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












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "419"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f106562%2fthe-clique-vs-independent-set-problem%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








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


















draft saved

draft discarded




















































Thanks for contributing an answer to Computer Science Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f106562%2fthe-clique-vs-independent-set-problem%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Knooppunt Holsloot

Image
Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
Read more

Altaar (religie)

Image
Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
Read more

Gregoriusmis

Image
Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
Read more