1. The Brick Wall: NP-Completeness Christopher King Joshua Greenspan

2. 2 History Before 1950s computers will solve anything 1950s & 1960s: The wall −Computers can’t solve basic problems. Today: The wall still stands John von Neumann, 1950

3. 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) −2 n or worse −Every time ‘n’ increases by 1, processing doubles −n! −Every time ‘n’ increases by 1, processing takes ‘n’ times longer

4. 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!=2.6x10 32 −30! Compute (1 trillion schedules)/sec = 30 years −31! = 930 years

5. 5 Jackhammer through the wall Faster Computers & Massive Parallelization Turn every molecule in the universe into processor: −10 60 processors Every processor: 1,000 times faster than current −10 15 schedules/second Compute a schedule with 63 welds in 1000 years 64 welds: 64,000 years or 63 more universes Even improving algorithm to 2 n would allow 240 welds Exponential Explosion of Complexity

6. 6 Avoiding the wall n!, or even 2 n wont work −n 2 needed, or even n k Greedy Algorithm? −Start at arbitrary point −Take optimal path A

7. 7 Avoiding the wall n!, or even 2 n wont work −n 2 needed, or even n k Greedy Algorithm? −Start at arbitrary point −Take optimal path A – D −18 −Total = 18

8. 8 Avoiding the wall n!, or even 2 n wont work −n 2 needed, or even n k Greedy Algorithm? −Start at arbitrary point −Take optimal path A – D – E −18 + 14 −Total = 32

9. 9 Avoiding the wall n!, or even 2 n wont work −n 2 needed, or even n k Greedy Algorithm? −Start at arbitrary point −Take optimal path A – D – E – C −18 + 14 + 14 −Total = 46

10. 10 Avoiding the wall n!, or even 2 n wont work −n 2 needed, or even n k Greedy Algorithm? −Start at arbitrary point −Take optimal path A – D – E – C – B −18 + 14 + 14 + 22 −Total = 68

11. 11 Avoiding the wall n!, or even 2 n wont work −n 2 needed, or even n k Greedy Algorithm? −Start at arbitrary point −Take optimal path A – D – E – C – B – A −18 + 14 + 14 + 22 + 25 −Total = 93

12. 12 Avoiding the wall n!, or even 2 n wont work −n 2 needed, or even n k 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. 13 Avoiding the wall For scheduling problem −No n 2, n 100, or n k algorithm has ever been found −No n k will ever be found For all NP-complete problems −No n 2, n 100, or n k algorithm has ever been found −No n k 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. 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. 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. 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. 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(2 n )

18. 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. 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. 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. 21 Proving NP-C Prove a problem is NP-C −Prove NP −Reduce problem to a known NP-C −n 4 reduces to n*n*n*n

22. 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. 23 Questions?