The Traveling Salesman Problem
in Theory & Practice
Lecture 6: Exploiting Geometry
25 February 2014
David S. Johnson
dstiflerj@gmail.com
http://davidsjohnson.net
Seeley Mudd 523, Tuesdays and Fridays
Outline
1. The k-d tree data structure
2. Exploiting k-d trees to speed up geometric tour
construction heuristics
K-d trees and the TSP
References
•
J.L. Bentley, “Multidimensional binary search trees used for
associative searching,’’ Comm. ACM 9 (1973), 509-517.
•
J.H. Friedman, J.L. Bentley, R.A. Finkel, “An algorithm for finding
best matches in logarithmic expected time,” ACM Trans. Math.
Software 3 (1977), 209-226.
•
J.L. Bentley, B.W. Weide, A.C. Yao, “Optimal expected-time
algorithms for closest point problems,” ACM Trans. Math. Software
6 (1980), 563-580.
•
J.L. Bentley, “K-d trees for semidynamic point sets,” Proc. 6th Ann.
ACM Symp on Computational Geometry, ACM, 1990, 187-197.
•
J.L. Bentley, “Experiments on traveling salesman heuristics,” Proc.
of 1st Ann. ACM-SIAM Symp. on Discrete Algorithms, 1990, 91-99.
•
J.L. Bentley, “Fast algorithms for geometric traveling salesman
problems,” ORSA J. Comput. 4 (1992), 387-411.
K-d trees [Bentley, 1975]
(Based on hierarchical partitioning, splitting at medians)
Stop partitioning is box contains k or fewer cities (typical k might be 5 or 8)
Data Elements
Array T: Permutation of Points derived from the partitioning
Array D: Array of Tree Vertices
Array H: H[i] = index of tree leaf vertex containing Point i
Point Structure:
Index of
Point
X-coord
Tree Vertex Structure:
Index L in
Array T of
First Point
Index U in
Array T of
Last Point
Widest
Dimension
Median Value
in Widest
Dimension
Leaf?
Y-coord
Deleted?
Implicit for Tree Vertex D[k]:
• Index of Median = ⎣(L+U)/2⎦
• Index of Parent = ⎣k/2⎦
• Index of Lower Child = 2k
• Index of Higher Child = 2k+1
Tree Vertex D[1]:
L=1
U=N
horizontal
0.71
0.48
Non-Leaf
0.60
Tree Vertex D[2]:
L=1
U=m
vertical
0.71
Non-Leaf
Tree Vertex D[3]:
0.48
(mth
vertex from left)
[L,U] is reordered so the city with median widestTdimension
contains coordinate
all the cities,
in position
arbitrary
order. and all cities
is in
⎣(L+U)/2⎦
to the left have widest dimension coords. ≤ the median.
L = m+1
U=N
vertical
0.60
Non-Leaf
Operations
• Construct k-d Tree: O(NlogN)
• Delete or Undelete Points (without rebalancing): O(1)
• Nearest Neighbor Search: Find undeleted point nearest to
a specified member of the point set: “Typically” O(logN)
• Fixed Radius Search: Return all undeleted points that are
within the specified distance to a specified member of the
point set: “Typically” O(logN)
• “Ball Search” – will be described later when we need it.
Finding nearest neighbor of point v
1.
Let the coordinates for v by (x,y).
2.
Use array H to find the index k of the tree vertex structure
for the leaf containing v.
3.
Find the nearest neighbor u of v among the cities in the the
interval of T specified by the tree vertex D[k].
4.
Let Δ = d(u,v). Any point (x’,y’) closer to v than u must have
x-Δ ≤ x’ ≤ x+Δ and y-Δ ≤ y’ ≤ y+Δ, so we only need consider
cities in leaves whose regions intersect this corresponding
square, which we can discover by going up the tree.
u
v
Finding nearest neighbor of point v:
The recursive search: Going up.
Initially, all four directions (up,down,left,right) are live. Set k to be the
index of the leaf that contains v.
Repeat until done:
1.
If k = 1 or all four directions are dead, we are done.
2.
Let D[k] be the current Tree Vertex.
3.
Consider the parent Tree Vertex D(⎣k/2⎦), and let D[k’] be the
sibling of D[k]. If D[k’] has already been explored, set k ←⎣k/2⎦
and continue.
4.
If the parent’s other child is in a dead direction, set k ⎣k/2⎦ and
continue.
5.
Test whether the parent’s median (in its specified direction) is
contained in the current square.
6.
1.
If yes, call “ExploreDown(k’)”.
2.
Otherwise, declare this direction dead.
Set k ←⎣k/2⎦ and continue.
Finding nearest neighbor of point v:
ExploreDown(k)
1.
If D[k] is a leaf, let w be the closest point to v in the list of points
for D[k].
1.
If d(v,w) < Δ, declare w to be the new nearest neighbor and set
Δ = d(v,w), thus shrinking the current square.
2.
Return.
2.
Otherwise, D[k] is an internal vertex, with at least one of its two
children potentially intersecting the current square. Let k’ be the
index of the child that lies to the “inside” direction of the median,
and let k’’ be the index of the other child.
3.
Call ExploreDown(k’).
4.
If the median lies entirely outside the current square, declare the
relevant direction dead and return.
5.
Otherwise, call ExploreDown(k’’) and then return.
Fixed Radius Search
Implemented just like Nearest Neighbor Search, except that Δ is
fixed and part of the query, and we return all vertices in the
encountered leaf nodes that satisfy the distance criterion.
Nearest Neighbor in “O(NlogN)”
1. Construct a k-d tree on cities.
2. Pick a starting city cπ(1) and delete it from the kd-tree.
3. While there remains an unvisited city, let the current last
city be cπ(i).
• Delete cπ(i) from the kd-tree, and use a Nearest
Neighbor Search to find the nearest unvisited
(undeleted) city to cπ(i). Call that city cπ(i+1) and declare
it the new “current last city”.
4. Add an edge from cπ(N) to cπ(i).
Greedy (Lazily) in “O(NlogN)”
Initialization
•
Construct a k-d tree on the set C of cities. Let G be a graph with vertex
set C and (initially) no edges. We will maintain the property that the kdtree contains only vertices with degree 0 or 1 in G. We will also maintain,
for each degree-1 city c, the identity tail(c) of the other end of the path
containing c.
•
For each city c, perform a Nearest Neighbor Search to identify its
nearest neighbor c’ and add the triple 〈c,c’,dist(c,c’)〉 to a priority
queue, sorted by increasing value of the third component.
In what follows, we shall say that an edge {c,c’} is “legal” if adding it to the
current graph G does not create
a) A vertex of degree 3 or
b) A cycle of length less than N.
Greedy (Lazily) in “O(NlogN)”
Choosing the next edge to add
While not yet a Hamilton Path:
•
While we haven’t yet selected a champion edge,
– Extract the minimum 〈c,c’,d(c,c’)〉 triple from the priority queue and
let Δ = d(c,c’).
– While, {c,c’} is not legal,
• If c’ = tail(c), we temporarily delete c’ from the kd-tree and do a
Nearest Neighbor Search to find a new nearest neighbor c’ for c.
– Once the we have obtained a legal {c,c’}, undelete all the cities deleted
during the interior while loop.
– If d(c,c’) ≤ Δ, declare {c,c’} to be the champion; otherwise, insert the
triple 〈c,c’d(c,c’)〉 into the priority queue -- there is no champion yet.
Greedy (Lazily) in “O(NlogN)”
Handling the Chosen Legal Edge
•
Add {c,c’} to G.
•
If either c or c’ now has degree 2, delete it (permanently) from the k-dtree. We also have to update some tail( ) values:
– If both c and c’ had degree-0 previously, set tail(c) = c’ and tail(c’) = c.
– If c was previously degree-1 and c’ was degree 0, set tail(c’) = tail(c)
and tail(tail(c)) = c’.
– If c’ was previously degree-1 and c was degree 0, set tail(c) = tail(c’)
and tail(tail(c’)) = c.
– If both c and c’ were degree-1, set tail (tail(c)) = tail(c’) and
tail(tail(c’)) = tail (c).
tail(c)
c
c’
tail(c’)
Once we have constructed a Hamilton path, we get a tour by adding an edge
between and the two endpoints of the path.
Savings Heuristic in “O(NlogN)”
•
c’ Greedy, except that the
Our implementation mimics our lazy approach to
triples in the priority queue
c will be 〈c,c’,s(c,c’)〉, the third component being
the savings from the shortcut, not the distance between c and c’. (Also, the
hub city c1 does not take part in any triple.)
•
The key difference is in computing the greatest savings for a given pair of
cities c and c’, which equals d(c1,c) + d(c1,c’) – d(c,c’).
•
Nominally, we want to find, for each c, that c’ which yields the biggest savings,
and insert the corresponding triple into the priority queue. However, this is
c1
not strictly necessary.
•
Call a city c’ “good” for c if d(c1,c’) ≤ d(c1,c) and otherwise call it “bad”
•
CLAIM: For a given c, it is not harmful to ignore bad candidates for c’.
•
PROOF: Suppose c’ is not good for c, in which case d(c1,c’) > d(c1,c). If the
edge {c,c’} provides the biggest savings overall, then it will also provides the
biggest savings for c’, and since c is good for c’ in this case, the triple involving
the edge {c,c’} will still be in the priority queue, as part of the triple 〈
c’,c,s(c’,c)〉, even under the restriction to only good candidates
Savings Heuristic in “O(NlogN)”
•
Construct a standard k-d tree as we did for Greedy.
•
For each city c, perform a modified Nearest Neighbor search to identify a good
neighbor c’ for c (if any exist) that yields the maximum savings, and add the
triple 〈c,c’,s(c,c’)〉 to the priority queue. The search proceeds as follows.
•
If there are any good cities in the leaf bucket containing c, let S be the
maximum savings for any such city, and let c’ be that city. Otherwise, let c’ be
c and set S to -∞.
•
CLAIM: If a good city c’’ for c yields savings S’’ > S, then we must have
d(c,c’’) ≤ 2d(c1,c) – S.
•
PROOF: By definition we have S’’ = d(c1,c’’) + d(c1,c) – d(c,c’’).
Hence d(c,c’’) = d(c1,c’’) + d(c1,c) – S’’ < 2d(c1,c) – S, since the fact that c’’ is good
for c implies d(c1,c’’) ≤ d(c1,c) .
•
So, from now on mimic Nearest Neighbor Search with d(c,c’) < 2d(c1,c) – S
playing the role of d(c,c’) < Δ,.
•
The rest of the implementation closely mimics that for Greedy, with our
modified Nearest Neighbor Search replacing the original.
Nearest Addition in “O(NlogN)”
Initialization
•
Construct a standard k-d tree as we did for Greedy.
•
Perform Nearest Neighbor Searches for each city c to find its nearest
neighbor n(c).
•
Use linear search to find a city c with minimum d(c,n(c)). Let c’ = n(c).
•
Let our initial tour consist of the two cities c and c’, delete c and c’ from
the k-d tree, compute the new nearest neighbors n(c) and n(c’) for c and c’,
and add the triples 〈c,n(c),d(c,n(c))〉 and 〈c’,n(c’),d(c,n(c’))〉 to a new
priority queue.
•
Inductively, we shall assume that the cities in the tour are deleted from
the k-d tree, while the non-tour cities remain undeleted. In addition, the
priority queue contains entries for every city in the current tour (and only
those), with the triples satisfying the property that, for each triple
〈
c,c’,d(c,c’)〉, if c’ is not in the current tour, then it is the nearest non-tour
city to c. Finally, we store the tour in a doubly-linked list, with each city
linked to its two tour neighbors, so that the two edges involving a given
vertex can be found in constant time.
Nearest Addition in “O(NlogN)”
Doing the Insertions
While we have not yet constructed a tour on all the cities,
•
While we have not yet found a valid candidate for the next city to be
inserted,
– Extract the minimum 〈c,c’,d(c,c’)〉 from the priority queue.
– If city c’ is currently in the tour, perform a Nearest Neighbor
Search to identify the nearest non-tour city c’’ to c, and insert the
triple 〈c,c’’,d(c,c’’)〉 into the priority queue.
•
Let {c,c1} an {c,c2} be the two tour edges involving c.
•
Determine into which of the two edges c’ can be more cheaply inserted
and perform that insertion.
•
Delete c’ from the k-d tree, compute its nearest non-tour neighbor c’’,
and insert 〈c’,c’’,d(c’,c’’)〉 into the priority queue.
Note: A simplified variant on this implementation will construct
MSTs in time “O(Nlog(N))”.
Nearest Insertion in “O(NlogN)”
• The choice of how to start and which city to insert next is the
same as for Nearest Addition.
• The new complexity is that, instead of just two choices of where
to insert the new city, now ALL tour edges are potential insertion
candidates.
• That is going to be complicated, so let’s start with something a bit
simpler and almost as good, an algorithm due to [Bentley, 1992]
that is intermediate between Nearest Addition and Nearest
Insertion:
• Nearest Augmented Addition (NA+): Choose the city c to be
inserted as in Nearest Addition, and let c’ be the nearest tour city
to it. As candidate insertion edges, consider all tour edges having
an endpoint c’’ with d(c,c’’) ≤ 2d(c,c’).
Nearest Augmented Addition in “O(NlogN)”
Finding the Best Insertion
• We maintain an additional k-d tree, this one containing all the
cities, but with all the cities initially marked “deleted” and only
undeleted when they are added to the tour.
• To find the candidate edges for city c with nearest tour neighbor
c’, we do a fixed radius search in our second kd-tree for c with
radius 2d(c,c’).
• For each city c’’ found in this search, we determine the cost of
inserting c into each of the tour edges with endpoint c’’, and insert
c into the best edge found, over all c’’ returned by the fixed
radius search.
Nearest Insertion in “O(NlogN)”
Finding the Best Insertion
• We exploit a theorem from [Bentley, 1992]:
• Best Insertion Theorem: Let c’ be the nearest tour neighbor of
non-tour city c, and for any tour city b, let longer(b) be the length
of the longer tour edge incident on b. The best tour edge into
which to insert c must have an endpoint e satisfying one of the
two following restrictions:
a) d(c,e) ≤ 1.5d(c,c’), or
b) d(c,e) ≤ 1.5longer(e).
• To exploit this, we need to use the “Ball search” operation
mentioned earlier, which we will now explain. We will prove the
theorem later.
Nearest Insertion in “O(NlogN)”
Finding the Best Insertion
Ball Search
•
Suppose we have a positive distance rad(c’) associated with each undeleted
city c’. Given a city c, return all those undeleted c’ such that d(c,c’) ≤
rad(c’).
[We wish to implement this in an environment where cities can be undeleted and the values
of rad(c’) can change dynamically.]
•
The basic idea is to store, with each leaf vertex, a list of all the undeleted
cities c’ that are within distance rad(c’) of some city in the leaf vertex.
•
To construct these lists, perform Fixed-Radius searches for each
undeleted c’, with radius set to rad(c’).
•
We can update the value of rad(c’) for a given c’ by performing a FixedRadius search for c’ with the larger of the before and after values for
rad(c’), and either adding or removing c’ from the leaf lists, as appropriate.
Nearest Insertion in “O(NlogN)”
Finding the Best Insertion
To set up our method for finding the best insertion, we maintain the property
that the undeleted cities in our k-d tree are precisely the tour cities, and that
for each such city c’’, we have rad(c’’) = 1.5longer(c’’). This involves at most
three Fixed Radius searches after each insertion: one for the inserted vertex
and one for each of its new tour neighbors.
By the Best Insertion Theorem, we can then find all potential candidates for
the best place to insert c by
1.
Doing a Fixed Radius search for c with radius 1.5d(c,c’), where c’ is
(already identified) city in the tour that is nearest to c.
2.
Doing a Ball Search for c.
The best edge will be a tour edge incident on one of our candidates, by the
Theorem.
With luck, this will yield far fewer candidates than there are cities in the
current tour, at least when the latter number number is large.
Proof of Best Insertion Theorem
Recall: c is the city to be inserted and c’ is the nearest tour city to c.
We need to show the the best place to insert c into the tour is next to a city e
satisfying one of
a)
d(c,e) ≤ 1.5d(c,c’), or
b) d(c,e) ≤ 1.5longer(e).
Observation: Denote the cost of inserting a vertex c into a tour edge {x,y},
which by definition is d(c,x) + d(c,y) - d(x,y), as cost(c,x-y). Then
cost(c,x-y) ≤ 2d(c,x).
Proof: The triangle inequality implies d(c,y) ≤ d(c,x) + d(x,y). Thus we have
cost(c,x-y) ≤ d(c,x) + (d(c,x) + d(x,y)) –d (x,y) = 2d(c,x).
We exploit this to prove the Best Insertion Theorem by contradiction. Assume
that {f,g} yields the minimum value of cost(c,x-y) for {x,y} a tour edge and c the
city we wish to insert, but neither (a) nor (b) holds when e is either f or g.
Case 1 [Long {f,g}]: d(f,g) > d(c,c’).
Let c’’ be either tour neighbor of c’.
c’’
Assume without loss of generality that d(f,g) = 1,
and hence both longer(f) and longer(g) are at
least 1.
c’
Since this is the “Long” case, this means that
d(c,c’) < 1.
By assumption we have thus have both d(c,f) >
1.5 and d(c,g) > 1.5.
By our observation, cost(c,c’-c’’) ≤ 2d(c,c’) < 2.
By definition, cost(c,f-g) = d(c,f) + d(c,g) – d(f,g)
> 1.5 + 1.5 – 1 = 2.
So cost(c,f-g) > cost(c,c’-c’’), a contradiction.
<1
c
> 1.5
> 1.5
g
f
1.5
1
1.5
Case 2 [Short {f,g}]: d(f,g) ≤ d(c,c’).
Assume without loss of generality that d(c,c’) = 1.
c’’
Then, since this is the “Short” case, we must have
that d(f,g) ≤ 1.
c’
1
By assumption we have thus have both d(c,f) > 1.5
and d(c,g) > 1.5.
c
1.5
By our observation, cost(c,c’-c’’) ≤ 2d(c,c’) = 2.
> 1.5
By definition, cost(c,f-g) = d(c,f) + d(c,g) – d(f,g) >
1.5 + 1.5 – 1 = 2.
> 1.5
So cost(c,f-g) > cost(c,c’-c’’), a contradiction.
f
≤1
g
Cheapest Insertion in “O(NlogN)”
•
Construct a standard k-d tree as we did for Greedy.
•
The intitial 2-city tour is constructed just as in Nearest Addition.
•
For subsequent insertions, we will maintain a (lazily evalutated) priority
queue whose entries are triples 〈{a,b},c,d(a,c)+d(b,c)〉 for {a,b} a tour
edge and c a non-tour city that yields the minimum value for Δ = d(a,c) +
d(b,c).
•
We determine the next insertion by repeatedly extracting the triple 〈
{a,b},c,Δ〉 with the minimum Δ from the priority queue, discarding it if
{a,b} is not a tour edge, updating and reinserting it if {a,b} remains a tour
edge but c is now in the tour, and otherwise stopping. Once we have
stopped we know that the current triple denotes the cheapest insertion,
and we perform it.
•
To compute the currently correct triple for a tour edge {a,b}, we
proceed as described on the next slide. (This computation is needed
when {a,b} first becomes a tour edge, or when a triple for {a,b} is
extracted from the priority queue and, although {a,b} remains a tour
edge, c is now in the tour.)
Cheapest Insertion in “O(NlogN)”
Finding the cheapest c for {a,b}
1)
First, use a Nearest Neighbor search to find a nearest neighbor c of a,
with ties broken in favor of the nearest neighbor that minimizes d(a,c) +
d(b,c).
2)
Next, do a fixed-radius search from b with radius d(b,c), and among the
returned cities (which must include c itself) find a city e that yields the
minimum value for d(b,e) + d(a,e).
Claim: The city e thus found yields the cheapest possible insertion for {a,b}.
Proof: Suppose not, and some city f yields a cheaper insertion. We thus
must have d(a,f) + d(b,f) < d(b,e) + d(a,e) ≤ d(a,c) + d(b,c). By step (1), we
must have d(a,f) ≥ d(a,c), which implies we must have d(b,f) < d(b,c). But this
means that f would have been considered in step (2), and so we cannot have
d(b,f) + d(a,f) < d(b,e) + d(a,e), a contradiction.
Convex Hull Cheapest Insertion in “O(NlogN)”
The same as Cheapest Insertion, except for the choice of starting tour.
Convex Hull Greatest Angle Insertion in “O(NlogN)”
A bit more complicated…
Figure from [JBMR94]
RI, RA, RA+ “O(NlogN)”
1.
Build k-d tree on cities, with all cities marked deleted (no
priority queue needed).
2. Pick two random cities c and c’ to start the tree, and mark them
undeleted.
3. While not yet done
1)
Pick a random non-tour city c.
2)
Do a Nearest Neighbor search from c to find its nearest tour neighbor c’.
3)
We then use the method for choosing the place to insert that is used by
NI, NA, or NA+, respectively.
4)
Insert c’ into the chosen edge and mark it as undeleted.
FI, FA, and FA+ in “O(NlogN)”
As usual, we start by constructing a k-d tree on all the cities, with all cities
deleted except those in the initial tour.
Then, for each city c not in our initial tour, we compute its nearest tour neighbor
c’, and enter the triple 〈c,c’,d(c,c’)〉 into a priority queue, sorted by decreasing
(rather than increasing) order of the third component.
To find the next city to insert:
1.
Extract the maximum element 〈c,c’,d(c,c’)〉 in the priority queue.
2.
Perform a Nearest Neighbor Search on c, and if it finds a tour city c’’ other
than c’, insert 〈c,c’’,d(c,c’’)〉 and go back to extracting.
3.
Otherwise, we choose to insert vertex c.
We then use the method for choosing the place to insert as used by NI, NA, or
NA+, respectively.
FI, FA, and FA+ in “O(NlogN)”
Hidden Challenge: Constructing the initial 2-city tour.
How do we identify the two most distant cities in O(NlogN) time?
Answer: It can be done.
Suppose C is the minimum-diameter circle that contains all our
cities. (This can be constructed in linear time [N. Megiddo, “Linear-time
algorithms for linear programming in R3 and related problems,” SIAM J. Comput. 12
(1983) 759-756]
.)
Claim: There is a pair {a,b} of most-distant cities, both of which lie
on circle C.
Proof by Picture: [On succeeding slides]
Slide C as far right as possible, until it hits a city c →
There are no
cities inside
this region:
Any such city
x would have
d(x,b) > d(a,b),
contradicting
the maximality
of d(a,b).
c
a
Circle of
radius d(a,b)
with center b.
C
b
{c,b} is also a maximum distance pair, and c is on the circle.
So we may assume there is a maximum distance pair {a,b} with
city a on the minimum-diameter circle containing all the cities.
there is a of
maximum-distance
pair
{a,b}
one, buta.
Case 1. Assume
The intersections
A and C are in the
half
of with
C containing
not both, endpoints on circle C.
A
C
a
Circle A of
radius d(a,b)
with center a.
b
There are no
cities in this
closed region:
Any such city
x would have
d(x,a) ≥ d(a,b),
contradicting
either the
maximality of
d(a,b) or the
assumption
that no max
distance pair
has both
endpoints on C.
Then we can slide C left by a small ε and it will still contain all cities,
but now none will be on its border. Thus, we can shrink C and still
enclose all the cities – a contradiction.
Case 2. The intersections of A and C are in the half of C not containing a.
A
x
C
a
Circle A of
radius d(a,b)
with center a.
b
All cities must
lie in this
region
There are no
cities in this
closed region:
Any such city
x would have
d(x,a) ≥ d(a,b),
contradicting
either the
maximality of
d(a,b) or the
assumption
that no max
distance pair
has both
endpoints on C.
Let x be the city on C that is closest to b,
and let X be a circle of radius d(a,b) centered at X.
All cities must lie in the darkest region (or on its boundaries).
A
x
C
a
Split C by a diameter that
intersects A between b and
the intersection of A and C
closest to x.
b
Note that both a and x must
lie on the same side of this
diameter, and somewhat
distant from it.
Now we can slide C by a small ε
in the direction indicated, and
all cities will remain inside, but
now none will be on the
boundary and we can shrink C
and still enclose them all.
Finding the farthest pair in time O(NlogN)
So now we know we can restrict attention to
cities on the minimum-diameter enclosing circle C.
Sort these cities in order around the circle in
time O(NlogN).
C
x
For a given boundary city a, we can now find the
boundary city b that is most distant from a in
constant time:
Construct the diameter of C that has c as an
endpoint, and let x be the intersection of that
diameter with the other side of the circle.
a
City b is the boundary city closest to x.
Thus the total time for finding a farthest pair is O(N) to find the smallest
enclosing circle, O(NlogN) to sort the cities on the circle, and O(N) to find the
farthest city for each boundary city and then determine the farthest over all.
Total time is hence O(NlogN), as hoped.
b
Farthest Insertion in “O(NlogN)”
• Dirty Little Secret:
• We don’t do what I just told you.
• In Jon Bentley’s Farthest Insertion code (which is used for our
results) the Initial Tour actually consists of the first city in the
input plus the city most distant from it.
• Similarly, his implementations of Nearest Addition, etc., all
start with a tour consisting of the first city in the input plus its
nearest neighbor.
• Advantages: Simpler programming.
• Disadvantages: Results depend on the order in which the input
cities are presented.
Double MST in “O(NlogN)”
•
Create k-d tree: O(NlogN)
•
Build MST using a priority queue and choosing cities to add (and where)
in the same way we chose vertices to add in Nearest Addition.
•
Duplicate the spanning tree: O(N)
•
Find Euler Tour: O(N)
•
Perform “smart shortcuts”: O(N)
Approximate Christofides in “O(NlogN)”
•
Create k-d tree: O(NlogN)
•
Build MST using a priority queue and choosing cities it add (and where)
in the same way we chose vertices to add in Nearest Addition.
•
Add a “greedy” matching on the odd-degree vertices using Nearest
Neighbor searches.
•
Find Euler Tour: O(N)
•
Perform “smart shortcuts”: O(N)
Random Euclidean Performance for N = 1,000,000
(% Excess over HK Bound)
Algorithm
Smart-Shortcut Christofides
%
9.8
Approximate Smart-Shortcut Christofides
11.2
Savings
12.2
Farthest Insertion
13.5
Greedy
14.2
Classic-Shortcut Christofides
14.5
Random Insertion
15.2
Implementation Breakdown
Algorithm
Data structures
Searches/City
Other Algs
Nearest Neighbor
1 kd
1 nn
Greedy
1 kd, 1 pq, tail
1+ nn
Savings
1 kd, 1 pq, tail
1+ nn*
Nearest Addition
1 kd, 1 pq
1+ nn
Nearest Addition+
2 kd, 1 pq
1+ nn, 1 fr
Nearest Insertion
2 kd, 1 pq
1+ nn, 1 fr, 1 ball
Cheapest Insertion
1 kd, 1 pq
2+ nn*, 2+ fr
Convex Hull, Cheapest Insertion
1 kd, 1 pq
2+ nn*, 2+ fr
Random Addition
1 kd
1 nn
Random Addition+
1 kd
1 nn, 1 fr
Random insertion
1 kd
1 nn, 1 fr, 1 ball
Farthest Addition
1 kd, 1 pq
2+ nn
Farthest Addition+
1 kd, 1 pq
2+ nn, 1 fr
Farthest Insertion
1 kd, 1 pq
2+ nn, 1 fr, 1 ball
Double MST
1 kd
1+ nn
euler tour
Approximate Christofides
1 kd
2+ nn
euler tour
convex hull
Running Times, Random Euclidean Instance
with N = 1,000,000
Normalized to a 500Mhz DEC Alpha Processor in a Compaq ES40 with 2 Gb RAM
Algorithm
secs.
Read Instance
1.0
Strip
2.8
Spacefilling Curve
%
Algorithm
secs.
%
Farthest Addition
148.7
43.1
30.1
Farthest Addition+
175.2
13.6
3.0
35.1
Random Insertion
202.9
15.1
Nearest Neighbor
25.6
23.3
Convex Hull, Cheapest Insertion
248.7
21.9
Random Addition
36.1
40.5
Cheapest Insertion
266.1
22.1
Random Addition+
60.0
15.5
Farthest Insertion
316.4
13.5
Greedy [Bentley]
60.4
14.2
Smart-Shortcut Christofides
422.9
9.8
Nearest Addition
78.1
32.6
Greedy [Johnson-McGeoch]
89.0
14.3
Nearest Addition+
95.5
27.1
2-Opt [Johnson-McGeoch]
196.0
4.7
Savings
99.6
12.1
3-Opt [Johnson-McGeoch]
224.3
2.9
Smart-Shortcut Double MST
101.4
40.0
Lin-Kernighan [Johnson-McGeoch]
574.5
2.0
Approximate Christofides
130.3
11.2
Lin-Kernighan [NYYY-6]
287.9
1.7
Nearest Insertion
130.6
27.0
Lin-Kernighan [NYYY-12]
507.6
1.5