程序代写代做代考 data structure algorithm compiler 03.key

03.key

http://people.eng.unimelb.edu.au/tobym

@tobycmurray

toby.murray@unimelb.edu.au

DMD 8.17 (Level 8, Doug McDonell Bldg)

Toby Murray

COMP90038  
Algorithms and Complexity

Lecture 3: Growth Rate and Algorithm Efficiency
(with thanks to Harald Søndergaard)

Update
• Compulsory Quizzes (first one closes Tuesday Week 3)

• Tutorials start this week

• Background knowledge catch-up tutorials:
• Weeks 2 and 3
• Thursday 1-2pm and 2:15-3:15pm 

Alice Hoy, Room 101

• Consultation Hours

• Discussion Board

Algorithm Efficiency

Why is one more efficient than the other?

What does “efficient” even mean?

How can we talk about these things precisely?

Two algorithms for computing gcd:

Linear Search Example

function find(A,x,n)
j ← 0
while j < n if A[j] = x return j j ← j+1 return -1 Y: Let’s trace the execution of find(Y,7,7) j: 0 A[j] n: 7x: 7A: Y j: 1j: 2j: 3j: 4 (returns 4) Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Linear Search Example 5 function find(A,x,n) j ← 0 while j < n if A[j] = x return j j ← j+1 return -1 How many times does the loop run to find 7? 5. How many times does the loop run to find 6? 1. How many times does the loop run to find 99? 7. (the length of the array) Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Assessing Algorithm “Efficiency” • Resources consumed: time and space • We want to assess efficiency as a function of input size • Mathematical vs empirical assessment • Average case vs worst case • Knowledge about input peculiarities may affect the choice of algorithm • The right choice of algorithm may also depend on the programming language used for implementation 6 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Running Time Dependencies • There are many things that a program’s running time depends on: 1.Complexity of the algorithms used 2.Input to the program 3.Underlying machine, including memory architecture 4.Language/compiler/operating system • Since we want to compare algorithms we ignore (3) and (4); just consider units of time • Use a natural number n to quantify (2)—size of the input • Express (1) as a function of n 7 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Linear Search Example 8 function find(A,x,n) j ← 0 while j < n if A[j] = x return j j ← j+1 return -1 How should we measure the size, n,   of the input to this algorithm? n = the length of the array How should we quantify the cost to run this algorithm? roughly, number of times the loop runs   (later in this lecture we will be more precise) Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Linear Search Example 9 function find(A,x,n) j ← 0 while j < n if A[j] = x return j j ← j+1 return -1 What is the worst case input? an array that doesn’t contain the item, x, we are searching for Worst case time complexity: n (since the loop runs n times in that case) Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Linear Search Example 10 function find(A,x,n) j ← 0 while j < n if A[j] = x return j j ← j+1 return -1 What is the best case input? an array that has the item, x, we are searching for in the first position Best case time complexity: 1 (since the loop runs once in that case) Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Estimating Time Consumption • Number of loop iterations is not a good estimate of running time. • Better is to identify the algorithm’s basic operation and how many times it is performed • If c is the cost of a basic operation and g(n) is the number of times the operation is performed for input size n,    then running time t(n) ≈ c ⋅ g(n) 11 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Linear Search Example 12 function find(A,x,n) j ← 0 while j < n if A[j] = x return j j ← j+1 return -1 What is the basic operation here? the comparison A[j] = x Rule of thumb: the most expensive operation executed each time in the inner-most loop of the program Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Examples: Input Size and Basic Operation 13 Problem Size Measure Basic Operation Search in a list   of n items n Key comparison Multiply two matrices of floats Matrix size  (rows x columns) Float multiplication Compute an log n Float multiplication Graph problem Number of nodes and edges Visiting a node Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Best, Average and Worst Case • The running time t(n) may well depend on more than just n • Worse case: analysis makes the most pessimistic assumptions about the input • Best case: analysis makes the most optimistic assumptions about the input • Average case: analysis aims to find the expected running time across all possible input of size n  (Note: not an average of the worst and best cases) • Amortised analysis takes context of running an algorithm into account, calculates cost spread over many runs. Used for “self-organising” data structures that adapt to their usage 14 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Large Input is what Matters • Small input does not properly stress an algorithm             for small values of m and n, their cost is similar     • Only as we let m and n grow large do we witness (big) differences in performance. 15 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Guessing Game Example • Guess which number I am thinking of, between 1 and n (inclusive). I will tell you if it is higher or lower than each guess. 16 1 10050 Wrong. My number is higher than 50. 51 75 Wrong. My number is lower than 75. We are halving the search space each time. Basic operation: guessing a number. (Worse case) complexity: log n Copyright University of Melbourne 2016, provided under Creative Commons Attribution License The Tyranny of Growth Rate 17 n log2 n n n log2 n n2 n3 2n n! 101 3 101 3 ·101 102 103 103 4 · 106 102 7 102 7 · 102 104 106 1030 9 · 10157 103 10 103 1 · 104 106 109 - - 1030 is 1,000 times the number of nano-seconds since the Big Bang. At a rate of a trillion (1012) operations per second, executing 2100   operations would take a computer in the order of 1010 years. That is more than the estimated age of the Earth Copyright University of Melbourne 2016, provided under Creative Commons Attribution License The Tyranny of Growth Rate 18 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Functions Often Met in   Algorithm Classification • 1: Running time independent of input • log n: typical for “divide an conquer” solutions, for example lookup in a balanced search tree • Linear (n): When each input must be processed once • n log n: Each input element processed once and processing involves other elements too, for example, sorting. • n2, n3: Quadratic, cubic. Processing all pairs (triples) of elements. • 2n: Exponential. Processing all subsets of elements. 19 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Asymptotic Analysis • We are interested in the growth rate of functions • Ignore constant factors • Ignore small input sizes 20 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Asymptotics • f(n) ≺ g(n) iff lim = 0 • That is, g approaches infinity faster than f • 1 ≺ log n ≺ nε ≺ nc ≺ nlog n ≺ cn ≺ nn  where 0 < ε < 1 < c • In asymptotic analysis, think big! • e.g., log n ≺ n0.0001, even though for n = 10100, 100 > 1.023.
• Try it for n = 101000000

n→∞
f(n)
g(n)

Big-Oh Notation
• O(g(n)) denotes the set of functions that grow no

faster than g, asymptotically.

• Formal definition: We write  
 
t(n) ∈ O(g(n))  
 
when, for some c and n0  
 
n > n0 ⇒ t(n) < c ·g(n) • For example: 1 + 2 + … + n ∈ O(n2) 22 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Big-Oh: What t(n) ∈ O(g(n)) Means 23 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Big-Oh Pitfalls • Levitin’s notation t(n) ∈ O(g(n)) is meaningful, but not standard. • Other authors use t(n) = O(g(n)) for the same thing. • As O provides an upper bound, it is correct to say both 3n ∈ O(n2) and 3n ∈ O(n) (so you can see why using ‘=’ is confusing); the latter, 3n ∈ O(n), is of course more precise and useful. • Note that c and n0 may be large. 24 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License Big-Omega and Big-Theta • Big Omega: Ω(g(n)) denotes the set of functions that grow no slower than g, asymptotically, so Ω is for lower bounds. • t(n) ∈ Ω(g(n)) iff n >n0 ⇒ t(n) > c ·g(n),  

for some n0 and c.
• Big Theta: Θ is for exact order of growth.

• t(n) ∈ Θ(g(n)) iff t(n) ∈ O(g(n)) and t(n) ∈ Ω(g(n)).

Big-Omega: What t(n) ∈ Ω(g(n)) Means

Big-Theta: What t(n) ∈ Θ(g(n)) Means

Establishing Growth Rate

• We can use the definition of O directly.  
 
t(n) ∈ O(g(n)) iff: n > n0 ⇒ t(n) < c ·g(n) • Exercise: use this to show that   1 + 2 + … + n ∈ O(n2) • Also show that:   17n2 + 85n + 1024 ∈ O(n2) 28 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License 1 + 2 + … + n ∈ O(n2) 29 1 + 2 + … + n < c · n2 Find some c and n0 such that, for all n > n0

1 + 2 + … + n

= n(n+1)
2

=
n2 + n

2
< n2 + n (for n > 0)

< n2 + n2 (for n > 1)

= 2n2 Choose n0 = 1, c = 2

17n2 + 85n + 1024 ∈ O(n2)

17n2 + 85n + 1024 < c · n2 Find some c and n0 such that, for all n > n0

Guess c = 18
17n2 + 85n + 1024 < 18n2 Need to prove: 85n + 1024 < n2i.e. Guess n0 = 1024 85·1024 + 1024 < 1024·1024 Check if: 85n0+ 1024 < n02 i.e. 86·1024 < 1024·1024 Clearly true. Choose c = 18, n0 = 1024 Copyright University of Melbourne 2016, provided under Creative Commons Attribution License 17n2 + 85n + 1024 ∈ O(n2) 31 17n2 + 85n + 1024 < c · n2 Find some c and n0 such that, for all n > n0

Alternative:

17n2 + 85n + 1024

Let c = 17 + 85 + 1024

Choose c = 17 + 85 + 1024, n0 = 1

< 17n2 + 85n2 + 1024n2 (for n > 1)
= (17 + 85 + 1024)n2

Of course, this works for any polynomial.

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