Arrange people at round table so that everyone knows the two people next to them
$begingroup$
Each of the guests know:
a) more than half of the guests
b) at least half of the guests.
Prove that in both of these cases it is possible to arrange them to sit around a round table so that everyone knows the two people next to them.
I believe that if we prove b) then we have at the same time proven a) as well. Can anyone give me a hint? I've tried drawing, but I'm not sure how to formally prove it. I was considering relationship properties, such as symmetry and transition, but couldn't work it out. Thanks in advance.
discrete-mathematics relations equivalence-relations
asked Dec 9 '18 at 9:02
ponikoliponikoli
416
$endgroup$
add a comment |
$begingroup$
Each of the guests know:
a) more than half of the guests
b) at least half of the guests.
Prove that in both of these cases it is possible to arrange them to sit around a round table so that everyone knows the two people next to them.
I believe that if we prove b) then we have at the same time proven a) as well. Can anyone give me a hint? I've tried drawing, but I'm not sure how to formally prove it. I was considering relationship properties, such as symmetry and transition, but couldn't work it out. Thanks in advance.
discrete-mathematics relations equivalence-relations
asked Dec 9 '18 at 9:02
ponikoliponikoli
416
$endgroup$
add a comment |
$begingroup$
Each of the guests know:
a) more than half of the guests
b) at least half of the guests.
Prove that in both of these cases it is possible to arrange them to sit around a round table so that everyone knows the two people next to them.
I believe that if we prove b) then we have at the same time proven a) as well. Can anyone give me a hint? I've tried drawing, but I'm not sure how to formally prove it. I was considering relationship properties, such as symmetry and transition, but couldn't work it out. Thanks in advance.
discrete-mathematics relations equivalence-relations
asked Dec 9 '18 at 9:02
ponikoliponikoli
416
$endgroup$
Each of the guests know:
a) more than half of the guests
b) at least half of the guests.
Prove that in both of these cases it is possible to arrange them to sit around a round table so that everyone knows the two people next to them.
I believe that if we prove b) then we have at the same time proven a) as well. Can anyone give me a hint? I've tried drawing, but I'm not sure how to formally prove it. I was considering relationship properties, such as symmetry and transition, but couldn't work it out. Thanks in advance.
discrete-mathematics relations equivalence-relations
discrete-mathematics relations equivalence-relations
asked Dec 9 '18 at 9:02
ponikoliponikoli
416
asked Dec 9 '18 at 9:02
ponikoliponikoli
416
edited Dec 11 '18 at 16:39
ponikoli
asked Dec 9 '18 at 9:02
ponikoliponikoli
416
asked Dec 9 '18 at 9:02
ponikoliponikoli
416
asked Dec 9 '18 at 9:02
ponikoliponikoli
416
416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Construct a graph with $n$ vertices representing the people and connect two vertices if the two people they represent know each other.
For a), the degree of each vertex if greater than $frac{n}{2}$, By dirac's theorem, there is a Hamiltonian cycle. And this implies we can arrange the people in a circle so that each person knows the ones sitting next to them. Same for b). I think.
Or consider first a random arrangement of the people around the table, Suppose a neighboring pair $(A,B)$ is a hostile couple with $B$ sitting to the right of $A$, then if we can find a neighboring pair $(A'B')$ with $B'$ sitting to the right of $A'$ and $B'$ is friend with $B$ and $A'$ is a friend of $A$. we can then swap $B$ with $A'$ and that will reduce the number of hostile neighboring couples. So it remains to show $(A'B')$ exists. Well $A$ has at least $n$ friends sitting to his right, and there are $n$ sits to the right of frinds of $A$. $B$ has at most $n-1$ enemy, So there is a friend of $A$,$A'$, with $B'$ sitting right to him, a friend of $B$. Done?
answered Dec 9 '18 at 16:55
nafhgoodnafhgood
1,805422
$endgroup$
$begingroup$
Is it possible to explain this in any other way, because we never used Dirac's theorem nor Hamiltonian cycle in our course?
$endgroup$
– ponikoli
Dec 11 '18 at 16:27
$begingroup$
It is the "Ambassadors at a Round Table" problem!!
$endgroup$
– nafhgood
Dec 11 '18 at 17:15
add a comment |
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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032178%2farrange-people-at-round-table-so-that-everyone-knows-the-two-people-next-to-them%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
$begingroup$
Construct a graph with $n$ vertices representing the people and connect two vertices if the two people they represent know each other.
For a), the degree of each vertex if greater than $frac{n}{2}$, By dirac's theorem, there is a Hamiltonian cycle. And this implies we can arrange the people in a circle so that each person knows the ones sitting next to them. Same for b). I think.
Or consider first a random arrangement of the people around the table, Suppose a neighboring pair $(A,B)$ is a hostile couple with $B$ sitting to the right of $A$, then if we can find a neighboring pair $(A'B')$ with $B'$ sitting to the right of $A'$ and $B'$ is friend with $B$ and $A'$ is a friend of $A$. we can then swap $B$ with $A'$ and that will reduce the number of hostile neighboring couples. So it remains to show $(A'B')$ exists. Well $A$ has at least $n$ friends sitting to his right, and there are $n$ sits to the right of frinds of $A$. $B$ has at most $n-1$ enemy, So there is a friend of $A$,$A'$, with $B'$ sitting right to him, a friend of $B$. Done?
answered Dec 9 '18 at 16:55
nafhgoodnafhgood
1,805422
$endgroup$
$begingroup$
Is it possible to explain this in any other way, because we never used Dirac's theorem nor Hamiltonian cycle in our course?
$endgroup$
– ponikoli
Dec 11 '18 at 16:27
$begingroup$
It is the "Ambassadors at a Round Table" problem!!
$endgroup$
– nafhgood
Dec 11 '18 at 17:15
add a comment |
$begingroup$
Construct a graph with $n$ vertices representing the people and connect two vertices if the two people they represent know each other.
For a), the degree of each vertex if greater than $frac{n}{2}$, By dirac's theorem, there is a Hamiltonian cycle. And this implies we can arrange the people in a circle so that each person knows the ones sitting next to them. Same for b). I think.
Or consider first a random arrangement of the people around the table, Suppose a neighboring pair $(A,B)$ is a hostile couple with $B$ sitting to the right of $A$, then if we can find a neighboring pair $(A'B')$ with $B'$ sitting to the right of $A'$ and $B'$ is friend with $B$ and $A'$ is a friend of $A$. we can then swap $B$ with $A'$ and that will reduce the number of hostile neighboring couples. So it remains to show $(A'B')$ exists. Well $A$ has at least $n$ friends sitting to his right, and there are $n$ sits to the right of frinds of $A$. $B$ has at most $n-1$ enemy, So there is a friend of $A$,$A'$, with $B'$ sitting right to him, a friend of $B$. Done?
answered Dec 9 '18 at 16:55
nafhgoodnafhgood
1,805422
$endgroup$
$begingroup$
Is it possible to explain this in any other way, because we never used Dirac's theorem nor Hamiltonian cycle in our course?
$endgroup$
– ponikoli
Dec 11 '18 at 16:27
$begingroup$
It is the "Ambassadors at a Round Table" problem!!
$endgroup$
– nafhgood
Dec 11 '18 at 17:15
add a comment |
$begingroup$
Construct a graph with $n$ vertices representing the people and connect two vertices if the two people they represent know each other.
For a), the degree of each vertex if greater than $frac{n}{2}$, By dirac's theorem, there is a Hamiltonian cycle. And this implies we can arrange the people in a circle so that each person knows the ones sitting next to them. Same for b). I think.
Or consider first a random arrangement of the people around the table, Suppose a neighboring pair $(A,B)$ is a hostile couple with $B$ sitting to the right of $A$, then if we can find a neighboring pair $(A'B')$ with $B'$ sitting to the right of $A'$ and $B'$ is friend with $B$ and $A'$ is a friend of $A$. we can then swap $B$ with $A'$ and that will reduce the number of hostile neighboring couples. So it remains to show $(A'B')$ exists. Well $A$ has at least $n$ friends sitting to his right, and there are $n$ sits to the right of frinds of $A$. $B$ has at most $n-1$ enemy, So there is a friend of $A$,$A'$, with $B'$ sitting right to him, a friend of $B$. Done?
answered Dec 9 '18 at 16:55
nafhgoodnafhgood
1,805422
$endgroup$
Construct a graph with $n$ vertices representing the people and connect two vertices if the two people they represent know each other.
For a), the degree of each vertex if greater than $frac{n}{2}$, By dirac's theorem, there is a Hamiltonian cycle. And this implies we can arrange the people in a circle so that each person knows the ones sitting next to them. Same for b). I think.
Or consider first a random arrangement of the people around the table, Suppose a neighboring pair $(A,B)$ is a hostile couple with $B$ sitting to the right of $A$, then if we can find a neighboring pair $(A'B')$ with $B'$ sitting to the right of $A'$ and $B'$ is friend with $B$ and $A'$ is a friend of $A$. we can then swap $B$ with $A'$ and that will reduce the number of hostile neighboring couples. So it remains to show $(A'B')$ exists. Well $A$ has at least $n$ friends sitting to his right, and there are $n$ sits to the right of frinds of $A$. $B$ has at most $n-1$ enemy, So there is a friend of $A$,$A'$, with $B'$ sitting right to him, a friend of $B$. Done?
answered Dec 9 '18 at 16:55
nafhgoodnafhgood
1,805422
edited Dec 11 '18 at 17:13
answered Dec 9 '18 at 16:55
nafhgoodnafhgood
1,805422
answered Dec 9 '18 at 16:55
nafhgoodnafhgood
1,805422
answered Dec 9 '18 at 16:55
nafhgoodnafhgood
1,805422
1,805422
$begingroup$
Is it possible to explain this in any other way, because we never used Dirac's theorem nor Hamiltonian cycle in our course?
$endgroup$
– ponikoli
Dec 11 '18 at 16:27
$begingroup$
It is the "Ambassadors at a Round Table" problem!!
$endgroup$
– nafhgood
Dec 11 '18 at 17:15
add a comment |
$begingroup$
Is it possible to explain this in any other way, because we never used Dirac's theorem nor Hamiltonian cycle in our course?
$endgroup$
– ponikoli
Dec 11 '18 at 16:27
$begingroup$
It is the "Ambassadors at a Round Table" problem!!
$endgroup$
– nafhgood
Dec 11 '18 at 17:15
$begingroup$
Is it possible to explain this in any other way, because we never used Dirac's theorem nor Hamiltonian cycle in our course?
$endgroup$
– ponikoli
Dec 11 '18 at 16:27
$begingroup$
Is it possible to explain this in any other way, because we never used Dirac's theorem nor Hamiltonian cycle in our course?
$endgroup$
– ponikoli
Dec 11 '18 at 16:27
$begingroup$
It is the "Ambassadors at a Round Table" problem!!
$endgroup$
– nafhgood
Dec 11 '18 at 17:15
$begingroup$
It is the "Ambassadors at a Round Table" problem!!
$endgroup$
– nafhgood
Dec 11 '18 at 17:15
add a comment |
Thanks for contributing an answer to Mathematics 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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032178%2farrange-people-at-round-table-so-that-everyone-knows-the-two-people-next-to-them%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
