David Evans cs.virginia/evans

David Evanshttp://www.cs.virginia.edu/evans

CS588: Security and PrivacyUniversity of VirginiaComputer Science

Lecture 15:Complexity Notes

This is selected from CS200 lectures 16,30, 37, 38 fromhttp://www.cs.virginia.edu/cs200/lectures/

http://www.cs.virginia.edu/cs200/lectures/

CS588 Spring 2005 2

What does really mean?•O(x) – it is no more than x work

(upper bound)(x) – work scales as x (tight

bound)(x) – it is at least x work

(lower bound)If O(x) and (x) are true,

then (x) is true.

CS588 Spring 2005 3

Meaning of O (“big Oh”)

f(x) is O (g (x)) means:There is a positive constant

c such that c * f(x) < g(x)

for all but a finite number of x values.

CS588 Spring 2005 4

O Examplesf(x) is O (g (x)) means:

There is a positive constant c such that c * f(x) < g(x)

for all but a finite number of x values.

x is O (x2)? Yes, c = 1 works fine.

10x is O (x)? Yes, c = .09 works fine.

x2 is O (x)? No, no matter what c we pick,cx2 > x for big enough x

CS588 Spring 2005 5

Lower Bound: (Omega)f(x) is (g (x)) means:

There is a positive constant c such that

c * f(x) > g(x) for all but a finite number of

x values. Difference from O – this was <

CS588 Spring 2005 6

f(x) is (g (x)) means:There is a positive constant c such that

c * f(x) > g(x) for all but a finite number of x values.

f(x) is O (g (x)) means:There is a positive constant c such that

c * f(x) < g(x) for all but a finite number of x values.

Examples

• x is (x)– Yes, pick c = 2

• 10x is (x)– Yes, pick c = 1

• Is x2 (x)?– Yes!

• x is O(x)– Yes, pick c = .5

• 10x is O(x)– Yes, pick c = .09

• x2 is not O(x)

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Tight Bound: (Theta)

f(x) is (g (x)) iff:

f(x) is O (g (x)) and f(x) is (g (x))

CS588 Spring 2005 8

Examples• 10x is (x)

– Yes, since 10x is (x) and 10x is O(x)•Doesn’t matter that you choose different

c values for each part; they are independent

• x2 is/is not (x)?– No, since x2 is not O (x)

• x is/is not (x2)?– No, since x2 is not (x)

CS588 Spring 2005 9

Measuring Work

• When we say a procedure or a problem is O(f(n)) what are we counting?

• Need a model computer to count steps

CS588 Spring 2005 10

How should we model a Computer?

Apollo Guidance Computer (1969)

Colossus (1944)

IBM 5100 (1975)

Cray-1 (1976)


Modeling Computers• Input

– Without it, we can’t describe a problem

• Output– Without it, we can’t get an answer

• Processing– Need some way of getting from the input

to the output

• Memory– Need to keep track of what we are doing


Modeling Input

Engelbart’s mouse and keypad

Punch Cards

Altair BASIC Paper Tape, 1976


Simplest Input

• Non-interactive: like punch cards and paper tape

• One-dimensional: just a single tape of values, pointer to one square on tape0 0 1 1 0 0 1 0 0 0

How long should the tape be?

Infinitely long! We are modeling a computer, not building one. Our model should not have silly practical limitations (like a real computer does).


Modeling Output

• Blinking lights are cool, but hard to model

• Output is what is written on the tape at the end of a computation

Connection Machine CM-5, 1993


Modeling Processing• Evaluation Rules

– Given an input on our tape, how do we evaluate to produce the output

• What do we need:– Read what is on the tape at the current

square– Move the tape one square in either

direction– Write into the current square

0 0 1 1 0 0 1 0 0 0

Is that enough to model a computer?


Modeling Processing

• Read, write and move is not enough• We also need to keep track of what

we are doing:– How do we know whether to read, write

or move at each step?– How do we know when we’re done?

• What do we need for this?


Finite State Machines

1Start 2

HALT

10

1#

0


Hmmm…maybe we don’t need those infinite tapes

after all?

1Start 2

HALT

(not a paren

)#

not a paren

)

ERROR

What if thenext input symbolis ( in state 2?


How many states do we need?

1Start 2

HALT

(not a paren

)#

not a paren

)

ERROR

3

not a paren

(

)

4(

)

not a paren

(


Finite State Machine

• There are lots of things we can’t compute with only a finite number of states

• Solutions:– Infinite State Machine

• Hard to describe and draw

– Add a tape to the Finite State Machine


FSM + Infinite Tape• Start:

– FSM in Start State– Input on Infinite Tape– Pointer to start of input

• Move:– Read one input symbol from tape– Follow transition rule from current state

• To next state• Write symbol on tape, and move L or R one square

• Finish:– Transition to halt state


Matching Parentheses• Repeat until halt:

– Find the leftmost )•If you don’t find one, the parentheses match,

write a 1 at the tape head and halt.

– Replace it with an X– Look left for the first (

•If you find it, replace it with an X (they matched)•If you don’t find it, the parentheses didn’t match

– end write a 0 at the tape head and halt


Matching Parentheses

1Start

HALT

Input: )Write: XMove: L

), X, L

2: look for (

X, X, RX, X, L

(, X, R

#, 0, ##, 1, #

(, (, R

Will this report thecorrect result for (()?


Matching Parentheses

1Start

HALT

), X, L

2: look for (

(, (, R

(, X, R

#, 0, ##, 1, #

X, X, L

X, X, R

#, #, L

3: look for (X, X, L

#, 1, #(, 0, #


Turing Machine• Alan Turing, On computable numbers:

With an application to the Entscheidungsproblem, 1936– Turing developed the machine

abstraction to show the halting problem really leads to a contradiction

– Our informal argument, depended on assuming we could do if and everything else except halts?


Describing Turing Machinesz z z z z z z z z z z z z z z zz z z z

TuringMachine ::= < Alphabet, Tape, FSM >Alphabet ::= { Symbol* }Tape ::= < LeftSide, Current, RightSide >OneSquare ::= Symbol | #Current ::= OneSquareLeftSide ::= [ Square* ]RightSide ::= [ Square* ]

Everything to left of LeftSide is #.Everything to right of RightSide is #.

1

Start

HALT

), X, L

2: look for (

#, 1, -

), #, R

(, #, L

(, X, R

#, 0, -

Finite State Machine


Describing Finite State Machines

TuringMachine ::= < Alphabet, Tape, FSM >FSM ::= < States, TransitionRules, InitialState, HaltingStates >States ::= { StateName* }InitialState ::= StateName must be element of StatesHaltingStates ::= { StateName* } all must be elements of StatesTransitionRules ::= { TransitionRule* }TransitionRule ::= < StateName, ;; Current State OneSquare, ;; Current square StateName, ;; Next State OneSquare, ;; Write on tape Direction > ;; Move tapeDirection ::= L, R, #

1Start

HALT

), X, L

2: look for (

), #, R (, #, L

(, X, R

#, 0, ##, 1, #

Transition Rule is a procedure:StateName X OneSquare StateName X OneSquare X Direction


Example Turing

Machine

TuringMachine ::= < Alphabet, Tape, FSM >FSM ::= < States, TransitionRules, InitialState, HaltingStates >Alphabet ::= { (, ), X }States ::= { 1, 2, HALT }InitialState ::= 1HaltingStates ::= { HALT }TransitionRules ::= { < 1, ), 2, X, L >,

< 1, #, HALT, 1, # >, < 1, ), #, R >, < 2, (, 1, X, R >,

< 2, #, HALT, 0, # >, < 2, ), #, L >,}

1Start

HALT

), X, L

2: look for (

), #, R (, #, L

(, X, R

#, 0, ##, 1, #


Enumerating Turing Machines

• Now that we’ve decided how to describe Turing Machines, we can number them

• TM-5023582376 = balancing parens• TM-57239683 = even number of 1s• TM-3523796834721038296738259873 = Photomosaic

Program• TM-3672349872381692309875823987609823712347823 = WindowsXP

Not the real numbers – they would be much

bigger!


Complexity Class P

Class P: problems that can be solved in polynomial time by a deterministic TM.

O (nk) for some constant k.

Easy problems like sorting, making a photomosaic using duplicate tiles, simulating the universe are all in P.


Deterministic Computing

• All our computing models so far are deterministic: you know exactly what to do every step– TM: only one transition rule can match

any machine configuration, only one way to follow that transition rule

– Lambda Calculus: always -reduce the outermost redex


Nondeterministic Computing• Allows computer to try different options at

each step, and then use whichever one works best

• Make a Finite State Machine non-deterministic: doesn’t change computing time

• Make a Turing Machine non-deterministic: might change computing time

• No one knows whether it does or not•This is the biggest open problem in Computer

Science


Making FSM’s Nondeterministic

1 2 30 1

0 0

1

HALT

#


Nondeterministic FSM

1 2 3

HALT

#

0, 1

0 1

Input Read Possible States

0 { 1, 2 }

00 { 1, 2 }

001 { 1, 3 }

0010 { 1, 2 }

00101 { 1, 3 }

00101# { HALT }


Power of NFSM

• Can NFSMs computing anything FSMs cannot compute?

• Can NFSMs compute faster than FSMs?

No – you can always convert an NFSM to a FSM bymaking each possible set of states in the NFSM into onestate in the new FSM. (Number of states may increase as 2s)

No – both read input one letter at a time, and transitionautomatically to the next state. (Both are always (n) wheren is the number of symbols in the input.)


Nondeterministic TMs

• Multiple transitions possible at every step

• Follow all possible steps:– Each possible execution maintains its

own state and its own infinite tape– If any possible execution halts (in an

accepting state), the NTM halts, and the tape output of that execution is the NTM output


Complexity ClassesClass P: problems that can be solved in polynomial time by a deterministic TM.

O (nk) for some constant k.Easy problems like simulating the universe are all in P.

Class NP: problems that can be solved in polynomial time by a nondeterministic TM

Hard problems like the pegboard puzzle sorting are in NP (as well as all problems in P).


Problem Classes

PNP

Decidable

Undecidable

Sorting: (n log n)

Simulating Universe: O(n3)

Cracker Barrel: O(2n) and (n)

Fill tape with 2n *s: (2n)

Halting Problem: ()


P = NP?• Is there a polynomial-time solution to

the “hardest” problems in NP?• No one knows the answer!• The most famous unsolved problem in

computer science and math• Listed first on Millennium Prize

Problems– win $1M if you can solve it – (also an automatic A+ in this course)


If P NP:

PNP

Decidable

Undecidable

Sorting: (n log n)




Halting Problem: ()


If P = NP:

PNP

Decidable

Undecidable

Sorting: (n log n)




Halting Problem: ()


Smileys Problem

Input: n square tilesOutput: Arrangement of the tiles in a square, where the colors and shapes match up, or “no, its impossible”.

Thanks to Peggy Reed for making the Smiley Puzzles!


How much work is the Smiley’s Problem?

• Upper bound: (O)O (n!)Try all possible permutations

• Lower bound: () (n)Must at least look at every tile

• Tight bound: ()

No one knows!


NP Problems

• Can be solved by just trying all possible answers until we find one that is right

• Easy to quickly check if an answer is right – Checking an answer is in P

• The smileys problem is in NPWe can easily try n! different answers

We can quickly check if a guess is correct (check all n tiles)


Is the Smiley’s Problem in P?

No one knows!

We can’t find a O(nk) solution.We can’t prove one doesn’t exist.


This makes a huge difference!

1

100

10000

1E+06

1E+08

1E+10

1E+12

1E+14

1E+16

1E+18

1E+20

1E+22

1E+24

1E+26

1E+28

1E+30

2 4 8 16 32 64 128

n! 2n

n2

n log n

today

2032

time since “Big Bang”

log-log scale

Solving a large smileys problem either takes a few seconds, or more time than the universe has been in existence. But, no one knows which for sure!


Who cares about Smiley puzzles?

If we had a fast (polynomial time) procedure to solve the smiley puzzle, we would also have a fast procedure to solve the 3/stone/apple/tower puzzle:

3


3SAT Smiley

Step 1: Transform into smileys

Step 2: Solve (using our fast smiley puzzle solving procedure)

Step 3: Invert transform (back into 3SAT problem


The Real 3SAT Problem(also can be quickly

transformed into the Smileys Puzzle)


Propositional GrammarSentence ::= Clause

Sentence Rule: Evaluates to value of Clause

Clause ::= Clause1 Clause2

Or Rule: Evaluates to true if either clause is true

Clause ::= Clause1 Clause2

And Rule: Evaluates to true iff both clauses are true


Propositional GrammarClause ::= Clause

Not Rule: Evaluates to the opposite value of clause (true false)

Clause ::= ( Clause )Group Rule: Evaluates to value of clause.

Clause ::= NameName Rule: Evaluates to value associated with Name.


PropositionExample

Sentence ::= ClauseClause ::= Clause1 Clause2 (or)

Clause ::= Clause1 Clause2 (and)

Clause ::= Clause (not)

Clause ::= ( Clause )Clause ::= Name

a (b c) b c


The Satisfiability Problem (SAT)

• Input: a sentence in propositional grammar

• Output: Either a mapping from names to values that satisfies the input sentence or no way (meaning there is no possible assignment that satisfies the input sentence)


SAT Example

SAT (a (b c) b c) { a: true, b: false, c: true } { a: true, b: true, c: false }

SAT (a a) no way






The 3SAT Problem• Input: a sentence in propositional

grammar, where each clause is a disjunction of 3 names which may be negated.

• Output: Either a mapping from names to values that satisfies the input sentence or no way (meaning there is no possible assignment that satisfies the input sentence)


3SAT / SAT

Is 3SAT easier or harder than SAT?

It is definitely not harder than SAT, since all 3SAT problemsare also SAT problems. Some SAT problems are not 3SAT problems.


3SAT Example

3SAT ( (a b c) (a b d)

(a b d) (b c d ) ) { a: true, b: false, c: false, d: false}






3SAT Smiley• Like 3/stone/apple/tower puzzle, we

can convert every 3SAT problem into a Smiley Puzzle problem!

• Transformation is more complicated, but still polynomial time.

• So, if we have a fast (P) solution to Smiley Puzzle, we have a fast solution to 3SAT also!


NP Complete• Cook and Levin proved that 3SAT was NP-

Complete (1971)• A problem is NP-complete if it is as hard

as the hardest problem in NP• If 3SAT can be transformed into a different

problem in polynomial time, than that problem must also be NP-complete.

• Either all NP-complete problems are tractable (in P) or none of them are!


NP-Complete Problems• Easy way to solve by trying all possible

guesses• If given the “yes” answer, quick (in P) way to

check if it is right– Solution to puzzle (see if it looks right)– Assignments of values to names (evaluate logical

proposition in linear time)

• If given the “no” answer, no quick way to check if it is right– No solution (can’t tell there isn’t one)– No way (can’t tell there isn’t one)


Traveling Salesperson Problem

– Input: a graph of cities and roads with distance connecting them and a minimum total distant

– Output: either a path that visits each with a cost less than the minimum, or “no”.

• If given a path, easy to check if it visits every city with less than minimum distance traveled


Graph Coloring Problem– Input: a graph of nodes with edges

connecting them and a minimum number of colors

– Output: either a coloring of the nodes such that no connected nodes have the same color, or “no”.

If given a coloring, easy to check if it no connected nodes have the same color, and the number of colors used.


Pegboard Problem- Input: a configuration of n pegs on a

cracker barrel style pegboard- Output: if there is a sequence of jumps that leaves a single peg, output that sequence of jumps. Otherwise, output false.If given the sequence of jumps, easy (O(n)) to check it is correct. If not, hard to know if there is a solution. Proof that variant of this

problem is NP-Complete isattached to today’s notes.


Minesweeper Consistency Problem

– Input: a position of n squares in the game Minesweeper– Output: either a assignment of bombs to squares, or “no”.

• If given a bomb assignment, easy to check if it is consistent.


Drug Discovery Problem– Input: a set of proteins,

a desired 3D shape– Output: a sequence of

proteins that produces the shape (or impossible)

Note: US Drug sales = $200B/year

If given a sequence, easy (not really) to check if sequence has the right shape.

Caffeine


Complexity and Cryptography

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