15-451 11/11/04
* Approximation Algorithms Reminder: quiz next Thurs.
Up to and including today
(old quiz on web page)
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NP-completeness recap:
Going back to network flow and linear programming, one thing that
makes these problems important is that you can use them to represent a
lot of problems (i.e., you can reduce a lot of problems to them). One
way to think about NP-completeness is that a problem like 3-SAT is so
expressive, it can represent *any* problem in NP. And for this
reason, we don't believe we will ever be able to find an efficient
algorithm to solve it.
Any more questions about NP-completeness?
APPROXIMATION ALGORITHMS
========================
General Question: Maybe we can't hope to always get the best solution,
but can we at least guarantee to get a "pretty good" solution? E.g.,
can we guarantee to find a solution that's within 10% of optimal? Or
how about within a factor of 2 of optimal? Or, anything non-trivial?
Interesting thing: even though all NP-complete problems are equivalent
in terms of difficulty of finding optimal solution, the hardness or
easyness of getting a good approximation varies all over the map.
VERTEX COVER
============
- GIVEN: a graph G. GOAL: find the smallest set of vertices such that
every edge is incident to (touches) at least one vertex in the set.
- Example: +----+----+
| | |
+----+----+
- Can think of as: what is the fewest # of guards we need to place
in museum to cover all the corridors.
- This problem is NP-hard. But it turns out we can get within a
factor of 2.
Let's start with some strategies that *don't* work.
Strawman #1: Pick arbitrary vertex with at least one uncovered edge
incident, put into cover, and repeat. What would be a bad example?
[A: how about a star graph]
Strawman #2: how about picking the vertex that covers the *most*
uncovered edges. Turns out this doesn't work either. [make bipartite
graph where opt is size t, but this alg finds one of size t +
floor(t/2) + floor(t/3) + floor(t/4) + ... + 1. This is Theta(t log
t). Best examples are with t=6 or t=12.]
How to get factor of 2. Two algorithms:
Alg1:
Pick arbitrary edge. We know any VC must have at least 1
endpt, so lets take both. Then throw out all edges covered
and repeat. Keep going until no uncovered edges left. What
we've found in the end is a matching (a set of edges no two of
which share an endpoint) that's "maximal" (meaning that you can't add
any more edges to it and keep it a matching). This means if
we take all endpoints, we have a VC. So, if we picked k
edges, our VC has size 2k. But, any VC must have size at least k since
you need to have at least one endpoint of each edge (and,
these edges don't touch, so these are k *different* vertices).
Here's another 2-approximation algorithm for Vertex Cover:
Alg2:
Step1: Solve a *fractional* version of the problem. Have a variable
x_i for each vertex. Constraint 0<= x_i <= 1. Each edge should be
"fractionally covered": for each edge (i,j) we want x_i+x_j >= 1.
Then our goal is to minimize sum_i x_i. We can solve this using
linear programming.
E.g., triangle-graph. E.g., star-graph.
Step2: now pick each vertex i such that x_i >= 1/2.
Claim 1: this is a VC. Why?
Claim 2: The size of this VC is at most twice the size of the
optimal VC. Why? Explanation in two parts: First, notice that
it's at most twice the value of the *fractional* solution we found
(because it's no worse than doubling and then rounding down).
Then, notice that the size of the optimal fractional solution is <=
the size of the optimal integral solution, i.e., the optimal VC,
because the optimal VC *is* a legal fractional solution.
Interesting fact: nobody knows any algorithm with approximation
ratio 1.9. Best known is 2 - O(1/sqrt(log n)), which is 2 - o(1).
Current best hardness result: Hastad shows 7/6 is NP-hard.
Recently improved to 1.361 by Dinur and Safra. Beating 2-epsilon
has been related to some other open problems, but not known to be
NP-hard.
SET-COVER
---------
On hwk5, you looked at the set-cover problem:
Given a domain X of n points, and m subsets S_1, S_2, ..., S_m of
these points. Goal: find the fewest number of these subsets needed to
cover all the points.
As you showed, set-cover is NP-hard. However, there's a simple
algorithm that gets a ratio of ln(n):
Greedy Algorithm: Pick the set that covers the most points. Throw out
all the points covered. Repeat.
What's an example where this *doesn't* find the best solution?
Theorem: If the optimal solution uses k sets, the greedy algorithm
finds a solution with at most k*ln(n) sets.
Proof: Since the optimal solution uses k sets, there must some
set that covers at least a 1/k fraction of the
points. Therefore, after the first iteration of the algorithm,
there are at most n(1 - 1/k) points left. After the second,
there are at most n(1 - 1/k)^2 points left, etc. After k
rounds, there are at most n(1 - 1/k)^k < n*(1/e) points left.
After k*ln(n) rounds, there are < n*(1/e)^{ln n} = 1 points
left, which means we must be done.
In fact, it's been proven that unless everything in NP can be solved in time
n^{O(loglog n)}, then you can't get better than (1-epsilon)*ln(n) [Feige].
MAX-SAT: Given a CNF formula (like in SAT), try to maximize the
number of clauses satisfied.
To make things cleaner, let's assume we have reduced each clause [so,
(x or x or y) would become just (x or y), and (x or not(x)) would be
removed]
Let m = # of clauses in the formula, m_1 = # clauses of size 1, m_2 =
# clauses of size 2, etc. So, m = m_1 + m_2 + ...
Claim: There exists an assignment that satisfies at least
sum_k m_k(1 - 1/2^k) clauses. Furthermore, there is an efficient
algorithm to find one.
Corollary: if every clause has size 3 (this is sometimes called the
MAX-exactly-3-SAT problem), then we can satisfy at least m(7/8) clauses.
So, this is for sure a 7/8-approximation, since the optimal solution
satisfies at most m clauses.
Interesting fact: getting a 7/8 + epsilon approximation for any
constant epsilon (like .001) is NP-hard.
Proof of claim: Consider a random assignment to the variables. For a
clause of size k, the chance it is *not* satisfied is 1/2^k. So, the
expected # of clauses satisfied is sum_k m_k(1 - 1/2^k). Since a
random assignment satisfies this many in expectation, surely an
assignment that does this well must exist. Also, if we just keep
trying random assignments, and it's not hard to show that whp, pretty
soon we'll find one that's as least as good as this expectation.
How about a deterministic algorithm? Here's what we can do. Look at
x_1: for each of the two possible settings (0 or 1) we can calculate
the expected number of clauses satisfied if we were to go with that
setting, and then set the rest of the variables randomly. (It is just
the number of clauses already satisfied plus sum_k m_k(1-1/2^k),
where m_k is the number of clauses of size k in the ``formula to
go''.) Fix x_1 to the setting that gives us a larger expectation.
Now go on to x_2 and do the same thing, setting it to the value with
the highest expectation-to-go, and then x_3 and so on. The point is
that since we always pick the setting whose expectation-to-go is
larger, this expectation-to-go never decreases (since our current
expectation is the average of the ones we get by setting the next
variable to 0 or 1).
This is called the ``conditional expectation'' method. The algorithm
itself is completely deterministic --- in fact we could rewrite
it to get rid of any hint of randomization by just viewing sum_k
m_k(1-1/2^k) as a way of weighting the clauses to favor the small
ones, but our motivation was based on the randomized method.
In general, the area of approximation algorithms and approximation
hardness is a major area of algorithms research. Occupies a good
fraction of major algorithms conferences.