Defining the connectivity of a graph












1












$begingroup$


Consider a simple binary undirected graph, which adjacency matrix is $A = {a_{i,j}} in {0, 1}^{N times N}.$



Suppose that all vertices of such graph have at least one neighbors, i.e.



$$k_i = sum_{j=1}^N{a_{i,j}} geq 1.$$



How do we call this graph?



Reading the definition of connectivity, I did not find anything about this particular case.



I would say that it is "minimally connected".



Is there a proper nomenclature for these graphs?






graph-theory network







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$endgroup$












  • $begingroup$
    What is a "binary" graph?
    $endgroup$
    – richarddedekind
    Dec 8 '18 at 3:41






  • 1




    $begingroup$
    Binary stands for "non-weighted", i.e. the entries of $A$ are in the set ${0, 1}$, not in $mathbb{R}^+$.
    $endgroup$
    – the_candyman
    Dec 8 '18 at 11:42
















1












$begingroup$


Consider a simple binary undirected graph, which adjacency matrix is $A = {a_{i,j}} in {0, 1}^{N times N}.$



Suppose that all vertices of such graph have at least one neighbors, i.e.



$$k_i = sum_{j=1}^N{a_{i,j}} geq 1.$$



How do we call this graph?



Reading the definition of connectivity, I did not find anything about this particular case.



I would say that it is "minimally connected".



Is there a proper nomenclature for these graphs?






graph-theory network







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is a "binary" graph?
    $endgroup$
    – richarddedekind
    Dec 8 '18 at 3:41






  • 1




    $begingroup$
    Binary stands for "non-weighted", i.e. the entries of $A$ are in the set ${0, 1}$, not in $mathbb{R}^+$.
    $endgroup$
    – the_candyman
    Dec 8 '18 at 11:42














1












1








1


2



$begingroup$


Consider a simple binary undirected graph, which adjacency matrix is $A = {a_{i,j}} in {0, 1}^{N times N}.$



Suppose that all vertices of such graph have at least one neighbors, i.e.



$$k_i = sum_{j=1}^N{a_{i,j}} geq 1.$$



How do we call this graph?



Reading the definition of connectivity, I did not find anything about this particular case.



I would say that it is "minimally connected".



Is there a proper nomenclature for these graphs?






graph-theory network







share|cite|improve this question









$endgroup$




Consider a simple binary undirected graph, which adjacency matrix is $A = {a_{i,j}} in {0, 1}^{N times N}.$



Suppose that all vertices of such graph have at least one neighbors, i.e.



$$k_i = sum_{j=1}^N{a_{i,j}} geq 1.$$



How do we call this graph?



Reading the definition of connectivity, I did not find anything about this particular case.



I would say that it is "minimally connected".



Is there a proper nomenclature for these graphs?






graph-theory network




graph-theory network






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 19:22









the_candymanthe_candyman

8,84632045




8,84632045












  • $begingroup$
    What is a "binary" graph?
    $endgroup$
    – richarddedekind
    Dec 8 '18 at 3:41






  • 1




    $begingroup$
    Binary stands for "non-weighted", i.e. the entries of $A$ are in the set ${0, 1}$, not in $mathbb{R}^+$.
    $endgroup$
    – the_candyman
    Dec 8 '18 at 11:42


















  • $begingroup$
    What is a "binary" graph?
    $endgroup$
    – richarddedekind
    Dec 8 '18 at 3:41






  • 1




    $begingroup$
    Binary stands for "non-weighted", i.e. the entries of $A$ are in the set ${0, 1}$, not in $mathbb{R}^+$.
    $endgroup$
    – the_candyman
    Dec 8 '18 at 11:42
















$begingroup$
What is a "binary" graph?
$endgroup$
– richarddedekind
Dec 8 '18 at 3:41




$begingroup$
What is a "binary" graph?
$endgroup$
– richarddedekind
Dec 8 '18 at 3:41




1




1




$begingroup$
Binary stands for "non-weighted", i.e. the entries of $A$ are in the set ${0, 1}$, not in $mathbb{R}^+$.
$endgroup$
– the_candyman
Dec 8 '18 at 11:42




$begingroup$
Binary stands for "non-weighted", i.e. the entries of $A$ are in the set ${0, 1}$, not in $mathbb{R}^+$.
$endgroup$
– the_candyman
Dec 8 '18 at 11:42










1 Answer
1






active

oldest

votes


















4












$begingroup$

Requiring that every vertex have positive degree is not sufficient to make the graph connected. You may describe your graph as having no isolated vertices.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    this makes sense. thanks a lot.
    $endgroup$
    – the_candyman
    Dec 8 '18 at 11:42











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Requiring that every vertex have positive degree is not sufficient to make the graph connected. You may describe your graph as having no isolated vertices.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    this makes sense. thanks a lot.
    $endgroup$
    – the_candyman
    Dec 8 '18 at 11:42
















4












$begingroup$

Requiring that every vertex have positive degree is not sufficient to make the graph connected. You may describe your graph as having no isolated vertices.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    this makes sense. thanks a lot.
    $endgroup$
    – the_candyman
    Dec 8 '18 at 11:42














4












4








4





$begingroup$

Requiring that every vertex have positive degree is not sufficient to make the graph connected. You may describe your graph as having no isolated vertices.






share|cite|improve this answer









$endgroup$



Requiring that every vertex have positive degree is not sufficient to make the graph connected. You may describe your graph as having no isolated vertices.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 19:28









Zach LangleyZach Langley

9731019




9731019












  • $begingroup$
    this makes sense. thanks a lot.
    $endgroup$
    – the_candyman
    Dec 8 '18 at 11:42


















  • $begingroup$
    this makes sense. thanks a lot.
    $endgroup$
    – the_candyman
    Dec 8 '18 at 11:42
















$begingroup$
this makes sense. thanks a lot.
$endgroup$
– the_candyman
Dec 8 '18 at 11:42




$begingroup$
this makes sense. thanks a lot.
$endgroup$
– the_candyman
Dec 8 '18 at 11:42


















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