The Clique vs. Independent Set Problem












2












$begingroup$


Suppose you have an undirected graph $G = (V, E)$, known to both Alice and Bob, Alice gets an independent set of $G$. Bob gets a Clique $B ⊆ V$.



Is there any algorithm in $O(log^2 n)$ bits that finds whether
$ A ∩ B = Ø $?



This is a well known communication complexity problem called CIS problem that was described by Yannakakis.





  • Lecture notes; the claim is Theorem 3

  • Link to Nisan & Kushilevitz's textbook


I'm not sure why and how does this work exactly. and which part of the $n/2$ vertices are reduced by both players.



P.S. I came to a conclusion that an independent and a clique can intersect in at most one vertex.






algorithms graphs communication-complexity







share|cite|improve this question











$endgroup$












  • $begingroup$
    The lecture notes contain a complete proof.
    $endgroup$
    – Yuval Filmus
    13 hours ago










  • $begingroup$
    I did not understand the algorithm completely to understand the proof itself.
    $endgroup$
    – Jay
    12 hours ago
















2












$begingroup$


Suppose you have an undirected graph $G = (V, E)$, known to both Alice and Bob, Alice gets an independent set of $G$. Bob gets a Clique $B ⊆ V$.



Is there any algorithm in $O(log^2 n)$ bits that finds whether
$ A ∩ B = Ø $?



This is a well known communication complexity problem called CIS problem that was described by Yannakakis.





  • Lecture notes; the claim is Theorem 3

  • Link to Nisan & Kushilevitz's textbook


I'm not sure why and how does this work exactly. and which part of the $n/2$ vertices are reduced by both players.



P.S. I came to a conclusion that an independent and a clique can intersect in at most one vertex.






algorithms graphs communication-complexity







share|cite|improve this question











$endgroup$












  • $begingroup$
    The lecture notes contain a complete proof.
    $endgroup$
    – Yuval Filmus
    13 hours ago










  • $begingroup$
    I did not understand the algorithm completely to understand the proof itself.
    $endgroup$
    – Jay
    12 hours ago














2












2








2





$begingroup$


Suppose you have an undirected graph $G = (V, E)$, known to both Alice and Bob, Alice gets an independent set of $G$. Bob gets a Clique $B ⊆ V$.



Is there any algorithm in $O(log^2 n)$ bits that finds whether
$ A ∩ B = Ø $?



This is a well known communication complexity problem called CIS problem that was described by Yannakakis.





  • Lecture notes; the claim is Theorem 3

  • Link to Nisan & Kushilevitz's textbook


I'm not sure why and how does this work exactly. and which part of the $n/2$ vertices are reduced by both players.



P.S. I came to a conclusion that an independent and a clique can intersect in at most one vertex.






algorithms graphs communication-complexity







share|cite|improve this question











$endgroup$




Suppose you have an undirected graph $G = (V, E)$, known to both Alice and Bob, Alice gets an independent set of $G$. Bob gets a Clique $B ⊆ V$.



Is there any algorithm in $O(log^2 n)$ bits that finds whether
$ A ∩ B = Ø $?



This is a well known communication complexity problem called CIS problem that was described by Yannakakis.





  • Lecture notes; the claim is Theorem 3

  • Link to Nisan & Kushilevitz's textbook


I'm not sure why and how does this work exactly. and which part of the $n/2$ vertices are reduced by both players.



P.S. I came to a conclusion that an independent and a clique can intersect in at most one vertex.






algorithms graphs communication-complexity




algorithms graphs communication-complexity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 12 hours ago









Yuval Filmus

196k15184349




196k15184349










asked 13 hours ago









JayJay

1415




1415












  • $begingroup$
    The lecture notes contain a complete proof.
    $endgroup$
    – Yuval Filmus
    13 hours ago










  • $begingroup$
    I did not understand the algorithm completely to understand the proof itself.
    $endgroup$
    – Jay
    12 hours ago


















  • $begingroup$
    The lecture notes contain a complete proof.
    $endgroup$
    – Yuval Filmus
    13 hours ago










  • $begingroup$
    I did not understand the algorithm completely to understand the proof itself.
    $endgroup$
    – Jay
    12 hours ago
















$begingroup$
The lecture notes contain a complete proof.
$endgroup$
– Yuval Filmus
13 hours ago




$begingroup$
The lecture notes contain a complete proof.
$endgroup$
– Yuval Filmus
13 hours ago












$begingroup$
I did not understand the algorithm completely to understand the proof itself.
$endgroup$
– Jay
12 hours ago




$begingroup$
I did not understand the algorithm completely to understand the proof itself.
$endgroup$
– Jay
12 hours ago










1 Answer
1






active

oldest

votes


















4












$begingroup$

The two players construct a sequence $V_0 supset V_1 supset cdots supset V_m$ of sets of vertices such that:





  1. $V_0$ consists of all vertices in the graph.


  2. $|V_{i+1}| leq (|V_i|+1)/2$.


  3. $V_i supseteq C cap I$.


The players stop once $|V_m| leq 1$. At this point they can answer the question using $O(1)$ communication.



At round $i$, the players know $V_{i-1}$, and wish to construct $V_i$. They act as follows:




  • If $C cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ neighbors in $V_{i-1}$, then Alice sends Bob one such vertex $v$, and both players set $V_i$ to be this set of neighbors, together with $v$ (this is valid since $C cap I subseteq C subseteq V_i$). Otherwise, she sends $bot$.


  • If Alice sent $bot$ and $I cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ non-neighbors in $V_{i-1}$, then Bob sends Alice one such vertex $v$, and both players set $V_i$ to be this set of non-neighbors, together with $v$ (this is valid since $C cap I subseteq I subseteq V_i$). Otherwise, he sends Alice $bot$.


  • If both players sent $bot$, then $C cap I = emptyset$. Indeed, if $v in C cap I$, then $v$ has at least $|V_{i-1}|/2$ neighbors and at least $|V_{i-1}|/2$ non-neighbors inside $|V_{i-1}|$, whereas the number of potential neighbors and non-neighbors is just $|V_{i-1}|-1$. Therefore the players can abort and conclude that $C cap I = emptyset$.



Each round takes $O(log n)$ bits of communication, and there are $O(log n)$ rounds, for a total of $O(log^2 n)$ bits of communication.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How does the number of vertices are resuced by factor of 2 - this leads to the $O(log(n))$ rounds?
    $endgroup$
    – Jay
    11 hours ago












  • $begingroup$
    I'm sorry, I can't explain it any better than what I wrote.
    $endgroup$
    – Yuval Filmus
    11 hours ago










  • $begingroup$
    Thanks a lot Yuval, I’ll try to figure it out.
    $endgroup$
    – Jay
    11 hours ago












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

The two players construct a sequence $V_0 supset V_1 supset cdots supset V_m$ of sets of vertices such that:





  1. $V_0$ consists of all vertices in the graph.


  2. $|V_{i+1}| leq (|V_i|+1)/2$.


  3. $V_i supseteq C cap I$.


The players stop once $|V_m| leq 1$. At this point they can answer the question using $O(1)$ communication.



At round $i$, the players know $V_{i-1}$, and wish to construct $V_i$. They act as follows:




  • If $C cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ neighbors in $V_{i-1}$, then Alice sends Bob one such vertex $v$, and both players set $V_i$ to be this set of neighbors, together with $v$ (this is valid since $C cap I subseteq C subseteq V_i$). Otherwise, she sends $bot$.


  • If Alice sent $bot$ and $I cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ non-neighbors in $V_{i-1}$, then Bob sends Alice one such vertex $v$, and both players set $V_i$ to be this set of non-neighbors, together with $v$ (this is valid since $C cap I subseteq I subseteq V_i$). Otherwise, he sends Alice $bot$.


  • If both players sent $bot$, then $C cap I = emptyset$. Indeed, if $v in C cap I$, then $v$ has at least $|V_{i-1}|/2$ neighbors and at least $|V_{i-1}|/2$ non-neighbors inside $|V_{i-1}|$, whereas the number of potential neighbors and non-neighbors is just $|V_{i-1}|-1$. Therefore the players can abort and conclude that $C cap I = emptyset$.



Each round takes $O(log n)$ bits of communication, and there are $O(log n)$ rounds, for a total of $O(log^2 n)$ bits of communication.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How does the number of vertices are resuced by factor of 2 - this leads to the $O(log(n))$ rounds?
    $endgroup$
    – Jay
    11 hours ago












  • $begingroup$
    I'm sorry, I can't explain it any better than what I wrote.
    $endgroup$
    – Yuval Filmus
    11 hours ago










  • $begingroup$
    Thanks a lot Yuval, I’ll try to figure it out.
    $endgroup$
    – Jay
    11 hours ago
















4












$begingroup$

The two players construct a sequence $V_0 supset V_1 supset cdots supset V_m$ of sets of vertices such that:





  1. $V_0$ consists of all vertices in the graph.


  2. $|V_{i+1}| leq (|V_i|+1)/2$.


  3. $V_i supseteq C cap I$.


The players stop once $|V_m| leq 1$. At this point they can answer the question using $O(1)$ communication.



At round $i$, the players know $V_{i-1}$, and wish to construct $V_i$. They act as follows:




  • If $C cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ neighbors in $V_{i-1}$, then Alice sends Bob one such vertex $v$, and both players set $V_i$ to be this set of neighbors, together with $v$ (this is valid since $C cap I subseteq C subseteq V_i$). Otherwise, she sends $bot$.


  • If Alice sent $bot$ and $I cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ non-neighbors in $V_{i-1}$, then Bob sends Alice one such vertex $v$, and both players set $V_i$ to be this set of non-neighbors, together with $v$ (this is valid since $C cap I subseteq I subseteq V_i$). Otherwise, he sends Alice $bot$.


  • If both players sent $bot$, then $C cap I = emptyset$. Indeed, if $v in C cap I$, then $v$ has at least $|V_{i-1}|/2$ neighbors and at least $|V_{i-1}|/2$ non-neighbors inside $|V_{i-1}|$, whereas the number of potential neighbors and non-neighbors is just $|V_{i-1}|-1$. Therefore the players can abort and conclude that $C cap I = emptyset$.



Each round takes $O(log n)$ bits of communication, and there are $O(log n)$ rounds, for a total of $O(log^2 n)$ bits of communication.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How does the number of vertices are resuced by factor of 2 - this leads to the $O(log(n))$ rounds?
    $endgroup$
    – Jay
    11 hours ago












  • $begingroup$
    I'm sorry, I can't explain it any better than what I wrote.
    $endgroup$
    – Yuval Filmus
    11 hours ago










  • $begingroup$
    Thanks a lot Yuval, I’ll try to figure it out.
    $endgroup$
    – Jay
    11 hours ago














4












4








4





$begingroup$

The two players construct a sequence $V_0 supset V_1 supset cdots supset V_m$ of sets of vertices such that:





  1. $V_0$ consists of all vertices in the graph.


  2. $|V_{i+1}| leq (|V_i|+1)/2$.


  3. $V_i supseteq C cap I$.


The players stop once $|V_m| leq 1$. At this point they can answer the question using $O(1)$ communication.



At round $i$, the players know $V_{i-1}$, and wish to construct $V_i$. They act as follows:




  • If $C cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ neighbors in $V_{i-1}$, then Alice sends Bob one such vertex $v$, and both players set $V_i$ to be this set of neighbors, together with $v$ (this is valid since $C cap I subseteq C subseteq V_i$). Otherwise, she sends $bot$.


  • If Alice sent $bot$ and $I cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ non-neighbors in $V_{i-1}$, then Bob sends Alice one such vertex $v$, and both players set $V_i$ to be this set of non-neighbors, together with $v$ (this is valid since $C cap I subseteq I subseteq V_i$). Otherwise, he sends Alice $bot$.


  • If both players sent $bot$, then $C cap I = emptyset$. Indeed, if $v in C cap I$, then $v$ has at least $|V_{i-1}|/2$ neighbors and at least $|V_{i-1}|/2$ non-neighbors inside $|V_{i-1}|$, whereas the number of potential neighbors and non-neighbors is just $|V_{i-1}|-1$. Therefore the players can abort and conclude that $C cap I = emptyset$.



Each round takes $O(log n)$ bits of communication, and there are $O(log n)$ rounds, for a total of $O(log^2 n)$ bits of communication.






share|cite|improve this answer











$endgroup$



The two players construct a sequence $V_0 supset V_1 supset cdots supset V_m$ of sets of vertices such that:





  1. $V_0$ consists of all vertices in the graph.


  2. $|V_{i+1}| leq (|V_i|+1)/2$.


  3. $V_i supseteq C cap I$.


The players stop once $|V_m| leq 1$. At this point they can answer the question using $O(1)$ communication.



At round $i$, the players know $V_{i-1}$, and wish to construct $V_i$. They act as follows:




  • If $C cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ neighbors in $V_{i-1}$, then Alice sends Bob one such vertex $v$, and both players set $V_i$ to be this set of neighbors, together with $v$ (this is valid since $C cap I subseteq C subseteq V_i$). Otherwise, she sends $bot$.


  • If Alice sent $bot$ and $I cap V_{i-1}$ contains a vertex $v$ with fewer than $|V_{i-1}|/2$ non-neighbors in $V_{i-1}$, then Bob sends Alice one such vertex $v$, and both players set $V_i$ to be this set of non-neighbors, together with $v$ (this is valid since $C cap I subseteq I subseteq V_i$). Otherwise, he sends Alice $bot$.


  • If both players sent $bot$, then $C cap I = emptyset$. Indeed, if $v in C cap I$, then $v$ has at least $|V_{i-1}|/2$ neighbors and at least $|V_{i-1}|/2$ non-neighbors inside $|V_{i-1}|$, whereas the number of potential neighbors and non-neighbors is just $|V_{i-1}|-1$. Therefore the players can abort and conclude that $C cap I = emptyset$.



Each round takes $O(log n)$ bits of communication, and there are $O(log n)$ rounds, for a total of $O(log^2 n)$ bits of communication.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 11 hours ago

























answered 12 hours ago









Yuval FilmusYuval Filmus

196k15184349




196k15184349












  • $begingroup$
    How does the number of vertices are resuced by factor of 2 - this leads to the $O(log(n))$ rounds?
    $endgroup$
    – Jay
    11 hours ago












  • $begingroup$
    I'm sorry, I can't explain it any better than what I wrote.
    $endgroup$
    – Yuval Filmus
    11 hours ago










  • $begingroup$
    Thanks a lot Yuval, I’ll try to figure it out.
    $endgroup$
    – Jay
    11 hours ago


















  • $begingroup$
    How does the number of vertices are resuced by factor of 2 - this leads to the $O(log(n))$ rounds?
    $endgroup$
    – Jay
    11 hours ago












  • $begingroup$
    I'm sorry, I can't explain it any better than what I wrote.
    $endgroup$
    – Yuval Filmus
    11 hours ago










  • $begingroup$
    Thanks a lot Yuval, I’ll try to figure it out.
    $endgroup$
    – Jay
    11 hours ago
















$begingroup$
How does the number of vertices are resuced by factor of 2 - this leads to the $O(log(n))$ rounds?
$endgroup$
– Jay
11 hours ago






$begingroup$
How does the number of vertices are resuced by factor of 2 - this leads to the $O(log(n))$ rounds?
$endgroup$
– Jay
11 hours ago














$begingroup$
I'm sorry, I can't explain it any better than what I wrote.
$endgroup$
– Yuval Filmus
11 hours ago




$begingroup$
I'm sorry, I can't explain it any better than what I wrote.
$endgroup$
– Yuval Filmus
11 hours ago












$begingroup$
Thanks a lot Yuval, I’ll try to figure it out.
$endgroup$
– Jay
11 hours ago




$begingroup$
Thanks a lot Yuval, I’ll try to figure it out.
$endgroup$
– Jay
11 hours ago


















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