the brick wall: np-completeness christopher king joshua greenspan
Post on 20-Dec-2015
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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
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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
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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
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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
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Avoiding the wall
• n!, or even 2n wont work− n2 needed, or even nk
• Greedy Algorithm?− Start at arbitrary point
− Take optimal path
• A
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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]
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