Presentation on theme: "Chapter 11 Limitations of Algorithm Power. Lower Bounds Lower bound: an estimate on a minimum amount of work needed to solve a given problem Examples:"— Presentation transcript:
Lower Bounds Lower bound: an estimate on a minimum amount of work needed to solve a given problem Examples: b number of comparisons needed to find the largest element in a set of n numbers b number of comparisons needed to sort an array of size n b number of comparisons necessary for searching in a sorted array b number of multiplications needed to multiply two n-by-n matrices
Lower Bounds (cont.) b Lower bound can be an exact countan exact count an efficiency class ( )an efficiency class ( ) b Tight lower bound: there exists an algorithm with the same efficiency as the lower bound Problem Lower boundTightness Problem Lower boundTightness sorting (nlog n) yes sorting (nlog n) yes searching in a sorted array (log n) yes searching in a sorted array (log n) yes element uniqueness (nlog n) yes element uniqueness (nlog n) yes n-digit integer multiplication (n) unknown n-digit integer multiplication (n) unknown multiplication of n-by-n matrices (n 2 ) unknown multiplication of n-by-n matrices (n 2 ) unknown
Methods for Establishing Lower Bounds b trivial lower bounds b information-theoretic arguments (decision trees) b adversary arguments b problem reduction
Trivial Lower Bounds Trivial lower bounds: based on counting the number of items that must be processed in input and generated as output Examples b finding max element b polynomial evaluation b sorting b element uniqueness b Hamiltonian circuit existence Conclusions b may and may not be useful b be careful in deciding how many elements must be processed
Decision Trees Decision tree — a convenient model of algorithms involving comparisons in which: b internal nodes represent comparisons b leaves represent outcomes Decision tree for 3-element insertion sort
Decision Trees and Sorting Algorithms b Any comparison-based sorting algorithm can be represented by a decision tree b Number of leaves (outcomes) n! b Height of binary tree with n! leaves log 2 n! b Minimum number of comparisons in the worst case log 2 n! for any comparison-based sorting algorithm b log 2 n! n log 2 n b This lower bound is tight (mergesort)
Adversary Arguments Adversary argument: a method of proving a lower bound by playing role of adversary that makes algorithm work the hardest by adjusting input Example 1: “Guessing” a number between 1 and n with yes/no questions questions Adversary: Puts the number in a larger of the two subsets generated by last question Example 2: Merging two sorted lists of size n a 1 < a 2 < … < a n and b 1 < b 2 < … < b n a 1 < a 2 < … < a n and b 1 < b 2 < … < b n Adversary: a i < b j iff i < j Output b 1 < a 1 < b 2 < a 2 < … < b n < a n requires 2n-1 comparisons of adjacent elements
Lower Bounds by Problem Reduction Idea: If problem P is at least as hard as problem Q, then a lower bound for Q is also a lower bound for P. bound for Q is also a lower bound for P. Hence, find problem Q with a known lower bound that can be reduced to problem P in question. Hence, find problem Q with a known lower bound that can be reduced to problem P in question. Example: P is finding MST for n points in Cartesian plane Q is element uniqueness problem (known to be in (nlogn))
11.3 P, NP, and NP-complete Problems b An algorithm solves a problem in polynomial time if its worst-case time efficiency belongs to O(p(n)) Where p(n) is a polynomial of the problem’s input size nWhere p(n) is a polynomial of the problem’s input size n b Problems that can be solved in polynomial time are called tractable b Problems that cannot be solved in polynomial time are called intractable. Cannot solve intractable problems in a reasonable length of timeCannot solve intractable problems in a reasonable length of time
25 city Traveling Salesperson Problem b There are 25! different possible paths to be considered. That is approximately 1.5 x 10 25 different paths.That is approximately 1.5 x 10 25 different paths. b Suppose the computer can analyze 10,000,000, or 10 7, paths per second. The number of seconds required to check all possible paths is about 1.5 x 10 25 /10 7, or about 1.5 x 10 18 seconds.The number of seconds required to check all possible paths is about 1.5 x 10 25 /10 7, or about 1.5 x 10 18 seconds. That’s roughly 10 12 years: about a trillion years.That’s roughly 10 12 years: about a trillion years. b This would not be a feasible algorithm.
The Halting Problem b Turing – 1936 b Given a computer program and an input to it, determine whether the program will halt on that input or continue working indefinitely on it. b Assume that A is an algorithm that solves the halting problem: A(P, I) = 1 if P halts on input I, 0 otherwise b Consider P as an input to itself and use the output of A for pair (P,P) to construct a program Q as follows: b Q(P) halts if A(P,P) = 0 (if program P does not halt on input P) b Q(P) does not halt if A(P, P) = 1 b Then substituting Q for P we obtain b Q(Q) halts if A(Q, Q) = 0 i.e. if program Q does not halt on input Q b Q(Q) does not halt if A(Q, Q) = 1 i.e. if program Q halts on input Q b This is a contradiction because neither of the two outcomes for program Q is possible. QED
Classifying Problem Complexity Is the problem tractable, i.e., is there a polynomial-time (O(p(n)) algorithm that solves it? Possible answers: b yes (give examples) b no because it’s been proved that no algorithm exists at all (e.g., Turing’s halting problem)because it’s been proved that no algorithm exists at all (e.g., Turing’s halting problem) because it’s been be proved that any algorithm takes exponential timebecause it’s been be proved that any algorithm takes exponential time b unknown
Problem Types: Optimization and Decision b Optimization problem: find a solution that maximizes or minimizes some objective function b Decision problem: answer yes/no to a question Many problems have decision and optimization versions. E.g.: traveling salesman problem b optimization: find Hamiltonian cycle of minimum length b decision: find Hamiltonian cycle of length m Decision problems are more convenient for formal investigation of their complexity.
Class P P: the class of decision problems that are solvable in O(p(n)) time, where p(n) is a polynomial of problem’s input size n Examples: b searching b element uniqueness b graph connectivity b graph acyclicity b primality testing (finally proved in 2002)
Class NP NP (nondeterministic polynomial): class of decision problems whose proposed solutions can be verified in polynomial time = solvable by a nondeterministic polynomial algorithm A nondeterministic polynomial algorithm is an abstract two-stage procedure that: b generates a random string purported to solve the problem b checks whether this solution is correct in polynomial time By definition, it solves the problem if it’s capable of generating and verifying a solution on one of its tries Why this definition? b led to development of the rich theory called “computational complexity”
Example: CNF satisfiability Problem: Is a boolean expression in its conjunctive normal form (CNF) satisfiable, i.e., are there values of its variables that makes it true? This problem is in NP. Nondeterministic algorithm: b Guess truth assignment b Substitute the values into the CNF formula to see if it evaluates to true Example: ( A | ¬B | ¬C ) & ( A | B ) & ( ¬B | ¬D | E ) & ( ¬D | ¬E ) Truth assignments: A B C D E 0 0 0 0 0...... 1 1 1 1 1 Checking phase: O(n)
What problems are in NP? b Hamiltonian circuit existence b Partition problem: Is it possible to partition a set of n integers into two disjoint subsets with the same sum? b Decision versions of TSP, knapsack problem, graph coloring, and many other combinatorial optimization problems. (Few exceptions include: MST, shortest paths) b All the problems in P can also be solved in this manner (but no guessing is necessary), so we have: P NP P NP b Big question: P = NP ?
NP-Complete Problems A decision problem D is NP-complete if it’s as hard as any problem in NP, i.e., b D is in NP b every problem in NP is polynomial-time reducible to D Cook’s theorem (1971): CNF-sat is NP-complete
NP-Complete Problems (cont.) Other NP-complete problems obtained through polynomial- time reductions from a known NP-complete problem Examples: TSP, knapsack, partition, graph-coloring and hundreds of other problems of combinatorial nature
P = NP ? Dilemma Revisited b P = NP would imply that every problem in NP, including all NP-complete problems, could be solved in polynomial time b If a polynomial-time algorithm for just one NP-complete problem is discovered, then every problem in NP can be solved in polynomial time, i.e., P = NP b Most but not all researchers believe that P NP, i.e. P is a proper subset of NP