Presented By:
Dahan Yakir
Sepetnitsky Vitali
Introduction
The Traveling Salesman (TSP) problem
 Given n points
 For each pair {i, j} of distinct points, a cost ci,j
 A shortest path is desired
 Shortest path = a closed path that
 Visits each point exactly once, denote a path like that, as
salesman tour
 The sum of all distances along the path is minimal (the
path have a minimal cost)
 This problem is NP-hard
Introduction
Euclidian TSP + PTAS
 In Euclidian TSP the points are in
(i.e. in 2D the
input is a set of n points in the plane) and the costs
d
2

xi  x j  

are Euclidian distances: ci , j   
i 1

d
1
2
 The distances satisfy triangle inequality
 Even 2D Euclidian TSP is NP-hard
 PTAS or Polynomial Time-approximation Scheme is a
polynomial-time algorithm which finds an
(1 + ε)-approximation to the optimal solution for ε > 0
 The run-time is polynomial by n
Introduction
The algorithm we will present
 We will show that Euclidian TSP has a PTAS
 The algorithm was given by Sanjeev Arora at 1996 and





improved several times through 1996-1998
A randomized algorithm will be presented
The PTAS, we will describe, is for points on the plane
(d = 2) but there is an extension to an arbitrary d
The algorithm computes an (1 + ε)-approximation to
the optimal solution in no(1∕ε) time
The algorithm is based on dynamic programming
The algorithm can be de-randomized, but then the
run-time should be increased by a factor of n4
Step 1
Bounding Box of a TSP instance
 The smallest axis parallel
square containing the
instance
 Let L be the side length of
the square and assume
w.l.o.g. that L is power of 2,
i.e. L = 2k
 We will call L the size of the
box
 We can scale all distances
such that L = 4n2 and thus:
k = 2+2∙log2n = O(log n)
Step 2
Perturbation of a TSP instance
 First, we perform a simple
“perturbation” of the
given TSP instance, I, and
create a new instance I’
 Define a unit grid on the
bounding box and
relocate each point to the
nearest grid point
 The maximum distance
from a point to a nearest
grid point is at most √2/2
L
Step 2
Error analysis of the Perturbation
 Approximate the error
we get by the described
perturbation
 At least two points are on
opposite edges of the
square, therefore each
solution cost is ≥ 2L
 For a given solution cost
SOL, the perturbation of
the instance increases the
solution length by √2 for
each point
L
Step 2
Error analysis of the Perturbation (Cont.)
 The perturbation of the
instance increases the
solution length by n√2 for all
the points
 Therefore if we denote the
cost of the solution after a
perturbation was made on it,
as SOL’ , we have:
(SOL’ – SOL)∕
SOL
= 1∕4√2n
≤ n√2∕2L = n√2∕8n2 =
L
Step 2
Error analysis of the Perturbation (Cont.)
 Assume we are given an
(1 + ε)-approximation of OPT
to the perturbed instance I’
of the TSP problem
 The corresponding solution
to the instance I (the points
are shifted to the original
place) is an (1 + ε + 1∕4√2n)approximation of OPT
 For sufficiently large n, we
can adjust ε to reach the
desired error
L
Step 3
Dissection of the Bounding Box
 A recursive partitioning
into four smaller squares
 An L × L square is divided
into four L∕2 × L∕2 squares
and so on
 We view the partitioning as
a 4-ary tree T , whose root
is the bounding box, the
four L∕2 × L∕2 squares are its
children and so on
L
L∕
2
Step 3
Dissection of the Bounding Box (Cont.)
 A useful square is a square
represented by a node in T
 The dissection is continued
until we obtain squares with
size ≤ 1 (therefore each one
has at most one point in its
interior)
 The tree depth is O(log L)
Step 3
Dissection of the Bounding Box (Cont.)
 Assign levels to the nodes of
T
 The root is at level 0, its
children at level 1 and so on
 The squares represented by
nodes of T get levels
accordingly
 Squares at level i have
dimensions L∕2i × L∕2i
Level 1
square
Level 0
square
Step 3
Dissection of the Bounding Box (Cont.)
 The basic dissection of the bounding box can be seen
as a 4-ary tree with depth O(k) = O(log L)
 At each level i there are 4i squares (i = 0 to k)
 The number of the nodes in the tree is therefore:
1 + 4 + 42 + … + 4k = (4k + 1 - 1) / (4 - 1) =
= O(4k) = O(42 + 2 log n) = O(n4)
The bounding box
Level 1 squares
…
…
…
Level 2 squares
Step 3
Dissection of the Bounding Box (Cont.)
 Now we assign levels for the
lines from which the
dissection is constructed
 Level i lines create the level
i squares
 Line at level i forms an edge
for useful squares at levels
i, i + 1, …
Level 1
square
Level 2
square
Level 3
square
Step 4
Portals
 A special set of points
defined for each line
 The solution is allowed to
cross a grid line only at a
portal
 The portals are
equidistant points - L∕(2im)
apart at level i line
 Each useful square has
at most 4m portals
on its sides and corners
portal
L∕ i
(2 m)
Step 4
Portals (Cont.)
 m is a power of 2 and
belongs to [k∕ε , 2 ∙ k∕ε]
 Therefore m = O(log n∕ε)
 A portal in a level i square
is a portal of a contained
level i + 1 square as
well
portal
L∕ i
(2 m)
Definitions
Well behaved tours
 A tour τ is well behaved with respect to the basic
dissection (WB) if it is a tour through all the n points
and any subset of the portals
 The tour is allowed to visit portals multiple times
 The tour must be non-self-intersecting
 A tour τ is well behaved with respect to the basic
dissection with limited crossings (WBLC) if it is a well
behaved tour (as defined above) and, in addition, it
visits each portal at most twice
Lemma
Well behaved tours (Cont.)
Any well behaved tour τ can be transformed to a
WBLC tour, whose length is at most the length of τ
Points of the proof:
 Given a well behaved tour τ, we will transform it to a
WBLC tour τ’
 For each portal in τ, located on line l, which is used
more than twice, we will perform a “short-cutting” on
the two sides of τ
 Then, self intersections (if created) will also be
removed from τ’ by short-cutting
Lemma – Proof
Stage 1: Making each portal’s crossings # to be ≤ 2
 Let τ be a well defined tour which crosses a portal p,
located on line l, α times (α > 2)
 There are two cases:
 α is odd
 α is even
 A “short-cutting” will be performed for each case and
make τ cross p at most twice, without increasing its
length
 The correctness will be guaranteed by triangle
inequality
Lemma – Proof (Cont.)
Stage 1, Case 1: Crossings # is odd
In this case we transform the tour to cross l only once:
a
l
c'
b
a
a'
c'
b
a'
p
c
l
p
b'
c
b'
Lemma – Proof (Cont.)
Stage 1, Case 2: Crossings # is even
In this case we transform the tour to cross l only twice:
c'
a
d
d
p
b
a'
l
c
p
b
a'
l
b'
d'
c'
a
c
b'
d'
Lemma – Proof (Cont.)
Stage 2: Eliminating intersections
 If the previous stage introduced self-intersections,
they can be removed by the following way:
a
c'
a
a’
c
c'
a’
c
Conclusion
The remaining parts of the algorithm
 After lemma 1 was proved we know that each WB tour
τ can be transformed to a WBLC tour τ’, such that
|τ’| ≤ |τ|
 We can show that there is a WB tour τ that is short
enough, i.e. |τ| ≤ (1 + ε) ∙ OPT
 From lemma 1 we know that there exists a WBLC tour
τ’ that is short enough too  |τ’| ≤ (1 + ε) ∙ OPT
 We will obtain all the WBLC tours and take the
shortest one τ’’ (the one with minimal sum of
distances), therefore |τ’’| ≤ |τ’|  |τ’’| ≤ (1 + ε) ∙ OPT
Definition
valid visit
For each useful square S, we
define a valid visit as an
assignment of crosses for
each portal on S, such that
c 1
the sum of all portal crosses
is even. In addition S’s
portals are divided to pairs, c‘, d 2
that are connected by some
valid WBLC tour
0
d‘, e 2
1
b'
0
1 b
1
a'
2 a, e’
0
0
0
Stage 5
Computing the optimal WBLC
The following steps:
 We will build a table, that for each square in the
dissection, contains the cost of each valid visit. The table
will be built up the 4-ary dissection tree
 The calculation uses dynamic programming
Stage 5
Computing the optimal WBLC (Cont.)
 We look only on WBLC tours which means that each




portal can be used (= crossed) 0, 1 or 2 times
Denote some WBLC tour by τ
The total number of times, τ can enter or exit a useful
square S is at most 8m
The part of τ inside S, can be viewed as at most 4m
paths, covering all the points inside S
First, we calculate the total number of possibilities to
define the number of crossings for each portal on S
sides
Stage 5
Computing the optimal WBLC (Cont.)
 The total number of possible assignments is
34m = 24m∙log23
 m in the range [O(log n)∕ε , 2 ∙ O(log n) ∕ε]
 Therefore 24 ∙ O(log n)∕ε ∙ (log 3) ≤ 34m ≤ 28 ∙ O(log n)∕ε ∙ (log 3)
4 log 3) ∕ε
≤ 34m ≤ n(8 log 3) ∕ε  34m = nO(1∕ε)
 Among all the possibilities of portals crossings, we
retain only those that involve an even sum of crossings
(allowing to enter and exit S)
 Once we defined the number of crossings for each
portal on S sides, not every possible pairing is allowed
because of the restriction of the self-intersection
 Thus we have n(
Stage 5
Computing the optimal WBLC (Cont.)
 Count the number of valid
pairings, for a given
assignment of portal
crossings
 Let’s assume that the sum of
portal usages is 2r (should
be even)
 We will prove that the
number of valid pairings is
the rth Catalan number
 Let’s look on a valid pairing:
A useful square S
Stage 5
Computing the optimal WBLC (Cont.)
Now, if we view the sides of S as a straight line, we can see
that a valid pairing can be transformed to a sequence of
balanced parentheses (a bijection can be created):
A valid pairing
An invalid pairing
Stage 5
Computing the optimal WBLC (Cont.)
 The number of balanced arrangements of 2r parentheses is
the
rth
Catalan number,
Cr=(1/r+1)∙
2r
r
( )
 It is known for Catalan numbers that Cr < 4r = 22r
 Since each visit on the portals of S crosses the portals at
most 8m times, we get that the number of valid pairings for
a given assignment of portal usages is bounded by:
28m = 28 ∙ O(k/ε) = 28 ∙ (O(log n)/ε) = nO(1∕ε)
 Thus we get:
1
 nO( ∕ε) possibilities for portal usages
1
 nO( ∕ε) valid pairings for each portal usage
 Total number of valid visits in S is nO(1∕ε) ∙ nO(1∕ε) = nO(1∕ε)
Dynamic programming
Computing the optimal WBLC: definition of the table
 Let’s build the following table in order to compute
optimal WBLC cost (for each useful square):
Minimum
Costs
Level 0
Level 1
…
Square 1
Square 1 Square 2 Square 3 Square 4
…
valid visit 1
valid visit 2
valid visit 3
…
 # of columns = # of squares in the 4-ary tree = O(n4)
 # of rows = # of valid visits for a square = nO(1∕ε)
 Total # of entries in the table = n4 ∙ nO(1∕ε) = nO(4 + 1∕ε)
Definition
Computing the optimal WBLC: consistency
We say that a valid visit of square S1 at level i is consistent
with a valid visit of its parent square S, if all common portals
get the same portal usages and internal pairings
Dynamic programming (Cont.)
Computing the optimal WBLC: the algorithm
The table is filled up the dissection tree T:
]
1. The level k is computed by calculating the minimal
cost for each valid visit in the squares at this level
2. For each level i, i = k – 1 downto 0
2.1. For each square S at level i
2.1.1. For each valid visit V in S
2.1.1.1. For each possibility of portal usages on the
two lines which divide S (to 4 squares)
2.1.1.1.1. For each valid pairing of all the portals
consistent with the visit V portals usage
Dynamic programming (Cont.)
Computing the optimal WBLC: the algorithm (Cont.)
 If the pairing is consistent, it gives rise to valid
visits in the four squares internal to S (whose
minimal WBLC costs were calculated)
2.1.1.1.1.1. Sum the optimal lengths of the
appropriate visits in the children squares
2.1.1.1.2 Compute the minimum cost among the
costs for this internal portal usage
2.1.1.2. Compute the minimum cost among the
costs for all internal portal usages and store
the returned cost in an appropriate entry
3. Finally, take the minimum cost of the level 0 column
Dynamic programming (Cont.)
Computing the optimal WBLC: runtime
1
 We have seen before that the table has nO( ∕ε) entries
 In order to calculate each entry, which represents a valid
visit V on a square S we need to consider at most all the
possibilities for internal portal usages
 The number of internal portals of S is at most 4m + 1
 Therefore the number of possibilities for internal portal
usages is at most 34m + 1 = nO(1∕ε) (since we have seen that
34m = nO(1∕ε))
 The total number of the internal portals along with the
external portals is 8m – 3 at most and therefore the total
number of pairings is also bounded by nO(1∕ε)
Dynamic programming (Cont.)
Computing the optimal WBLC: runtime (Cont.)
 Therefore, in total we should consider nO(1∕ε) ∙ nO(1∕ε) = nO(1∕ε)
valid pairings for each square. That is because we have
nO(1∕ε) valid pairings for the internal portals, for a single
usage assignment of them, and nO(1∕ε) usage assignments.
 Finally, the total cost for a single entry calculation is nO(1∕ε)
 The total runtime of the algorithm is:
#entries ∙ nO(1∕ε) = nO(1∕ε)
Correctness: We compute all the available WBLCs costs,
since each WBLC tour of a square S at level i can be divided
to 4 squares at level i + 1, whose minimal WBLCs costs were
calculated for each valid visit
WB tour vs. Optimal tour
How an optimal WB tour differs from OPT
 First, consider an optimal WB tour and let’s calculate
how much it can differ from OPT’s cost
 Convert OPT to WB and see how much its length
increases by this conversion
L/2im
An edge of
the changed
tour
Portal
Portal
An edge of
the optimal
tour
WB tour vs. Optimal tour
How an optimal WB tour differs from OPT (Cont.)
 We can easily see that for each crossing of a dissection
line, |OPT| increases by at most L/2im
 We can notice that the increase depends on the line’s
level. i.e. the lower the level, the greater the increase
L/2im
An edge of
the changed
tour
Portal
Portal
An edge of
the optimal
tour
Is it sufficient?
How an optimal WB tour differs from OPT
 Is it enough? Is an optimal WB tour always an
(1 + ε)-approximation for the optimal TSP solution
 The answer is NO!!!
 We will show an instance of the Euclidian TSP
problem, for which the process of transforming an
optimal tour to be WB, increases its length by a fixed
constant
 Therefore the algorithm will fail if ε is chosen to be
below this bound
Counterexample
How an optimal WB tour differs from OPT (Cont.)
 Let’s denote L/2im as x
 An Euclidian TCP
instance is shown on the
right, along with the
optimal TCP solution
 We can easily obtain:
4x
8x
L
8x
OPT 
4
 3x    4 x 
2
2
 2y  L
8x
4x
2x
x
y
Counterexample (Cont.)
How an optimal WB tour differs from OPT (Cont.)
 Now let’s modify the tour
to be WB and call the
new tour SOL
 Again, we can easily
obtain:
SOL 
  3x 2   2 x 2 

4


2
2

x

x


 2y  L
4x
8x
L
8x
8x
4x
2x
x
y
Counterexample (Cont.)
How an optimal WB tour differs from OPT (Cont.)
 Since y ≤ L∕2 we obtain
|SOL| ∕
|OPT| ≥
1.0015
 Hence, the described
algorithm fails for a
selection of an error
parameter ε < 0.0015
 To bypass this problem
we will randomize the
algorithm and
generalize the concept
of the dissection
4x
8x
L
8x
8x
4x
2x
x
y
Definition
(a,b)-Shifted Dissection of the Bounding Box
 Let’s define L2 different dissections of the bounding




box which are shifts of the basic dissection
Choose randomly two integers a and b such that
0 ≤ a, b < L
Shift each vertical line from its original location x to
(x + a) mod L
Shift each horizontal line from its original location y
to (y + b) mod L
We call the result (a,b)-shifted dissection
Randomization
(a,b)-Shifted Dissection of the Bounding Box (Cont).
Here we a shifting of the level-1 lines is illustrated
b
a
Randomization (Cont.)
(a,b)-Shifted Dissection analysis
 By performing the shifting, any specific line has a
random level. For instance, the level of the middle
vertical line in the counter-example is not necessarily
1, but depends on a: it is the level of the
[(L∕2 + a) mod L]-line in the original dissection
 There are 21 lines at the first level, 22 lines at the
second, …, 2k lines at level k (2i lines at random level i)
 Thus, we have 2k+1 - 2 total # of lines in a dissection
 The probability to randomly choose line with level i is:
2i
2i
2i
Pr[level  i]  k 1


2  2 2L  2 L
Randomization (Cont.)
Distance between portals on a random line
 We showed that when a line at level i is crossed by the
path τ, the maximal increase in the process of making
τ a WB tour is x(i) = L / 2im
 Now, we can define X as a random variable which
describes the possibilities of length increases by a
single grid-line cross
 Thus, the expected value of the increase for a
randomly chosen line is:
k
i
 m  2
2 L
k 
EX   p  i   x  i     i 

m
i 1
i 1 L 2 m
k
k
k


Randomization (Cont.)
Number of crossings for a random dissection
 Now let’s bound the number of crossings of dissection
lines
 Let τ be an optimal tour and let N(τ) be the total
number of dissection lines crossings of τ (intersection
is counted twice). Denote: |τ| = OPT.
 We claim that: N(τ) ≤ √2 ∙ OPT (will be proved later)
 Therefore, if we denote by Y the total increase in the
path cost, we have EY = N(τ) ∙ EX ≤ √2 ∙ OPT ∙ ε
EX
 Markov’s inequality says that: Pr  X  a  
a
Randomization (Cont.)
Finalizing the algorithm
 Thus, we have:
EY
1
Pr Y  2 2    OPT  

2 2    OPT 2
 This leads us to the following corollary:
Pick a and b uniformly at random from [0, L). Then,
the probability that there is a WB tour of length
(1+2√2ε)-approximation is greater or equal to 0.5
Randomization (Cont.)
Finalizing the algorithm (Cont.)
Three remarks:
 The 2√2 prefix can be removed by updating ε to
ε’ > 0 which can be calculated by solving the
equation 2√2ε’ = ε
 We can reduce the probability that the total increase
in length is to high, by repeating the procedure for c
different (a,b)-shifted dissections and choosing the
shortest tour
 We can derandomize the algorithm by trying all the
possible L4 = O(n4) dissection and, again, choosing
the shortest tour
Randomization (Cont.)
Proving N(τ) ≤ √2 ∙ OPT
 Consider a segment C (of
τ) from a point (x1,y1) to a
point (x2,y2), remember
that each one is located
on a grid point
 Let D be the Euclidian
distance between the
points (clearly D ≥ 1)
 Denote:
 Dx = |x1 - x2|
 Dy = |y1 – y2|
(x2,y2)
D
C
(x1,y1)
Dx=3
Dy=4
Randomization (Cont.)
Proving N(τ) ≤ √2 ∙ OPT (Cont.)
 C crosses vertical grid





lines Dx - 1 times
C crosses horizontal grid
D
lines Dy - 1 times
D2 = Dx2+ Dy2
Dx2+ Dy2 ≥ (Dx+ Dy)2 / 2
Thus we have:
2D2 ≥ (Dx+ Dy)2 
(x1,y1)
√2D ≥ Dx+ Dy ≥ N(C)
Adding up we obtain the
desired result: N(τ) ≤ √2 OPT
(x2,y2)
C
Dx=3
Dy=4
Summary
PTAS for Euclidian TSP (in R2)
Given a TSP instance, perform:
1. Find the bounding box
2. Perturbation of the instance
3. Create an (a,b)-shifted dissection of the bounding
box for some randomly chosen integers 0 ≤ a, b < L
4. Calculate an optimal WBLC tour τ, using dynamic
programming
5. With probability of at least 1∕2 , τ is an (1 + ε)approximation to the optimal solution
6. Repeat 3-5 until the required probability is obtained
References
 V. V. Vazirani, Approximation Algorithms, Springer,
2001.
 Sanjeev Arora, Polynomial-time Approximation
Schemes for Euclidean TSP and other Geometric
Problems, Journal of the ACM 45(5) 753–782, 1998.