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CS880: Approximations Algorithms Scribe: Dave Andrzejewski Lecturer: Shuchi Chawla Topic: Metric multiway cut and multicut, integrality gap Date: 3/15/
The lecture further explores the use of cut metrics, with applications to the multiway cut and multicut problems. Also, the idea of an expander graph is introduced and applied to deriving the integrality gap between an optimal LP solution and the optimal corresponding integral solution.
16.1 Metric multiway cut
16.1.1 Problem setup and LP
GIVEN: a graph G = (V, E) and a set T ⊂ V of terminals. DO: find the minimum weight cut separating every pair of terminals ti, tj ∈ T from one another.
Just as in the previous lecture, we will formulate this cut problem as a metric LP. We enforce the separation of all pairs of terminals by requiring that our metric assign them distance ≥ 1.
min
(u,v)∈E
cuvd(u, v) obj fcn
d is a metric d(ti, tj ) ≥ 1 ∀ti, tj ∈ T
where cuv is the cost of the edge between vertices u and v.
16.1.2 Rounding
Once we’ve found an optimal solution to the metric LP above, we need to transform our metric to a cut metric, which will yield a valid multiway cut. As usual, we use the fact that LP ∗^ ≤ OP T.
To do this analysis, it will be useful to consider a set of interesting physical analogies for the quantities in our problem.
- edges → pipes
- ce → cross-sectional area of pipe e
- de → = length of pipe e
- B(ti, r) → = ball of radius r, centered at ti
- f (ti, r, e) → = fraction of edge e covered by ball B(ti, r)
- V ol(B(ti, r)) → = total pipe volume enclosed by ball B(ti, r)
- Area′(B(ti, r)) → = the total pipe cross section area on the surface of ball B(ti, r)
- Area(B(ti, r)) → = the total cost (ce) of the edges crossing the ball B(ti, r)
Note that Area 6 = Area′^ in general, because the surface of the ball may not be perpendicular to the pipe (Figure 16.1.1). In fact, Area′^ ≥ Area. The expressions for f ,V ol and Area are as below. For f , edge e = (u, v), and f is simply 1 or 0 if both or neither of (u, v) are contained in the ball, with the expression below covering the more interesting case where u ∈ B and v /∈ B.
f (ti, r, e) =
r − d(ti, u) d(ti, v) − d(ti, u)
V ol(B(ti, r)) =
e∈E
f (ti, r, e)dece (16.1.2)
E′^ = {(u, v) s.t. |B ∩ u, v| = 1} (16.1.3) Area(B(ti, r)) =
(u,v)∈E′
cuv (16.1.4)
Area′(B(ti, r)) =
d dr V ol(B(ti, r)) =
(u,v)∈E′
cuv
d(u, v) d(ti, v) − d(ti, u)
These quantities will provide an intuitive framework with which to analyze our rounding scheme.
Algorithm
- ∀i, pick ri = arg minr∈[0, 1 /2] Area(B(ti, r))
- let Ci be the cut associated with that radius
- pick the k − 1 minimum weight cuts from C 1 , C 2 , ..., Ck
Since each pair of terminals (ti, tj ) must satisfy d(ti, tj ) ≥ 1 in the original LP solution, making our cuts with at a radius r ≤ 1 /2 around each terminal will clearly yield a valid set of separating cuts.
We now derive the approximation factor of our resulting scheme.
Lemma 16.1.1 Area(ti, ri) ≤ 2 V ol(ti, 1 /2) ∀i
Proof: As the radius of a ball increases, the volume enclosed by the ball will grow proportionally to the surface areas of all pipes currently cut by the surface of the ball.
If all cut pipes were perpendicular to the surface of the expanding ball the volume growth would be exactly equal to the current surface area, but since this is not neccessarily the case (see 16.1.1) Area lower bounds the rate of volume growth (Area′).
Area(ti, r) ≤ d dr
V ol(ti, r)
pair. As in previous lectures, this formulation would have the disadvantage of exponentially many constraints, but could be approached using the ellipsoid method. The ellipsoid method requires an efficient ’separation oracle’ to reveal which constraint is violated by any proposed solution. For this problem we could use the results of the polytime all-pairs shortest path algorithm for this purpose.)
16.2.2 Rounding
Given an optimal LP solution, how can we round to a valid cut, and how can we analyze this scheme? The method applied to the multiway cut problem is no longer directly applicable, because the fact that we now only separate pairs of terminal nodes means that the B centered at each terminal are no longer guaranteed to be disjoint.
The key idea to overcome this is to only charge the area to the sub-ball entirely enclosed by a given cut. Once the edges are charged to that cut, they are then removed from the graph, potentially changing area and volume calculations for subsequent steps. Crucially, these modifications ensure that the volumes remain disjoint.
Our derivation will involve dividing by V ol(si, 0) at some point, which would be an undefined divide-by-zero operation. We avoid this by redefining volume slightly. Call F the total volume of the graph. Then redefine volume by assigning volume F/k to each terminal si.
V ol′(si, r) = F/k +
e
fecede
Now the total volume of the graph has doubled, so all previous volume lemmas still hold, with both sides multiplied by 2. The initial volume of a ball is then F/k, and its final volume must be ≤ F/k + F.
Algorithm
- for each i, pick minimum r such that Area(si, r) ≤ αV ol(si, r)
- make the cut Ci at that r, remove those edges from the graph
- repeat for next i
We pick the α value to be 2 ln (k + 1).
Lemma 16.2.1 ri ≤ 1 / 2 ∀i
Proof: As before, we know that Area(si, r) ≤ (^) drd V ol(si, r).
Now assume ri ≥ 1 /2. Then we must have that
d dr
V ol(si, r) > αV ol(si, r) ∀r < 1 / 2
because otherwise we would have picked one of those smaller r.
We then manipulate the equation, integrating from r = 0 to r = 1/2.
dV ol(si, r) > αV ol(si, r) dr 1 V ol(si, r)
dV ol(si, r) > α dr ∫ (^1) / 2
0
dV ol(si, r) V ol(si, r)
0
α dr
ln
V ol(si, 1 /2) V ol(si, 0)
α/ 2
Recall that the maximum possible value of V ol is now F + F/k, and that V ol(si, 0) is now defined to be F/k. Substitute these values in to get another inequality.
ln (1 + k) = ln
F + F/k F/k
ln
V ol(si, 1 /2) V ol(si, 0)
α/ 2
This gives us α < 2 ln (1 + k). Since we have chosen α = 2 ln (1 + k), we have derived a contradic- tion.
Lemma 16.2.
i V ol(si, ri)^ ≤^2 LP^
∗ = 2F
Proof: This is acheived by construction. Our modified disjoint volumes now sum to no more than 2F. Since LP ∗^ is equal to the volume of the full graph F , this is clearly true.
These lemmas will be used in our full derivation of the algoirthm approximation factor.
Theorem 16.2.
i |Ci|^ =^
i Area(si, ri)^ ≤^2 αLP^ ∗ (^) ≤ 2 αOP T
This theorem simply follows from the combination of the previous lemmas and definitions. Plugging our chosen α = 2 ln (1 + k) in shows that this algorithm achieves an 4 log (1 + k)-approximation [4].
Multicut is known to be APX-hard [1], meaning that it cannot be approximated within every constant factor.
Furthermore, if Unique Games Conjecture is true, it cannot even be approximated within any constant factor. A stronger version the Unique Games Conjecture further implies that it cannot be approximated with a factor Ω(log log n) either [3].
16.3 Integrality gap analysis
To derive an integrality gap for a given problem, we find a problem instance in which the optimal integral solution is significantly worse than the optimal fractional solution. The factor by which the integral solution is worse is known as the integrality gap. The existence such a gap bounds the performance of approximation algorithms based on LP rounding, since we cannot approximate LP ∗^ at a factor better than the integrality gap. Our specific approach will be to use a special mathematical stucture known as an expander graph to construct a problem instance with this property.
