The Cinderella game on holes and anti-holes. Introduction - - PowerPoint PPT Presentation

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The Cinderella game on holes and anti-holes. Introduction - - PowerPoint PPT Presentation

Nov 21, 2023 35 likes •258 views

The Cinderella game on holes and anti-holes. Introduction Definitions Example game Marijke H.L. Bodlaender 1 Cor A.J. Hurkens 2 The game on general graphs Gerhard J. Woeginger 2 The game on holes Conjectures 22 June 2011 1 Department of

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Introduction Definitions Example game The game on general graphs The game on holes Conjectures

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The Cinderella game on holes and anti-holes.

Marijke H.L. Bodlaender1 Cor A.J. Hurkens2 Gerhard J. Woeginger2 22 June 2011

1Department of Information and Computing Sciences, Universiteit

Utrecht, The Netherlands

2Department of Mathematics and Computer Science, TU Eindhoven,

The Netherlands

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Introduction Definitions Example game The game on general graphs The game on holes Conjectures

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Introduction Definitions Example game The game on general graphs The game on holes Conjectures

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Introduction Definitions Example game The game on general graphs The game on holes Conjectures

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Introduction to the game

◮ Proposed problem for the International Mathematical

Olympiad

◮ We study variant where water arrives in rounds and the

game board is an undirected graph

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The game

◮ Game played on undirected simple graph G = (V , E) ◮ Every vertex v contains a bucket ◮ Every edge [u, v] ∈ E indicates an incompatibility ◮ In every round the Stepmother distributes a liter of water

in the buckets

◮ Cinderella empties the buckets in an independent set ◮ Stepmother tries to reach an overflow ◮ Cinderella wants to avoid an overflow

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Some definitions and notation

◮ bucket′(G): infimum of all bucket sizes Cinderella needs

to win

◮ bucket(G) : bucket′(G) − 1 is the bucket number of G ◮ GREEDY: empty maximum weight independent set every

turn

◮ g-bucket(G): bucket number of G when Cinderella uses a

GREEDY strategy

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Definitions

◮ x = (xv) v ∈ V where xv is the contents of bucket v at the

start of a round

◮ x(S) =

v∈S xv

◮ yv the contents of v after the Stepmother moved ◮ χ(G) the chromatic number of G ◮ ω(G) the clique number of G ◮ For S ⊆ V we write χ(S) ◮ H k = 1 + 1

2 + 1 3 + 1 4 + . . . . . . + 1 k

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Example with bucket size 1.6

◮ x1 = x2 = x3 = 0

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Example with bucket size 1.6

◮ y1 = y2 = y3 = 1/3

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Example with bucket size 1.6

◮ x1 = 1/3 ◮ x2 = x3 = 0

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Introduction Definitions Example game The game on general graphs The game on holes Conjectures

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Example with bucket size 1.6

◮ y1 = y2 = 2/3 ◮ y3 = 0

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Example with bucket size 1.6

◮ x1 = 2/3 ◮ x2 = x3 = 0

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Example with bucket size 1.6

◮ y1 = 5/3 > 1.6 ◮ y2 = y3 = 0

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Results in this paper

◮ g-bucket(G) ≤ H χ(G) − 1 ◮ bucket(G) ≥ H ω(G) − 1 ◮ bucket(G) = H ω(G) − 1

∀ graphs on n ≤ 6 vertices

◮ bucket(C2m+1) = 1 ◮ g-bucket(C2m+1) = 1 + 1

m · 2−m

◮ g-bucket

  • C2m+1
  • ≤ H m − 1/(2m)

◮ g-bucket

  • C2m+1
  • ≥ H m − 1 + m2−3m+1

2m2(m−1)

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Upper bound on general graphs

Theorem

Every graph G = (V , E) satisfies g-bucket(G) ≤ H χ(G) − 1

Proof.

GREEDY maintains the following system of invariants x(S) < χ(S)·( 1 + H χ(G) − 1 − H χ(S) ) for all sets S ⊆ V (1) Apply (1) to S = {v} to show xv < H χ(G) − 1

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Upper bound for GREEDY on general graphs continued

If χ(S) = χ(G), then y(S) ≤ y(V − I) ≤ χ(G) − 1 χ(G) y(V ) ≤ χ(G) − 1 χ(G) (x(V ) + 1) < χ(G) − 1 Assume that χ(S) < χ(G) observe that y(S) ≤ χ(S) · y(I) (2)

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Upper bound for GREEDY on general graphs continued

Furthermore x(S ∪ I) < ( χ(S) + 1 ) · ( 1 + H χ(G) − 1 − H χ(S) + 1 ) (3) Applying (2) and (3) we derive y(S) ≤ χ(S) χ(S) + 1 (y(S) + y(I)) ≤ χ(S) χ(S) + 1 (x(S ∪ I) + 1) < χ(S) ·

  • 1 + H χ(G) − 1 − H χ(S) + 1 +

1 χ(S) + 1

  • =

χ(S) · (1 + H χ(G) − 1 − H χ(S))

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Lower bound on general graphs

Theorem

Every graph G = (V , E) satisfies bucket(G) ≥ H ω(G) − 1 Let ω(G) = n Define a strategy for the Stepmother:

◮ Play game on the the largest clique, K ◮ At the first phase:

  • Fill repeatedly all buckets in K to the same level
  • This converges to 1 − ǫ

◮ In second phase

  • In r-th round fill n − r fullest buckets to the same level
  • At the end of round n − 2 at least one bucket contains

H n − 1 − ǫ

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Perfect graphs

Theorem

Every perfect graph G has bucket(G) = g-bucket(G) = H ω(G) − 1

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GREEDY odd holes: Upper bound

Theorem

The odd cycle C2m+1 has g-bucket(C2m+1) ≤ 1 + 1

m · 2−m

Proof(Upper bound).

GREEDY maintains the following invariants

2m+1

  • i=1

xi < m + 1 m

k+2t−1

  • i=k

xi < 1 + 1 m · 2t−m for 1 ≤ k ≤ 2m + 1, 1 ≤ t ≤ m xk < 1 + 1 m · 2−m for 1 ≤ k ≤ 2m + 1

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GREEDY odd holes: Lower bound

Theorem

The odd cycle C2m+1 has g-bucket(C2m+1) ≥ 1 + 1

m · 2−m

Proof.

First phase: Fill repeatedly all buckets to the same level Second phase:

B1 B2 B3 . . . B2m−3 B2m−2 B2m−1 B2m SL

1 m 1 m 1 m

. . .

1 m 1 m 1 m 1 m

CL

1 m

. . .

1 m 1 m 1 m

SL

1 m 1 m

. . . α1 α1 α1 α1 CL

1 m

. . . α1 α1 . . . Bk+1 Bk+2 . . . B2m−k−1 B2m−k B2m−k+1 B2m−k+2 CL

1 m

. . .

1 m

αk−1 αk−1 SL

1 m 1 m

. . . αk αk αk αk CL

1 m

. . . αk αk

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GREEDY odd holes: Lower bound (continued)

Third phase: . . . Bm+1 Bm+2 . . . CL αm−1 αm−1 SL αm−1 + 1

2

αm−1 + 1

2

CL αm−1 + 1

2

The alpha values solve to αk = 1 2m

  • k + 1 + 2−k

⇒ αm−1 + 1 2 = 1 + 1 m · 2−m

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Possible future research

Conjecture

Every graph G satisfies bucket(G) = H ω(G) − 1

Conjecture

A graph G is perfect, if and only if bucket(G) = g-bucket(G)

Conjecture

The difference between g-bucket(G) and bucket(G) is bounded by an absolute constant (that does not depend on G)