Transcript Lecture 10, 1 April 2014

The Traveling Salesman Problem
in Theory & Practice
Lecture 10: The Cutting Plane and Branch-and-Bound
Approaches to TSP Optimization
1 April 2014
David S. Johnson
[email protected]
http://davidsjohnson.net
Seeley Mudd 523, Tuesdays and Fridays
Outline
1. Cutting Planes
2. Branch and Bound
3. Student Presentation by Arka Bhattacharya, on
“An O(log n/loglog n) randomized approximation
algorithm for the ATSP,” by Asadpour, Goemans,
Madry, Gharan, and Saberi, SODA 2010.
Optimization, The Early Days
[Applgate, Bixby, Chvátal, & Cook, The Traveling Salesman Problem, 2006]
In the 1940’s and 50’s, even 10-city instances were state-of-the-art.
• Computers were much slower than today, if you had access to one at all,
and compiler optimization almost non-existent.
• 9!/2 = 181,440 was just too daunting, so even pruned exhaustive search
over permutations seemed infeasible.
First breakthrough was
[Dantzig, Fulkerson, & Johnson, “Solution of a large-scale travelingsalesman problem”, Operations Research 2 (1954), 393-410],
which solved a 49-city problem, generated by picking one city from each of
the then 48 U.S. states, plus Washington DC, and using the roadmap
distances between them.
(Actually, they deleted 7 East-coast cities, solved the resulting 42-city
instance, and, lucky for them, the optimal tour contained a link from
Washington DC to Boston, and the shortest path between those two
contained all the deleted cities.)
The Cutting Plane Approach
Solve the edge-based integer programming formulation for the TSP, as
follows:
1.
Start by solving a weak linear programming relaxation.
2.
While the LP solution is not a tour,
a.
Identify a valid inequality that holds for all tours but not the
current solution (a “cutting plane” or “cut” for short) .
b.
Add it to the formulation and re-solve.
Dantzig, Ford, and Fulkerson started with
Minimize Σedexe, subject to
Σxϵe xe= 2 for all cities x.
where de is the length of edge e.
Note: This can be solved in O(N3) time by b-matching techniques.
Dantzig et al. solved it using the Simplex algorithm.
By Hand.
The Cutting Plane Approach, Illustrated
Hyperplane
perpendicular to
the vector of
edge lengths
Optimal Tour
-- Points in RN(N-1)/2 corresponding to a tour.
Optimal Tour is a point on the convex hull of all tours.
Facet
Note: Subtour
constraints are
all facets.
Unfortunately, the LP relaxation of the TSP, especially when we
To improve
it, add more
(“cuts”)
omit the subtour
constraints,
can constraints
be a very poor
approximation
to the convex hull of tours.
Digression: Solving the Degree-2 Problem
in O(N3) Time
w1
w4
w2
w3
Graphical TSP Instance
Replace each city by a pair of cities.
First Idea
Doesn’t work – can get multiple copies of the same original edge.
Replace each edge by a four edges, one joining each pair of a copy of one endpoint
to a copy of the other, all with the original weight.
Compute a standard weighted matching.
Doesn’t work – can get multiple copies of the same original edge.
If you don’t take the
weighted link, then
you must take its
neighbors.
Hence you cannot
cover any
representatives of
the original edge’s
endpoint with this
gadget.
Second Idea
w1
If you do take the
weighted link, then
you must take edges
covering
representatives of
both of the original
edge’s endpoints.
Replace each edge by a gadget like this, with all new edges having weight
zero except the middle one, which gets the weight of the original edge.
Compute a standard weighted matching.
The resulting matching corresponds to a set of edges in
our original graph that makes every city have degree 2.
Finding Violated Constraints
First Choice: Subtour-elimination (subtour) constraints.
Σe={u,v},uϵS,v∉S xe ≥ 2 for S a proper subset of the cities.
How to find a violated one:
Consider the graph consisting of edges e with xe > 0.
Determine its connected components (takes linear time).
If the whole graph is not connected, we get a violated constraint for
each connected component.
This is unfortunately not enough – the graph containing all the nonzero edges can be highly connected and still violate a subtour
constraint.
Danzig et al. found violated subtour constraints by hand.
We can do it in O(N4) time:
Finding Violated Subtour Constraints
•
Let G be a graph with the set C of all cities as vertices, and an edge joining
cities a and b if and only if x{a,b} > 0, that edge having weight equal to x{a,b}.
•
Pick a city c.
•
If there is a violated subtour constraint, c will be on one side of it and some
other city c’ will be on the other.
•
Thus there must be some a set S of cities containing c but not c’, such that
the total weight of edges linking S and C-S less than 2.
•
But by the maxflow-mincut theorem, this means that the maximum flow from
c to c’, where an edge’s capacity equals its weight, will be less than 2.
•
Such a maxflow problem can be solved in time O(N3).
•
To find a violated subtour constraint (if any exists) we need only solve such
problems for all pairs (c,c’), c’ ∈ S-{c}, for a total of O(N4) time.
Digression: Computing the HK bound in
Polynomial Time (by Linear Programming)
• Input data:
– Number N of cities
– Edge lengths d{a,b} for each pair of cities.
– Selected city c0.
• Variables:
– x{a,b} for each pair {a,b} of cities (from the LP
relaxation of the TSP).
– f(a,b,c’) for each ordered pair (a,b) of distinct cities and
each city c’ ≠ c0.
• Goal: Minimize
Σ{a,b}d{a,b}x{a,b}
• Subject to the following constraints.
Constraints of the HK Linear Program
• Variable range constraints: 0 ≤ z ≤ 1, for all variables z.
• Capacity constraints: f(a,b,c’) ≤ x{a,b}, for all triples (a,b,c’).
• Flow conservation constraints:
Σbf(b,a,c’) = Σbf(a,b,c’),
for all c’ ≠ c0 and all a ∉ {c0,c’}.
• MinCut constraints:
Σbf(b,c’,c’) ≥ 2, for all c’ ≠ c0.
O(N3) total variables and constraints
Back to [DFJ54]:
More Cutting Planes Needed
After reaching a solution that satisfied all degree-2 and
subtour constraints, there were still fractional variable values.
Picture from [ABCC06]
Solid edges have xe = 1, dashed edges have xe = ½.
Another Facet Class: Comb Inequalities
H
T1
T2 T3
Tt-1
Tt
Teeth Ti are disjoint, t ≥ 3 is odd, all teeth contain at
least one city in H and one not in H.
H
T1
T2
T3
Tt-1
Tt
• For Y the handle or a tooth, let δx(Y) be the total value of the
edge variables xe for edges with one endpoint in Y and one
outside.
• By the subtour inequalities, we must have δx(Y) ≥ 2 for each
such Y. δx(Y) also must be even, which is exploited to prove
the following comb inequality, which holds whenever x
corresponds to a tour:
δx(H) +
t
∑δx(Ti)
i=1
≥ 3t+1
H
T1
T2
T3
Tt-1
Tt
Theorem: If x is a 0-1 vector of length n(n-1)/2 corresponding to a tour (with x[i] = 1
if and only if edge ei is in the tour) and the sets H and Ti are as described, then
δx(H) +
t
∑δx(Ti)
≥ 3t+1.
i=1
Proof: If δx(Ti) = 2, then the tour corresponding to x must contain a Hamilton path
through the cities in Ti. That path must have at least one edge between a city in H
and one not in H, which will be counted in both δx(Ti) and δx(H).
If δx(Ti) > 2, then we must have δx(Ti) ≥ 4.
Suppose that the number of teeth with the latter property is k. Then our sum is at
least 3t + k, which is at least 3t + 1 if k > 0.
If k = 0, then δx(H) ≥ t. Since t is odd, this means δx(H) ≥ t+1.
In both cases we have that our overall sum is at least 3t+1. QED
Back to [DFJ54]:
Are there any violated combs?
T1
H
T2
T3
Picture from [ABCC06]
δx(H) = 1 + 1 + 1 = 3
δx(T1) = 1 + ½ + ½ = 2
δx(T2) = ½ + ½ + ½ + ½ = 2
δx(T3) = 1 + ½ + ½ = 2
δx(H) +
t
∑δx(Ti) = 9
i=1
< 3t+1 = 10.
New LP Solution
Are there any more violated combs?
H
T3
T2
T1
Picture from [ABCC06]
δx(H) = 1 + 1 + ½ + ½ = 3
δx(T1) = 1 + ½ + ½ = 2
δx (T2) = ½ + ½ + ½ + ½ = 2
δx(T3) = ½ + ½ + ½ + ½ = 2
δx(H) +
t
∑δx(Ti) = 9
< 3t+1 = 10.
i=1
With this extra constraint, the LP yields an optimal tour.
Finding Violated Combs
No general polynomial-time algorithm known.
Not known to be NP-hard either.
Many fast heuristics (See [ABCC06] for details and references).
Some polynomial-time solvable special cases:
•
Combs in which all teeth contain exactly two cities.
– These are called “blossom inequalities.”
– Takes time O(N3).
– See [Padberg & Rao, “Odd minimum cuts and b-matchings,” Math. of O.R. 7
(1982), 67-80].
•
Combs with t teeth for fixed t.
– See [R. Carr, “Separating clique trees and bipartition inequalities having a fixed
number of handles and teeth in polynomial time,” Math. of O.R. 22 (1997), 257265].
– Takes time O(N2t) which is not practical even for combs with 3 teeth (and even
this assumes that your value of x already satisfies all the subtour constraints).
Generalizations of Combs I: The Clique Tree
•
A non-empty collection of pairwise-disjoint “handles” H1, H2, …, Hh.
•
A set of at least three pairwise-disjoint “teeth” T1, T2, …, Tt.
•
These sets satisfying:
– Each tooth contains at least on city that is not in any handle.
– Each handle intersects an odd number of teeth, with that number being at
least 3.
– The graph G, with a vertex for each handle and each tooth, and an edge
between to vertices if the corresponding sets intersect, is a tree.
Clique Tree Inequality:
h
t
i=1
j=1
∑δx(Hi) + ∑δx(Tj)
≥ 2h + 3t - 1
These inequalities all correspond to TSP facets [Grötschel & Pullyblank, 1986].
A violated one can be found in polynomial time for fixed h and t [Carr,1997].
Generalizations of Combs II: Stars and Paths
•
Start with a set of common-tooth combs {Hi,T1,T2,…,Tk}, 1 ≤ i ≤ k, where
– Hi ⊂ Hi+1, 1 ≤ i < k, and
– Hi+1 - Hi - ∪Tj = ϕ.
•
Call a sequence {p, p+1, …, p+k} ⊆ {1, 2, …, h} a “handle interval” for tooth
Tj if it is a maximal such sequence with the following property:
Hp∩Tj = Hp+1∩Tj = … = Hp+k∩Tj
If each city in
a tooth can
only be used in
one comb, we
will need two
copies of the
teeth T1, T2,
and T4, and
three copies
of T5.
H3
H2
H1
{1}, {2,3} for T1
{1,2}, {3} for T2
{1}, {2}, {3} for T3
{1}, {2,3} for T4
{1,2,3} for T5
T1
T2
T3
T4
T5
Stars and Paths, Continued
•
Now suppose we have αi copies of handle Hi and βj copies of tooth Tj.
•
We get “Star Inequality” if, for each tooth Tj and each handle interval
I for Tj,
∑αi ≤ βj.
i∈I
•
We have a “Path Inequality” if, for each tooth Tj and each handle
interval I for Tj,
∑αi = βj.
i∈I
•
In both cases, for valid tours, we have
h
t
h
t
i=1
j=1
i=1
j=1
∑αiδx(Hi) + ∑βjδx(Tj) ≥ (t+1)∑αi + 2∑βj
•
Path Inequalities are facet-inducing [Naddef & Pochet, 2001].
More Cutting Planes
•
•
•
•
Domino Parity Cuts [Letchford, 2000]
Bipartition Cuts [Carr, 1997].
Local Cuts [ABCC, 2006].
…
But the number of facets of the TSP polytope grows
exponentially in N, so the cutting plane approach, by
itself, runs out of gas quite quickly.
Another idea is needed.
Branch & Bound for the TSP
Assume we have an algorithm ALB that computes a lower bound on the TSP length
when certain edges are fixed (forced to be in the tour, or forced not to be in the
tour), and which, for some subproblems, may produce a tour as well.
•
Start with an initial heuristic-created “champion” tour TUB, an upper bound UB =
length(TUB) on the optimal tour length, and a single “live” subproblem in which
no edge is fixed.
•
While there is a live subproblem, pick one, say subproblem P, and apply
algorithm ALB to it.
– If LB < UB
1.
Pick an edge e that is unfixed in P and create two new subproblems as its children, one
with e forced to be in the tour, and one with e forced not to be in the tour.
2.
If algorithm ALB produced a tour T, and length(T) < UB
1.
Set UB = length(T) and TUB = T.
2.
Delete all subproblems with current LB ≥ UB, as well as their children and their ancestors that
no longer have any live children.
– Otherwise (LB ≥ UB), delete subproblem P and all its ancestors that no longer have
live children.
•
Halt. Our current champion is an optimal tour.
UB = 97
Initial LP, UB = 100, LB = 90
X{a,b} = 0
X{a,b} = 1
LB = 92
X{c,d} = 0
LB = 92
LB = 93
X{c,d} = 1
LB = 100
X{a,c} = 0
X{a,c} = 1
LB = 98
LB = 97
New Opt = 97
X{e,a} = 0
LB = 101
X{e,a} = 1
LB = 100
Opt = 97
First Effective Implementation
•
A branch-and-bound scheme of Held & Karp that used their iterative
one-tree-based approach to approximating the LP relaxation of the TSP
(described last week) for computing lower bounds.
[Held & Karp, ‘‘The traveling-salesman problem and minimum spanning trees:
Part II,’’ Math. Programming 1 (1971), 6-25].
•
Beat Dantzig-Fulkerson-Johnson’s record of 49 cities, set 17 years
earlier, by solving two 64-city instances (one random Euclidean instance,
and one that merged the 42-city reduced instance of Dantzig et al. with
22 random Euclidean cities).
•
Unlike Dantzig et al., who solved their instance by hand, this one was
solved by computer (programmed Linda Ibrahim at UC Berkeley).
•
Subsequent work along this line pushed the record up to 67.
[Camerini, Fratta, & Maffioli, “On improving relaxation methods by modified
gradient techniques,” Math. Programming Study 3 (1975), 26-34].
•
But again we seem to be hitting limits and another idea is needed.
Branch & Cut
• Combine Branch-and-bound with Cutting Planes.
• More details next time.