Mapping the cycle graph into the real line
$begingroup$
I am trying to work on the following exercise.
Suppose $f: (C_n, d_n) to (mathbf{R}, |cdot|)$ is a map of the cycle graph $C_n$ (with nodes labelled, $1, 2, dots, n$) with the shortest path metric $d_n$ into $mathbf{R}$ with the usual metric $|cdot|$. Show that if $f$ is non-expansive, which means $|f(i) - f(j)| leq d_n(i,j)$ for all $i,j in {1, dots, n}$, then there exist two nodes $i,j$ for which
$$frac{d_n(i, j)}{|f(i) - f(j)|} = Omega(n).$$
(By the way $f(n) = Omega(n)$ means that for all $n$ big enough, $f(n) geq Cn$, where $C > 0$ is a universal constant.)
I know if you look at an equilateral triangle on the graph, then you can show that the distances contract by at least a constant factor, but I don't know how to go from this to th result, or if this is even helpful.
geometry discrete-mathematics graph-theory metric-spaces
asked Oct 19 '18 at 21:58
Drew BradyDrew Brady
682315
$endgroup$
add a comment |
$begingroup$
I am trying to work on the following exercise.
Suppose $f: (C_n, d_n) to (mathbf{R}, |cdot|)$ is a map of the cycle graph $C_n$ (with nodes labelled, $1, 2, dots, n$) with the shortest path metric $d_n$ into $mathbf{R}$ with the usual metric $|cdot|$. Show that if $f$ is non-expansive, which means $|f(i) - f(j)| leq d_n(i,j)$ for all $i,j in {1, dots, n}$, then there exist two nodes $i,j$ for which
$$frac{d_n(i, j)}{|f(i) - f(j)|} = Omega(n).$$
(By the way $f(n) = Omega(n)$ means that for all $n$ big enough, $f(n) geq Cn$, where $C > 0$ is a universal constant.)
I know if you look at an equilateral triangle on the graph, then you can show that the distances contract by at least a constant factor, but I don't know how to go from this to th result, or if this is even helpful.
geometry discrete-mathematics graph-theory metric-spaces
asked Oct 19 '18 at 21:58
Drew BradyDrew Brady
682315
$endgroup$
add a comment |
$begingroup$
I am trying to work on the following exercise.
Suppose $f: (C_n, d_n) to (mathbf{R}, |cdot|)$ is a map of the cycle graph $C_n$ (with nodes labelled, $1, 2, dots, n$) with the shortest path metric $d_n$ into $mathbf{R}$ with the usual metric $|cdot|$. Show that if $f$ is non-expansive, which means $|f(i) - f(j)| leq d_n(i,j)$ for all $i,j in {1, dots, n}$, then there exist two nodes $i,j$ for which
$$frac{d_n(i, j)}{|f(i) - f(j)|} = Omega(n).$$
(By the way $f(n) = Omega(n)$ means that for all $n$ big enough, $f(n) geq Cn$, where $C > 0$ is a universal constant.)
I know if you look at an equilateral triangle on the graph, then you can show that the distances contract by at least a constant factor, but I don't know how to go from this to th result, or if this is even helpful.
geometry discrete-mathematics graph-theory metric-spaces
asked Oct 19 '18 at 21:58
Drew BradyDrew Brady
682315
$endgroup$
I am trying to work on the following exercise.
Suppose $f: (C_n, d_n) to (mathbf{R}, |cdot|)$ is a map of the cycle graph $C_n$ (with nodes labelled, $1, 2, dots, n$) with the shortest path metric $d_n$ into $mathbf{R}$ with the usual metric $|cdot|$. Show that if $f$ is non-expansive, which means $|f(i) - f(j)| leq d_n(i,j)$ for all $i,j in {1, dots, n}$, then there exist two nodes $i,j$ for which
$$frac{d_n(i, j)}{|f(i) - f(j)|} = Omega(n).$$
(By the way $f(n) = Omega(n)$ means that for all $n$ big enough, $f(n) geq Cn$, where $C > 0$ is a universal constant.)
I know if you look at an equilateral triangle on the graph, then you can show that the distances contract by at least a constant factor, but I don't know how to go from this to th result, or if this is even helpful.
geometry discrete-mathematics graph-theory metric-spaces
geometry discrete-mathematics graph-theory metric-spaces
asked Oct 19 '18 at 21:58
Drew BradyDrew Brady
682315
asked Oct 19 '18 at 21:58
Drew BradyDrew Brady
682315
asked Oct 19 '18 at 21:58
Drew BradyDrew Brady
682315
asked Oct 19 '18 at 21:58
Drew BradyDrew Brady
682315
asked Oct 19 '18 at 21:58
Drew BradyDrew Brady
682315
682315
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Put $I={1,dots,n}$ and $n’in {1,dots,lfloor n/2rfloor}$. We prove a stronger claim: there exist $i,jin I$ such that $d_n(i,j)=n’$ and $|f(i)-f(j)|le 1$. For each $i,jin I$ let $i+’j$ equals $i+j$, if $i+jle n$, and equals $i+j-n$, otherwise. Remark that $d_n(i, i+’n’)=n’$ for each $i$. Put $I_1={iin I:f(i)le f(i+’n’)}$ and $I_2={iin I:f(i)ge f(i+n’)}$. Then $I_1cup I_2$ is a cover of $I$ by its two non-empty subsets. It is easy to see that there exists $iin I$ such that $iin I_1$ and $i+1in I_2$ or $iin I_2$ and $i+1in I_1$. Assume that $iin I_1$ and $i+1in I_2$. Then $f(i)le f(i+'n’)$ and $f(i+1)ge f(i+1+’n’)$. If $f(i)<f(i+'n’)-1$ and $f(i+1)>f(i+1+’n’)+1$ then $f(i+1+’n’)<f(i+1)-1le f(i)< f(i+'n’)-1$, a contradiction. The case $iin I_2$ and $i+1in I_1$ is considered similarly.
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
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%2f2962656%2fmapping-the-cycle-graph-into-the-real-line%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
$begingroup$
Put $I={1,dots,n}$ and $n’in {1,dots,lfloor n/2rfloor}$. We prove a stronger claim: there exist $i,jin I$ such that $d_n(i,j)=n’$ and $|f(i)-f(j)|le 1$. For each $i,jin I$ let $i+’j$ equals $i+j$, if $i+jle n$, and equals $i+j-n$, otherwise. Remark that $d_n(i, i+’n’)=n’$ for each $i$. Put $I_1={iin I:f(i)le f(i+’n’)}$ and $I_2={iin I:f(i)ge f(i+n’)}$. Then $I_1cup I_2$ is a cover of $I$ by its two non-empty subsets. It is easy to see that there exists $iin I$ such that $iin I_1$ and $i+1in I_2$ or $iin I_2$ and $i+1in I_1$. Assume that $iin I_1$ and $i+1in I_2$. Then $f(i)le f(i+'n’)$ and $f(i+1)ge f(i+1+’n’)$. If $f(i)<f(i+'n’)-1$ and $f(i+1)>f(i+1+’n’)+1$ then $f(i+1+’n’)<f(i+1)-1le f(i)< f(i+'n’)-1$, a contradiction. The case $iin I_2$ and $i+1in I_1$ is considered similarly.
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
add a comment |
$begingroup$
Put $I={1,dots,n}$ and $n’in {1,dots,lfloor n/2rfloor}$. We prove a stronger claim: there exist $i,jin I$ such that $d_n(i,j)=n’$ and $|f(i)-f(j)|le 1$. For each $i,jin I$ let $i+’j$ equals $i+j$, if $i+jle n$, and equals $i+j-n$, otherwise. Remark that $d_n(i, i+’n’)=n’$ for each $i$. Put $I_1={iin I:f(i)le f(i+’n’)}$ and $I_2={iin I:f(i)ge f(i+n’)}$. Then $I_1cup I_2$ is a cover of $I$ by its two non-empty subsets. It is easy to see that there exists $iin I$ such that $iin I_1$ and $i+1in I_2$ or $iin I_2$ and $i+1in I_1$. Assume that $iin I_1$ and $i+1in I_2$. Then $f(i)le f(i+'n’)$ and $f(i+1)ge f(i+1+’n’)$. If $f(i)<f(i+'n’)-1$ and $f(i+1)>f(i+1+’n’)+1$ then $f(i+1+’n’)<f(i+1)-1le f(i)< f(i+'n’)-1$, a contradiction. The case $iin I_2$ and $i+1in I_1$ is considered similarly.
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
add a comment |
$begingroup$
Put $I={1,dots,n}$ and $n’in {1,dots,lfloor n/2rfloor}$. We prove a stronger claim: there exist $i,jin I$ such that $d_n(i,j)=n’$ and $|f(i)-f(j)|le 1$. For each $i,jin I$ let $i+’j$ equals $i+j$, if $i+jle n$, and equals $i+j-n$, otherwise. Remark that $d_n(i, i+’n’)=n’$ for each $i$. Put $I_1={iin I:f(i)le f(i+’n’)}$ and $I_2={iin I:f(i)ge f(i+n’)}$. Then $I_1cup I_2$ is a cover of $I$ by its two non-empty subsets. It is easy to see that there exists $iin I$ such that $iin I_1$ and $i+1in I_2$ or $iin I_2$ and $i+1in I_1$. Assume that $iin I_1$ and $i+1in I_2$. Then $f(i)le f(i+'n’)$ and $f(i+1)ge f(i+1+’n’)$. If $f(i)<f(i+'n’)-1$ and $f(i+1)>f(i+1+’n’)+1$ then $f(i+1+’n’)<f(i+1)-1le f(i)< f(i+'n’)-1$, a contradiction. The case $iin I_2$ and $i+1in I_1$ is considered similarly.
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.9k32282
$endgroup$
Put $I={1,dots,n}$ and $n’in {1,dots,lfloor n/2rfloor}$. We prove a stronger claim: there exist $i,jin I$ such that $d_n(i,j)=n’$ and $|f(i)-f(j)|le 1$. For each $i,jin I$ let $i+’j$ equals $i+j$, if $i+jle n$, and equals $i+j-n$, otherwise. Remark that $d_n(i, i+’n’)=n’$ for each $i$. Put $I_1={iin I:f(i)le f(i+’n’)}$ and $I_2={iin I:f(i)ge f(i+n’)}$. Then $I_1cup I_2$ is a cover of $I$ by its two non-empty subsets. It is easy to see that there exists $iin I$ such that $iin I_1$ and $i+1in I_2$ or $iin I_2$ and $i+1in I_1$. Assume that $iin I_1$ and $i+1in I_2$. Then $f(i)le f(i+'n’)$ and $f(i+1)ge f(i+1+’n’)$. If $f(i)<f(i+'n’)-1$ and $f(i+1)>f(i+1+’n’)+1$ then $f(i+1+’n’)<f(i+1)-1le f(i)< f(i+'n’)-1$, a contradiction. The case $iin I_2$ and $i+1in I_1$ is considered similarly.
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.9k32282
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.9k32282
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.9k32282
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.9k32282
39.9k32282
add a comment |
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%2f2962656%2fmapping-the-cycle-graph-into-the-real-line%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
