Prove that the even cycles are 2-list-colorable.












1












$begingroup$


I need to prove the statement in the title. It's very easy to show that with a particular 2-list-assignment, we can have a proper 2-list-coloring (e.g. with a greedy coloring algorithm). I'm just not sure if that's enough to show that the statement is true for ALL 2-list-assignments.



Here are the relevant definitions.




8.4.24. Definition. For each vertex $v$ in a graph $G$, let $L(v)$ denote a list of colors available at $v$. A list coloring or choice function is a proper coloring $f$ such that $f(v) in L(v)$ for all $v$. A graph is $k$-choosable or list $k$-colorable if every assignment of $k$-element lists to the vertices permits a proper list coloring. The list chromatic number, choice number, or choosability $chi_l(G)$ is the minimum $k$ such that $G$ is $k$-choosable.







proof-verification graph-theory coloring







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I need to prove the statement in the title. It's very easy to show that with a particular 2-list-assignment, we can have a proper 2-list-coloring (e.g. with a greedy coloring algorithm). I'm just not sure if that's enough to show that the statement is true for ALL 2-list-assignments.



    Here are the relevant definitions.




    8.4.24. Definition. For each vertex $v$ in a graph $G$, let $L(v)$ denote a list of colors available at $v$. A list coloring or choice function is a proper coloring $f$ such that $f(v) in L(v)$ for all $v$. A graph is $k$-choosable or list $k$-colorable if every assignment of $k$-element lists to the vertices permits a proper list coloring. The list chromatic number, choice number, or choosability $chi_l(G)$ is the minimum $k$ such that $G$ is $k$-choosable.







    proof-verification graph-theory coloring







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I need to prove the statement in the title. It's very easy to show that with a particular 2-list-assignment, we can have a proper 2-list-coloring (e.g. with a greedy coloring algorithm). I'm just not sure if that's enough to show that the statement is true for ALL 2-list-assignments.



      Here are the relevant definitions.




      8.4.24. Definition. For each vertex $v$ in a graph $G$, let $L(v)$ denote a list of colors available at $v$. A list coloring or choice function is a proper coloring $f$ such that $f(v) in L(v)$ for all $v$. A graph is $k$-choosable or list $k$-colorable if every assignment of $k$-element lists to the vertices permits a proper list coloring. The list chromatic number, choice number, or choosability $chi_l(G)$ is the minimum $k$ such that $G$ is $k$-choosable.







      proof-verification graph-theory coloring







      share|cite|improve this question











      $endgroup$




      I need to prove the statement in the title. It's very easy to show that with a particular 2-list-assignment, we can have a proper 2-list-coloring (e.g. with a greedy coloring algorithm). I'm just not sure if that's enough to show that the statement is true for ALL 2-list-assignments.



      Here are the relevant definitions.




      8.4.24. Definition. For each vertex $v$ in a graph $G$, let $L(v)$ denote a list of colors available at $v$. A list coloring or choice function is a proper coloring $f$ such that $f(v) in L(v)$ for all $v$. A graph is $k$-choosable or list $k$-colorable if every assignment of $k$-element lists to the vertices permits a proper list coloring. The list chromatic number, choice number, or choosability $chi_l(G)$ is the minimum $k$ such that $G$ is $k$-choosable.







      proof-verification graph-theory coloring




      proof-verification graph-theory coloring






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 20:08









      Misha Lavrov

      46.2k656107




      46.2k656107










      asked Dec 9 '18 at 17:01









      ensbanaensbana

      256213




      256213






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          There are actually some cases in which a greedy coloring algorithm will not work. For instance, suppose you have a $4$-cycle with the following color lists:




          1. Red and blue

          2. Red and green

          3. Green and yellow

          4. Red and yellow


          A greedy algorithm will color vertex 1 red, then vertex 2 green, then vertex 3 yellow, and then get stuck: vertex 4 is adjacent to vertices 1 and 3, which are red and yellow respectively, so there is no way to color it.



          It's possible to get this greedy algorithm "unstuck" by starting appropriately. If we began by coloring vertex 1 blue and then used the greedy algorithm, we would not run into the problem. (It's easy to see that we can get stuck only on the last vertex we color.) So you need a strategy for picking how to start, to avoid getting stuck on the last vertex.



          Here's one possible strategy. (I urge you to think about it yourself before looking at the spoiler, because now you know what the problem is with the greedy strategy - and have the chance to figure out yourself how to avoid that problem.)




          Pick two adjacent vertices $i, i+1$ which don't have the same color lists. Then color $i+1$ with a color not available for $i$ and proceed to color $i+2, i+3, dots$ greedily. Each vertex other than the last vertex (vertex $i$) can be colored without problems. Vertex $i$ is guaranteed not to conflict with vertex $i+1$, so it also can be colored: at most one of its options is ruled out by the color of vertex $i-1$.



          This works in all cases except when all vertices have the same color lists, in which case we're back in the world of ordinary 2-coloring, where we know what happens.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the answer. I've thought about your suggestions and come up with this idea, which is more or less similar to yours. I'd still like to hear your opinion though: instead of picking two vertices with the property like you stated, can we just start with any vertex, and look at the vertex "behind" it (i.e. the end of the cycle)? Let $v_1$ and $v_{2n}$ be those two vertices, respectively. If $L(v_1) cap L(v_{2n}) = 1$, then we color $v_1$ with the color not available to $v_{2n}$ and then proceed, otherwise we just apply the greedy algorithm like usuals.
            $endgroup$
            – ensbana
            Dec 9 '18 at 20:38






          • 1




            $begingroup$
            In the other case $|L(v_1) cap L(v_{2n})| = 2$, it's possible that later on you'll have two choices, and only one of them is correct - so just applying the greedy algorithm won't work. Say, you have two choices for $L(v_{2n-1})$, and one of them rules out the remaining option on $L(v_{2n})$. Or you have two choices for $L(v_{2n-2})$, one of which forces the color of $v_{2n-1}$ to be the color that rules out the remaining option on $L(v_{2n})$.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:46






          • 1




            $begingroup$
            My suggestion is to start at a vertex with choices, so that you can make a choice at the very beginning that solves the problem at the last vertex, and then proceed greedily. You can also give an argument that waits to make such a choice at a vertex other than the first vertex, but I think that's trickier.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:47











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          1 Answer
          1






          active

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          active

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          active

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          2












          $begingroup$

          There are actually some cases in which a greedy coloring algorithm will not work. For instance, suppose you have a $4$-cycle with the following color lists:




          1. Red and blue

          2. Red and green

          3. Green and yellow

          4. Red and yellow


          A greedy algorithm will color vertex 1 red, then vertex 2 green, then vertex 3 yellow, and then get stuck: vertex 4 is adjacent to vertices 1 and 3, which are red and yellow respectively, so there is no way to color it.



          It's possible to get this greedy algorithm "unstuck" by starting appropriately. If we began by coloring vertex 1 blue and then used the greedy algorithm, we would not run into the problem. (It's easy to see that we can get stuck only on the last vertex we color.) So you need a strategy for picking how to start, to avoid getting stuck on the last vertex.



          Here's one possible strategy. (I urge you to think about it yourself before looking at the spoiler, because now you know what the problem is with the greedy strategy - and have the chance to figure out yourself how to avoid that problem.)




          Pick two adjacent vertices $i, i+1$ which don't have the same color lists. Then color $i+1$ with a color not available for $i$ and proceed to color $i+2, i+3, dots$ greedily. Each vertex other than the last vertex (vertex $i$) can be colored without problems. Vertex $i$ is guaranteed not to conflict with vertex $i+1$, so it also can be colored: at most one of its options is ruled out by the color of vertex $i-1$.



          This works in all cases except when all vertices have the same color lists, in which case we're back in the world of ordinary 2-coloring, where we know what happens.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the answer. I've thought about your suggestions and come up with this idea, which is more or less similar to yours. I'd still like to hear your opinion though: instead of picking two vertices with the property like you stated, can we just start with any vertex, and look at the vertex "behind" it (i.e. the end of the cycle)? Let $v_1$ and $v_{2n}$ be those two vertices, respectively. If $L(v_1) cap L(v_{2n}) = 1$, then we color $v_1$ with the color not available to $v_{2n}$ and then proceed, otherwise we just apply the greedy algorithm like usuals.
            $endgroup$
            – ensbana
            Dec 9 '18 at 20:38






          • 1




            $begingroup$
            In the other case $|L(v_1) cap L(v_{2n})| = 2$, it's possible that later on you'll have two choices, and only one of them is correct - so just applying the greedy algorithm won't work. Say, you have two choices for $L(v_{2n-1})$, and one of them rules out the remaining option on $L(v_{2n})$. Or you have two choices for $L(v_{2n-2})$, one of which forces the color of $v_{2n-1}$ to be the color that rules out the remaining option on $L(v_{2n})$.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:46






          • 1




            $begingroup$
            My suggestion is to start at a vertex with choices, so that you can make a choice at the very beginning that solves the problem at the last vertex, and then proceed greedily. You can also give an argument that waits to make such a choice at a vertex other than the first vertex, but I think that's trickier.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:47
















          2












          $begingroup$

          There are actually some cases in which a greedy coloring algorithm will not work. For instance, suppose you have a $4$-cycle with the following color lists:




          1. Red and blue

          2. Red and green

          3. Green and yellow

          4. Red and yellow


          A greedy algorithm will color vertex 1 red, then vertex 2 green, then vertex 3 yellow, and then get stuck: vertex 4 is adjacent to vertices 1 and 3, which are red and yellow respectively, so there is no way to color it.



          It's possible to get this greedy algorithm "unstuck" by starting appropriately. If we began by coloring vertex 1 blue and then used the greedy algorithm, we would not run into the problem. (It's easy to see that we can get stuck only on the last vertex we color.) So you need a strategy for picking how to start, to avoid getting stuck on the last vertex.



          Here's one possible strategy. (I urge you to think about it yourself before looking at the spoiler, because now you know what the problem is with the greedy strategy - and have the chance to figure out yourself how to avoid that problem.)




          Pick two adjacent vertices $i, i+1$ which don't have the same color lists. Then color $i+1$ with a color not available for $i$ and proceed to color $i+2, i+3, dots$ greedily. Each vertex other than the last vertex (vertex $i$) can be colored without problems. Vertex $i$ is guaranteed not to conflict with vertex $i+1$, so it also can be colored: at most one of its options is ruled out by the color of vertex $i-1$.



          This works in all cases except when all vertices have the same color lists, in which case we're back in the world of ordinary 2-coloring, where we know what happens.







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for the answer. I've thought about your suggestions and come up with this idea, which is more or less similar to yours. I'd still like to hear your opinion though: instead of picking two vertices with the property like you stated, can we just start with any vertex, and look at the vertex "behind" it (i.e. the end of the cycle)? Let $v_1$ and $v_{2n}$ be those two vertices, respectively. If $L(v_1) cap L(v_{2n}) = 1$, then we color $v_1$ with the color not available to $v_{2n}$ and then proceed, otherwise we just apply the greedy algorithm like usuals.
            $endgroup$
            – ensbana
            Dec 9 '18 at 20:38






          • 1




            $begingroup$
            In the other case $|L(v_1) cap L(v_{2n})| = 2$, it's possible that later on you'll have two choices, and only one of them is correct - so just applying the greedy algorithm won't work. Say, you have two choices for $L(v_{2n-1})$, and one of them rules out the remaining option on $L(v_{2n})$. Or you have two choices for $L(v_{2n-2})$, one of which forces the color of $v_{2n-1}$ to be the color that rules out the remaining option on $L(v_{2n})$.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:46






          • 1




            $begingroup$
            My suggestion is to start at a vertex with choices, so that you can make a choice at the very beginning that solves the problem at the last vertex, and then proceed greedily. You can also give an argument that waits to make such a choice at a vertex other than the first vertex, but I think that's trickier.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:47














          2












          2








          2





          $begingroup$

          There are actually some cases in which a greedy coloring algorithm will not work. For instance, suppose you have a $4$-cycle with the following color lists:




          1. Red and blue

          2. Red and green

          3. Green and yellow

          4. Red and yellow


          A greedy algorithm will color vertex 1 red, then vertex 2 green, then vertex 3 yellow, and then get stuck: vertex 4 is adjacent to vertices 1 and 3, which are red and yellow respectively, so there is no way to color it.



          It's possible to get this greedy algorithm "unstuck" by starting appropriately. If we began by coloring vertex 1 blue and then used the greedy algorithm, we would not run into the problem. (It's easy to see that we can get stuck only on the last vertex we color.) So you need a strategy for picking how to start, to avoid getting stuck on the last vertex.



          Here's one possible strategy. (I urge you to think about it yourself before looking at the spoiler, because now you know what the problem is with the greedy strategy - and have the chance to figure out yourself how to avoid that problem.)




          Pick two adjacent vertices $i, i+1$ which don't have the same color lists. Then color $i+1$ with a color not available for $i$ and proceed to color $i+2, i+3, dots$ greedily. Each vertex other than the last vertex (vertex $i$) can be colored without problems. Vertex $i$ is guaranteed not to conflict with vertex $i+1$, so it also can be colored: at most one of its options is ruled out by the color of vertex $i-1$.



          This works in all cases except when all vertices have the same color lists, in which case we're back in the world of ordinary 2-coloring, where we know what happens.







          share|cite|improve this answer









          $endgroup$



          There are actually some cases in which a greedy coloring algorithm will not work. For instance, suppose you have a $4$-cycle with the following color lists:




          1. Red and blue

          2. Red and green

          3. Green and yellow

          4. Red and yellow


          A greedy algorithm will color vertex 1 red, then vertex 2 green, then vertex 3 yellow, and then get stuck: vertex 4 is adjacent to vertices 1 and 3, which are red and yellow respectively, so there is no way to color it.



          It's possible to get this greedy algorithm "unstuck" by starting appropriately. If we began by coloring vertex 1 blue and then used the greedy algorithm, we would not run into the problem. (It's easy to see that we can get stuck only on the last vertex we color.) So you need a strategy for picking how to start, to avoid getting stuck on the last vertex.



          Here's one possible strategy. (I urge you to think about it yourself before looking at the spoiler, because now you know what the problem is with the greedy strategy - and have the chance to figure out yourself how to avoid that problem.)




          Pick two adjacent vertices $i, i+1$ which don't have the same color lists. Then color $i+1$ with a color not available for $i$ and proceed to color $i+2, i+3, dots$ greedily. Each vertex other than the last vertex (vertex $i$) can be colored without problems. Vertex $i$ is guaranteed not to conflict with vertex $i+1$, so it also can be colored: at most one of its options is ruled out by the color of vertex $i-1$.



          This works in all cases except when all vertices have the same color lists, in which case we're back in the world of ordinary 2-coloring, where we know what happens.








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 19:53









          Misha LavrovMisha Lavrov

          46.2k656107




          46.2k656107












          • $begingroup$
            Thanks for the answer. I've thought about your suggestions and come up with this idea, which is more or less similar to yours. I'd still like to hear your opinion though: instead of picking two vertices with the property like you stated, can we just start with any vertex, and look at the vertex "behind" it (i.e. the end of the cycle)? Let $v_1$ and $v_{2n}$ be those two vertices, respectively. If $L(v_1) cap L(v_{2n}) = 1$, then we color $v_1$ with the color not available to $v_{2n}$ and then proceed, otherwise we just apply the greedy algorithm like usuals.
            $endgroup$
            – ensbana
            Dec 9 '18 at 20:38






          • 1




            $begingroup$
            In the other case $|L(v_1) cap L(v_{2n})| = 2$, it's possible that later on you'll have two choices, and only one of them is correct - so just applying the greedy algorithm won't work. Say, you have two choices for $L(v_{2n-1})$, and one of them rules out the remaining option on $L(v_{2n})$. Or you have two choices for $L(v_{2n-2})$, one of which forces the color of $v_{2n-1}$ to be the color that rules out the remaining option on $L(v_{2n})$.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:46






          • 1




            $begingroup$
            My suggestion is to start at a vertex with choices, so that you can make a choice at the very beginning that solves the problem at the last vertex, and then proceed greedily. You can also give an argument that waits to make such a choice at a vertex other than the first vertex, but I think that's trickier.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:47


















          • $begingroup$
            Thanks for the answer. I've thought about your suggestions and come up with this idea, which is more or less similar to yours. I'd still like to hear your opinion though: instead of picking two vertices with the property like you stated, can we just start with any vertex, and look at the vertex "behind" it (i.e. the end of the cycle)? Let $v_1$ and $v_{2n}$ be those two vertices, respectively. If $L(v_1) cap L(v_{2n}) = 1$, then we color $v_1$ with the color not available to $v_{2n}$ and then proceed, otherwise we just apply the greedy algorithm like usuals.
            $endgroup$
            – ensbana
            Dec 9 '18 at 20:38






          • 1




            $begingroup$
            In the other case $|L(v_1) cap L(v_{2n})| = 2$, it's possible that later on you'll have two choices, and only one of them is correct - so just applying the greedy algorithm won't work. Say, you have two choices for $L(v_{2n-1})$, and one of them rules out the remaining option on $L(v_{2n})$. Or you have two choices for $L(v_{2n-2})$, one of which forces the color of $v_{2n-1}$ to be the color that rules out the remaining option on $L(v_{2n})$.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:46






          • 1




            $begingroup$
            My suggestion is to start at a vertex with choices, so that you can make a choice at the very beginning that solves the problem at the last vertex, and then proceed greedily. You can also give an argument that waits to make such a choice at a vertex other than the first vertex, but I think that's trickier.
            $endgroup$
            – Misha Lavrov
            Dec 9 '18 at 20:47
















          $begingroup$
          Thanks for the answer. I've thought about your suggestions and come up with this idea, which is more or less similar to yours. I'd still like to hear your opinion though: instead of picking two vertices with the property like you stated, can we just start with any vertex, and look at the vertex "behind" it (i.e. the end of the cycle)? Let $v_1$ and $v_{2n}$ be those two vertices, respectively. If $L(v_1) cap L(v_{2n}) = 1$, then we color $v_1$ with the color not available to $v_{2n}$ and then proceed, otherwise we just apply the greedy algorithm like usuals.
          $endgroup$
          – ensbana
          Dec 9 '18 at 20:38




          $begingroup$
          Thanks for the answer. I've thought about your suggestions and come up with this idea, which is more or less similar to yours. I'd still like to hear your opinion though: instead of picking two vertices with the property like you stated, can we just start with any vertex, and look at the vertex "behind" it (i.e. the end of the cycle)? Let $v_1$ and $v_{2n}$ be those two vertices, respectively. If $L(v_1) cap L(v_{2n}) = 1$, then we color $v_1$ with the color not available to $v_{2n}$ and then proceed, otherwise we just apply the greedy algorithm like usuals.
          $endgroup$
          – ensbana
          Dec 9 '18 at 20:38




          1




          1




          $begingroup$
          In the other case $|L(v_1) cap L(v_{2n})| = 2$, it's possible that later on you'll have two choices, and only one of them is correct - so just applying the greedy algorithm won't work. Say, you have two choices for $L(v_{2n-1})$, and one of them rules out the remaining option on $L(v_{2n})$. Or you have two choices for $L(v_{2n-2})$, one of which forces the color of $v_{2n-1}$ to be the color that rules out the remaining option on $L(v_{2n})$.
          $endgroup$
          – Misha Lavrov
          Dec 9 '18 at 20:46




          $begingroup$
          In the other case $|L(v_1) cap L(v_{2n})| = 2$, it's possible that later on you'll have two choices, and only one of them is correct - so just applying the greedy algorithm won't work. Say, you have two choices for $L(v_{2n-1})$, and one of them rules out the remaining option on $L(v_{2n})$. Or you have two choices for $L(v_{2n-2})$, one of which forces the color of $v_{2n-1}$ to be the color that rules out the remaining option on $L(v_{2n})$.
          $endgroup$
          – Misha Lavrov
          Dec 9 '18 at 20:46




          1




          1




          $begingroup$
          My suggestion is to start at a vertex with choices, so that you can make a choice at the very beginning that solves the problem at the last vertex, and then proceed greedily. You can also give an argument that waits to make such a choice at a vertex other than the first vertex, but I think that's trickier.
          $endgroup$
          – Misha Lavrov
          Dec 9 '18 at 20:47




          $begingroup$
          My suggestion is to start at a vertex with choices, so that you can make a choice at the very beginning that solves the problem at the last vertex, and then proceed greedily. You can also give an argument that waits to make such a choice at a vertex other than the first vertex, but I think that's trickier.
          $endgroup$
          – Misha Lavrov
          Dec 9 '18 at 20:47


















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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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