How many distinct ways are there to traverse the edges of an N-dimensional hypercube?












2














I have an N-dimensional hypercube, and would like to traverse the edges so that I reach each vertex exactly once. One path that satisfies this property is known as the Gray code, but there are many others. How many such paths are there? Paths that can be mapped to each other by the symmetries of the hypercube may or may not be counted as distinct, depending on your preference.






combinatorics graph-theory







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  • 2




    The name for what you're asking about is "Hamiltonian path." The short answer to your question is "a lot." According to this paper (ftp.inf.ethz.ch/doc/tech-reports/6xx/601.pdf), the number of Hamiltonian cycles on the $n$-dimensional hypercube $Q_n$ grows superexponentially; there are $6$ cycles (all automorphic) for $n = 3$, but for $n = 7$, the best known upper bound is about $2.5 times 10^{67}$. Note that every Hamiltonian cycle on $Q_n$ yields $2^n$ distinct paths by removing one edge of the cycle, and there are paths that can't be completed into cycles.
    – Connor Harris
    Nov 29 '18 at 15:14












  • You could try a more-or-less brute-force computer search, but I don't think this would be tractable beyond $n = 5$.
    – Connor Harris
    Nov 29 '18 at 15:15












  • @ConnorHarris If only an upper bound is known, feel free to post that as an answer and I will accept it. Unless of course someone decides to announce their new result here. ;)
    – Display Name
    Nov 29 '18 at 15:42












  • Well, $n!$ is an immediate lower bound; let $Q=u^0u^1ldots u^{2^n}$ be a Hamiltonian path on $H_n$ and let $pi$ be any permutation in $S_n$. Then let $pi(Q) = u^0v^1 ldots v^{2^n}$ be the Hamiltonian path specified as follows: let $k$ be such that $u^j$ and $u^{j+1}$ differ precisely in the $k$-th bit. Then $v^j$ and $v^{j+1}$ differ precisely in the $pi(k)$-th bit.
    – Mike
    Nov 29 '18 at 16:10








  • 1




    See OEIS sequences A003043 and A091299.
    – Robert Israel
    Nov 29 '18 at 18:31


















2














I have an N-dimensional hypercube, and would like to traverse the edges so that I reach each vertex exactly once. One path that satisfies this property is known as the Gray code, but there are many others. How many such paths are there? Paths that can be mapped to each other by the symmetries of the hypercube may or may not be counted as distinct, depending on your preference.






combinatorics graph-theory







share|cite|improve this question


















  • 2




    The name for what you're asking about is "Hamiltonian path." The short answer to your question is "a lot." According to this paper (ftp.inf.ethz.ch/doc/tech-reports/6xx/601.pdf), the number of Hamiltonian cycles on the $n$-dimensional hypercube $Q_n$ grows superexponentially; there are $6$ cycles (all automorphic) for $n = 3$, but for $n = 7$, the best known upper bound is about $2.5 times 10^{67}$. Note that every Hamiltonian cycle on $Q_n$ yields $2^n$ distinct paths by removing one edge of the cycle, and there are paths that can't be completed into cycles.
    – Connor Harris
    Nov 29 '18 at 15:14












  • You could try a more-or-less brute-force computer search, but I don't think this would be tractable beyond $n = 5$.
    – Connor Harris
    Nov 29 '18 at 15:15












  • @ConnorHarris If only an upper bound is known, feel free to post that as an answer and I will accept it. Unless of course someone decides to announce their new result here. ;)
    – Display Name
    Nov 29 '18 at 15:42












  • Well, $n!$ is an immediate lower bound; let $Q=u^0u^1ldots u^{2^n}$ be a Hamiltonian path on $H_n$ and let $pi$ be any permutation in $S_n$. Then let $pi(Q) = u^0v^1 ldots v^{2^n}$ be the Hamiltonian path specified as follows: let $k$ be such that $u^j$ and $u^{j+1}$ differ precisely in the $k$-th bit. Then $v^j$ and $v^{j+1}$ differ precisely in the $pi(k)$-th bit.
    – Mike
    Nov 29 '18 at 16:10








  • 1




    See OEIS sequences A003043 and A091299.
    – Robert Israel
    Nov 29 '18 at 18:31
















2












2








2







I have an N-dimensional hypercube, and would like to traverse the edges so that I reach each vertex exactly once. One path that satisfies this property is known as the Gray code, but there are many others. How many such paths are there? Paths that can be mapped to each other by the symmetries of the hypercube may or may not be counted as distinct, depending on your preference.






combinatorics graph-theory







share|cite|improve this question













I have an N-dimensional hypercube, and would like to traverse the edges so that I reach each vertex exactly once. One path that satisfies this property is known as the Gray code, but there are many others. How many such paths are there? Paths that can be mapped to each other by the symmetries of the hypercube may or may not be counted as distinct, depending on your preference.






combinatorics graph-theory




combinatorics graph-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 14:49









Display NameDisplay Name

2008




2008








  • 2




    The name for what you're asking about is "Hamiltonian path." The short answer to your question is "a lot." According to this paper (ftp.inf.ethz.ch/doc/tech-reports/6xx/601.pdf), the number of Hamiltonian cycles on the $n$-dimensional hypercube $Q_n$ grows superexponentially; there are $6$ cycles (all automorphic) for $n = 3$, but for $n = 7$, the best known upper bound is about $2.5 times 10^{67}$. Note that every Hamiltonian cycle on $Q_n$ yields $2^n$ distinct paths by removing one edge of the cycle, and there are paths that can't be completed into cycles.
    – Connor Harris
    Nov 29 '18 at 15:14












  • You could try a more-or-less brute-force computer search, but I don't think this would be tractable beyond $n = 5$.
    – Connor Harris
    Nov 29 '18 at 15:15












  • @ConnorHarris If only an upper bound is known, feel free to post that as an answer and I will accept it. Unless of course someone decides to announce their new result here. ;)
    – Display Name
    Nov 29 '18 at 15:42












  • Well, $n!$ is an immediate lower bound; let $Q=u^0u^1ldots u^{2^n}$ be a Hamiltonian path on $H_n$ and let $pi$ be any permutation in $S_n$. Then let $pi(Q) = u^0v^1 ldots v^{2^n}$ be the Hamiltonian path specified as follows: let $k$ be such that $u^j$ and $u^{j+1}$ differ precisely in the $k$-th bit. Then $v^j$ and $v^{j+1}$ differ precisely in the $pi(k)$-th bit.
    – Mike
    Nov 29 '18 at 16:10








  • 1




    See OEIS sequences A003043 and A091299.
    – Robert Israel
    Nov 29 '18 at 18:31
















  • 2




    The name for what you're asking about is "Hamiltonian path." The short answer to your question is "a lot." According to this paper (ftp.inf.ethz.ch/doc/tech-reports/6xx/601.pdf), the number of Hamiltonian cycles on the $n$-dimensional hypercube $Q_n$ grows superexponentially; there are $6$ cycles (all automorphic) for $n = 3$, but for $n = 7$, the best known upper bound is about $2.5 times 10^{67}$. Note that every Hamiltonian cycle on $Q_n$ yields $2^n$ distinct paths by removing one edge of the cycle, and there are paths that can't be completed into cycles.
    – Connor Harris
    Nov 29 '18 at 15:14












  • You could try a more-or-less brute-force computer search, but I don't think this would be tractable beyond $n = 5$.
    – Connor Harris
    Nov 29 '18 at 15:15












  • @ConnorHarris If only an upper bound is known, feel free to post that as an answer and I will accept it. Unless of course someone decides to announce their new result here. ;)
    – Display Name
    Nov 29 '18 at 15:42












  • Well, $n!$ is an immediate lower bound; let $Q=u^0u^1ldots u^{2^n}$ be a Hamiltonian path on $H_n$ and let $pi$ be any permutation in $S_n$. Then let $pi(Q) = u^0v^1 ldots v^{2^n}$ be the Hamiltonian path specified as follows: let $k$ be such that $u^j$ and $u^{j+1}$ differ precisely in the $k$-th bit. Then $v^j$ and $v^{j+1}$ differ precisely in the $pi(k)$-th bit.
    – Mike
    Nov 29 '18 at 16:10








  • 1




    See OEIS sequences A003043 and A091299.
    – Robert Israel
    Nov 29 '18 at 18:31










2




2




The name for what you're asking about is "Hamiltonian path." The short answer to your question is "a lot." According to this paper (ftp.inf.ethz.ch/doc/tech-reports/6xx/601.pdf), the number of Hamiltonian cycles on the $n$-dimensional hypercube $Q_n$ grows superexponentially; there are $6$ cycles (all automorphic) for $n = 3$, but for $n = 7$, the best known upper bound is about $2.5 times 10^{67}$. Note that every Hamiltonian cycle on $Q_n$ yields $2^n$ distinct paths by removing one edge of the cycle, and there are paths that can't be completed into cycles.
– Connor Harris
Nov 29 '18 at 15:14






The name for what you're asking about is "Hamiltonian path." The short answer to your question is "a lot." According to this paper (ftp.inf.ethz.ch/doc/tech-reports/6xx/601.pdf), the number of Hamiltonian cycles on the $n$-dimensional hypercube $Q_n$ grows superexponentially; there are $6$ cycles (all automorphic) for $n = 3$, but for $n = 7$, the best known upper bound is about $2.5 times 10^{67}$. Note that every Hamiltonian cycle on $Q_n$ yields $2^n$ distinct paths by removing one edge of the cycle, and there are paths that can't be completed into cycles.
– Connor Harris
Nov 29 '18 at 15:14














You could try a more-or-less brute-force computer search, but I don't think this would be tractable beyond $n = 5$.
– Connor Harris
Nov 29 '18 at 15:15






You could try a more-or-less brute-force computer search, but I don't think this would be tractable beyond $n = 5$.
– Connor Harris
Nov 29 '18 at 15:15














@ConnorHarris If only an upper bound is known, feel free to post that as an answer and I will accept it. Unless of course someone decides to announce their new result here. ;)
– Display Name
Nov 29 '18 at 15:42






@ConnorHarris If only an upper bound is known, feel free to post that as an answer and I will accept it. Unless of course someone decides to announce their new result here. ;)
– Display Name
Nov 29 '18 at 15:42














Well, $n!$ is an immediate lower bound; let $Q=u^0u^1ldots u^{2^n}$ be a Hamiltonian path on $H_n$ and let $pi$ be any permutation in $S_n$. Then let $pi(Q) = u^0v^1 ldots v^{2^n}$ be the Hamiltonian path specified as follows: let $k$ be such that $u^j$ and $u^{j+1}$ differ precisely in the $k$-th bit. Then $v^j$ and $v^{j+1}$ differ precisely in the $pi(k)$-th bit.
– Mike
Nov 29 '18 at 16:10






Well, $n!$ is an immediate lower bound; let $Q=u^0u^1ldots u^{2^n}$ be a Hamiltonian path on $H_n$ and let $pi$ be any permutation in $S_n$. Then let $pi(Q) = u^0v^1 ldots v^{2^n}$ be the Hamiltonian path specified as follows: let $k$ be such that $u^j$ and $u^{j+1}$ differ precisely in the $k$-th bit. Then $v^j$ and $v^{j+1}$ differ precisely in the $pi(k)$-th bit.
– Mike
Nov 29 '18 at 16:10






1




1




See OEIS sequences A003043 and A091299.
– Robert Israel
Nov 29 '18 at 18:31






See OEIS sequences A003043 and A091299.
– Robert Israel
Nov 29 '18 at 18:31












1 Answer
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This really is a sketch of a lower bound--but is too long for the comments



The lower bound for the number $X_n$ of Hamiltonian cycles in $H_n$ is at least something like $2^{2^n}$. In fact, $X_{n+1} geq 2^{2^n}X_n$.



Indeed, let $Q = v^0v^1ldots v^{2^n}$ be a Hamiltonian cycle in $H_{n}$. We will construct $2^{2^{n}}$ Hamilonian cycles in $H_{n+1}$ from $Q$.




  1. Let $w^0w^1 ldots w^{2^n}$ be vertices such that (a) the first $n$ bits of $w^i$ are the same as the first $n$ bits of $v^i$, but the $n+1$-th bit of $w^i$ is free (for $i=0,1,2, ldots, 2^n-1$). So $2^{2^n}$ ways to chose the $w^i$s.


  2. Let $H^i_2$ be the 4-vertex Hamiltonian path in the 4-vertex hypercube ${v^i,v^{i+1},v^i+ {bf{e}}_{n+1}, v^{i+1}+{bf{e}}_{n+1} }$ starting from $w^i$ and ending at $w^{i+1}$, where ${bf{e}}_l$ is the element where the $l$-th coordinate is 1 and the rest are all 0. [Note that $w^i in {v^i,v^i+ {bf{e}}_{n+1}}$, and $w^{i+1} in {v^{i+1},v^{i+1}+ {bf{e}}_{n+1}}$]


  3. Then let $Q$ be the Hamiltonian cycle consisting of the $H^i_2$s concatonated. This specifies $2^{2^n}$ distinct Hamiltoninan cycles in $H_{n+1}$ from one Hamiltonian cycle $Q$ in $H_n$. In fact, if $X_n$ is the number of Hamiltoninan cycles in $H_n$ then $X_{n+1} geq 2^{2^n} X_n$.




The (trivial) upper bound $U_n$ for the number of Hamiltonian cycles in $H_n$ is $n^{2^n} = 2^{2^n log n}$ (given the $i$-th vertex of the Hamiltonian cycle there are only at most $n$ choices for the $(i+1)$-th vertex), so letting $L_n$ be the lower bound derived above, we note that $log log U_n le log log L_n + log log n$.






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    This really is a sketch of a lower bound--but is too long for the comments



    The lower bound for the number $X_n$ of Hamiltonian cycles in $H_n$ is at least something like $2^{2^n}$. In fact, $X_{n+1} geq 2^{2^n}X_n$.



    Indeed, let $Q = v^0v^1ldots v^{2^n}$ be a Hamiltonian cycle in $H_{n}$. We will construct $2^{2^{n}}$ Hamilonian cycles in $H_{n+1}$ from $Q$.




    1. Let $w^0w^1 ldots w^{2^n}$ be vertices such that (a) the first $n$ bits of $w^i$ are the same as the first $n$ bits of $v^i$, but the $n+1$-th bit of $w^i$ is free (for $i=0,1,2, ldots, 2^n-1$). So $2^{2^n}$ ways to chose the $w^i$s.


    2. Let $H^i_2$ be the 4-vertex Hamiltonian path in the 4-vertex hypercube ${v^i,v^{i+1},v^i+ {bf{e}}_{n+1}, v^{i+1}+{bf{e}}_{n+1} }$ starting from $w^i$ and ending at $w^{i+1}$, where ${bf{e}}_l$ is the element where the $l$-th coordinate is 1 and the rest are all 0. [Note that $w^i in {v^i,v^i+ {bf{e}}_{n+1}}$, and $w^{i+1} in {v^{i+1},v^{i+1}+ {bf{e}}_{n+1}}$]


    3. Then let $Q$ be the Hamiltonian cycle consisting of the $H^i_2$s concatonated. This specifies $2^{2^n}$ distinct Hamiltoninan cycles in $H_{n+1}$ from one Hamiltonian cycle $Q$ in $H_n$. In fact, if $X_n$ is the number of Hamiltoninan cycles in $H_n$ then $X_{n+1} geq 2^{2^n} X_n$.




    The (trivial) upper bound $U_n$ for the number of Hamiltonian cycles in $H_n$ is $n^{2^n} = 2^{2^n log n}$ (given the $i$-th vertex of the Hamiltonian cycle there are only at most $n$ choices for the $(i+1)$-th vertex), so letting $L_n$ be the lower bound derived above, we note that $log log U_n le log log L_n + log log n$.






    share|cite|improve this answer




























      1














      This really is a sketch of a lower bound--but is too long for the comments



      The lower bound for the number $X_n$ of Hamiltonian cycles in $H_n$ is at least something like $2^{2^n}$. In fact, $X_{n+1} geq 2^{2^n}X_n$.



      Indeed, let $Q = v^0v^1ldots v^{2^n}$ be a Hamiltonian cycle in $H_{n}$. We will construct $2^{2^{n}}$ Hamilonian cycles in $H_{n+1}$ from $Q$.




      1. Let $w^0w^1 ldots w^{2^n}$ be vertices such that (a) the first $n$ bits of $w^i$ are the same as the first $n$ bits of $v^i$, but the $n+1$-th bit of $w^i$ is free (for $i=0,1,2, ldots, 2^n-1$). So $2^{2^n}$ ways to chose the $w^i$s.


      2. Let $H^i_2$ be the 4-vertex Hamiltonian path in the 4-vertex hypercube ${v^i,v^{i+1},v^i+ {bf{e}}_{n+1}, v^{i+1}+{bf{e}}_{n+1} }$ starting from $w^i$ and ending at $w^{i+1}$, where ${bf{e}}_l$ is the element where the $l$-th coordinate is 1 and the rest are all 0. [Note that $w^i in {v^i,v^i+ {bf{e}}_{n+1}}$, and $w^{i+1} in {v^{i+1},v^{i+1}+ {bf{e}}_{n+1}}$]


      3. Then let $Q$ be the Hamiltonian cycle consisting of the $H^i_2$s concatonated. This specifies $2^{2^n}$ distinct Hamiltoninan cycles in $H_{n+1}$ from one Hamiltonian cycle $Q$ in $H_n$. In fact, if $X_n$ is the number of Hamiltoninan cycles in $H_n$ then $X_{n+1} geq 2^{2^n} X_n$.




      The (trivial) upper bound $U_n$ for the number of Hamiltonian cycles in $H_n$ is $n^{2^n} = 2^{2^n log n}$ (given the $i$-th vertex of the Hamiltonian cycle there are only at most $n$ choices for the $(i+1)$-th vertex), so letting $L_n$ be the lower bound derived above, we note that $log log U_n le log log L_n + log log n$.






      share|cite|improve this answer


























        1












        1








        1






        This really is a sketch of a lower bound--but is too long for the comments



        The lower bound for the number $X_n$ of Hamiltonian cycles in $H_n$ is at least something like $2^{2^n}$. In fact, $X_{n+1} geq 2^{2^n}X_n$.



        Indeed, let $Q = v^0v^1ldots v^{2^n}$ be a Hamiltonian cycle in $H_{n}$. We will construct $2^{2^{n}}$ Hamilonian cycles in $H_{n+1}$ from $Q$.




        1. Let $w^0w^1 ldots w^{2^n}$ be vertices such that (a) the first $n$ bits of $w^i$ are the same as the first $n$ bits of $v^i$, but the $n+1$-th bit of $w^i$ is free (for $i=0,1,2, ldots, 2^n-1$). So $2^{2^n}$ ways to chose the $w^i$s.


        2. Let $H^i_2$ be the 4-vertex Hamiltonian path in the 4-vertex hypercube ${v^i,v^{i+1},v^i+ {bf{e}}_{n+1}, v^{i+1}+{bf{e}}_{n+1} }$ starting from $w^i$ and ending at $w^{i+1}$, where ${bf{e}}_l$ is the element where the $l$-th coordinate is 1 and the rest are all 0. [Note that $w^i in {v^i,v^i+ {bf{e}}_{n+1}}$, and $w^{i+1} in {v^{i+1},v^{i+1}+ {bf{e}}_{n+1}}$]


        3. Then let $Q$ be the Hamiltonian cycle consisting of the $H^i_2$s concatonated. This specifies $2^{2^n}$ distinct Hamiltoninan cycles in $H_{n+1}$ from one Hamiltonian cycle $Q$ in $H_n$. In fact, if $X_n$ is the number of Hamiltoninan cycles in $H_n$ then $X_{n+1} geq 2^{2^n} X_n$.




        The (trivial) upper bound $U_n$ for the number of Hamiltonian cycles in $H_n$ is $n^{2^n} = 2^{2^n log n}$ (given the $i$-th vertex of the Hamiltonian cycle there are only at most $n$ choices for the $(i+1)$-th vertex), so letting $L_n$ be the lower bound derived above, we note that $log log U_n le log log L_n + log log n$.






        share|cite|improve this answer














        This really is a sketch of a lower bound--but is too long for the comments



        The lower bound for the number $X_n$ of Hamiltonian cycles in $H_n$ is at least something like $2^{2^n}$. In fact, $X_{n+1} geq 2^{2^n}X_n$.



        Indeed, let $Q = v^0v^1ldots v^{2^n}$ be a Hamiltonian cycle in $H_{n}$. We will construct $2^{2^{n}}$ Hamilonian cycles in $H_{n+1}$ from $Q$.




        1. Let $w^0w^1 ldots w^{2^n}$ be vertices such that (a) the first $n$ bits of $w^i$ are the same as the first $n$ bits of $v^i$, but the $n+1$-th bit of $w^i$ is free (for $i=0,1,2, ldots, 2^n-1$). So $2^{2^n}$ ways to chose the $w^i$s.


        2. Let $H^i_2$ be the 4-vertex Hamiltonian path in the 4-vertex hypercube ${v^i,v^{i+1},v^i+ {bf{e}}_{n+1}, v^{i+1}+{bf{e}}_{n+1} }$ starting from $w^i$ and ending at $w^{i+1}$, where ${bf{e}}_l$ is the element where the $l$-th coordinate is 1 and the rest are all 0. [Note that $w^i in {v^i,v^i+ {bf{e}}_{n+1}}$, and $w^{i+1} in {v^{i+1},v^{i+1}+ {bf{e}}_{n+1}}$]


        3. Then let $Q$ be the Hamiltonian cycle consisting of the $H^i_2$s concatonated. This specifies $2^{2^n}$ distinct Hamiltoninan cycles in $H_{n+1}$ from one Hamiltonian cycle $Q$ in $H_n$. In fact, if $X_n$ is the number of Hamiltoninan cycles in $H_n$ then $X_{n+1} geq 2^{2^n} X_n$.




        The (trivial) upper bound $U_n$ for the number of Hamiltonian cycles in $H_n$ is $n^{2^n} = 2^{2^n log n}$ (given the $i$-th vertex of the Hamiltonian cycle there are only at most $n$ choices for the $(i+1)$-th vertex), so letting $L_n$ be the lower bound derived above, we note that $log log U_n le log log L_n + log log n$.







        share|cite|improve this answer














        share|cite|improve this answer



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        edited Nov 29 '18 at 19:48

























        answered Nov 29 '18 at 16:50









        MikeMike

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            Willebadessen

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            Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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            Residenzschloss Arolsen

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            Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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