Let $ngeq 1$ and B a group of elements $(i _1,i_2,…,i_n)$ so $i_j in {-1,1}$. Let’s assume $...












1












$begingroup$


Let $ngeq 1$ and B a group of elements $(i _1,i_2,…,i_n)$ so $i_j in {-1,1}$. Let’s assume $ |B|>dfrac{2^{n+1}}{n}$. Show that exists 3 diffrent elements in B so every 2 of them differ from each other in exactly 2 coordinates( 3 points in B are vertexes of a Equilateral triangle with length √8)



Tried to solve with pigeon hole principle but it doesn't really work.I really not sure what to even try to use in the question. If someone have any leads how to try and solve it will be really helpful






combinatorics discrete-mathematics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please use MathJax in future. I suggest you edit the question too.
    $endgroup$
    – Shaun
    Jan 1 at 14:36






  • 1




    $begingroup$
    @IdanDaniel can you explain < מ+! > or rewrite it? And please check if $(מ+!)/n$ is the right exponent. It was not clear to me.
    $endgroup$
    – user376343
    Jan 1 at 14:44










  • $begingroup$
    sorry, i translated it from another language. edited it
    $endgroup$
    – Idan Daniel
    Jan 1 at 14:50










  • $begingroup$
    The title still needs editing, @IdanDaniel.
    $endgroup$
    – Shaun
    Jan 1 at 15:06










  • $begingroup$
    now i edited the title too
    $endgroup$
    – Idan Daniel
    Jan 1 at 15:31
















1












$begingroup$


Let $ngeq 1$ and B a group of elements $(i _1,i_2,…,i_n)$ so $i_j in {-1,1}$. Let’s assume $ |B|>dfrac{2^{n+1}}{n}$. Show that exists 3 diffrent elements in B so every 2 of them differ from each other in exactly 2 coordinates( 3 points in B are vertexes of a Equilateral triangle with length √8)



Tried to solve with pigeon hole principle but it doesn't really work.I really not sure what to even try to use in the question. If someone have any leads how to try and solve it will be really helpful






combinatorics discrete-mathematics







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please use MathJax in future. I suggest you edit the question too.
    $endgroup$
    – Shaun
    Jan 1 at 14:36






  • 1




    $begingroup$
    @IdanDaniel can you explain < מ+! > or rewrite it? And please check if $(מ+!)/n$ is the right exponent. It was not clear to me.
    $endgroup$
    – user376343
    Jan 1 at 14:44










  • $begingroup$
    sorry, i translated it from another language. edited it
    $endgroup$
    – Idan Daniel
    Jan 1 at 14:50










  • $begingroup$
    The title still needs editing, @IdanDaniel.
    $endgroup$
    – Shaun
    Jan 1 at 15:06










  • $begingroup$
    now i edited the title too
    $endgroup$
    – Idan Daniel
    Jan 1 at 15:31














1












1








1


1



$begingroup$


Let $ngeq 1$ and B a group of elements $(i _1,i_2,…,i_n)$ so $i_j in {-1,1}$. Let’s assume $ |B|>dfrac{2^{n+1}}{n}$. Show that exists 3 diffrent elements in B so every 2 of them differ from each other in exactly 2 coordinates( 3 points in B are vertexes of a Equilateral triangle with length √8)



Tried to solve with pigeon hole principle but it doesn't really work.I really not sure what to even try to use in the question. If someone have any leads how to try and solve it will be really helpful






combinatorics discrete-mathematics







share|cite|improve this question











$endgroup$




Let $ngeq 1$ and B a group of elements $(i _1,i_2,…,i_n)$ so $i_j in {-1,1}$. Let’s assume $ |B|>dfrac{2^{n+1}}{n}$. Show that exists 3 diffrent elements in B so every 2 of them differ from each other in exactly 2 coordinates( 3 points in B are vertexes of a Equilateral triangle with length √8)



Tried to solve with pigeon hole principle but it doesn't really work.I really not sure what to even try to use in the question. If someone have any leads how to try and solve it will be really helpful






combinatorics discrete-mathematics




combinatorics discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 16:12









Martin Sleziak

44.8k10119272




44.8k10119272










asked Jan 1 at 14:31









Idan DanielIdan Daniel

347




347












  • $begingroup$
    Please use MathJax in future. I suggest you edit the question too.
    $endgroup$
    – Shaun
    Jan 1 at 14:36






  • 1




    $begingroup$
    @IdanDaniel can you explain < מ+! > or rewrite it? And please check if $(מ+!)/n$ is the right exponent. It was not clear to me.
    $endgroup$
    – user376343
    Jan 1 at 14:44










  • $begingroup$
    sorry, i translated it from another language. edited it
    $endgroup$
    – Idan Daniel
    Jan 1 at 14:50










  • $begingroup$
    The title still needs editing, @IdanDaniel.
    $endgroup$
    – Shaun
    Jan 1 at 15:06










  • $begingroup$
    now i edited the title too
    $endgroup$
    – Idan Daniel
    Jan 1 at 15:31


















  • $begingroup$
    Please use MathJax in future. I suggest you edit the question too.
    $endgroup$
    – Shaun
    Jan 1 at 14:36






  • 1




    $begingroup$
    @IdanDaniel can you explain < מ+! > or rewrite it? And please check if $(מ+!)/n$ is the right exponent. It was not clear to me.
    $endgroup$
    – user376343
    Jan 1 at 14:44










  • $begingroup$
    sorry, i translated it from another language. edited it
    $endgroup$
    – Idan Daniel
    Jan 1 at 14:50










  • $begingroup$
    The title still needs editing, @IdanDaniel.
    $endgroup$
    – Shaun
    Jan 1 at 15:06










  • $begingroup$
    now i edited the title too
    $endgroup$
    – Idan Daniel
    Jan 1 at 15:31
















$begingroup$
Please use MathJax in future. I suggest you edit the question too.
$endgroup$
– Shaun
Jan 1 at 14:36




$begingroup$
Please use MathJax in future. I suggest you edit the question too.
$endgroup$
– Shaun
Jan 1 at 14:36




1




1




$begingroup$
@IdanDaniel can you explain < מ+! > or rewrite it? And please check if $(מ+!)/n$ is the right exponent. It was not clear to me.
$endgroup$
– user376343
Jan 1 at 14:44




$begingroup$
@IdanDaniel can you explain < מ+! > or rewrite it? And please check if $(מ+!)/n$ is the right exponent. It was not clear to me.
$endgroup$
– user376343
Jan 1 at 14:44












$begingroup$
sorry, i translated it from another language. edited it
$endgroup$
– Idan Daniel
Jan 1 at 14:50




$begingroup$
sorry, i translated it from another language. edited it
$endgroup$
– Idan Daniel
Jan 1 at 14:50












$begingroup$
The title still needs editing, @IdanDaniel.
$endgroup$
– Shaun
Jan 1 at 15:06




$begingroup$
The title still needs editing, @IdanDaniel.
$endgroup$
– Shaun
Jan 1 at 15:06












$begingroup$
now i edited the title too
$endgroup$
– Idan Daniel
Jan 1 at 15:31




$begingroup$
now i edited the title too
$endgroup$
– Idan Daniel
Jan 1 at 15:31










1 Answer
1






active

oldest

votes


















0












$begingroup$

It can be solved using double counting method and pigeonhole principle. Define for $v=(v_i)_{i=1}^n$ and $ w=(w_i)_{i=1}^n$ in $V:={-1,1}^n$,
$$
d(v,w) = |{i ;|;v_i neq w_i}|.
$$ That is, $d(v,w)$ is the number of coordinates $i$ such that $v_i neq w_i$. We will show there exist $vin V$ and distinct $w_i in B$, $i=1,2,3$ such that $d(v,w_i) = 1$ for $i=1,2,3$. Then it follows $d(w_i, w_j) = 2$ for all $ineq j$ as we wanted.



Proof: Let us define $$X= {(v,w)in Vtimes B;|;d(v,w) =1}.
$$ We will count $|X|$ using double counting method. First, note that
$$
|X| = sum_{(v,w)in X}1 = sum_{vin V} |{win B;|; d(v,w) = 1}|.
$$ On the other hand, we have
$$
|X| = sum_{w in B} |{vin V ;|; d(v,w) =1}| =sum_{w in B} n = n|B|>2^{n+1}.
$$ Since $|V| = 2^n$, this gives us
$$
frac{1}{|V|}sum_{vin V} |{win B;|; d(v,w) = 1}|>2.
$$ By pigeonhole principle, it follows that there exists $vin V$ such that
$$
|{win B;|; d(v,w) = 1}|ge 3.
$$ Thus, there exist distinct $w_i, i=1,2,3$ in $B$ such that $d(v,w_i) = 1$. This proves the claim.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for the detailed proof! it exactly what i needed
    $endgroup$
    – Idan Daniel
    Jan 1 at 18:51










  • $begingroup$
    I'm glad it helped :)
    $endgroup$
    – Song
    Jan 1 at 18:54










  • $begingroup$
    For each $w$, $v$ for which $d(v,w)=1$ is obtained by changing sign of $j$-th coordinate $w_j$ of $w$ and leaving the rest. We can take $j = 1,2,ldots, n$. Thus there are exactly $n$ number of $v$'s for which $d(v,w)=1$.
    $endgroup$
    – Song
    Jan 1 at 19:08










  • $begingroup$
    got it, thanks!
    $endgroup$
    – Idan Daniel
    Jan 1 at 19:53











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1 Answer
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1 Answer
1






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active

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active

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0












$begingroup$

It can be solved using double counting method and pigeonhole principle. Define for $v=(v_i)_{i=1}^n$ and $ w=(w_i)_{i=1}^n$ in $V:={-1,1}^n$,
$$
d(v,w) = |{i ;|;v_i neq w_i}|.
$$ That is, $d(v,w)$ is the number of coordinates $i$ such that $v_i neq w_i$. We will show there exist $vin V$ and distinct $w_i in B$, $i=1,2,3$ such that $d(v,w_i) = 1$ for $i=1,2,3$. Then it follows $d(w_i, w_j) = 2$ for all $ineq j$ as we wanted.



Proof: Let us define $$X= {(v,w)in Vtimes B;|;d(v,w) =1}.
$$ We will count $|X|$ using double counting method. First, note that
$$
|X| = sum_{(v,w)in X}1 = sum_{vin V} |{win B;|; d(v,w) = 1}|.
$$ On the other hand, we have
$$
|X| = sum_{w in B} |{vin V ;|; d(v,w) =1}| =sum_{w in B} n = n|B|>2^{n+1}.
$$ Since $|V| = 2^n$, this gives us
$$
frac{1}{|V|}sum_{vin V} |{win B;|; d(v,w) = 1}|>2.
$$ By pigeonhole principle, it follows that there exists $vin V$ such that
$$
|{win B;|; d(v,w) = 1}|ge 3.
$$ Thus, there exist distinct $w_i, i=1,2,3$ in $B$ such that $d(v,w_i) = 1$. This proves the claim.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for the detailed proof! it exactly what i needed
    $endgroup$
    – Idan Daniel
    Jan 1 at 18:51










  • $begingroup$
    I'm glad it helped :)
    $endgroup$
    – Song
    Jan 1 at 18:54










  • $begingroup$
    For each $w$, $v$ for which $d(v,w)=1$ is obtained by changing sign of $j$-th coordinate $w_j$ of $w$ and leaving the rest. We can take $j = 1,2,ldots, n$. Thus there are exactly $n$ number of $v$'s for which $d(v,w)=1$.
    $endgroup$
    – Song
    Jan 1 at 19:08










  • $begingroup$
    got it, thanks!
    $endgroup$
    – Idan Daniel
    Jan 1 at 19:53
















0












$begingroup$

It can be solved using double counting method and pigeonhole principle. Define for $v=(v_i)_{i=1}^n$ and $ w=(w_i)_{i=1}^n$ in $V:={-1,1}^n$,
$$
d(v,w) = |{i ;|;v_i neq w_i}|.
$$ That is, $d(v,w)$ is the number of coordinates $i$ such that $v_i neq w_i$. We will show there exist $vin V$ and distinct $w_i in B$, $i=1,2,3$ such that $d(v,w_i) = 1$ for $i=1,2,3$. Then it follows $d(w_i, w_j) = 2$ for all $ineq j$ as we wanted.



Proof: Let us define $$X= {(v,w)in Vtimes B;|;d(v,w) =1}.
$$ We will count $|X|$ using double counting method. First, note that
$$
|X| = sum_{(v,w)in X}1 = sum_{vin V} |{win B;|; d(v,w) = 1}|.
$$ On the other hand, we have
$$
|X| = sum_{w in B} |{vin V ;|; d(v,w) =1}| =sum_{w in B} n = n|B|>2^{n+1}.
$$ Since $|V| = 2^n$, this gives us
$$
frac{1}{|V|}sum_{vin V} |{win B;|; d(v,w) = 1}|>2.
$$ By pigeonhole principle, it follows that there exists $vin V$ such that
$$
|{win B;|; d(v,w) = 1}|ge 3.
$$ Thus, there exist distinct $w_i, i=1,2,3$ in $B$ such that $d(v,w_i) = 1$. This proves the claim.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot for the detailed proof! it exactly what i needed
    $endgroup$
    – Idan Daniel
    Jan 1 at 18:51










  • $begingroup$
    I'm glad it helped :)
    $endgroup$
    – Song
    Jan 1 at 18:54










  • $begingroup$
    For each $w$, $v$ for which $d(v,w)=1$ is obtained by changing sign of $j$-th coordinate $w_j$ of $w$ and leaving the rest. We can take $j = 1,2,ldots, n$. Thus there are exactly $n$ number of $v$'s for which $d(v,w)=1$.
    $endgroup$
    – Song
    Jan 1 at 19:08










  • $begingroup$
    got it, thanks!
    $endgroup$
    – Idan Daniel
    Jan 1 at 19:53














0












0








0





$begingroup$

It can be solved using double counting method and pigeonhole principle. Define for $v=(v_i)_{i=1}^n$ and $ w=(w_i)_{i=1}^n$ in $V:={-1,1}^n$,
$$
d(v,w) = |{i ;|;v_i neq w_i}|.
$$ That is, $d(v,w)$ is the number of coordinates $i$ such that $v_i neq w_i$. We will show there exist $vin V$ and distinct $w_i in B$, $i=1,2,3$ such that $d(v,w_i) = 1$ for $i=1,2,3$. Then it follows $d(w_i, w_j) = 2$ for all $ineq j$ as we wanted.



Proof: Let us define $$X= {(v,w)in Vtimes B;|;d(v,w) =1}.
$$ We will count $|X|$ using double counting method. First, note that
$$
|X| = sum_{(v,w)in X}1 = sum_{vin V} |{win B;|; d(v,w) = 1}|.
$$ On the other hand, we have
$$
|X| = sum_{w in B} |{vin V ;|; d(v,w) =1}| =sum_{w in B} n = n|B|>2^{n+1}.
$$ Since $|V| = 2^n$, this gives us
$$
frac{1}{|V|}sum_{vin V} |{win B;|; d(v,w) = 1}|>2.
$$ By pigeonhole principle, it follows that there exists $vin V$ such that
$$
|{win B;|; d(v,w) = 1}|ge 3.
$$ Thus, there exist distinct $w_i, i=1,2,3$ in $B$ such that $d(v,w_i) = 1$. This proves the claim.






share|cite|improve this answer











$endgroup$



It can be solved using double counting method and pigeonhole principle. Define for $v=(v_i)_{i=1}^n$ and $ w=(w_i)_{i=1}^n$ in $V:={-1,1}^n$,
$$
d(v,w) = |{i ;|;v_i neq w_i}|.
$$ That is, $d(v,w)$ is the number of coordinates $i$ such that $v_i neq w_i$. We will show there exist $vin V$ and distinct $w_i in B$, $i=1,2,3$ such that $d(v,w_i) = 1$ for $i=1,2,3$. Then it follows $d(w_i, w_j) = 2$ for all $ineq j$ as we wanted.



Proof: Let us define $$X= {(v,w)in Vtimes B;|;d(v,w) =1}.
$$ We will count $|X|$ using double counting method. First, note that
$$
|X| = sum_{(v,w)in X}1 = sum_{vin V} |{win B;|; d(v,w) = 1}|.
$$ On the other hand, we have
$$
|X| = sum_{w in B} |{vin V ;|; d(v,w) =1}| =sum_{w in B} n = n|B|>2^{n+1}.
$$ Since $|V| = 2^n$, this gives us
$$
frac{1}{|V|}sum_{vin V} |{win B;|; d(v,w) = 1}|>2.
$$ By pigeonhole principle, it follows that there exists $vin V$ such that
$$
|{win B;|; d(v,w) = 1}|ge 3.
$$ Thus, there exist distinct $w_i, i=1,2,3$ in $B$ such that $d(v,w_i) = 1$. This proves the claim.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 1 at 18:22

























answered Jan 1 at 17:22









SongSong

15.9k1739




15.9k1739












  • $begingroup$
    Thanks a lot for the detailed proof! it exactly what i needed
    $endgroup$
    – Idan Daniel
    Jan 1 at 18:51










  • $begingroup$
    I'm glad it helped :)
    $endgroup$
    – Song
    Jan 1 at 18:54










  • $begingroup$
    For each $w$, $v$ for which $d(v,w)=1$ is obtained by changing sign of $j$-th coordinate $w_j$ of $w$ and leaving the rest. We can take $j = 1,2,ldots, n$. Thus there are exactly $n$ number of $v$'s for which $d(v,w)=1$.
    $endgroup$
    – Song
    Jan 1 at 19:08










  • $begingroup$
    got it, thanks!
    $endgroup$
    – Idan Daniel
    Jan 1 at 19:53


















  • $begingroup$
    Thanks a lot for the detailed proof! it exactly what i needed
    $endgroup$
    – Idan Daniel
    Jan 1 at 18:51










  • $begingroup$
    I'm glad it helped :)
    $endgroup$
    – Song
    Jan 1 at 18:54










  • $begingroup$
    For each $w$, $v$ for which $d(v,w)=1$ is obtained by changing sign of $j$-th coordinate $w_j$ of $w$ and leaving the rest. We can take $j = 1,2,ldots, n$. Thus there are exactly $n$ number of $v$'s for which $d(v,w)=1$.
    $endgroup$
    – Song
    Jan 1 at 19:08










  • $begingroup$
    got it, thanks!
    $endgroup$
    – Idan Daniel
    Jan 1 at 19:53
















$begingroup$
Thanks a lot for the detailed proof! it exactly what i needed
$endgroup$
– Idan Daniel
Jan 1 at 18:51




$begingroup$
Thanks a lot for the detailed proof! it exactly what i needed
$endgroup$
– Idan Daniel
Jan 1 at 18:51












$begingroup$
I'm glad it helped :)
$endgroup$
– Song
Jan 1 at 18:54




$begingroup$
I'm glad it helped :)
$endgroup$
– Song
Jan 1 at 18:54












$begingroup$
For each $w$, $v$ for which $d(v,w)=1$ is obtained by changing sign of $j$-th coordinate $w_j$ of $w$ and leaving the rest. We can take $j = 1,2,ldots, n$. Thus there are exactly $n$ number of $v$'s for which $d(v,w)=1$.
$endgroup$
– Song
Jan 1 at 19:08




$begingroup$
For each $w$, $v$ for which $d(v,w)=1$ is obtained by changing sign of $j$-th coordinate $w_j$ of $w$ and leaving the rest. We can take $j = 1,2,ldots, n$. Thus there are exactly $n$ number of $v$'s for which $d(v,w)=1$.
$endgroup$
– Song
Jan 1 at 19:08












$begingroup$
got it, thanks!
$endgroup$
– Idan Daniel
Jan 1 at 19:53




$begingroup$
got it, thanks!
$endgroup$
– Idan Daniel
Jan 1 at 19:53


















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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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