CSC236 Week 3236/larry/3/lec03.pdfProblem Set 1 due Oct 4 Make sure to read “Submission...

CSC236 Week 3Larry Zhang

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Announcements

● Problem Set 1 due Oct 4

● Make sure to read “Submission Instructions” on the handout and the course website.○ submit both tex and pdf

○ make sure you can login to MarkUs

○ create a group and submit one PDF for each group

○ late submissions are not accepted

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NEW TOPIC

Asymptotic NotationsBig-Oh, Big-Omega, Big-Theta

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Review: Asymptotic Notations

● Algorithm Runtime: we describe it in terms of the number of steps as a function of input size

○ Like n², nlog(n), n, sqrt(n), ...

● Asymptotic notations are for describing the growth rate of functions.

○ constant factors don’t matter

○ only the highest-order term matters4

ConcepTest

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Big-Oh, Big-Omega, Big-Theta

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O( f(n) ): The set of functions that grows no faster than f(n)● asymptotic upper-bound on growth rate

Ω( f(n) ): The set of functions that grows no slower than f(n)● asymptotic lower-bound on growth rate

Θ( f(n) ): The set of functions that grows no faster and no slower than f(n)

● asymptotic tight-bound on growth rate

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growth rate ranking of typical functions

grow slowly

grow fast

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The formal mathematical definition of big-Oh

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n0

Chicken grows slower than turkey, or

chicken size is in O(turkey size).

What it really means:

● Baby chicken might be larger than baby turkey at the beginning.

● But after certain “breakpoint”, the chicken size will be surpassed by the turkey size.

● From the breakpoint on, the chicken size will always be smaller than the turkey size.

Definition of f(n) is in O(g(n))

Let f(n) and g(n) be two functions of n. By “f(n)’s growth rate is upper-bounded by g(n)”, we mean the following:

● f(n) is bounded from above by g(n) in some way, something like f(n) ⩽ g(n)

● constant factors don’t matter, so we’re okay with f(n) being bounded by g(n) multiplied by some constant factor, i.e., f(n) ⩽ cg(n)

● we only care about large inputs, so we just need the above inequality to be true when n is larger than some “breakpoint” n₀

So the formal definition of “f(n) is in O(g(n))” is the following

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Definition of big-Oh

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cg(n)

f(n)

n₀

Beyond the breakpoint n₀, f(n) is upper-bounded by cg(n), where c is some wisely chosen constant multiplier.

Side Note

Both ways below are fine

● f(n) ∈ O(g(n)), i.e., f(n) is in O(g(n))

● f(n) = O(g(n)), i.e., f(n) is O(g(n))

Both means the same thing, while the latter is a slight abuse of notation.

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Knowing the definition, now we can write proofs for big-Oh.

The key is finding n₀ and c

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Example 1

Prove that 100n + 10000 is in O(n²)

Need to find the n₀ and c such that 100n + 10000 can be upper-bounded by n² multiplied by some c.

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large

small

c n²

100n + 10000

under-estimate and simplify

over-estimate and simplify

Pick n₀ = 1

100n² + 10000# coz n >= 1

100n² + 10000n²# coz n >= 1

10100n²

Pick c = 10100

Write up the proofProof:

Choose n₀=1, c = 10100,

then for all n >= n₀,

100n + 10000 <= 100n² + 10000n² # because n >= 1

= 10100n²

= cn²

Therefore by definition of big-Oh, 100n + 10000 is in O(n²)

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Proof that 100n + 10000 is in O(n²)

Quick note

The choice of n₀ and c is not unique.

There can be many (actually, infinitely many) different combinations of n₀ and c that would make the proof work.

It depends on what inequalities you use while doing the upper/lower-bounding.

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Example 2

Prove that 5n² - 3n + 20 is in O(n²)

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large

small

c n²

5n² - 3n + 20

under-estimate and simplify

over-estimate and simplify

Pick n₀ = 1

Pick c = 25

5n² + 20# remove a negative term

5n² + 20n²# coz n >= 1

25n²

Takeaway

over-estimate and simplify

under-estimate and simplify

large

small

₀≥

➔

◆ ≥➔

◆ ≥

➔

◆ ≤➔

◆ ≤20

The formal mathematical definition of big-Omega

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Definition of big-Omega

The only difference

Example 3

Prove that 2n³ - 7n + 1 = Ω(n³)

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large

small

2n³ - 7n + 1

c n³

under-estimate and simplify

over-estimate and simplify

Prove that 2n³ - 7n + 1 = Ω(n³)

n³ + n³ - 7n + 1

Pick n₀ = 3(coz we want n³ - 7n > 0)

n³ + 1# n³ - 7n > 0# coz n >= 3

n³

Pick c = 1

Write up the proofProof:

Choose n₀= 3, c = 1,

then for all n >= n₀,

2n³ - 7n + 1 = n³ + (n³ - 7n) + 1

>= n³ + 1 # because n >= 3

>= n³ = cn³

Therefore by definition of big-Omega, 2n³ - 7n + 1 is in Ω(n³)

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Prove that 2n³ - 7n + 1 is in Ω(n³)

Takeaway

Additional trick learned

● Splitting a higher order term● Choose n₀ to however large you need it to be

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n³ + n³ - 7n + 1

The formal mathematical definition of big-Theta

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Definition of big-Theta

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In other words, if you want to prove big-Theta, just prove both big-Oh and big-Omega, separately.

Summary of Asymptotic Notations

● We use functions to describe algorithm runtime○ Number of steps as a function of input size

● Big-Oh/Omega/Theta are used for describing function growth rate

● A proof for big-Oh and big-Omega is basically a chain of inequality.○ Choose the n₀ and c that makes the chain work.

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Handout, Exercise 1

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Analyzing Algorithm RuntimeUsing the Asymptotic Notations

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Runtime Analysis● We can use the asymptotic notations to describe algorithm runtimes.

● We will first focus on iterative algorithms

● The strategy: sum of the work done by all the iterations of the algorithm, then determine a Θ bound.

● In this course, we focus on worst-case analysis

○ What is the maximum number of operations used by the algorithm for input size n?

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Example

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● total = total + 1 occurs at least as often as any other operation

● So, we can count the number of times that this operation occurs, and use that as an asymptotic bound on the algorithm. (Why?)

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Bounding the sum

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