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Computation, Computers, and Programs Cook's Theoremhttp://www.cs.caltech.edu/courses/cs20/a/ December 2, 2002 1

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Term summary

• Finite automata– Deterministice FA– Nondeterministic FA, with e-transitions– Regular expressions– Equivalence of Regex, eNFA, NFA, DFA– Pumping Lemma


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Determistic Finite Automata

• A finite automaton is a 5-tuple (Q,Σ, δ, q0, F)

– Q is a finite set of states

– Σ is an alphabet

– δ : Q× Σ→ Q is a transition function

– q0 ∈ Q is the initial state

– F ⊆ Q is a set of final or accepting states


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Transition diagrams

Q = {q0, q1, q2, q3}Σ = {0,1}δ =

State Symbol Goto

q0 0 q2

q0 1 q1

q1 0 q3

q1 1 q0

q2 0 q0

q2 1 q3

q3 0 q1

q3 1 q2

q0 = q0

F = {q0}

1

1

0 0

1

1

0 0

q0 q1

q2 q3


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A finite automaton

0 1 1 0 0 1 0 1

FiniteControl

Read-only tape

Tape head

1. In state q a. read a symbol c b. move the tape head right b. goto state delta(q, c)2. Accept iff the FA is in a final state after reading the last symbol


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Regex->eNFA

• Prove by structural induction: induction on the size of the regular expression

• Base cases:– Show the empty RE has an e-NFA– Show the RE has an e-NFA

– Show the a (symbol) RE has an e-NFA

• Induction:– Show (xy) has an e-NFA– Show (x + y) has an e-NFA– Show x* has an e-NFA


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Choice

ε

ε ε

ε

Machine 1

Machine 2

Start state Any (originally)final state


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Equivalence classes of RM

L(M)

Equivalence classes of RM


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Myhill-Nerode

Theorem (The Myhill-Nerode theorem). The followingthree statements are equivalent:

• The set L ⊆ Σ∗ is accepted by a FA.

• The set L is the union of some of the equivalenceclasses of a right-invariant equivalence relation withfinite index.

• Let equivalence relation RL be defined as xRLy iffxz ∈ L� yz ∈ L. Then RL has finite index.


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Regular languages

• Intuition: if a FA accepts a string that is “long enough,” it must repeat a state– But it can’t remember that the state was repeated– So it can be forced to repeat the state over and over

q0 qmqj=qk

a1,...,aj

aj+1,...,ak

ak+1,...,am


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Pumping lemma

Lemma (the Pumping Lemma)

• Let L be a regular set.

• There is a constant n s.t. for any z where |z| ≥ n,then z can be written z = uvw, where

– |uv| ≤ n– |v| ≥ 1

– For all i ≥ 0, uviw ∈ L– n is bounded by |Q|


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Term summary

• Context-free languages– Context-free grammars– Grammar simplification– Pushdown automata– Equivalence of CFGs and PDAs


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Sentence diagramming; derivation trees

FRUIT FLIES LIKE A BANANA

adjective noun verb noun

noun-phrase

verb-phrase

noun-phrase

sentence


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CFG: formal definition

• A CFG is a four-tuple (V , T , P, S)

– V is a finite set of nonterminals

– T is a finite set of terminals (V and T are dis-joint)

– P is a finite set of productions

– S is a nonterminal called the start symbol


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Ambiguity

e ::= e + e | e * e | NUMBER

e

e e

e e

1 2

3

+

*

e

ee

e e

2 3

1

*

+

(1 * 2) + 3 1 * (2 + 3)

Leftmost derivation Rightmost derivation


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Ambiguity

• A leftmost-derivation is a derivation in which a production is always applied to the leftmost symbol– A rightmost derivation applies to the rightmost symbol

• In general, a string may have multiple left and rightmost derivations

• A grammar in which some word has two parse trees is said to be ambiguous


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Simplification

• A nonterminal A isuseless iff– S * xAy * xzy– Otherwise, it is useless


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Eliminating epsilon-productions

Theorem All ε-productions can be eliminated, exceppossibly a production of the form S → ε

Definition A nonterminal A is nullable if A→∗G ε

Algorithm (finding nullable nonterminals)

• Base: if A→ ε then A is nullable

• Step: if A → α1 · · ·αn and α1 · · ·αn are all nulable, then A is nullable


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PDA machine

1 + 2 * 3 + 4 ;

FiniteControl

Read-only tape

Tape head

1. In state q a. read a symbol c, stack symbol Z b. move the tape head right b. goto state delta(q, c, Z).1 d. replace Z with delta(q, c, Z).2--or: a. goto state delta(q, ε, Z).1 b. replace Z with delta(q, c, Z).22. Accept iff the FA is in a final state after reading the last symbol

Stack

2

+1

Z


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PDA: formal definition

A (nondetermistic) PDA is a 7-tuple (Q,Σ, Γ , δ, q0, Z0, F)

• Q is a set of states

• Σ is an input alphabet

• Γ is a stack alphabet

• q0 is the start state

• Z0 is a stack symbol called the start symbol

• F ⊆ Q is a set of final states

• δ : Q× (Σ∪ {ε})× Γ → 2Q×Γ∗)


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Balanced parentheses

M = ({q1}, {), (}, {S, (}, δ, S, {})

δ(q1, (, S) = {(q1, ()}δ(q1, (, () = {(q1, ()}δ(q1, ), () = {(q1, ε)}δ(q1, ε, S) = {(q1, ε)}


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Empty stack vs final state

• Final state PDA -> empty stack PDA– Simulate: whenever the PDA reaches a final state, empty

the stack

• Empty stack PDA -> final state PDA– Add a special marker $ at the bottom of the stack– Add transitions delta(q, epsilon, S) qf for some new

state qf in F


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Compiling a CFG to a PDA

Theorem If G = (V , T , P, S), then there is a PDA Mwhere L(G) = L(M) (empty stack)


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Building the PDA from the GNF

To build the PDA for a grammar G = (V , T , P , S) in GNF

• Each production has the formA→ a1 . . . anX1 . . . Xm

• Define M = ({q}, T , V , δ, q, S, {})• Let δ(q,a,A) contains (q, γ) iff A→ aγ ∈ P


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Building a CFG from a PDA

Theorem IfM is a PDA, then there is a CFGG = (V , T , P , S)s.t. L(M) = L(G)

• Let M = (Q,Σ, Γ , δ, q0, Z0, F)

• Let V = {[q,A,p] | q,p ∈ Q∧A ∈ Γ} ∪ S• Productions

– S → [q0, Z0, q] for each q ∈ Q– [q,A, qm+1]→ a[q1, B1, q2] · · · [qm,Bm,qm+1]

for each

(q1, B1 . . . Bm) ∈ δ(q,a,A)


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Pumping lemma for CFL

Lemma Let L be a CFL. Then there is a constant n suchthat for any string z ∈ L where |z| ≥ n, then z =uvxyz and

• |vx| ≥ 1

• |vwx| ≤ n• uviwxiy ∈ L for i ≥ 0


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Pumping

A

A

A

A

A

S

S

vu y

vu y

w x v w x

x

Pump


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Ogden’s lemma

Ogden's Lemma Let L be a CFL. Then there is a constantn, then for any z ∈ L we can mark n or more positionsof z = uvwxy , such that:

• v and x have at least one marked symbol

• vwx has at most n marked symbols

• uviwxiy ∈ L for i ≥ 0


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Building the path

Constructed pathb

b

b

P

* * * ****

Branch points

Marked symbols


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Term summary: Turing Machines

• Turing Machines– Executions– Diagonalization– The halting problem– Rice’s theorem


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Turing machines

0 1 1 0 0 b b b

FiniteControl

Infinite read-write tape

Tape head

1. In state q a. read a symbol c b. print a new symbol c' c. move the tape head one position left or right d. goto state delta(q, c)2. Accept iff the TM ever enters a final state

Blanks


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Turing machine formal definition

A Turing Machine is a 7-tuple M = (Q,Σ, Γ , δ, s, b, F)

• Q is a finite set of states,

• Γ is a finite set of tape symbols,

• b ∈ Γ is the blank symbol,

• Σ ⊆ Γ − {b} is the set of input symbols,

• δ : Q × Γ → Q × Γ × {L,R} is a partial transitionfunction,

• s ∈ Q is the start state,

• F ⊆ Q is the set of final states


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Instantaneous descriptions

• An instantaneous description ID (σ ) isα1qα2, where

– q is the current state of the TM,

– α1α2 ∈ Γ∗ if the contents of the tape (to thelast non-blank symbol)

– The current symbol is the first symbol of α2


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Universal Turing Machines

Next, build a universal Turing machine U that accepts

L(U) ≡ {M#x | x ∈ L(M)}

• U checks that M#x is a valid description

• Copy M to one tape, and x to another tape

• Use a scratch tape to represents current state, headposition, etc.

• Simulate M : for each move:

– Figure out δ function from M

– Move the head,

– modify the tape x


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Another diagonalization argument

Place a * if M halts on w

Input string w0 1 2 3 4 5 ...

M0

M1

M2

M3

M4...

*

*

*

*

*

*

*

*

*

*

*

*

*

*

*

Machine MTuring Machine K = { M#w | M halts on w}


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Halting problem

• Problem: determine entries in the matrix• Find a recursive machine K, given M#x

– K halts and accepts if M accepts x– K halts and rejects if M does not terminate

• Is there such a machine?• Suppose so, then build a new machine N, that,

given x, runs K on Mx#x and:– Accepts if K rejects– Loops forever of K accepts

• If K exists, then N does too; let N = My

– N#y halts and accepts if N#y does not halt– N#y loops forever if N#y halts


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Halting problem

N(x) = if K(Mw#w) accepts then loop else halt

0 1 2 3 4 5 ...

M0

M1

M2

M3

M4...

*

*

*

*

*

*

*

*

*

*

*

*

*

Turing Machine K = { M#w | M halts on w}

N * * ?


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Rice’s theorem

• A property P is a set of r.e. languages (so it is a set containing sets of strings)

• For any r.e. language L, we say P(L) is true iff L is a member of P

• Theorem: Any nontrivial property of the r.e. languages is undecidable– Nontrivial means: the property is neither always true

nor always false


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Proof of Rice’s theorem

• Consider a property P– Assume P({}) = false, otherwise invert P

• Since P is nontrivial, there is a language L in P– Let ML be a TM accepting L

• To determine if M halts on input w, build M’– Ignore the input, and simulate M on w– If M halts, then start ML on the input string– So P(M’) iff M halts on w

w

x

Halts

MML yes/no


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Term summary: alternate models

• Primitive and partial recursive programs– For-programs– While-programs– Godel numbering– The recursion theorem– Rice’s theorem


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Primitive recursion

Functions N tuples → N tuplesThe following functions are primitive recursive:

• s(x) = x + 1 (the successor function)

• z(x) = 0 (the zero function)

• πmi (x1, . . . , xm) = xi (projection)

• For any f : Nm → Nm, and g1, . . . , gm : Nk → N,the composition f(g1(x), . . . , gm(x)) is primitiverecursive


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For-programs

• Variables x,y, z over N

• Simple assignments:

– x ← y

– x ← s(y)

– x ← 0

• Induction: if p,q are for-programs, so are:

– p;q (sequential composition)

– if x = y then p else q (conditional)

– fory do p done (for-loop)


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For programs and primitive recursion

Theorem The for-programs define exactly the primitive-recursive functions.

Encoding p.r. functions with for-loops. By inductionon the size of the p.r. function.

f(0, x) = h(x)f(s(y), x) = g(y,x, f (y,x))


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Partial recursive functions (Godel)

• To capture r.e. computations, we need more• A partial recursive computation includes the

prim-rec computations, plus unbounded minimization


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Unbounded minimization

• Find the least y s.t. g(x,y) = 0

f(x) = µy.g(x,y) = 0

where g is primitive recursive

• Alternatively:

f (x) = µy.(g(x,y) = 0∧g(x, z) is defined for z ≤ y)

where g does not have to be primitive recursive


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While-programs

The while programs are defined inductively:

• All for-programs are while-programs

• Add whilex ≠ y do p done, where p is a while-program


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Godel numbering

Gödel numbering:

• There is a universal function U(〈i, x〉) = ϕi(x)

• For anym,n there is a total recursive function Smn(a partial recursive function defined on all inputs),s.t. for any i and x1, . . . , xn, y1, . . . , ym

ϕSmn (i,x1,...,xn)(y1, . . . , ym) = ϕi(x1, . . . , xn,y1, . . . , ym)

An indexing is acceptable iff it satisfies these two prop-erties.


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Recursion theorem

Theorem (Fixed-point theorem)For all total recursive functions f : N → N, there is anindex i s.t. ϕi = ϕf(i).

ProofTake any v ∈ N, and consider a recursive function thatdoes the following on x:

1. Compute ϕv(v)

2. If it halts, apply f to get f (ϕv(v))

3. Use it as an index: compute ϕf(ϕv(v))(x)


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Rice’s theorem

Theorem (Rice's theorem)Every non-trivial property of the partial recursive func-tions is undecidable. That is, there is no total recursivefunction p : N→ {0,1} s.t.

• p(k) ≠ p(l)

• For any i, j, if ϕi =ϕj , then p(i) = p(j)

Proof Suppose p exists. Define

f(i) ={k if p(i) = p(l)l if p(i) = p(k)

f is total, and it has no fixpoint.


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Term summary: lambda calculus

• Syntax• Substitution• Reduction


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Lambda calculus

• A starting point for reasoning about functions• The foundation of most functional programming

languages, including the Lisp and ML languages

v ranges over a countable number of variables

e ::= v (variables)| e1 e2 (function application)| λv.e (function abstraction)


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Single-step evaluation

• A single-step reduction is just a substitution• Called “beta-reduction”

(λv.e1) e2 →β e1[e2/v]


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Term summary: logic

• Propositional logic• Predicate calculus• Godel’s incompleteness theorem (Rice’s theorem

in disguise)


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Defining the syntax

Start with a countable set of propositional letters P,Q,R, . . .Define the propositions inductively:

• (true) is a proposition

• ⊥ (false) is a proposition

• Any propositional letter is a proposition

• If A is a proposition, so it ¬A (negation)

• If A and B are propositions, so are

– A∧ B (conjunction)

– A∨ B (disjunction)

– A⇒ B (implication)


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Standard syntax definition

Propositions:

e ::= | ⊥| P,Q,R, . . .| ¬e| e∧ e| e∨ e| e ⇒ e


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Rule table

Γ � ∆1, ,∆2 Γ1,⊥, Γ2 � ∆

Γ � ∆1, α,∆2 Γ � ∆1, β,∆2

Γ � ∆1, α∧ β,∆2

Γ1, α, β, Γ2 � ∆Γ1, α∧ β, Γ2 � ∆

Γ � ∆1, α, β,∆2

Γ � ∆1, α∨ β,∆2

Γ1, α, Γ2 � ∆ Γ1, β, Γ2 � ∆Γ1, α∨ β, Γ2 � ∆

Γ , α � ∆1, β,∆2

Γ � ∆1, α⇒ β,∆2

Γ1, β, Γ2 � ∆ Γ1, Γ2 � α,∆Γ1, α⇒ β, Γ2 � ∆


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Pierce’s law

A � B,A 4

� A⇒ B,A 3 A � A 5

(A⇒ B) ⇒ A � A 2

� ((A ⇒ B)⇒ A)⇒ A 1


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Formalizing arithmetic

The language of arithmetic:

e ::= i | v | e+ e | · · · | e/eP ::= e = e | e < e | · · ·

| ¬P| P ∧ P | P ∨ P | P ⇒ P| ∃v.P[v] | ∀v.P[v]

Example:

∀x1.∃x2.∀x3.∃x4.(x3 + x2 ≤ (1+ x1)∗ x4)


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Indexing the arithmetic formulas

Define �ϕ� is the Gödel index of formula ϕDefine the normal way:

• �xi� = [0 :: i]

• �e1 + e2 = 1 :: �e1�@�e2�

• �e1 < e2� = 10 :: �e1�@�e2�

• �P1 ∧ P2� = 20 :: �P1�@�P2�

• · · ·


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Fixpoint theorem for arithmetic

Theorem (Fixpoint theorem)For any formula ψ(y) with free variable y , there is asentence ϕ with no free variables s.t. ϕ� ψ(�ϕ�).


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Godel’s Incompleteness Theorem

Theorem (Gödel's Incompletness theorem)There is a formula that is not provable.Proof

• Define a formula unprovable(�ϕ�) iff ϕ is notprovable.

• This is a formula in arithmetic

ψ(y) ≡ unprovable(y) ≡ unprovable(�ϕy�)

• Consider, for some ϕ, the formula

ϕ� unprovable(�ϕ�)


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Term summary: complexity

• Time & space bounded machines• NP computations

– SAT


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Space-bounded TMs

• Input tape is read-only• The number of storage

tapes is an arbitrary constant k

• M is DSPACE(T(n)) if:– M is deterministic– For any input of length n,

M scans at most T(n) cells on any storage tape

• NSPACE(T(n)) is the nondeterministic bound

Read-only input

FiniteControl

Storage tapes


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Time-bounded TMs

• All tapes are 2-way infinite

• M is DTIME(T(n)) if– M is deterministic– For any input of length n,

M takes at most T(n) steps

• M is NTIME(T(n))– Nondeterministic case

FiniteControl

Storage tapes

R/W inputput inpR/W


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Asymptotic complexity: big-Oh notation

Let f ,g : N→ N

• f is O(g) if

∃c ∈ N.∞∀n ∈ N.f (n) ≤ c · g(n)

∞∀ means ``for all but finitely many.'' Asympoti-cally, f grows no faster than g within a constantmultiple.


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CNF satisfiability

• A propositional formula is in conjunctive normalform (CNF) if it is a conjunction of disjunctions ofliterals.

• A literal is a propositional letter, or the negationof a propositional letter.

• Every propositional formula can be represented inCNF.

(A∨¬B ∨ C)∧ (D ∨¬B ∨¬C)∧ (¬A∨ B ∨ E ∨¬F)...∧ (E ∨ F ∨¬D ∨A∨¬B)


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Satisfiability phase transition (Selman)

• Using modern algorithms, plot how long it takes to solve a SAT formula as ratio of clauses/literals

1 4.3 10

t

clauses/literal

% clauses satisfiable

CPU time


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Term summary

• Graph theory– Undirected, directed graphs– Storngly-connected components– Solving 2SAT in linear time


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Formal definition of graphs

• A graph G is defined as a pair G = (V , E) where

– V is a set of vertices

– E is a set of edges (vi, vj), . . .

• n = |V | is the size of the graph

• |E| is the number of edges


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Path example

A simple path

A simple cycle

A non-simple path


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ILO

GY


Graph components

A connected graph

A disconnected graph

Articulation point

V1 V2


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


DFS: Finalize

2

2

2

2

0

1 3

45

6

7

0

1 3

45

6

7

0

1 3

45

6

7

0

1 3

45

6

7


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


DFS backedges

2

0

1 3

45

6

7Back edges


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


Compute low values

3 2

1

4

5 7

6

8

low[3] = 1

low[2] = 1

low[1] = 1

low[4] = 3

low[5] = 4

low[6] = 6

low[7] = 7 low[8] = 6

x = min{low(w) | w is an immediate descendent of v}y = min{z | z is reachable by a back-edge from v}

low(v) = min(x,y)


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


Term summary

• Reductions– 3SAT Clique– 3SAT Coloring

• Cook’s theorem– NP SAT


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


Conflict graph for 3SAT

x

¬x¬y x

¬y

yC1

C1

C2

C2

C3

C3

(x ∨ y) ∧ (¬x ∨ ¬y) ∧ (x ∨ ¬y)


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


SAT/clique example

x

¬x¬y x

¬y

yC1

C1

C2

C2

C3

C3

(x ∨ y) ∧ (¬x ∨ ¬y) ∧ (x ∨ ¬y)

Satisfying assignment:(Both cases): (x ∧ ¬y)


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


More NP problems: graph coloring

Definition Consider an undirected graph G = (V , E),and a finite set C . A coloring is a map ϕ : V → C suchthat ϕ(u) ≠ ϕ(v) for (u,v) ∈ E. Given G and a con-stant k, the k-coloring problem is to determine whetherthere exists a coloring using no more than k colors.

Reduction We show that coloring is hard by showingSAT ≤pm 3-colorability.

• Use the colors R, G, B.

• Build a triangle for each literal.


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


Unsatisfiable clause

¬x y ¬z w

Conflict

Suppose top row is entirely red

G G G G

G G


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


Satisfiable clause

¬x y ¬z w

Suppose at least one is green

G G G G

G G


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


Cook’s theorem

Theorem If R ∈ NP then R ≤pm SAT

Preparation

• Let M be the TM for R

• Assume the depth of the configuration tree is atmost N = |x|k for some constant k

• Assume thatM uses a single tape, delimited at theleft by a special symbol �

• Also, assume when M accepts, it erases its input,and moves all the way left

• Note, M can scan at most N tape cells


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


Boolean variables

• Qqi : at time i, the machine is in state q

• Hij : at time i, the tape head is at position j

• Saij : at time i, the symbol in cell j is a


CALI

FOR

NIA

INSTITUTE OF TECH

NO

ILO

GY


Boolean variables

S00Qq00

Q : at time i, the machine is in state qqi

00

S : at time i, tape cell j contains symbol a aij

Sa101 Sa2

02 SaN0N

H

S10Qq11 Sa1

11 Sa212 SaN

1N

11H

SN0QqNN

N0

S bN1 S b

N2 S bNN

H

S20Qq22 S x

21 Sa222 SaN

2N

22H

ijH : at time i, the tape head is at cell j

summary - Caltech Computingcourses.cms.caltech.edu/cs20/a/slides/summary/summary.pdf · If this...

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Transcript of summary - Caltech Computingcourses.cms.caltech.edu/cs20/a/slides/summary/summary.pdf · If this...