complexity ©d.moshkovitz 1 paths on the reasonability of finding paths in graphs

Complexity©D.Moshkovitz

1

PathsOn the Reasonability of Finding Paths in Graphs


2

Introduction

• Objectives:– To introduce more graph theory

problems.

• Overview:– Hamiltonian paths– Eulerian paths


3

Hamiltonian Path

Instance: a directed graph G=(V,E) and two vertices s,tV.

Problem: To decide if there exists a path from s to t, which goes through each node once.


4

Can You Find One Here?

s t


5

HAMPATH is in NP

• Non-deterministically choose v1,…,vnV (n=|V|).

• For any 1i,jn, verify vivj.

• Verify s=v1 and t=vn.

• For any 1in, verify (vi,vi+1)E.


6

HAMPATH is NP-Complete

Proof: We’ll show 3SATpHAMPATH.

SIP 262-267

..)..(.......)..(..

s

t

. . .

p


7

Representing Variables

For any variable xi,

Diamond

. . .


8

Representing Clauses

• For any clause ci,


9

High-Level Structure

. . .

x1

x2

.

.

.

xl

c1

c2

c3

.

.

.

ck

. . .

. . .

s

. . .

. . .

t


10

The Internal Structure of the Diamonds

k pairs of internal nodesk pairs of internal nodesk pairs of internal nodesk pairs of internal nodes

Diamond

. . .


11

Connecting Clauses to Variables

If the clause cj contains the literal xi,

j’th pair

xi

cj

. . .


12

Connecting Clauses to Variables

If the clause cj contains the literal xi,

j’th pair

xi

cj

. . .


13

Construction Completed

• This concludes the construction.• Its size is polynomial in the size of

the formula (Check!).

• We proceed to prove correctness.


14

Completeness

• Assume there exists a satisfying assignment for the 3CNF formula.

• Let us demonstrate a Hamiltonian tour on the graph.


15

Assignment Path

If the variable xi is assigned TRUE,

. . .


16

Assignment Path

If the variable xi is assigned FALSE,

. . .


17

Covering the ClausesIf the clause cj is satisfied due to the literal

xi,

xi

cj

. . .

j’th pair


18

. . .

Covering the ClausesIf the clause cj is satisfied due to the literal

xi,

xi

cj

j’th pair


19

Soundness

• Assume there exists a Hamiltonian tour on the graph,

• Let us construct a satisfying assignment for the 3CNF formula.


20

Observation

• Every Hamiltonian path must contain either the entire right path or the entire left path of each diamond.

. . .


21

Path Assignment

• For each variable, decide on its truth value,

• According to the direction of the path in its diamond.

. . .


22

x1

x2

.

.

.

xl

. . .

. . .

s

. . .

. . .

t

Observation

cj

A Hamiltonian path cannot jump from inside one

diamond to another

Proof Idea: Assume it can and

observe this node


23

Satisfaction

• Note that because of the observation

• And the fact the path goes through all clause nodes,

• All clauses are satisfied


24

The Punch Line

HAMPATH is NP-Complete.


25

Food for Thought…

Is HAMCYCLE (Whether there exists a simple cycle which goes through all

nodes) NP-Complete?


26

Food for Thought…

What about finding Hamiltonian paths in undirected graphs?


27

Similar Problem

• Children usually like this riddle:

• Can you draw the following shape in one line?


28

Eulerian Path

Instance: an undirected graph G=(V,E) and two vertices stV.

Problem: To decide if there exists a path from s to t, which goes through each edge exactly once.


29

Can You Find One Here?

s t


30

Eulerian Paths Vs. Hamiltonian Paths

• What do you make of this problem?

• Is it reducible to HAMPATH?• Or vice-versa?

?


31


32

The Euler Theorem

Theorem: A connected graph has an Eulerian path from s to t iff

1. s and t’s degrees are odd.2. the degree of all other

vertices is even.


33

First Direction

• Assume there is an Eulerian path from s to t in the graph.

• Let us prove all degrees, but s and t’s, are odd.


34

Analyzing Degrees

• Observe the path. It passes through all edges.• For internal vertices, it gets out every time it gets

in, thus adding 2 to the degree each time.• Except it first gets out of s and finally gets in to t.


35

Second Direction

• Assume all degrees, but s and t’s, are even.

• Let us describe an algorithm which finds an Eulerian path from s to t.


36

Walks• Start at s,• While there are new edges

– Arbitrarily choose such,– And walk on it.


37

Observation

Claim: Such a walk must get stuck at t.

Proof: After leaving s, t is the only vertex whose degree is odd. Thus we can get out of every other vertex we visit.


38

Were All Edges Visited?

Not necessarily! Check out this example:

s t


39

Observation

Claim: If there are edges not visited, than there exists such edge which “hits” a visited vertex.

Proof: By graph connectivity.

st


40

Observation

Claim: A walk that starts at the visited vertex must get stuck at that vertex itself.

Proof: After leaving the visited vertex it remains the only vertex the number of new edges hitting it is odd.


41

Putting the Pieces Together

• We can add the new walk to the old one and so cover at least one more edge.

• Now repeat this process if there are more uncovered edges.…

s t

… …


42

The Punch Line

There is a polynomial time algorithm for the Eulerian path problem.


43

Food for Thought…

How would you efficiently implement the Euler algorithm?


44

Summary

NP

P

NPC

EULERPATHHAMPATH


45

Summary

• In this lecture we’ve examined few problems related to paths in graphs.

• We’ve shown that finding a path which visits all vertices (Hamiltonian) is NP-Hard.

• While we can efficiently find paths which visit all edges (Eulerian).

complexity ©d.moshkovitz 1 paths on the reasonability of finding paths in graphs

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