Algorithmic Game Theory and Internet Computing

Algorithmic Game Theoryand Internet Computing

Vijay V. Vazirani

Georgia Tech

Dispelling an Old Myth

about an Ancient Algorithm

Central to the Theory of Algorithms

Kuhn, 1955: Primal-dual paradigm

Edmonds, 1965: Definition of P

Valiant, 1979: Definition of #P

Jerrum, Valiant & Vazirani, 1986: Equivalance

of random generation and approximate counting

Matching in game theory and economics

Gale & Shapley, 1962: Stable marriage

Shapley & Shubik, 1971: Matching markets

Matching in game theory and economics

Gale & Shapley, 1962: Stable marriage

Shapley & Shubik, 1971: Matching markets

Mehta, Saberi, Vazirani & Vazirani, 2005:

Adwords market

Maximum matching: a matching

of maximum cardinality in graph

U V

alternating path

U V

augmenting path: alternating

path between 2 unmatched

vertices.

alternating path

U V

alternating path

augmenting path: alternating

path between 2 unmatched

vertices.

Maximum matching algorithm

M ← ∅;While ∃ an augmenting path w.r.t. M do: Find one, say p;

M ← M ⊕p;end;output M;

Maximum matching algorithm

M ← ∅;While ∃ an augmenting path w.r.t. M do: Find one, say p;


O(n) iterations, O(m)/iteration, so O(mn) algorithm.

Shortest augmenting paths

M ← ∅;While ∃ an augmenting path w.r.t. M do: Find a shortest one, say p;


Disjoint.Therefore aug paths w.r.t. M.

Maximal set.

M

Shortest augmenting paths

M ← ∅;While ∃ an augmenting path w.r.t. M do: Find a maximal set of disjoint mimimum length paths;

AugmentM;end;output M;

phase

Hopcroft & Karp, 1973:

Theorem: O n( ) phases suffice

in general graphs.

Theorem: In bipartite graphs:

O(m) per phase⇒ O m n( ) algorithm.

Micali & Vazirani, 1980:

Theorem: In general graphs:

O(m) per phase ⇒ O m n( ) algorithm

Micali & Vazirani, 1980:

Theorem: In general graphs:

O(m) per phase ⇒ O m n( ) algorithm,

using Gabow & Tarjan, 1985:linear time set-union on RAM.

In G =(U, V, E) w.r.t. current matching M define

level(v): length of shortest alternating path

from an unmatched U vertex to v.

Find, via an alternating BFS, starting from

all unmatched vertices in U.

At search level i: level i vertices find

level i+1 vertices.

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U

V

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a b

{a, b}

Set ofpredecessors

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General Graphs

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G =(V, E). W.r.t. matching M define:

evenlevel(v): length of min even alternating path

from an unmatched vertex to v.

oddlevel(v): length of min odd alternating path

from an unmatched vertex to v.

minlevel(v): smaller of e(v) and o(v)

maxlevel(v): larger of e(v) and o(v)

Qualitative difference

BFS honesty:

Minimum length alternating paths in bipartite

graphs are BFS honest.

f vu


BFS honesty:

Minimum length alternating paths in bipartite

graphs are BFS honest.

Not so in non-bipartite graphs!

f vu


BFS honesty:

Not so in non-bipartite graphs!

Because shortest f to u path has v on it at

opposite parity.

f vu

v

a

b

Central issue

Need to find longer and longer alternating

paths to v in order to find min alternating

paths to other vertices.

Central issue

Need to find longer and longer alternating

paths to v in order to find min alternating

paths to other vertices.

Should require exponential time!

Saving grace

Incredible structure of minimum length

alternating paths.

Will give a purely graph-theoretic definition

of blossoms from this perspective.

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Will mark predecessors

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bridge

base

prop: an edge (u,v) s.t. u is a predecessor of v,

i.e., u assigns minlevel to v.

bridge: an edge that is not a prop.

Assigns maxlevels.

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base

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WANT

Find “blossom” by identifying “base”

i.e., highest bottleneck w.r.t. predecessors.

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WANT



If a nested blossom is encountered on the way, should skip to its base.

New algorithmic procedure:Double Depth First Search

In a layered graph,assume bridge (a,b)

a and b: special vertices on highest layer

(layers correspond to minlevels)

Edges can skip layers (nested blossoms)

Accomplish WANT.

In time: O(no. of edges above “base”)

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WANT



If a nested blossom is encountered on the way, should skip to its base.

If no bottleneck, then reach 2 unmatched

vertices – min aug path found!

GOT MORE …

bridge = (a, b)

1). Mark each v ∈B as red or blue.2). For each red v, ∃ a red path from a to v.3). Also∃ a red path from a to base of B.

GOT MORE …

bridge = (a, b)

1). Mark each v ∈B as red or blue.2). For each red v, ∃ a red path from a to v.3). Also∃ a red path from a to base of B or to red unmatched vertex.

WANT



If a nested blossom encountered on the way,

should skip to its base.

If no bottleneck, then reach 2 unmatched


WANT




should skip to its base – mark blossom so

this can be done efficiently. If no bottleneck, then reach 2 unmatched


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tenacity(v) = evenlevel(v) + oddlevel(v)

= Length of min alternating

walk (may be self-intersecting) between

unmatched vertices on which v occurs once.

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Let tenacity(v) = t. Consider the highest vertex of

tenacity > t on each minlevel(v) and maxlevel(v)

path.

v



path.

v v v



path.

v

v v v

vv

The base of a vertex

Theorem: Let v be a vertex of tenacity t.

Consider the highest vertex of tenacity > t

on each minlevel(v) and maxlevel(v) path.

There is a ! such vertex.

Denoted: base(v)

Blossom

Let tenacity(b) > t.

The blossom of tenacity t with base b

is defined to be

Bb,t = v|tenacity(v)≤t and base(v)=b{ }

v

base(v)

base(base(v))

base3(v)

=base2 (v)

Theorem: base(v), base2 (v), ... occur on

every evenlevel(v) and oddlevel(v) path.

tenacity(u,v) = Length of min alternating

walk (may be self-intersecting) between

unmatched vertices on which (u,v) occurs

once.

(u,v) unmatched ⇒tenacity(u,v) = evenlevel(u) + evenlevel(v)+1

(u,v) matched ⇒tenacity(u,v) = oddlevel(u) + oddlevel(v)+1

u v

u v

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Support of a bridge

Let (u, v) = bridge of tenacity t

support(u, v) = vertices of tenacity t found

in DDFS from (u, v).

Bridge (u, v) is responsible for assigning

maxlevels to these vertices.

Main ideas in algorithm

DDFS

Precise synchronization of events

At search level i:

MIN: level i vertices reach, via alternating BFS,

all vertices having minlevel of i+1.

MAX: Run DDFS on each bridge of tenacity 2i+1

to find maxlevels of all vertices

of tenacity 2i+1.

Algorithm: intertwined BSF & DDFS

1315

19

At search level i:





of tenacity 2i+1.


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tenacity(bridge) = 13Will do DDFS at

search level 6.

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After DDFS atsearch level 6.

At search level i:





of tenacity 2i+1.


Are we done?

Must show: Every vertex having tenacity 2i+1 lies in the support of a bridge of tenacity 2i+1

Must show: Every vertex having tenacity 2i+1 lies in the support of a bridge of tenacity 2i+1

Theorem: On any maxlevel(v) path

∃! bridge of tenacity = tenacity(v), and v lies it its support.

The agents that assign levels

minlevel: props

maxlevel: bridges

WANT




should skip to its base – mark blossom so

this can be done efficiently. If no bottleneck, then reach 2 unmatched


Computing base*(v)

Using set-union algorithm of Tarjan, 1975.

Gabow & Tarjan, 1985: Linear time on RAM.

Open: Path compression suffices

(by charging edges of blossoms)

GOT MORE …

bridge = (a, b)

1). Mark each v ∈B as red or blue.2). For each red v, ∃ a red path from a to v.3). Also∃ a red path from a to base of B or to red unmatched vertex.

a b

v

Find via two DFSs:

a to v path in red subgraph,

b to "base" path in blue subgraph.

Q: How will next path be found in

“half-eaten” blossom?

Theorem: base(v), base2 (v), ... occur on

every evenlevel(v) and oddlevel(v) path.

Q: How will next path be found in

“half-eaten” blossom?

A: Since base*(v) is removed, no part of

“half-eaten” blossom lies on a disjoint

min augmenting path!

Single thought?!

Algorithmic Game Theory and Internet Computing

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