Connectivity and minimum degree games

Prologue The Connectivity Game The Minimum Degree Game Epilogue

The Connectivity and Minimum Degree Games

Dalya Gartzman

Tel Aviv University

Based on notes by Michael Krivelevich

Dalya Gartzman Tel Aviv University



What Are We Playing?

CG

sequential games with perfectinformation

PG

each move consists of claiming apreviously-unclaimed position

MBG

the board is a set V , andH ⊂ P(V ) are the winning sets





CG


PG


MBG






CG


PG


MBG






CG


PG


MBG






CG


PG


MBG





Tic-tac-toe as a Maker-Breaker Game

Maker Vs. Breaker

maker=X,breaker=O

Maker Wins

notice the a-symmetry!





Maker Vs. Breaker

maker=X,breaker=O

Maker Wins






Maker Vs. Breaker

maker=X,breaker=O

Maker Wins






Maker Vs. Breaker

maker=X,breaker=O

Maker Wins





History and Motivation

A Mathematician’s Apology

”Nothing I have ever done isof the slightest practical use”

BUT THEN CAME RSA

So instead of asking useless questions about the usefulness ofMathematics, let‘s just enjoy the fact that we may play gamesduring working time :)







BUT THEN CAME RSA








BUT THEN CAME RSA








BUT THEN CAME RSA








BUT THEN CAME RSA





Introducing The Connectivity Game

Rules of the Game

I Maker starts by claiming one edge

I Breaker continues by claiming b ≥ 1 edges

I If Maker claimed a spanning tree - she wins

I If Breaker created a cut - he wins

I Game ends anyway when we are out of edges

What is the Critical Bias?

I When b = 1 Maker plays greedily and wins after n − 1 moves

I When b = n − 1 Breaker wins on the first round

I What is the critical bias bC?




Introducing The Connectivity GameRules of the Game


































































































































Probabilistic Intuition

The Random (1 : b) Game

By the end of the game RandomMaker occupies a random graph

with⌈

(n2)1+b

⌉≈ n2

2b edges.

What is brandC ?

THM 1.1 (Erdos − Renyi)

Consider the random graph G (n,M), and set 0 < ε 1.If M = (1− ε) n ln n

2 then G (n,M) is not connected WHP,

while if M = (1 + ε) n ln n2 then G (n,M) is connected WHP.

Corollary 1.2

brandC ≈ n2

2M ≈n

ln n







with⌈

(n2)1+b

⌉≈ n2

2b edges.

What is brandC ?





Corollary 1.2

brandC ≈ n2

2M ≈n

ln n







with⌈

(n2)1+b

⌉≈ n2

2b edges.

What is brandC ?





Corollary 1.2

brandC ≈ n2

2M ≈n

ln n







with⌈

(n2)1+b

⌉≈ n2

2b edges.

What is brandC ?





Corollary 1.2

brandC ≈ n2

2M ≈n

ln n







with⌈

(n2)1+b

⌉≈ n2

2b edges.

What is brandC ?





Corollary 1.2

brandC ≈ n2

2M ≈n

ln n







with⌈

(n2)1+b

⌉≈ n2

2b edges.

What is brandC ?





Corollary 1.2

brandC ≈ n2

2M ≈n

ln n







with⌈

(n2)1+b

⌉≈ n2

2b edges.

What is brandC ?





Corollary 1.2

brandC ≈ n2

2M ≈n

ln n




Proof of THM 1.1

Consider G (n, p) with p(n) = M

(n2)(≈ G (n,M))

When p = (1 + ε) ln nn

Remember: G is connected iff there exists no subset S ⊂ V (G )such that G has no edges between S and S\V .So we get (details ahead):

E [# of cuts in G (n, p)]n→∞−−−→ 0

From Markov, this is also an upper bound on the probability thatG (n, p) has at least one cut, and hence G is asymptotically almostsurely connected.




Proof of THM 1.1


(n2)(≈ G (n,M))



E [# of cuts in G (n, p)]n→∞−−−→ 0





Proof of THM 1.1


(n2)(≈ G (n,M))



E [# of cuts in G (n, p)]n→∞−−−→ 0





Proof of THM 1.1


(n2)(≈ G (n,M))


Remember: G is connected iff there exists no subset S ⊂ V (G )such that G has no edges between S and S\V .

So we get (details ahead):

E [# of cuts in G (n, p)]n→∞−−−→ 0





Proof of THM 1.1


(n2)(≈ G (n,M))



E [# of cuts in G (n, p)]n→∞−−−→ 0





Proof of THM 1.1


(n2)(≈ G (n,M))



E [# of cuts in G (n, p)]n→∞−−−→ 0





Proof of THM 1.1 (calculations for the connected case)

By linearity of expectation we get that E [# of cuts in G (n, p)] is

n/2∑i=1

(n

i

)(1− p)i(n−i) ≤

n/2∑i=1

(n

i

)e−pi(n−i)

≤

√n∑

i=1

(ne−p(n−i)

)i+

n/2∑i=√n+1

2ne−p√n(n−

√n)

≤

√n∑

i=1

(e−ε ln n(1+o(1))

)i+

n/2∑i=√n+1

e−Ω(ln n√n)

and both terms tend to zero when n tends to infinity, as desired.






n/2∑i=1

(n

i

)(1− p)i(n−i) ≤

n/2∑i=1

(n

i

)e−pi(n−i)

≤

√n∑

i=1

(ne−p(n−i)

)i+

n/2∑i=√n+1

2ne−p√n(n−

√n)

≤

√n∑

i=1

(e−ε ln n(1+o(1))

)i+

n/2∑i=√n+1

e−Ω(ln n√n)







n/2∑i=1

(n

i

)(1− p)i(n−i) ≤

n/2∑i=1

(n

i

)e−pi(n−i)

≤

√n∑

i=1

(ne−p(n−i)

)i+

n/2∑i=√n+1

2ne−p√n(n−

√n)

≤

√n∑

i=1

(e−ε ln n(1+o(1))

)i+

n/2∑i=√n+1

e−Ω(ln n√n)







n/2∑i=1

(n

i

)(1− p)i(n−i) ≤

n/2∑i=1

(n

i

)e−pi(n−i)

≤

√n∑

i=1

(ne−p(n−i)

)i+

n/2∑i=√n+1

2ne−p√n(n−

√n)

≤

√n∑

i=1

(e−ε ln n(1+o(1))

)i+

n/2∑i=√n+1

e−Ω(ln n√n)

and both terms tend to zero when n tends to infinity, as desired.Dalya Gartzman Tel Aviv University



Proof of THM 1.1 (cont.)

When p = (1− ε) ln nn

Remember: if G has an isolated vertex, then it is surelydisconnected.By linearity of expectation we get

E [# of isolated vertices in G (n, p)] = n(1− p)n−1 ≈ ne−pn

Since p = (1− ε) ln nn we get ne−pn = nε →∞.

Apply Chevishev‘s inequality to get (details ahead):

Pr(∣∣∣# of isolated vertices in G (n, p)− nε

∣∣∣ ≥ n2ε/3)→ 0

meaning the probability that there are no isolated vertices, tends tozero when n tends to infinity, as desired.











∣∣∣ ≥ n2ε/3)→ 0







Remember: if G has an isolated vertex, then it is surelydisconnected.

By linearity of expectation we get





∣∣∣ ≥ n2ε/3)→ 0












∣∣∣ ≥ n2ε/3)→ 0












∣∣∣ ≥ n2ε/3)→ 0












∣∣∣ ≥ n2ε/3)→ 0












∣∣∣ ≥ n2ε/3)→ 0





Proof of THM 1.1 (calculations for the disconnected case)

Let Z be the number of isolated vertices in G (n, p), and let’scompute its variance:

Var(Z ) = E[Z 2]− E2 [Z ]

= n(1− p)n−1 + n(n − 1)(1− p)2n−3 − n2(1− p)2n−2

= n(1− p)n−1(1− (1− p)n−2

)+ n2(1− p)2n−2

(p

1−p

)≈ nε

(1− nε−1

)+ n2ε (1−ε) ln n

n−(1−ε) ln n

From Chebyshev’s inequality we get:

Pr(|Z − E[Z ]| ≥ n

2ε3

)≤ Var(Z)

n4ε3≈ n−

ε3 − n−1+ 2ε

3 + n2ε3

(1−ε) ln nn−(1−ε) ln n

All three terms tend to zero as n tends to infinity, giving thedesired result.






Var(Z ) = E[Z 2]− E2 [Z ]

= n(1− p)n−1 + n(n − 1)(1− p)2n−3 − n2(1− p)2n−2

= n(1− p)n−1(1− (1− p)n−2

)+ n2(1− p)2n−2

(p

1−p

)≈ nε

(1− nε−1

)+ n2ε (1−ε) ln n

n−(1−ε) ln n


Pr(|Z − E[Z ]| ≥ n

2ε3

)≤ Var(Z)

n4ε3≈ n−

ε3 − n−1+ 2ε

3 + n2ε3








Var(Z ) = E[Z 2]− E2 [Z ]

= n(1− p)n−1 + n(n − 1)(1− p)2n−3 − n2(1− p)2n−2

= n(1− p)n−1(1− (1− p)n−2

)+ n2(1− p)2n−2

(p

1−p

)≈ nε

(1− nε−1

)+ n2ε (1−ε) ln n

n−(1−ε) ln n


Pr(|Z − E[Z ]| ≥ n

2ε3

)≤ Var(Z)

n4ε3≈ n−

ε3 − n−1+ 2ε

3 + n2ε3








Var(Z ) = E[Z 2]− E2 [Z ]

= n(1− p)n−1 + n(n − 1)(1− p)2n−3 − n2(1− p)2n−2

= n(1− p)n−1(1− (1− p)n−2

)+ n2(1− p)2n−2

(p

1−p

)

≈ nε(1− nε−1

)+ n2ε (1−ε) ln n

n−(1−ε) ln n


Pr(|Z − E[Z ]| ≥ n

2ε3

)≤ Var(Z)

n4ε3≈ n−

ε3 − n−1+ 2ε

3 + n2ε3








Var(Z ) = E[Z 2]− E2 [Z ]

= n(1− p)n−1 + n(n − 1)(1− p)2n−3 − n2(1− p)2n−2

= n(1− p)n−1(1− (1− p)n−2

)+ n2(1− p)2n−2

(p

1−p

)≈ nε

(1− nε−1

)+ n2ε (1−ε) ln n

n−(1−ε) ln n


Pr(|Z − E[Z ]| ≥ n

2ε3

)≤ Var(Z)

n4ε3≈ n−

ε3 − n−1+ 2ε

3 + n2ε3








Var(Z ) = E[Z 2]− E2 [Z ]

= n(1− p)n−1 + n(n − 1)(1− p)2n−3 − n2(1− p)2n−2

= n(1− p)n−1(1− (1− p)n−2

)+ n2(1− p)2n−2

(p

1−p

)≈ nε

(1− nε−1

)+ n2ε (1−ε) ln n

n−(1−ε) ln n


Pr(|Z − E[Z ]| ≥ n

2ε3

)≤ Var(Z)

n4ε3≈ n−

ε3 − n−1+ 2ε

3 + n2ε3








Var(Z ) = E[Z 2]− E2 [Z ]

= n(1− p)n−1 + n(n − 1)(1− p)2n−3 − n2(1− p)2n−2

= n(1− p)n−1(1− (1− p)n−2

)+ n2(1− p)2n−2

(p

1−p

)≈ nε

(1− nε−1

)+ n2ε (1−ε) ln n

n−(1−ε) ln n


Pr(|Z − E[Z ]| ≥ n

2ε3

)≤ Var(Z)

n4ε3≈ n−

ε3 − n−1+ 2ε

3 + n2ε3






Asymptotic Probabilistic Intuition

Does this Intuition hold?

Yes.On our next segment we will show that the threshold biases in therandom game, and in the clever game, are equal.

Do all graph properties behave in this manner?

No.I.e. take diameter-2 game, where brand

g = n3/2

2√

ln nwhile bg = 1!

In fact, one of the major questions you may ask in this regard, iswhether this connection appears, and under what circumstances.









g = n3/2

2√

ln nwhile bg = 1!







Yes.

On our next segment we will show that the threshold biases in therandom game, and in the clever game, are equal.



g = n3/2

2√

ln nwhile bg = 1!










g = n3/2

2√

ln nwhile bg = 1!










g = n3/2

2√

ln nwhile bg = 1!










g = n3/2

2√

ln nwhile bg = 1!










g = n3/2

2√

ln nwhile bg = 1!






So should I play randomly?

No. Play clever.ASI gives us an estimate on the outcome of the Clever Game bysimulating a Random Game.Also, the Clever Game ends much faster...

API and AI

Using a Monte-Carlo algorithm, a machine can choose its moveswisely, by simulating random games resulting from each possiblemove, and act according to the one that gives the best winningchances.This technique allowed an unprecedented breakthrough in AI forgames.







API and AI







No. Play clever.

ASI gives us an estimate on the outcome of the Clever Game bysimulating a Random Game.Also, the Clever Game ends much faster...

API and AI







No. Play clever.ASI gives us an estimate on the outcome of the Clever Game bysimulating a Random Game.

Also, the Clever Game ends much faster...

API and AI








API and AI








API and AI








API and AI

Using a Monte-Carlo algorithm, a machine can choose its moveswisely, by simulating random games resulting from each possiblemove, and act according to the one that gives the best winningchances.

This technique allowed an unprecedented breakthrough in AI forgames.







API and AI





Cutting to the Chase

THM 1.3 (Gebauer − Szabo)

For every ε > 0 there is an n0 = n0(ε) such that for every n ≥ n0,Maker can build a spanning tree of Kn while playing against aBreaker who plays with bias (1− ε) n

ln n .In particular bC = (1 + o(1)) n

ln n .

(We only show she lower bound today)

The Plan

Maker will abandon the intuitive approach of building a spanningtree, and adopt a strategy of ”breaking” into every cut.Succeeding with this will imply connectivity, and maintaining acycle free graph will imply completing a spanning tree.








ln n .


The Plan









ln n .


The Plan









ln n .


The Plan

Maker will abandon the intuitive approach of building a spanningtree, and adopt a strategy of ”breaking” into every cut.

Succeeding with this will imply connectivity, and maintaining acycle free graph will imply completing a spanning tree.








ln n .


The Plan





Proof of THM 1.3 - Setting

I Let b=(1− ε) nln n .

I component = a connected component on Maker‘s graph

I C (v) = the connected component of vertex v

I The degree of a vertex v (or deg(v)) = the degree of v inBreaker‘s graph

I A dangerous component = a component that contains atmost 2b vertices

I A non-dangerous component is connected to the rest of thegraph by at least 2b(n − 2b) > (n − 1)b edges
































































Proof of THM 1.3 - Maker’s Strategy

Goal: Maintain an active vertex vC in each component C .Tool: define a danger function

dang(v) =

deg(v) if v is in a dangerous component0 otherwise

MS

I At first all the vertices are active

I On each round, pick an active vertex vC with maximal danger

I Connect vC to an arbitrary component C ′

I Deactivate vC and assign vC∪C ′ = vC ′





Goal: Maintain an active vertex vC in each component C .

Tool: define a danger function

dang(v) =


MS










dang(v) =


MS










dang(v) =


MS










dang(v) =


MS










dang(v) =


MS










dang(v) =


MS










dang(v) =


MS








Proof of THM 1.3 - Proof of Maker’s win

Assume by contradiction that Breaker has a winning strategyagainst MS .

Let g − 1 be the winning round, and let (K ,V \K ) be the cut thatbreaker occupied. Assume WLOG that |K | ≤ |V \ K |, and letvg ∈ K be an arbitrary active vertex.

Notice: |K | ≤ 2b, dang(vg ) ≥ n − 2b.

Let Mi ,Bi be Maker’s and Breaker’s moves on round i .Let vi be the vertex Maker deactivated on round i , and letJi = vi+1, ..., vg, J = J0.














Let g − 1 be the winning round,

and let (K ,V \K ) be the cut thatbreaker occupied. Assume WLOG that |K | ≤ |V \ K |, and letvg ∈ K be an arbitrary active vertex.








Let g − 1 be the winning round, and let (K ,V \K ) be the cut thatbreaker occupied.

Assume WLOG that |K | ≤ |V \ K |, and letvg ∈ K be an arbitrary active vertex.








Let g − 1 be the winning round, and let (K ,V \K ) be the cut thatbreaker occupied. Assume WLOG that |K | ≤ |V \ K |,

and letvg ∈ K be an arbitrary active vertex.






























Proof of THM 1.3 - Proof of Maker’s win (cont.)

Let

dang (Mi ) =

∑v∈Ji−1

dang(v)

|Ji−1|dang (Bi ) =

∑v∈Ji dang(v)

|Ji |

Notice: dang (Mg ) = dang (vg ) ≥ n − 2b, dang (M1) = 0.

Goal: show the average danger cannot change so drastically (from0 to almost n) in less than n rounds, to get a contradiction.





Let

dang (Mi ) =

∑v∈Ji−1

dang(v)


∑v∈Ji dang(v)

|Ji |







Let

dang (Mi ) =

∑v∈Ji−1

dang(v)


∑v∈Ji dang(v)

|Ji |







Let

dang (Mi ) =

∑v∈Ji−1

dang(v)


∑v∈Ji dang(v)

|Ji |






Proof of THM 1.3 - Change during Maker’s move

Lemma 1.3.1

For every i , 1 ≤ i ≤ g − 1, we get

dang (Mi ) ≥ dang (Bi ) .

Proof of Lemma 1.3.1: Follows immediately from MS .





Lemma 1.3.1








Lemma 1.3.1







Proof of THM 1.3 - Change during Breaker’s move

Lemma 1.3.2


dang (Bi ) ≥ dang (Mi+1)− b + a(i)− a(i − 1)

|Ji |− 1,

where a(i) is the number of edges Breaker occupied within Jiduring the first i rounds.

Proof of Lemma 1.3.2:Let edouble be the number of edges Breaker adds to Ji during Bi .

So we get dang (Bi ) ≥ dang (Mi+1)− b+edouble|Ji | .

To finish, notice that the number of edges taken by Breaker untilBi−1 in Ji = a(i)− edouble , but is also ≥ a(i − 1)− |Ji |.





Lemma 1.3.2



|Ji |− 1,









Lemma 1.3.2



|Ji |− 1,


Proof of Lemma 1.3.2:

Let edouble be the number of edges Breaker adds to Ji during Bi .







Lemma 1.3.2



|Ji |− 1,









Lemma 1.3.2



|Ji |− 1,









Lemma 1.3.2



|Ji |− 1,








Proof of THM 1.3 - Almost the end

Let k = nln n , and notice that during the first g − k − 1 rounds of

the game we surely get dang (Bi ) ≥ dang (Mi+1)− 2b|Ji | .

This, together with the application of Lemma 1.3.2 on the last krounds of the game, gives (details ahead)

0 = dang (M1) ≥ · · · ≥ n − b(ln n + ln ln n + 3)− n

ln n,

whereas if the game ended after less than k rounds, we get (detailsahead)

0 = dang (M1) ≥ · · · ≥ n − b(ln n + 3)− n

ln n.

On both cases, b must be at least (1 + o(1)) nln n to satisfy these

inequalities, contradicting our assumption, and finishing the proof.









ln n,


0 = dang (M1) ≥ · · · ≥ n − b(ln n + 3)− n

ln n.











ln n,


0 = dang (M1) ≥ · · · ≥ n − b(ln n + 3)− n

ln n.











ln n,


0 = dang (M1) ≥ · · · ≥ n − b(ln n + 3)− n

ln n.











ln n,


0 = dang (M1) ≥ · · · ≥ n − b(ln n + 3)− n

ln n.











ln n,


0 = dang (M1) ≥ · · · ≥ n − b(ln n + 3)− n

ln n.











ln n,


0 = dang (M1) ≥ · · · ≥ n − b(ln n + 3)− n

ln n.


inequalities, contradicting our assumption, and finishing the proof.Dalya Gartzman Tel Aviv University



Proof of THM 1.3 - calculations for the long game version

0 = dang (M1) ≥ dang (B1) ≥ dang (M2)− 2bg−1 ≥ · · ·

≥ dang (Mg )− b+a(g−1)−a(g−2)|Jg−1| − · · · − b+a(g−k)−a(g−k−1)

|Jg−k | − k

− 2bk+1 − · · · −

2bg−1

♠≥ dang (Mg )− b

1 −b2 − · · · −

bk −

a(g−1)1 − k − 2b

k+1 − · · · −2bg−1

≥ (n − 2b)− b(1 + ln k)− k − 2b(ln g − ln k)

≥ n − b(ln n + ln ln n + 3)− nln n

Where in ♠ we used the fact that a(g − `)(

1

|Jg−`+1| −1

|Jg−`|

)≥ 0

for all `, and that a(g − 1) = 0 since Jg−1 contains only onevertex, and hence no edges.







|Jg−k | − k

− 2bk+1 − · · · −

2bg−1


1 −b2 − · · · −

bk −

a(g−1)1 − k − 2b

k+1 − · · · −2bg−1

≥ (n − 2b)− b(1 + ln k)− k − 2b(ln g − ln k)



1

|Jg−`+1| −1

|Jg−`|

)≥ 0








|Jg−k | − k

− 2bk+1 − · · · −

2bg−1


1 −b2 − · · · −

bk −

a(g−1)1 − k − 2b

k+1 − · · · −2bg−1

≥ (n − 2b)− b(1 + ln k)− k − 2b(ln g − ln k)



1

|Jg−`+1| −1

|Jg−`|

)≥ 0








|Jg−k | − k

− 2bk+1 − · · · −

2bg−1


1 −b2 − · · · −

bk −

a(g−1)1 − k − 2b

k+1 − · · · −2bg−1

≥ (n − 2b)− b(1 + ln k)− k − 2b(ln g − ln k)



1

|Jg−`+1| −1

|Jg−`|

)≥ 0








|Jg−k | − k

− 2bk+1 − · · · −

2bg−1


1 −b2 − · · · −

bk −

a(g−1)1 − k − 2b

k+1 − · · · −2bg−1

≥ (n − 2b)− b(1 + ln k)− k − 2b(ln g − ln k)



1

|Jg−`+1| −1

|Jg−`|

)≥ 0








|Jg−k | − k

− 2bk+1 − · · · −

2bg−1


1 −b2 − · · · −

bk −

a(g−1)1 − k − 2b

k+1 − · · · −2bg−1

≥ (n − 2b)− b(1 + ln k)− k − 2b(ln g − ln k)



1

|Jg−`+1| −1

|Jg−`|

)≥ 0








|Jg−k | − k

− 2bk+1 − · · · −

2bg−1


1 −b2 − · · · −

bk −

a(g−1)1 − k − 2b

k+1 − · · · −2bg−1

≥ (n − 2b)− b(1 + ln k)− k − 2b(ln g − ln k)



1

|Jg−`+1| −1

|Jg−`|

)≥ 0





Proof of THM 1.3 - calculations for the short game version

0 = dang (M1) ≥ dang (B1) ≥ dang (M2)− b+a(1)−a(0)|J1| − 1 ≥ · · ·

≥ dang (Mg )− b+a(g−1)−a(g−2)|Jg−1| − · · · − b+a(1)−a(0)

|J1| − g

♣≥ dang (Mg )− b

1 −b2 − · · · −

bg−1 −

a(g−1)1 − g

≥ (n − 2b)− b(1 + ln g)− g

≥ n − b(ln n + 3)− nln n

Where in ♣ we used the same argument as before.







|J1| − g


1 −b2 − · · · −

bg−1 −

a(g−1)1 − g

≥ (n − 2b)− b(1 + ln g)− g

≥ n − b(ln n + 3)− nln n








|J1| − g


1 −b2 − · · · −

bg−1 −

a(g−1)1 − g

≥ (n − 2b)− b(1 + ln g)− g

≥ n − b(ln n + 3)− nln n








|J1| − g


1 −b2 − · · · −

bg−1 −

a(g−1)1 − g

≥ (n − 2b)− b(1 + ln g)− g

≥ n − b(ln n + 3)− nln n








|J1| − g


1 −b2 − · · · −

bg−1 −

a(g−1)1 − g

≥ (n − 2b)− b(1 + ln g)− g

≥ n − b(ln n + 3)− nln n








|J1| − g


1 −b2 − · · · −

bg−1 −

a(g−1)1 − g

≥ (n − 2b)− b(1 + ln g)− g

≥ n − b(ln n + 3)− nln n








|J1| − g


1 −b2 − · · · −

bg−1 −

a(g−1)1 − g

≥ (n − 2b)− b(1 + ln g)− g

≥ n − b(ln n + 3)− nln n





Introducing The Minimum Degree Game

Goal

To win Dc , Maker needs to obtain the edges of a spanningsubgraph of Kn of minimum degree c .

What is brandDc

?


Consider the random graph G (n, p), and set 0 < ε 1.if p = (1− ε) ln n

n then δ (G (n, p)) < c WHP,

while if p = (1 + ε) ln nn then δ (G (n, p)) ≥ c WHP.

(THM 2.1 can be proven in a similar way as THM 1.1)We will now show that Dc satisfies API.





Goal


What is brandDc

?










Goal


What is brandDc

?










Goal


What is brandDc

?










Goal


What is brandDc

?










Goal


What is brandDc

?





(THM 2.1 can be proven in a similar way as THM 1.1)

We will now show that Dc satisfies API.





Goal


What is brandDc

?









Setting the stage for proving bDc


bDc = (1 + o(1)) nln n .

(Again, we only show the lower bound)

I Set b = (1− ε) nln n

I Let dM(v), dB(v) be the degrees of vertex v in Maker’s, andBreaker’s graphs, respectively

I A vertex v is dangerous if dM(v) < c






bDc = (1 + o(1)) nln n .










bDc = (1 + o(1)) nln n .










bDc = (1 + o(1)) nln n .










bDc = (1 + o(1)) nln n .










bDc = (1 + o(1)) nln n .








Proof of THM 2.2

MS

In round i , pick an arbitrary dangerous vertex vi with the largestdanger, and occupy an arbitrary edge incident to vi .This action is called easing vi .

Maker is going to have a hard time if some vertex v satisfies

dM(v) + dn−1−dB(v)−dM(v)b+1 e < c

⇒ Make damage control and be sure that every v satisfies

dB(v)− b · dM(v) ≤ n − 1− c(b + 1)

⇒ Let the danger function for the game be

dang(v) = dB(v)− 2b · dM(v)




Proof of THM 2.2

MS





dB(v)− b · dM(v) ≤ n − 1− c(b + 1)






Proof of THM 2.2

MS

In round i , pick an arbitrary dangerous vertex vi with the largestdanger, and occupy an arbitrary edge incident to vi .

This action is called easing vi .




dB(v)− b · dM(v) ≤ n − 1− c(b + 1)






Proof of THM 2.2

MS





dB(v)− b · dM(v) ≤ n − 1− c(b + 1)






Proof of THM 2.2

MS





dB(v)− b · dM(v) ≤ n − 1− c(b + 1)






Proof of THM 2.2

MS





dB(v)− b · dM(v) ≤ n − 1− c(b + 1)






Proof of THM 2.2

MS





dB(v)− b · dM(v) ≤ n − 1− c(b + 1)






Proof of THM 2.2 - cont.


I Let g − 1 be Breaker’s winning round. Then by this time heoccupied at least n − c edges incident to some vertex vg .

I Maker is easing a vertex every turn, hence g − 1 < cn.I Since dM (vg ) < c at the end fo the game, we get

dang (vg ) > n − c − 2bc = (1− o(1)) n.I Let vi be the vertex that Maker eased on her i-th turn, and

define the set Ji = vi+1, ..., vg.

Define dang (Mi ) , dang (Bi ) as before, and get:dang (Mg ) = dang(vg ) ≥ (1− o(1))n , dang (M1) = 0,a difference we will now show to be impossible.
















I Let g − 1 be Breaker’s winning round.

Then by this time heoccupied at least n − c edges incident to some vertex vg .





















I Maker is easing a vertex every turn, hence g − 1 < cn.

I Since dM (vg ) < c at the end fo the game, we getdang (vg ) > n − c − 2bc = (1− o(1)) n.

I Let vi be the vertex that Maker eased on her i-th turn, anddefine the set Ji = vi+1, ..., vg.









dang (vg ) > n − c − 2bc = (1− o(1)) n.

I Let vi be the vertex that Maker eased on her i-th turn, anddefine the set Ji = vi+1, ..., vg.


























Lemma 2.2.1



If Ji = Ji−1, then also

dang (Mi ) ≥ dang (Bi ) +2b

|Ji |.






Lemma 2.2.1





|Ji |.






Lemma 2.2.1





|Ji |.






Lemma 2.2.1





|Ji |.






Lemma 2.2.1





|Ji |.






Lemma 2.2.2

For every i , 1 ≤ i ≤ g − 1, we get the two lower bounds

dang (Bi ) ≥ dang (Mi+1)− 2b

|Ji |(1)


|Ji |− 1 (2)

where a(i) is the number of edges Breaker occupies within Jiduring the first i rounds.

Proof of Lemma 2.2.2: Exactly the same arguments as on theConnectivity game.





Lemma 2.2.2



|Ji |(1)


|Ji |− 1 (2)







Lemma 2.2.2



|Ji |(1)


|Ji |− 1 (2)







Lemma 2.2.2



|Ji |(1)


|Ji |− 1 (2)






Proof of THM 2.2 - Change between Maker’s moves

Corollary 2.2.3


dang (Mi ) ≥ dang (Mi+1) , provided Ji = Ji−1 (3)

dang (Mi ) ≥ dang (Mi+1)−min

2b

|Ji |,b + a(i)− a(i − 1)

|Ji |− 1

(4)


So the interesting situation is when Ji 6= Ji+1, then the dangermight grow.





Corollary 2.2.3




2b

|Ji |,b + a(i)− a(i − 1)

|Ji |− 1

(4)







Corollary 2.2.3




2b

|Ji |,b + a(i)− a(i − 1)

|Ji |− 1

(4)







Corollary 2.2.3




2b

|Ji |,b + a(i)− a(i − 1)

|Ji |− 1

(4)







Let 1 ≤ i1 ≤ ... ≤ ir ≤ g − 1 the indices where Ji 6= Ji−1.Notice: |Jir−`| = `+ 1, |Jir | = |Jg−1| = 1, |Ji1−1| = |J0| = r + 1.Let k = n

ln n . Applying the estimates on Corollary 2.2.3 we get(details ahead):If r ≥ k

0 = dang (M1) ≥ · · · ≥ n − b(ln n + ln ln n + 2c + 1)− c − nln n ,

and if r < k

0 = dang (M1) ≥ · · · ≥ n − b(2c + ln r −+1)− c − nln n .







Let 1 ≤ i1 ≤ ... ≤ ir ≤ g − 1 the indices where Ji 6= Ji−1.

Notice: |Jir−`| = `+ 1, |Jir | = |Jg−1| = 1, |Ji1−1| = |J0| = r + 1.Let k = n



and if r < k








Let 1 ≤ i1 ≤ ... ≤ ir ≤ g − 1 the indices where Ji 6= Ji−1.Notice: |Jir−`| = `+ 1, |Jir | = |Jg−1| = 1, |Ji1−1| = |J0| = r + 1.

Let k = nln n . Applying the estimates on Corollary 2.2.3 we get

(details ahead):If r ≥ k


and if r < k









ln n . Applying the estimates on Corollary 2.2.3 we get(details ahead):

If r ≥ k


and if r < k











and if r < k











and if r < k











and if r < k








0 = dang (M1) ≥ dang (Mi1) ≥ dang (Mi2)− 2b

|Ji1 |≥ · · ·

≥ dang (Mg )− b+a(ir )−a(ir−1)|Jir |

− · · · − b+a(ir−k+1)−a(ir−k+1−1)∣∣∣Jir−k+1

∣∣∣ − k

− 2b∣∣∣Jir−k

∣∣∣ − · · · − 2b

|Ji1 |♠≥ dang (Mg )− b

1 −b2 − · · · −

bk −

a(ir )1 − k − 2b

k+1 − · · · −2bg−1

≥ (n − c − 2bc)− b(1 + ln k)− k − 2b(ln n − ln k)

≥ n − b(ln n + ln ln n + 2c + 1)− c − nln n

Where in ♠ we used the fact that a(ir−` − 1) ≥ a(ir−`−1) for all `(since Jir−j−1

= Jir−j−1), and that a(ir ) = 0 since Jg−1 containsonly one vertex, and hence no edges.






|Ji1 |≥ · · ·


− · · · − b+a(ir−k+1)−a(ir−k+1−1)∣∣∣Jir−k+1

∣∣∣ − k


∣∣∣ − · · · − 2b


1 −b2 − · · · −

bk −

a(ir )1 − k − 2b

k+1 − · · · −2bg−1










|Ji1 |≥ · · ·


− · · · − b+a(ir−k+1)−a(ir−k+1−1)∣∣∣Jir−k+1

∣∣∣ − k


∣∣∣ − · · · − 2b

|Ji1 |


1 −b2 − · · · −

bk −

a(ir )1 − k − 2b

k+1 − · · · −2bg−1










|Ji1 |≥ · · ·


− · · · − b+a(ir−k+1)−a(ir−k+1−1)∣∣∣Jir−k+1

∣∣∣ − k


∣∣∣ − · · · − 2b


1 −b2 − · · · −

bk −

a(ir )1 − k − 2b

k+1 − · · · −2bg−1










|Ji1 |≥ · · ·


− · · · − b+a(ir−k+1)−a(ir−k+1−1)∣∣∣Jir−k+1

∣∣∣ − k


∣∣∣ − · · · − 2b


1 −b2 − · · · −

bk −

a(ir )1 − k − 2b

k+1 − · · · −2bg−1










|Ji1 |≥ · · ·


− · · · − b+a(ir−k+1)−a(ir−k+1−1)∣∣∣Jir−k+1

∣∣∣ − k


∣∣∣ − · · · − 2b


1 −b2 − · · · −

bk −

a(ir )1 − k − 2b

k+1 − · · · −2bg−1










|Ji1 |≥ · · ·


− · · · − b+a(ir−k+1)−a(ir−k+1−1)∣∣∣Jir−k+1

∣∣∣ − k


∣∣∣ − · · · − 2b


1 −b2 − · · · −

bk −

a(ir )1 − k − 2b

k+1 − · · · −2bg−1









0 = dang (M1) ≥ dang (Mi1) ≥ dang (Mi2)− b+a(i2)−a(i1)

|Ji1 |− 1 ≥ · · ·


− · · · − b+a(i2)−a(i1)

|Ji1 |− r


1 −b2 − · · · −

br −

a(ir )1 − r

≥ (n − c − 2bc)− b(1 + ln r)− r

≥ n − b(2c + ln r −+1)− c − nln n







|Ji1 |− 1 ≥ · · ·


− · · · − b+a(i2)−a(i1)

|Ji1 |− r


1 −b2 − · · · −

br −

a(ir )1 − r

≥ (n − c − 2bc)− b(1 + ln r)− r

≥ n − b(2c + ln r −+1)− c − nln n







|Ji1 |− 1 ≥ · · ·


− · · · − b+a(i2)−a(i1)

|Ji1 |− r


1 −b2 − · · · −

br −

a(ir )1 − r

≥ (n − c − 2bc)− b(1 + ln r)− r

≥ n − b(2c + ln r −+1)− c − nln n







|Ji1 |− 1 ≥ · · ·


− · · · − b+a(i2)−a(i1)

|Ji1 |− r


1 −b2 − · · · −

br −

a(ir )1 − r

≥ (n − c − 2bc)− b(1 + ln r)− r

≥ n − b(2c + ln r −+1)− c − nln n







|Ji1 |− 1 ≥ · · ·


− · · · − b+a(i2)−a(i1)

|Ji1 |− r


1 −b2 − · · · −

br −

a(ir )1 − r

≥ (n − c − 2bc)− b(1 + ln r)− r

≥ n − b(2c + ln r −+1)− c − nln n







|Ji1 |− 1 ≥ · · ·


− · · · − b+a(i2)−a(i1)

|Ji1 |− r


1 −b2 − · · · −

br −

a(ir )1 − r

≥ (n − c − 2bc)− b(1 + ln r)− r

≥ n − b(2c + ln r −+1)− c − nln n







|Ji1 |− 1 ≥ · · ·


− · · · − b+a(i2)−a(i1)

|Ji1 |− r


1 −b2 − · · · −

br −

a(ir )1 − r

≥ (n − c − 2bc)− b(1 + ln r)− r

≥ n − b(2c + ln r −+1)− c − nln n





Where do we go from here?

I We know the properties of connectivity and minimum degree1 appear with high correlation in random graphs.Open conjecture: for all n large enough, bC = bD1 .

I What can be a strong criteria for a property to satisfy API?

I Next time - Hemiltonicity game.
































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Connectivity and minimum degree games