Luca, since a year has passed, you probably have researched your own
answer. I'm answering some of your questions here just for the
record. I review some Lagrangian-relaxation algorithms for the
problems you mention, and sketch the connection to learning (in
particular, following expert advice). I don't comment here on SDP
algorithms.
Note that the particular algorithms that you mention do not run in
nearly linear time. (There is a nearly linear-time algorithm for
explicitly given packing or covering problems. See
Beating Simplex for Fractional Packing and Covering Linear Programs.)
The algorithms you have in mind typically have variants that run in a
nearly linear number of iterations, but each iteration typically
requires at least linear time as well. I discuss some of these
algorithms below.
Some useful functions
Before we start, here are some functions that we will use in the proof sketches.
(If you are interested in the algorithms, but not the proof details, you can skip ahead.)
For any vector $y$, define $\mbox{Lmax}(y)$ to be $\ln \sum_i \exp(y_i)$.
This function is an upper bound on $\max_i y_i$:
$$\max_i y_i ~\le~ \mbox{Lmax}(y) ~\le~ \max_i y_i + \ln m.$$
Analogously, define $\mbox{Lmin}(y)$ to be $-\mbox{Lmax}(-y)$,
a lower bound on $\min_i y_i$.
For convenience in what follows,
we use $g(y)$ to denote the gradient $\nabla \mbox{Lmin}(y)$ of Lmin.
We use $G(y)$ to denote the gradient $\nabla \mbox{Lmax}(y)$ of Lmax.
Explicitly, $g_i(y)$ is $\exp(-y_i)/\sum_{i'} \exp(-y_{i'})$
while $G_i(y)$ is $\exp(y_i)/\sum_{i'} \exp(y_{i'})$.
Lmin and Lmax are smooth in the following sense:
for any vectors $d\in[0,\varepsilon]^n$ and $y\in R^n$,
$$\mbox{Lmin}(y+d) ~\ge~ \mbox{Lmin}(y)
~+~ (1-O(\varepsilon))\, d \cdot g(y)$$
and
$$\mbox{Lmax}(y+d) ~\le~ \mbox{Lmax}(y)
~+~ (1+O(\varepsilon))\, d \cdot G(y).$$
Note that both of the gradients have 1-norm equal to 1:
$|G(y)| = |g(y)| = 1$. (Throughout we use $|z|$ to denote the 1-norm.)
Note also that, for a matrix $A$,
the gradient of the function $x\mapsto \mbox{Lmin}(Ax)$ with respect to $x$
is (by the chain rule) $(g(Ax))^T A$.
More explicitly, the partial derivative of the function with respect to $x_j$
is $\sum_i A_{ij} \exp(-A_i x) / \sum_i \exp(-A_i x)$.
Likewise, the partial derivative of Lmax$(Ax)$
with respect to $x_j$ is $\sum_i A_{ij} \exp(A_i x)/\sum_i \exp(A_i x)$.
Fractional Set Cover
Fix a Set-Cover instance.
Let $A$ denote the element/set incidence matrix.
Thus, $A_{es} = 1$ if $e\in s$, else 0,
and $A_e x$ is the extent that fractional cover $x$ covers element $e$.
The LP is $\min\{ |x| : A x \ge 1; x \ge 0\}$.
Given $\varepsilon\in (0,1)$, the algorithm is
- Initialize all $x_s = 0$. Let $N=\log(n)/\varepsilon$.
Repeat until $\min_e A_e x \ge N$:
2.1. Choose $s$ maximizing the partial derivative of Lmin$(Ax)$ w.r.t. $x_s$.
(Explicitly, choose $s$ maximizing $\sum_{e\in s} \exp(-\sum_{s'\ni e} x_{s'})$.)
2.2. Increase $x_s$ by $\varepsilon$.
Return $x/\min_{e} A_e x$.
The algorithm returns a $(1+O(\varepsilon))$ approximate solution
in $O(|x^*|\log(n)/\varepsilon^2)$ iterations, where $n$ is the number of elements
and $x^*$ is the optimal fractional set cover (trivially $|x^*|\le n$).
(A similar algorithm appears in the paper Chandra mentioned.
Vertex Cover is of course a special case.)
(Remark: Note that the iteration bound does not depend on the number
of sets, just the number of elements. Thus, the algorithm can be used
with an implicitly defined set system, as long as, given weights on the
elements, one can efficiently find a set of maximum (or near-maximum) total weight.
This kind of oracle is the same as the separation oracle required to apply
the ellipsoid algorithm to the dual problem. For packing problems such
as set packing, you need an oracle that, given weights on the elements,
returns a set minimizing the total weight. For problems such as
multi-commodity flow, you might, for example, need to find a path
minimizing the sum of some given edge weights.)
Here's a sketch of the proof of the performance guarantee.
In each iteration, the partial derivative w.r.t. the chosen $s$
is at least $1/|x^*|$, where $x^*$ is the optimal fractional set cover.
(To see why, recall that the gradient of Lmin$(Ax)$ with respect to $x$ is $(g(Ax))^T A$.
If we were to choose a set $s'$ at random from the distribution $x^*/|x^*|$,
the expected value of the partial derivative with respect to $x_{s'}$
would thus be $(g(Ax))^T \,A x^*/|x^*|$.
Since $Ax^* \ge 1$, this is at least $|g(Ax)|/|x^*|$.
Since $|g(Ax)|=1$, this is at least $1/|x^*|$.
Thus, there must exist some $s$ giving partial derivative at least $1/|x^*|$.
Since the algorithm chooses $x_s$ in each iteration to maximize
the partial derivative, it achieves a partial derivative of at least $1/|x^*|$.)
Then, the step size $\varepsilon$ is chosen just small enough so that
no coordinate of $A x$ increases by more than $\varepsilon$.
Thus, because of the smoothness of Lmin, increasing $x_s$
to $x_s+\varepsilon$ increases $\mbox{Lmin}(Ax)$ by at least
$(1-O(\varepsilon))\varepsilon/|x^*|$.
In this way, the algorithm maintains the invariant
$$\mbox{Lmin}(Ax) \ge (1-O(\varepsilon)) |x|/|x^*| - \ln n.$$
(Note that Lmin$(\overline 0)$ equals $\ln n$.)
At termination, in the invariant, the $\ln n$ term is $O(\varepsilon)$ times the left-hand side,
so by calculation one gets $\min_e A_e x \ge (1-O(\varepsilon)) |x|/|x^*|$.
After the normalization in the last line of the algorithm,
this implies $|x| \le (1+O(\varepsilon))|x^*|$.
FWIW, the inequalities involved in proving the invariant are essentially
the same as those involved in proving the Chernoff bound.
(In fact, this algorithm can be derived by applying the
method of conditional probabilities to a randomized-rounding
scheme that repeatedly samples sets from the distribution $x^*/|x^*|$
(with replacement), increasing $x_s$ for each sampled set $s$.
This derandomization of that gives the algorithm:
the underlying invariant is just that the pessimistic estimator stays below 1.
The exponential penalties in the pessimistic estimator come from the
using the Chernoff bound in the analysis of the rounding scheme.
This basic idea is explained further in the paper Chandra mentioned.)
Fractional Weighted Set Cover (and general fractional Covering)
To handle problems such as Weighted Set Cover efficiently,
we modify the algorithm to use non-uniform increments
(an idea due to Garg and Konemann).
The LP is $\min\{ c\cdot x : (\forall e) \sum_{s\ni e} x_s \ge 1\}$,
where $e$ ranges over the elements, $s$ ranges over the sets,
and all variables are non-negative.
To present the algorithm, first rewrite the problem as a general covering problem.
Let $A_{es} = 1/c_s$ for $e\in s$ and $A_{es} = 0$ otherwise.
Then (with a change of variables, scaling each $x_s$ by $c_s$),
the LP is $\min\{ |x| : A x \ge 1; x \ge 0\}$,
which we can view as a general covering LP.
Here is the algorithm:
Initialize all $x_s = 0$. Let $N=\log(n)/\varepsilon$.
Repeat until all covering constraints have been deleted:
2.1. Choose $s$ maximizing the partial derivative of Lmin$(Ax)$ w.r.t. $x_s$.
(Explicitly, choose $s$ maximizing $\sum_{e\in s} \exp({-\sum_{s'\ni e} x_{s'}})/c_s$.)
2.2. Increase $x_s$ by $\delta$, where $\delta$ is chosen maximally such that, for every remaining covering constraint $e$, the increase in $A_e \cdot x$ is at most $\varepsilon$.
2.3 Delete all covering constraints $e$ such that $A_e\cdot x \ge N$.
Return $x/\min_e A_e\cdot x$.
The algorithm returns a $(1+O(\varepsilon))$-approximate solution
in $O(n\log(n)/\varepsilon^2)$ iterations, where $n$ is the number of covering constraints.
(Each iteration increases some remaining $A_e x$ by $\varepsilon$; this can happen
only $N/\varepsilon$ times to a constraint before it is deleted.)
The proof of correctness is via essentially the same invariant as for Set Cover.
Weighted Vertex Cover is a special case.
Maximum Fractional Bipartite Matching
Given a graph $G=(U,W,E)$,
the natural LP for the problem is $\max\{|x| : \forall v.\, \sum_{e\ni v} x_e \le 1\}$.
In matrix representation, this is a packing LP $\max\{|x| : Ax \le 1; x \ge 0\}$
with 0-1 coefficients ($A_{ve} = 1$ if $v\in e$).
Such problems do not require non-uniform increments, so a simple algorithm analogous to the unweighted Set Cover algorithm (but for packing) will do:
- Initialize all $x_e = 0$. Let $N=\log(n)/\varepsilon$.
While $A x < N$:
2.1. Choose $e$ minimizing the partial derivative of Lmax$(Ax)$ w.r.t. $x_e$.
(Explicitly, choose $e$ to minimize $\sum_{v\in e} \exp(\sum_{e'\ni v} x_{e'})$.)
2.2. Increase $x_e$ by $\varepsilon$.
Return $x/\max_{v} A_v x$.
The algorithm returns a $(1-O(\varepsilon))$-approximate solution in $O(n\log(n)/\varepsilon^2)$ iterations.
(This is because each iteration increases $|x|$ by $\varepsilon$,
and finally, before normalization, $|x| = O(N n)$.)
Just for fun, here is a curious alternative algorithm for Perfect Bipartite Matching.
Recall that $G=(U,W,E)$. Let $n=|U|=|W|$.
- Initialize all $x_e = 0$. Let $N=4\ln(n)/\varepsilon$.
Repeat $n\,N$ times:
2.1. Choose $u$ uniformly at random from $U$.
2.2. Choose $w$ such that $(u,w)\in E$ minimizing $\sum_{e\ni w} x_e$.
2.3. Increase $x_{uw}$ by $\varepsilon$.
Return $x/N$.
If $G$ has a perfect matching, the algorithm returns
an $x$ such that $|x|=n$, and, with high probability,
for all vertices $u\in U$, $1-O(\varepsilon) \le \sum_{e\ni u} x_e \le 1+O(\varepsilon)$,
and for all vertices $w\in W$, $\sum_{e\ni w} x_e \le 1+O(\varepsilon)$.
If you are interested in the details of the proof, please ask...
Mixed Packing and Covering
You may have asked about bipartite matching hoping for an example
of a mixed packing and covering problem, that is, one of the form
$$\exists x?~ Px \le p; Cx \ge c; x \ge 0.$$
Here's one algorithm for such problems.
First, normalize so that $p=\overline 1$ and $c=\overline 1$.
Let $m$ be the number of constraints (rows in $P$ plus rows in $C$).
- Initialize all $x_j = 0$. Let $N=2\ln(m)/\varepsilon$.
While $P x < N$:
2.1. Choose $j$ so that the partial derivative of Lmax$(Px)$ with respect to $x_j$ is at most the partial derivative of Lmin$(Cx)$ with respect to $x_j$. (Explicitly, choose $j$ such that $$\frac{\sum_i P_{ij} \exp(P_i x)}{\sum_{i}\exp(P_i x)} \le \frac{\sum_i C_{ij} \exp(-C_i x)}{\sum_{i}\exp(-C_i x)}.)$$
2.2. Increase $x_j$ by $\delta$, where $\delta$ is chosen maximally such that no constraint $P_i x$ or remaining constraint $C_i x$ increases by more than $\varepsilon$.
2.3. Delete all covering constraints $i$ such that $C_i x \ge N$.
Return $x/\max_i P_i x$.
Assuming the given problem is feasible, the algorithm
returns an $x$ such that $Px\le 1$
and $Cx\ge 1-O(\varepsilon)$.
The number of iterations is $O(m\ln(m)/\varepsilon^2)$,
because each iteration increases some constraint by $\varepsilon$,
and this can happen for each constraint at most $N$ times.
The proof of correctness is via the invariant
$$\mbox{Lmax}(Px) \le 2\ln(m) + (1+O(\varepsilon)) \mbox{Lmin}(Cx).$$
The invariant implies
$$\max Px \le 2\ln(m) + (1+O(\varepsilon)) \min Cx.$$
At termination the left-hand side is $\Omega(\log(m)/\varepsilon)$,
proving the performance guarantee.
In Step 2.1, the desired $j$ must exist as long as the original problem is feasible.
(This is because, for any feasible $x^*$, and any $x$,
if we were to choose a random $j'$ from the distribution $x^*/|x^*|$,
the expected value of the partial derivative of Lmax$(Px)$ with respect to $x_{j'}$
would be at most $1/|x^*|$ (see the previous proof sketch for Set Cover).
Likewise, the expected value of the partial derivative of Lmin$(Cx)$ with respect to $x_{j'}$
would be at least $1/|x^*|$.
Thus, there is an $j$ such that the partial derivative of Lmax$(Px)$ with respect to $x_{j'}$
is at most the partial derivative of Lmin$(Cx)$.)
Then the invariant is maintained in each iteration because,
by the choice of $x_j$ and $\delta$,
and the smoothness of Lmin and Lmax,
increasing $x_j$ to $x_j+\delta$
increases Lmax$(Px)$
by at most $1+O(\varepsilon)$
times the increase in Lmin$(Cx)$.
Learning (following experts / boosting)
One reference for understanding this connection is
Adaptive game playing using multiplicative weights, by Freund and Schapire.
Here is a quick summary to give the technical idea.
Consider the following repeated game. In each round $t$:
- You choose a probability distribution $p^t$ on $[n]$ (the $n$ so-called experts).
- Knowing $p^t$, the adversary then chooses a payoff vector $a^t\in [0,1]^{n}$.
- You receive payoff $p^t\cdot a^t$ for the round.
The game stops after some number of rounds.
Your goal is to minimize your regret
in comparison to any single expert (i.e., pure strategy) $i$.
That is, your goal is to minimize $(\max_i \sum_t a^t_i) - \sum_t p^t\cdot a^t$.
Fix any $\varepsilon>0$.
Let vector $y^t$ denote $\varepsilon \sum_{s \le t} a^s$, that is,
$\varepsilon$ times the vector sum of the payoff vectors up to time $t$.
Recall that $G(y)$ is the gradient of Lmax$(y)$.
Here's the basic strategy we will analyze:
On round $t$, choose $p^t$ to be $G(y^{t-1})$.
By inspection, this gives you payoff $a^t \cdot G(y^{t-1})$ in round $t$.
Because of the smoothness property of $F$,
$$\mbox{Lmax}(y^t) \le \mbox{Lmax}(y^{t-1}) + (1+O(\varepsilon)) \varepsilon a^t \cdot G(y^{t-1}).$$
That is, in each round, $\mbox{Lmax}(y^t)$ can't increase by more than $\varepsilon(1+O(\varepsilon))$ times your payoff.
Since $\mbox{Lmax}(\overline 0) = \ln n$,
this maintains the invariant that $\mbox{Lmax}(y^t)$
is at most your total payoff times $\varepsilon(1+O(\varepsilon)$,
plus $\ln(n)$.
On the other hand, your regret in comparison to the best expert $i$
is $\max_i \sum_t a^t_i$, i.e., $\varepsilon^{-1} \max_i y^t_i$,
which is in turn at most $\varepsilon^{-1} \mbox{Lmax}(y^t)$.
Thus, your regret is at most $\varepsilon^{-1} \ln(n)$, plus $O(\varepsilon)$ times your total payoff.
Remark: I think, as Freund and Schapire point out, a "boosting" algorithm (in learning theory) is also implicit in this analysis. See their paper for more details.
Minimizing total payoff
You can derive a similar strategy for the setting where the goal is to minimize,
rather than maximize, the total payoff.
Your regret, which you still want to minimize, is $\sum_t p^t\cdot a^t - \min_i a^t_i$.
In that case, the corresponding strategy is to choose $p^t$ to be the gradient of $\mbox{Lmin}(y^t)$.
With this strategy your regret is again at most
$\varepsilon^{-1} \ln n$ plus $O(\varepsilon)$ times your total payoff.
Connection to Lagrangian-relaxation algorithms
To see the connection to Lagrangian-relaxation algorithms, fix a Set-Cover instance.
Consider the latter type of game (with the goal of minimizing payoff),
where the experts correspond to the elements $e$ of your set system.
In each round, choose the probability distribution $p^t$
to be the gradient of Lmin$(y^t)$ as above,
and have the adversary choose the payoff vector $a^t$ as a function of $p^t$ as follows:
choose the set $s^t$ maximizing $\sum_{e\in s} p^t_e$,
then let $a^t_e = 1$ if $e\in s^t$, and $a^t_e = 0$ otherwise.
Given the correct stopping condition (discussed below),
this process gives you exactly the Set-Cover algorithm discussed at the start.
The performance guarantee of the algorithm follows from the regret bound as follows.
Let $X_s$ be the number of times the adversary chose set $s$ during the play.
Let $x^*$ be the optimal fractional set cover.
Let $T=|X_s|$ be the number of rounds played.
The regret bound implies
$$\textstyle \sum_t a^t\cdot p^t \le \varepsilon^{-1}\ln(m) + \min_e \sum_t a_e^t.$$
Using the definition of $a^t$, the $t$th payoff
(the $t$th term in the sum on the left)
equals $\sum_{e\in s^t} p^t_e$.
The adversary chose $s^t$ to minimize this payoff.
If the adversary had instead chosen $s^t$ randomly
from the distribution $x^*/|x^*|$, the expectation of
the payoff would have been
$$ \sum_s \frac{x^*_s}{|x^*|} \sum_{e\in s} p^t_e
~=~
\frac{1}{|x^*|} \sum_e p^t_e \sum_{s\ni e} x^*_s
~\ge~
\frac{1}{|x^*|} \sum_e p^t_e
~=~
\frac{1}{|x^*|}.$$
(Above we use that $\sum_{s\ni e} x^*_s \ge 1$ for all $e$, and $|p^t| = 1$.)
Since each payoff is at least $1/|x^*|$, the regret bound implies
$$ \frac{T}{|x^*|} \le \varepsilon^{-1}\ln(m) + \min_e \sum_t a_e^t.$$
By the definition of $X$,
we have $|X| = T$ (each round chooses one set),
and $\sum_t a_e^t = \sum_e [e\in s^t] = \sum_{s\ni e} X_s$,
giving
$$\frac{|X|}{|x^*|} \le \varepsilon^{-1}\ln(m) +\min_e \sum_{s\ni e} X_s.$$
We make the process stop when $\min_e \sum_{s\ni e} X_s = \Omega(\varepsilon^{-2}\ln m)$,
so then (rearranging terms)
$$\frac{|X|}{\min_e \sum_{s\ni e} X_s}~ \le~ (1+O(\varepsilon)|x^*|.$$
That is, normalizing $X$ gives a fractional set cover
of size at most $(1+O(\varepsilon))$ times optimum.
Remark: In a sense, this learning theory interpretation
generalizes the algorithmic interpretation. However, some
of the algorithmic techniques necessary for efficiency
(such as non-uniform increments and dropping satisfied
covering constraints) don't seem to carry over into the
learning theory setting naturally. Likewise, algorithms for
mixed packing and covering LPs (e.g. these)
don't seem to have natural analogues in the learning-theory setting.
