Given the Voronoi diagram of some points on a line, determine the points












5














Definition:



Assume that a set of points $P={p_1,dots,p_n}$ on a line is given. The Voronoi diagram of $P$ is a set of points $V(P)={x_1,dots,x_{n-1}}$ such that $x_i$ is the midpoint of $p_i p_{i+1}$.



Question:





  1. Suppose you are given a set $X={x_1,dots,x_{n-1}}$. How can we determine whether or not $X$ is a one-dimensional Voronoi diagram of a set of points?

  2. If $X$ is indeed a one-dimensional Voronoi diagram, How can we determine $P$?




My try:



I believe that it suffices for the points of $X$ to be on the same line in order to be a one-dimensional Voronoi diagram. Also, for the second part, I have an idea. We know that:



$$x_1 = frac{p_1+p_2}{2} implies p_2=2x_1-p_1$$
$$x_2 = frac{p_2+p_3}{2} implies p_3=2x_2-(2x_1-p_1)$$
$$dots$$
$$x_{n-1} = frac{p_{n-1}+p_n}{2} implies p_n=2x_{n-1}-p_{n-1}$$



So if we determine $p_1$, we get every other point of $P$ according to these equations. But I'm stuck at determining $p_1$.










share|cite|improve this question
























  • Regarding question 1, I don’t think every set of points on a line can be a Voronoi diagram. For example consider two points ‘close’ to each other, then a large gap to another two points that are ‘close’ to each other. There must exist some relationship between the sequential gaps I’m guessing..
    – T. Ford
    Dec 12 '18 at 8:27










  • For the general problem, see the paper Recognizing Dirichlet tessellations.
    – lhf
    Dec 12 '18 at 10:23


















5














Definition:



Assume that a set of points $P={p_1,dots,p_n}$ on a line is given. The Voronoi diagram of $P$ is a set of points $V(P)={x_1,dots,x_{n-1}}$ such that $x_i$ is the midpoint of $p_i p_{i+1}$.



Question:





  1. Suppose you are given a set $X={x_1,dots,x_{n-1}}$. How can we determine whether or not $X$ is a one-dimensional Voronoi diagram of a set of points?

  2. If $X$ is indeed a one-dimensional Voronoi diagram, How can we determine $P$?




My try:



I believe that it suffices for the points of $X$ to be on the same line in order to be a one-dimensional Voronoi diagram. Also, for the second part, I have an idea. We know that:



$$x_1 = frac{p_1+p_2}{2} implies p_2=2x_1-p_1$$
$$x_2 = frac{p_2+p_3}{2} implies p_3=2x_2-(2x_1-p_1)$$
$$dots$$
$$x_{n-1} = frac{p_{n-1}+p_n}{2} implies p_n=2x_{n-1}-p_{n-1}$$



So if we determine $p_1$, we get every other point of $P$ according to these equations. But I'm stuck at determining $p_1$.










share|cite|improve this question
























  • Regarding question 1, I don’t think every set of points on a line can be a Voronoi diagram. For example consider two points ‘close’ to each other, then a large gap to another two points that are ‘close’ to each other. There must exist some relationship between the sequential gaps I’m guessing..
    – T. Ford
    Dec 12 '18 at 8:27










  • For the general problem, see the paper Recognizing Dirichlet tessellations.
    – lhf
    Dec 12 '18 at 10:23
















5












5








5


1





Definition:



Assume that a set of points $P={p_1,dots,p_n}$ on a line is given. The Voronoi diagram of $P$ is a set of points $V(P)={x_1,dots,x_{n-1}}$ such that $x_i$ is the midpoint of $p_i p_{i+1}$.



Question:





  1. Suppose you are given a set $X={x_1,dots,x_{n-1}}$. How can we determine whether or not $X$ is a one-dimensional Voronoi diagram of a set of points?

  2. If $X$ is indeed a one-dimensional Voronoi diagram, How can we determine $P$?




My try:



I believe that it suffices for the points of $X$ to be on the same line in order to be a one-dimensional Voronoi diagram. Also, for the second part, I have an idea. We know that:



$$x_1 = frac{p_1+p_2}{2} implies p_2=2x_1-p_1$$
$$x_2 = frac{p_2+p_3}{2} implies p_3=2x_2-(2x_1-p_1)$$
$$dots$$
$$x_{n-1} = frac{p_{n-1}+p_n}{2} implies p_n=2x_{n-1}-p_{n-1}$$



So if we determine $p_1$, we get every other point of $P$ according to these equations. But I'm stuck at determining $p_1$.










share|cite|improve this question















Definition:



Assume that a set of points $P={p_1,dots,p_n}$ on a line is given. The Voronoi diagram of $P$ is a set of points $V(P)={x_1,dots,x_{n-1}}$ such that $x_i$ is the midpoint of $p_i p_{i+1}$.



Question:





  1. Suppose you are given a set $X={x_1,dots,x_{n-1}}$. How can we determine whether or not $X$ is a one-dimensional Voronoi diagram of a set of points?

  2. If $X$ is indeed a one-dimensional Voronoi diagram, How can we determine $P$?




My try:



I believe that it suffices for the points of $X$ to be on the same line in order to be a one-dimensional Voronoi diagram. Also, for the second part, I have an idea. We know that:



$$x_1 = frac{p_1+p_2}{2} implies p_2=2x_1-p_1$$
$$x_2 = frac{p_2+p_3}{2} implies p_3=2x_2-(2x_1-p_1)$$
$$dots$$
$$x_{n-1} = frac{p_{n-1}+p_n}{2} implies p_n=2x_{n-1}-p_{n-1}$$



So if we determine $p_1$, we get every other point of $P$ according to these equations. But I'm stuck at determining $p_1$.







computational-geometry voronoi-diagram






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edited Dec 12 '18 at 8:07







Arman Malekzadeh

















asked Dec 12 '18 at 7:53









Arman MalekzadehArman Malekzadeh

1,7951028




1,7951028












  • Regarding question 1, I don’t think every set of points on a line can be a Voronoi diagram. For example consider two points ‘close’ to each other, then a large gap to another two points that are ‘close’ to each other. There must exist some relationship between the sequential gaps I’m guessing..
    – T. Ford
    Dec 12 '18 at 8:27










  • For the general problem, see the paper Recognizing Dirichlet tessellations.
    – lhf
    Dec 12 '18 at 10:23




















  • Regarding question 1, I don’t think every set of points on a line can be a Voronoi diagram. For example consider two points ‘close’ to each other, then a large gap to another two points that are ‘close’ to each other. There must exist some relationship between the sequential gaps I’m guessing..
    – T. Ford
    Dec 12 '18 at 8:27










  • For the general problem, see the paper Recognizing Dirichlet tessellations.
    – lhf
    Dec 12 '18 at 10:23


















Regarding question 1, I don’t think every set of points on a line can be a Voronoi diagram. For example consider two points ‘close’ to each other, then a large gap to another two points that are ‘close’ to each other. There must exist some relationship between the sequential gaps I’m guessing..
– T. Ford
Dec 12 '18 at 8:27




Regarding question 1, I don’t think every set of points on a line can be a Voronoi diagram. For example consider two points ‘close’ to each other, then a large gap to another two points that are ‘close’ to each other. There must exist some relationship between the sequential gaps I’m guessing..
– T. Ford
Dec 12 '18 at 8:27












For the general problem, see the paper Recognizing Dirichlet tessellations.
– lhf
Dec 12 '18 at 10:23






For the general problem, see the paper Recognizing Dirichlet tessellations.
– lhf
Dec 12 '18 at 10:23












2 Answers
2






active

oldest

votes


















1














Every point $x_i$ is equidistant of two points $p_i$ and $p_{i+1}$. Let's call that distance $d_i$. We also know that $p_{i+1} - p_i = (p_{i+1} - x_{i+1}) + (x_{i+1} - p_i) = d_{i+1} + d_i$. The $d_i$ are distances so they're all positive.



For a given set $X$ with $n$ elements, you have $n-1$ linear equations and $n$ inequalities (all the distances must be positive). One way to solve it is to compute your vector $(d_1,...d_n)$ as a function of $d_1$ and check whether there's a $d_1$ for which all the $d_i$ are positive.



All the $d_i$ are linear in $d_1$, so your inequalities will translate to linear inequalities on $d_1$ of the form $(-1)^{b_i}d_1 le c_i$, which may or may not have a solution depending on the specific values.



Edit: Computing the actual inequalities.



We're trying to express all the $d_i$ as a function of $d_1$.



$$d_2 + d_1 = (p_2 - p_1) Leftrightarrow d_2 = (p_2 - p_1) - d_1$$
$$d_3 + d_2 = (p_3 - p_2) Leftrightarrow d_3 = (p_3 - p_2) - d_2 = (p_3 - p_2) - (p_2 - p_1) + d_1$$
$$...$$
$$d_i = sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) + (-1)^{i-1} d_1$$



So: $$d_i ge 0 Leftrightarrow (-1)^{i-1} d_1 ge -sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) $$



Your set is a Voronoi diagram if and only if there's a $d_1$ that satisfies all these inequalities.






share|cite|improve this answer























  • Can you provide an example?
    – Arman Malekzadeh
    Dec 12 '18 at 9:30










  • I'll compute the actual inequalities as soon as I have some time.
    – RcnSc
    Dec 12 '18 at 9:42










  • @ArmanMalekzadeh Let me know if something is unclear
    – RcnSc
    Dec 12 '18 at 10:14










  • Thank you... I was just wondering if the solution was unique. It seems it's not :)
    – Arman Malekzadeh
    Dec 12 '18 at 10:28



















2














For (2), knowing $V(P)$ is not sufficient to determine $P$. For example, if $V(P) = {5,15}$ then $P$ could be ${0,10,20}$ or ${1,9,21}$ or ${2,8,22}$ etc.



Difficulty is that $V(P)$ gives you $n-1$ linear equations in $n$ unknowns, so system is underdetermined. Additional constraint that $p_{k+1} ge p_k$ is not sufficient to give a unique solution.






share|cite|improve this answer























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    2 Answers
    2






    active

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    Every point $x_i$ is equidistant of two points $p_i$ and $p_{i+1}$. Let's call that distance $d_i$. We also know that $p_{i+1} - p_i = (p_{i+1} - x_{i+1}) + (x_{i+1} - p_i) = d_{i+1} + d_i$. The $d_i$ are distances so they're all positive.



    For a given set $X$ with $n$ elements, you have $n-1$ linear equations and $n$ inequalities (all the distances must be positive). One way to solve it is to compute your vector $(d_1,...d_n)$ as a function of $d_1$ and check whether there's a $d_1$ for which all the $d_i$ are positive.



    All the $d_i$ are linear in $d_1$, so your inequalities will translate to linear inequalities on $d_1$ of the form $(-1)^{b_i}d_1 le c_i$, which may or may not have a solution depending on the specific values.



    Edit: Computing the actual inequalities.



    We're trying to express all the $d_i$ as a function of $d_1$.



    $$d_2 + d_1 = (p_2 - p_1) Leftrightarrow d_2 = (p_2 - p_1) - d_1$$
    $$d_3 + d_2 = (p_3 - p_2) Leftrightarrow d_3 = (p_3 - p_2) - d_2 = (p_3 - p_2) - (p_2 - p_1) + d_1$$
    $$...$$
    $$d_i = sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) + (-1)^{i-1} d_1$$



    So: $$d_i ge 0 Leftrightarrow (-1)^{i-1} d_1 ge -sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) $$



    Your set is a Voronoi diagram if and only if there's a $d_1$ that satisfies all these inequalities.






    share|cite|improve this answer























    • Can you provide an example?
      – Arman Malekzadeh
      Dec 12 '18 at 9:30










    • I'll compute the actual inequalities as soon as I have some time.
      – RcnSc
      Dec 12 '18 at 9:42










    • @ArmanMalekzadeh Let me know if something is unclear
      – RcnSc
      Dec 12 '18 at 10:14










    • Thank you... I was just wondering if the solution was unique. It seems it's not :)
      – Arman Malekzadeh
      Dec 12 '18 at 10:28
















    1














    Every point $x_i$ is equidistant of two points $p_i$ and $p_{i+1}$. Let's call that distance $d_i$. We also know that $p_{i+1} - p_i = (p_{i+1} - x_{i+1}) + (x_{i+1} - p_i) = d_{i+1} + d_i$. The $d_i$ are distances so they're all positive.



    For a given set $X$ with $n$ elements, you have $n-1$ linear equations and $n$ inequalities (all the distances must be positive). One way to solve it is to compute your vector $(d_1,...d_n)$ as a function of $d_1$ and check whether there's a $d_1$ for which all the $d_i$ are positive.



    All the $d_i$ are linear in $d_1$, so your inequalities will translate to linear inequalities on $d_1$ of the form $(-1)^{b_i}d_1 le c_i$, which may or may not have a solution depending on the specific values.



    Edit: Computing the actual inequalities.



    We're trying to express all the $d_i$ as a function of $d_1$.



    $$d_2 + d_1 = (p_2 - p_1) Leftrightarrow d_2 = (p_2 - p_1) - d_1$$
    $$d_3 + d_2 = (p_3 - p_2) Leftrightarrow d_3 = (p_3 - p_2) - d_2 = (p_3 - p_2) - (p_2 - p_1) + d_1$$
    $$...$$
    $$d_i = sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) + (-1)^{i-1} d_1$$



    So: $$d_i ge 0 Leftrightarrow (-1)^{i-1} d_1 ge -sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) $$



    Your set is a Voronoi diagram if and only if there's a $d_1$ that satisfies all these inequalities.






    share|cite|improve this answer























    • Can you provide an example?
      – Arman Malekzadeh
      Dec 12 '18 at 9:30










    • I'll compute the actual inequalities as soon as I have some time.
      – RcnSc
      Dec 12 '18 at 9:42










    • @ArmanMalekzadeh Let me know if something is unclear
      – RcnSc
      Dec 12 '18 at 10:14










    • Thank you... I was just wondering if the solution was unique. It seems it's not :)
      – Arman Malekzadeh
      Dec 12 '18 at 10:28














    1












    1








    1






    Every point $x_i$ is equidistant of two points $p_i$ and $p_{i+1}$. Let's call that distance $d_i$. We also know that $p_{i+1} - p_i = (p_{i+1} - x_{i+1}) + (x_{i+1} - p_i) = d_{i+1} + d_i$. The $d_i$ are distances so they're all positive.



    For a given set $X$ with $n$ elements, you have $n-1$ linear equations and $n$ inequalities (all the distances must be positive). One way to solve it is to compute your vector $(d_1,...d_n)$ as a function of $d_1$ and check whether there's a $d_1$ for which all the $d_i$ are positive.



    All the $d_i$ are linear in $d_1$, so your inequalities will translate to linear inequalities on $d_1$ of the form $(-1)^{b_i}d_1 le c_i$, which may or may not have a solution depending on the specific values.



    Edit: Computing the actual inequalities.



    We're trying to express all the $d_i$ as a function of $d_1$.



    $$d_2 + d_1 = (p_2 - p_1) Leftrightarrow d_2 = (p_2 - p_1) - d_1$$
    $$d_3 + d_2 = (p_3 - p_2) Leftrightarrow d_3 = (p_3 - p_2) - d_2 = (p_3 - p_2) - (p_2 - p_1) + d_1$$
    $$...$$
    $$d_i = sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) + (-1)^{i-1} d_1$$



    So: $$d_i ge 0 Leftrightarrow (-1)^{i-1} d_1 ge -sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) $$



    Your set is a Voronoi diagram if and only if there's a $d_1$ that satisfies all these inequalities.






    share|cite|improve this answer














    Every point $x_i$ is equidistant of two points $p_i$ and $p_{i+1}$. Let's call that distance $d_i$. We also know that $p_{i+1} - p_i = (p_{i+1} - x_{i+1}) + (x_{i+1} - p_i) = d_{i+1} + d_i$. The $d_i$ are distances so they're all positive.



    For a given set $X$ with $n$ elements, you have $n-1$ linear equations and $n$ inequalities (all the distances must be positive). One way to solve it is to compute your vector $(d_1,...d_n)$ as a function of $d_1$ and check whether there's a $d_1$ for which all the $d_i$ are positive.



    All the $d_i$ are linear in $d_1$, so your inequalities will translate to linear inequalities on $d_1$ of the form $(-1)^{b_i}d_1 le c_i$, which may or may not have a solution depending on the specific values.



    Edit: Computing the actual inequalities.



    We're trying to express all the $d_i$ as a function of $d_1$.



    $$d_2 + d_1 = (p_2 - p_1) Leftrightarrow d_2 = (p_2 - p_1) - d_1$$
    $$d_3 + d_2 = (p_3 - p_2) Leftrightarrow d_3 = (p_3 - p_2) - d_2 = (p_3 - p_2) - (p_2 - p_1) + d_1$$
    $$...$$
    $$d_i = sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) + (-1)^{i-1} d_1$$



    So: $$d_i ge 0 Leftrightarrow (-1)^{i-1} d_1 ge -sum_{j = 0}^{i-1} (-1)^j(p_{i-j} - p_{i-j-1}) $$



    Your set is a Voronoi diagram if and only if there's a $d_1$ that satisfies all these inequalities.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 12 '18 at 10:19

























    answered Dec 12 '18 at 9:19









    RcnScRcnSc

    1014




    1014












    • Can you provide an example?
      – Arman Malekzadeh
      Dec 12 '18 at 9:30










    • I'll compute the actual inequalities as soon as I have some time.
      – RcnSc
      Dec 12 '18 at 9:42










    • @ArmanMalekzadeh Let me know if something is unclear
      – RcnSc
      Dec 12 '18 at 10:14










    • Thank you... I was just wondering if the solution was unique. It seems it's not :)
      – Arman Malekzadeh
      Dec 12 '18 at 10:28


















    • Can you provide an example?
      – Arman Malekzadeh
      Dec 12 '18 at 9:30










    • I'll compute the actual inequalities as soon as I have some time.
      – RcnSc
      Dec 12 '18 at 9:42










    • @ArmanMalekzadeh Let me know if something is unclear
      – RcnSc
      Dec 12 '18 at 10:14










    • Thank you... I was just wondering if the solution was unique. It seems it's not :)
      – Arman Malekzadeh
      Dec 12 '18 at 10:28
















    Can you provide an example?
    – Arman Malekzadeh
    Dec 12 '18 at 9:30




    Can you provide an example?
    – Arman Malekzadeh
    Dec 12 '18 at 9:30












    I'll compute the actual inequalities as soon as I have some time.
    – RcnSc
    Dec 12 '18 at 9:42




    I'll compute the actual inequalities as soon as I have some time.
    – RcnSc
    Dec 12 '18 at 9:42












    @ArmanMalekzadeh Let me know if something is unclear
    – RcnSc
    Dec 12 '18 at 10:14




    @ArmanMalekzadeh Let me know if something is unclear
    – RcnSc
    Dec 12 '18 at 10:14












    Thank you... I was just wondering if the solution was unique. It seems it's not :)
    – Arman Malekzadeh
    Dec 12 '18 at 10:28




    Thank you... I was just wondering if the solution was unique. It seems it's not :)
    – Arman Malekzadeh
    Dec 12 '18 at 10:28











    2














    For (2), knowing $V(P)$ is not sufficient to determine $P$. For example, if $V(P) = {5,15}$ then $P$ could be ${0,10,20}$ or ${1,9,21}$ or ${2,8,22}$ etc.



    Difficulty is that $V(P)$ gives you $n-1$ linear equations in $n$ unknowns, so system is underdetermined. Additional constraint that $p_{k+1} ge p_k$ is not sufficient to give a unique solution.






    share|cite|improve this answer




























      2














      For (2), knowing $V(P)$ is not sufficient to determine $P$. For example, if $V(P) = {5,15}$ then $P$ could be ${0,10,20}$ or ${1,9,21}$ or ${2,8,22}$ etc.



      Difficulty is that $V(P)$ gives you $n-1$ linear equations in $n$ unknowns, so system is underdetermined. Additional constraint that $p_{k+1} ge p_k$ is not sufficient to give a unique solution.






      share|cite|improve this answer


























        2












        2








        2






        For (2), knowing $V(P)$ is not sufficient to determine $P$. For example, if $V(P) = {5,15}$ then $P$ could be ${0,10,20}$ or ${1,9,21}$ or ${2,8,22}$ etc.



        Difficulty is that $V(P)$ gives you $n-1$ linear equations in $n$ unknowns, so system is underdetermined. Additional constraint that $p_{k+1} ge p_k$ is not sufficient to give a unique solution.






        share|cite|improve this answer














        For (2), knowing $V(P)$ is not sufficient to determine $P$. For example, if $V(P) = {5,15}$ then $P$ could be ${0,10,20}$ or ${1,9,21}$ or ${2,8,22}$ etc.



        Difficulty is that $V(P)$ gives you $n-1$ linear equations in $n$ unknowns, so system is underdetermined. Additional constraint that $p_{k+1} ge p_k$ is not sufficient to give a unique solution.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 12 '18 at 11:16

























        answered Dec 12 '18 at 10:23









        gandalf61gandalf61

        7,921624




        7,921624






























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