# COSC 3100 Fundamentals of Analysis of Algorithm Efficiency

COSC 3100 Fundamentals of Analysis of Algorithm Efficiency Instructor: Tanvir Outline (Ch 2) Input size, Orders of growth, Worst-case, Best-case, Average-case (Sec 2.1) Asymptotic notations and Basic efficiency classes (Sec 2.2) Analysis of Nonrecursive Algorithms

(Sec 2.3) Analysis of Recursive Algoithms (Sec 2.4) Example: Computing n-th Fibonacci Number (Sec 2.5) Empirical Analysis, Algorithm Visualization (Sec 2.6, 2.7) Analysis of Algoithms An investigation of an algorithms efficiency with respect to two resources: Running time (time efficiency or time complexity)

Memory space (space efficiency or space complexity) Analysis Framework Measuring an inputs size Measuring Running Time Orders of Growth (of algorithms efficiency function) Worst-case, Best-case, and Average-case efficiency Measuring Input Size Efficiency is defined as a function of input size Input size depends on the problem Example 1: what is the input size for sorting n numbers?

Example 2: what is the input size for evaluating Example 3: what is the input size for multiplying two nn matrices? Units of measuring running time Measure running time in milliseconds, seconds, etc. Depends on which computer Count the number of times each operation is executed Difficult and unnecessary Count the number of times an algorithms basic operation is

executed Measure running time in terms of # of basic operations Basic operation: the operation that contributes the most to the total running time of an algorithm Usually the most time consuming operation in the algorithms innermost loop Input size and basic operation examples Measure of input size

Basic operation Search for a key in a list of n items # of items in the list Key comparison Add two nn matrices Dimensions of the matrices, n

Addition Polynomial evaluation Order of the polynomial Multiplication Problem Theoretical Analysis of Time Efficiency Count the number of times the algorithms basic operation is executed on inputs of size n: C(n)

Input size T(n) cop C(n) Running time Execution time for basic operation Ignore cop, Focus on orders of growth # of times basic op. is executed

If C(n) = , how much longer it runs if the input size is doubled? Orders of Growth Why do we care about the order of growth of an algorithms efficiency function, i.e., the total number of basic operations? Euclid Consecutive s Integer Checking 3 13

gcd(31415, 14142) 10 14142 gcd(218922995834555169026, 135301852344706746049) 97 > 1020 gcd(60, 24) How fast efficiency function grows

as n gets larger and larger Orders of Growth (contd.) n lgn n nlg n n2 n3 2n

n! 10 3.3 10 3.310 102 103 103

3.6106 102 6.6 102 6.6102 104 106 1.3103 9.3101

0 57 103 10 103 10103 106 109

104 13 104 13104 108 1012 105 17 105

17105 1010 1015 106 20 106 20106 1012

1018 Orders of Growth (contd.) Plots of growth Consider only the leading term Ignore the constant coefficients Worst, Best, Average Cases Efficiency depends on input size n For some algorithms, efficiency depends on the type of input Example: Sequential Search Given a list of n elements and a search key k, find if k is in the list Scan list, compare elements with k

until either found a match (success), or list is exhausted (failure) Sequential Search Algorithm ALGORITHM SequentialSearch(A[0..n-1], k) //Input: A[0..n-1] and k //Output: Index of first match or -1 if no match is //found i <- 0 while i < n and A[i] k do i <- i+1 if i < n return i //A[i] = k else return -1

Different cases Worst case efficiency Efficiency (# of times the basic op. will be executed) for the worst case input of size n Runs longest among all possible inputs of size n Best case efficiency Runs fastest among all possible inputs of size n Average case efficiency Efficiency for a typical/random input of size n NOT the average of worst and best cases How do we find average case efficiency?

Average Case of Sequential Search Two assumptions: Probability of successful search is p (0 p 1) Search key can be at any index with equal probability (uniform distribution) Cavg(n) = Expected # of comparisons = Expected # of comparisons for success + Expected # of comparisons if k is not in the list Summary of Analysis Framework Time and space efficiencies are functions of input size

Time efficiency is # of times basic operation is executed Space efficiency is # of extra memory units consumed Efficiencies of some algorithms depend on type of input: requiring worst, best, average case analysis Focus is on order of growth of running time (or extra memory units) as input size goes to infinity Asymptotic Growth Rate 3 notations used to compare orders of growth of an algorithms basic operation count O(g(n)): Set of functions that grow no faster than g(n)

(g(n)): Set of functions that grow at least as fast as g(n) (g(n)): Set of functions that grow at the same rate as g(n) O(big oh)-Notation c g(n) t(n) Doesnt matter n n0 t(n) O(g(n))

O-Notation (contd.) Definition: A function t(n) is said to be in O(g(n)), denoted t(n) O(g(n)), if t(n) is bounded above by some positive constant multiple of g(n) for sufficiently large n. If we can find +ve constants c and n0 such that: t(n) c g(n) for all n n0 O-Notation (contd.) Is 100n+5 O(n2) ? Is 2n+1 O(2n) ? Is 22n O(2n) ? Is n(n-1) O(n2) ?

(big omega)-Notation t(n) c g(n) Doesnt matter n n0 t(n) (g(n)) -Notation (contd.) Definition: A function t(n) is said to be in (g(n)) denoted t(n) (g(n)), if t(n) is bounded below by

some positive constant multiple of g(n) for all sufficiently large n. If we can find +ve constants c and n0 such that t(n) c g(n) for all n n0 -Notation (contd.) Is n3 (n2) ? Is 100n+5 (n2) ? Is n(n-1) (n2) ? (big theta)-Notation c1 g(n) t(n) c2 g(n) Doesnt

matter n n0 t(n) (g(n)) -Notation (contd.) Definition: A function t(n) is said to be in (g(n)) denoted t(n) (g(n)), if t(n) is bounded both above and below by some positive constant multiples of g(n) for all sufficiently large n. If we can find +ve constants c1, c2, and n0 such that

c2g(n) t(n) c1g(n) for all n n0 -Notation (contd.) Is n(n-1) (n2) ? Is n2+sin(n) (n2) ? Is an2+bn+c (n2) for a > 0? Is (n+a)b (nb) for b > 0 ? O, , and () (g(n)): functions that grow at least as fast as g(n) g(n) (=) (g(n)): functions that grow at the same rate as g(n)

() O(g(n)): functions that grow no faster that g(n) Theorem If t1(n) O(g1(n)) and t2(n) O(g2(n)), then t1(n) + t2(n) O(max{g1(n),g2(n)}) Analogous assertions are true for and notations. Implication: if sorting makes no more than n2 comparisons and then binary search makes no more than log2n comparisons, then efficiency is O(max{n2, log2n}) = O(n2) Theorem (contd.)

If t1(n) O(g1(n)) and t2(n) O(g2(n)), then t1(n) + t2(n) O(max{g1(n),g2(n)}) t1(n) c1g1(n) for n n01 and t2(n) c2g2(n) for n n02 t1(n) + t2(n) c1g1(n) + c2g2(n) for n max { n01, n02 } t1(n) + t2(n) max{c1, c2} g1(n) + max{c1, c2} g2(n) for n max { n01, n02 } t1(n) + t2(n) 2max{c1, for c2}nmax{g max{n 1(n),01, n02} c g2(n) }, n0

Using Limits for Comparing Orders of Growth = 0 implies that t(n) grows slower than g(n) c implies that t(n) grows at the same order as g(n) implies that t(n) grows faster than g(n) 1. First two cases (0 and c) means t(n) O(g(n)) 2. Last two cases (c and ) means t(n) (g(n)) 3. The second case (c) means t(n) (g(n)) Taking Limits LHpitals rule: if limits of both

t(n) and g(n) are as n goes to , Stirlings formula: For large n, we have n! Stirlings Formula n! = n (n-1) (n-2) 1 ln(n!) = n! Trapezoidal rule: 1 234 n Add to both sides, Exponentiate both sides, n! C nne-nn1/2

=C Taking Limits (contd.) Compare orders of growth of 22n and 3n Compare orders of growth of lgn and Compare orders of growth of n! and 2n Establish Order of Growth of Basic Operation Count Method 1: Using Limits Method 2: Using Theorem Method 3: Using definitions of O-, -, and -notations.

Basic Efficiency Classes Class Name Comments 1 constant May be in best cases lgn logarithmic

Halving problem size at each iteration n linear Scan a list of size n nlgn linearithmic Divide and conquer algorithms, e.g., mergesort

n2 quadratic Two embedded loops, e.g., selection sort n3 cubic Three embedded loops, e.g., matrix multiplication 2n

exponential All subsets of nelements set n! factorial All permutations of an n-elements set Important Summation Formulas Analysis of Nonrecursive

Algorithms ALGORITHM MaxElement(A[0..n-1]) //Determines largest element maxval <- A[0] Input size: n for i <- 1 to n-1 doBasic operation: > or maxval maxval <- A[i] return maxval C(n) = (n) of Nonrecursive algorithms Decide on a parameter indicating an inputs size Identify the basic operation Does C(n) depends only on n or does it

also depend on type of input? Set up sum expressing the # of times basic operation is executed. Find closed form for the sum or at least establish its order of growth Analysis of Nonrecursive (contd.) ALGORITHM UniqueElements(A[0..n-1]) //Determines whether all elements are //distinct for i <- 0 to n-2 do for j <- i+1 to n-1 do if A[i] = A[j] return false return true Input size: n

Basic operation: A[i] = A[j] Does C(n) depend on type of input? UniqueElements (contd.) for i <- 0 to n-2 do for j <- i+1 to n-1 do if A[i] = A[j] return false return true Cworst(n) = Why Cworst(n) is better than saying Cworst(n) (contd.) ALGORITHM MatrixMultiplication(A[0..n-1, 0..n-1], B[0..n-1, 0..n-1])

//Output: C = AB T(n) cmM(n)+caA(n) for i <- 0 to n-1 do = cmn3+can3 for j <- 0 to n-1 do C[i, j] = 0.0 = (cm+ca)n3 for k <- 0 to n-1 do C[i, j] = C[i, j] + A[i, k]B[k, j] return C Input size: n Basic operation: M(n) = (contd.) ALGORITHM Binary(n)

// Output: # of digits in // representation of n count <- 1 while n > 1 do count <- count+1 n

Analysis of Nonrecursive (contd.) ALGORITHM Mystery(n) S <- 0 for i <- 1 to n do S <- S + ii What does this algorithm compute? return S What is the basic operation? How many times is the basic operation executed? Whats its efficiency class? Can you improve it further, or Can you prove that no improvement is possible? Analysis of Nonrecursive

(contd.) ALGORITHM Secret(A[0..n-1]) r <- A[0] What does this algorithm compute? s <- A[0] for i <- 1 to n-1 do What is the basic operation? if A[i] < r How many times is the basic operation executed? r <- A[i] if A[i] > s Whats its efficiency class? s <- A[i] return s-r Can you improve it further, Or Could you prove that no improvement is possible?

Analysis of Recursive Algorithms ALGORITHM F(n) // Output: n! Input size: n Basic operation: if n = 0 return 1 M(n) = M(n-1) + 1 for n > 0 else to compute F(n-1) return nF(n-1) to multiply n and F(n-1)

Analysis of Recursive (contd.) M(n) = M(n-1) + 1 Recurrence relation M(0) = 0 By Backward substitution M(n) = M(n-1) + 1 = M(n-2)+ 1 + 1 = = M(n-n) + 1 + + 1 = 0 + n = n n 1s We could compute it nonrecursively, saves function call overhead Recursive Decide on input size parameter

Identify the basic operation Does C(n) depends also on input type? Set up a recurrence relation Solve the recurrence or, at least establish the order of growth of its solution Analysis of Recursive (contd.): Tower of Hanoi Tower of Hanoi (contd.) M(n) = M(n-1) + 1 + M(n-1) M(1) = 1 1

Tower of Hanoi (contd.) ALGORITHM ToH(n, s, m, d) M(1) = 1 if n = 1 print move from s to d else M(n-1) ToH(n-1, s, d, m) print move from s to d ToH(n-1, m, s, d) M(n-1) M(1) Tower of Hanoi (contd.) M(n) = 2M(n-1) + 1 for n > 1

M(1) = 1 M(n) = 2M(n-1) + 1 [Backward substitution] = 2[ 2M(n-2)+1] + 1 = 22M(n-2)+2+1 = 22[2M(n-3)+1]+2+1 = 23M(n-3)+ 22+2+1 = 2n-1M(n-(n-1))+2n-2+2n-3++22+2+1 = 2n-1+ 2n-2+2n-3++22+2+1 = = 2n-1 M(n) (2n) Be careful! Recursive algorithm may look simple But might easily be exponential in complexity. Tower of Hanoi (contd.) Recursion tree (# of function calls) n

n-1 n-2 n-2 n-2 n-2 2 1

n-1 2 2 1 1 1 1 C(n) = = 2n-1 2 1

1 1 Analysis of Recursive (contd.) ALGORITHM BinRec(n) //Output: # of binary digits in ns //binary representation if n = 1 Input size: n = 2 Basic operation: + return 1 else return BinRec()+1 k

A(n) = A(2k) = A(2k-1) + 1 = [A(2k-2) + 1] + 1 = = A(2k-k)+k = A(1) + k = k = lg(n) A(n) (lg(n)) Analysis of Recursive (contd.) Maximum how many slices of pizza can a person obtain by making n straight cuts with a pizza knife? L0 = 1 L2 = 4

L1 = 2 1 1, L3 = 7 Bonus problem in HW2 2 2, 3 4, difference 7, Ln = Ln-1 + n for n > 0

L0 = 1 EXAMPLE: nth Fibonacci Number 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, F(n) = F(n-1) + F(n-2) for n > 1 F(0) = 0, F(1) = 1 Lets try backward substitution F(n) = F(n-2)+F(n-3)+F(n-3)+F(n-4) = F(n-2)+2F(n-3)+F(n-4) = F(n-3)+3F(n-4)+3F(n-5)+F(n-6) nth Fibonacci (contd.) ax(n)+bx(n-1)+cx(n-2) = 0 Homogeneous linear second-order recurrence with constant coefficients

Guess: x(n) = rn arn+brn-1+crn-2 = 0 Characteristic equation ar2+br+c = 0 THEOREM: If r1, r2 are two real distinct roots of characteristic equation then x(n) = rr1n+rr2n where r, r are arbitrary constants nth Fibonacci (contd.) F(n)-F(n-1)-F(n-2) =0 r2-r-1 = 0 => r1,2 = F(n) F(0) F(1)

F(n) = r+r = r+r= 0 => r+r=0 = r+r= 1 == where and = - nth Fibonacci (contd.) F(n) = where 1.61803 and 0.61803 A(n) = A(n-1)+A(n-2)+1 F(n) (n) A(0) = 0, A(1) = 0 ALGORITHM F(n) [A(n)+1] [A(n-1)+1]-[A(n-2)+1]=0 Let B(n) = A(n)+1 if n 1

B(n) B(n-1)-B(n-2)=0 return n B(0)=1, B(1)=1 else Notice, B(n) = F(n+1) = B(n)-1=F(n+1)-1 return F(n-1)+F(n-2) A(n) = -1 A(n) (n) Basic op: + nth Fibonacci (contd.) F(5) F(3) F(4)

F(3) F(2) F(1) F(2) F(1) F(1) F(1) F(2) F(0) F(1) F(0)

F(0) Only n-1 additions, (n)! ALGORITHM Fib(n) F[0] <- 0 F[1] <- 1 for i <- 2 to n do F[i] <- F[i-1]+F[i-2] return F[n] ALGORITHM Fib(n) f <- 0 fnext <- 1 for i <- 2 to n do tmp <- fnext fnext <- fnext+f f <- tmp

return fnext Empirical Analysis Mathematical Analysis Advantages Machine and input independence Disadvantages Average case analysis is hard Empirical Analysis Advantages Applicable to any algorithm Disadvantages Machine and input dependency

Empirical Analysis (contd.) General Plan Understand Experiments purpose: What is the efficiency class? Compare two algorithms for same problem Efficiency metric: Operation count vs. time unit Characteristic of input sample (range, size, etc.) Write a program Generate sample inputs Run on sample inputs and record data Analyze data Empirical Analysis

(contd.) Insert counters into the program to count C(n) Or, time the program, (tfinish-tstart), in Java System.currentTimeMillis() Typically not very accurate, may vary Remedy: make several measurements and take mean or median May give 0 time, remedy: run in a loop and take mean Empirical Analysis (contd.) Increase sample size systematically, like 1000, 2000, 3000, , 10000 Impact is easier to analyze

Better is to generate random sizes within desired range Because, odd sizes may affect Pseudorandom number generators In Java Math.random() gives uniform random number in [0, 1) If you need number in [min, max]: min + (int)( Math.random() * ( (max-min) + 1 ) ) Empirical Analysis (contd.) Pseudorandom in [min, max] Math.random() * (max-min) gives [0, maxmin) min + Math.random() * (max-min) gives [min, max)

(int)[Math.random() * (max-min+1)] gives [0, max-min] min+ (int)[Math.random() * (max-min+1)] gives [min, max] To get [5, 10] (int)Math.random()*(10-5+1) gives [0, 10-5] or [0, 5] 5+ (int)Math.random()*(10-5+1) gives [5, 10] Empirical Analysis (contd.) Profiling: find out most time-consuming portion, Eclipse and Netbeans have it Tabulate data n

M(n) g(n) M(n)/g(n) If M(n)/g(n) converges to a +ve constant, we know M(n) (g(n)) Or look at M(2n)/M(n); e.g., if M(n) (lgn) then M(2n)/M(n) will be low Empirical Analysis (contd.) Scatterplot count/time

count/time M(n) can lgM(n) nlga + lgc n n lgn n count/time nlgn or n2 n

Algorithm Visualization (contd.) Sequence of still pictures or animation Sorting Out Sorting on YouTube Nice animation of 9 sorting algorithms Summary Time and Space Efficiency C(n): Count of # of times the basic operation is executed for input of size n C(n) may depend on type of input and then we need worst, average, and best case analysis Usually order of growth (O, , ) is all that

matters: logarithmic, linear, linearithmic, quadratic, cubic, and exponential Input size, basic op., worst case?, sum or recurrence, solve it We run on computers for empirical analysis Homework 2 DUE ON FEBRUARY 07 (TUESDAY) IN CLASS

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