My Math Forum  

Go Back   My Math Forum > Science Forums > Computer Science

Computer Science Computer Science Forum


Reply
 
LinkBack Thread Tools Display Modes
May 9th, 2015, 03:47 AM   #1
Member
 
Joined: Nov 2013

Posts: 62
Thanks: 0

Question Complexity problem regarding Graph Isomorphism.

$A,B$ are symmetric matrices of graphs $G,H$ respectively. For $x \in G$, consider the graph $(G-x)$, it has vertices adjacent to $x$ and vertices non-adjacent to $x$.


$ A_x$ is the symmetric matrix of the graph $(G-x)$, where $C$ is the symmetric matrix of the graph created by vertices of $(G-x)$ which are adjacent to $x$ and $D$ is the symmetric matrix of graph the created by vertices of $(G-x)$ which are not adjacent to $x$.

$ A_x =
\left( \begin{array}{cc}
C & E \\
E^{T} & D\\
\end{array} \right) $


It should be noted, that-

1. Interchanging/swapping any two rows(or columns) of $C$ does not affect matrix $D$ (and vice versa).
2. Any change in $C$ or $D$ or both $C$ and $D$ changes matrix $E$.



If $G \simeq H $ then there exists $u \in H$ so that, $(G-x)\simeq (H-u)$, the matrix of $(H-u)$ is $B_u$, where $ B_u= \left( \begin{array}{cc}
S & R \\
R^{T} & Q\\
\end{array} \right)$

Consider the situation where $S \neq C,D \neq Q$. Since $(G-x)\simeq (H-u)$ , there exists a permutation matrix $P$, so that $PQ=D$.

Once $Q$ is arranged exactly like $D$ then $C$ can be arranged in linear/ polynomial time using the arrangement of rows of $E$ because after $Q=D$, the rows of $E$ and $R$ would be the same but arranged in different order. So, arrange the order of rows of $R$ like $Q$, it will make $S = C$.




One of the $S,Q$ has vertices $<\frac{n}{2}$ where $ n=|(G-x)|=|(H-u)|$, consider the smallest matrix to find the permutation matrix $P$.

So, to determine whether $(G-x)\simeq (H-u)$ is true or not, there are $n$ vertices to check. Let the time complexity to find this permutation matrix $P$ be $K(n/2)$. So, the complexity would be:

$n*K(n/2)$ (omitting the time complexity of rearranging $E$ as it has less growth than $K$)

But recursion of above method can be used to minimize the complexity of $K(n/2)$ for smallest matrix, say, $Q$, in that case, complexity would be

$n*n/2* K(n/4)$
Using recursion, this gives a bound $\frac{n^{log_2(n)}}{2^{\sum log_2(n)}}$.


Is this bound correct?? If not, what is the correct complexity of the above algorithm?

Last edited by skipjack; May 9th, 2015 at 07:48 AM.
jim198810 is offline  
 
Reply

  My Math Forum > Science Forums > Computer Science

Tags
complexity, graph, isomorphism, problem, regrading



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Graph(regular) Isomorphism in n^(O(log2(n))) . jim198810 Computer Science 8 May 9th, 2015 03:54 AM
Graph Isomorphism Testing mt2 Applied Math 0 October 18th, 2013 05:52 AM
Complexity of Problem UnOriginal Applied Math 2 August 26th, 2009 07:05 AM
complexity problem baxy7 Applied Math 14 April 20th, 2009 02:29 PM
isomorphism problem Ujjwal Linear Algebra 1 November 8th, 2008 05:48 PM





Copyright © 2018 My Math Forum. All rights reserved.