- By
**livia** - Follow User

- 123 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'GAMES AND COMPLEXITY' - livia

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### GAMES AND COMPLEXITY

Part 7

The

Evasiveness

Game

Peter van Emde Boas

ILLC-FNWI-Univ. of Amsterdam

© Games Workshop

© Games Workshop

References and slides available at: http://staff.science.uva.nl/~peter/teaching/gac08.html

Testing Graph Properties

Has an edge ? Yes

Has all edges ? No

Connected ? Yes

Has a Cycle ? No

Even number of edges ? Yes

.....

B

A

C

E

D

F

G

Graph: n nodes and e edges

Size Adjacency List representation: n + e

Size Adjacency Matrix representation: n(n-1)/2 (undirected)

and n2 in the directed case; n2-n when self loops are excluded.

Empirical observation around 1972

Testing graph properties algorithmically can for many of

these properties be performed in linear time O(n+e) , provided

the Adjacency list representation is used.

If the Adjacency Matrix representation is used (edge probe model)

in the worst case all edges must be probed and the time required

becomes W(n2) !

This issue became disturbing when for many important graph

properties linear time algorithms based on Depth First Search

were discovered.

Examples: connectivity, planarity, ....

Conjecture: this is no accident....

Edge probe model

Adversory model for property testing:

2 person game between seeker (Thorgrim) and hider (Urgat) .

Thorgrim identifies an edge and asks for its presence;

Urgat gives yes-no answer.

Thorgrim wins the game if he can decide the property before

all edges have been asked; Urgat wins if the property only is

decided when the complete graph is known.

Note that Urgat is allowed to construct the graph during the game:

Adaptive strategy.

The property is called Evasive when Urgat has a winning strategy

in this game (Evasiveness Game)

an ancient artifact

a sheet from my 1974 presentation on this topic…

Excluding Trivialities

Property 0 : The graph has n nodes

This property doesn’t depend on the presence or absense of

edges: Trivial property

Property 1 : The graph contains a self loop : (there are only n loops)

Property 2 : Edge (A,B) is present

This property depends on the individuality of nodes and is

therefore not invariant under renaming nodes:

not a Graph property

Restriction: Non Trivial graph properties which

don’t involve loops only !

Non trivial: property doesn’t depend on the number of nodes only

Graph property: Property is preserved under renaming nodes.

Rosenberg Conjecture

In 1973 Rosenberg conjectured that all non trivial Graph properties

are Evasive. But Aandara soon presented a directed Counterexample:

Property: The graph contains a sink

Sink

G’

Sink: all incoming edges are present

no outgoing edge is present

O(n)-time Sink test

Ask for edge (X,Y)

If present conclude that X is not a sink (has outgoing edge)

If absent conclude that Y is not a sink (misses incoming edge)

Always ask for an edge connecting two candidate sinks; regardless the

answer given one candidate sink is removed.

After n-1 probes only one candidate sink S remains;

asking for its remaining incoming and outgoing edges

(at most 2n-3 of which are still unprobed) decides whether S

is a true sink.

Note that the property is independent of edges within the subgraph G’

Also the property can de disrupted both by adding and removing edges

(Non monotone property).

The Scorpion Example

O(n) probe example for a property on undirected graphs

(invented by M. Best 1974).

t

s

b

The scorpion has a tail t , a body

b connected via s and n-3 other

nodes. The tail t has degree 1, node

s has degree 2 , and the body b has

degree n-2 . Connections amongst the

n-3 other nodes are irrelevant.

O(n) Scorpion test

Idea: each node has two body-tickets and two tail-tickets.

For each incident edge which is granted (refused) a tail-ticket

(body-ticket) is destroyed.

A dual candidate has still tickets remainin of both sorts.

A body (tail) candidate has ony body (tail) tickets remaining.

A node having lost all its tickets is moved to the irrelevant part.

Strategy: questions between dual candiates destroy two tickets

regardless the answer. If no more dual candidates remain

questions between candidate bodies and candidate tails will

destroy at least one ticket.

O(n) Scorpion test

In a first stage all n questions along a Hamiltonian path are probed.

This will destroy 2n tickets. The remaining nodes are divided in

3 classes: body candidates, tail candiates and dual candidates.

The dual candidates, which are the endpoints of chains of connected

edges along the cycle, always can be matched by non-cycle edges

and hence after at most n/2 more probes no dual candidates remain.

In a second stage edges are probed connecting remaining candidate

tails and candidate bodies. The remaining number of tickets being

at most 2n this stage will terminate after at most 2n probes, resulting

in a state where t candiate tails remain and b candidate bodies, and

all t.b edges connecting them have been asked. At the same time

from the connecting edges at most t have been granted and at most

b have been refused, showing that t.b < t + b .

The only solutions for this inequality are t=1, b=1 or t = b = 2 .

O(n) Scorpion test

Third stage:

t=1 : ask all edges for the unique candidate tail; this identifies s .

Validate that s has degree 2 and so identify the bodyb . Next

validate that b indeed has degree n-2. This will require at most 3n

probes.

b=1 : ask all edges for the unique candidate body ; this identifies t.

Next proceed as above, except for the final validation of b. Again

at most 3n probes.

t=b=2 : s must be one of the two candidate bodies. Asking their

remaining edges will determine who is who. This identifies t as well.

Finalize by asking all edges for the remaining tail candidate.

At most 3n probes are required.

Total number of probes < n + n/2 + 2n + 3n < 7n .

Monotonicity

Upward monotonicity: the property is not disturbed by adding edges

Downward monotonicity: the property is not disturbed by removing

edges.

Property Q is upward monotone iff the complementrary property Q*

is downward monotone.

Q*(G) iff notQ(G)

Upward monotone non trivial : notQ(En) & Q(Kn) ; EG. is connected

Downward monotone non trivial: Q(En) & notQ(Kn) ; EG. is acyclic

En= empty graph on n nodes ; Kn = complete graph on n nodes .

Aandara Rosenberg Conjecture

Based on the “has a sink” counterexample Aandara and Rosenberg

revised the conjecture:

Any Non-trivial (Upwards or downwards) Monotone Graph property

is Evasive (Sharp version of AR-conjecture)

Any Non-trivial (Upwards or downwards) Monotone Graph property

requires in the worst case W(n2) probes in the edge probe model

(original version of AR-conjecture)

The original AR-conjecture was proven by Rivest and Vuillemin in

1974; The sharpened version is still open, but holds for many

special cases. This result suffices for establishing the superiority of

the Adjacency List Representation for Algorithmic purposes....

Lattice: powerset lattice on the

set of m edges in Kn;

m = n(n-1)/2

Property : subset of Lattice

Monotone : upward(downward)

closed

Non trivial : neither empty nor full

Graph property : invariant under

lattice automorphisms

induced by permuting

nodes

Permutations in Sn induce

permutations in Sm

n = 3 is atypical since then

n = m ....

Y

N

AC?

BC?

Y

N

Y

N

BC?

AC?

AC?

BC?

Y

Y

N

N

Y

N

N

Y

Algorithm / Strategy of ThorgrimDecision tree for the property “is connected” for the case n = 3

The purple boxes correspond to positive results.

Y

N

AC?

BC?

Y

N

Y

N

Thorgrim wins

Thorgrim wins

AC?

BC?

Y

N

Y

N

Urgat wins

If the answer no longer depends on unprobed edges some questions

can be omitted; but in the worst case all edges must be probed;

Thorgrim has no better strategy; Urgat can win by giving one of the

two edges probed first...

Relation Tree - Lattice

Each node in the decision tree represents a set T of edges given

and a set R of edges refused in this position of the game

That node also corresponds to the sublattice of all graphs with an

set of edges including T and disjoint with R : G(T,R)

The depth of this node equals d = #T + #R

The game terminates if this entire set G(T,R) is included in the

property or is disjoint with the property

# G(T,R) = 2 m-#T-#R = 2 m-d .

A sufficient condition for Evasiveness

The number of graphs having the property equals:

S2 m-d(v)

v is terminal with G(T(v),R(v))

having the property

Corollary: If the number of graphs having the property is odd

then the property is Evasive

Proof: in the above sum one of the terms must be odd, which

requires d(v) = m indicating that all edges have been

probed.

Urgat’s winning strategy: give that answer which keeps the number

of consistent graphs with the property odd.

Example: the graph is non-empty.

Evasiveness based on Strategy

Property: the graph contains more than k edges

Winning strategy for Urgat: answer “yes” on the first k questions

and reject all subsequent ones.

As long as there exists an unprobed edge, any subsequent

“yes” aswer will produce a graph having the property

while the graph actually constructed doesn’t have it....

Question: for which properties will such a “simple” strategy work ?

Other examples: “does contain a cycle”, planarity

General condition for this Strategy

Theorem: Let Q be a (upward monotone) property such that any

non-maximal non- Q graph G can be extended in at least two

different ways to a non- Q graph:

if there is an f G such that G {f} Q then there is also

an f’ f , f’ G such that G {f’} Q

Then Q is evasive .

Strategy for Urgat: Say yes to all edges, unless adding the edge asked

for produces a graph in Q .

Analysis

We denote a graph by its edge-set. Assume that Thorgrim has a

winning strategy.

By the end of the game the set of all edges is divided in three parts:

E : edges given by Urgat

F : edges refused by Urgat

H : edges not yet asked for.

By construction the graph E doens’t have the property. Also H .

E H doesn’t have the property either since Thorgrim has won.

Let x H . Then E H \ {x} is a non-maximal non-Qgraph so there

exists an y x with E H \ {x} {y} Q . So also E {y} Q

However y F , i.e. y was refused by Urgat which doesn’t

conform to his strategy. Contradiction.

Examples

The graph contains a cycle... The non-maximal non-examples

are disconnected collections of trees. Except for the trivial

case of two nodes, there exist always multiple edges which

can be added.

Planarity: Maximal planar graphs embedded in a sphere have

triangual faces only. Removing an edge (a,b) will create a

face which is a quadrangle, whose other diagonal (c,d) can

be added, unless it is already present, in which case

either only 4 nodes are present and the property is

trivial, or looking at the two faces adjacent to this edge

(c,d) another edge which can be added will be found.

A dual result

Theorem: Let Q be a (upward monotone) property such that any

non-minimal Q graph G can be reduced in at least two

different ways to a Q graph:

if there is an f G such that G \ {f} Q then there is also

an f’ f , f’ G such that G \ {f’} Q

Then Q is evasive .

Strategy for Urgat: Say “no” to all edges, unless refusing the edge

asked for produces a graph which can’t be extended to a

graph in Q .

Analysis

We denote a graph by its edge-set. Assume that Thorgrim has a

winning strategy.

The game will terminate with Urgat granting an edge x and producing

a set of edges E such that the graph E has the property;

otherwise Urgat has violated his strategy. To prove that this

only happens when the last edge is asked for. So assume

that there still exists an unprobed edge y.

Since E {y} is a non-minimal graph with the property there exists an

edge z y E such that E {y}\ {z} has the property. But then

Urgat shouldn’t have granted edge z . (this argument applies

regardless whether x = z or not....

Example: the graph is connected ; non-minimal connected graphs

consist of a spanning tree and at least one more edge

creating a cycle.

Algebra 101

Group: non empty set A with operation A x A - -> A

written a.b , ab or a+b (additive notation)

Properties:

Associativity: (ab)c = a(bc) a+(b+c) = (a+b)+c

Neutral element: ea = ae = a a + 0 = 0 + a = a

Inverse: a-1 a = aa-1 = e a + (-a) = (-a) + a = 0

Potential Extra property

Commutativity: ab = baa + b = b + a

Additive notation for Commutative (Abelian) groups only.

Algebra 101

Properties:

Uniqueness of neutral element: e = e.e’ = e’

Uniqueness of inverse a* = a*.(a.a#) = (a*.a).a# = a#

Division: ac = bc implies a = b

Inverse of product: (ab)-1 = b-1 a-1

Inverse of inverse: (a-1)-1 = a

Typical examples:

Groups from arithmetic: Z+ , Q+ , R+ , Z/nZ, Q*, R*

Transformation groups, Automorphism groups,

Permutation groups Sn

Algebra 101

Homomorphismf : A - -> B is a mapping preserving

operations:

f(ab) = f(a).f(b) and f(a-1) = f(a)-1

Isomorphism : bijective Homomorphism

Automorphism: Isomorphism from group onto itself

Conjugation: fx(a) = xax-1 ; inner Automorphism

(trivial for Abelian groups)

Caley Representation: a group is isomorph to a group

of transformations on itself:

a - -> [ ta : x - -> x.a ]

Algebra 101

SubgroupB A : non empty subset closed under operations

(product & inverse): x,y B - -> x.y-1 B

Coset relation: x ~B y iff x.y-1 B

This is an equivalence relation:

Reflexive: x.x-1 = e B

Symmetry: x.y-1 B - -> (x.y-1)-1 = (y-1)-1.x-1 = y.x-1 B

Transitivity: x.y-1 , y.z-1 B - -> x.y-1.y.z-1 = x.z-1 B

The equivalence classes are called cosets Bx ;

Bx = By <- -> x.y-1 B

Right multiplication with x maps B bijectively on the coset Bx

Finite Groups

Order finite group: #A:=number of elements inA

Index [A:B] :=number of cosets Bx in A

Theorem [Lagrange]: #B . [A:B] = #A

Corrolary: #B and [A:B] are divisors of #A

Order element o(a) := order k cyclic subgroup <a> generated by a :

{a, a2, a3, ... , ak = e }

Hence o(a) divides #A ; moreover <a> is isomorph with Z/(o(a)Z)

and in particular <a> is Abelian.

Groups working on structures

Group actionG on set X : map P: G x X - -> X

Conditions: P(e,x) = x , P(a,p(b,x)) = P(ab, x)

Preserving a set : Y X : {a G | P(a,y) Y iff y Y }

Fixing a set Y X : {a G | P(a,y) = y for all y Y }

Stabilizer of a pointx : Gx {a G | P(a,x) = x}

Orbit of point x : G(x) {P(g,x) | g G }

There is a 1-1 correspondence between points in the orbit of x

and cosets relative Gx

Size orbit #G(x) and order Gx both divide the order #G

Groups working on structures

Transitive Group action: Orbit of some (equivalently all points)

is the entire set X.

Consequence: all the subgroups fixing some point are conjugate:

If fx = y then fGx f-1 is the subgroup fixing y.

Moreover #X is a divisor of #G.

The Caley representation is an example of a transitive group action.

Rings and Fields

Ring: set R with two operations + and .

Properties: (R,+,0) is an Abelian group

Multiplication . is associative a(bc) = (ab)c

Distributivity laws: a(b+c) = ab + ac , (a+b)c = ab + ac

Potential extra properties:

neutral element1 : 1a = a1 = a

Inverses for a 0 : a-1 a = aa-1 = 1

Unit: element having an inverse

Commutativity of multiplication : ab = ba

Field: Commutative Ring with 1 and inverses for alla 0

Rings and Fields

Examples: Z, Q, R, Z/nZ

Polynomials over a commutative ring: R[X]

R[X] := { r0 + r1X + r2X2 + .... + rkXk | k N , ri R }

Addition: coefficientwise: ( riXi )+( smXm) = (ri+si)Xi

Multiplication: ( riXi ) ( smXm) = k( i+m=krism)Xk

Evaluation: R[X] x R - -> R : ( riXi ) (v) := rivi

for fixed v this is a Ring homomorphism.

Ring homomorphism f: R - -> R’ extend to polynomial rings

f : R[X] - -> R’[X] by f(( riXi ) ) = ( f(ri)Xi )

Specific example : Z[X] - -> Z/pZ[X]

Finite Rings and Fields

The units of Z/nZ are the residue classes a with (a,n) = 1

grace to the solvability of ax + ny = 1, i.e. ax 1 (mod n)

In the special case that n = p a prime number, all rfesidue

classes a 0 satisfy the condition (a,p) = 1

Hence Z/pZ is a finite Field, denoted Fp .

In Commutative rings of characteristicp like Fp and Fp[X] raising

to the power p is a ring homomorphism (Frobenius)

(xy)p = xp.yp grace to associativity and commutativity

(x+y)p = Si=0...p (pi) xi y(p-i) (Binomium of Newton)

Since (pi) is divisible by p for j = 1 ... p-1 only the first and last term

remain : (x+y)p = xp + yp

Pairable sets

Two sets A,B X form a pair when X D Y := (X \ Y) (Y \ X)

consists of a single point :

Y is the result of adding to or removing from X a single point.

A family of sets is pairable if it is the disjoint union of a

collection of pairs

A Graph property Q (regarded as a collection of sets of edges) is

called Pairable when both the property and its complement

are pairable

Pairable Sets

Theorem: Non evasive graph properties are Pairable.

Proof: the property and its complement both are the disjoint union

of the sets G(T,R) , which are pairable when there are unprobed

edges...

Moreover Urgat has a winning strategy for unpairable sets

by always giving an answer such that the remaining part of

the property (or its complement) remains unpairable.

Application

Let Q be an upward monotone property.

Let m[i] be the number of graphs having the property Q with

i edges. Let j be the number of edges in a minimal graph

with the property.

Theorem: Ifm[j] > m[j+1]then the property is evasive.

Proof: Non-evasive - -> pairable. Now graphs with j edges can

only be paired with graphs of j+1 edges, and there are

not enough of those graphs....

Example: The graph contains at least m edges for m > n(n-1)/4

Intervals

Let A B be subsets of a set X with #(B \ A) = k

Then the set <A:B> :={C | A C B } is a k-interval

A k-interval contains 2k sets

A k-interval is pairable

Generalization of the previous necessary condition:

Theorem: If for some propertyThorgrimhas a winning strategy

where the game always terminates with at leastk

unprobed edges remaining, then both the property and

its complement are the disjoint union of a collection of

k-intervals

In this situation the number of graphs having the property is

divisible by 2k

Strictly Pairable sets

- Inductive definition of a strictly pairable subset Y P(X)
- (the powerset of X):
- a) A k-interval with k > 0 is strictly pairable
- b) If A and B are strictly pairable, and there exists an element
- z X such that every set a A contains z and no set b B
- contains z , then A B is strictly pairable.
- c) That’s all folks...

Necessary and Sufficient Condition

Theorem: Property Q , regarded as a subset of P( E ) with

E = set of edges in Kn is non-evasive iff both Q and its

complement are strictly pairable.

Proof: If Q is non-evasive then Thorgrim has a winning strategy

represented by a decision tree. Leaves in this tree

correspond to k-intervals for k > 0 and internal nodes

are labeled with an edge being asked for which separates

the subsets represented by the two successor nodes, thus

satsifying the inductive clause of the definition....

Conversily: from a certificate for strictly pairability of Q

a decision tree can be constructed which proves that

Thorgrim has a winning strategy. (minor details omitted)

Generating Function

Property Q on graphs on n nodes ; m = n(n-1) ( m= n(n-1)/2 )

number of edges in complete (undirected) graph

Generating function for this property :

HQ(X) := SA Q X#A = Si=0..m qiXi

where qi := #{ A | A Q and #A = i }

This generating function is a polynomial of degree at most m

If Kn has the property the degree is exactly m

If En doen’t have the property the constant term vanishes.

Pairing and Generating function

The contribution to a pair (A , A {e}) to the generating function

equals X#A ( 1+X )

Corollary : If the property Q is pairable then (1 + X) divides the

Generating function HQ(X)

Since non-evasive properties are pairable this yields yet another

sufficient condition for evasiveness

Theorem: if (1 + X)doesn’t divideHQ(X)thenQis evasive

Further consequences

Substituting arguments for X in theGenerating function HQ(X)

yields additional necessary conditions for evasiveness:

X = 1 : this yields the (already observed) consequence that the

number of graphs with an evasive propery must be even

X = -1 : the alternating sum condition: the number of graphs having

the property with an even number of edges equals the

number of graphs having the property with an odd

number of edges: AQ :=HQ(-1) := SA Q (-1)#A = Si=0..m qi(-1)i

Since divisibility of polynomials is preserved under taking the

coefficients modpwe also obtain the conditions:

(1 + X) |HQ(X) mod p and HQ(-1) = 0 mod p

Survey of conditions

Qis evasive<= =>

Qis strictly pairable = =>

Qand its complement are pairable= =>

Qis pairable = =>

(1 + X) |HQ(X) = =>

(1 + X) |HQ(X) mod p

Except for the first one the converses of these implications are

not true.

Counterexample

A property in P(0..3) such that both the property and its complement

are pairable but which is not strictly pairable

(monotone but not invariant under transitive permutation group)

Counterexample

Property on P(0..2) ; the complement is pairable, the property

is not.

The generating function 1 + X3 is divisible by 1 + X

Counting can be sufficient

Property: “the graph is starlike” : either it has no edges, or

all edges share a vertex.

Theorem: “the graph is starlike” is evasive forn > 2.

Proof: all graphs with 0 or 1 edges are starlike. If 2 or more

edges are present then the shared vertex is uniquely

determined.

Hence the alternating sum can be evaluated like:

AQ = 1 – n(n-1)/2 + n. Si=2..n-1 (n-1i) (-1)i =

= 1 – n(n-1)/2 + n( -1 + (n-1)) + n. Si=0..n-1 (n-1i) (-1)i =

= 1 + n(n-3)/2= (n-1)(n-2)/2 which is positive for n > 2 .

Counting can be sufficient

Theorem: “the graph contains no pair of disjoint edges”

is evasive forn > 3.

Proof: A graph without disjoint edges is either starlike or it is

a triangle. In order to evaluate the alternating sum

it suffices to add the contribution of triangles

-n(n-1)(n-2)/6 to the previous result:

Hence the alternating sum can be evaluated to:

AQ = (n-1)(n-2)/2- n(n-1)(n-2)/6 = -(n-1)(n-2)(n-3)/6

which is negative for n > 3 .

Inclusion & Exclusion

# (A B) = #A + #B - #(AB)

# (A B C) = #A + #B + #C - #(AB) - #(AC) - #(BC) + #(ABC)

In general: let F be a collection of sets covering a set X : X = F

Then # X = - S F’ F (-1) #F’ #(F’)

Proof: if element x X is covered by r members of F the

contribution of x to the RHS equals Si=1..r (ri)(-1) (i+1) = 1

Inclusion & Exclusion

In general: let F be a collection of sets covering a set X : X = F

Let w : X - -> R be some weight function, and let W(Y) := Sy Y w(y)

Then W( X ) = - S F’ F (-1) #F’ W(F’)

Proof: if element x X is covered by r members of F the

contribution of x to the RHS equals Si=1..r w(x)(ri)(-1) (i+1) = w(x)

The weights can be elements in some arbitrary Abelian Group

Use of In- and Exclusion

QP(X) an upward monotone property

MQ := collection of minimal sets in P(X) having the property

For A MQ let U(A) be the interval <A:X> ;

the family { U(A) | A MQ } covers Q

The weight of a set w(B) = (-1)#B

Now AQ = W(Q) = S F’ MQ (-1) (#F’+1) W({U(A)|A F’}) =

= S F’ MQ (-1) (#F’+1) W( U(F’) )

Use of In- and Exclusion

Now AQ= S F’ MQ (-1) (#F’+1) W( U(F’) )

In this sum contributions of families F’ which don’t cover X vanish;

U(F’) is a proper interval which contributes weight zero

to the alternating sum.

Hence AQ= S F’ MQ,F’ = X (-1) (#F’+1) W( U(F’) ) =

= S F’ MQ,F’ = X (-1) (#F’+1) W( { X } ) =

= S F’ MQ,F’ = X (-1) (#F’+1) (-1) #X

Use of In- and Exclusion

AQ= S F’ MQ,F’ = X (-1) (#F’+1) (-1) #X

The factor (-1) #X being constant, the evaluation reduces to counting an

alternating sum of coverings of X by minimal Q-sets.

Application: a star is a node with all connecting edges present.

The property “contains a star” is evasive.

Proof: Evidently the stars are the minimal graphs with the property.

A covering of all edges by stars requires all stars except one

to be present (if stars on nodes x and y are missing you

loose the edge (x,y) ).

There exist therefore n non maximal coverings by n-1 stars

and a single covering by all stars.

Hence the alternating sum equals ± (n – 1) 0

First edge is irrelevant.

Assume that Q is non-evasive and that Thorgrim has a winning

strategy. Thorgrim probes edge (a,b) and regardless

whether Urgat grants this edge or refuses it, Thorgrim

can win the game.

Without loss of generality we cas assume that the first edge probed

is the edge (1,2) .

Q[T,R] := the residual property on the set of unprobed edges

when edges in T are granted and edges in R are refused.

NB! Edges in T count for the size of a graph inQ[T,R]but not for

the degree of its monomial in generating function...

Q[T,R] has a generating function of itself. As long as the residual

property is non-evasive all the necessary conditions shown

previously must hold for this residual generating function.

Extra factor (1 + X)

Assume that Qis a non-trivial non-evasive string property on P(A)

which is invariant under a group operating transitive on A

Consider the generating functions for the residual properties:

FQ[{a},](X) and FQ[,{a}](X) ; Let #A = m

These two polynomials are independent of the choice of a .

So are its coefficients counting graphs in these residual properties.

Let qi := #{ B | B Q and #B = i }

q*i := #{ B | B Q[{a},] and #B = i }

q**i := #{ B | B Q[,{a}] and #B = i }

Now mq*i = iqi = # { <a,B> | aB Q and #B = i }

Andmq**i = (m-i)qi = # { <a,B> | aB Q and #B = i }

Extra factor (1 + X)

Now mq*i = iqi = # { <a,B> | aB Q and #B = i }

Andmq**i = (m-i)qi = # { <a,B> | aB Q and #B = i }

The generating function FQ[{a},](X) satisfies:

mX. FQ[{a},](X) = Si=1..m X.mq*i.X(i-1) =

= X Si=1..m i.qi X(i-1) = X (d/dX) FQ (X)

If the property is non-evasive, so is the residual property Q[{a},].

So (1+X) divides both FQ (X) and its derivative.

Then (1+X)2 must divide FQ (X) with theconsequence that

# Q = FQ (1) is divisible by 4 .

Lemma of Vera Pless

Assume string property Q on a set A with #A = pk for some

prime number p is invariant under a transitive group

G acting on A.

Let O(B) be the orbit of a subset B A .

Then either O(B) is a singleton, which implies B = or B = A,

or p | #O(B)

Proof: Consider a matrix M indexed by rows C O(B) and

columns a A , where M[C,a] := if a C then 1 else 0 fi .

The size of this matrix is w by pk , where 1 w = #O(B) .

Each column contains an equal number of 1’s say f .

Each row contains an equal number of 1’s say h = #B .

So the total number of 1’s in the matrix equals:

f.pk = h.w . Since 0 < h < pknot all k factors p in this

product can be included in h , so at least one factor p

divides w . QED.

Lemma of Vera Pless

Picture:

pk

1,1,1,0,1,0,....

0,1,1,0,1,1,....

....

....

0,0,1,0,0,1,....

C O(B)

h 1’s /row

w

f.pk = h.w

a A

Remark: The result would be trivial if we

know that #G is a power of p ,

but that’s not assumed.

Group theory tells us that such a group exists

f 1’s/column

Using group actions

A usual Q is a string property on P(A) , and G is a p-group

acting on A.

Subsets B A either are invariant under G or they generate a

non trivial orbit O(B) . But then p | #O(B) .

Let Q<G> be the property of both having Q and being invariant

under the action of G

Then HQ (X) = HQ<G> (X) mod p

Proof: write HQ (X) as the sum of the contributions of invariant sets

in Q and sets with a non trivial orbit. The latter contribution

vanishes since all non-trivial orbits have a size divisible by p.

the first contribution equals HQ<G> (X) .

Application

Theorem: non-trivial string properties Q on a set A with #A = pk

and Q and A Q preserved under a group acting

transitively on A are evasive.

Proof: By the lemma of Vera Pless the sizes of non-trivial orbits

are divisible by p . The unique invariant subsets are A and

the empty set .

Hence HQ (X) = HQ<G> (X) = X #A mod p

and this polynomial is not divisible by (1+X)

Remark: This doesn’t resolve the AR conjecture, since the number

of edges in a graph: n(n-1)/2 or n(n-1) is never a prime power

except for n=2 (directed) and n=3 (undirected).

Application

The property “contains a Hamiltonial Cycle” is evasive for

graphs of prime order p .

Proof: Consider the group generated by a cyclic permutation

of the p nodes (this group is isomorphic to Z/pZ).

All invariant nonempty graphs are Hamiltonian, since

the oribit of a single edge already gives a Hamiltonian

cycle.

Hence mod p the Generating Function for the complement

of the property equals the constant 1 .

Remark: same proof works for the property “strongly connected”

Application

The property “contains a directed cycle” is evasive for all orders.

Proof: Consider the action of the 2 element group generated by

permuting the first two nodes {1,2}

An invariant graph either includes both edges (1,2) and

(2,1) and thus a directed cycle, or the graph is obtained

from a directed graph of n-1 nodes by doubling the first

node, and will contain a directed cycle iff this n-1 node

graph contains one.

Hence if #Q(n) denotes the number of directed graphs without the

property, we have #Q(n) = #Q(n-1) mod 2

Since #Q(2) = 1 it follows that #Q(n) is odd for all n .

Rivest & Vuillemin proof

The key problem for our algebraic approach is that the number

of edges in a graph is never a prime power when n > 3 .

For the Andaara Rosenberg Conjecture it suffices that a large

subset of the edges needs to be probed. Hence it suffices to show

that Urgat can force that the game develops into a subgame for

a residual property for which the conditions for applying the

Pless lemma hold.

So we will grant Urgat before the game starts a move where he

exposes a subset of all edges granting some and refusing others.

If the resulting residual property is evasive then this shows that

a large fraction of the edges must be probed in the standard game.

Building Graphs

Let U and V be graphs with edge sets E and F

U + V := the disjoint union of U and V without connecting edges

U * V := the disjoint union of U and V with all connecting edges

U * V

U

V

U + V

A fundamental series of graphs

D[k,0] := Emwhere m = 2k

D[k,1] := sum of 2k-1copies of K2

D[k,i] := sum of 2k-icopiesof Kmwhere m = 2i

D[k,k] := Kmwhere m = 2k

D[4,0]

D[4,1]

D[4,2]

D[4,4]

Properties

The number of nodes in D[k,i] = 2k independent of i

D[k-1,i] + D[k-1,i] = D[k,i]

D[k-1,i] * D[k-1,i] D[k,i+1]

There exists a group preserving D[k,i+1] which acts

transitive on the 2(2k-2) cross edges in D[k,i+1]

For an upward monotone property Q there must exist an i

such that D[k,i] doesn’t have the property while D[k,i+1]

has it. (this follows since D[k,0] is the empty graph and

D[k,k] is the complete graph).

A good residual property

Let i be chosen as above. Take U = V = D[k-1,i]

Then we have: U + V doesn’t have the property, U * V has it, and

there exists a group preserving the set of cross edges W and

acting transitive on it. # W = 2(2k-2)

Consider the residual property Q* on P(W) :

Q*( B ) iff adding B to U + V yields a subgraph of U * V havingQ

Theorem: Q*onP(W)is evasive

Proof: all the conditions for the application of the Pless lemma hold.

Corollary: For graphs on n nodes with n = 2k the AR conjecture holds

with constant > 1/4

The case for general n

Consider for a given n the monotone non trivial graph property

which is the easiest to determine; maximally non-evasive.

W[n] := the number of probes required for deciding this property.

AR conjecture < == > W[n] = W(n2) < == >

W[n] > c.n2 for some constant c > 0 .

We have shown that W[n] > n2/4 if n is a power of 2

If we can show that W[n] > W[n-1] this would sufffice for proving

the AR conjecture with c = 1/16 . But this monotonicity is not

required. Rives and Vuillemin show instead that:

W[n] > min ( W[n-1] , m2/4 ) where m is the largest power of 2

below n . This will also prove AR conjecture with c = 1/16

More residual properties

Let Q be a non-trivial upward monotone graph property,

n arbitrary, and let m be the largest power of 2 below n.

Case distinction:

Case 1: the graph Kn-1 + K1 has the property:

Now Urgat refuses before the game all edges connected to node n

which yields a residual monotone non trivial graph property on

n-1 node graphs.

So in this case at least W[n-1] probes are required for the

residual property.

More residual properties

Case 2: the graph En-1 * K1 has not the property:

Now Urgat before the game grants all edges connected to node n

and again a residual monotone non-trivial graph property on

n-1 node graphs results.

So again W[n-1] probes are required.

Case 1:Kn-1 + K1 has the property

Case 2:En-1 * K1 has not the property

B

m/2

m/2

r

r

A

A

C

C

m/2

m/2

More residual propertiesCase 3: the graph En-1 * K1 has the property and Kn-1 + K1 has not

Now take r = n-m and decompose the nodes in three sets:

A,B and C with #A = #B = m/2 and #C = r

Before the game Urgat will grant all edges within A C and

refuse all edges within B or connecting B and C

B

m/2

m/2

r

r

A

A

C

C

m/2

m/2

The unprobed edges are the edges connecting A and B

The smallest possible graph is now Q := (Km/2 * Kr ) + Em/2

The largest possible graph is now R := Km/2 * (Kr + Em/2 )

Since Q Kn-1 + K1 the graph Q has not the property

Since En-1 * K1 R the graph R has the property

So the property induces a non trivial upward monotone string

property on the unprobed edges.

B

m/2

m/2

r

r

A

A

C

C

m/2

m/2

The number of unprobed edges equals m/2.m/2 = m2/4

Since the full permutation groups on the sets A , B and C preserve

the graphs Q and R there exists a transitive group preserving

this residual property

Hence the Pless Lemma applies: the residual property is evasive

and at least m2/4 probes are necessary.

Combining these three cases one obtains W[n] > min ( W[n-1] , m2/4 )

Other Results

Illies Counterexample (1978):

Illies produces an example of an evasive non-trivial string

property in P({0,…,11})

The property is non-monotone, but invariant under a transitive

group acting on {0,…,11} (more specifically a cyclic rotation).

More examples of this sort have been given by Lutz (1999).

Kleitman & Kwiatkowski(1980) improve the constant from the

Rivest-Vuillemin proof to 1/9 .

Other results

Kahn, Saks & Sturtevant(1984) introduce a necessary condition

for non-evasiveness based on Simplicial Complexes and

techniques from Algebraic Topology.

This yields a further improvement of the constant to 1/4 .

More importantly the technique allows to prove full evasiveness

for graphs with a number of nodes being a prime power.

Chakrabarti, Khot and Shi (2002) use this technique to prove

evasiveness for general properties based on containing some fixed

subgraph H for arithmetic progressions of the order n .

This raises the constant for such properties to 1/2 .

Further results

The question for the minimal number of probes required

is also meaningful for nondeterministic, probabilistic, or,

more recently, Quantum Computing.

Nondeterministic results have been obtained by Babai & Nisan.

Probabilistic results have been obtained by Yao, King & Hainal.

See Classroom notes by Lovasz.

Download Presentation

Connecting to Server..