Show any two edges in a 2-connected graph lie on a cycle











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So I found some proofs on any two vertices would lie on a cycle, but stuck on dealing with edges. We can say any two edges are connected, but does that just imply they will be on a common cycle?






discrete-mathematics graph-theory connectedness graph-connectivity







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  • Which kind of $2$-connectedness are you referring to: $2$-edge-connected or $2$-vertex-connected? I would guess that $2$-edge-connected is what you are aiming for, but I can't say for sure.
    – Batominovski
    2 days ago












  • @Batominovski Here I'm referring to 2-vertex-connected.
    – Thomas
    2 days ago















up vote
1
down vote

favorite
1












So I found some proofs on any two vertices would lie on a cycle, but stuck on dealing with edges. We can say any two edges are connected, but does that just imply they will be on a common cycle?






discrete-mathematics graph-theory connectedness graph-connectivity







share|cite|improve this question
























  • Which kind of $2$-connectedness are you referring to: $2$-edge-connected or $2$-vertex-connected? I would guess that $2$-edge-connected is what you are aiming for, but I can't say for sure.
    – Batominovski
    2 days ago












  • @Batominovski Here I'm referring to 2-vertex-connected.
    – Thomas
    2 days ago













up vote
1
down vote

favorite
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up vote
1
down vote

favorite
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1





So I found some proofs on any two vertices would lie on a cycle, but stuck on dealing with edges. We can say any two edges are connected, but does that just imply they will be on a common cycle?






discrete-mathematics graph-theory connectedness graph-connectivity







share|cite|improve this question















So I found some proofs on any two vertices would lie on a cycle, but stuck on dealing with edges. We can say any two edges are connected, but does that just imply they will be on a common cycle?






discrete-mathematics graph-theory connectedness graph-connectivity




discrete-mathematics graph-theory connectedness graph-connectivity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited 2 days ago









Batominovski

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31.5k23187










asked 2 days ago









Thomas

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896












  • Which kind of $2$-connectedness are you referring to: $2$-edge-connected or $2$-vertex-connected? I would guess that $2$-edge-connected is what you are aiming for, but I can't say for sure.
    – Batominovski
    2 days ago












  • @Batominovski Here I'm referring to 2-vertex-connected.
    – Thomas
    2 days ago


















  • Which kind of $2$-connectedness are you referring to: $2$-edge-connected or $2$-vertex-connected? I would guess that $2$-edge-connected is what you are aiming for, but I can't say for sure.
    – Batominovski
    2 days ago












  • @Batominovski Here I'm referring to 2-vertex-connected.
    – Thomas
    2 days ago
















Which kind of $2$-connectedness are you referring to: $2$-edge-connected or $2$-vertex-connected? I would guess that $2$-edge-connected is what you are aiming for, but I can't say for sure.
– Batominovski
2 days ago






Which kind of $2$-connectedness are you referring to: $2$-edge-connected or $2$-vertex-connected? I would guess that $2$-edge-connected is what you are aiming for, but I can't say for sure.
– Batominovski
2 days ago














@Batominovski Here I'm referring to 2-vertex-connected.
– Thomas
2 days ago




@Batominovski Here I'm referring to 2-vertex-connected.
– Thomas
2 days ago










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Let $G = (V, E)$ graph, $|V| ge 3$. Suppose that there exists an edge $e = (u, v) in E$ such that $e$ does not lie on a cycle. Then every path from $u$ to $v$ must contain $e$ (1). If $u$ or $v$ are a leaf, then removing the vertex that connects the leaf from the rest of the graph makes the leaf disconnected (!). Otherwise, define the following partition of $G$ without $e$:



$$G_u = (V_u, E_u) quad quad G_v = (V_v, E_v)$$



where $V_u = {w in V |$ there is a path from $w$ to $u$ that does not contain $e}$, $E_u = {(w, t) in E | w, t in V_u}$ and $V_v$, $E_v$ are defined analogously (2). If we remove $u$ or $v$ from the graph, then we are also removing $e$. Note, however, that if we remove $e$, then there is no path to go from any $w in G_u$ to any $t in G_v$. In other words, the graph becomes disconnected (!).



Notes:



(1) Suppose that there exists a path from $u$ to $v$ that does not contain $e$. Then we can join this path to $e$ to form a cycle, but by doing so we are showing that $e$ lies on a cycle (!).



(2) It is easy to show that $G_u$ and $G_v$ are disjoint.






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    Let $G = (V, E)$ graph, $|V| ge 3$. Suppose that there exists an edge $e = (u, v) in E$ such that $e$ does not lie on a cycle. Then every path from $u$ to $v$ must contain $e$ (1). If $u$ or $v$ are a leaf, then removing the vertex that connects the leaf from the rest of the graph makes the leaf disconnected (!). Otherwise, define the following partition of $G$ without $e$:



    $$G_u = (V_u, E_u) quad quad G_v = (V_v, E_v)$$



    where $V_u = {w in V |$ there is a path from $w$ to $u$ that does not contain $e}$, $E_u = {(w, t) in E | w, t in V_u}$ and $V_v$, $E_v$ are defined analogously (2). If we remove $u$ or $v$ from the graph, then we are also removing $e$. Note, however, that if we remove $e$, then there is no path to go from any $w in G_u$ to any $t in G_v$. In other words, the graph becomes disconnected (!).



    Notes:



    (1) Suppose that there exists a path from $u$ to $v$ that does not contain $e$. Then we can join this path to $e$ to form a cycle, but by doing so we are showing that $e$ lies on a cycle (!).



    (2) It is easy to show that $G_u$ and $G_v$ are disjoint.






    share|cite|improve this answer



























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      Let $G = (V, E)$ graph, $|V| ge 3$. Suppose that there exists an edge $e = (u, v) in E$ such that $e$ does not lie on a cycle. Then every path from $u$ to $v$ must contain $e$ (1). If $u$ or $v$ are a leaf, then removing the vertex that connects the leaf from the rest of the graph makes the leaf disconnected (!). Otherwise, define the following partition of $G$ without $e$:



      $$G_u = (V_u, E_u) quad quad G_v = (V_v, E_v)$$



      where $V_u = {w in V |$ there is a path from $w$ to $u$ that does not contain $e}$, $E_u = {(w, t) in E | w, t in V_u}$ and $V_v$, $E_v$ are defined analogously (2). If we remove $u$ or $v$ from the graph, then we are also removing $e$. Note, however, that if we remove $e$, then there is no path to go from any $w in G_u$ to any $t in G_v$. In other words, the graph becomes disconnected (!).



      Notes:



      (1) Suppose that there exists a path from $u$ to $v$ that does not contain $e$. Then we can join this path to $e$ to form a cycle, but by doing so we are showing that $e$ lies on a cycle (!).



      (2) It is easy to show that $G_u$ and $G_v$ are disjoint.






      share|cite|improve this answer

























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        up vote
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        Let $G = (V, E)$ graph, $|V| ge 3$. Suppose that there exists an edge $e = (u, v) in E$ such that $e$ does not lie on a cycle. Then every path from $u$ to $v$ must contain $e$ (1). If $u$ or $v$ are a leaf, then removing the vertex that connects the leaf from the rest of the graph makes the leaf disconnected (!). Otherwise, define the following partition of $G$ without $e$:



        $$G_u = (V_u, E_u) quad quad G_v = (V_v, E_v)$$



        where $V_u = {w in V |$ there is a path from $w$ to $u$ that does not contain $e}$, $E_u = {(w, t) in E | w, t in V_u}$ and $V_v$, $E_v$ are defined analogously (2). If we remove $u$ or $v$ from the graph, then we are also removing $e$. Note, however, that if we remove $e$, then there is no path to go from any $w in G_u$ to any $t in G_v$. In other words, the graph becomes disconnected (!).



        Notes:



        (1) Suppose that there exists a path from $u$ to $v$ that does not contain $e$. Then we can join this path to $e$ to form a cycle, but by doing so we are showing that $e$ lies on a cycle (!).



        (2) It is easy to show that $G_u$ and $G_v$ are disjoint.






        share|cite|improve this answer














        Let $G = (V, E)$ graph, $|V| ge 3$. Suppose that there exists an edge $e = (u, v) in E$ such that $e$ does not lie on a cycle. Then every path from $u$ to $v$ must contain $e$ (1). If $u$ or $v$ are a leaf, then removing the vertex that connects the leaf from the rest of the graph makes the leaf disconnected (!). Otherwise, define the following partition of $G$ without $e$:



        $$G_u = (V_u, E_u) quad quad G_v = (V_v, E_v)$$



        where $V_u = {w in V |$ there is a path from $w$ to $u$ that does not contain $e}$, $E_u = {(w, t) in E | w, t in V_u}$ and $V_v$, $E_v$ are defined analogously (2). If we remove $u$ or $v$ from the graph, then we are also removing $e$. Note, however, that if we remove $e$, then there is no path to go from any $w in G_u$ to any $t in G_v$. In other words, the graph becomes disconnected (!).



        Notes:



        (1) Suppose that there exists a path from $u$ to $v$ that does not contain $e$. Then we can join this path to $e$ to form a cycle, but by doing so we are showing that $e$ lies on a cycle (!).



        (2) It is easy to show that $G_u$ and $G_v$ are disjoint.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








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