Construction of graph with degrees $d$ and $(d + 1)$
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Let $n = a + b$ and $d$ be non-negative integers such that: $ad + b(d + 1)$ is even and $(d + 1) leq (n - 1)$. Does there exist a graph with $n$ vertices such that $a$ of them have degree $d$ and $b$ of them have degree $(d + 1)$? Is there an explicit construction?
graph-theory
asked Nov 17 at 16:09
user404944
738212
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Let $n = a + b$ and $d$ be non-negative integers such that: $ad + b(d + 1)$ is even and $(d + 1) leq (n - 1)$. Does there exist a graph with $n$ vertices such that $a$ of them have degree $d$ and $b$ of them have degree $(d + 1)$? Is there an explicit construction?
graph-theory
asked Nov 17 at 16:09
user404944
738212
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $n = a + b$ and $d$ be non-negative integers such that: $ad + b(d + 1)$ is even and $(d + 1) leq (n - 1)$. Does there exist a graph with $n$ vertices such that $a$ of them have degree $d$ and $b$ of them have degree $(d + 1)$? Is there an explicit construction?
graph-theory
asked Nov 17 at 16:09
user404944
738212
Let $n = a + b$ and $d$ be non-negative integers such that: $ad + b(d + 1)$ is even and $(d + 1) leq (n - 1)$. Does there exist a graph with $n$ vertices such that $a$ of them have degree $d$ and $b$ of them have degree $(d + 1)$? Is there an explicit construction?
graph-theory
graph-theory
asked Nov 17 at 16:09
user404944
738212
asked Nov 17 at 16:09
user404944
738212
asked Nov 17 at 16:09
user404944
738212
asked Nov 17 at 16:09
user404944
738212
asked Nov 17 at 16:09
user404944
738212
738212
add a comment |
add a comment |
1 Answer
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This is not the answer but maybe that can give some idea:
Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.
proof:
Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.
answered Nov 19 at 0:51
mathnoob
73411
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is not the answer but maybe that can give some idea:
Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.
proof:
Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.
answered Nov 19 at 0:51
mathnoob
73411
add a comment |
up vote
1
down vote
This is not the answer but maybe that can give some idea:
Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.
proof:
Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.
answered Nov 19 at 0:51
mathnoob
73411
add a comment |
up vote
1
down vote
up vote
1
down vote
This is not the answer but maybe that can give some idea:
Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.
proof:
Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.
answered Nov 19 at 0:51
mathnoob
73411
This is not the answer but maybe that can give some idea:
Fix an integer $kgeq3$. There exist a $(2k-1)$-vertex $(k-2)$-edge-connected simple graph $H_k=(V_k,E_k)$ with $V_k={y_1,y_2,...,y_{k+1},z_{1},z_{2},...,z_{k-2}}$, where all vertices $y_i$ have degree $k$ and all vertices $z_j$ have degree $(k-1)$.
proof:
Start with a k-vertex complete graph on the vertices ${y_1,...,y_k}$, plus a $(k-2)$-vertex complete graph on the vertices ${z_1,...,z_{k-2}}$. Next place an edge from the vertex $y_{k+1}$ to the vertices $y_{k-1}$ and $y_{k}$, then $(k-2)$ edges of the form $y_jz_j$ and $(k-2)$ edges of the form $y_{k+1}z_j$.
answered Nov 19 at 0:51
mathnoob
73411
answered Nov 19 at 0:51
mathnoob
73411
answered Nov 19 at 0:51
mathnoob
73411
answered Nov 19 at 0:51
mathnoob
73411
73411
add a comment |
add a comment |
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