Skip to main content

Posts

Showing posts from November, 2016

DESIGN TM FOR {a^n b^n c^n | n>=1}

Concept:- General_idea: click_me L={a^n b^n c^n | n>=1} w={abc,aabbcc,aaabbbccc,....} a a a b b b c c c  X a a Y b b Z c c X X a Y Y b Z Z c Step_1: Replace a by X and go right side. TM tape can move any side. It has infinite memory in both side. And it can remember previous states details. Step_2: Replace b by Y and go right side of tape. Step_3: Replace c by Z and turn left till X .  Step_4: After seeing X(capital X) replace next a by X and do same loop. Step_5: X portions finished. Step_6: Move to Y states. Denote it for one move first and after that notate self loops. Goto final state Z in self loop. Finally stop turing machine. Success. Next step: How to add self loop states? Ok listen me... For that we are taking two steps X a a Y b b Z c c X X a Y Y b Z Z c  Just understand in between symbols. And make it as self loops. If any doubt please post comments..I can explain it. Small editing...This one is right answ
Post a Comment
Read more

Show that a tree has either one or two centers.

pf : No need to consider any trees on fewer than 3 vertices (WHY?). If the radius of the tree on n ≥ 3 vertices is one, then there is a vertex with eccentricity 1, which means it is connected to all other points. Claim: this center is unique. Assume otherwise, then another point is connected to all other points as well, including the known center. Taking an edge from this point to the known center, another edge out to a third distinct point, and finally an edge back to this point gives a cycle. That contradicts that this graph is a tree. Now let T be a tree with radius r > 1. Remove all vertices of degree 1 and their associated edges. This creates a graph of radius r − 1, which we can assume by induction has either 1 or 2 centers, which would obviously be the centers T. Reference: [1] 9.1 Introduction to Trees ,Available: click_me 
Post a Comment
Read more