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CSC401 – Analysis of Algorithms Lecture Notes 2 Asymptotic Analysis Objectives: Mathematics foundation for algorithm analysis Amortization analysis techniques Case studies Asymptotic notions: Big-Oh, Big-Omega, Little-oh, little-omega, Big-Theta Summations Summation: b f (i) f (a) f (a 1) f (a 2) f (b) i a 1 a n1 a 1 a a a Geometric summation: 1 a i 0 1 i 2 n a 1 a a a 1 a i 0 n i 2 n n Natural summation: i 1 2 3 n i 0 n(n 1) 2 1 1 1 1 Harmonic number: 1 ln n 2 3 n i 0 i Split summation: n k n i 0 i 1 i k 1 f (i) f (i) f (i) 2 Logarithms and Exponents Logarithms – properties of logarithms: logb(xy) = logbx + logby logb (x/y) = logbx - logby logbxa = alogbx logba = logxa/logxb Exponents – properties of exponentials: a(b+c) = aba c abc = (ab)c ab /ac = a(b-c) b = a logab bc = a c*logab Floor and ceiling functions – – x the largest integer less than or equal to x x the smallest integer greater than or equal to x 3 Proof Techniques By Example -- counterexample Contrapositive – Principle: To justify “if p is true, then q is true”, show “if q is not true, then p is not true”. – Example: To justify “if ab is odd, a is odd or b is even”, assume “’a is odd or b is odd’ is not true”, then “a is even and b is odd”, then “ab is even”, then “’ab is odd’ is not true” Contradiction – Principle: To justify “p is true”, show “if p is not true, then there exists a contradiction”. – Example: To justify “if ab is odd, then a is odd or b is even”, let “ab is odd”, and assume “’a is odd or b is even’ is not true”, then “a is even and b is odd”, thus “ab is even” which is a contradiction to “ab is odd”. 4 Proof by Induction Induction -- To justify S(n) for n>=n0 – Principle Base cases: Justify S(n) is true for n0<=n<=n1 Assumption: Assume S(n) is true for n=N>=n1; or Assume S(n) is true for n1<=n<=N Induction: Justify S(n) is true for n=N+1 – Example: Question: Fibonacci sequence is defined as F(1)=1, F(2)=1, and F(n)=F(n-1)+F(n-2) for n>2. Justify F(n)<2^n for all n>=1 Proof by induction where n0=1. Let n1=2. – Base cases: For n0<=n<=n1, n=1 or n=2. If n=1, F(1)=1<2=2^1. If n=2, F(2)=2<4=2^2. It holds. – Assumption: Assume F(n)<2^n for n=N>=2. – Induction: For n=N+1, F(N+1)=F(N)+F(N1)<2^N+2^(N-1)<2 2^N=2^(N+1). It holds. 5 Proof by Loop Invariants Principle: – To prove a statement S about a loop is correct, define S in terms of a series of smaller statements S0, S1, …, Sk, where The initial claim S0 is true before the loop begins If Si-1 is true before iteration i begins, then show that Si is true after iteration i is over The final statement Sk is true Example: Consider the algorithm arrayMax – Statement S: max is the maximum number when finished – A series of smaller statements: Si: max is the maximum Algorithm arrayMax(A, n) in the first i+1 elements max A[0] of the array for i 1 to n 1 do S0 is true before the loop if A[i] max then If Si is true, then easy to max A[i] return max show Si+1 is true 6 S=Sn-1 is also true Basic Probability Sample space: the set of all possible outcomes from some experiment Probability space: a sample space S together with a probability function that maps subsets of S to real numbers between 0 and 1 Event: Each subset of A of S called an event Properties of the probability function Pr – – – – Pr(Ø)=0 Pr(S)=1 0<=Pr(A)<=1 for any subset A of S If A and B are subsets of S and AB=, then Pr(AB)=Pr(A)+Pr(B) Independence – A and B are independent if Pr(AB)=Pr(A)Pr(B) – A1, A2, …, An are mutually independent if Pr(A1A2… An)=Pr(A1)Pr(A2)…Pr(An) 7 Basic Probability Conditional probability – The conditional probability that A occurs, given B, is defined as: Pr(A|B)=Pr(AB)/Pr(B), assuming Pr(B)>0 Random variables – Intuitively, Variables whose values depend on the outcomes of some experiment – Formally, a function X that maps outcomes from some sample space S to real numbers – Indicator random variable: a variable that maps outcomes to either 0 or 1 Expectation:a random variable has random values – Intuitively, average value of a random variable – Formally, expected value of a random variable X is defined as E(X)=xxPr(X=x) – Properties: Linearity: E(X+Y) = E(X) + E(Y) Independence: If X and Y are independent, that is, Pr(X=x|Y=y)=Pr(X=x), then E(XY)=E(X)E(Y) 8 Computing Prefix Averages We further illustrate asymptotic analysis with two algorithms for prefix averages The i-th prefix average of an array X is average of the first (i 1) elements of X: A[i] (X[0] X[1] … X[i])/(i+1) Computing the array A of prefix averages of another array X has applications to financial analysis 35 30 X A 25 20 15 10 5 0 1 2 3 4 5 6 7 9 Prefix Averages (Quadratic) The following algorithm computes prefix averages in quadratic time by applying the definition Algorithm prefixAverages1(X, n) Input array X of n integers Output array A of prefix averages of X #operations A new array of n integers n for i 0 to n 1 do n s X[0] n for j 1 to i do 1 2 … (n 1) s s X[j] 1 2 … (n 1) A[i] s / (i 1) n return A 1 10 Arithmetic Progression The running time of prefixAverages1 is O(1 2 … n) The sum of the first n integers is n(n 1) / 2 – There is a simple visual proof of this fact Thus, algorithm prefixAverages1 runs in O(n2) time 7 6 5 4 3 2 1 0 1 2 3 4 5 6 11 Prefix Averages (Linear) The following algorithm computes prefix averages in linear time by keeping a running sum Algorithm prefixAverages2(X, n) Input array X of n integers Output array A of prefix averages of X A new array of n integers s0 for i 0 to n 1 do s s X[i] A[i] s / (i 1) return A #operations n 1 n n n 1 Algorithm prefixAverages2 runs in O(n) time 12 Amortization Amortization – Typical data structure supports a wide variety of operations for accessing and updating the elements – Each operation takes a varying amount of running time – Rather than focusing on each operation – Consider the interactions between all the operations by studying the running time of a series of these operations – Average the operations’ running time Amortized running time – The amortized running time of an operation within a series of operations is defined as the worst-case running time of the series of operations divided by the number of operations – Some operations may have much higher actual running time than its amortized running time, while some others 13 have much lower The Clearable Table Data Structure The clearable table – An ADT Storing a table of elements Being accessing by their index in the table – Two methods: add(e) -- add an element e to the next available cell in the table clear() -- empty the table by removing all elements Consider a series of operations (add and clear) performed on a clearable table S – Each add takes O(1) – Each clear takes O(n) – Thus, a series of operations takes O(n2), because it may consist of only clears 14 Amortization Analysis Theorem: – A series of n operations on an initially empty clearable table implemented with an array takes O(n) time Proof: – Let M0, M1, …, Mn-1 be the series of operations performed on S, where k operations are clear – Let Mi0, Mi1, …, Mik-1 be the k clear operations within the series, and others be the (n-k) add operations – Define i-1=-1,Mij takes ij-ij-1, because at most ij-ij-1-1 elements are added by add operations between Mij-1 and Mij – The total time of the series is: (n-k) + ∑k-1j=0(ij-ij-1) = n-k + (ik-1 - i-1) <= 2n-k Total time is O(n) Amortized time is O(1) 15 Accounting Method The method – Use a scheme of credits and debits: each operation pays an amount of cyber-dollar – Some operations overpay --> credits – Some operations underpay --> debits – Keep the balance at any time at least 0 Example: the clearable table – Each operation pays two cyber-dollars – add always overpays one dollar -- one credit – clear may underpay a variety of dollars the underpaid amount equals the number of add operations since last clear - 2 – Thus, the balance is at least 0 – So the total cost is 2n -- may have credits 16 Potential Functions Based on energy model – associate a value with the structure, Ø, representing the current energy state – each operation contributes to Ø a certain amount of energy t’ and consumes a varying amount of energy t – Ø0 -- the initial energy, Øi -- the energy after the i-th operation – for the i-th operation ti -- the actual running time t’i -- the amortized running time ti = t’i + Øi-1 - Øi – Overall: T=∑(ti), T’=∑(t’i) – The total actual time T = T’ + Ø0 - Øn – As long as Ø0 <= Øn, T<=T’ 17 Potential Functions Example -- The clearable table – the current energy state Ø is defined as the number of elements in the table, thus Ø>=0 – each operation contributes to Ø t’=2 – Ø0 = 0 – Øi = Øi-1 + 1, if the i-th operation is add add: t = t’+ Øi-1 - Øi = 2 - 1 = 1 – Øi = 0, if the i-th operation is clear clear: t =t’+ Øi-1 = 2 + Øi-1 – Overall: T=∑ (t), T’=∑(t’) – The total actual time T = T’ + Ø0 - Øn – Because Ø0 = 0 <= Øn, T<=T’ =2n 18 Extendable Array ADT: Extendable array is an array with extendable size. One of its methods is add to add an element to the array. If the array is not full, the element is added to the first available cell. When the array is full, the add method performs the following – – – – Allocate a new array with double size Copy elements from the old array to the new array Add the element to the first available cell in the new array Replace the old array with the new array Question: What is the amortization time of add? – Two situations of add: add -- O(1) Extend and add -- (n) – Amortization analysis: Each add deposits 3 dollars and spends 1 dollar Each extension spends k dollars from the size k to 2k (copy) Amortization time of add is O(1) 19 Relatives of Big-Oh big-Omega – f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) c•g(n) for n n0 big-Theta – f(n) is (g(n)) if there are constants c’ > 0 and c’’ > 0 and an integer constant n0 1 such that c’•g(n) f(n) c’’•g(n) for n n0 little-oh – f(n) is o(g(n)) if, for any constant c > 0, there is an integer constant n0 0 such that f(n) c•g(n) for n n0 little-omega – f(n) is (g(n)) if, for any constant c > 0, there is an integer constant n0 0 such that f(n) c•g(n) for n n0 20 Intuition for Asymptotic Notation Big-Oh – f(n) is O(g(n)) if f(n) is asymptotically less than or equal to g(n) big-Omega – f(n) is (g(n)) if f(n) is asymptotically greater than or equal to g(n) big-Theta – f(n) is (g(n)) if f(n) is asymptotically equal to g(n) little-oh – f(n) is o(g(n)) if f(n) is asymptotically strictly less than g(n) little-omega – f(n) is (g(n)) if is asymptotically strictly greater than g(n) 21 Example Uses of the Relatives of Big-Oh 5n2 is (n2) f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) c•g(n) for n n0 let c = 5 and n0 = 1 5n2 is (n) f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) c•g(n) for n n0 let c = 1 and n0 = 1 5n2 is (n) f(n) is (g(n)) if, for any constant c > 0, there is an integer constant n0 0 such that f(n) c•g(n) for n n0 need 5n02 c•n0 given c, the n0 that satisfies this is n0 c/5 0 22