Time Complexity

p2.

Measuring Complexity Worst-case analysis Average-case analysis

Define M: a det. Turing machine that halts on all input The running time or time complexity of M is the

function f: NN, where f(n) is the maximum number of steps that M uses on any input of length n

If f(n) is the running time of M, we say M runs in time f(n) and that M is an f(n) time Turing machine.

p3.

Big-O and small-o Notation Def:

Def:

Def:

0 0

f(n)=O(g(n)), if there exists positive integersc and n so that for every n n , f(n) cg(n)

. ( )( ) ( ), ( ) ( )

eg f n n n n

f n O n f n O n

3 2

4 35 2 22 6

( ), : . ( ) ( ( )) lim

( ) Say that if

n

f nf g N R f n o g n

g n

0

. ( ), ( ), log log ( log ) eg n o n n o n n n o n n 2 3

: a functionTIME(t(n)): time complexity class

L is a language decided by a O(t(n)) time= L Turing machine

t N N

p4.

Complexity relationships among models

( ))2

Thm: t(n): a function n Every t(n) time multitape Turing machine has an equivalent O(t time single-tape Turing machine.

n

O(t(n))

Thm: t(n): a function n Every t(n) time non deterministic single-tape TM has an equivalent 2 time deterministic single-tape TM.

p5.

The class PP is the class of languages that are decidable in polynomial time on a deterministic single-tape TM

In other words, ( )k

k

P TIME n

p6.

, , is a directed graph tha

has a directed path from sto t

GPATH G s t

s t

:Thm PATH P

, | , are relatively primeRELPRIME x y x y :Thm RELPRIME P

p7.

The Class NP Hamiltonian path in a directed graph G is a directed

path that goes through each node once

Note: It is not hard to find an exponential time algorithm for the

HAMPATH problem The HAMPATH problem can be verified in polynomial time

G is a directed graph with aHAMPATH= , ,

Hamiltonian path from s to tG s t

s t

p8.

Not all problem can be verified in poly. time.

Def: A verifier for a language A is an algorithm V, whereA = { | V accepts < ,c> for some string c} A polynomial time verifier runs in polynomial time in the

length of A language A is polynomially verifiable if it has a polynomial

time verifier. A verifier uses additional information c to verify that a string

is member of A C: certificate or proof of membership in A

, for integers, p,q >1COMPOSITES x x pq

12653 = 123 x 111

verifiable in poly. time

p9.

Def: NP is the class of languages that have polynomial time verifiersNP: Non-deterministic Polynomial time

Def:

Thm: A language is in NP iff it is decided by some non-deterministic polynomial time Turing machine

Pf:

is a language decided by ( ( ))

a O(t(n)) time non-det TM

LNTIME t n L

k

Let V be the polynomial time verifier for A that exists by t

" ": Let A NP and show that A is decided by a polynomial

time

he

definition of NP.

Assume V runs in time n with input of

NTM N

si

.

ze n

p10.

k

Construct N:

N = "On input of length n

1. Nondeterministically guess the string c of length n

2. Run V on input < , c>

3. If V accepts, ACCEPT; otherwise, REJECT."

" ": Assume that A is decided by a poly. time NTM

N and construct a poly time verifier V as follows.

"On input < ,c>, where and c are strings:

treating each symbol of c 1. Simulate N o an , s a

V

description of the non-det choice to make at each

step.

2. If this branch of N's accepts, ACCEPT; otherwise

REJECT"

p11.

Cor:

Examples of problem in NP Def: clique is a graph, where every two nodes are

connected by an edge A k-clique is a clique that contains k nodes

( )k

k

NP NTIME n

A graph of 5-clique

p12.

Def:

Thm: CLIQUE is in NPpf :

is an undirected graph,

with a k-clique

GQUE GC kLI

" On input <G,k>, where G is a graph:

1. Nondeterministically guess a subset c of k nodes of G

2. Test whether G contains all edges connecting node in

C

3. If yes, ACCEP

N

T; otherwise, REJECT."

p13.

Def:

Thm: SUBSET-SUM is in NPPf:

1

1 i

{ ,...., } and for some,

{y ,...., } , we have yk

k

S x xSUBSET SUM S t

y S t

1 l

: <{ 4, 11, 16, 21, 27 } , 25 > SUBSET-SUM

4+

Note: S and {y ,...., y } can be

21

mu

= 25

ltisets

Eg

"On input <S,t>"

1. Non-deterministrically guess a subset c of the

numbers in S

2. Test whether c is a collection of numbers

that sum to t

3. If both pass, ACCE

N

PT; otherwise, REJECT."

p14.

P = the class of languages where membership can be decided quickly

NP= the class of languages where membership can be verified quickly

BIG Question!! P = NP?

p15.

NP-completeness

The satisfiability problem is to test whether a Boolean formula is satisfiable

_ _

( ) ( ) Boolean formula

a boolean formula is satisfiable if some assigment

of 0's and 1's to the variables makes the formula

evaluate to 1

x y x z

{ | is a satisfiable Boolean formula}SAT

p16.

Thm: Cook-Levin Theorem

Def:

SAT P iff P=NP

* *: is a polynomial time computable

function if some polynomial time Turing Machine

M exists that halts with just ( ) on its tape, when

started on any input w.

f

f

p17.

Def:

p

* *

Language A is polynomial time mapping reducible

to language B, A B, if a polynomial time

computable function : exists, where

for every ,

A ( )

f

f B

The function is called the polynomial time reduction

of A to B

f

A B

p18.

Thm:

Pf: pIf A B and B P, then A P

Let M be the polynomial time algorithm deciding

B and be the polynomial time reduction from

A to B

N="On input :

1. comput ( )

2. Run M on input and ( ) output whatever M

outputs.

f

f

f

"

p19.

Def: A language B is NP-complete if it satisfies two condition:1. B is in NP

2. Every A in NP is polynomial time reducible to B(with only 2, it is called NP-hard)

Thm:

Pf:

If B is NP-complete and B P, then P=NP

Follows from the definition of polynomial time reducibility

p20.

Thm:

Pf:

pIf B is NP-complete and B C for C in NP, then

C is NP-complete

p p p

B is NP-complete, every language in NP is polynomial

time reducible to B

Need to show that every A in NP is polynomial time

reducible to C

Hence every lan

A B, B C A

guage in NP s po

C

i

lynomial time reducible

to C

p21.

Def

__

__ __

1 2 3 4

__ __ __

1 2 3 4 5 3 6

1

( )

literal: a boolean variable or a negated boolean

variable or

clause:

conjunctive n

( )

ormal form: (cnf )

3cnf-formula:

( ) ( )

(

x x

x x x x

x x x x x x x

x x

__ __ __

2 3 3 5 6 3 6 4) ( ) ( )x x x x x x x

3SAT={< | is a satisfiable 3cnf-formula}

p22.

Thm: 3SAT is polynomial time reducible to CLIQUEPf:

1x

2x

2x

1x

2x

__

1x__

2x

__

1x

__

2x

1 1 2 1 2 2

1

1

1

2 2

1 2 2 2

Let be a formula with k clauses:

( ) ( ) ...

( ) ( )

)

( )

(

,

k k k

x

a b c a b

x

f

c a b c

x x x x x x x

G k

Need to explain

1. f is poly. time reduction

2. is satisfiable

iff <G,k> CLIQUE

p23.

Cook-Levin Theorem ： SAT is NP-complete.Proof ：

1> SAT NP ：A NTM can guess an assignment to a given formula and accept if the assignment satisfies . 2> For any A NP and show that A is polynomial time reducible to SAT.

Let N be a non-deterministic TM that decides A in nk time for some constant k.

A tableau for N on w is an nkXnk table whose rows are the configurations of a branch of the

Computation of N on input w.

###

#

q0 w1 nww2 #

##

#

start configuration

2nd configuration

window

nkth configuration

kn

kn

cell

A tableau is accepting if any row of the tableau is accepting configuration.

Every accepting tableau for N on w corresponds to a computation branch of N on w.

* Determine whether N accepts w is equivalent to determine whether an accepting tableau for N on w exists.

f ： polynomial time reduction from A to SAT.On input w, the reduction produces a formula .Let Q and be the state set and the tape alphabet of N.Let For and each we have a variable such variables. {#}.C Q knji ,1 ,Cs .,, sjix kn2~

If , it means cell[i,j] contains an s. Design so that a satisfying assignment to the variable does correspond to an accepting

tableau

for N on w.

1,, sjix

cell start move accept

(1)

, , , , , ,,1 , k

x x xcell i j s i j s i j ts C s t Ci j n

s t

si

j

(a)

#,,1,1,,3,1,2,1,4,1,3,1,2,1#,1,11210 kkn nnxnwnwwqstart xxxxxxxx

Ensure that the first row of the table is the starting configuration of N on w.

(3) Guarantees that an accepting configuration occurs in the tableau.accept

acceptk qjinji

accept x ,,,1

(4) This part is more complicated !:moveguarantee that each row of the table corresponds to a configuration that legally follows the preceding

Row’s configuration according to N’s rules.

)},,,{(),( 11 Rbqaq

)},,{(),( 22 Rbqaq

)}.,,(),,,{(),( 221 RaqLcqbq

###

b a

b

b

q2

b

q1

a

q2

b a

c a

a

a

a

a

ca q1 bq2 a

(b)a q1 b

q2a a

(c)a a q1

ba a

(d)

a# b a# b

(b)a b a

q2a b

(c)b b b

bc b

legal windows

Examples of illegal windows ：

(a)

aa b aa a

(b)a q1 bq1 aa

(c)a a

bq1

q2 q2

Claim ： If the top row of the table is the start configuration and every window in the table is legal, each row

of the table is a configuration that legally follows the preceding one.

##

b a

b q2

q1

a

b a

c a

a

a

##

)legal is winodwj)(i, the(,1 knji

move

)legal is winodwj)(i, the()(

65432161

,1,1,1,,1,1,,1,,,,1,...,

ajiajiajiajiajiajiaa

xxxxxx

is a legal window

a1 a3

a5

a2

a4 a6

i+1ii-1

jj+1

Size of ： fixed. and Non depends ~Let ),( 2 lCnO k Total number of variables .)( 2knO

).(: , )(: 2 kstart

kcell nOnO

).(:, 2kmoveaccept nO Thus the reduction is poly.

p27.

Cor ：3SAT is NP-complete.

Cor ：CLIQUE is NP-complete.

Proof ：

3SAT is in NP.

:3SATSAT p

)()()( 43214321 aazzaaaaaa

3 )( 321 laaaa l

1 2 1 1 3 2 2 4 3 3 1( ) ( ) ( ) ( )l l la a z z a a z a z z a a

How about )?( 21 aa

p28.

Vertex cover of G ：

If G is an undirected graph, a vertex cover of G is a subset of the

nodes where every edge of G touches one of those nodes.

1 4

3

{2,3} is a vertex cover.

{1,3} is not a vertex cover.

VERTEX-COVER={<G,k>|G is an undirected graph that has a k-node vertex cover}.

Thm ：

VERTEX-COVER is NP-complete.

Proof ：

VERTEX-COVER is in NP.

3 .pSAT VERTEX COVER

)()()( 221221211 xxxxxxxxx

1x 1x

1x

1x 2x

1x

2x 2x

2x 2x

1x

2x 2x

Variable gadgets

Clause gadgets

Need to show that is satisfiable iff G has a vertex cover with k nodes.

Let have m variables and l clauses. Let k=m+2l.(1) One true variable from each variable gadget.

two node from each clause gadget.

(2) Each of the 3 edges connecting the variable gadgets with each clause gadget is covered.

p30.

SUBSET-SUM ：

1 1 l i{ , | { ,..., } and for some { ,...,y } S, y }. kSUBSET SUM S t S x x y t

Thm ：

SUBSET-SUM is NP-complete.

Proof ：SUBSET-SUM is in NP.

SUMSUBSETSAT p 3Let be a boolean formula with variables x1,…,xl and clauses c1,…,ck. kccc ...21

., tS

1y1z2y2z

1x

2x

3y3z

3x

lylz

lx

1g1h2g2h

1c

2c

kgkh

kc

1 2 3 4 ….. l1 0 0 0 ….. 01 0 0 0 ….. 0

1 0 0 ….. 01 0 0 ….. 0

1 0 ….. 01 0 ….. 0

11

C1 C2 ….. Ck

1 0 ….. 0 0 0 ….. 0 0 1 ….. 0 1 0 ….. 0 1 1 ….. 0 0 0 ….. 1

0 0 ….. 0 0 0 ….. 01 0 ….. 01 0 ….. 0

1 ….. 01 ….. 0

11

t 1 1 1 1 ….. 1 3 3 ….. 3

iy 1 0 ….. 0

il

iz 1 0 ….. 0

jc

….. 1 ….….. 0 ….

clause cj contains xi

Suppose is satisfiable.(1)

Construct a subset S as follows.

If xi is assigned TRUE, select yi

else select zi. I.e. for each I, we select either yi or zi.

Last k digits add up between 1 and 3.

Select enough of g and h numbers to make each of the last k digits up to 3.

(2)Suppose a subset of S sums to t.We construct a satisfying assignment to .

If the subset contains yi , we assigned xi TRUE

else assign xi FALSE.

This assignment satisfied ! Why??

The table has size )(~).)(( 22 nOlkO

p33.

MAX-SAT:Given a boolean formula , and an integer k, is there a

truthassignment that satisfies at least k clauses?

Thm: MAX-SAT is NP-completePf:

SAT p MAX-SAT

= c1 c2 … cm

k = m Thm: DOUBLE-SAT:

Given a boolean formula , are there at least 2 truthassignment for ?

DOUBLE-SAT is NP-complete

p34.

Def: Hamiltonian path in a directed graph G is a directed path that goes through each node once.

HAMPATH:Given a directed graph and its 2 vertices s and t, is

there a Hamiltonian path from s to t?

Thm: HAMPATH is NP-completePf: (long…later)

3SAT p HAMPATH

UHAMPATH:Given a undirected graph and its 2 vertices s and t, is

therea Hamiltonian path from s to t?

p35.

Proof: (HAMPATH is NPC) The following is a 3cnf-formula with k clauses:=(a1∨b1∨c1) ∧ (a2∨b2∨c2) ∧ …∧ (ak∨bk∨ck)

Each a,b,c is a literal xi or , and x1,…,xl are l variables of .

xi : ci :

ix

p36.

Proof conti.s

c1

1x

c2

2x

t

cl

lx

p37.

Proof conti.

1c 2c

13 k

jc

jc

jc

jc

ij xc contains

ij xc contains

p38.

Proof conti.

If there is a satisfying assignment, “select exactly one of the literals” in a clause and assign it TRUE.

If there is a truth assignment, then there exists an Hamiltonian path from s to t.

) trueis when (

zagzig

ix

)false is when (

zigzag

ix

p39.

Proof conti. If the Hamiltonian path is normal, that is, it

goes through the diamonds in order from the top one to the bottom one, we can easily get the satisfying assignment. Because each clause node appears on the path, by

observing the diamond at which the detour to it is taken, we may determine which of the literals in the corresponding clause is TRUE.

If the path zigzag, we assign corresponding variable TRUEIf the path zagzig, we assign FALSE.

p40.

Proof conti. We show that a Hamiltonian path must be

normal. c

1a 3a2a

1b 2b

at. points aat th

node availableonly theis a because

a fromenter cannot path the:2 casein

nodes thesefrom

elsewhere goespath thebecause

cor a fromenter cannot path the: 1 casein

acontain cannot path thecase,both In

seperator is a :2 case

seperator a is a :1 case

2

3

3

1

2

3

2

p41.

Thm: UHAMPATH is NP-completePf:

HAMPATH p UHAMPATH

directed G undirected G’u V(G)vertices of G’: u uin, umid, uout

s sout ; t tin

edges of G’: u v E(G) uin – umid – uout – vin – vmid – vout

if s u1 u2 … uk t is a Hamiltonian path in G,

then sout – u1in – u1

mid – u1out – u2

in – u2mid – u2

out – …

– ukin – uk

mid – ukout – tin is an undirected

Hamiltonian path in G’

su

v

t

vmid

sout

uin

umiduout

vin

vout

tin

p42.

Undirected Hamiltonian Cycle Thm:

Undirected Hamiltonian Cycle is NP-comlpletePf: UHAMPATH p UHAMCYCLE

Traveling Salesman Problem: (TSP)Given an integer n2, an nn distance matrix of some

cities,and an integer B0, is there a tour that visits every cityexactly once and returns to the starting city by traveling within distance B?

Thm: TSP is NP-completePf: UHAMCYCLE p TSP

p43.

Longest Cycle:Given a graph and integer k, is there a cycle, with no repeated nodes, of length at least k?

Thm: LONGEST CYCLE is NP-complete

SUBGRAPH ISOMORPHISM:Given 2 undirected graphs G and H, is G a subgraph of

H? Thm:

SUBGRAPH ISOMORPHISM is NP-complete

p44.

Coping with NP-completeness:Approximation algorithms: Let x be an instance of an optimization problem Opt(x): the optimum solution of x A: a poly. time algorithm for x : positive real numberIf A satisfies

For all x, then we say A is an -approximation algorithm

)(

|)()(|

xopt

xAxopt

p45.

Eg. The following is a 1-approximation algorithm for the

vertex cover problem.G: C={2,3}

Algorithm: 1. C :=;2. while there is an edge [u,v] in G add u, v to C and delete them from

GLet be the optimum vertex cover

c={ , , ,…, }

1

23

4

c

1|ˆ|

|ˆ|

|ˆ|

||||ˆ||

|ˆ| |||ˆ| 2

|||ˆ|

c

c

c

cc

cccc

c

p46.

Polynomial Time Approximation Scheme(PTAS)We say that an approximation scheme is PTAS, if for

any fixed >0, the scheme runs in time polynomial in the size n

of itsinput instance.

Ratio bound: Inapproximable problem:

If there is no -approximation algorithm for them withhowever large , unless P=NP

)(),max(*

*ne

c

c

c

c

p47.

Thm: TSP is inapproximable unless P=NPPf:

Let G be a graph with n nodes.UHAMCYCLE p TSP

If G has a Hamiltonian cycle, then the optimum cost of a tour is n; otherwise if G has no Hamiltonian cycle, then the optimum cost of a tour > n(1+)

o/w 2

),( if 1

if 0

nGvv

ji

d jiij

1

1

2+3

p48.

If TSP had an -approximation algorithm A, then we would be able to tell whether G has a Hamiltonian cycle.Run A for TSP:i> if the returned cost n(1+)+1: No cycleii> if the returned cost n(1+): has a cycleUnless P=NP, TSP is inapproximable.

p49.

Backtracking and Branch&Bound:A={S0}

while A is not empty dochoose a subproblem S and delete it from A choose a way of branching out of S , say to subproblems

S1, S2,…, Sr

for each subproblem Si in this list do

if test(Si) returns “solution found”

else if test(Si) returns “?” then add Si to A

Return “no solution”

p50.

Eg.

)()()()()( zyxxzzyyxzyx

zzyzyzyzyy )()( )()(

yyzz

x = T x = F

y = T y = F z = T z = F

p51.

Branch&Bound algorithm:A:={S0} , best_so_far:=

while A is not empty dochoose a subproblem S and delete it from Achoose a way of branching out of S , say to subproblems

S1, S2,…, Sr

for each subproblem Si in this list do

if |Si|=1 then update best_so_far

else if lowerbound(Si) < best_so_far then add Si to A

Return best_so_far

p52.

Local improvement algorithm:S= initial solutionwhile there is a solution S’ such that

S’ is a neighbor of S and cost(S’) < cost(S)do S:=S’

Return S

p53.

Simulated annealing:S= initial solution; T:=T0

repeatgenerate a random solution S’ such that

S’ is a neighbor of S and let = cost(S’)cost(S)if 0 then S:=S’

else S:=S’ with probability e-/T

update (T)until T=0Return the best solution seen!

T: temperature