Skip to main content

13/11/2012

Clique in an undirected graph is a subgraph, wherein every 2nodes are connected by an edge.

Clique Problem is to determine whether a graph contains a clique of specified size.

Satisfiability Problem is to test whether a boolean formula is satisfiable. Boolean formula is an expression involving boolean variable & operations. 

Halting Problem is the problem for determining whether a turing m/c halts(by accepting/by rejecting) on a given input.

Decidable  is a problem whose language is recursive. 

A problem is undecidable if there exist no algorithm that takes as input an instance of the problem and determine whether the answer to that instance is "yes" or "no".

Halting Problem is undecidable theorem and proof's scanned copy will upload next day. 

Church's Thesis: - What function cant be computed by turing machines? This is very surprising. It is believed that there are no functions that can be defined by humans,whose calculation can be described by any well defined mathematical algorithm that people can be taught to perform,that cant be computed by TMs. The TM is believed to be the ultimate calculating mechanism. 
Church actually said that any machine that can do certain list of operations will be able to perform all concievable algorithms. he tied together what logicians had called recursive functions and computable functions. TMs can do all that Church's asked, so they are possible model of the universal algorithm machine Church described.

Listen students......
Chomsky Classification/Hierarchy is not equal to Chomsky Normal Form. I will upload some notes about that next day.

Comments

  1. I had posted the same things in Our students google groups corner...

    MESCE CSE A '10 - '14

    ReplyDelete

Post a Comment

Popular posts from this blog

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...

Design a TM(Turing Machine) , L={a^nb^n | n>=1}

Design View: By JFLAP Software Concept :  aaabbb Case1: XaaYbb Case2:XXaYYb Case3:XXXYYY Case 1: Read first 'a' and replace it by 'X' Then read first 'b' and replace it as 'Y' What i am doing is .....Meaning of a^n b^n is " Equal no: of a's must followed by equal no: of b's " In Case 1 : I read first 'a' and First 'b'. Before reading Second 'b'....Should read Second 'a' .... Then continue the steps.....Finally 'B' reads....Machine will stop... Transitions concept will continue ......

Counter Machines

Counter machine has the same structure as the multi-stack machine  but in place of each stack is a counter. Counters hold any non-negative integer,but we can only distinguish between zero and nonzero counters. That's the move of the counter machine depends on its state,input symbol and which if any of the counters are zero. In one move,the counter machine can a)change state b)Add or subtract 1 from any of its counters,independently. However a counter is not allowed to become negative,so it cant subtract from a counter that is currently 0. Counter Machine may also be regarded as a restricted multi-stack machine. The restrictions are as follows, a)There are only two stack symbols,which we shall refer to as Z0(the bottom of stack marker) and X. b)Z0 is initially on each stack. c)We may replace Z0 only by a string of the form X^iz0 for some    i >=0. d)We may replace X only by X^i for some i >= 0. That's Z0 appears   only on the bottom of each stack and...