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

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Define the <math>\text{Core} =\text{intersection of all } T \text{ where } C(T)=1</math>.  
 
Define the <math>\text{Core} =\text{intersection of all } T \text{ where } C(T)=1</math>.  
  
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The empty set is denoted as <math>\varnothing</math>, <math>\cap</math> denotes intersection, and <math>\cup</math> denotes union. T1<T2 means T1 is a subset of T2 but not =T2.
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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.
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Let <math>Sn={n}</math> are one-element subsets.
 
  
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Let mCk denote m choose k = <math>\frac{m!}{k!(m-k)!}</math>
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Let <math>m_{c_k}=\dbinom{m}{k} = \frac{m!}{k!(m-k)!}</math> denote m choose k.
  
  
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(Case I) <math>f(\null)=1</math>. Then for distinct m and k, <math>f(Sm+Sk)=f(Sm)f(Sk)</math>, meaning only if Sm and Sk are both blue is their union blue. Namely <math>C(Sm+Sk)=C(Sm)C(Sk).</math>
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(Case I) <math>f(\varnothing)=1</math>. Then for distinct m and k, <math>f(S_m \cup S_k)=f(S_m)f(S_k)</math>, meaning only if <math>S_m</math> and <math>S_k</math> are both blue is their union blue. Namely <math>C(S_m \cup S_k)=C(S_m)C(S_k).</math>
  
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Similarly, for distinct m,n,k, f(Sm \cup Sk \cup Sn)=f(Sm \cup Sk)f(Sn), C(Sm \cup Sk \cup Sn)=C(Sm)C(Sk)C(Sn). This procedure of determination continues to S. Therefore, if <math>T={a_1,a_2, \cdots a_k}</math>, then <math>C(T)=C(Sa1)C(Sa2)...C(Sak)</math>. 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.  
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Similarly, for distinct <math>m,n,k</math>, <math>f(S_m \cup S_k \cup Sn)=f(S_m \cup S_k)f(S_n)</math>, <math>C(S_m \cup S_k \cup S_n)=C(S_m)C(S_k)C(S_n)</math>. This procedure of determination continues to <math>S</math>. Therefore, if <math>T=\{a_1,a_2, \cdots a_k\}</math>, then <math>C(T)=C(S_{a_1})C(S_{a_2}) \cdots C(S_{a_k})</math>. All colorings thus determined by the free colors chosen for subsets of one single elements satisfy the condition.  There are <math>2^n</math> colorings in this case.  
  
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(Case II.)  f(<math>\varnothing</math>)=0.  
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(Case II.)  <math>f(\varnothing)=0</math>.
  
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(Case II.1)  Core=<math>\varnothing</math>. Then either (II.1.1) there exist two nonintersecting subsets A and B, <math>C(A)=C(B)=1</math>, but f<math>(A)f(B)=0</math> which is a contradiction, or (II.1.2) all subsets has C(T)=0, which is easily confirmed to satisfy the condition <math>f(T1)f(T2)=f(T1 \cap T2)f(T1 \cup T2)</math>.  There is one coloring in this case.
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(Case II.1)  <math>\text{Core}=\varnothing</math>. Then either (II.1.1) there exist two nonintersecting subsets A and B, <math>C(A)=C(B)=1</math>, but f<math>(A)f(B)=0</math>, a contradiction, or (II.1.2) all subsets has <math>C(T)=0</math>, which is easily confirmed to satisfy the condition <math>f(T_1)f(T_2)=f(T_1 \cap T_2)f(T_1 \cup T_2)</math>.  There is one coloring in this case.
  
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(Case II.2) Core= a subset of 1 element. WOLG, C(S1)=1. Then f(S1)=1, and subsets containing element 1 may be colored Blue.  
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(Case II.2) Core = a subset of 1 element. WLOG, <math>C(S_1)=1</math>. Then <math>f(S_1)=1</math>, and subsets containing element 1 may be colored blue. <math>f(S_1 \cup S_m)f(S_1\cup S_n)=f(S_1 \cup S_m \cup S_n)</math>, namely <math>C(S_1 \cup S_m \cup S_n)=C(S_m \cup S_1)C(S_n \cup S_1)</math>. Now S_1 functions as the <math>\varnothing</math> in case I, with <math>n-1</math> elements to combine into a base of <math>n-1</math> two-element sets, and all the other subsets are determined. There are <math>2^{n-1}</math> colorings for each choice of core. However, there are nC1 = n such cores. Hence altogether there are <math>n2^{n-1}</math> colorings in this case.
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<math>f(S1 \cup Sm)f(S1\cup Sn)=f(S1 \cup Sm \cup Sn)</math>, namely C(S1 \cup Sm \cup Sn)=C(Sm \cup S1)C(Sn \cup S1). Now S1 functions as the <math>\varnothing</math> in case I, with n-1 elements to combine into a base of n-1 2-element sets, and all the other subsets are determined. There are 2^(n-1) legit colorings for each choice of core. But there are nC1 (i.e. n choose 1) = n such cores. Hence altogether there are <math>n2^(n-1)</math> colorings in this case.
 
  
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(Case II.3) Core = a subset of 2 elements. WLOG, C(S1+S2)=1. Only subsets containing the core may be colored Blue. With the same reasoning as in the preceding case, there are <math>(nC2)2^(n-2)</math> colorings.
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(Case II.3) Core = a subset of 2 elements. WLOG, let <math>C(S_1 \cup S_2)=1</math>. Only subsets containing the core may be colored blue. With the same reasoning as in the preceding case, there are <math>(nC2)2^{n-2}</math> colorings.
  
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... (Case II.n+1) Core = S. Then C(S)=1, with all other subsets <math>C(T)=0</math>, there is <math>1=(nCn)2^0</math>
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<math>\dots</math>
  
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Combining all the cases, <math>1+[1+(\dbinom{n}{1})2^{n-1}+(\dbinom{n}{2})2^{n-2}+\cdot \cdot \cdot +(\dbinom{n}{n})2^0]=1+3^n</math> is the total number of colorings satisfying the given condition.
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(Case II.n+1) Core = S. Then <math>C(S)=1</math>, with all other subsets <math>C(T)=0</math>, there is <math>1=\dbinom{n}{n}2^0</math>
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Combining all the cases, we have <math>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}</math> colorings. sponsored by ALLEN

Latest revision as of 04:55, 14 April 2023

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) $f(\varnothing)=1$. 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. sponsored by ALLEN

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