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‎.gitignore

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## html
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*.html
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*.pyc
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## Intermediate documents:
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*.dvi
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*-converted-to.*

‎4.do.txt

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========= Lecture 4: Reductions, Cook-Levin Theorem and NP-Completeness =========
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The notes are mostly from Section 7.4, 7.5 in the Sipser book.
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======= Polynomial Time Reductions =======
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Last lecture, we showed that for some problems, the search problem can be solved in polytime,
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if the corresponding desciion problem can be solved in polytime. In this lecture,
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we will show that many desicion problems can be solved by solving one particular
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desciion problem called $3$-SAT.
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For this we need first define a reduction. The following problems were defined previously
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o $3$-SAT = $\{ \phi : \phi \text{ is a 3CNF formula that is satisfiable} \}$.
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o CLIQUE = $\{ (G,k) : G \text{ has a clique of size } k \}.$
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We will say that $f$ is a (poly time) reduction from $3$-SAT to CLIQUE, if it maps $3$-CNF formulas $\phi$
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to a tuple $(G,k)$ where $G$ is a graph and $k$ is a number such that
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o $f$ can be computed by a polynomial time TM.
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o $\phi$ is satisfiable if and only if $G$ has a $k$ clique.
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If we have such an $f$, any polynomial time algorithm for CLIQUE can be used to
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design a polynomial time algorithm for $3$-SAT.
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Verify that: If we have such an $f$, give a polynomial time algorithm for $3$-SAT,
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assuming there is a polynomial time algorithm for CLIQUE.
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The algorithm for computing $f$ is as follows:
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o For every clause in $\phi$, put $3$ new verticies correponding to each literal in the clause.
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o Put all edges in the graph except:
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o between the 3 veritices correponding to the same clause.
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o $x_i$ an $\bar x_i$ for all $i$.
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o Set $k$ to be equal to the number of clauses.
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Verify the following:
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o $f$ is polynomial time.
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o Show that if $\phi$ is satisfiable $G$ has a $k$ CLIQUE.
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o Show that if $G$ has a $k$ CLIQUE then $\phi$ is satisfiable.
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Such a reduction says that CLIQUE is a harder problem than $3$-SAT, because an algo for
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CLIQUE gives an algo for $3$-SAT and we dont know if the reverse is True. Hence it is denote as
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$$ \text{CLIQUE} \geq_p 3\text{-SAT}.$$
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======= Cook-Levin Theorem =======
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The Cook-Levin Theorem tells that the reverse reduction also exists. In fact,
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it states that any language in NP can be reduced to SAT
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__Cook-Levin Theorem.__
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For any language $L \in NP$, SAT $\geq_p $ L.
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For doing this, for any language $L$, that has a nondeterministic polynomial time TM,
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we need to come up with a polynomial time reduction, which satisfies the conditions,
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given in the previous section.
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See Theorem 7.37 for the proof of Cook-Levin Theorem.
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!bu-problem
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Show that $3$-SAT $\geq_p$ SAT.
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!eu-problem
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======= NP Compeleness / Hardness =======
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Vertex Cover 3SAT Reduction.
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!bu-problem
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Solve Problem 7.21 in Sipser book
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!eu-problem
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!bu-problem
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Solve Problem 7.23 in Sipser book
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!eu-problem
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!bu-problem
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Solve Problem 7.27 in Sipser book
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!eu-problem

‎5.do.txt

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========= Lecture 5: More Time Complexity =========
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======= NP-completeness of Vertex Cover and Subset Sum =======
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See Section 7.5 in Sipser 2nd Edition
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======= EXP and Time Heirachy Theorem =======
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EXPTIME is the set of languages for which there is a poly time determinitic
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TM that runs in $2^{n^k}$ for some integer $k$.
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Clearly $\text{P} \subseteq \text{NP} \subseteq \text{EXPTIME}$.
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As we discussed, the question if P=NP or P $\subsetneq$ NP is an open problem.
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But what about P $\subsetneq$ EXP?
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This is easy to prove using diagonalization. Let DTIME($f(n)$), be the
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set of languages that can be decided in time $f(n)$ by a determinitic TM.
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Note that P $\subseteq$ DTIME($2^n$). We will design a language that is
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different from all languages in DTIME($2^n$) by can be decided in EXPTIME.
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We define the language by giving a TM $D$. $D$ takes encodings of TMs as
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input $\langle M \rangle$. D simulates M on $\langle M \rangle$ for $2^n$ steps.
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If M halts in $2^n$ steps, it gives the opposite answer. Otherwise it rejects.
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Due to
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the overheads involved in simulating a TM, the TM $D$ will take more time
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than $2^n$, however halts in less than $2^{n^2}$ steps. Hence the
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language decided by $D$ is in EXPTIME.
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But can the language decided by $D$ be in DTIME($2^n$). That is their
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another TM $D'$ that decided this language in $2^n$ steps. If so
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what is the output of $D$ on the input $\langle D' \rangle$.
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You can see that this results in a contradiction and hence the
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language decided by $D$ is not in P. So P $\subsetneq$ EXPTIME.
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The question of whether NP = EXPTIME is again open.
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======= coNP and Map of Complexity classes =======
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The complement of a language $L$ is
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$$ \bar L = \{ x : x \not in L \}.$$
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Let coNP $= \{ L : \bar L \in \text{NP} \}$.
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Another way of defining coNP is: it is the set of languages $L$ for which
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there is a determinitic TM $M$ that takes 2 inputs $x,y$ such that
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o If $x \in L$ then for every $y$, $M(x,y)$ accepts.
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p If $x \notin L$ then there exist a $y$, for which $M(x,y)$ rejects.
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!bu-problem
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Show that these two definitions gives the same set of languages.
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!eu-problem
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coNP is a set of languages for which there is a certificate for non membership.
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Similar to NP-complete, we can define:
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$$\text{coNP-complete} = \{ L \in \text{coNP} : \forall L' \in \text{coNP}, L' \leq_p L \}$$
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!bu-problem
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Show that UNSAT = $\{ \phi : \phi \text{ is a boolean formula that is unsatisfiable} \}$,
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is coNP-complete
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(a formula is unstatisfiable when for all assignments to the variable, the formula evaluates to False).
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!eu-problem
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Note that P = coP.
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Map of complexity classes
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‎6.do.txt

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========= Lecture 5: Space Complexity =========
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======= Space Complexity classes and some Relationships =======
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Let SPACE$(f(n))$ be the set of languages that can be decided in space $f(n)$,
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then
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$$\text{TIME}(f(n)) \subseteq \text{SPACE}(f(n)) \subseteq \text{TIME}(2^{cf(n)}).$$
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Let coSPACE$(f(n))$ be the complements of languages in SPACE$(f(n))$, then
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$$\text{coSPACE}(f(n)) = \text{SPACE}(f(n)).$$
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Let L be the set of languages that can be decided in space $c\log n$ for some constant $c$,
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and PSPACE the set of languages that can be deciding in polynomial space. Then
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$$ \text{L} \subsetneq \text{PSPACE}.$$
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!bu-problem
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Show that
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o $\text{L} = \text{coL} \subseteq \text{P} \subset \text{PSPACE} \subseteq \text{EXPTIME}.$
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o $\text{L} \subsetneq \text{PSPACE}.$
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!eu-problem
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======= Non Determinism and Space : NL-complete, Savith's Theorem =======
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Just like we defined NP, we can define NL as the set of languages that
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is decided by a non deterministic log-space TM. Whether L = NL is again
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a well known open problem like P = NP. We can do reductions in NL.
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As we will see later $NL \subseteq P$. Therefore any two languages
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in NL are polynomial time reducible (in fact a polytime machine solve the problem
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itself). Hence for defining NL-complete langauges, we use log-space reductions
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that runs in $O(\log n)$ space (denoted by $\leq_l$). That is
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$$\text{NL-complete} = \{ R : R \text{ is a language such that for all other languages R' } \in L, R' \leq_l R \}.$$
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!bu-problem
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Show that
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o $\text{REACHABILITY} \in \text{NL-complete}.$
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o $\text{NL} \subseteq \text{P}.$
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!eu-problem
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Actually all languages is NL can be decided by a deterministic $O(\log^2 n)$ space TM.
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This theorm is known as Savitch's Theorem.
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Savitch's theorem says that $\text{NPSPACE}(f(n)) \subseteq \text{SPACE}(f^2(n))$.
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See proof in proof of Theorem 8.5 in Sipser book.
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======= Overview of next part of course =======
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Streaming algorithms for Well paranthesised expressions.
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Lowerbound using communication complexity.
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======= Readings =======
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Chapter 8, Introduction to Theory of Computation by Micheal Sipser, Edition 2.
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Chapter 4, Computational Complexity: A Modern Approach
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Sanjeev Arora and Boaz Barak
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http://theory.cs.princeton.edu/complexity/book.pdf
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‎7.do.txt

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========= Lecture 5: Space Complexity =========
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======= Space Complexity classes and some Relationships =======
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Let SPACE$(f(n))$ be the set of languages that can be decided in space $f(n)$,
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then
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$$\text{TIME}(f(n)) \subseteq \text{SPACE}(f(n)) \subseteq \text{TIME}(2^{cf(n)}).$$
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Let coSPACE$(f(n))$ be the complements of languages in SPACE$(f(n))$, then
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$$\text{coSPACE}(f(n)) = \text{SPACE}(f(n)).$$
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Let L be the set of languages that can be decided in space $c\log n$ for some constant $c$,
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and PSPACE the set of languages that can be deciding in polynomial space. Then
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$$ \text{L} \subsetneq \text{PSPACE}.$$
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!bu-problem
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Show that
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o $\text{L} = \text{coL} \subseteq \text{P} \subset \text{PSPACE} \subseteq \text{EXPTIME}.$
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o $\text{L} \subsetneq \text{PSPACE}.$
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!eu-problem
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======= Non Determinism and Space : NL-complete, Savith's Theorem =======
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Just like we defined NP, we can define NL as the set of languages that
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is decided by a non deterministic log-space TM. Whether L = NL is again
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a well known open problem like P = NP. We can do reductions in NL.
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As we will see later $NL \subseteq P$. Therefore any two languages
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in NL are polynomial time reducible (in fact a polytime machine solve the problem
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itself). Hence for defining NL-complete langauges, we use log-space reductions
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that runs in $O(\log n)$ space (denoted by $\leq_l$). That is
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$$\text{NL-complete} = \{ R : R \text{ is a language such that for all other languages R' } \in L, R' \leq_l R \}.$$
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!bu-problem
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Show that
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o $\text{REACHABILITY} \in \text{NL-complete}.$
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o $\text{NL} \subseteq \text{P}.$
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!eu-problem
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Actually all languages is NL can be decided by a deterministic $O(\log^2 n)$ space TM.
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This theorm is known as Savitch's Theorem.
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Savitch's theorem says that $\text{NPSPACE}(f(n)) \subseteq \text{SPACE}(f^2(n))$.
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See proof in proof of Theorem 8.5 in Sipser book.
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======= Immerman-Szelepcsenyi Theorem =======
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In time complexity, we didnt have a proof of NP = coNP. This remains an open problem.
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However in space complexity, we know that NL = coNL. This theorem is known as the
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Immerman-Szelepcsenyi Theorem.
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See Section 8.6 in Sipser book.
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======= Readings =======
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Chapter 8, Introduction to Theory of Computation by Micheal Sipser, Edition 2.
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Chapter 4, Computational Complexity: A Modern Approach
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Sanjeev Arora and Boaz Barak
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http://theory.cs.princeton.edu/complexity/book.pdf
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‎full.do.txt

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# #include "1.do.txt"
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# #include "2.do.txt"
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# #include "3.do.txt"
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# #include "4.do.txt"
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# #include "5.do.txt"
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# #include "6.do.txt"
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