## Obtaining laplacian of a graph

### Deepa Maheshvare (view profile)

on 16 Dec 2018
Latest activity Commented on by Christine Tobler

### Christine Tobler (view profile)

on 19 Dec 2018
The Neumann Laplacian of a simple graph(G) can be formed from the commands degree(G) and adjacency(G), L= D- A .
Could someone suggest how Dirichlet laplacian can be obtained?

Christine Tobler

### Christine Tobler (view profile)

on 18 Dec 2018
Based on formula (1.10) in that paper, that would be L = 2*d*I + (2*d*I - D) - A, with I the identity and d a scalar depending on the dimensionality of the graph, wouldn't it?
Deepa Maheshvare

### Deepa Maheshvare (view profile)

on 19 Dec 2018
I tried the above formula for an 1 D graph with 5 nodes.
d = 1,
L = 2*d*I + (2*d*I - D) - A,
gives,
3 -1 0 0 0
-1 2 -1 0 0
0 -1 2 -1 0
0 0 -1 2 -1
0 0 0 -1 3
whereas, the pseudo dirichlet 2*d*I-A (1.9)gives
2 -1 0 0 0
-1 2 -1 0 0
0 -1 2 -1 0
0 0 -1 2 -1
0 0 0 -1 2
which matches with the laplacian computed using centered difference formula for the second derivative operator with dirichlet boundary condition.
Also, I am not sure how the dimensionality of any given graph can be determined.
Any suggestions?
Christine Tobler

### Christine Tobler (view profile)

on 19 Dec 2018
A graph doesn't have an inherent dimensionality, this would have to be based on the construction of the graph. Perhaps the linked paper has more information.