Number of 'Tri - Coloured Triangles' in a random graph.
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I am attempting some practice questions where no answers are given. Would it be possible to verify my solution for this particular problem?
Let RBG be a coloured random graph with $n$ vertices ${1, 2, · · · , n}$, where every
pair of vertices is connected either by a red edge (with probability $p$) or by a blue edge (with
probability $q$) or by a green edge (with probability $r$), such that $p + q + r = 1$. (Note that all
edges are present, but each edge carries exactly one of three different colours - red, blue or green)
A set of three vertices ${i, j, k$} is said to form a “tri-coloured triangle” if
the triangle formed by ${i, j, k}$ has exactly one red edge, exactly one blue edge and exactly
one green edge. Let $N_3$ denote the number of tri-coloured triangles in the coloured random
graph RBG. Calculate $E(N_3)$
My approach is as follows
Given a triangle, there are $3! = 6$ ways of colouring the triangle such that each edge has a distinct colour.
Therefore, $P(text{A triangle is tri - coloured}) = 6pqr$.
There are ${n}choose{3}$ ways of choosing 3 vertices in the random graph.
Therefore, $E(N_3) =$ ${n}choose{3}$ $6pqr = n(n-1)(n-2)pqr$
probability graph-theory expected-value
asked Dec 2 '18 at 4:50
dzldzl
348214
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add a comment |
$begingroup$
I am attempting some practice questions where no answers are given. Would it be possible to verify my solution for this particular problem?
Let RBG be a coloured random graph with $n$ vertices ${1, 2, · · · , n}$, where every
pair of vertices is connected either by a red edge (with probability $p$) or by a blue edge (with
probability $q$) or by a green edge (with probability $r$), such that $p + q + r = 1$. (Note that all
edges are present, but each edge carries exactly one of three different colours - red, blue or green)
A set of three vertices ${i, j, k$} is said to form a “tri-coloured triangle” if
the triangle formed by ${i, j, k}$ has exactly one red edge, exactly one blue edge and exactly
one green edge. Let $N_3$ denote the number of tri-coloured triangles in the coloured random
graph RBG. Calculate $E(N_3)$
My approach is as follows
Given a triangle, there are $3! = 6$ ways of colouring the triangle such that each edge has a distinct colour.
Therefore, $P(text{A triangle is tri - coloured}) = 6pqr$.
There are ${n}choose{3}$ ways of choosing 3 vertices in the random graph.
Therefore, $E(N_3) =$ ${n}choose{3}$ $6pqr = n(n-1)(n-2)pqr$
probability graph-theory expected-value
asked Dec 2 '18 at 4:50
dzldzl
348214
$endgroup$
$begingroup$
Perfect, except for five missing full stops.
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– bof
Dec 2 '18 at 4:55
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Thank you, @bof ! :)
$endgroup$
– dzl
Dec 3 '18 at 5:27
add a comment |
$begingroup$
I am attempting some practice questions where no answers are given. Would it be possible to verify my solution for this particular problem?
Let RBG be a coloured random graph with $n$ vertices ${1, 2, · · · , n}$, where every
pair of vertices is connected either by a red edge (with probability $p$) or by a blue edge (with
probability $q$) or by a green edge (with probability $r$), such that $p + q + r = 1$. (Note that all
edges are present, but each edge carries exactly one of three different colours - red, blue or green)
A set of three vertices ${i, j, k$} is said to form a “tri-coloured triangle” if
the triangle formed by ${i, j, k}$ has exactly one red edge, exactly one blue edge and exactly
one green edge. Let $N_3$ denote the number of tri-coloured triangles in the coloured random
graph RBG. Calculate $E(N_3)$
My approach is as follows
Given a triangle, there are $3! = 6$ ways of colouring the triangle such that each edge has a distinct colour.
Therefore, $P(text{A triangle is tri - coloured}) = 6pqr$.
There are ${n}choose{3}$ ways of choosing 3 vertices in the random graph.
Therefore, $E(N_3) =$ ${n}choose{3}$ $6pqr = n(n-1)(n-2)pqr$
probability graph-theory expected-value
asked Dec 2 '18 at 4:50
dzldzl
348214
$endgroup$
I am attempting some practice questions where no answers are given. Would it be possible to verify my solution for this particular problem?
Let RBG be a coloured random graph with $n$ vertices ${1, 2, · · · , n}$, where every
pair of vertices is connected either by a red edge (with probability $p$) or by a blue edge (with
probability $q$) or by a green edge (with probability $r$), such that $p + q + r = 1$. (Note that all
edges are present, but each edge carries exactly one of three different colours - red, blue or green)
A set of three vertices ${i, j, k$} is said to form a “tri-coloured triangle” if
the triangle formed by ${i, j, k}$ has exactly one red edge, exactly one blue edge and exactly
one green edge. Let $N_3$ denote the number of tri-coloured triangles in the coloured random
graph RBG. Calculate $E(N_3)$
My approach is as follows
Given a triangle, there are $3! = 6$ ways of colouring the triangle such that each edge has a distinct colour.
Therefore, $P(text{A triangle is tri - coloured}) = 6pqr$.
There are ${n}choose{3}$ ways of choosing 3 vertices in the random graph.
Therefore, $E(N_3) =$ ${n}choose{3}$ $6pqr = n(n-1)(n-2)pqr$
probability graph-theory expected-value
probability graph-theory expected-value
asked Dec 2 '18 at 4:50
dzldzl
348214
asked Dec 2 '18 at 4:50
dzldzl
348214
asked Dec 2 '18 at 4:50
dzldzl
348214
asked Dec 2 '18 at 4:50
dzldzl
348214
asked Dec 2 '18 at 4:50
dzldzl
348214
348214
$begingroup$
Perfect, except for five missing full stops.
$endgroup$
– bof
Dec 2 '18 at 4:55
$begingroup$
Thank you, @bof ! :)
$endgroup$
– dzl
Dec 3 '18 at 5:27
add a comment |
$begingroup$
Perfect, except for five missing full stops.
$endgroup$
– bof
Dec 2 '18 at 4:55
$begingroup$
Thank you, @bof ! :)
$endgroup$
– dzl
Dec 3 '18 at 5:27
$begingroup$
Perfect, except for five missing full stops.
$endgroup$
– bof
Dec 2 '18 at 4:55
$begingroup$
Perfect, except for five missing full stops.
$endgroup$
– bof
Dec 2 '18 at 4:55
$begingroup$
Thank you, @bof ! :)
$endgroup$
– dzl
Dec 3 '18 at 5:27
$begingroup$
Thank you, @bof ! :)
$endgroup$
– dzl
Dec 3 '18 at 5:27
add a comment |
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$begingroup$
Perfect, except for five missing full stops.
$endgroup$
– bof
Dec 2 '18 at 4:55
$begingroup$
Thank you, @bof ! :)
$endgroup$
– dzl
Dec 3 '18 at 5:27