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89
On
the Spacefilling Curve Heuristic
for the Euclidean Traveling
Salesman Problem
Dimitris Bertsimas
Sloan School of Management
Massachusetts Institute of Technology
Michelangelo Grigni
of Mathematics
Massachusetts Institute of Technology
Department
M.T
Sloan School Working Paper No. 2105-88
MASSACHUSETTS INSTITUTE
OF TECHNOLOGY
NOV
1 4 2000
LIBRARIES
On
the Spacefilling Curve Heuristic for the
Euclidean Traveling Salesman Problem
Dimitris Bertsimas*
Sloan School of Management,
Rm. E53-359
Massachusetts Institute of Technology
Cambridge,
MA
02139
Michelangelo Grigm*
Department
of
Mathematics
Massachusetts Institute of Technology
Cambridge,
December
MA
18,
02139
1988
Abstract
Bartholdi and Platzman
[3]
proposed the spacefilling curve heuristic
for the
Euclidean Traveling Salesman Problem and proved that their heuristic returns
a tour within an
worst-case ratio
O(lgn)
is
factor of optimal length.
in fact
0(1).
showing the O(lgn) upper bound
"This research
In this note
we
They conjectured that the
exhibit a counterexample
is tight.
Foundation under grant ECS-
is
partially supported by the National Science
is
based upon work supported under a National Science Foundation Graduate
8717970.
T
This material
Fellowship.
Introduction
1
Bartholdi and Platzman
proposed a heuristic
[3]
man Problem (ETSP) based
continuous map from the unit
<f>
as a total linear ordering
may be
Their curve
on a spacefilling curve.
means
<$ on the
x
if
—
<
y
points of the unit square
<
—
x'
fall in
standard position with vertices
same
First rotate
at (0.0), (1,0),
denoted by an arrow). In the following we
means
<
<
1
x'
+
points from the lower
all
subtriangle
is
left
y')
left
and we apply a recursive
(for this
into
transformation to
our figures this orientation
in
is
refer to this triangle as the 'unit triangle".
<^
In figure 1(b) this
(x',y').
we
Now
the same subtriangle.
fall in
similar to the original unit triangle, so
we may
where d
this (oriented)
recurse. After
t
iterations
are considering has a hypotenuse of length
n the scale of the original unit square), hence the process
halt after 0(\og(l/d)) iterations,
,
subtriangle precede points from the upper right
of taking subtriangles, the subtriangle
(j
1
In figure 1(a),
and enlarge the triangle
and (1,1)
the unit triangle, then (x,y)
y' in
subtriangle. Otherwise, both points
20 -0/2
define
This ordering
.
<4> (x',y').
triangle,
be uniquely defined, we have to orient the triangles;
y
we
the order.
first in
then (x,y)
y',
the
ordering on points in such a triangle.
+
[0, l]
2
points from the lower right triangle precede points from the upper
all
Otherwise, both points
triangle.
x
a uniformly
defined by a recursive procedure; for any two distinct points (x, y) and (x
In the unit square,
If
is
interval to the unit square. For our purposes
the procedure will eventually decide which point comes
this
Euclidean Traveling Sales-
for the
is
is
guaranteed to
the Euclidean distance between the two
input points (x, y) and (x',y').
Now we
[3]:
point.
As remarked
in [4], this heuristic is
pair of k-bit fractions, a 2fc-bit sorting key
(t is
S
given a set
of n points in the unit square,
the points in the order $ defined above, and finish the tour by returning to the
visit
first
describe the heuristic of
really
an inverse
4>~
l
(x,y) where
interval onto the unit square).
and then sorted (by radix-sort
input
<f>
is
very
may be computed
t
O(kn)
all
the keys
bit operations,
in
i.e.
input as a
(x, y)
in O(fc) bit
defined as a continuous
Given n such points,
[1]) in
Given point
fast.
operations
map from the unit
may be computed
time linear
in
the
size.
Let L^(5) be the length of the tour produced by this heuristic, and
the length of the optimal tour. In
[5]
they proved that L '(S)/L {S)
further conjecture that the worst case ratio
by exhibiting a simple
set
Sn
m
4
is
of n points with
in fact
0(1).
We
=
)
L'(S) be
O(lgn). They
refute this conjecture
L '(Sn)/L"(Sn = 0(lgn).
<i
let
A
2
0(lgn)
Example
Consider n points uniformly spaced along the
line
from (1/3,1/3) to (1,1/3). Pre
the set of points
cisely, define
Sn = l(x it yi) = I- +
This definition of
Sn
is
2i_-
2
for
'
3
2n
-J
1
<i <n\
(1)
convenient for our proof, but in fact any reasonably uniform
—
distribution on the line y
1/3 would
points in the order of increasing
i,
suffice.
The optimal
tour on
and has length L*(S„) < 4/3.
It
Sn
now
visits
the
suffices to
show:
Theorem
length
2.1
=
For n
k
traversing
,
Sn
in the
order of curve o produces a tour of
L*{Sn ) > 2k 19.
Proof: The basic idea
visits
2
is
Sn and observe
,
to recursively
decompose the order that the
that at each level there are significantly long 'jumps'.
Let Ok be the length of the path produced by traversing
<f>
(note at
< //(Sn)
spacefilling curve
since
Sn
i*(5n ) counts the extra edge used
in
the order of curve
to close the path into
a cycle). Let bk be the length of the path traversed by the heuristic on the following
similar set of
7?
=
2
k
points:
H^H3
5
+
3-^'3J
(a)
Figure
1:
In
both the square
points of subtriangle
2.
for
^^"l
(2)
(b)
(a)
and the triangle
(b), points of subtriangle
1
precede
By
the recursive definition of
hence
in 5„;
We now
and
let 7\,
Sn
of
bk
=
the points of S'n are visited in the
<p.
lie in
as those
a k /2.
derive a recursion for a k
T2l
same order
T4
J3,
.
denote the subtriangles
T2
subtriangles 7\,
and
,
T
In figure 2, let
T3
in
Hence
.
denote the entire unit triangle,
the order visited by
ak
sum
the
is
The
<f>.
points
of the path lengths
within each subtriangle, together with the lengths of the jumps between consecutive
subtriangles.
The n/4
ak
.
4>
is
points of
Sn
form a half-scale image of S'n
T\
f]
Similarly the n/A points of
reversible
Sn f]T2 form
contributing another term of b k _ 2 /2 to a k
where
j 12 is
and j 2 3
bk _ 2
of
+a
fc
Sn / 2
_i/2
jump
,
+ J12
+J23
we need
to
same
,
4
.
Since
as the path in T\.
Finally the n/2 points of
=
(afc_2
+
a*-i)/2
last point in
know the
T2
first
Sn f)T3 form
+ .7i2 + j 2A
first
point in
to the first point in
and
a
Hence
.
the last point in T] to the
jump from the
lengths
.
the
is
contributing a k _i/2 to a k
jump from
the length of the
estimate these
in
image
the length of the
is
contributing b k _ 2 /2 to
a reversed half-scale image of S'n
points within that triangle), the length of their path
aA
,
4
'reversing the arrow' of a triangle exactly reverses the order of
(i.e.
half-scale reversed
,
last
points of
S'n
T2
.
T3 To
.
visited
each subtriangle.
Lemma
order
<f>
Figure
2.2 For n
is
2:
(xi,yi),
The
=
first
k
,
and the
k
>
1,
the
first
T2 and
Sn (and by similarity
where p = (n/2) + 1.
point in
last point is (x p ,y p )
set 5i 6 in the unit triangle
reversed copy of S'4 in
point
2
T
decomposes into a copy of S'4
a reversed copy of Sg in
,
and the ninth point
last
S'n ) under
T3
.
Curve
(marked by arrowheads).
<j>
visits
in 7\,
a
the leftmost
Proof: Let the points of Sn be indexed as
<
k
For k
2.
The
in T\.
>
we again use
3,
Sn
points in
form a half-scale image of S'n
The
last
point of
is
(Z(n/2)+l,y(n/2)+l) of
Now we
<i<
1
and inductively the
,
find the
Sn
first
point
first
T3 The
the last point in
is
Sn
T
is
inspection for
the
points in
.
=
among them
Sn
point
first
T3
{x 1 ,y l ).
is
(~)T3 are a half-
(x(n/2)+i>y(n/2)+i),
1
< <
*
corresponds to the
5n / 2 would be
point visited in
in
By
n/4}, with identical indices. They
reversed, the last point visited in
Inductively the
-
T
?/,),
,
Since the order
Sn / 2
in
4
t ,
point of
first
image of Sn / 2 label them {(-r^y-)
scale reversed
in
Sn
,
The
figure 2.
are {(z
f] Ji
the definition (1).
in
(x'1 .y[);
n
/2J-
first point
this
is
point
.
jumps distances ju and
and last points of Sn in
j'23-
For &
>
3,
we may apply the lemma
T]5 T2, and T3 The last point in T\ has
— n/8 + 1, and the first point in T2 (since it is reversed) has index 3n/8.
index
= n/2, and the first point in T3 has index
Similarly the last point in Ti has index
3?>/4 + 1. Hence j 12 = (n/4 - l)(2/3n) = 1/6 - 2/3??, j 23 = {n/4 + l)(2/3n) =
to the find the
first
.
i
i
1/6
+
2/3n, so for k
>
3
we have the simple
a*
From the base
Finally
cases a !
we have I* (5
,
ri
=
-(«i-i +0A-2J
Bertsimas
[2]
applies the spacefilling curve heuristic to the Euclidean Probabalistic
and a probability
For a tour r on S,
points of
X
in
2fc/9+(4/27)(-l/2)*+5/27.
PTSP
p, for
is
defined as follows:
each point (i„j/,) G S.
then generated by including each point (x
Pi.
g-
)
Traveling Salesman Problem which
points
+
= 1/3, a 2 = 2/3 follows that a k =
= a k + 1/3 > 2fc/9 as claimed.
Application to the
3
recurrence:
let
T
to choose tour r to get
r
was chosen by the
we may show that
in
?/,)
L (X) be the length
the order followed in
is
t ,
T
E[L (X)]
r.
in
X
we are given
A random
a set
instance
X
5
of n
C S
is
independently with probability
of the tour generated by visiting the
Suppose tour a minimizes E[L"(X)}\ the goal
close to
spacefilling heuristic.
E[L° (X)). Let L*{X) be the length when
Then by a
similar analysis as in section 2
the worst case, E[L+(X)]/E[L"(X)]
=
0(lgn).
4
Concluding Remarks
Similar examples hold for other curves in the unit square,
in
particular the Hilbert
curve (figure 3(a)) and 'zig-zag' curve (figure 3(b)) mentioned in
curve, take n points along the diagonal line x
points uniformly along the horizontal line y
curve
<f>
+
=
y
=
1/2.
[4].
For the Hilbert
2/3. For the zig-zag curve, take
Numerical experiments with the
strongly suggest that n points uniformly spaced on a 'random' line across the
unit square have expected tour length £'[£'*]
(a)
(b)
=
0(lgn).
rithms. Addison- Wesley, 1974.
[2]
D. Bertsimas. Probabilistic Combinatorial Optimization Problems. Ph.D thesis,
Technical report 194 (1988), Operations Research Center, Massachusetts
Insti-
tute of Technology, Cambridge, Mass.
[3]
J.
J.
Bartholdi. L. K. Platzman.
An 0(N log N)
Planar Traveling Salesman
Heuristic Based on Spacefilling Curves. Operations Research Letters, 1 (1982),
121-125.
[4]
J.
J.
Bartholdi, L. K. Platzman. Heuristics Based on Spacefilling Curves for
Combinatorial Problems
in
Euclidean Space. Management Science, 34. no.
3,
(19S8), 291-305.
[5]
L. K.
Platzman,
J. J.
Bartholdi. Spacefilling Curves and Planar Traveling Sales-
man Problem. PDRC technical
19S3 (to appear in JACM).
report 83-02, Georgia Institute of Technology.
Mar
Date Due
?nor
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