Lecture 12 - 3-D Matching Problem

Report
MCS 312: NP Completeness
and Approximation algorithms
Instructor
Neelima Gupta
[email protected]
Table of Contents
• 3D Matching
• Exact Cover
3-D Matching Problem
Problem Definition
Given
• Disjoint sets X, Y, Z, each of size n
• A set T XxYxZ of ordered triples
To Find If there exists a set of n triples in T so that
each element of XYZ is contained in exactly
one of these triples.
3-D Matching Problem
Claim 3-D Matching is NP Complete
Proof
Part I 3-D Matching belongs to NP
Given
• An instance of 3-D Matching
• A certificate containing a collection of
triples T’ T
It can clearly be verified in polynomial time that each
element in X Y  Z belongs to exactly one of the triples in
T’.
3-D Matching Problem
Part II
3-D Matching is NP-hard
We prove the above by showing that
3-SAT ≤p 3-D Matching
We adopt the following approach
 We first design gadgets that encode the independent
c h o i c e s i n v o l v e d i n t h e t r u t h
assignment to each variable
 We then add gadgets to encode constraints
imposed by clauses.
Given: A 3-SAT instance with n variables and k clauses
3-D Matching Problem
Basic gadget associated with variable xi
 Core of the gadget as Ai = {ai1,ai2,….,ai,2k}
 Tips of the gadget Bi = {bi1,bi2,….,bi,2k}
 A triple tij = (aij,ai,j+1,bij) for all j = 1, 2k, where addition is
modulo 2k.
 A triple tij is even if j is even, and odd otherwise.
Analogous for tip bij.
• Red nodes are the core elements
• Blue nodes are tips
Enhancing the variable gadget to model the clauses
and the constraints on possible assignments imposed
by them
For clause C j , clause gadget
Two core elements
 For each of the three terms
If term = xi, define a
If term = x i ’, define a
is defined as follows:
Pj = {pj, pj’}.
in C j do the following
triple (pj, pj’,bi,2j).
triple (p j , p j ’,b i , 2 j - 1 ).
3-D Matching Problem
We understand this enhancement through
an example. Consider the following clause
C1= x1’ v x2 v x3’
Interpretation: The matching should select the
even tips of the first gadget, or odd tips of the
second gadget, or even tips the third gadget.
We first create gadgets for the three variables
3-D Matching Problem
x1
x2
x3
3-D Matching Problem
We now model the assignment constraints
with a clause gadget defined as follows:
A set of 2 core elements P1 = {p1, p1’}
contained in the following 3 triples:
• (p1, p1’,b1,1)
• (p1, p1’,b2,2)
• (p1, p1’,b3,1)
3-D Matching Problem
C1
3-D Matching Problem
How do we encode the idea that xi can either be set to 0
or 1 ?
Either use all the even triples in gadget i, or all odd
ones !
• Selecting all even triples corresponds to setting xi = 0. In
this case the odd tips would remain free for use.
• Selecting all odd triples corresponds to xi = 1. In this
case, the even tips would remain free for use.
Enhancing the variable gadget to model the clauses
and the constraints on possible assignments imposed
by them
For clause C j , clause gadget is defined as follows:
Two core elements Pj = {pj, pj’}.
 For each of the three terms in C j do the following
If term = xi, define a triple (pj, pj’,bi,2j).
If term = x i ’, define a triple (p j , p j ’,b i , 2 j - 1 ).
Reserving the (2j)th and (2j-1)th tips of each variable
gadget for clause gadget j ensures that the clause
gadgets never compete for free tips.
Since only the 3 triples defined corresponding to the 3
terms of Cj cover Pj, one of these must be used. This
enforces the clause constraint.
Cleanup Gadgets
We introduce (n-1)k cleanup gadgets !
The rth cleanup gadget consists of 2 core
elements Qr = {qr, qr’}
Create a triple (qr, qr’,b) for every tip b in
variable gadgets.
3-D Matching Problem
Define X,Y and Z as follows:
X consists of
1. all aij with j even ….total number nk
2. The set of all pj ….. total number k
3. The set of all qi ….. total number (n – 1) k
Y consists of
1. all aij with j odd ….total number nk
2. The set of all pj’ ….. total number k
3. The set of all qi’….. total number (n – 1) k
Z consists of
1. Set of all tips bij …. ….. total number 2 n k
Can you see that
• X, Y and Z are of same size….what’s the size?
• Each triple consists of one element from each of X,Y,
and Z.
3-D Matching Problem
Y consists of
1. all aij with j odd
2. The set of all pj’
3. The set of all qi’
Z consists of
1. Set of all tips bij
Can you see that
•
X, Y and Z are of same size
•
Each triple consists of one element from each of
X,Y, and Z.
3-D Matching Problem
We understand the above reduction with the help
of an example !
Consider an instance of 3-SAT having
4 variables : x1,x2 x3,x4
3 clauses C1 : (x1 v x2 v x3’)
C2 : (x2’ v x3 v x4)
C3 : (x1’ v x2 v x4’)
Variable gadget for x1 has the following components
 Core A1 = {a11,a12,….,a1,6}
 Tips B1 = {b11,b12,….,b1,6}
 A triple t1j = (a1j,a1,j+1,b1j) for all j, where addition
is modulo 6.
Tip 1
Tip 6
Tip 2
Tip 5
Tip 3
Tip 4
3-D Matching Problem
Gadgets for the remaining variables can
similarly be drawn.
We now bring the clause gadgets into
picture
C1 : (x1 v x2 v x3’)
p1
p1 ’
x1
x2
x3
C2 : (x2’ v x3 v x4)
x2
x3
p2
p2 ’
x4
C3 : (x1’ v x2 v x4’)
x1
p3
p3 ’
x2
x4
3-D Matching Problem
Part I Consider the following satisfying
assignment for the given 3-SAT instance
x1 = 1
x2 = 0
x3 = 1
x4 = 0
We now try to find a matching.
We select the odd triples of gadget for x1 since
x1 = 1. This leaves the even tips free for use
Tip 1
Tip 6
Tip 2
Tip 5
Tip 3
Tip 4
We select the even triples of gadget for x2 since
x 2 = 0. This leaves the odd tips free for use
Tip 1
Tip 6
Tip 2
Tip 5
Tip 3
Tip 4
We select the odd triples of gadget for x3 since
x3 = 1. This leaves the even tips free for use
Tip 1
Tip 6
Tip 2
Tip 5
Tip 3
Tip 4
We select the even triples of gadget for x4 since
x 4 = 0. This leaves the odd tips free for use
Tip 1
Tip 6
Tip 2
Tip 5
Tip 3
Tip 4
C1 : (x1 v x2 v x3’)
p1
p1 ’
x1
x2
x3
C2 : (x2’ v x3 v x4)
x2
x3
p2
p2 ’
x4
C3 : (x1’ v x2 v x4’)
x1
p3
p3 ’
x2
x4
3-D Matching Problem
Claim: 3-SAT is satisfiable if and only if
there exists a perfect 3-D Matching
Part I
Given
A satisfying assignment for the
3-SAT instance
To Prove There exists a perfect 3-D
Matching
Recap
Define X,Y and Z as follows:
X consists of
1. all aij with j even ….total number nk
2. The set of all pj ….. total number k
3. The set of all qi ….. total number (n – 1) k
Y consists of
1. all aij with j odd ….total number nk
2. The set of all pj’ ….. total number k
3. The set of all qi’….. total number (n – 1) k
Z consists of
1. Set of all tips bij …. ….. total number 2 n k
Can you see that
• X, Y and Z are of same size….what’s the size?
• Each triple consists of one element from each of X,Y,
and Z.
3-D Matching Problem
Proof
Make the corresponding choices of
odd/even for each variable gadget based on
the given assignment.
 This leaves at least one free tip for each clause
gadget.
 Using the cleanup gadgets, we cover all the core
elements in the variable, clause and cleanup
gadgets, and all the tips too !
3-D Matching Problem
Part II
Given A perfect 3-D Matching in the
instance constructed.
Can we claim that this matching cannot
possibly cover some even and some odd
triples in any variable gadget, i.e., either all
even or all odd triples are chosen ?
3-D Matching Problem
Yes !
Any two consecutive triples share one of the
core elements ! Therefore, any perfect
matching cannot pick consecutive triples!
• If even triples have been chosen, set xi = 0
• If odd triples have been chosen, set xi = 1
3-D Matching Problem
Have the clauses been satisfied too ?
Yes !
Since the core elements of every clause gadget
have been covered, the following holds true for all
clauses: At least one of the three variable gadgets
corresponding to a term in any clause Cj must
have made the correct even/odd decision.
Big Thanks to Sonika Arora

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