COSC 3100 Fundamentals of Analysis

of Algorithm EfficiencyInstructor: 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 algorithm’s

efficiency with respect to two resources:– Running time (time efficiency or time

complexity)– Memory space (space efficiency or

space complexity)

Analysis Framework• Measuring an input’s size• Measuring Running Time• Orders of Growth (of algorithm’s

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 n×n 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 algorithm’s “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 algorithm’s innermost loop

Input size and basic operation examples

Problem 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 n×n matrices

Dimensions of the matrices, n

Addition

Polynomial evaluation

Order of the polynomial

Multiplication

Theoretical Analysis of Time Efficiency

• Count the number of times the algorithm’s basic operation is executed on inputs of size n: C(n)

T(n) ≈ cop × C(n)

Execution time for basic operation

# of times basic op. is executedRunning time

Input size Ignore cop,Focus on orders of growth

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 algorithm’s efficiency function, i.e., the total number of basic operations? Euclid’

sConsecutive

Integer Checking

gcd(60, 24) 3 13gcd(31415, 14142) 10 14142gcd(218922995834555169026, 135301852344706746049)

97 > 1020

How fast efficiency function grows as n gets larger and larger…

Orders of Growth (contd.)n lgn n n×lg

nn2 n3 2n n!

10 3.3 10 3.3×10 102 103 103 3.6×106

102 6.6 102 6.6×102 104 106 1.3×1030

9.3×10157

103 10 103 10×103 106 109

104 13 104 13×104 108 1012

105 17 105 17×105 1010 1015

106 20 106 20×106 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 <- 0while i < n and A[i] ≠ k do

i <- i+1if i < n

return i //A[i] = kelse

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 algorithm’s basic operation count– O(g(n)): Set of functions that grow no

faster than g(n)– Ω(g(n)): Set of functions that grow at

least as fast as g(n)– Θ(g(n)): Set of functions that grow at

the same rate as g(n)

O(big oh)-Notation

Doesn’tmatter

nn0

c × g(n)t(n)

t(n) є O(g(n))

O-Notation (contd.)• Definition: A function t(n) is said

to be in O(g(n)), denoted t(n) є 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 є O(n2) ?

• Is 2n+1 є O(2n) ? Is 22n є O(2n) ?

• Is n(n-1) є O(n2) ?

Ω(big omega)-Notation

Doesn’tmatter

n0

t(n)c × g(n)

n

t(n) є Ω(g(n))

Ω-Notation (contd.)• Definition: A function t(n) is said

to be in Ω(g(n)) denoted t(n) є Ω(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

Ω-Notation (contd.)

• Is n3 є Ω(n2) ?

• Is 100n+5 є Ω(n2) ?

• Is n(n-1) є Ω(n2) ?

Θ(big theta)-Notation

Doesn’tmatter

n0n

t(n) є Θ(g(n))

c1 × g(n)

c2 × g(n)t(n)

Θ-Notation (contd.)• Definition: A function t(n) is said to

be in Θ(g(n)) denoted t(n) є Θ(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

c2×g(n) ≤ t(n) ≤ c1×g(n) for all n ≥ n0

Θ-Notation (contd.)• Is n(n-1) є Θ(n2) ?

• Is n2+sin(n) є Θ(n2) ?

• Is an2+bn+c є Θ(n2) for a > 0?

• Is (n+a)b Θ(nb) for b > 0 ?

O, Θ, and Ω

g(n)

(≥) Ω(g(n)): functions that grow at least as fast as g(n)

(=) Θ(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) є O(g1(n)) and t2(n) є O(g2(n)),

then t1(n) + t2(n) є O(maxg1(n),g2(n))

• Analogous assertions are true for Ω and Θ notations.

• Implication: if sorting makes no more than n2 comparisons and then binary search makes no more than log2n comparisons, then efficiency is O(maxn2, log2n) = O(n2)

Theorem (contd.)• If t1(n) є O(g1(n)) and t2(n) є

O(g2(n)), then t1(n) + t2(n) є O(maxg1(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) ≤ maxc1, c2 g1(n) + maxc1, c2 g2(n) for n ≥ max n01, n02 t1(n) + t2(n) ≤ 2maxc1, c2 maxg1(n), g2(n) , c

for n ≥ maxn01, n02n0

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) є O(g(n))

2. Last two cases (c and ∞) means t(n) є Ω(g(n))

3. The second case (c) means t(n) є Θ(g(n))

Taking Limits• L’Hôpital’s rule: if limits of both

t(n) and g(n) are ∞ as n goes to ∞,

• Stirling’s formula: For large n, we have n! ≈

Stirling’s Formula• n! = n (n-1) (n-2) … 1• ln(n!) =

• Trapezoidal rule: • Add to both sides,• • Exponentiate both sides, n! ≈ C nne-nn1/2

= C

n! ≈

…

1 2 3 4 n

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

Summary of How to Establish Order of Growth of Basic Operation Count

• Method 1: Using Limits• Method 2: Using Theorem• Method 3: Using definitions of O-,

Ω-, and Θ-notations.

Basic Efficiency ClassesClass Name Comments

1 constant May be in best caseslgn logarithmic Halving problem size

at each iterationn linear Scan a list of size n

n×lgn linearithmic Divide and conquer algorithms, e.g.,

mergesortn2 quadratic Two embedded loops,

e.g., selection sortn3 cubic Three embedded

loops, e.g., matrix multiplication

2n exponential All subsets of n-elements 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 elementmaxval <- A[0]for i <- 1 to n-1 do

if A[i] > maxvalmaxval <- A[i]

return maxval

Input size: nBasic operation: > or <-

C(n) = є Θ(n)

General Plan for Analysis of Nonrecursive

algorithms• Decide on a parameter indicating an

input’s 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 //distinctfor i <- 0 to n-2 do

for j <- i+1 to n-1 doif A[i] = A[j]

return falsereturn true

Input size: nBasic 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 doif A[i] = A[j]

return falsereturn true

Cworst(n) =

Why Cworst(n) is better thansaying Cworst(n)

Analysis of Nonrecursive (contd.)

ALGORITHM MatrixMultiplication(A[0..n-1, 0..n-1], B[0..n-1, 0..n-1])//Output: C = ABfor i <- 0 to n-1 do

for j <- 0 to n-1 doC[i, j] = 0.0for k <- 0 to n-1 do

C[i, j] = C[i, j] + A[i, k]×B[k, j]return C

Input size: nBasic operation: ×M(n) =

T(n) ≈ cmM(n)+caA(n) = cmn3+can3

= (cm+ca)n3

Analysis of Nonrecursive (contd.)

ALGORITHM Binary(n)// Output: # of digits in binary // representation of ncount <- 1while n > 1 do

count <- count+1n <-

return count

Input size: nBasic operation: C(n) = ?

Each iteration halves nLet m be such that 2m ≤ n < 2m+1

Take lg: m ≤ lg(n) < m+1, som = and m+1 = +1 є Θ(lg(n))

Analysis of Nonrecursive (contd.)

ALGORITHM Mystery(n)S <- 0for i <- 1 to n do

S <- S + i×ireturn S What does this algorithm compute?

What is the basic operation?

How many times is the basic operation executed?

What’s its efficiency class?Can you improve it further, orCan you prove that no improvementis possible?

Analysis of Nonrecursive (contd.)

ALGORITHM Secret(A[0..n-1])r <- A[0]s <- A[0]for i <- 1 to n-1 do

if A[i] < rr <- A[i]

if A[i] > ss <- A[i]

return s-r

What does this algorithm compute?

What is the basic operation?

How many times is the basic operation executed?

What’s its efficiency class?

Can you improve it further, Or Could you prove that no improvement is possible?

Analysis of Recursive Algorithms

ALGORITHM F(n)// Output: n!if n = 0

return 1else

return n×F(n-1)

Input size: nBasic operation: ×

M(n) = M(n-1) + 1 for n > 0

to compute F(n-1) to multiply

n and F(n-1)

Analysis of Recursive (contd.)

M(n) = M(n-1) + 1M(0) = 0

M(n) = M(n-1) + 1 = M(n-2)+ 1 + 1 = … …= M(n-n) + 1 + … + 1 = 0 + n = n

We could compute it nonrecursively, saves function call overhead

Recurrence relation

n 1’s

By Backward substitution

General Plan for 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-1)

1

M(n-1)

M(n) = M(n-1) + 1 + M(n-1)M(1) = 1

Tower of Hanoi (contd.)ALGORITHM ToH(n, s, m, d)if n = 1

print “move from s to d”else

ToH(n-1, s, d, m)print “move from s to d”ToH(n-1, m, s, d)

M(1) = 1

M(n-1)

M(1)

M(n-1)

Tower of Hanoi (contd.)M(n) = 2M(n-1) + 1 for n > 1M(1) = 1M(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) є Θ(2n)Be careful! Recursive algorithm may look simpleBut might easily be exponential in complexity.

Tower of Hanoi (contd.)• Recursion tree (# of function calls)

n

n-1 n-1

n-2 n-2 n-2 n-2

… … …2

112

11

211

211

C(n) = = 2n-1

Analysis of Recursive (contd.)

ALGORITHM BinRec(n)//Output: # of binary digits in n’s //binary representationif n = 1

return 1else

return BinRec()+1

Input size: n = 2k

Basic operation: +

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) є Θ(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 L1 = 2 L2 = 4

L3 = 7

1, 2, 4, 7, …1 2 3

Ln = Ln-1 + n for n > 0L0 = 1

difference

Bonus problem in HW2

EXAMPLE: nth Fibonacci Number

• 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …F(n) = F(n-1) + F(n-2) for n > 1F(0) = 0, F(1) = 1Let’s 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) = 0Homogeneous linear second-order recurrence with constant coefficientsGuess: x(n) = rn

arn+brn-1+crn-2 = 0ar2+br+c = 0 THEOREM: If r1, r2 are two real distinct roots of characteristic equation then x(n) = αr1

n+βr2n where α, β are arbitrary constants

Characteristic equation

nth Fibonacci (contd.)F(n)-F(n-1)-F(n-2) = 0r2-r-1 = 0 => r1,2 = F(n) = α+βF(0) = α+β= 0 => α+β=0F(1) = α+β= 1 F(n) = =

where and = -

nth Fibonacci (contd.)• F(n) = where ≈ 1.61803 and ≈ -

0.61803• F(n) є Θ(φn) ALGORITHM F(n)if n ≤ 1

return nelse

return F(n-1)+F(n-2)

Basic op: +

A(n) = A(n-1)+A(n-2)+1A(0) = 0, A(1) = 0

[A(n)+1] –[A(n-1)+1]-[A(n-2)+1]=0Let B(n) = A(n)+1

B(n) –B(n-1)-B(n-2)=0B(0)=1, B(1)=1

Notice, B(n) = F(n+1)A(n) = B(n)-1=F(n+1)-1= -1

A(n) є Θ(φn)

nth Fibonacci (contd.)F(5)

F(4) F(3)

F(2) F(1)

F(0)F(1)

F(2)F(3)

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

F(0)F(1)Only n-1 additions, Θ(n)!

ALGORITHM Fib(n)F[0] <- 0F[1] <- 1for i <- 2 to n do

F[i] <- F[i-1]+F[i-2]return F[n]

ALGORITHM Fib(n)f <- 0fnext <- 1for i <- 2 to n do

tmp <- fnextfnext <- fnext+ff <- 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 Experiment’s 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, max-

min)– 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

• If M(n)/g(n) converges to a +ve constant, we know M(n) є Θ(g(n))

• Or look at M(2n)/M(n); e.g., if M(n) є Θ(lgn) then M(2n)/M(n) will be low

n M(n) g(n) M(n)/g(n)

Empirical Analysis (contd.)

• Scatterplot

n

count/time

n

count/time

n

count/timelgn

n

nlgn or n2

M(n) ≈ can

lgM(n) ≈ nlga + lgc

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, Ω, Θ) 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