Tengyu Ma
Xiaoming Sun Huacheng Yu
Institute for
Institute for
Institute for
Interdisciplinary
Advanced Study,
Interdisciplinary
Information Sciences Tsinghua University Information Sciences
Tsinghua University
Tsinghua University

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
3 cooperative players
each is assigned a hat of color
red or blue
each can only see others’ hat
guess own color or pass
players win if: at least one
correct and no wrong guess
goal : to maximize winning
probability
pass

strategy1: only a pre-specified
player guesses randomly
winning prob. =
pass

1
2
strategy2: if other two have
same color, guess the
opposite, otherwise pass.
winning prob. =
3

4
is optimal
3
4

𝒏 cooperative players:
◦ coordinate a strategy initially

assigned a blue or red hat
◦ uniformly and independently


guess a color or pass
winning condition:
◦ at least 𝒌 correct guesses and no
wrong guess

goal: to maximize winning prob.



𝑘 = 1 case is well studied by [?], [?]..
Observation 1: randomized strategy does
not help
Observation 2: related to the minimum
𝑘-dominating set of 0,1 𝑛
Definition: A 𝑘-dominating set for a graph
𝐺 = 𝑉, 𝐸 is a subset 𝐷 of 𝑉, such that every
vertex not in 𝐷 has at least 𝑘 neighbors in 𝐷
pass
win!
pass
pass
pass
lose
pass
win!
pass
pass
lose
pass
0,1,0,1,1,1 ∈ 0,1
winning point
6
1,1,0,1,1,1 ∈ 0,1
losing point
6
pass
win!
pass
pass
lose
pass
winning point has at least 𝑘 losing points
as neighbors

all losing points → 𝐷 ⊂ 0,1
𝑛
◦ 𝐷 is 𝑘-dominating set of 0,1
◦ winning prob. = 1 −


𝐷
2𝑛
𝑛
reduction can be done vice versa
by counting argument:
◦ 𝐷 ≥
𝑘
2𝑛
𝑛+𝑘
◦ ⇒ winning prob. ≤
𝑛
𝑛+𝑘

Theorem:
◦ There exists a 𝑘-dominating set 𝐷 of size
𝑘
2𝑛 ,
𝑛+𝑘
𝑘
2𝑛
𝑛+𝑘
as long as
is an integer, for
large enough 𝑛 (𝑛 ≥ Ω(𝑘2𝑘 ) ).
◦ It follows that there exists a strategy of
the hat guessing games with winning
𝑛
prob.
𝑛+𝑘

theorem is not true for small 𝑛
◦ example: 𝑛, 𝑘 = 5,3 , (11,5)
0,1
𝑛
∖𝐷
𝐷
each has 𝑘
neighbors
in 𝐷
each has 𝑛
neighbors
in 0,1 𝑛 ∖ 𝐷
𝑉1
each has 𝑑1
neighbors
in 𝑉1
𝑉2
each has 𝑑2
neighbors
in 𝑉2

𝑘-DS of 0,1 n → 𝑘, 𝑛 − 𝑘 -RP of 0,1
possible (𝑑1 , 𝑑2 )-RP of 0,1 𝑛 :

possible 𝑘-DS corresponds to the case

𝑛
◦ the parameters are of the following form
𝑑1 = 𝑘, 𝑑2 = 2𝑠 − 𝑘, 𝑛 = 2𝑠 − 𝑟
𝑑1 = 𝑘, 𝑑2 = 2𝑠 − 𝑘, 𝑛 = 2𝑠 − 𝑘

easy case
𝑑1 = 1, 𝑑2 = 2𝑠 − 1, 𝑛 = 2𝑠 − 1

hard case
𝑑1 = 𝑘, 𝑑2 = 2𝑠 − 𝑘, 𝑛 = 2𝑠 − 𝑟, 𝑟 = 2, … , 𝑘


from the cases 𝑑1 , 𝑑2 , 𝑛 = (1,2𝑠 − 1,2𝑠 −
1) to 𝑘, 2𝑠 − 𝑘, 2𝑠 − 1 -- nontrivial, [?]
from 𝑘, 2𝑠 − 𝑘, 2𝑠 − 𝑟
to 𝑘, 2𝑠+1 − 1,2𝑠+1 − 𝑟 + 1


solve the case 3,5,6 from 3,1,3
given 3,1 -RP of 0,1 3 , (𝑉1 , 𝑉2 ) :
111
011
𝑉1
101
001
𝑉2
110
010
000
100





now construct (3,5)-partition for 0,1
for each 𝑝 = (𝑝1 , 𝑝2 , 𝑝3 ) ∈ 𝑉2
𝐸𝑝 : sys. of equations over 0,1 6
𝑥1 + 𝑥2 = 𝑝1
𝑥3 + 𝑥4 = 𝑝2
𝑥5 + 𝑥6 = 𝑝3
6
𝑠𝑜𝑙 𝐸𝑝 , the collection of solutions of 𝐸𝑝
◦ 𝐸𝑝 is an independent set

0,1
6
= 𝑠𝑜𝑙 𝐸000 ∪ 𝑠𝑜𝑙 𝐸001 ∪ … ∪ 𝑠𝑜𝑙(𝐸111 )
𝑠𝑜𝑙(𝐸011 )
𝑠𝑜𝑙(𝐸001 )
𝑠𝑜𝑙(𝐸000 )
𝑠𝑜𝑙(𝐸101 )
𝑉1
𝑉2
𝑠𝑜𝑙(𝐸010 )
𝑠𝑜𝑙(𝐸000 )
𝑠𝑜𝑙(𝐸110 )
𝑠𝑜𝑙(𝐸100 )



find a perfect matching in 𝑉2
cut each black set by an additional eqn.
for 𝑠𝑜𝑙 𝐸011 and 𝑠𝑜𝑙 𝐸010 use eqn.:
𝑥1 + 𝑥3 + 𝑥5 + 𝑥6 = 0
𝑠𝑜𝑙(𝐸011 )
𝑠𝑜𝑙(𝐸001 )
𝑠𝑜𝑙(𝐸000 )
𝑠𝑜𝑙(𝐸101 )
6 = 2 * the
index of the
different bit
𝑉1
𝑉2
𝑠𝑜𝑙(𝐸010 )
𝑠𝑜𝑙(𝐸000 )
𝑠𝑜𝑙(𝐸110 )
𝑠𝑜𝑙(𝐸100 )



find a perfect matching in 𝑉2
cut each black set by an additional eqn.
for 𝑠𝑜𝑙 𝐸001 and 𝑠𝑜𝑙 𝐸101 use eqn.:
𝑥1 + 𝑥3 + 𝑥5 + 𝑥2 = 0
𝑠𝑜𝑙(𝐸011 )
𝑠𝑜𝑙(𝐸001 )
𝑠𝑜𝑙(𝐸000 )
𝑠𝑜𝑙(𝐸101 )
2 = 2 * the
index of the
different bit
𝑉1
𝑉2
𝑠𝑜𝑙(𝐸010 )
𝑠𝑜𝑙(𝐸000 )
𝑠𝑜𝑙(𝐸110 )
𝑠𝑜𝑙(𝐸100 )

all the grey points → 𝑉2′ ⊂ 0,1 6 , 𝑉1′ = 𝑉2′ 𝐶 .
◦ 𝑉1′ , 𝑉2′ is a 3,5 -RP of 0,1

6
this idea is extendable to general cases
𝑠𝑜𝑙(𝐸011 )
𝑠𝑜𝑙(𝐸001 )
𝑠𝑜𝑙(𝐸000 )
𝑠𝑜𝑙(𝐸101 )
𝑉1
𝑉2
𝑠𝑜𝑙(𝐸010 )
𝑠𝑜𝑙(𝐸000 )
𝑠𝑜𝑙(𝐸110 )
𝑠𝑜𝑙(𝐸100 )

Main contribution:
◦ foy any odd 𝑘, and 0 < 𝑐 ≤ 𝑘, when 𝑠 ≥ 𝑐 +
log 𝑘 − 1, there exists a 𝑘, 2𝑠 − 𝑘 -regular
𝑠 −𝑐
2
partition of 0,1
.
◦ particularly, it follows that for large
enough 𝑛, there exists 𝑘-dominating set
of size
𝑘
2𝑛 ,
𝑛+𝑘
as long as
𝑘
2𝑛
𝑛+𝑘
is integer.