added lec 3

Author: geevi

2.do.txt

@@ -4,7 +4,7 @@
 ======= Russell's Paradox and Diagonalization ======= 
 
 *There is a village where the barbers shave only those people who 
-do not shave by themselves. Now will a does a barber shave himself?*
+do not shave by themselves. Now does a barber shave himself?*
 
 If YES, then he cannot shave himself. If NO, then he can shave himself.
 
@@ -63,6 +63,21 @@ input $x$. It is just another TM having its own transition function, alphabets a
 such that it can accept a TM encoding $\langle M \rangle$ as in part of the input, and the transition function
 is defined such that it simulates the running of $M$ on a string $x$.
 
+__Configuration.__ An important concept for doing simulation is the configuration of a TM. Lets 
+stick to 1-tape TMs for now. The "state" of the algorithm (not the TM state) is
+really consist of the the TM's state, the contents of its tape as well as the position
+of the tape heads. If a TM is at state $\omega_i$, has the string 101101111 in 
+its tape and the tape head is at the 5th position, then its current configuration is 
+1011 $\omega_i$ 01111. Note that this is a string over the alphabet $\mathcal{T} \cup \Omega$. 
+The computation of TM is a graph in the space of all TM configurations. If the TM 
+halts, it is a path which ends in a configuration having an accept or reject state.
+If it loops, there will be a cycle in this graph. 
+
+ 
+A universal TM, goes over a configuration string, then goes over the encoding of the delta function,
+finds out which rule in the delta function to use, and can update the configuration,
+to reflect the next configuration of the TM begin simulated.
+
 
 ======= Halting Problem ======= 
 

3.do.txt

@@ -3,10 +3,49 @@
 
 ======= Non Deterministic Turing Machines ======= 
 
-
-
-
-
-
-
-======= NP and Search Problems ======= 
+Last lecture we saw that TM's can be encoded as strings and simulated by a
+universal TM using the configurations. We also saw that simulation of a TM,
+is essentialy tracing a path in the configuration space. But in the TMs that 
+we defined, the out degree of any node (a configuration) 
+in the graph (in the configuration space) is one. 
+
+Similar to the Nondeterministic finite automata, we can also define TMs 
+with delta rules that results in multiple next states 
+(called Nondeterministic Turing Machines or NTMs). Then the delta rules
+will be of the form $\delta:\Omega\times \mathcal{T} \rightarrow \mathcal{P}(\Omega\times \mathcal{T}\times \{\langle, \rangle, -\})$.
+($\mathcal{P}$ denotes the power set). So the configuration graph of a 
+NTM can have out degree greater than $1$. However the it still has to be a finite 
+number since the size of $\mathcal{P}(\Omega\times \mathcal{T}\times \{\langle, \rangle, -\})$ is 
+finite. Following the different paths, an NTM could accept, reject or keep looping.
+So we need to define what is meant by deciding a language by an NTM.
+
+An NTM is said to decide a language (a descision problem) $L$ iff,
+for all strings in the language, there exists one path in the configuration 
+space that results in accept state.  For strings not in the language, all 
+paths in the configuration space should result in reject state.
+
+We know that NFA (Nondeterministic Finite automata) can always be converted to 
+a deterministic one. However for Pushdown Automaton this coverstion is not possible
+always. We will see that for TMs, this conversion can always be done. That is the 
+set of languages that can be decided by TMs does not change by allowing 
+nondeterminism.
+
+Recall that UTM simulated a deterministic TM, by tracing the path in the configuration
+space. But for NTMs, the graph in the configuration space is a tree. A simple 
+idea is for a UTM to do a graph traversal. DFS might be a bad idea, because
+some of the paths go into infinite loops. Hence it can to BFS. The first time,
+it finds that the NTM has reached the accept state, the UTM can also accept.
+If it never finds an accept state, the simulating TM rejects.
+
+======= NP and Search Problems ======= 
+
+As we disscussed earlier, an NTM can take different paths in the configuration space.
+The length of the path is essentialy the number of steps. Now we will define 
+the worst case running time for a NTM. 
+
+For a language $L$, on inputs of size $n$, the worst case running time of an NTM 
+is the length of the longest path in the configuration space on any of the inputs 
+of size $n$.
+
+NP or Nondeterministic Polynomial time  is the class of descision problems for which,
+there is an NTM which decides it in worst case polynomial time.