The Brick Wall:NP-Completeness

Christopher King

Joshua Greenspan

2

History

• Before 1950s computers will solve anything

• 1950s & 1960s: The wall− Computers can’t solve basic

problems.

• Today: The wall still standsJohn von Neumann, 1950

3

The Wall

• Many fundamentally important problems can’t be solved in tractable time− Engineering− Operations research− VLSI chip design− Database management− Etc.

• NP Complete (Nondeterministic Polynomial time)− 2n or worse

− Every time ‘n’ increases by 1, processing doubles

− n!− Every time ‘n’ increases by 1, processing takes ‘n’ times longer

4

Scheduling Example

Industry: Welding Schedule• Manufacturing-line welding• Fastest path = Line speed• Finding fastest path:

− Tabulate all possible paths− Compute path times− Choose best path

• All possible paths = n! (n = #welds)− 3!=6, 4!=24, 5!=120, 6!=720, 30!=2.6x1032

− 30! Compute (1 trillion schedules)/sec = 30 years− 31! = 930 years

5

Jackhammer through the wall

Faster Computers & Massive Parallelization• Turn every molecule in the universe into processor:

− 1060 processors

• Every processor: 1,000 times faster than current− 1015 schedules/second

• Compute a schedule with 63 welds in 1000 years• 64 welds: 64,000 years or 63 more universes

• Even improving algorithm to 2n would allow 240 welds

• Exponential Explosion of Complexity

6

Avoiding the wall

• n!, or even 2n wont work− n2 needed, or even nk

• Greedy Algorithm?− Start at arbitrary point

− Take optimal path

• A

7

Avoiding the wall

• n!, or even 2n wont work− n2 needed, or even nk

• Greedy Algorithm?− Start at arbitrary point

− Take optimal path

• A – D− 18

− Total = 18

8

Avoiding the wall

• n!, or even 2n wont work− n2 needed, or even nk

• Greedy Algorithm?− Start at arbitrary point

− Take optimal path

• A – D – E− 18 + 14

− Total = 32

9

Avoiding the wall

• n!, or even 2n wont work− n2 needed, or even nk

• Greedy Algorithm?− Start at arbitrary point

− Take optimal path

• A – D – E – C− 18 + 14 + 14

− Total = 46

10

Avoiding the wall

• n!, or even 2n wont work− n2 needed, or even nk

• Greedy Algorithm?− Start at arbitrary point

− Take optimal path

• A – D – E – C – B− 18 + 14 + 14 + 22

− Total = 68

11

Avoiding the wall

• n!, or even 2n wont work− n2 needed, or even nk

• Greedy Algorithm?− Start at arbitrary point

− Take optimal path

• A – D – E – C – B – A− 18 + 14 + 14 + 22 + 25

− Total = 93

12

Avoiding the wall

• n!, or even 2n wont work− n2 needed, or even nk

• Greedy Algorithm?− Start at arbitrary point

− Take optimal path

• A – D – E – C – B – A− 18 + 14 + 14 + 22 + 25 = 93

• A – D – C – E – B – A− 18 + 15 + +14 +18 +25 = 90

13

Avoiding the wall

• For scheduling problem− No n2, n100, or nk algorithm has ever been found

− No nk will ever be found

• For all NP-complete problems− No n2, n100, or nk algorithm has ever been found

− No nk will ever be found

• Can good solutions be made efficiently?− For non-Euclidean, no such guarantee

− For scheduling problem using Euclidean distances− Polynomial can be found to produce 50% optimal

− Many other NP-C problems can be approximated efficiently

14

Bin packing problem

• Goal: Store telephones of different sizes in bins

• Optimal solution: Use as few bins as possible

• First Fit Algorithm− Place each phone in first bin it fits in

− No bin except last can be less then half full

− This means: FF requires no more then 2x optimal + 1

− Proof: If two bins are less then half full

−You can combine the contents of these bins into 1

15

Efficient approximation guarantee

• Better algorithms can be found− First-Fit Decreasing - Put the biggest phone in the

lowest bin it fits in

− guarantee within 22% of optimal

− proof is complex− Shows ratio of bins

produced by FFD

to optimal is 11/9

Example of worst case

16

Probabilistic Results

• Guarantees can be much worse than actual

• Consider where you have infinite phones with sizes between 0 to bin-size− Always find two phones to fit in a bin

• Actually holds for more general cases using a finite number of phones of even size distribution− High probability that two phones will fit bin

• Proved that under uniform distribution− Constant number of bins are wasted

• Combining locally optimal solutions can make NP-C approximations efficient

• Not all NP-C can be solved using locally optimal solutions

17

Satisfiability (SAT) problem

• Important because first to be shown NP-C

• Used to prove many other problems are NP-C

• Boolean equation in Conjunctive Normal Form− Equation is an “AND” of clauses− Clause is an “OR” of literals− Literals are True or false

• Clause is true if at least one literal is true

• Equation is satisfiable (true) if every clause is true

• Problem: Determine if given formula is satisfiable

• Known solution is O(2n)

)()()()( 643653132321 xxxxxxxxxxxx

18

SAT Problem

• Since it is either satisfiable or not satisfiable − No approximations

− No locally optimal solutions

• SAT allows research into NP-C problems• Special classes of SAT can be solved efficiently

− Restrict to 2 literals, time to solve proportional to size of formula

− Restrict so only 1 non-negated literal, also proportional to size of formula

• Problems with 3 literals per clause are those of NP-C problems

19

Probability of Satisfiability

• It is trivial to find satisfiabilities probabilistically− Given M clauses and N literals− Formula is likely satisfiable if M/N < 4.2− Near 100% not-satisfiable if M/N > 4.3

• It is non-trivial to find satisfiability with certainty− Known special classes are likely only with M/N < 1− Other efficient solutions are likely (Near 100%)

if M/N < 3.003− No known efficiently solution when M/N > 4.3

20

Other NP Problems

• NP− All decision problems where ‘yes’ answers

have simple proofs− Ex. Integer factorization

• NP Complete (NPC)− Most challenging NP problems− Proving 1 NP-C=P proves all NP=P− Proving 1 NP-C≠P proves all NP-C≠P− Ex. Boolean

• NP Easy− At most as hard as NP

• NP Hard− At least as hard as NP− May not be NP

• Ex. (NP-C) Subset Sum, Traveling Salesperson

• Ex. (! NP-C) Halting problem

21

Proving NP-C

• Prove a problem is NP-C− Prove NP− Reduce problem to a known NP-C− n4 reduces to n*n*n*n

22

Does NP≠P

• It has yet to be shown if NP problems can be solved in Polynomial Time− Clay Mathematics Institute (CMI): $1,000,000

• If NP-C≠P for one problem: The same for all

• If NP-C=P: Solutions to all CMI problems

• If NP-C=P: All life’s mysteries will be solved, artificially intelligent beings will emerge, and they will enslave humanity [Stross, 2000]

23

Questions?