Chapter 6. Relaxation

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Chapter 6. Relaxation

• (1) Superstring

Ding-Zhu Du
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Whats relaxation?
x
Min f(x)
x in O
xA
f(x) lt opt min f(x) lt f(xA)
x in O
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Whats relaxation?
x
max f(x)
x in O
xA
f(x) lt opt max f(x) lt f(xA)
x in O
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Traveling Salesman Problem
(Given a distance table for a complete graph)
relaxed
tour ? spanning graph ?
minimum spanning graph
minimum spanning tree ? tour
add 0.5 opt
So, the p.r. 1.5
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Directed TSP with triangular inequality
(Given a distance table for a complete digraph
without loop)
relaxed
Directed tour ? 1-factor (assignment)
? minimum assignment
? directed tour
a directed tour connecting all cycles in 1-factor
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TSP in Digraph
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Superstring
Given n strings s1, , sn, find a shortest string
s containing every si as a substring.
ov(s1,s2) maximum overlapping of s1 and s2
s1 suf(s1,s2) ov(s1,s2)
s1
ov(s1,s2)
suf(s1,s2)
s2
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Superstring
Directed TSP
vertices s1, , sn, sn1 e
distance d(s1,s2) suf(si,sj)
Minimum assignment
r1
r2
rtsn1e
ct
c1
c2
d(ci) total distance of cycle ci.
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cycle (r1,r2, , rt)
d(cycle (r1,r2, , rt)) d(r1,r2)
d(rt-2,rt-1) d(rt-1,rt) d(rt,r1)
suf(r1,r2) suf(rt-2,rt-1) rt-1
superstring ltr1,r2,,rt-1gt
S ri - ov(r1,r2) - - ov(rt-2,rt-1)
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ltr1,r2, , rt-1gt
r1
ov(r1,r2)
r2
rt-1
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assume ltr1,r2, , r(t-1)gt is opt superstring
ltr1,r2, , rt-1gt is obtained by Greedy
Algorithm on overlapping

ov(r1,r2) ov(r(t-2),r(t-1))
lt 2(ov(r1,r2) ov(rt-2,rt-1))
ltr1,r2, , rt-1gt
0.5(ov(r1,r2) ov(r(t-2),r(t-1)))
lt opt
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Greedy Algorithm
S ? s1, , sn,
while S gt 1 do
choose s and s from S to maximize ov(s,s) and
set S ? (S s,s) U suf(s,s)s,
output unique string in S.
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ltr1,r2, , rt-1gt
S ri - ov(r1,r2) - - ov(rt-2,rt-1)
lt S ri - 0.5(ov(r1,r2)
ov(r(t-2),r(t-1)))
lt ltr1,r2, , r(t-1)gt
0.5(ov(r1,r2) ov(r(t-2),r(t-1)))
lt opt
(opt superstring for s1, , sn)
0.5(ov(r1,r2) ov(r(t-2),r(t-1)))
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Lemma. ov(ri,rj) lt d(ci) d(cj)
If it is true, then
ov(r1,r2) ov(r(t-2),r(t-1)) lt 2 S
d(ci) lt 2 opt
Therefore, we obtain a 3-approximation
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3-approximation
Step 1. Compute a minimum assignment c1, , ct.
Step 2. Choose a vertex ri from ci. But, ct
sn1e
Step 3. Find an ordering (r1, , rt-1) by greedy
algorithm on overlapping.
Step 4. Merging c1, , ct into a Hamiltonian
cycle based on the ordering (r1,
, rt-1, rt).
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Lemma. ov(ri,rj) lt d(ci) d(cj)
ov(ri,rj) gt d(ci) d(cj).
Proof
Assume
We want to show ?(ci) ?(cj)
where ?(c) is a root of c.
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Root
For a nonempty string x, ?(x) is the shortest
string y such that
m
x y
for some integer m gt 0.
For a set c of strings, ?(c) is a string y such
that every string in c is a substring of y
for some m.
m

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Property of nonempty string root
?(u) ?(v)
uv vu
Proof. (by induction on uv)
Case 1. u v. We have u v.
Case 2. u gt v.
Then u vw.
So, vwv vvw. Hence, wv vw.
Since v ? e, wv lt uv.
By induction hypothesis, ?(w)?(v). Hence, ?(u)
?(vw) ?(v).
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1st Property of ?(c) for string set c
There exists a cycle c connecting all strings in
c Such that d(c) ?(c).
?(c)
?(c)
s1
d(s1,s2)
s2
d(s2,s3)
s3
s1
d(s3,s1)
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2nd Property of ?(c) for string set c
For any permutation (s1, , sk) of strings in c,
suf(s1, s2) suf(s2, s3) suf(sk-1, sk)suf(sk,
s1)
is a ?(c).
s1
s2
s3
s1
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3rd Property of ?(c) for string set c
Any substring of ?(c)?(c) with length ?(c) is a
root of c.
?(c)
?(c)
?(c)
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Lemma. ov(ri,rj) lt d(ci) d(cj)
ov(ri,rj) gt d(ci) d(cj).
Proof
Assume
We want to show ?(ci) ?(cj)
If this is true, then cycles ci and cj can be
merged Into one c with ?(c) ?(ci) ?(cj),
Contradicting the minimum of assignment.
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Lemma. ov(ri,rj) lt d(ci) d(cj)
ov(ri,rj) gt d(ci) d(cj).
Proof
Assume
We want to show ?(ci) ?(cj)
ri
u
v
d(cj)
d(ci)
v
u
rj
d(cj)
d(ci)
uv vu !
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Thanks, End