Theory of Parallel and Distributed Systems (WS2016/17)

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(WS2016/17)

Kapitel 3 More Algorithms

Walter Unger

Lehrstuhl für Informatik 1

11:52 Uhr, den 30. Januar 2017

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Colorings I Eulerian cycle Matchings Colorings II MIS Coloring III S.

3 Inhaltsverzeichnis Walter Unger 30.1.2017 11:52 WS2016/17 Z

Inhalt I

1 Colorings I Cycle Trees

2 Eulerian cycle Introduction Introduction The algorithm

3 Matchings Introduction Algorithm Running times

4 Colorings II Bipartite graphs

∆ +1 Coloring any Graph

5 MIS Introduction MIS and Coloring Small Domination Set Randomized Distributed MIS

6 Coloring III Outerplanar graphs

7 Simulations Simple simulations Simple simulations Simple simulations Advanced simulations

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Colorings

Coloring Problem

Given undirected graphG= (V,E)andk∈N. Compute [exists?] Functionc :V 7→ {1,· · ·,k}with:

∀{a,b} ∈E :c(a)6=c(b).

Coloring number (chromatic index) ofG:

χ(G) :=min{{k| ∃c:V 7→ {1,· · ·,k} | ∀{a,b} ∈E:c(a)6=c(b)}.

Coloring problem is NP-complete.

LetG=Cn, i.e.G= ({v0,· · ·,vn−1},{{vi,v(i+1)modn} |06i <n}).

Then we haveχ(Cn)63 andχ(C2·n)62 (χ(C2·n+1) =3).

We do not have a nice order on the nodes:

letπ(i)be a permutation LetG=Cn, i.e.

G = ({v0,· · ·,vn−1},{{vπ(i),vπ((i+1)modn)} |06i<n}).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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Colorings

Coloring Problem

∀{a,b} ∈E :c(a)6=c(b).

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3:2 Cycle 1/7 Walter Unger 30.1.2017 11:52 WS2016/17 Z

Parallel Coloring Algorithm of (on) a cycle (Idea)

A processorPi works onv_π(i−1)for some permutationπ.

RegisterRi holdsπ(i−1).

RegisterNi holdsπ(i).

In registerCi will be the color ofvR_i. InitializeCi withi.

Reduce step by step the number of colors.

We will use the colors{0,1,· · ·,n}.

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Parallel Coloring Algorithm of (on) a cycle (Idea)

Programm: color-cycle

for allPi+1 where 06i <ndo in parallel π(i−1)→Ri

π(i)→Ni

c=i c→Ci

repeatdlog^∗(n)e+2times CN_i →c⁰

minimalk with:((ck)%2)6= ((c⁰k)%2).

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

At the start we are usingncolors.

Within each color-reduction will the coloring stay correct.

Within each color reduction will thc coloring number be reduced fromx to log(x) +O(1).

Afterdlog^∗(n)ereductions steps will be the coloring numbers65.

A second reduction of colors will follow now:

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Parallel Coloring Algorithm of (on) a cycle (Idea)

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Last Steps

The rows holdc and the columns holdc⁰. The entries in the table hold the newc.

000 001 010 011 100 101 110 111

0 000 1 001 2 010 3 011 4 100 5 101 6 110 7 111

We only have the colors 000,001,010,011,100,101 (65).

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Last Steps

000 001 010 011 100 101 110 111

0 000 000 010 000 100 000 010 000

1 001 001 001 010 001 100 001 010

2 010 011 000 000 011 000 100 000

3 011 001 011 001 001 011 001 100

4 100 101 000 010 000 000 010 000

5 101 001 101 001 010 001 001 010

6 110 011 000 101 000 011 000 000

7 111 001 011 001 101 001 011 001

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Last Steps

000 001 010 011 100 101 110 111

0 000 0 2 0 4 0 2 0

1 001 1 1 2 1 4 1 2

2 010 3 0 0 3 0 4 0

3 011 1 3 1 1 3 1 4

4 100 5 0 2 0 0 2 0

5 101 1 5 1 2 1 1 2

6 110 3 0 5 0 3 0 0

7 111 1 3 1 5 1 3 1

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Last Steps

000 001 010 011 100 101 110 111

0 000 0 2 0 4 0 2 0

1 001 1 1 2 1 4 1 2

2 010 3 0 0 3 0 4 0

3 011 1 3 1 1 3 1 4

4 100 5 0 2 0 0 2 0

5 101 1 5 1 2 1 1 2

6 110 3 0 5 0 3 0 0

7 111 1 3 1 5 1 3 1

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

ifc=r then CN_i →c⁰ c⁰→Ci

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a cycle (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Coloring a Cycle

Theorem:

A cycle withnnodes could be colored withnprocessors in timeO(log^∗n)with at most 3 colors.

Proof: see above.

Theorem:

A cycle ofnprocessors may color itself in timeO(log^∗n)with at most 3 colors.

Proof: see above.

Theorem:

A cycle ofnprocessors needs at leastΩ(log^∗n)time to color itself with at most 3 colors.

Proof: see V4.

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Coloring a Cycle

Theorem:

Proof: see above.

Theorem:

Proof: see above.

Theorem:

Proof: see V4.

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Coloring a Cycle

Theorem:

Proof: see above.

Theorem:

Proof: see above.

Theorem:

Proof: see V4.

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Coloring a Cycle

Theorem:

Proof: see above.

Theorem:

Proof: see above.

Theorem:

Proof: see V4.

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Coloring a Tree

RegisterNi holdsπ(j−1)wherejis the father ofi. The father of the rootr isr.

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3:8 Trees 2/8 Walter Unger 30.1.2017 11:52 WS2016/17 Z

Coloring a Tree

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Coloring a Tree

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Coloring a Tree

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Coloring a Tree

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Coloring a Tree

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Coloring a Tree

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Coloring a Tree

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Parallel Coloring Algorithm of (on) a tree (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

π(i)→Ni

c=i c→Ci

c=2·k+ ((ck)%2).

c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

Within each color reduction will the coloring number be reduced fromx to log(x) +O(1).

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Parallel Coloring Algorithm of (on) a tree (Idea)

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Parallel Coloring Algorithm of (on) a tree (Idea)

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Parallel Coloring Algorithm of (on) a tree (Idea)

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Parallel Coloring Algorithm of (on) a tree (Idea)

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Parallel Coloring Algorithm of (on) a tree (Idea)

Programm: color-tree

π(i)→Ni

c=i andc→Ci

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

Programm: color-tree

π(i)→Ni

c=i andc→Ci

repeatdlog^∗(n)e+2times, ifRi 6=Ni

CN_i →c⁰

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

π(j−1)→Ni withj is father ofi c=i andc→Ci

c=2·k+ ((ck)%2).

c→Ci

ifRi =Ni thenc=min({0,1} \Ri elsec=CN_i

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Parallel Coloring Algorithm of (on) a tree (Idea)

c=2·k+ ((ck)%2).

c→Ci

forr:=5downto3do:

CN_i →c⁰⁰

c:=min({0,1,2} \ {c⁰,c⁰⁰}) c→Ci

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Coloring a Tree

Theorem:

A tree withnnodes could be colored with nprocessors in timeO(log^∗n)with at most 3 colors.

Proof: see above.

Theorem:

A tree ofnprocessors may color itself in timeO(log^∗n)with at most 3 colors.

Proof: see above.

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Coloring a Tree

Theorem:

A tree withnnodes could be colored with nprocessors in timeO(log^∗n)with at most 3 colors.

Proof: see above.

Theorem:

A tree ofnprocessors may color itself in timeO(log^∗n)with at most 3 colors.

Proof: see above.

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Eulerian cycle

Definition:

A graphG= (V,E)is called Eulerian, iff there exists a cycle which visits each edge precisely once.

Theorem

A non-directed graphG= (V,E)is Eulerian G is connected and

each node ofG has even degree.

Theorem

A directed graphG= (V,E)is Eulerian G is strong connected and

each node as as many incoming edges as outgoing ones.

Problem: Compute Eulerian cycle on Eulerian graphs.

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3:14 Introduction 2/4 Walter Unger 30.1.2017 11:52 WS2016/17 Z

Eulerian cycle

Definition:

Theorem

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Eulerian cycle

Definition:

Theorem

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Eulerian cycle

Definition:

Theorem

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Idea

Non Parallel:

Start with a free edge and follow free/unused edges till a cycle is closed.

Repeat till all edges are is some cycle.

Join pairs of cycles into a single one.

Repeat till just one cycle remains.

IfG is non-directed, then make a directed version ofG. Compute a cover of cycles.

Compute an additional cycle which meets each cycle precisely once.

Uses these to compute a cycle forG

Delete some edges to get a Eulerian cycle forG.

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Idea

Non Parallel:

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Idea

Non Parallel:

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Idea

Non Parallel:

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Idea

Non Parallel:

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Idea

Non Parallel:

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Idea

Non Parallel:

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Idea

Non Parallel:

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Idea

Non Parallel:

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Idea

Non Parallel:

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Change a non-directed Graph into a directed one

G containmnon-directed edges.

Substitute each non-directed edge with two directed ones:

{i,j}becomes(i,j)and(j,i).

Define a successor for each edge:

The neighbors ofv are:v0,v1,· · ·,vd−1. Then define for alli:

Succ((vi,v)) := (v,v(i+1)modd)und Succ((v(i+1)modd,v)) := (v,vi).

Each directed edge is in precisely one cycle (defined bySucc).

For each cycleC exists one cycleC⁰, which consists the reverse edges.

We will now delete one of the two cyclesC orC⁰.

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Change a non-directed Graph into a directed one

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