Let $ngeq 1$ and B a group of elements $(i _1,i_2,…,i_n)$ so $i_j in {-1,1}$. Let’s assume $...
$begingroup$
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
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Jan 1 at 14:31
Idan DanielIdan Daniel
347
$endgroup$
add a comment |
$begingroup$
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
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Jan 1 at 14:31
Idan DanielIdan Daniel
347
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Jan 1 at 14:31
Idan DanielIdan Daniel
347
$endgroup$
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
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Jan 1 at 14:31
Idan DanielIdan Daniel
347
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
asked Jan 1 at 14:31
Idan DanielIdan Daniel
347
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
edited Jan 1 at 16:12
Martin Sleziak
44.8k10119272
44.8k10119272
asked Jan 1 at 14:31
Idan DanielIdan Daniel
347
asked Jan 1 at 14:31
Idan DanielIdan Daniel
347
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 1 at 17:22
SongSong
15.9k1739
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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.
answered Jan 1 at 17:22
SongSong
15.9k1739
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 1 at 17:22
SongSong
15.9k1739
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 1 at 17:22
SongSong
15.9k1739
$endgroup$
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.
answered Jan 1 at 17:22
SongSong
15.9k1739
edited Jan 1 at 18:22
answered Jan 1 at 17:22
SongSong
15.9k1739
answered Jan 1 at 17:22
SongSong
15.9k1739
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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