1/6/20161 CS 3343: Analysis of Algorithms Lecture 2: Asymptotic Notations.

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CS 3343: Analysis of Algorithms

Lecture 2: Asymptotic Notations

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Outline

• Review of last lecture

• Order of growth

• Asymptotic notations– Big O, big Ω, Θ

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How to express algorithms?

Nature language (e.g. English)

Pseudocode

Real programming languages

Increasing precision

Ease of expression

Describe the ideas of an algorithm in nature language.Use pseudocode to clarify sufficiently tricky details of the algorithm.

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Insertion Sort

InsertionSort(A, n) {for j = 2 to n {

}

}1 j

sorted

1. Find position i in A[1..j-1] such that A[i] ≤ A[j] < A[i+1]2. Insert A[j] between A[i] and A[i+1]

▷ Pre condition: A[1..j-1] is sorted

▷ Post condition: A[1..j] is sorted

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key = A[j];i = j - 1;while (i > 0) and (A[i] > key)

{A[i+1] = A[i];i = i – 1;

}A[i+1] = key

}}

Insertion Sort

1 i j

Keysorted

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Use loop invariants to prove the correctness of loops

• A loop invariant (LI) is a formal statement about the variables in your program which holds true throughout the loop

• Proof by induction– Initialization: the LI is true prior to the 1st iteration– Maintenance: if the LI is true before the jth iteration, it

remains true before the (j+1)th iteration– Termination: when the loop terminates, the invariant

shows the correctness of the algorithm

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Loop invariants and correctness of insertion sort

• Claim: at the start of each iteration of the for loop, the subarray consists of the elements originally in A[1..j-1] but in sorted order.

• Proof: by induction

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Prove correctness using loop invariantsInsertionSort(A, n) {for j = 2 to n {

key = A[j];i = j - 1;▷Insert A[j] into the sorted sequence A[1..j-1]

while (i > 0) and (A[i] > key) {A[i+1] = A[i];i = i – 1;

}A[i+1] = key

}}

Loop invariant: at the start of each iteration of the for loop, the subarray A[1..j-1] consists of the elements originally in A[1..j-1] but in sorted order.

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InitializationInsertionSort(A, n) {for j = 2 to n {



}A[i+1] = key

}}

Subarray A[1] is sorted. So loop invariant is true before the

loop starts.

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MaintenanceInsertionSort(A, n) {for j = 2 to n {



}A[i+1] = key

}}

Assume loop variant is true prior to iteration j

1 i j

Keysorted

Loop variant will be true before iteration j+1

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TerminationInsertionSort(A, n) {for j = 2 to n {



}A[i+1] = key

}} 1 j=n+1

Sorted

Upon termination, A[1..n] contains all the original

elements of A in sorted order.

n

The algorithm is correct!

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Efficiency

• Correctness alone is not sufficient– Brute-force algorithms exist for most problems– E.g. use permutation to sort– Problem: too slow!

• How to measure efficiency?– Accurate running time is not a good measure– It depends on input, computer, and

implementation, etc.

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Machine-independent

• A generic uniprocessor random-access machine (RAM) model– No concurrent operations– Each simple operation (e.g. +, -, =, *, if, for)

takes 1 step.• Loops and subroutine calls are not simple

operations.

– All memory equally expensive to access• Constant word size• Unless we are explicitly manipulating bits

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Analysis of insertion Sort


key = A[j]i = j - 1;while (i > 0) and (A[i] > key) {

A[i+1] = A[i]i = i - 1

}A[i+1] = key

}

}

How many times will this line execute?

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key = A[j]i = j - 1;while (i > 0) and (A[i] > key) {

A[i+1] = A[i]i = i - 1

}A[i+1] = key

}

}

How many times will this line execute?

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Analysis of insertion SortStatement cost time__

InsertionSort(A, n) {

for j = 2 to n { c1 n

key = A[j] c2 (n-1)

i = j - 1; c3 (n-1)

while (i > 0) and (A[i] > key) { c4 S

A[i+1] = A[i] c5 (S-(n-1))

i = i - 1 c6 (S-(n-1))

} 0

A[i+1] = key c7 (n-1)

} 0

}

S = t2 + t3 + … + tn where tj is number of while expression evaluations for the jth for loop iteration

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Analyzing Insertion Sort• T(n) = c1n + c2(n-1) + c3(n-1) + c4S + c5(S - (n-1)) + c6(S - (n-1)) + c7(n-1)

= c8S + c9n + c10

• What can S be?– Best case -- inner loop body never executed

• tj = 1 S = n - 1 • T(n) = an + b is a linear function

– Worst case -- inner loop body executed for all previous elements

• tj = j S = 2 + 3 + … + n = n(n+1)/2 - 1• T(n) = an2 + bn + c is a quadratic function

– Average case• Can assume that on average, we have to insert A[j] into the

middle of A[1..j-1], so tj = j/2• S ≈ n(n+1)/4• T(n) is still a quadratic function

Θ (n)

Θ (n2)

Θ (n2)

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Analysis of insertion SortStatement cost time__

InsertionSort(A, n) {


key = A[j] c2 (n-1)

i = j - 1; c3 (n-1)


A[i+1] = A[i] c5 (S-(n-1))

i = i - 1 c6 (S-(n-1))

} 0

A[i+1] = key c7 (n-1)

} 0

}What are the basic operations (most executed lines)?

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Statement cost time__InsertionSort(A, n) {


key = A[j] c2 (n-1)

i = j - 1; c3 (n-1)


A[i+1] = A[i] c5 (S-(n-1))

i = i - 1 c6 (S-(n-1))

} 0

A[i+1] = key c7 (n-1)

} 0

}


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Statement cost time__InsertionSort(A, n) {


key = A[j] c2 (n-1)

i = j - 1; c3 (n-1)


A[i+1] = A[i] c5 (S-(n-1))

i = i - 1 c6 (S-(n-1))

} 0

A[i+1] = key c7 (n-1)

} 0

}


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What can S be?

• S = j=2..n tj

• Best case:• Worst case:• Average case:

1 i j

sorted

Inner loop stops when A[i] <= key, or i = 0

Key

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Best case

• Array already sorted• S = j=2..n tj

• tj = 1 for all j• S = n-1 T(n) = Θ (n)

1 i j

sorted Key

Inner loop stops when A[i] <= key, or i = 0

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Worst case

• Array originally in reverse order sorted• S = j=2..n tj

• tj = j• S = j=2..n j = 2 + 3 + … + n = (n-1) (n+2) / 2 = Θ (n2)

1 i j

sorted

Inner loop stops when A[i] <= key

Key

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Average case

• Array in random order• S = j=2..n tj

• tj = j / 2 on average• S = j=2..n j/2 = ½ j=2..n j = (n-1) (n+2) / 4 = Θ (n2)

1 i j

sorted

Inner loop stops when A[i] <= key

Key

What if we use binary search? Answer: still Θ(n2)

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Asymptotic Analysis

• Running time depends on the size of the input– Larger array takes more time to sort– T(n): the time taken on input with size n– Look at growth of T(n) as n→∞.

“Asymptotic Analysis”• Size of input is generally defined as the

number of input elements– In some cases may be tricky

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Asymptotic Analysis

• Ignore actual and abstract statement costs• Order of growth is the interesting measure:

– Highest-order term is what counts• As the input size grows larger it is the high order term that

dominates

0 50 100 150 2000

1

2

3

4x 10

4

n

T(n

)

100 * n

n2

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Comparison of functions

log2n n nlog2n n2 n3 2n n!

10 3.3 10 33 102 103 103 106

102 6.6 102 660 104 106 1030 10158

103 10 103 104 106 109

104 13 104 105 108 1012

105 17 105 106 1010 1015

106 20 106 107 1012 1018

For a super computer that does 1 trillion operations per second, it will be longer than 1 billion years

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Order of growth

1 << log2n << n << nlog2n << n2 << n3 << 2n << n!

(We are slightly abusing of the “<<“ sign. It means a smaller order of growth).

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Exact analysis is hard!

• Worst-case and average-case are difficult to deal with precisely, because the details are very complicated

It may be easier to talk about upper and lower bounds of the function.

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Asymptotic notations

• O: Big-Oh

• Ω: Big-Omega

• Θ: Theta

• o: Small-oh

• ω: Small-omega

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Big O

• Informally, O (g(n)) is the set of all functions with a smaller or same order of growth as g(n), within a constant multiple

• If we say f(n) is in O(g(n)), it means that g(n) is an asymptotic upper bound of f(n)– Intuitively, it is like f(n) ≤ g(n)

• What is O(n2)?– The set of all functions that grow slower than

or in the same order as n2

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So: n O(n2)

n2 O(n2)

1000n O(n2)

n2 + n O(n2)

100n2 + n O(n2)

But:1/1000 n3 O(n2)

Intuitively, O is like ≤

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small o

• Informally, o (g(n)) is the set of all functions with a strictly smaller growth as g(n), within a constant multiple

• What is o(n2)?– The set of all functions that grow slower than n2

So: 1000n o(n2)

But: n2 o(n2)

Intuitively, o is like <

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Big Ω

• Informally, Ω (g(n)) is the set of all functions with a larger or same order of growth as g(n), within a constant multiple

• f(n) Ω(g(n)) means g(n) is an asymptotic lower bound of f(n)– Intuitively, it is like g(n) ≤ f(n)

So: n2 Ω(n) 1/1000 n2 Ω(n)

But:1000 n Ω(n2)

Intuitively, Ω is like ≥

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small ω

• Informally, ω (g(n)) is the set of all functions with a larger order of growth as g(n), within a constant multiple

So: n2 ω(n)

1/1000 n2 ω(n)

n2 ω(n2)

Intuitively, ω is like >

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Theta (Θ)

• Informally, Θ (g(n)) is the set of all functions with the same order of growth as g(n), within a constant multiple

• f(n) Θ(g(n)) means g(n) is an asymptotically tight bound of f(n)– Intuitively, it is like f(n) = g(n)

• What is Θ(n2)?– The set of all functions that grow in the same

order as n2

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So: n2 Θ(n2) n2 + n Θ(n2) 100n2 + n Θ(n2)

100n2 + log2n Θ(n2)

But:nlog2n Θ(n2)1000n Θ(n2)1/1000 n3 Θ(n2)

Intuitively, Θ is like =

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Tricky cases

• How about sqrt(n) and log2 n?

• How about log2 n and log10 n

• How about 2n and 3n

• How about 3n and n!?

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Big-Oh

• Definition: O(g(n)) = {f(n): positive constants c and n0

such that 0 ≤ f(n) ≤ cg(n) n≥n0}

• lim n→∞ g(n)/f(n) > 0 (if the limit exists.)

• Abuse of notation (for convenience):f(n) = O(g(n)) actually means f(n) O(g(n))

There existFor all

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Big-Oh

• Claim: f(n) = 3n2 + 10n + 5 O(n2)• Proof by definition

(Hint: to prove this claim by definition, we need to find some positive constants c and n0 such that f(n) <= cn2 for all n ≥ n0.) (Note: you just need to find one concrete example of c and n0 satisfying the condition, but it needs to be correct for all n ≥ n0. So do not try to plug in a concrete value of n and show the inequality holds.)Proof:3n2 + 10n + 5 3n2 + 10n2 + 5, n ≥ 1

3n2 + 10n2 + 5n2, n ≥ 1 18 n2, n ≥ 1

If we let c = 18 and n0 = 1, we have f(n) c n2, n ≥ n0. Therefore by definition, f(n) = O(n2).

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Big-Omega

• Definition: Ω(g(n)) = {f(n): positive constants c and n0

such that 0 ≤ cg(n) ≤ f(n) n≥n0}

• lim n→∞ f(n)/g(n) > 0 (if the limit exists.)

• Abuse of notation (for convenience):f(n) = Ω(g(n)) actually means f(n) Ω(g(n))

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Big-Omega

• Claim: f(n) = n2 / 10 = Ω(n)

• Proof by definition: f(n) = n2 / 10, g(n) = n

Need to find a c and a no to satisfy the definition of f(n) Ω(g(n)), i.e., f(n) ≥ cg(n) for n ≥ n0

Proof:

n ≤ n2 / 10 when n ≥ 10

If we let c = 1 and n0 = 10, we have f(n) ≥ cn, n ≥ n0. Therefore, by definition, n2 / 10 = Ω(n).

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Theta

• Definition:– Θ(g(n)) = {f(n): positive constants c1, c2,

and n0 such that 0 c1 g(n) f(n) c2 g(n), n n0 }

• f(n) = O(g(n)) and f(n) = Ω(g(n))

• Abuse of notation (for convenience):f(n) = Θ(g(n)) actually means f(n) Θ(g(n))

Θ(1) means constant time.

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Theta

• Claim: f(n) = 2n2 + n = Θ (n2)

• Proof by definition:– Need to find the three constants c1, c2, and n0

such that

c1n2 ≤ 2n2+n ≤ c2n2 for all n ≥ n0

– A simple solution is c1 = 2, c2 = 3, and n0 = 1

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More Examples

• Prove n2 + 3n + lg n is in O(n2)• Need to find c and n0 such that

n2 + 3n + lg n <= cn2 for n ≥ n0

• Proof:n2 + 3n + lg n <= n2 + 3n2 + n for n ≥ 1

<= n2 + 3n2 + n2 for n ≥ 1 <= 5n2 for n ≥ 1

Therefore by definition n2 + 3n + lg n O(n2).(Alternatively: n2 + 3n + lg n <= n2 + n2 + n2 for n ≥ 10

<= 3n2 for n ≥ 10)

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More Examples

• Prove n2 + 3n + lg n is in Ω(n2)

• Want to find c and n0 such that

n2 + 3n + lg n >= cn2 for n ≥ n0

n2 + 3n + lg n >= n2 for n ≥ 1

n2 + 3n + lg n = O(n2) and n2 + 3n + lg n = Ω (n2)

=> n2 + 3n + lg n = Θ(n2)

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O, Ω, and Θ

The definitions imply a constant n0 beyond which they aresatisfied. We do not care about small values of n.

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Using limits to compare orders of growth

0• lim f(n) / g(n) = c > 0

∞n→∞

f(n) o(g(n))

f(n) Θ (g(n))

f(n) ω (g(n))

f(n) O(g(n))

f(n) Ω(g(n))

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logarithms

• compare log2n and log10n

• logab = logcb / logca

• log2n = log10n / log102 ~ 3.3 log10n

• Therefore lim(log2n / log10 n) = 3.3

• log2n = Θ (log10n)

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• Compare 2n and 3n

• lim 2n / 3n = lim(2/3)n = 0

• Therefore, 2n o(3n), and 3n ω(2n)

• How about 2n and 2n+1?

2n / 2n+1 = ½, therefore 2n = Θ (2n+1)

n→∞ n→∞

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L’ Hopital’s rule

• You can apply this transformation as many times as you want, as long as the condition holds

lim f(n) / g(n) = lim f(n)’ / g(n)’n→∞ n→∞

Condition:

If both lim f(n) and lim g(n) are or 0

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• Compare n0.5 and log n

• lim n0.5 / log n = ?

• (n0.5)’ = 0.5 n-0.5

• (log n)’ = 1 / n• lim (n-0.5 / 1/n) = lim(n0.5) = ∞• Therefore, log n o(n0.5)• In fact, log n o(nε), for any ε > 0

n→∞

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Stirling’s formula

nnn

ene

nnn

2/122!

!n nn en 2/1(constant)

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• Compare 2n and n!

• Therefore, 2n = o(n!)

• Compare nn and n!

• Therefore, nn = ω(n!)

• How about log (n!)?

n

nnn

n

nnn e

nnc

e

nncn

2lim

2lim

2

!lim

0limlim!

lim nnnn

n

nnn e

nc

en

nnc

n

n

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)log(loglog)!log( 2/1 nnn

n

enCe

nncn

)log(

log2

1)log

2(log

2

log2

1log

nn

nnnn

nn

C

nnnnC

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More advanced dominance ranking

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Asymptotic notations

• O: Big-Oh

• Ω: Big-Omega

• Θ: Theta

• o: Small-oh

• ω: Small-omega

• Intuitively:

O is like o is like <

is like is like >

is like =

1/6/20161 CS 3343: Analysis of Algorithms Lecture 2: Asymptotic Notations.

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