lecture 8: computational complexity

Lecture 8: Computational Complexity

虞台文大同大學資工所智慧型多媒體研究室

Content

Time-Bounded Turing Machine Rate of Growth Functions Time-Bound Simulations P and NP NP –Completeness Bounded Halting Problems Some Important NP-Complete Problems

Time-Bounded Turing Machine

大同大學資工所智慧型多媒體研究室

Computational Effectiveness for Solvable Problems

Not Solvable Solvable

Can such a problem, e.g., TSP,

be solved in any practical sense?

Yields in t Steps tM├

C C0M├ iff C C

C C1tM├ iff

C such that

C CtM├ and C C M├

Time-Bounded TM

Let T: N N be a function. Let L be a

language, and let be a k-tape TM

with 0 . We say that M decides L in time T if

the following holds:

, , ,M K s

1. w L

2. w L

, # #, #, , # , # #, #, , #s w h Y tM├ for some t T(|w|);

, # #, #, , # , # #, #, , #s w h N tM├ for some t T(|w|);

TIME(T)

We say that L is decidable in time T if there is some k-tape TM (k > 0) that decides L in time T.

The class of all languages decidable in time T is denoted by

( )TIME T

Absolute Time Bound (2n + 4)

Y/N ##

• |w|+1 writing steps

• |w|+3 head-moving steps

erase w (write ‘#’)write ‘Y’ or ‘N’

( ) 2 4T n n

Rate ofGrowth Functions

Time-Bound Notations

Upper Bound

Lower Bound

Exact Bound

( )f O g

, :f g N N

( )f g

0 0st , ( ) i( ) fc n f n cg n n n

1 2 0 1 2 0, , ( ) (st if) ( )c c n c g n f n c g n n n

Time-Bound Notations

Upper Bound

Lower Bound

Exact Bound

( )f O g

, :f g N N

( )f g

0 0st , ( ) i( ) fc n f n cg n n n

1 2 0 1 2 0, , ( ) (st if) ( )c c n c g n f n c g n n n

( ) ( ) an ( )df g f O g f g

( ) 0 d

f n a n a

( ) ( )

b) and ) ( ) ( )

f n O n f n n

Lemma0

( ) 0 d

f n a n a

b) and

( ) ( )

f n O n f n n

Pf)For a, we need to prove and .( ) ( )df n O n ( )dn O f n

( ) ( )df n O n

( ) ( ) | | | |d d

j dj j

f n f n a n n a

( )dn O f n

0( ) if dn cf n n n

0( ) 0 if dcf n n n n

Therefore, we have to find c (exercise) such that

11( ) ( 1) 0d d d

d dcf n n ca n ca n

Lemma0

( ) 0 d

f n a n a

b) and

( ) ( )

f n O n f n n

Pf)1( ) ( )df n O n

1 ( )d dn O n

Suppose that nd+1 = O(nd), Then, nnd = nd+1 cnd for n n0.But for any c, when n > c, nd+1 cnd.

Since f = O(nd), we only need to show that nd = O(nd+1), which is trivially satisfied because nd nd+1 for all n 0.

Theorem

( 0. ) d

f n a n a

Let with Then,

( ) ( ), 1.nf n O r r

Theorem0

( 0. ) d

f n a n a

Let with Then,

( ) ( ), 1.nf n O r r

Pf) Because f = O(nd), we only need to prove that nd = O(rn), i.e.,

0 d nn cr n n f or0

0( ) 0n kdn k cr k f or

0 0 00 0

1 2( )

d d n n n kn k n

n n n k

00 0 0

1 1 11 1 1

dnn n n k

k terms

0 0n k n kcr cr r

Theorem0

( 0. ) d

f n a n a

Let with Then,

( ) ( ), 1.nf n O r r

Pf) Because f = O(nd), we only need to prove that nd = O(rn), i.e.,

0 d nn cr n n f or0

0( ) 0n kdn k cr k f or

0 00 0

1( ) 1

d d n k n kn k nn

cr cr r

Let 00dncr n

n 0 1/

Hence, nd = O(rn).

Time-Bound Simulations

Time-Complexities on TMs

The complexities to solve problems

on different TMs are different.

Theorem 1

1 22-way standardinfinite 1-way

tape tape

L TIME T L TIME T

where 2 1( ) 6 ( ) 3 8.T n T n n

Theorem 2

1 2-tape standard

1-waytape

L TIME T L TIME T

22 1 1( ) 4 ( ) (4 4 3) ( ) 5 15.T n T n n k T n n

Conclusions

A polynomial algorithm of any TM can be simulated using some polynomial algorithms of other TM’s.

In the following, we consider two classes of problems:– Polynomial algorithms (P)– Exponential algorithms (NP)

P and NP

Definition (P )

( ) : 0dTIME n d P

That is, P denotes the class of all languages that are decided by TM’s in some polynomial time bounds.

Example: (Integer Programming)

, 1, 2, ,n

ij j ij

a x b i m

have solution in 0 and 1?

Matrix formulation:

ij m nA a

Given A and b, is there any binary vector x such that

1 0 2 3

0 3 2 2

Encode: ( )A II I III I IIc c II IIIc c cc ( )b IIII IIccI

( ) ( ) has a binary soluti@ onB A b Ax b Language

?BP No answer!

( ) ( ) has a binary soluti@ onB A b Ax b Language

?BP No answer!

( ) ( ) ( ) is a binary solut@ ion to $B A b x x Ax b

Language

?BP Yes, it is.

3B TIME nFact:

Definition

:T N N*0 0, L

( , , , )M K s

M accepts L in nondeterministic time T if the following holds

For all ,*0w

( , # #) ( , )w L s w h u v iff tM├

for some u, v *, , and t T(|w|).

We say that L is acceptable in nondeterministic time T if there is an NTM that accepts L in nondeterministic time T.

Definition

( )NTIME TThe class of languages acceptable in nondeterministic time T.

What is the difference between TIME(T) and NTIME(T)?

Definition (NP )

( ) : 0dNTIME n d NP

That is, NP denotes the class of all languages that are accepted by NTM’s in some nondeterministic polynomial time bounds.

Definition (Step Counting Function)

f : NN is a step-counting function if a k-tape Turing machine M such that for all w*, M halts in exactly f(|w|) steps on input w, i.e.,

1 1 1( , # #, , #) ( , , , )k k ks w h u a v u a v (| |)Mf w├

for some u1, …, uk, v1, …, vk * and a1, …, ak.

Exercise

Given d >0, there exists a polynomial p of

degree d being a step counting function.

Theorem 3

Let L be a language accepted in time T1 by

a NTM M1=(K1, 1, 1, s1) where T1 is a step

counting function.

A k2-tape DTM M2=(K2, 2, 2, s2) decides L in time T2 where

1 ( )2 2( ) ( 1, and 1)T nT n r k r

See text for the proof.

Polynomial Balanced Languages

Let L *$* be a language, where $.

Then, L is said to be polynomial balanced

if there exists a polynomial p such that

| | | | .$x Ly y p x only if

Theorem 4

Let L * be a language, $ and ||

* \ $ ,L L L andPL NP iff L’ *$* such that

where L’ is a polynomial balanced language.

Theorem 4

Pf) “”

* \ $ .L L L andP

L NP iff L’ *$* such that

where L’ is a p.b.l.

L’ is a p.b.l. Polynomial p s.t. x$yL’ only if |y| p(|x|).

L’P k–tape DTM M’ that decides L’ in time q (a polynomial).

To prove L NP, we construct an NTM M based on M’ to accept L as follows:

M’’:

M: >$(1)

M’’

Theorem 4

Pf) “”

* \ $ .L L L andP

L’ is a p.b.l. Polynomial p s.t. x$yL’ only if |y| p(|x|).

L’P k–tape DTM M’ that decides L’ in time q (a polynomial).

To prove L NP, we construct an NTM M based on M’ to accept L as follows:

M’’:

M: >$(1)

M’’

M can be accept L in nondeterministic time p + q.

Hence, L NP.

M can be accept L in nondeterministic time p + q.

Hence, L NP.

Theorem 4

Pf) “”

* \ $ .L L L andP

We shall show that L’ = {x$Y(x) : x L, |Y(x)| p(|x|)},

where p is a polynomial, such that L’P, i.e.,

a DTM M’ can decides L’ in polynomial time.

Y(x)=? M’ =?

* \ $ .L L L andP

Theorem 4

Pf) “”

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

a DTM M’can decides L’ in polynomial time.

Y(x)=? M’ =?

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?Since LNP, an NTM M

that accept L in

nondeterministic time q (a

polynomial).That is, given xL, we have

such that0 1 tC C C├M ├M ├M

0 ( , # #)C s x( , )tC h w u | |t q x

Let be the encoding function that encodes TM’s and their memory configurations in {I,

c}*.Let Y(x) be the set of all strings

0( )@ ( )@ @ ( )ty M C C

where M is the NTM that accepts L.

* \ $ .L L L andP

Theorem 4

Pf) “”

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?0( )@ ( )@ @ ( )ty M C C

( ) ( ) #@'t

y M C s

constant to be discussed

q(|x|)

| | | |jC q x x input length

accept in time q

(| |) | |jC M q x x

* \ $ .L L L andP

Theorem 4

Pf) “”

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?

( ) ( ) #@'t

y M C s

q(|x|)

( ) (| |) 2 (| |) | | (| |)M q x q x x q x

0( )@ ( )@ @ ( )ty M C C

* \ $ .L L L andP

Theorem 4

Pf) “”

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?

( ) ( ) #@'t

y M C s

( ) (| |) 2 (| |) | | (| |)M q x q x x q x

Polynomial of |x|

0( )@ ( )@ @ ( )ty M C C

* \ $ .L L L andP

Theorem 4

Pf) “”

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?

( ) ( ) #@'t

y M C s

( ) (| |) 2 (| |) | | (| |)M q x q x x q x

Let $ , ( )L x y x L y Y x

0( )@ ( )@ @ ( )ty M C C

L’ is polynomial balanced.

* \ $ .L L L andP

Theorem 4

Pf) “”

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?

We shall show that

L’= {x$Y(x) : x L, |Y(x)| p(|x|)},

Y(x)=? M’ =?

( ) ( ) #@'t

y M C s

( ) (| |) 2 (| |) | | (| |)M q x q x x q x

Let $ , ( )L x y x L y Y x

0( )@ ( )@ @ ( )ty M C C

L’ is polynomial balanced.

* \ $ .L L L andP

Theorem 4

Pf) “” $ , ( )L x y x L y Y x We now show that L’P by designing a DTM M’ which decides L’ as follows:

1. Check the syntax of the input string

To verify the following things:a) Is Cj a configuration of M?b) C0 = (s, #x#)?c) Ct = (h, uv)?

If rejected, print out ‘N’ then halt. This steps requires O(n) time.2. Check Cj├M Cj+1, j= 0, …, t1 by consulting (M). If rejected print out ‘

N’ and then halt. This steps requires O(n2).3. Therefore, L’P.

0$ ( )@ ( )@ @ ( )tw x M C C

Example (Integer Programming BNP )

( ) ( ) has a binary soluti@ onB A b Ax b

| | (| |)| |

Polynomial Balanced

$ ( ) ( ) ( @ )w O ww

A b x B

3( )B TIME n P BNP

Example (Integer Programming BNP )

3( )B TIME n P BNP

?BPNo answer

Is P = NP ?

How to show?

P NPShow NP Pand

Trivial (why?) DifficultNo answer, now.

Example (TSPNP )

[ ]ijD d Distance matrix

Find the shortest tour t.

Example (TSPNP )

Find the shortest tour t.

:[1, 2, , ] [1,2, , ]t n n t is a bijection E.g.

,1 7 6 4 5 3 8

1 2 3 4 5 6 7 8

It is not language decision problem. Rather, it is a problem of function evaluation.

Example (TSPNP )

TSPF Find the shortest tour t.

TSP Denote the cost of t as D(t).

@ ( )@ a tour with ( )n bTSP I D I t D t b

TSP’

( )@ ( )@ ( )$n bTSP I D I bt D t

Polynomial balanced

Exercises

Show that

1. TSPNP.

2. TSPP a polynomial algorithm for TSPF.

3. TSP’P.

Remark

TSPP P = NP

NP-Completeness

Definition: Polynomial-Time Computable

* *1 2:f is said to be computed in time T if

a k-tape DTM s.t.

, # #, #, , # , # ( ) #, #, , #s x h f x tM├

with t T(|x|).

Furthermore, if T(|x|) is a polynomial, then f is said to be polynomial-time computable.

Definition: Polynomial-Time Reduction

be polynomial-time computable.* *1 2:

Then L1 is said to be polynomial-time reducible to L2 iff

Let and be languages,*1 1L *

1 2( )x L x L

More on NP-Completeness

1. LNP..2. .p rL L L NP

NP-Complete

p.r.p.r.

NP-Hard

NP-Complete

p.r.p.r.

NP-Hard

Important Theorem

Let L be an NP-complete language.

Then, L P NP P.NP

NP-Complete

Pf)“”

“”

L P NP P.

L P P NP.

trivial

Remark

NP-Complete

If one can decide any NP-complete language in polynomial time, then all languages in NP can be decided effectively.

Surely, the guy will win Turing Award.

Bounded Halting Problems

Bounded Halting Problem N0

The NTM accepts input( ) ( )

in or fewer steps@ tN M w I

Halting problem is not solvable.

Is Bounded Halting problem is solvable?

Theorem: N0 is NP-Complete

01. N NP

.2. .p rL L N NP

01. N NP

Define

( ) ( ) ( ) ( ) ( )

such that

(a) ( , , , ) an NTM

(b) a configuration of

(c) ( , # #), ( , )

@ @ @ @

M w I C C C

C s w C h u v

in or fewer step@

stN M I

01. N NP

Define

( ) ( ) ( ) ( ) ( )

such that

(a) ( , , , ) an NTM

(b) a configuration of

(c) ( , # #), ( , )

M I C C C

C s w C h u v

polynomial balanceN

P 0N NP

Exercise

in or fewer step@

stN M I

.2. .p rL L N NPFor any LNP, find L.

• LNP: NTM

: polynomial

to accept

• Let | |( ) ( ) ( )@ LP wLL w w IM

• Fact: L(w) is polynomial computable.

• w L L(w) N0.

The Language0N̂

The NTM accepts inp

utˆ ( ) ( ) in or fewer steps

u uM u

in or fewer steps@ tN M u I

Theorem: is NP-Complete0N̂

The NTM accepts inp

u uM u

0. ˆ1 N NP. .

0 0ˆ2. p rN N

Exercise

Find .

Theorem: is NP-Complete0N̂ 0

ˆ ( ) ( ) 2 | | stepsuN M u

0 @( ) ( ) steps.tN M u I t

.0: ˆ p rN N Find

( ) ( )@ tw M u I ( ) ( ) ( )w M u

. .p r 0

ˆ( )w N

ˆ ( ) ( ) 2 | | stepsuN M u

.0: ˆ p rN N Find

( ) ( )@ tw M u I ( ) ( ) ( )w M u

. .p r 0

ˆ( )w N

Case 1. ( ) ( ) 2 | | 2, @ tw M u I t u

2 2| |#t uu u #

: M L RM #

t + 2 2|u| + 2 + t

= 2t + 4 2|u| =2|u’|

ˆ ( ) ( ) 2 | | stepsuN M u

.0: ˆ p rN N Find

( ) ( )@ tw M u I ( ) ( ) ( )w M u

. .p r 0

ˆ( )w N

Case 1. ( ) ( ) 2 | | 2, @ tw M u I t u

2 2| |#t uu u #

: M L RM #

t + 2 2|u| + 2 + t

= 2t + 4 2|u|

Halt within t steps.

Halt within 2|u’|

steps.

Halt within 2|u’|

steps.

=2|u’|

ˆ ( ) ( ) 2 | | stepsuN M u

.0: ˆ p rN N Find

( ) ( )@ tw M u I ( ) ( ) ( )w M u

. .p r 0

ˆ( )w N

Case 2. ( ) ( ) 2 | | 2, @ tw M u I t u

u u 2| |: u tM MR t +2|u| t =2|u| =2|u’|

ˆ ( ) ( ) 2 | | stepsuN M u

.0: ˆ p rN N Find

( ) ( )@ tw M u I ( ) ( ) ( )w M u

. .p r 0

ˆ( )w N

Case 2. ( ) ( ) 2 | | 2, @ tw M u I t u

u u 2| |: u tM MR t +2|u| t =2|u| =2|u’|

Halt within t steps.

Halt within 2|u’|

steps.

Halt within 2|u’|

steps.

ˆ ( ) ( ) 2 | | stepsuN M u

.0: ˆ p rN N Find

( ) ( )@ tw M u I ( ) ( ) ( )w M u

. .p r 0

ˆ( )w N

Hence,

is not in the form of ( ) ( )@ ( )

( ) ( ) is in the form of ( ) ( )@

I w M u Iw

M u w M u I

See previous slides

Theorem: is NP-Complete0N̂

0. ˆ1 N NP. .

0 0ˆ2. p rN N

0ˆ ( ) ( ) 2 | | stepsuN M u

Known NP-complete

Summary of the proof:

Special Bounded Halting Problem N2

The NTM accepts inp

u uM u

( , , , ), and( )

( , #) ( , ) wi@

th t M K s

N M Is h u v t t

0ˆ ( ) ( )N M u

0 ( ) ( )@ tN M u I

( , , , ), and( )

( , #) ( , ) wi@

th t M K s

N M Is h u v t t

0N0N̂. .p r LNP

2Np.r.

Pf)21. N NP

2. . .p r

obvious.

0ˆ ( ) ( )N M u

0 ( ) ( )@ tN M u I

( , , , ), and( )

( , #) ( , ) wi@

th t M K s

N M Is h u v t t

0N0N̂. .p r LNP

2Np.r.

Pf)21. N NP

2. . .p r

obvious.

0 ( ) ( )@ tN M u I 2 ( )@ tN M I :( ) ( )@ tw M u I

0N0N̂. .p r LNP

2Np.r.

2. . .p r

0N̂. .p r LNP

2Np.r.p.r.

2. . .p r

( ) ( )@ tw M I

0w N 2( )w N

0 ( ) ( )@ tN M u I 2 ( )@ tN M I :( ) ( )@ tw M u I

0N0N̂. .p r LNP

2Np.r.

2. . .p r

0N̂. .p r LNP

2Np.r.p.r.

2. . .p r

( ) ( )@ tw M I

0w N 2( )w N

(1) (2) (: | |)u u uM R R RMuR Let

2|u| 1 t

2 | | 1t t u

0 ( ) ( )@ tN M u I 2 ( )@ tN M I :( ) ( )@ tw M u I

0N0N̂. .p r LNP

2Np.r.

2. . .p r

0N̂. .p r LNP

2Np.r.p.r.

2. . .p r

( ) ( )@ tw M I

0w N 2( )w N

(1) (2) (: | |)u u uM R R RMuR Let

2|u| 1 t

2 | | 1t t u

M halts on u in t stepsM halts on u in t steps

M ’ halts on in t ’ steps

0 ( ) ( )@ tN M u I 2 ( )@ tN M I :( ) ( )@ tw M u I ( ) ( )@ tw M I

0w N 2( )w N

(1) (2) (: | |)u u uM R R RMuR

2|u| 1 t

2 | | 1t t u

2| | 1

if is not in form of ( ) ( )( )

( ) if is in form of ( )@(@ )

I w M u Iw

M I w M u I

Summary

0N0N̂. .p r LNP

2Np.r.

. .p r

Some Important NP-Complete

Problems

Bounded Tiling Problem (BT)

0( , , , )D d H VD

Tiling System:

D: a set of finite number of tiles.

d0D: the tile place at the origin.

HDD: the Horizontal tiling rule.

VDD: the Vertical tiling rule.

0( , , , )D d H VD

Given D and s > 0, an s s tiling by D is a function

:[0,1, , 1] [0,1, , 1]f s s D

such that

01. (0,0)f d

2. ( , ), ( 1, )f m n f m n H

3. ( , ), ( , 1)f m n f m n V 0 1 2 3 4 5

0( , , , )D d H VD

( ) there is an tiling by @ sBT I s s D D

BT is NP-Complete

Theorem: BT is NP-Complete

0( , , , )D d H VD

Pf)1. BT NP

2. 0N0N̂. .p r LNP

2Np.r.

. .p r

( ) ( ) is an tiling by $@ sBT I D D s s D Da p.b.l.P

BT. .p r

2. 0N0N̂. .p r LNP

2Np.r.

. .p r

BT. .p r

2. 0N0N̂. .p r LNP

2Np.r.

. .p r 0N0N̂. .p r LNP

2Np.r.p.r.

. .p r

BT. .p r

( )@ sBT I D 2 ( )@ tN M I :( )@ tw M I ( ) ( )@ sw I D

2w N ( )w BT

0( , , , ) ?D d H V D?s

0 1 t t+1

( )@ tw M I ( ) ( )@ sw I D

0 1 t t+1

( )@ tw M I ( ) ( )@ sw I D

0( , , , ) ?D d H V D

( , , , )M K s

1) For each a, create tiles

(a, k)

(a, k+1)

( )@ tw M I ( ) ( )@ sw I D

0( , , , ) ?D d H V D

( , , , )M K s

(a, k)

(a, k+1)

0 1 6 7

2 3 4 5

. ., 6e g t 2 8s t

(a, 0)

(a, 1)

(a, 2)

(a, 3)

(a, 4)

(a, 5)

(a, 6)

(a, 7)

( )@ tw M I ( ) ( )@ sw I D

0( , , , ) ?D d H V D

( , , , )M K s

2) For each (q, a, p, b) where b and q h, create tiles

(q, a, k)

(p, b, k+1)

0 1 6 7

2 3 4 5

. ., 6e g t 2 8s t

(q, a, 0)

(p, b, 1)(q, a, 1)

(p, b, 2)(q, a, 2)

(p, b, 3)(q, a, 3)

(p, b, 4)(q, a, 4)

(p, b, 5)(q, a, 5)

(p, b, 6)

Can’t reach the highest layer if q h.

( )@ tw M I ( ) ( )@ sw I D

0( , , , ) ?D d H V D

( , , , )M K s

3) For each (q, a, p, L) , create tiles

0 k t 0 1 6 7

2 3 4 5

. ., 6e g t 2 8s t

(q, a, k)

(a, k+1)p

(b, k)

(p, b, k+1)p

(q, a, 2)

(a, 3)

(b, 2)

(p, b, 3)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(b, 0)

(b, 1)

(b, 2)

(a, 0)

(a, 1)

(q, a, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

( )@ tw M I ( ) ( )@ sw I D

0( , , , ) ?D d H V D

( , , , )M K s

0 k t 0 1 6 7

2 3 4 5

. ., 6e g t 2 8s t

(q, a, k)

(a, k+1)p

(b, k)

(p, b, k+1)p

(q, a, 2)

(a, 3)

(b, 2)

(p, b, 3)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(b, 0)

(b, 1)

(b, 2)

(a, 0)

(a, 1)

(q, a, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(a, 3)

(a, 4)

(a, 5)

(a, 6)

(a, 7)

( )@ tw M I ( ) ( )@ sw I D

0( , , , ) ?D d H V D

( , , , )M K s

4) For each (q, a, p, R) , create tiles

0 k t 0 1 6 7

2 3 4 5

. ., 6e g t 2 8s t

(q, a, k)

(a, k+1)p

(b, k)

(p, b, k+1)p

(q, a, 2)

(a, 3)

(b, 2)

(p, b, 3)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(a, 0)

(a, 1)

(q, a, 2)

(b, 0)

(b, 1)

(b, 2)

(?, 0)

(?, 1)

(?, 2)

( )@ tw M I ( ) ( )@ sw I D

0( , , , ) ?D d H V D

( , , , )M K s

0 k t 0 1 6 7

2 3 4 5

. ., 6e g t 2 8s t

(q, a, k)

(a, k+1)p

(b, k)

(p, b, k+1)p

(q, a, 2)

(a, 3)

(b, 2)

(p, b, 3)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(a, 0)

(a, 1)

(q, a, 2)

(b, 0)

(b, 1)

(b, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(a, 3)

(a, 4)

(a, 5)

(a, 6)

(a, 7)

( )@ tw M I ( ) ( )@ sw I D

0( , , , ) ?D d H V D

( , , , )M K s

0 1 6 7

2 3 4 5

. ., 6e g t 2 8s t

(h, a, k)

(h, a, k+1)

(q, a, 2)

(a, 3)

(b, 2)

(h, b, 3)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 0)

(?, 1)

(?, 2)

(a, 0)

(a, 1)

(q, a, 2)

(b, 0)

(b, 1)

(b, 2)

(?, 0)

(?, 1)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 2)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 3)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 4)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 5)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(?, 6)

(?, 7)

(a, 3)

(a, 4)

(a, 5)

(a, 6)

(a, 7)

(h, a, k+1)

( )@ tw M I ( ) ( )@ sw I D

0( , , , ) ?D d H V D

( , , , )M K s

0 1 6 7

2 3 4 5

. ., 6e g t 2 8s t

(s, #, 0)

d0 = #

7) Base tile

(#, 0)

(s, #, 0)#

(#, 0)##

( )@ tw M I ( ) ( )@ sw I D

( , , , )M K s 2s t

(a, k)

(a, k+1)

2) For each (q, a, p, b) where b and q h, create tiles

(q, a, k)

(p, b, k+1)

(q, a, k)

(a, k+1)

(b, k)

(p, b, k+1)

p 0 k t

(q, a, k)

(a, k+1)

(b, k)

(p, b, k+1)

p 0 k t

(h, a, k)

(h, a, k+1)

(s, #, 0)

d0 = #

7) Base tile

(#, 0)

0( , , , )D d H VD

Pf)1. BT NP

2. 0N0N̂. .p r LNP

2Np.r.

. .p r

( ) ( ) is an tiling by $@ sBT I D D s s D Da p.b.l.P

BT. .p r

Integer Programming

BNP already known

Is B NP-complete?

Theorem: B is NP-Complete

0N0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT B. .p r

( )@ sBT I D

@( ) ( )B A b

0N0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT B. .p r

( )@ sBT I D :( )@ sw I D )( @) ( ( )w A b

w BT ( )w B

@( ) ( )B A b ( )@ sBT I D :( )@ sw I D )( @) ( ( )w A b

w BT ( )w B

( , )i j

kd1 is placed at ( , )

0 is not placed at ( , )k

d D i jx

d D i j

Define variables

where0 ,

0 | | 1

( , )i j

• Every place has exactly one tile

1 is placed at ( , )

d D i jx

d D i j

• Horizontal rules

• Vertical rules

• Place d0 at (0, 0)

0 , 1i j s

1, ,( , )

ijk i j ld d H

, 1,( , )

ijk i j ld d V

0 | | 1

000 1x

0 | | 1

( , )i j

1 is placed at ( , )

d D i jx

d D i j

• Vertical rules

0 , 1i j s

1, ,( , )

ijk i j ld d H

, 1,( , )

ijk i j ld d V

0 | | 1

000 1x

0 | | 1

• Vertical rules

0 , 1i j s

1, ,( , )

ijk i j ld d H

, 1,( , )

ijk i j ld d V

0 | | 1

000 1x

0 | | 1

Introducing slack variables

, {0,1}ijkijky z

1, ,( , )

ijk i j ld d H

ijkx yx

, 1,( , )

ijk i j ld d V

ijkx zx

• Vertical rules

0 , 1i j s

0 | | 1

000 1x

0 | | 1

Introducing slag variables

, {0,1}ijkijky z

1, ,( , )

ijk i j ld d H

ijkx yx

, 1,( , )

ijk i j ld d V

ijkx zx

• Vertical rules

0 , 1i j s

0 | | 1

000 1x

0 | | 1

Introducing slag variables

, {0,1}ijkijky z

1, ,( , )

ijk i j ld d H

ijkx yx

, 1,( , )

ijk i j ld d V

ijkx zx

The coefficient of each

variable is 0 or 1. The coefficient of each

variable is 0 or 1. The right-hand side of each

equality is 1.The right-hand side of each

equality is 1.

Language Br is NP-Complete

all entries of are in {0,1},

( ) ( ) all entries of are 1, and

has a binary solut

B A b b

Hamilton Cycle

Theorem: H is NP-Complete

( ) has a Hamilton CycleH G GPf)

1. H NP

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

B H. .p r

( ) has a Hamilton CycleH G G

A Hamilton cycle entering the box from a, it must leave the box from c.

A Hamilton cycle entering the box from d, it must leave the box from b.

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

B H. .p r

( )H G ( )@ ( )rB A b :( )@ ( )w A b ( ) ( )w G

rw B ( )w H

1 2 3 4

x x x x

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

B H. .p r

1 2 3 4

x x x x

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

B H. .p r

1 2 3 4

x x x x

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

B H. .p r

1 2 3 4

x x x x

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

B H. .p r

1 2 3 4

x x x x

( )H G ( )@ ( )rB A b :( )@ ( )w A b ( ) ( )w G

rw B ( )w H

Traveling Salesman Problem (TSP)

( )@ ( )@ ( )$n bTSP I D I bt D t

TSPNP already known

Is TSP NP-complete?

Theorem: TSP is NP-Complete

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H

TSPp.r.

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H

TSPp.r.

@ ( )@n bTSP I D I ( )H G :( )w G ( ) @ ( )@n bw I D I

w H ( )w TSP

( )w G ( ) @ ( )@n bw I D I

w H ( )w TSP

?ijD d

@ ( )@n bTSP I D I ( )H G :

?ijD d

@ ( )@n bH I D I ( )rB G :( )w G ( ) @ ( )@n bw I D I

w H ( )w TSP

G = (V, E)

0 if ,

if ,ij

i j Ed

?ijD d

@ ( )@n bH I D I ( )rB G :( )w G ( ) @ ( )@n bw I D I

w H ( )w TSP

G = (V, E)

0 if ,

if ,ij

i j Ed

There exists a Hamilton cycle in graph G=(V, E) if and only if there exists a zero-cost tour for the following TSP .

Satisfiability Problem (SAT)

: Boolean Variables; 1 2, ,x x

: the negation of xi; ix literals

A formula in the propositional calculus is an expression that can be constructed using literals and the operations (and) and (or). A satisfiability problem is to determine if a formula is true for any truth assignment of variables.

Example: 1 2 3 4x x x x

Normal Forms

Cook’s Theorem

SAT is NP-complete

SATP iff P =NP

Cook’s TheoremSAT is NP-complete

1. SAT NP

2.0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

SATp.r.

Cook’s Theorem0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

SATp.r.p.r.

( , )m n

Define literals:

is placed at ( , )

is not placed at ( , )k

true d D m nP

false d D m n

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

SATp.r.p.r.

0 1 ,0 1

m s d d Dn s d d

mnd mndP P

( , )m n

000 dP

1 20 10 1

,V mnm s d d H

1 20 10 1

,V mnm s d d V

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

SATp.r.p.r.

0 1 ,0 1

m s d d Dn s d d

mnd mndP P

( , )m n

000 dP

1 20 10 1

,V mnm s d d H

1 20 10 1

,V mnm s d d V

True if and only if there is no different tiles being placed at

each cell.

True if and only if there is no different tiles being placed at

each cell.

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

SATp.r.p.r.

0 1 ,0 1

m s d d Dn s d d

mnd mndP P

( , )m n

000 dP

1 20 10 1

,V mnm s d d H

1 20 10 1

,V mnm s d d V

True if and only if horizontal rules are

satisfied at each cell.

True if and only if horizontal rules are

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

SATp.r.p.r.

0 1 ,0 1

m s d d Dn s d d

mnd mndP P

( , )m n

000 dP

1 20 10 1

,V mnm s d d H

1 20 10 1

,V mnm s d d V

True if and only if vertical rules are

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

SATp.r.p.r.

0 1 ,0 1

m s d d Dn s d d

mnd mndP P

( , )m n

000 dP

1 20 10 1

,V mnm s d d H

1 20 10 1

,V mnm s d d V

True if an only if d0 is place at (0, 0).

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

SATp.r.p.r.

0 1 ,0 1

m s d d Dn s d d

mnd mndP P

( , )m n

000 dP

1 20 10 1

,V mnm s d d H

1 20 10 1

,V mnm s d d V

Summary

0N 0N̂. .p r LNP

2Np.r.

. .p r

. .p r BT . .p r

. .p r H TSP. .p r

p.r.SAT

lecture 8: computational complexity

Documents

on the computational complexity of maxsat central position...

computational physics (lecture 10)

컴퓨팅적 사고 (computational thinking) 수업...

výroková a predikátová logika - ikti.mff.cuni.cz ›...

1 cse 417: algorithms and computational complexity winter...

using complexity to help understand and manage public policy...

computational complexity. on the geometry of polynomials and...

computational physics (lecture 11)

computational movement analysis lecture 5: segmentation,...

linear rergression time complexity calculation · 2021. 3....

computational physics (lecture 13)

complexity theory csci 5403 lecture vii: diagonalization

חישוביות וסיבוכיות computability and...

computational physics (lecture 17)

an introduction to computational complexity edith elkind...

computational complexity of kernel-based density-ratio...

cs 350 algorithms and complexity - computer action...

optimization theory - keio...

unit 1a: computational complexity -...

modern computational statistics [1em] lecture 13:...