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Difference between revisions of "2015 USAMO Problems/Problem 3"

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The empty set is denoted as <math>\varnothing</math>, <math>\cap</math> denotes intersection, and <math>\cup</math> denotes union. Let <math>S_n=\{n\}</math> are one-element subsets.
 
The empty set is denoted as <math>\varnothing</math>, <math>\cap</math> denotes intersection, and <math>\cup</math> denotes union. Let <math>S_n=\{n\}</math> are one-element subsets.
  
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Let mCk = <math>\dbinom{m}{k} = \frac{m!}{k!(m-k)!}</math> denote m choose k.
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Let <math>m_{c_k}=\dbinom{m}{k} = \frac{m!}{k!(m-k)!}</math> denote m choose k.
  
  

Revision as of 10:08, 10 August 2020

Problem

Let $S = \{1, 2, ..., n\}$, where $n \ge 1$. Each of the $2^n$ subsets of $S$ is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set $T \subseteq S$, we then write $f(T)$ for the number of subsets of T that are blue.

Determine the number of colorings that satisfy the following condition: for any subsets $T_1$ and $T_2$ of $S$, \[f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2).\]

Solution

Define function: $C(T)=1$ if the set T is colored blue, and $C(T)=0$ if $T$ is colored red. Define the $\text{Core} =\text{intersection of all } T \text{ where } C(T)=1$.

The empty set is denoted as $\varnothing$, $\cap$ denotes intersection, and $\cup$ denotes union. Let $S_n=\{n\}$ are one-element subsets.

Let $m_{c_k}=\dbinom{m}{k} = \frac{m!}{k!(m-k)!}$ denote m choose k.


(Case I) $f(\null)=1$. Then for distinct m and k, $f(S_m \cup S_k)=f(S_m)f(S_k)$, meaning only if $S_m$ and $S_k$ are both blue is their union blue. Namely $C(S_m \cup S_k)=C(S_m)C(S_k).$

Similarly, for distinct $m,n,k$, $f(S_m \cup S_k \cup Sn)=f(S_m \cup S_k)f(S_n)$, $C(S_m \cup S_k \cup S_n)=C(S_m)C(S_k)C(S_n)$. This procedure of determination continues to $S$. Therefore, if $T=\{a_1,a_2, \cdots a_k\}$, then $C(T)=C(S_{a_1})C(S_{a_2}) \cdots C(S_{a_k})$. All colorings thus determined by the free colors chosen for subsets of one single elements satisfy the condition. There are $2^n$ colorings in this case.

(Case II.) $f(\varnothing)=0$.

(Case II.1) $\text{Core}=\varnothing$. Then either (II.1.1) there exist two nonintersecting subsets A and B, $C(A)=C(B)=1$, but f$(A)f(B)=0$, a contradiction, or (II.1.2) all subsets has $C(T)=0$, which is easily confirmed to satisfy the condition $f(T_1)f(T_2)=f(T_1 \cap T_2)f(T_1 \cup T_2)$. There is one coloring in this case.

(Case II.2) Core = a subset of 1 element. WLOG, C(S_1)=1. Then $f(S_1)=1$, and subsets containing element 1 may be colored Blue. $f(S_1 \cup S_m)f(S_1\cup S_n)=f(S_1 \cup S_m \cup S_n)$, namely $C(S_1 \cup S_m \cup S_n)=C(S_m \cup S_1)C(S_n \cup S_1)$. Now S_1 functions as the $\varnothing$ in case I, with $n-1$ elements to combine into a base of $n-1$ two-element sets, and all the other subsets are determined. There are $2^{n-1}$ colorings for each choice of core. However, there are nC1 = n such cores. Hence altogether there are $n2^{n-1}$ colorings in this case.

(Case II.3) Core = a subset of 2 elements. WLOG, let $C(S_1 \cup S_2)=1$. Only subsets containing the core may be colored blue. With the same reasoning as in the preceding case, there are $(nC2)2^{n-2}$ colorings.

$\dots$

(Case II.n+1) Core = S. Then $C(S)=1$, with all other subsets $C(T)=0$, there is $1=\dbinom{n}{n}2^0$

Combining all the cases, we have $1+\left[2^n+\dbinom{n}{1}2^{n-1}+\dbinom{n}{2}2^{n-2}+ \cdots + \dbinom{n}{n}2^0\right]=\boxed{1+3^n}$ colorings.

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