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
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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
39.8k32281
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1 Answer
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1 Answer
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active
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active
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$begingroup$
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
39.8k32281
$endgroup$
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
39.8k32281
$endgroup$
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
39.8k32281
$endgroup$
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
39.8k32281
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.8k32281
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.8k32281
answered Dec 4 '18 at 8:37
Alex RavskyAlex Ravsky
39.8k32281
39.8k32281
add a comment |
add a comment |
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