The threshold dimension of a graph Lucas Mol Joint work with - PowerPoint PPT Presentation

A N E XAMPLE Consider the graph G = K 1 , 5 .

A N E XAMPLE Consider the graph G = K 1 , 5 . β ( G ) = 4

A N E XAMPLE Consider the graph G = K 1 , 5 . β ( G ) = 4 Now add a couple of carefully chosen edges:

A N E XAMPLE Consider the graph G = K 1 , 5 . β ( G ) = 4 Now add a couple of carefully chosen edges:

A N E XAMPLE Consider the graph G = K 1 , 5 . β ( G ) = 4 Now add a couple of carefully chosen edges:

A N E XAMPLE Consider the graph G = K 1 , 5 . β ( G ) = 4 Now add a couple of carefully chosen edges:

A N E XAMPLE Consider the graph G = K 1 , 5 . β ( G ) = 4 Now add a couple of carefully chosen edges: [ 2 , 2 ] [ 1 , 2 ] [ 2 , 1 ] [ 1 , 1 ]

A N E XAMPLE Consider the graph G = K 1 , 5 . β ( G ) = 4 Now add a couple of carefully chosen edges: [ 2 , 2 ] [ 1 , 2 ] [ 2 , 1 ] τ ( G ) ≤ 2 [ 1 , 1 ]

A N E XAMPLE Consider the graph G = K 1 , 5 . β ( G ) = 4 Now add a couple of carefully chosen edges: [ 2 , 2 ] [ 1 , 2 ] [ 2 , 1 ] τ ( G ) = 2 [ 1 , 1 ]

P LAN M ETRIC D IMENSION T HRESHOLD D IMENSION B OUNDS E MBEDDINGS

T REES

T REES For every positive real number x , let d x denote the smallest positive integer such that x ≤ 2 d x + d x .

T REES For every positive real number x , let d x denote the smallest positive integer such that x ≤ 2 d x + d x . ◮ Note that d x < log 2 ( x ) .

T REES For every positive real number x , let d x denote the smallest positive integer such that x ≤ 2 d x + d x . ◮ Note that d x < log 2 ( x ) . Theorem (MMO 2019+): Let T be a tree of order n . Then τ ( T ) ≤ d n . Moreover, this bound is sharp.

T REES For every positive real number x , let d x denote the smallest positive integer such that x ≤ 2 d x + d x . ◮ Note that d x < log 2 ( x ) . Theorem (MMO 2019+): Let T be a tree of order n . Then τ ( T ) ≤ d n . Moreover, this bound is sharp. Sketch of Proof:

T REES For every positive real number x , let d x denote the smallest positive integer such that x ≤ 2 d x + d x . ◮ Note that d x < log 2 ( x ) . Theorem (MMO 2019+): Let T be a tree of order n . Then τ ( T ) ≤ d n . Moreover, this bound is sharp. Sketch of Proof: ◮ If β ( T ) ≤ d n , then we are done.

T REES For every positive real number x , let d x denote the smallest positive integer such that x ≤ 2 d x + d x . ◮ Note that d x < log 2 ( x ) . Theorem (MMO 2019+): Let T be a tree of order n . Then τ ( T ) ≤ d n . Moreover, this bound is sharp. Sketch of Proof: ◮ If β ( T ) ≤ d n , then we are done. ◮ Otherwise, it must be the case that T has at least d n leaves.

T REES For every positive real number x , let d x denote the smallest positive integer such that x ≤ 2 d x + d x . ◮ Note that d x < log 2 ( x ) . Theorem (MMO 2019+): Let T be a tree of order n . Then τ ( T ) ≤ d n . Moreover, this bound is sharp. Sketch of Proof: ◮ If β ( T ) ≤ d n , then we are done. ◮ Otherwise, it must be the case that T has at least d n leaves. ◮ Take any set W of d n leaves – this is going to be our resolving set.

T REES For every positive real number x , let d x denote the smallest positive integer such that x ≤ 2 d x + d x . ◮ Note that d x < log 2 ( x ) . Theorem (MMO 2019+): Let T be a tree of order n . Then τ ( T ) ≤ d n . Moreover, this bound is sharp. Sketch of Proof: ◮ If β ( T ) ≤ d n , then we are done. ◮ Otherwise, it must be the case that T has at least d n leaves. ◮ Take any set W of d n leaves – this is going to be our resolving set. ◮ Since n ≤ 2 d n + d n , there are at most 2 d n vertices outside of W .

T REES For every positive real number x , let d x denote the smallest positive integer such that x ≤ 2 d x + d x . ◮ Note that d x < log 2 ( x ) . Theorem (MMO 2019+): Let T be a tree of order n . Then τ ( T ) ≤ d n . Moreover, this bound is sharp. Sketch of Proof: ◮ If β ( T ) ≤ d n , then we are done. ◮ Otherwise, it must be the case that T has at least d n leaves. ◮ Take any set W of d n leaves – this is going to be our resolving set. ◮ Since n ≤ 2 d n + d n , there are at most 2 d n vertices outside of W . ◮ Attach each vertex not in W to a unique subset of W .

A B OUND IN TERMS OF THE C HROMATIC N UMBER

A B OUND IN TERMS OF THE C HROMATIC N UMBER Theorem (MMO 2019+): Let G be a graph of order n with chromatic number k . Then τ ( G ) < k ( d n / k + 2 ) < k (log 2 ( n / k ) + 2 ) .

A B OUND IN TERMS OF THE C HROMATIC N UMBER Theorem (MMO 2019+): Let G be a graph of order n with chromatic number k . Then τ ( G ) < k ( d n / k + 2 ) < k (log 2 ( n / k ) + 2 ) . Sketch of proof:

A B OUND IN TERMS OF THE C HROMATIC N UMBER Theorem (MMO 2019+): Let G be a graph of order n with chromatic number k . Then τ ( G ) < k ( d n / k + 2 ) < k (log 2 ( n / k ) + 2 ) . Sketch of proof: ◮ The vertices of G can be partitioned into k independent sets V 1 , . . . V k , say of orders n 1 , . . . , n k , respectively.

A B OUND IN TERMS OF THE C HROMATIC N UMBER Theorem (MMO 2019+): Let G be a graph of order n with chromatic number k . Then τ ( G ) < k ( d n / k + 2 ) < k (log 2 ( n / k ) + 2 ) . Sketch of proof: ◮ The vertices of G can be partitioned into k independent sets V 1 , . . . V k , say of orders n 1 , . . . , n k , respectively. ◮ Add all edges between these independent sets – we are now looking at K n 1 , n 2 ,..., n k .

A B OUND IN TERMS OF THE C HROMATIC N UMBER Theorem (MMO 2019+): Let G be a graph of order n with chromatic number k . Then τ ( G ) < k ( d n / k + 2 ) < k (log 2 ( n / k ) + 2 ) . Sketch of proof: ◮ The vertices of G can be partitioned into k independent sets V 1 , . . . V k , say of orders n 1 , . . . , n k , respectively. ◮ Add all edges between these independent sets – we are now looking at K n 1 , n 2 ,..., n k . ◮ For every i , take d n i vertices from V i – together, these will form a resolving set.

A B OUND IN TERMS OF THE C HROMATIC N UMBER Theorem (MMO 2019+): Let G be a graph of order n with chromatic number k . Then τ ( G ) < k ( d n / k + 2 ) < k (log 2 ( n / k ) + 2 ) . Sketch of proof: ◮ The vertices of G can be partitioned into k independent sets V 1 , . . . V k , say of orders n 1 , . . . , n k , respectively. ◮ Add all edges between these independent sets – we are now looking at K n 1 , n 2 ,..., n k . ◮ For every i , take d n i vertices from V i – together, these will form a resolving set. ◮ Use ideas like we did for trees.

A B OUND IN TERMS OF THE C HROMATIC N UMBER Theorem (MMO 2019+): Let G be a graph of order n with chromatic number k . Then τ ( G ) < k ( d n / k + 2 ) < k (log 2 ( n / k ) + 2 ) . Sketch of proof: ◮ The vertices of G can be partitioned into k independent sets V 1 , . . . V k , say of orders n 1 , . . . , n k , respectively. ◮ Add all edges between these independent sets – we are now looking at K n 1 , n 2 ,..., n k . ◮ For every i , take d n i vertices from V i – together, these will form a resolving set. ◮ Use ideas like we did for trees. ◮ Finally, show that the worst case is when the n i ’s are approximately equal.

P LAN M ETRIC D IMENSION T HRESHOLD D IMENSION B OUNDS E MBEDDINGS

E MBEDDINGS AND S TRONG P RODUCTS An embedding of G in H is an injective function φ : V ( G ) → V ( H ) satisfying xy ∈ E ( G ) ⇒ φ ( x ) φ ( y ) ∈ E ( H ) .

E MBEDDINGS AND S TRONG P RODUCTS An embedding of G in H is an injective function φ : V ( G ) → V ( H ) satisfying xy ∈ E ( G ) ⇒ φ ( x ) φ ( y ) ∈ E ( H ) . In other words, an embedding of G in H is an injective homomorphism from G to H .

E MBEDDINGS AND S TRONG P RODUCTS An embedding of G in H is an injective function φ : V ( G ) → V ( H ) satisfying xy ∈ E ( G ) ⇒ φ ( x ) φ ( y ) ∈ E ( H ) . In other words, an embedding of G in H is an injective homomorphism from G to H . G H

E MBEDDINGS AND S TRONG P RODUCTS An embedding of G in H is an injective function φ : V ( G ) → V ( H ) satisfying xy ∈ E ( G ) ⇒ φ ( x ) φ ( y ) ∈ E ( H ) . In other words, an embedding of G in H is an injective homomorphism from G to H . G H

S TRONG P RODUCTS

S TRONG P RODUCTS We will be concerned with embeddings of graphs in the strong product of a number of paths.

S TRONG P RODUCTS We will be concerned with embeddings of graphs in the strong product of a number of paths. ◮ The strong product of 2 paths is a 2-dimensional grid with diagonal edges in addition to horizontal and vertical ones.

S TRONG P RODUCTS We will be concerned with embeddings of graphs in the strong product of a number of paths. ◮ The strong product of 2 paths is a 2-dimensional grid with diagonal edges in addition to horizontal and vertical ones. 8 7 6 5 4 3 2 1 00 1 2 3 4 5 6 7 8

S TRONG P RODUCTS We will be concerned with embeddings of graphs in the strong product of a number of paths. ◮ The strong product of 2 paths is a 2-dimensional grid with diagonal edges in addition to horizontal and vertical ones. 8 7 6 5 4 3 2 1 00 1 2 3 4 5 6 7 8 ◮ The strong product of b paths is an analogous b -dimensional grid.

Lemma (MMO 2019+): If β ( G ) = b , then G can be embedded in the strong product of b paths.

Lemma (MMO 2019+): If β ( G ) = b , then G can be embedded in the strong product of b paths. Idea of proof:

Lemma (MMO 2019+): If β ( G ) = b , then G can be embedded in the strong product of b paths. Idea of proof: ◮ Let W = { w 1 , . . . , w b } be a resolving set of G .

Lemma (MMO 2019+): If β ( G ) = b , then G can be embedded in the strong product of b paths. Idea of proof: ◮ Let W = { w 1 , . . . , w b } be a resolving set of G . ◮ Define φ by φ ( x ) = [ d ( x , w 1 ) , d ( x , w 2 ) , . . . , d ( x , w k )] .

Lemma (MMO 2019+): If β ( G ) = b , then G can be embedded in the strong product of b paths. Idea of proof: ◮ Let W = { w 1 , . . . , w b } be a resolving set of G . ◮ Define φ by φ ( x ) = [ d ( x , w 1 ) , d ( x , w 2 ) , . . . , d ( x , w k )] . Example: 3 [ 0 , 3 ] [ 3 , 0 ] 2 1 00 [ 2 , 3 ] [ 1 , 2 ] [ 2 , 1 ] [ 3 , 2 ] 1 2 3

N OTATION ◮ For an embedding φ of G in H , let φ ( G ) denote the subgraph of H induced by the set φ ( V ( G )) .

N OTATION ◮ For an embedding φ of G in H , let φ ( G ) denote the subgraph of H induced by the set φ ( V ( G )) . G H

N OTATION ◮ For an embedding φ of G in H , let φ ( G ) denote the subgraph of H induced by the set φ ( V ( G )) . G H

N OTATION ◮ For an embedding φ of G in H , let φ ( G ) denote the subgraph of H induced by the set φ ( V ( G )) . φ ( G ) G

G OOD E MBEDDINGS

G OOD E MBEDDINGS ◮ Call an embedding φ of G in the strong product of b paths good if there is a set of vertices W = { w 1 , . . . , w b } such that for all vertices x of G , we have φ ( x ) = [ d φ ( G ) ( φ ( x ) , φ ( w 1 )) , . . . , d φ ( G ) ( φ ( x ) , φ ( w b ))] .

G OOD E MBEDDINGS ◮ Call an embedding φ of G in the strong product of b paths good if there is a set of vertices W = { w 1 , . . . , w b } such that for all vertices x of G , we have φ ( x ) = [ d φ ( G ) ( φ ( x ) , φ ( w 1 )) , . . . , d φ ( G ) ( φ ( x ) , φ ( w b ))] . ◮ Essentially, the coordinates of φ ( x ) are the distances from φ ( x ) to the vertices in the set φ ( W ) in the subgraph φ ( G ) .

G OOD E MBEDDINGS ◮ Call an embedding φ of G in the strong product of b paths good if there is a set of vertices W = { w 1 , . . . , w b } such that for all vertices x of G , we have φ ( x ) = [ d φ ( G ) ( φ ( x ) , φ ( w 1 )) , . . . , d φ ( G ) ( φ ( x ) , φ ( w b ))] . ◮ Essentially, the coordinates of φ ( x ) are the distances from φ ( x ) to the vertices in the set φ ( W ) in the subgraph φ ( G ) . ◮ The embeddings constructed in the previous lemma are the prototypical good embeddings.

G OOD E MBEDDINGS ◮ Call an embedding φ of G in the strong product of b paths good if there is a set of vertices W = { w 1 , . . . , w b } such that for all vertices x of G , we have φ ( x ) = [ d φ ( G ) ( φ ( x ) , φ ( w 1 )) , . . . , d φ ( G ) ( φ ( x ) , φ ( w b ))] . ◮ Essentially, the coordinates of φ ( x ) are the distances from φ ( x ) to the vertices in the set φ ( W ) in the subgraph φ ( G ) . ◮ The embeddings constructed in the previous lemma are the prototypical good embeddings. 3 [ 0 , 3 ] [ 3 , 0 ] 2 1 00 [ 2 , 3 ] [ 1 , 2 ] [ 2 , 1 ] [ 3 , 2 ] 1 2 3

M ORE G OOD E MBEDDINGS This tree has metric dimension 5, but has a good embedding in the strong product of only 2 paths. 6 5 4 3 2 1 00 1 2 3 4 5 6

M ORE G OOD E MBEDDINGS This tree has metric dimension 5, but has a good embedding in the strong product of only 2 paths. 6 5 4 3 2 1 00 1 2 3 4 5 6

M ORE G OOD E MBEDDINGS This tree has metric dimension 5, but has a good embedding in the strong product of only 2 paths. 6 5 4 3 2 1 00 1 2 3 4 5 6 This tree has threshold dimension 2.

A G EOMETRICAL C HARACTERIZATION Theorem (MMO 2019+): Let G be a graph. Then τ ( G ) = b if and only if b is the smallest number such that there is a good embedding of G in the strong product of b paths.

A G EOMETRICAL C HARACTERIZATION Theorem (MMO 2019+): Let G be a graph. Then τ ( G ) = b if and only if b is the smallest number such that there is a good embedding of G in the strong product of b paths. ◮ So the notion of threshold dimension corresponds in some way to our usual geometrical notion of dimension.

A G EOMETRICAL C HARACTERIZATION Theorem (MMO 2019+): Let G be a graph. Then τ ( G ) = b if and only if b is the smallest number such that there is a good embedding of G in the strong product of b paths. ◮ So the notion of threshold dimension corresponds in some way to our usual geometrical notion of dimension. ◮ We thought this was cool!

A G EOMETRICAL C HARACTERIZATION Theorem (MMO 2019+): Let G be a graph. Then τ ( G ) = b if and only if b is the smallest number such that there is a good embedding of G in the strong product of b paths. ◮ So the notion of threshold dimension corresponds in some way to our usual geometrical notion of dimension. ◮ We thought this was cool! ◮ Is it useful?

Question: Are there trees of arbitrarily large metric dimension whose threshold dimension is 2?