Regular Expression to NFA

Page 1 of 1

1 Replies - 5139 Views - Last Post: 26 November 2011 - 09:49 PM

Topic Sponsor:

#1 m-e-g-a-z  Icon User is offline

  • Winning
  • member icon


Reputation: 492
  • View blog
  • Posts: 1,448
  • Joined: 19-October 09

Regular Expression to NFA

Posted 26 November 2011 - 12:36 PM

I have been given the following regular expression ab*((b|d)|c*) and need to create a NFA for this expression. I have done it but i'm not sure if it is 100% correct. Any feedback would be much appreciated.


Posted Image

This post has been edited by m-e-g-a-z: 30 November 2011 - 08:53 PM


Is This A Good Question/Topic? 0
  • +

Replies To: Regular Expression to NFA

#2 Nikitin  Icon User is offline

  • D.I.C Regular

Reputation: 46
  • View blog
  • Posts: 256
  • Joined: 02-August 10

Re: Regular Expression to NFA

Posted 26 November 2011 - 09:49 PM

Are the 'e' symbols stand for epsilon transitions? Why so many of them? I don't see the requirement that the first character is 'a' being enforced on the graph, so it seems wrong to me.

Edit: Ah, looks like you meant to group the 'a' and 'b' together. In that case, the graph looks correct.

This post has been edited by Nikitin: 26 November 2011 - 09:54 PM

Was This Post Helpful? 0
  • +
  • -
Page 1 of 1