How to determine if a random graph with a fixed degree sequence has - PowerPoint PPT Presentation

How to determine if a random graph with a fixed degree sequence has a giant component Felix Joos, Guillem Perarnau, Dieter Rautenbach, Bruce Reed A walkthrough by Angus Southwell Monash University 1

Random graphs with a given degree sequence: G n , ❞ Definition Let ❞ = ( d 1 , . . . , d n ). Then let G ( ❞ ) be a uniformly chosen simple graph with labelled vertices { 1 , . . . , n } and degree sequence ❞ . The probability space of such graphs is G n , ❞ . • Pro: these graphs are much more like most real-world graphs than G ( n , p ). • Con: they are much more complicated to analyse. Example In G ( n , p ), P ( u ∼ v ) = p trivially. In G n , ❞ , P ( u ∼ v ) is not known in general. 2

Random graphs with a given degree sequence: G n , ❞ Definition Let ❞ = ( d 1 , . . . , d n ). Then let G ( ❞ ) be a uniformly chosen simple graph with labelled vertices { 1 , . . . , n } and degree sequence ❞ . The probability space of such graphs is G n , ❞ . • Pro: these graphs are much more like most real-world graphs than G ( n , p ). • Con: they are much more complicated to analyse. Example In G ( n , p ), P ( u ∼ v ) = p trivially. In G n , ❞ , P ( u ∼ v ) is not known in general. 2

Random graphs with a given degree sequence: G n , ❞ Definition Let ❞ = ( d 1 , . . . , d n ). Then let G ( ❞ ) be a uniformly chosen simple graph with labelled vertices { 1 , . . . , n } and degree sequence ❞ . The probability space of such graphs is G n , ❞ . • Pro: these graphs are much more like most real-world graphs than G ( n , p ). • Con: they are much more complicated to analyse. Example In G ( n , p ), P ( u ∼ v ) = p trivially. In G n , ❞ , P ( u ∼ v ) is not known in general. 2

Random graphs with a given degree sequence: G n , ❞ Definition Let ❞ = ( d 1 , . . . , d n ). Then let G ( ❞ ) be a uniformly chosen simple graph with labelled vertices { 1 , . . . , n } and degree sequence ❞ . The probability space of such graphs is G n , ❞ . • Pro: these graphs are much more like most real-world graphs than G ( n , p ). • Con: they are much more complicated to analyse. Example In G ( n , p ), P ( u ∼ v ) = p trivially. In G n , ❞ , P ( u ∼ v ) is not known in general. 2

Switchings v t v t w w ❙ t − 1 ❙ t − 1 y x y x A switching is an operation that takes G ( ❞ ) to G ′ ( ❞ ). They are used to find the probability of certain events occurring that we previously got via configuration model, such as: • probability of specific edges being present, • probability that a given undiscovered vertex is found at the next step, • probability of a giant component in the preprocessed vertices. 3

Switchings v t v t w w ❙ t − 1 ❙ t − 1 y x y x A switching is an operation that takes G ( ❞ ) to G ′ ( ❞ ). They are used to find the probability of certain events occurring that we previously got via configuration model, such as: • probability of specific edges being present, • probability that a given undiscovered vertex is found at the next step, • probability of a giant component in the preprocessed vertices. 3

Giant component problem Problem Given a random graph model, what is the distribution of the size of the largest connected component? Older results “Double jump” threshold for Erd˝ os–R´ enyi random graphs at around 1 2 n edges. • Below the threshold, all components are order O (log n ). • At the threshold, largest component has order Θ( n 2 / 3 ). • Above the threshold, largest component has order Θ( n ). 4

Giant component problem Problem Given a random graph model, what is the distribution of the size of the largest connected component? Older results “Double jump” threshold for Erd˝ os–R´ enyi random graphs at around 1 2 n edges. • Below the threshold, all components are order O (log n ). • At the threshold, largest component has order Θ( n 2 / 3 ). • Above the threshold, largest component has order Θ( n ). 4

Results for fixed degree sequences Theorem (Molloy and Reed (1995)) Let ❉ ( n ) be a “well behaved” degree sequence with max. degree 1 4 − ε . Then define at most n Q ( D ) := 1 � d ( j )( d ( j ) − 2) . n j ∈ [ n ] • If Q ( D ) < 0, then all components have size O (log n ). • If Q ( D ) > 0, then there exists a component with at least α n vertices and β n cycles for α, β > 0. 5

Proof sketch • Breadth first search on the graph. • Keep track of X t , the number of “half edges” in your component that can be explored still. v t w t ❙ t − 1 • Show E t − 1 [ X t − X t − 1 ] stays positive (or negative) for a sufficiently long time. 6

Proof sketch • Breadth first search on the graph. • Keep track of X t , the number of “half edges” in your component that can be explored still. v t w t ❙ t − 1 Figure 1: “Superman using his laser vision on the four-vertex empty graph, soon to be the three-vertex empty graph” - Tim 6

Proof sketch • Breadth first search on the graph. • Keep track of X t , the number of “half edges” in your component that can be explored still. v t w t ❙ t − 1 • Show E t − 1 [ X t − X t − 1 ] stays positive (or negative) for a sufficiently long time. 6

Limitations of MR result • Proven in configuration model rather than G n , ❞ . • Handling of large degree vertices is nonexistent. • Criterion does not extend to general degree sequences: Consider n = k 2 for large odd k , and ❞ = (1 , . . . , 1 , 2 k ). Then Q ( D ) ≈ 3, so we would expect a giant component according to the MR criterion. · · · 1 2 ( n − 1 − 2 k ) Figure 1: “ An evil spider that can pick between 1 and n toothpicks as its weapon” - Tim 7

Limitations of MR result • Proven in configuration model rather than G n , ❞ . • Handling of large degree vertices is nonexistent. • Criterion does not extend to general degree sequences: Consider n = k 2 for large odd k , and ❞ = (1 , . . . , 1 , 2 k ). Then Q ( D ) ≈ 3, so we would expect a giant component according to the MR criterion. · · · 1 2 ( n − 1 − 2 k ) Figure 1: “ An evil spider that can pick between 1 and n toothpicks as its weapon” - Tim 7

Limitations of MR result • Proven in configuration model rather than G n , ❞ . • Handling of large degree vertices is nonexistent. • Criterion does not extend to general degree sequences: Consider n = k 2 for large odd k , and ❞ = (1 , . . . , 1 , 2 k ). Then Q ( D ) ≈ 3, so we would expect a giant component according to the MR criterion. · · · 1 2 ( n − 1 − 2 k ) Figure 1: “ An evil spider that can pick between 1 and n toothpicks as its weapon” - Tim 7

Limitations of MR result • Proven in configuration model rather than G n , ❞ . • Handling of large degree vertices is nonexistent. • Criterion does not extend to general degree sequences: Consider n = k 2 for large odd k , and ❞ = (1 , . . . , 1 , 2 k ). Then Q ( D ) ≈ 3, so we would expect a giant component according to the MR criterion. · · · 1 2 ( n − 1 − 2 k ) Figure 1: “ An evil spider that can pick between 1 and n toothpicks as its weapon” - Tim 7

Limitations of MR result • Proven in configuration model rather than G n , ❞ . • Handling of large degree vertices is nonexistent. • Criterion does not extend to general degree sequences: Consider n = k 2 for large odd k , and ❞ = (1 , . . . , 1 , 2 k ). Then Q ( D ) ≈ 3, so we would expect a giant component according to the MR criterion. · · · 1 2 ( n − 1 − 2 k ) Figure 1: “ An evil spider that can pick between 1 and n toothpicks as its weapon” - Tim 7

The more comprehensive result Theorem (Joos et al. (2018)) For any function δ → 0 as n → ∞ , for every γ > 0, if R D ≤ δ M D , the probability that G ( D ) has a component of order at least γ n is o (1). If there exists an ε > 0 such that R D ≥ ε M D , then the probability that G ( D ) contains a component of size γ n for some γ > 0 is 1 − o (1). Big Small n 1 j D R D M D (+2 n 2 ) 8

The more comprehensive result Theorem (Joos et al. (2018)) For any function δ → 0 as n → ∞ , for every γ > 0, if R D ≤ δ M D , the probability that G ( D ) has a component of order at least γ n is o (1). If there exists an ε > 0 such that R D ≥ ε M D , then the probability that G ( D ) contains a component of size γ n for some γ > 0 is 1 − o (1). Big Small n 1 j D R D M D (+2 n 2 ) 8

Proof sketch • Breadth first search again. • Now with added preprocessing! P n j ∗ 1 j D D R D M (+2 n 2 ) • Suppression of degree 2 vertices. • Switchings used to work in the graph model to get edge probabilities. 9

Proof sketch • Breadth first search again. • Now with added preprocessing! P n j ∗ 1 j D D R D M (+2 n 2 ) • Suppression of degree 2 vertices. • Switchings used to work in the graph model to get edge probabilities. 9

Proof sketch • Breadth first search again. • Now with added preprocessing! P n j ∗ 1 j D D R D M (+2 n 2 ) • Suppression of degree 2 vertices. • Switchings used to work in the graph model to get edge probabilities. 9

Proof sketch • Breadth first search again. • Now with added preprocessing! P n j ∗ 1 j D D R D M (+2 n 2 ) • Suppression of degree 2 vertices. • Switchings used to work in the graph model to get edge probabilities. 9