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HMMT 二月 2017 · COMB 赛 · 第 9 题

HMMT February 2017 — COMB Round — Problem 9

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Let m be a positive integer, and let T denote the set of all subsets of { 1 , 2 , . . . , m } . Call a subset Sof T δ - good if for all s , s ∈ S , s 6 = s , | ∆( s , s ) | ≥ δm , where ∆ denotes symmetric difference (the
    1 2 1 2 1 2
    symmetric difference of two sets is the set of elements that is in exactly one of the two sets). Find the
    1024
    largest possible integer s such that there exists an integer m and a -good set of size s .
    2047
解析

英文解析

  1. Let m be a positive integer, and let T denote the set of all subsets of { 1 , 2 , . . . , m } . Call a subset Sof T δ - good if for all s , s ∈ S , s 6 = s , | ∆( s , s ) | ≥ δm , where ∆ denotes symmetric difference (the
    1 2 1 2 1 2
    symmetric difference of two sets is the set of elements that is in exactly one of the two sets). Find the
    1024
    largest possible integer s such that there exists an integer m and a -good set of size s .
    2047
    Proposed by: Yang Liu
    Answer: 2048
    ∑
    Let n = | S | . Let the sets in S be s , s , . . . , s . We bound the sum | ∆( s , s ) | in two ways.
    1 2 n i j
    1 ≤ i<j ≤ n
    On one hand, by the condition we have the obvious bound
    ( )
    ∑
    | ∆( s , s ) | ≥ δm.ni j
    1 ≤ i<j ≤ n 2
    On the other hand, for 1 ≤ i ≤ m , let t = |{ 1 ≤ j ≤ n : i ∈ s }| . Then it is clear thati j
    2 m
    ∑ ∑
    | ∆( s , s ) | = t ( n − t ) ≤ mni j k k
    1 ≤ i<j ≤ n k =14
    by AM-GM. Therefore, we get the bound
    ( )
    n n 2 δ2
    δm ≤ m ⇒ n ≤ = 2048 .
    2 4 2 δ − 1
    To give a construction with n = 2048, take m = 2047 . For the rest of this construction, we will beinterpreting the integers 1 , 2 , . . . , m as 11-digit integers in binary. Given this interpretation, define adot product x y of two positive integers 0 ≤ x, y ≤ m the following way. If x = ( x x . . . x ) , y =
    1 2 11 2
    ( y y . . . y ) in binary, then
    1 2 11 2
    ∑
    x y = x y (mod 2) .
    i i
    Now we can define the sets s , s , . . . , s . Define
    1 2 2048
    s = { 1 ≤ j ≤ m : ( i − 1) j = 1 . } .
    A computation shows that this construction works.i
    Some notes: here is the motivation behind the construction. We are treating the integers 0 , 1 , . . . , m asthe vector space V = F , and the sets s correspond to linear functionals f : V → F . In particular,11
    i i 2
    the function f : V → F is simply defined as f ( x ) = ( i − 1) x , which one can easily check to be 2
    i 2 ilinear. This construction corresponds to Hadamard matrices of size 2 .11