely. So there is only one Efron-Stein decomposition for
every probability space. So first we define this functions f subset t by
this formula so it's the expectation of f given xi's with i in t, and then we
use the inclusion -- exclusion formula to get this expression for fs.
Okay. Now let's apply the Efron-Stein decomposition to other problem. So
recall that we consider this random variable Z which is a number of satisfied
constraints, and we can just use these explicit formulas to compute the
Efron-Stein decomposition of Z.
And what we get -- okay, and -- okay. First let's know that Z is just the
sum of indicator functions of basic -- of basic ordering predicates of this
form x1 is less than x2 et cetera less than xk. And because the Efron-Stein
decomposition is linear, it's enough to compute the Efron-Stein decomposition
of just these predicates.
And here we just use those formulas and we get where the decomposition has
this form. So it's some constant times a sum over all permutation -permutations of the indexes 1 et cetera k, and here some polynomial p within
integer coefficients of degree at most k that depends on variables x1 et
cetera xk and times the indicator of the event that xi's ordered according to
pi.
So we get that the decomposition has this form. And because the Efron-Stein
decomposition is linear, we get that the decomposition of Z also has this
form.
Now, what is important here, that these polynomials P prime have integer
coefficients. And so the set of all functions of this form forms a discrete
lattice in a finite dimensional space. And because of that, if this term ZS
in the Efron-Stein decomposition is not identical equal to 0, then it must be
bounded away from 0. And specifically its variance should be at least some
number that depends on non k, some positive number. And to prove that we use
a very basic compactness argument.
Okay. Now we are ready to prove our theorem 1. So recall that we assume
that the instance depends on at least r variables and we want to prove that
the variance of Z is large. So we compute the Efron-Stein decomposition of
Z. And now know that each term ZS depends on at most k variables. Right?
Because each predicate depends on at most k variables.
There are four -- and since Z depends on at least r variables, there are four
that are at least r/k terms. Right? Each term contributes at least Bk to
the variance of Z. And so what we get that the variance of Z is at least
rBk/k. So it depends linear -- it's lower bounded by linear function of r.
And so this concludes the proof of theorem 1.
Now we want to show the proof, theorem 2. And the plan is as follows. So
what we want to prove is that if the variance of Z is large, then the optimum
value is at least average plus b square root r. And we show using a new
Bonami-type lemma that we prove, that we've proven, which I will show in the
next slide. Then the fourth moment of Z is upper bounded by some constant
that depends on non k times the variance of Z squared.
And from this it follows by result of Alon et al. that Z attains values
greater than the average plus b square root r with positive probability.
this result of Alon et al. is similar to the Markov inequality. So it's
not -- it's not difficult to show.
And
So let me now say a few words about this Bonami lemma. And so what we show
that -- I mean, let's first recall what the standard Bonami lemma for minus 1
random variables is. So the Bonami lemma says that if f is a polynomial of
degree at most k than the fourth moment of f is bounded by some number 9 to
the power of k times the second moment of f squared.
Now, what we -- we prove that if f -- so let's
from the segment 0, 1 to r, or this also works
let's assume that all terms in the Efron-Stein
most k. Then a very similar inequality holds,
consider that function f now
for any probability space, and
decomposition has degree at
but we also need to assume
additionally that this condition is true, that the expectation of the product
of n four terms is bounded by the product of the variance is times some
number.
And a condition like this is necessary. So result of this condition, this
lemma is not true. And we use this for ours -- in our problem we show that
this condition is true and we get that the fourth moment is bounded by the
second moment squared. So this is how we apply this lemma in our case.
>>: So if I pick a polynomial [inaudible] so if 1 [inaudible] on the
[inaudible].
>> Yury Makarychev:
Right.
>>: Does it -- like does -- is that -- does it have Efron-Stein
decomposition of the [inaudible]?
>> Yury Makarychev:
>>:
Yeah.
So extended by just [inaudible], right?
>> Yury Makarychev:
>>:
Right.
Oh, oh, oh.
By doing what?
By --
[inaudible]
>> Yury Makarychev:
>>:
In the extended...
By linearity.
>> Yury Makarychev:
By rounding it or by linear, it's kind of -Just -- just -So you consider the same polynomial.
>>: Yeah.
>> Yury Makarychev: It will have in the Efron-Stein decomposition, but it
will be different from ->>:
It won't be -- it won't have any k?
>> Yury Makarychev:
have degree K.
No, yeah, it should have the degree k, yeah.
>>: And what would the last condition mean in that case?
trivial?
>> Yury Makarychev:
true.
>>:
Okay.
It will
Or is that
If it's just in this function -- I mean, it should be
>> Yury Makarychev: So for kind of normal function it's true. So basically
it's a problem that the function is unbounded. If it's unbounded by some
constant, yeah, it should [inaudible]. And, yeah, I mean, it -- it will help
this small constant [inaudible]. Any questions?
So let me now tell you how we prove this Bonami lemma for the Efron-Stein
decomposition. And the idea is to reduce this problem to the standard Bonami
lemma. So we are going to define an auxiliary polynomial g, and we proceed
as follows. So let gs be just the variance of this term fs, and let chi s be
the character for the set that S. And so we define this polynomial g with
Fourier coefficients gs, and now we want to compare f and g.
And these distributions may be very different, but we want just to compare
the second and fourth moments. And know that since f has Efron decomposition
of degree at most k, g is a polynomial of degree at most K.
First of all, by our choice of Fourier coefficients, the variance of g is
equal to the sum of gs squared, and this is equal to the sum of variances of
terms fs, so this is equal to the variance of f. So these two functions have
the same variance.
And now what we want to -- so we want to upper bound the fourth moment of f.
Now we want to -- what we are going to prove, that the fourth moment of f is
bounded by the fourth moment of g times some constant c. And using the
standard Bonami lemma, this is less than some number times the variance of -sorry, this is the variance of g squared -- than the variance of g squared.
And this is equal to the -- again, to the same constant times the variance of
f squared. So this should be just f. Okay? So this is our plan.
And the only thing that we have to prove is this inequality, right? Then the
fourth moment of f is upper bounded by the fourth moment of g. And let's
just expand the fourth moment of f so the fourth moment of f is equal to the
sum over subsets S1, S2, S3, S4, the expectation of the product of FS1, FS2,
FS3, FS4, right? And the fourth moment of g is given by this formula. Okay?
Now we want to show that this inequality holds term by term so that we can
just write that this term for every set S1 S et cetera S4 is less than the
corresponding term here. And recall that this additional condition that we
require says that the product -- the expectation of the product of FS1, FS2,
FS3, FS4 is less than some constant c times the product of the variances of g
si squared. So basically we get that this coefficient is greater than this
expectation up to this constant C.
So now the only thing that we need to show is that this expectation is at
least 1. Now, and here we have a slight problem. So if some variable appear
doesn't appear in the sets S1, S2, S3, S4, then -- so, okay. What we have,
we have here just the product of variable Z, right, and we can write it as
variable Z1 to some power times variable, Z2 to some power and so on, and
these variables are independent. So we can consider each variable
separately.
Now, if some variable doesn't appear in the sets S1, S2, S3, S4 et al., then
of course it doesn't contribute to anything. If a variable appears one time,
then its expectation is 0. And then this expression is also 0. But it's not
hard to see that in this case, in this case if some index appears in these
sets only once, then this expectation is also 0. So that's not a problem.
If a variable appears two or four times, then the expectation of its square
and its fourth power is equal to 1, so we are in good shape. There is a
problem, however, if a variable appears three times here and three times
here. Because in this case this expectation is equal to 0 and this
expectation is not necessarily equal to 0.
However, it turns out that this is easy to fix. So basically instead of
considering plus minus 1 variables we consider a Bonami variable with 0 first
moment but with positive first, second, and third -- with positive 30 second
[inaudible] and fourth moments.
And specifically we just consider a random variable that takes values 3 and
minus 1 with probabilities 1 quarter and 3 quarters. This variable has
expectation 0, but it has positive first, second, and third moments. And so
instead of -- and we corresponding consider characters for this variable.
We also need to show kind of when [inaudible] of the standard Bonami lemma
for functions defined on this probability space, but it turns out that this
is very easy.
And then from that we get the result of this. We get this Bonami lemma for
the Efron-Stein decomposition. Any questions?
>>: That condition [inaudible] are there examples -- so what's an example
[inaudible]? And also that condition on the board [inaudible].
>> Yury Makarychev: This one, I mean, basically the function belongs to L2
but doesn't belong to -- I mean, if this function, right, so even if all
these terms are equal, the function belongs to L2 but doesn't belong to L4,
then this is infinite and this is unbounded. Any questions? Okay. And
using this Bonami lemma for the Efron-Stein decomposition we get theorem 2.
So this concludes the proof.
Now let me mention our other result. So we prove actually our result for a
wide class of constraint satisfaction problems defined on domain 0, 1, and it
calls for all CSPs that have certain structural properties, in particular it
calls for CSPs with constraints defined by linear equations with small
integer coefficients or with constraints defined by inequalities,
inequalities which are defined by a bounded degree polynomials with small
integer coefficients.
And just to give you some idea what these constraints might be, for instance,
we can have constraints of the form xi is greater than the average of xj and
xk, or say xi is closer to xj than to xk. Also all ordering CSPs are in this
class. They can be defined by trivial linear inequalities, and also all
standard assignment CSPs are in this class.
Okay. Let me summarize what we do. What we did. We proved the conjecture
of Gutin et al. and moreover we generalized it to a wide class of constraints
satisfaction problems. And additionally we prove the Bonami-type lemma for
the Efron-Stein decomposition. It would be interesting to see if it has more
applications in computer science. Thank you.
[applause]
>> Konstantin Makarychev:
Questions?
>>: Are those CSPs also approximate, like is it clear that on the last ones
that you mentioned that you could possibly could not get multiplicative?
>> Yury Makarychev: I mean, so this class is very generalized, as I describe
it. And in particular it includes all ordering CSPs. So yes. We can't
approximate them. But, I mean, we consider some subclasses, then they know
the difficulty. I mean, in particular or just standard CSPs belong to this
class of CSPs. And for them, for many of them, we do have good approximation
algorithms. Yes.
>>: It's funny because the [inaudible] is very simple. And could you expect
maybe to take more out of your analysis and get a better algorithm with
complexity maybe 2 to the t instead of 2 to the t squared?
>> Yury Makarychev:
I don't know.
But I think --
>>: You would be in case 1 and you would say, I don't know, something like -something more refined than just [inaudible].
>> Yury Makarychev:
I don't know.
>>:
Okay.
>>:
There's no obstruction.
>> Yury Makarychev: I'm not sure. I don't know. I think that might be
possible. I mean, of course, it might be possible to have some improved
approximation factor but improve not by a constant but by something else.
For instance, maybe it's possible to get the result of the form that we can
get one plus one over log n times average. Or maybe something along this
line.
>>:
So something like this is known for max acyclic subgraph?
>> Yury Makarychev:
For what?
>>:
The max acyclic subgraph.
>> Yury Makarychev: For the maximum acyclic subgraph, we can get average
plus OPT minus average divided by log n. So that's basically -- this is ->>:
How would this 1/2 be if it's --
>> Yury Makarychev: Yes. So this is
maximum acyclic subgraph, we can look
is our advantage over random, right?
log n approximation factor -- this is
an approximate -- I mean, so for the
at kind of a different objective, what
And then for this function, this is a
a log n approximation algorithm.
>>: Yeah, so is that -- but that's not exactly the approximation [inaudible]
ideal? So does it generalize to all the [inaudible] CSPs?
>> Yury Makarychev:
>>:
This result?
Yeah.
>> Yury Makarychev: No. I mean, that would be very interesting to get
something like that for other ordering CSPs. But that's not, no.
>>:
These techniques are very different from these ones or...
>> Yury Makarychev: For this result, the technique is [inaudible] technique.
Yeah. I mean, the technique is very different. So we don't -- we use here
[inaudible]. So we use here [inaudible]. Here we can't use [inaudible] as
is. And after that it's very different.
I mean, it's funny that here we also use Fourier analysis, but the Fourier
analysis is very different from this Fourier analysis. Any questions?
>> Konstantin Makarychev:
[applause]
More questions?
Let's thank the speaker.