Specialization of Erdos Gallai Theorem To Vertex Exclusions
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The Erdos Gallai theorem says that a degree sequence $d=(d_1,cdots,d_n)$ corresponds to a graph iff it satisfies:
$$sum_{i=1}^kd_ileq k(k-1)+sum_{i=k+1}^nmin(k,d_i).$$
Is there a specialization of this result to allow for additional exclusions on possible connections in the graph? Something like for each vertex $v_i$ there is an exclusion list of vertices which cannot connect to $v_i$? For example, $v_1$ cannot connect to $v_2,v_3$.
graph-theory
asked Dec 8 '18 at 1:20
Alex R.Alex R.
24.9k12452
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add a comment |
$begingroup$
The Erdos Gallai theorem says that a degree sequence $d=(d_1,cdots,d_n)$ corresponds to a graph iff it satisfies:
$$sum_{i=1}^kd_ileq k(k-1)+sum_{i=k+1}^nmin(k,d_i).$$
Is there a specialization of this result to allow for additional exclusions on possible connections in the graph? Something like for each vertex $v_i$ there is an exclusion list of vertices which cannot connect to $v_i$? For example, $v_1$ cannot connect to $v_2,v_3$.
graph-theory
asked Dec 8 '18 at 1:20
Alex R.Alex R.
24.9k12452
$endgroup$
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By vertex $v_i$ do you mean the vertex with degree $d_i$, or are you not given the correspondence between vertices (as relevant for the exclusion list) and their degrees?
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– Misha Lavrov
Dec 8 '18 at 1:44
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@Misha Lavrov: exactly so vertex $v_i$ has degree $d_i$.
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– Alex R.
Dec 8 '18 at 2:01
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How did Erdös figure that out!!? That guy's mind just baffles me. ¶ Do you mean, though, ∀k∊{1 ... n}? I would surmise that it does occasion that qualifier ... but I am not certain; and it would be a lovely finishing-touch if you would say explicitly whether it does.
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– AmbretteOrrisey
Dec 17 '18 at 8:16
add a comment |
$begingroup$
The Erdos Gallai theorem says that a degree sequence $d=(d_1,cdots,d_n)$ corresponds to a graph iff it satisfies:
$$sum_{i=1}^kd_ileq k(k-1)+sum_{i=k+1}^nmin(k,d_i).$$
Is there a specialization of this result to allow for additional exclusions on possible connections in the graph? Something like for each vertex $v_i$ there is an exclusion list of vertices which cannot connect to $v_i$? For example, $v_1$ cannot connect to $v_2,v_3$.
graph-theory
asked Dec 8 '18 at 1:20
Alex R.Alex R.
24.9k12452
$endgroup$
The Erdos Gallai theorem says that a degree sequence $d=(d_1,cdots,d_n)$ corresponds to a graph iff it satisfies:
$$sum_{i=1}^kd_ileq k(k-1)+sum_{i=k+1}^nmin(k,d_i).$$
Is there a specialization of this result to allow for additional exclusions on possible connections in the graph? Something like for each vertex $v_i$ there is an exclusion list of vertices which cannot connect to $v_i$? For example, $v_1$ cannot connect to $v_2,v_3$.
graph-theory
graph-theory
asked Dec 8 '18 at 1:20
Alex R.Alex R.
24.9k12452
asked Dec 8 '18 at 1:20
Alex R.Alex R.
24.9k12452
asked Dec 8 '18 at 1:20
Alex R.Alex R.
24.9k12452
asked Dec 8 '18 at 1:20
Alex R.Alex R.
24.9k12452
asked Dec 8 '18 at 1:20
Alex R.Alex R.
24.9k12452
24.9k12452
$begingroup$
By vertex $v_i$ do you mean the vertex with degree $d_i$, or are you not given the correspondence between vertices (as relevant for the exclusion list) and their degrees?
$endgroup$
– Misha Lavrov
Dec 8 '18 at 1:44
$begingroup$
@Misha Lavrov: exactly so vertex $v_i$ has degree $d_i$.
$endgroup$
– Alex R.
Dec 8 '18 at 2:01
$begingroup$
How did Erdös figure that out!!? That guy's mind just baffles me. ¶ Do you mean, though, ∀k∊{1 ... n}? I would surmise that it does occasion that qualifier ... but I am not certain; and it would be a lovely finishing-touch if you would say explicitly whether it does.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 8:16
add a comment |
$begingroup$
By vertex $v_i$ do you mean the vertex with degree $d_i$, or are you not given the correspondence between vertices (as relevant for the exclusion list) and their degrees?
$endgroup$
– Misha Lavrov
Dec 8 '18 at 1:44
$begingroup$
@Misha Lavrov: exactly so vertex $v_i$ has degree $d_i$.
$endgroup$
– Alex R.
Dec 8 '18 at 2:01
$begingroup$
How did Erdös figure that out!!? That guy's mind just baffles me. ¶ Do you mean, though, ∀k∊{1 ... n}? I would surmise that it does occasion that qualifier ... but I am not certain; and it would be a lovely finishing-touch if you would say explicitly whether it does.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 8:16
$begingroup$
By vertex $v_i$ do you mean the vertex with degree $d_i$, or are you not given the correspondence between vertices (as relevant for the exclusion list) and their degrees?
$endgroup$
– Misha Lavrov
Dec 8 '18 at 1:44
$begingroup$
By vertex $v_i$ do you mean the vertex with degree $d_i$, or are you not given the correspondence between vertices (as relevant for the exclusion list) and their degrees?
$endgroup$
– Misha Lavrov
Dec 8 '18 at 1:44
$begingroup$
@Misha Lavrov: exactly so vertex $v_i$ has degree $d_i$.
$endgroup$
– Alex R.
Dec 8 '18 at 2:01
$begingroup$
@Misha Lavrov: exactly so vertex $v_i$ has degree $d_i$.
$endgroup$
– Alex R.
Dec 8 '18 at 2:01
$begingroup$
How did Erdös figure that out!!? That guy's mind just baffles me. ¶ Do you mean, though, ∀k∊{1 ... n}? I would surmise that it does occasion that qualifier ... but I am not certain; and it would be a lovely finishing-touch if you would say explicitly whether it does.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 8:16
$begingroup$
How did Erdös figure that out!!? That guy's mind just baffles me. ¶ Do you mean, though, ∀k∊{1 ... n}? I would surmise that it does occasion that qualifier ... but I am not certain; and it would be a lovely finishing-touch if you would say explicitly whether it does.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 8:16
add a comment |
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$begingroup$
By vertex $v_i$ do you mean the vertex with degree $d_i$, or are you not given the correspondence between vertices (as relevant for the exclusion list) and their degrees?
$endgroup$
– Misha Lavrov
Dec 8 '18 at 1:44
$begingroup$
@Misha Lavrov: exactly so vertex $v_i$ has degree $d_i$.
$endgroup$
– Alex R.
Dec 8 '18 at 2:01
$begingroup$
How did Erdös figure that out!!? That guy's mind just baffles me. ¶ Do you mean, though, ∀k∊{1 ... n}? I would surmise that it does occasion that qualifier ... but I am not certain; and it would be a lovely finishing-touch if you would say explicitly whether it does.
$endgroup$
– AmbretteOrrisey
Dec 17 '18 at 8:16