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

45.1k10123277




45.1k10123277










asked Jan 1 at 14:31









Idan DanielIdan Daniel

397




397












  • $begingroup$
    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






active

oldest

votes









active

oldest

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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
















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

18.6k21651




18.6k21651












  • $begingroup$
    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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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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