How many edges a graph with 500 vertices and 19 components has? The 2019 Stack Overflow...
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How many edges a graph with 500 vertices and 19 components has?
The 2019 Stack Overflow Developer Survey Results Are InIf a graph with $n$ vertices and $n$ edges there must a cycle?prove that a connected graph with $n$ vertices has at least $n-1$ edgesNumbers of ways $k - 1$ edges to be added to $k$ connected components to make the graph connectedMinimum and maximum number of edges graph with 25 vertices and 6 connected components can haveNumber of connected components in a graph with n vertices and n-k edgesHamiltonian graph and connected componentsA connected graph with n vertices has at least n-1 edgesThe simple graph with $n$ vertices and every vertex has at least 2 edgesProve that a graph on $n$ vertices with $c$ components has at least $n-c$ edges.Minimum number of cycles of a graph with $n$ vertices, $m$ edges and $c$ components.A simple connected graph on 100 vertices has 102 edges. Show that…
$begingroup$
$G$ is a graph with no cycles, 500 vertices, and 19 connected components. How many edges it has?
any help/hint?
discrete-mathematics graph-theory
asked Mar 21 at 15:22
JackJack
1047
$endgroup$
add a comment |
$begingroup$
$G$ is a graph with no cycles, 500 vertices, and 19 connected components. How many edges it has?
any help/hint?
discrete-mathematics graph-theory
asked Mar 21 at 15:22
JackJack
1047
$endgroup$
1
$begingroup$
See for example this answer
$endgroup$
– mrf
Mar 21 at 15:32
add a comment |
$begingroup$
$G$ is a graph with no cycles, 500 vertices, and 19 connected components. How many edges it has?
any help/hint?
discrete-mathematics graph-theory
asked Mar 21 at 15:22
JackJack
1047
$endgroup$
$G$ is a graph with no cycles, 500 vertices, and 19 connected components. How many edges it has?
any help/hint?
discrete-mathematics graph-theory
discrete-mathematics graph-theory
asked Mar 21 at 15:22
JackJack
1047
asked Mar 21 at 15:22
JackJack
1047
asked Mar 21 at 15:22
JackJack
1047
asked Mar 21 at 15:22
JackJack
1047
asked Mar 21 at 15:22
JackJack
1047
1047
1
$begingroup$
See for example this answer
$endgroup$
– mrf
Mar 21 at 15:32
add a comment |
1
$begingroup$
See for example this answer
$endgroup$
– mrf
Mar 21 at 15:32
1
1
$begingroup$
See for example this answer
$endgroup$
– mrf
Mar 21 at 15:32
$begingroup$
See for example this answer
$endgroup$
– mrf
Mar 21 at 15:32
add a comment |
1 Answer
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$begingroup$
Each component is connected and does not contain circuits. Thus, each component is a tree. A tree with $n$ vertices has $n-1$ edges. Thus, the total number of edges is$$sum_{i=1}^{19}(n_i-1)=500-19=481$$where $n_i$ is the number of vertices in the $i^{th}$ component.
answered Mar 21 at 15:46
Shubham JohriShubham Johri
5,558818
$endgroup$
$begingroup$
The step that I can not understand logically that led me to ask the question is the one that passes from the formula you wrote to 500-19...can you confirm that we need to assume that there are exactly 19 components so even if we don't know the number of vertices of each component they are generally 500 and since we subtract 1 at every iteration at the and we subtract 19 from the total number of vertices?
$endgroup$
– Jack
Mar 21 at 15:56
$begingroup$
@Jack Yes, only the number of components matters. The $i^{th}$ component has $n_i$ vertices and $n_i-1$ edges, so you need to sum $n_i-1$ over all components. $500-19$ arises from the fact $sum(n_i-1)=sum n_i-sum1$.
$endgroup$
– Shubham Johri
Mar 21 at 16:02
$begingroup$
Thank you. All clear now
$endgroup$
– Jack
Mar 21 at 16:04
add a comment |
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$begingroup$
Each component is connected and does not contain circuits. Thus, each component is a tree. A tree with $n$ vertices has $n-1$ edges. Thus, the total number of edges is$$sum_{i=1}^{19}(n_i-1)=500-19=481$$where $n_i$ is the number of vertices in the $i^{th}$ component.
answered Mar 21 at 15:46
Shubham JohriShubham Johri
5,558818
$endgroup$
$begingroup$
The step that I can not understand logically that led me to ask the question is the one that passes from the formula you wrote to 500-19...can you confirm that we need to assume that there are exactly 19 components so even if we don't know the number of vertices of each component they are generally 500 and since we subtract 1 at every iteration at the and we subtract 19 from the total number of vertices?
$endgroup$
– Jack
Mar 21 at 15:56
$begingroup$
@Jack Yes, only the number of components matters. The $i^{th}$ component has $n_i$ vertices and $n_i-1$ edges, so you need to sum $n_i-1$ over all components. $500-19$ arises from the fact $sum(n_i-1)=sum n_i-sum1$.
$endgroup$
– Shubham Johri
Mar 21 at 16:02
$begingroup$
Thank you. All clear now
$endgroup$
– Jack
Mar 21 at 16:04
add a comment |
$begingroup$
Each component is connected and does not contain circuits. Thus, each component is a tree. A tree with $n$ vertices has $n-1$ edges. Thus, the total number of edges is$$sum_{i=1}^{19}(n_i-1)=500-19=481$$where $n_i$ is the number of vertices in the $i^{th}$ component.
answered Mar 21 at 15:46
Shubham JohriShubham Johri
5,558818
$endgroup$
$begingroup$
The step that I can not understand logically that led me to ask the question is the one that passes from the formula you wrote to 500-19...can you confirm that we need to assume that there are exactly 19 components so even if we don't know the number of vertices of each component they are generally 500 and since we subtract 1 at every iteration at the and we subtract 19 from the total number of vertices?
$endgroup$
– Jack
Mar 21 at 15:56
$begingroup$
@Jack Yes, only the number of components matters. The $i^{th}$ component has $n_i$ vertices and $n_i-1$ edges, so you need to sum $n_i-1$ over all components. $500-19$ arises from the fact $sum(n_i-1)=sum n_i-sum1$.
$endgroup$
– Shubham Johri
Mar 21 at 16:02
$begingroup$
Thank you. All clear now
$endgroup$
– Jack
Mar 21 at 16:04
add a comment |
$begingroup$
Each component is connected and does not contain circuits. Thus, each component is a tree. A tree with $n$ vertices has $n-1$ edges. Thus, the total number of edges is$$sum_{i=1}^{19}(n_i-1)=500-19=481$$where $n_i$ is the number of vertices in the $i^{th}$ component.
answered Mar 21 at 15:46
Shubham JohriShubham Johri
5,558818
$endgroup$
Each component is connected and does not contain circuits. Thus, each component is a tree. A tree with $n$ vertices has $n-1$ edges. Thus, the total number of edges is$$sum_{i=1}^{19}(n_i-1)=500-19=481$$where $n_i$ is the number of vertices in the $i^{th}$ component.
answered Mar 21 at 15:46
Shubham JohriShubham Johri
5,558818
answered Mar 21 at 15:46
Shubham JohriShubham Johri
5,558818
answered Mar 21 at 15:46
Shubham JohriShubham Johri
5,558818
answered Mar 21 at 15:46
Shubham JohriShubham Johri
5,558818
5,558818
$begingroup$
The step that I can not understand logically that led me to ask the question is the one that passes from the formula you wrote to 500-19...can you confirm that we need to assume that there are exactly 19 components so even if we don't know the number of vertices of each component they are generally 500 and since we subtract 1 at every iteration at the and we subtract 19 from the total number of vertices?
$endgroup$
– Jack
Mar 21 at 15:56
$begingroup$
@Jack Yes, only the number of components matters. The $i^{th}$ component has $n_i$ vertices and $n_i-1$ edges, so you need to sum $n_i-1$ over all components. $500-19$ arises from the fact $sum(n_i-1)=sum n_i-sum1$.
$endgroup$
– Shubham Johri
Mar 21 at 16:02
$begingroup$
Thank you. All clear now
$endgroup$
– Jack
Mar 21 at 16:04
add a comment |
$begingroup$
The step that I can not understand logically that led me to ask the question is the one that passes from the formula you wrote to 500-19...can you confirm that we need to assume that there are exactly 19 components so even if we don't know the number of vertices of each component they are generally 500 and since we subtract 1 at every iteration at the and we subtract 19 from the total number of vertices?
$endgroup$
– Jack
Mar 21 at 15:56
$begingroup$
@Jack Yes, only the number of components matters. The $i^{th}$ component has $n_i$ vertices and $n_i-1$ edges, so you need to sum $n_i-1$ over all components. $500-19$ arises from the fact $sum(n_i-1)=sum n_i-sum1$.
$endgroup$
– Shubham Johri
Mar 21 at 16:02
$begingroup$
Thank you. All clear now
$endgroup$
– Jack
Mar 21 at 16:04
$begingroup$
The step that I can not understand logically that led me to ask the question is the one that passes from the formula you wrote to 500-19...can you confirm that we need to assume that there are exactly 19 components so even if we don't know the number of vertices of each component they are generally 500 and since we subtract 1 at every iteration at the and we subtract 19 from the total number of vertices?
$endgroup$
– Jack
Mar 21 at 15:56
$begingroup$
The step that I can not understand logically that led me to ask the question is the one that passes from the formula you wrote to 500-19...can you confirm that we need to assume that there are exactly 19 components so even if we don't know the number of vertices of each component they are generally 500 and since we subtract 1 at every iteration at the and we subtract 19 from the total number of vertices?
$endgroup$
– Jack
Mar 21 at 15:56
$begingroup$
@Jack Yes, only the number of components matters. The $i^{th}$ component has $n_i$ vertices and $n_i-1$ edges, so you need to sum $n_i-1$ over all components. $500-19$ arises from the fact $sum(n_i-1)=sum n_i-sum1$.
$endgroup$
– Shubham Johri
Mar 21 at 16:02
$begingroup$
@Jack Yes, only the number of components matters. The $i^{th}$ component has $n_i$ vertices and $n_i-1$ edges, so you need to sum $n_i-1$ over all components. $500-19$ arises from the fact $sum(n_i-1)=sum n_i-sum1$.
$endgroup$
– Shubham Johri
Mar 21 at 16:02
$begingroup$
Thank you. All clear now
$endgroup$
– Jack
Mar 21 at 16:04
$begingroup$
Thank you. All clear now
$endgroup$
– Jack
Mar 21 at 16:04
add a comment |
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$begingroup$
See for example this answer
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– mrf
Mar 21 at 15:32