• 沒有找到結果。

# The Proof (continued)

N/A
N/A
Protected

Share "The Proof (continued)"

Copied!
73
0
0

(1)

### Notations

• Suppose M is a TM accepting L.

• Write L(M) = L.

– In particular, if M (x) = for all x, then L(M) = ∅.

• If M(x) is never “yes” nor  (as required by the deﬁnition of acceptance), we also let L(M ) = ∅.

(2)

### Nontrivial Properties of Sets in RE

• A property of the recursively enumerable languages can be deﬁned by the set C of all the recursively

enumerable languages that satisfy it.

– The property of finite recursively enumerable languages is

{ L : L = L(M) for a TM M, L is ﬁnite }.

• A property is trivial if C = RE or C = ∅.

– Answer to a trivial property is always “yes”or always

“no.”

(3)

### Nontrivial Properties of Sets in RE (concluded)

• Here is a trivial property (always yes): Does the TM accept a recursively enumerable language?a

• A property is nontrivial if C = RE and C = ∅.

– In other words, answer to a nontrivial property is

“yes” for some TMs and “no” for others.

• Here is a nontrivial property: Does the TM accept an empty language?b

• Up to now, all nontrivial properties (of recursively

enumerable languages) are undecidable (pp. 156–157).

• In fact, Rice’s theorem conﬁrms that.

aOr, L(M ) ∈ RE?

bOr, L(M ) = ∅?

(4)

### Rice’s Theorem

Theorem 13 (Rice, 1956) Suppose C = ∅ is a proper subset of the set of all recursively enumerable languages.a Then the question “L(M ) ∈ C?” is undecidable.

• Note that the input is a TM program M.

• Assume that ∅ ∈ C (otherwise, repeat the proof for the class of all recursively enumerable languages not in C).

• Let L ∈ C be accepted by TM ML (recall that C = ∅).

• Let MH accept the undecidable language H.

MH exists (p. 139).

aA nontrivial property, i.e.

(5)

### The Proof (continued)

• Construct machine Mx(y):

if MH(x) = “yes” then ML(y) else 

• On the next page, we will prove that

x ∈ H if and only if L(Mx) ∈ C. (1) – As a result, the halting problem is reduced to

deciding L(Mx) ∈ C.

– Hence L(Mx) ∈ C must be undecidable, and we are done.

(6)

### The Proof (concluded)

• Suppose x ∈ H, i.e., MH(x) = “yes.”

Mx(y) determines this, and it either accepts y or never halts, depending on whether y ∈ L.

– Hence L(Mx) = L ∈ C.

• Suppose MH(x) =.

Mx never halts.

L(Mx) = ∅ ∈ C.

(7)

• C must be arbitrary.

• The following Mx(y), though similar, will not work:

if ML(y) = “yes” then MH(x) else .

• Rice’s theorem is about properties of the languages accepted by Turing machines.

• It then says any nontrivial property is undecidable.

• Rice’s theorem is not about Turing machines

themselves, such as ”Does a TM contain 5 states?”

(8)

### Consequences of Rice’s Theorem

Corollary 14 The following properties of recursively enumerative sets are undecidable.

• Emptiness.

• Finiteness.

• Recursiveness.

• Σ.

• Regularity.a

• Context-freedom.b

aIs it a regular language?

bIs it a context-free language?

(9)

### Undecidability in Logic and Mathematics

• First-order logic is undecidable (answer to Hilbert’s (1928) Entscheidungsproblem).a

• Natural numbers with addition and multiplication is undecidable.b

• Rational numbers with addition and multiplication is undecidable.c

aChurch (1936).

bRosser (1937).

cRobinson (1948).

(10)

### Undecidability in Logic and Mathematics (concluded)

• Natural numbers with addition and equality is decidable and complete.a

• Elementary theory of groups is undecidable.b

aPresburger’s Master’s thesis (1928), his only work in logic. The direction was suggested by Tarski. Moj¯zesz Presburger (1904–1943) died in a concentration camp during World War II.

bTarski (1949).

(11)

(12)

(13)

## Boolean Logic

(14)

Christianity is either false or true.

— Girolamo Savonarola (1497) Both of us had said the very same thing.

Did we both speak the truth

—or one of us did

—or neither?

— Joseph Conrad (1857–1924), Lord Jim (1900)

(15)

### Boolean Logic

a

Boolean variables: x1, x2, . . ..

Literals: xi, ¬xi.

Boolean connectives: ∨, ∧, ¬.

Boolean expressions: Boolean variables, ¬φ (negation), φ1 ∨ φ2 (disjunction), φ1 ∧ φ2 (conjunction).

n

i=1 φi stands for φ1 ∨ φ2 ∨ · · · ∨ φn.

n

i=1 φi stands for φ1 ∧ φ2 ∧ · · · ∧ φn.

Implications: φ1 ⇒ φ2 is a shorthand for ¬φ1 ∨ φ2.

Biconditionals: φ1 ⇔ φ2 is a shorthand for 1 ⇒ φ2) ∧ (φ2 ⇒ φ1).

aGeorge Boole (1815–1864) in 1847.

(16)

### Truth Assignments

• A truth assignment T is a mapping from boolean variables to truth values true and false.

• A truth assignment is appropriate to boolean expression φ if it deﬁnes the truth value for every variable in φ.

{ x1 = true, x2 = false } is appropriate to x1 ∨ x2. { x2 = true, x3 = false } is not appropriate to

x1 ∨ x2.

(17)

### Satisfaction

• T |= φ means boolean expression φ is true under T ; in other words, T satisfies φ.

• φ1 and φ2 are equivalent, written φ1 ≡ φ2,

if for any truth assignment T appropriate to both of them, T |= φ1 if and only if T |= φ2.

(18)

### Truth Table

a

• Suppose φ has n boolean variables.

• A truth table contains 2n rows.

• Each row corresponds to one truth assignment of the n variables and records the truth value of φ under it.

• A truth table can be used to prove if two boolean expressions are equivalent.

– Just check if they give identical truth values under all appropriate truth assignments.

aPost (1921); Wittgenstein (1922). Here, 1 is used to denote true; 0 is used to denote false. This is called the standard representation (Beigel, 1993).

(19)

p q p ∧ q

0 0 0

0 1 0

1 0 0

1 1 1

(20)

p q p ∨ q

0 0 0

0 1 1

1 0 1

1 1 1

(21)

p ¬p

0 1

1 0

(22)

### p ⇒ q ≡ ¬q ⇒ ¬p

p q p ⇒ q ¬q ⇒ ¬p

0 0 1 1

0 1 1 1

1 0 0 0

1 1 1 1

(23)

### De Morgan’s Laws

a

• De Morgan’s laws state that

¬(φ1 ∧ φ2) ≡ ¬φ1 ∨ ¬φ2,

¬(φ1 ∨ φ2) ≡ ¬φ1 ∧ ¬φ2.

• Here is a proof of the ﬁrst law:

φ1 φ2 ¬(φ1 ∧ φ2) ¬φ1 ∨ ¬φ2

0 0 1 1

0 1 1 1

1 0 1 1

1 1 0 0

aAugustus DeMorgan (1806–1871) or William of Ockham (1288–

(24)

### Conjunctive Normal Forms

• A boolean expression φ is in conjunctive normal form (CNF) if

φ =

n i=1

Ci,

where each clause Ci is the disjunction of zero or more literals.a

– For example,

(x1 ∨ x2) ∧ (x1 ∨ ¬x2) ∧ (x2 ∨ x3).

• Convention: An empty CNF is satisﬁable, but a CNF containing an empty clause is not.

aImproved by Mr. Aufbu Huang (R95922070) on October 5, 2006.

(25)

### Disjunctive Normal Forms

• A boolean expression φ is in disjunctive normal form (DNF) if

φ =

n i=1

Di,

where each implicanta or simply term Di is the conjunction of zero or more literals.

– For example,

(x1 ∧ x2) ∨ (x1 ∧ ¬x2) ∨ (x2 ∧ x3).

aDi implies φ, thus the term.

(26)

### Clauses and Implicants

• The 

of clauses yields a clause.

– For example,

(x1 ∨ x2) ∨ (x1 ∨ ¬x2) ∨ (x2 ∨ x3)

= x1 ∨ x2 ∨ x1 ∨ ¬x2 ∨ x2 ∨ x3.

• The 

of implicants yields an implicant.

– For example,

(x1 ∧ x2) ∧ (x1 ∧ ¬x2) ∧ (x2 ∧ x3)

= x1 ∧ x2 ∧ x1 ∧ ¬x2 ∧ x2 ∧ x3.

(27)

Any Expression φ Can Be Converted into CNFs and DNFs φ = xj:

• This is trivially true.

φ = ¬φ1 and a CNF is sought:

• Turn φ1 into a DNF.

• Apply de Morgan’s laws to make a CNF for φ.

φ = ¬φ1 and a DNF is sought:

• Turn φ1 into a CNF.

• Apply de Morgan’s laws to make a DNF for φ.

(28)

Any Expression φ Can Be Converted into CNFs and DNFs (continued)

φ = φ1 ∨ φ2 and a DNF is sought:

• Make φ1 and φ2 DNFs.

φ = φ1 ∨ φ2 and a CNF is sought:

• Turn φ1 and φ2 into CNFs,a φ1 =

n1



i=1

Ai, φ2 =

n2



j=1

Bj.

• Set

φ =

n1



i=1 n2



j=1

(Ai ∨ Bj).

aCorrected by Mr. Chun-Jie Yang (R99922150) on November 9, 2010.

(29)

Any Expression φ Can Be Converted into CNFs and DNFs (concluded)

φ = φ1 ∧ φ2 and a CNF is sought:

• Make φ1 and φ2 CNFs.

φ = φ1 ∧ φ2 and a DNF is sought:

• Turn φ1 and φ2 into DNFs, φ1 =

n1



i=1

Ai, φ2 =

n2



j=1

Bj.

• Set

φ =

n1



i=1 n2



j=1

(Ai ∧ Bj).

(30)

An Example: Turn ¬((a ∧ y) ∨ (z ∨ w)) into a DNF

¬((a ∧ y) ∨ (z ∨ w))

¬(CNF∨CNF)

= ¬(((a) ∧ (y)) ∨ ((z ∨ w)))

¬(CNF)

= ¬((a ∨ z ∨ w) ∧ (y ∨ z ∨ w))

de Morgan

= ¬(a ∨ z ∨ w) ∨ ¬(y ∨ z ∨ w)

de Morgan

= (¬a ∧ ¬z ∧ ¬w) ∨ (¬y ∧ ¬z ∧ ¬w).

(31)

### Functional Completeness

• A set of logical connectives is called functionally

complete if every boolean expression is equivalent to one involving only these connectives.

• The set { ¬, ∨, ∧ } is functionally complete.

– Every boolean expression can be turned into a CNF, which involves only ¬, ∨, and ∧.

• The sets { ¬, ∨ } and { ¬, ∧ } are functionally complete.a – By the above result and de Morgan’s laws.

• { nand } and { nor } are functionally complete.b

aPost (1921).

bPeirce (c. 1880); Sheﬀer (1913).

(32)

### Satisfiability

• A boolean expression φ is satisfiable if there is a truth assignment T appropriate to it such that T |= φ.

• φ is valid or a tautology,a written |= φ, if T |= φ for all T appropriate to φ.

aWittgenstein (1922). Wittgenstein is one of the most important philosophers of all time. Russell (1919), “The importance of ‘tautology’

for a deﬁnition of mathematics was pointed out to me by my former pupil Ludwig Wittgenstein, who was working on the problem. I do not know whether he has solved it, or even whether he is alive or dead.”

“God has arrived,” the great economist Keynes (1883–1946) said of him on January 18, 1928, “I met him on the 5:15 train.”

(33)

### Satisfiability (concluded)

• φ is unsatisfiable or a contradiction if φ is false under all appropriate truth assignments.

– Or, equivalently, if ¬φ is valid (prove it).

• φ is a contingency if φ is neither a tautology nor a contradiction.

(34)

### Ludwig Wittgenstein (1889–1951)

Wittgenstein (1922),

“Whereof one cannot speak, thereof one must be silent.”

(35)

### satisfiability (sat)

• The length of a boolean expression is the length of the string encoding it.

• satisfiability (sat): Given a CNF φ, is it satisﬁable?

• Solvable in exponential time on a TM by the truth table method.

• Solvable in polynomial time on an NTM, hence in NP (p. 121).

• A most important problem in settling the “P = NP”? problem (p. 323).

(36)

### unsatisfiability (unsat or sat complement) and validity

• unsat (sat complement): Given a boolean expression φ, is it unsatisﬁable?

• validity: Given a boolean expression φ, is it valid?

φ is valid if and only if ¬φ is unsatisﬁable.

φ and ¬φ are basically of the same length.

– So unsat and validity have the same complexity.

• Both are solvable in exponential time on a TM by the truth table method.

(37)

### Relations among sat, unsat, and validity

Contingent

Valid Unsatisfiable

• The negation of an unsatisﬁable expression is a valid expression.

• None of the three problems—satisﬁability,

unsatisﬁability, validity—are known to be in P.

(38)

### Boolean Functions

• An n-ary boolean function is a function

f : { true, false }n → { true, false }.

• It can be represented by a truth table.

• There are 22n such boolean functions.

– We can assign true or false to f for each of the 2n truth assignments.

(39)

### Boolean Functions (continued)

Assignment Truth value 1 true or false 2 true or false

... ...

2n true or false

• A boolean expression expresses a boolean function.

– Think of its truth values under all possible truth assignments.

(40)

### Boolean Functions (continued)

• A boolean function expresses a boolean expression.



T |= φ, literal yi is true in “row” T (y1 ∧ · · · ∧ yn).a

∗ The implicant y1 ∧ · · · ∧ yn is called the minterm over { x1, . . . , xn } for T .

– The sizeb is ≤ n2n ≤ 22n.

– This DNF is optimal for the parity function, for example.c

aSimilar to programmable logic array. This is called the table lookup representation (Beigel, 1993).

bWe count only the literals here.

cDu & Ko (2000).

(41)

### Boolean Functions (continued)

x1 x2 f(x1, x2)

0 0 1

0 1 1

1 0 0

1 1 1

The corresponding boolean expression:

(¬x1 ∧ ¬x2) ∨ (¬x1 ∧ x2) ∨ (x1 ∧ x2).

(42)

### Boolean Functions (concluded)

Corollary 15 Every n-ary boolean function can be expressed by a boolean expression of size O(n2n).

• In general, the exponential length in n cannot be avoided (p. 212).

• The size of the truth table is also O(n2n).a

aThere are 2n n-bit strings.

(43)

### Boolean Circuits

• A boolean circuit is a graph C whose nodes are the gates.

• There are no cycles in C.

• All nodes have indegree (number of incoming edges) equal to 0, 1, or 2.

• Each gate has a sort from

{ true, false, ∨, ∧, ¬, x1, x2, . . . }.

– There are n + 5 sorts.

(44)

### Boolean Circuits (concluded)

• Gates with a sort from { true, false, x1, x2, . . . } are the inputs of C and have an indegree of zero.

• The output gate(s) has no outgoing edges.

• A boolean circuit computes a boolean function.

• A boolean function can be realized by infinitely many equivalent boolean circuits.

(45)

### Boolean Circuits and Expressions

• They are equivalent representations.

• One can construct one from the other:

¬ [L ¬

[L

[L ∨ [M

[L [M

[L ∧ [M

[L [M

(46)

### An Example

((x1 x2 ) (x3 x4)) (x3 x4))

x1 x2 x3 x4

• Circuits are more economical because of the possibility of “sharing.”

(47)

### circuit sat and circuit value

circuit sat: Given a circuit, is there a truth assignment such that the circuit outputs true?

• circuit sat ∈ NP: Guess a truth assignment and then evaluate the circuit.a

circuit value: The same as circuit sat except that the circuit has no variable gates.

• circuit value ∈ P: Evaluate the circuit from the input gates gradually towards the output gate.

aEssentially the same algorithm as the one on p. 121.

(48)

a

### Boolean Functions Need Exponential Circuits

b

Theorem 16 For any n ≥ 2, there is an n-ary boolean function f such that no boolean circuits with 2n/(2n) or fewer gates can compute it.

• There are 22n diﬀerent n-ary boolean functions (p. 202).

• So it suﬃces to prove that the number of boolean circuits with 2n/(2n) or fewer gates is less than 22n.

aCan be strengthened to “Almost all.”

bRiordan & Shannon (1942); Shannon (1949).

(49)

### The Proof (concluded)

• There are at most ((n + 5) × m2)m boolean circuits with m or fewer gates (see next page).

• But ((n + 5) × m2)m < 22n when m = 2n/(2n):

m log2((n + 5) × m2)

= 2n



1 log2 n+54n2 2n



< 2n for n ≥ 2.

(50)

(51)

### Claude Elwood Shannon (1916–2001)

Howard Gardner (1987), “[Shan- non’s master’s thesis is] possibly the most important, and also the most famous, master’s thesis of the cen- tury.”

(52)

• The lower bound 2n/(2n) is rather tight because an upper bound is n2n (p. 204).

• The proof counted the number of circuits.

– Some circuits may not be valid at all.

– Diﬀerent circuits may also compute the same function.

• Both are ﬁne because we only need an upper bound on the number of circuits.

• We do not need to consider the outgoing edges because they have been counted as incoming edges.a

aIf you prove the theorem by considering outgoing edges, the bound will not be good. (Try it!)

(53)

## Relations between Complexity Classes

(54)

It is, I own, not uncommon to be wrong in theory and right in practice.

— Edmund Burke (1729–1797), A Philosophical Enquiry into the Origin of Our Ideas of the Sublime and Beautiful (1757) The problem with QE is it works in practice, but it doesn’t work in theory.

— Ben Bernanke (2014)

(55)

### Proper (Complexity) Functions

• We say that f : N → N is a proper (complexity) function if the following hold:

f is nondecreasing.

– There is a k-string TM Mf such that Mf(x) = f(| x |) for any x.a

Mf halts after O(| x | + f(| x |)) steps.

Mf uses O(f (| x |)) space besides its input x.

• Mf’s behavior depends only on | x | not x’s contents.

• Mf’s running time is bounded by f (n).

aThe textbook calls “” the quasi-blank symbol. The use of Mf(x) will become clear in Proposition 17 (p. 222).

(56)

### Examples of Proper Functions

• Most “reasonable” functions are proper: c, log n, polynomials of n, 2n, √n , n!, etc.

• If f and g are proper, then so are f + g, fg, and 2g.a

• Nonproper functions when serving as the time bounds for complexity classes spoil “theory building.”

– For example, TIME(f (n)) = TIME(2f(n)) for some recursive function f (the gap theorem).b

• Only proper functions f will be used in TIME(f(n)), SPACE(f (n)), NTIME(f (n)), and NSPACE(f (n)).

aFor f (g(n)), we need to add f (n) ≥ n.

bTrakhtenbrot (1964); Borodin (1972). Theorem 7.3 on p. 145 of the textbook proves it.

(57)

### Precise Turing Machines

• A TM M is precise if there are functions f and g such that for every n ∈ N, for every x of length n, and for every computation path of M ,

M halts after precisely f(n) steps,a and

– All of its strings are of length preciselyb g(n) at halting.c

∗ Recall that if M is a TM with input and output, we exclude the ﬁrst and last strings.

• M can be deterministic or nondeterministic.

aFully time constructible (Hopcroft & Ullman, 1979).

bThis strong requirement does not seem needed later.

cFully space constructible (Hopcroft & Ullman, 1979).

(58)

### Precise TMs Are General

Proposition 17 Suppose a TMa M decides L within time (space) f (n), where f is proper. Then there is a precise TM M which decides L in time O(n + f (n)) (space O(f (n)), respectively).

• M on input x ﬁrst simulates the TM Mf associated with the proper function f on x.

• Mf’s output, of length f (| x |), will serve as a

“yardstick” or an “alarm clock.”

aIt can be deterministic or nondeterministic.

(59)

### The Proof (continued)

• Then M simulates M (x).

• M(x) halts when and only when the alarm clock runs out—even if M halts earlier.

• If f is a time bound:

– The simulation of each step of M on x is matched by advancing the cursor on the “clock” string.

– Because M stops at the moment the “clock” string is exhausted—even if M (x) stops earlier, it is precise.

– The time bound is therefore O(| x | + f(| x |)).

(60)

### The Proof (concluded)

• If f is a space bound (sketch):

M simulates M on the quasi-blanks of Mf’s output string.a

– The total space, not counting the input string, is O(f(n)).

– But we still need a way to make sure there is no inﬁnite loop even if M does not halt.b

aThis is to make sure the space bound is precise.

bSee the proof of Theorem 24 (p. 241).

(61)

### Important Complexity Classes

• We write expressions like nk to denote the union of all complexity classes, one for each value of k.

• For example,

NTIME(nk) =Δ 

j>0

NTIME(nj).

(62)

### Important Complexity Classes (concluded)

P =Δ TIME(nk), NP =Δ NTIME(nk), PSPACE =Δ SPACE(nk), NPSPACE =Δ NSPACE(nk),

E =Δ TIME(2kn), EXP =Δ TIME(2nk), NEXP =Δ NTIME(2nk),

L =Δ SPACE(log n), NL =Δ NSPACE(log n).

(63)

### Complements of Nondeterministic Classes

• Recall that the complement of L, or ¯L, is the language Σ − L.

– sat complement is the set of unsatisﬁable boolean expressions.

• R, RE, and coRE are distinct (p. 161).

– Again, coRE contains the complements of languages in RE, not languages that are not in RE.

• How about coC when C is a complexity class?

(64)

### The Co-Classes

• For any complexity class C, coC denotes the class { L : ¯L ∈ C }.

• Clearly, if C is a deterministic time or space complexity class, then C = coC.

– They are said to be closed under complement.

– A deterministic TM deciding L can be converted to one that decides ¯L within the same time or space bound by reversing the “yes” and “no” states.a

• Whether nondeterministic classes for time are closed under complement is not known (see p. 113).

aSee p. 158.

(65)

• As

coC = { L : ¯L ∈ C }, L ∈ C if and only if ¯L ∈ coC.

• But it is not true that L ∈ C if and only if L ∈ coC.

– coC is not deﬁned as ¯C.

• For example, suppose C = {{ 2, 4, 6, 8, 10, . . . }, . . . }.

• Then coC = {{ 1, 3, 5, 7, 9, . . . }, . . . }.

• But ¯C = 2{ 1,2,3,... } − {{ 2, 4, 6, 8, 10, . . . }, . . . }.

(66)

### The Quantified Halting Problem

• Let f(n) ≥ n be proper.

• Deﬁne

Hf =Δ { M; x : M accepts input x after at most f (| x |) steps }, where M is deterministic.

• Assume the input is binary as usual.

(67)

f

3

### )

• For each input M; x, we simulate M on x with an alarm clock of length f (| x |).

– Use the single-string simulator (p. 85), the universal TM (p. 137), and the linear speedup theorem (p. 95).

– Our simulator accepts M ; x if and only if M accepts x before the alarm clock runs out.

• From p. 92, the total running time is O(MkM2 f(n)2), where M is the length to encode each symbol or state of M and kM is M ’s number of strings.

• As MkM2 = O(n), the running time is O(f (n)3), where the constant is independent of M .

(68)

f

### ∈ TIME(f(n/2))

• Suppose TM MHf decides Hf in time f (n/2).

• Consider machine:

Df(M ) {

if MHf(M ; M ) = “yes”

then “no”;

else “yes”;

}

(69)

### The Proof (continued)

• MHf(M ; M ) runs in time f (2n+12 ) = f(n), where n = | M |.a

• By construction, Df(M ) runs in the same amount of time as MHf(M ; M ), i.e., f (n), where n = | M |.

aMr. Hsiao-Fei Liu (F92922019) and Mr. Hong-Lung Wang (F92922085) pointed out on October 6, 2004, that this estimation (and the text’s Lemma 7.2) forgets to include the time to write down M ; M .

(70)

### The Proof (concluded)

• First, suppose Df(Df) = “yes”.

• This implies

Df; Df ∈ Hf.

• Thus Df does not accept Df within time f (| Df |).

• But Df(Df) stops in time f (| Df |) with an answer.

• Hence Df(Df) = “no”, a contradiction

• Similarly, Df(Df) = “no” ⇒ Df(Df) = “yes.”

(71)

### The Time Hierarchy Theorem

Theorem 18 If f (n) ≥ n is proper, then

TIME(f (n))  TIME(f(2n + 1)3).

• The quantiﬁed halting problem makes it so.

Corollary 19 P  E.

• P ⊆ TIME(2n) because poly(n) ≤ 2n for n large enough.

• But by Theorem 18,

TIME (2n)  TIME

(22n+1)3

⊆ E.

• So P  E.

(72)

### The Space Hierarchy Theorem

Theorem 20 (Hennie & Stearns, 1966) If f (n) is proper, then

SPACE(f (n))  SPACE(f(n) log f(n)).

Corollary 21 L  PSPACE.

(73)

### Nondeterministic Time Hierarchy Theorems

Theorem 22 (Cook, 1973) NTIME(nr)  NTIME(ns) whenever 1 ≤ r < s.

Theorem 23 (Seiferas, Fischer, & Meyer, 1978) If T1(n) and T2(n) are proper, then

NTIME(T1(n))  NTIME(T2(n)) whenever T1(n + 1) = o(T2(n)).

[r]

The PROM is a combinational programmable logic device (PLD) – an integrated circuit with programmable gates divided into an AND array and an OR array to provide an

The design of a sequential circuit with flip-flops other than the D type flip-flop is complicated by the fact that the input equations for the circuit must be derived indirectly

Graph Algorithms Euler Circuit Hamilton Circuit.. for Sprout 2014 by Chin

Graph Algorithms Euler Circuit Hamilton Circuit.. for Sprout 2014 by Chin

 There is no Hamilton circuit in G2 (this can be seen by nothing that any circuit containing every vertex must contain the edge {a,b} twice), but G2 does have a Hamilton

 If SAT can be solved in deterministic polynomial time, then so can any NP problems  SAT ∈ NP-hard..  If A is an NP-hard problem and B can be reduced from A, then B is an

• If we know how to generate a solution, we can solve the corresponding decision problem. – If you can find a satisfying truth assignment efficiently, then sat is