CSC 325 Chapter 01, Spring 2012

Chapter 1 Analysis of Algorithms Algorithm Input Output An algorithm is a step-by-step procedure for solving a problem in a finite amount of time. Analysis of Algorithms 2 Running Time (§1.1) Most algorithms transform input objects into output objects. The running time of an algorithm typically grows with the input size. Average case time is often difficult to determine. We focus on the worst case running time.  Easier to analyze  Crucial to applications such as games, finance and robotics 0 20 40 60 80 100 120 Running Time 1000 2000 3000 4000 Input Size best case average case worst case Analysis of Algorithms 3 Limitations of Experiments It is necessary to implement the algorithm, which may be difficult Results may not be indicative of the running time on other inputs not included in the experiment. In order to compare two algorithms, the same hardware and software environments must be used Analysis of Algorithms 4 Theoretical Analysis Uses a high-level description of the algorithm instead of an implementation Characterizes running time as a function of the input size, n. Takes into account all possible inputs Allows us to evaluate the speed of an algorithm independent of the hardware/software environment Analysis of Algorithms 5 Pseudocode (§1.1) High-level description of an algorithm More structured than English prose Less detailed than a program Preferred notation for describing algorithms Hides program design issues Algorithm arrayMax(A, n) Input array A of n integers Output maximum element of A currentMax  A[0] for i  1 to n  1 do if A[i]  currentMax then currentMax  A[i] return currentMax Example: find max element of an array Analysis of Algorithms 6 Pseudocode Details Control flow  if … then … [else …]  while … do …  repeat … until …  for … do …  Indentation replaces braces Method declaration Algorithm method (arg [, arg…]) Input … Output … Method call var.method (arg [, arg…]) Return value return expression Expressions  Assignment (like  in Java)  Equality testing (like  in Java) n2 Superscripts and other mathematical formatting allowed Analysis of Algorithms 7 The Random Access Machine (RAM) Model A CPU An potentially unbounded bank of memory cells, each of which can hold an arbitrary number or character 012 Memory cells are numbered and accessing any cell in memory takes unit time. Analysis of Algorithms 8 Primitive Operations Basic computations performed by an algorithm Identifiable in pseudocode Largely independent from the programming language Exact definition not important (we will see why later) Assumed to take a constant amount of time in the RAM model Examples:  Evaluating an expression  Assigning a value to a variable  Indexing into an array  Calling a method  Returning from a method Analysis of Algorithms 9 Counting Primitive Operations (§1.1) By inspecting the pseudocode, we can determine the maximum number of primitive operations executed by an algorithm, as a function of the input size Algorithm arrayMax(A, n) # operations currentMax  A[0] 2 for i  1 to n  1 do 2 + n if A[i]  currentMax then 2(n  1) currentMax  A[i] 2(n  1) { increment counter i } 2(n  1) return currentMax 1 Total 7n  1 Analysis of Algorithms 10 Estimating Running Time Algorithm arrayMax executes 7n  1 primitive operations in the worst case. Define: a = Time taken by the fastest primitive operation b = Time taken by the slowest primitive operation Let T(n) be worst-case time of arrayMax. Then a (7n  1)  T(n)  b(7n  1) H h i i i b d d Analysis of Algorithms 11 Growth Rate of Running Time Changing the hardware/ software environment  Affects T(n) by a constant factor, but  Does not alter the growth rate of T(n) The linear growth rate of the running time T(n) is an intrinsic property of algorithm arrayMax Analysis of Algorithms 12 Growth Rates Growth rates of functions:  Linear  n  Quadratic  n2  Cubic  n3 In a log-log chart, the slope of the line corresponds to the growth rate of the function 1E+0 1E+2 1E+4 1E+6 1E+8 1E+10 1E+12 1E+14 1E+16 1E+18 1E+20 1E+22 1E+24 1E+26 1E+28 1E+30 1E+0 1E+2 1E+4 1E+6 1E+8 1E+10 n T (n ) Cubic Quadratic Linear Analysis of Algorithms 13 Constant Factors The growth rate is not affected by  constant factors or  lower-order terms Examples  102n + 105 is a linear function  105n2 + 108n is a quadratic function 1E+0 1E+2 1E+4 1E+6 1E+8 1E+10 1E+12 1E+14 1E+16 1E+18 1E+20 1E+22 1E+24 1E+26 1E+0 1E+2 1E+4 1E+6 1E+8 1E+10 n T (n ) Quadratic Quadratic Linear Linear Analysis of Algorithms 14 Big-Oh Notation (§1.2) Given functions f(n) and g(n), we say that f(n) is O(g(n)) if there are positive constants c and n0 such that f(n)  cg(n) for n  n0 Example: 2n + 10 is O(n)  2n + 10  cn  (c  2) n  10  n  10/(c  2)  Pick c = 3 and n0 = 10 1 10 100 1,000 10,000 1 10 100 1,000 n 3n 2n+ 10 n Analysis of Algorithms 15 Big-Oh Example Example: the function n2 is not O(n)  n2  cn  n  c  The above inequality cannot be satisfied since c must be a constant 1 10 100 1,000 10,000 100,000 1,000,000 1 10 100 1,000 n n^ 2 100n 10n n Analysis of Algorithms 16 More Big-Oh Examples 7n-2 7n-2 is O(n) need c > 0 and n0  1 such that 7n-2  c•n for n  n0 this is true for c = 7 and n0 = 1  3n3 + 20n2 + 5 3n3 + 20n2 + 5 is O(n3) need c > 0 and n0  1 such that 3n3 + 20n2 + 5  c•n3 for n  n0 this is true for c = 4 and n0 = 21  3 log n + log log n 3 log n + log log n is O(log n) need c > 0 and n0  1 such that 3 log n + log log n  c•log n for n  n0 this is true for c = 4 and n0 = 2 Analysis of Algorithms 17 Big-Oh and Growth Rate The big-Oh notation gives an upper bound on the growth rate of a function The statement “f(n) is O(g(n))” means that the growth rate of f(n) is no more than the growth rate of g(n) We can use the big-Oh notation to rank functions according to their growth rate f(n) is O(g(n)) g(n) is O(f(n)) g(n) grows more Yes No f(n) grows more No Yes Same growth Yes Yes Analysis of Algorithms 18 Big-Oh Rules If is f(n) a polynomial of degree d, then f(n) is O(nd), i.e., 1. Drop lower-order terms 2. Drop constant factors Use the smallest possible class of functions  Say “2n is O(n)” instead of “2n is O(n2)” Use the simplest expression of the class  Say “3n + 5 is O(n)” instead of “3n + 5 is O(3n)” Analysis of Algorithms 19 Asymptotic Algorithm Analysis The asymptotic analysis of an algorithm determines the running time in big-Oh notation To perform the asymptotic analysis  We find the worst-case number of primitive operations executed as a function of the input size  We express this function with big-Oh notation Example:  We determine that algorithm arrayMax executes at most 7n  1 primitive operations  We say that algorithm arrayMax “runs in O(n) time” Since constant factors and lower-order terms are eventually dropped anyhow, we can disregard them when counting primitive operations Analysis of Algorithms 20 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 0 5 10 15 20 25 30 35 1 2 3 4 5 6 7 X A Analysis of Algorithms 21 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 Analysis of Algorithms 22 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 0 1 2 3 4 5 6 7 1 2 3 4 5 6 Analysis of Algorithms 23 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 #operations A  new array of n integers n s  0 1 for i  0 to n  1 do n s  s + X[i] n A[i]  s / (i + 1) n return A 1 Algorithm prefixAverages2 runs in O(n) time Analysis of Algorithms 24 properties of logarithms: logb(xy) = logbx + logby logb (x/y) = logbx - logby logbxa = alogbx logba = logxa/logxb properties of exponentials: a(b+c) = aba c abc = (ab)c ab /ac = a(b-c) b = a logab bc = a c*logab Summations (Sec. 1.3.1) Logarithms and Exponents (Sec. 1.3.2) Proof techniques (Sec. 1.3.3) Basic probability (Sec. 1.3.4) Math you need to Review Analysis of Algorithms 25 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 Analysis of Algorithms 26 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) Analysis of Algorithms 27 Example Uses of the Relatives of Big-Oh 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 5n0 2  c•n0  given c, the n0 that satifies this is n0  c/5  0  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 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 (n2)