Mapping the cycle graph into the real line












1












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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







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$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question









      $endgroup$




      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






      share|cite|improve this question













      share|cite|improve this question











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      asked Oct 19 '18 at 21:58









      Drew BradyDrew Brady

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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.






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            1 Answer
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            active

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            active

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            0












            $begingroup$

            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.






            share|cite|improve this answer









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              0












              $begingroup$

              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.






              share|cite|improve this answer









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                0





                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$



                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 8:37









                Alex RavskyAlex Ravsky

                39.9k32282




                39.9k32282






























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